Jekyll2020-07-02T05:24:42+00:00https://nivent.github.io/feed.xmlThoughts of a ProgrammerNiven Achenjang's Personal WebsiteCharacteristic Classes2020-07-02T00:00:00+00:002020-07-02T00:00:00+00:00https://nivent.github.io/blog/char-class<p>For a long time I’ve been telling myself that I should try and get a solid understanding of characteristic classes, so why not do that now?</p>
<p>Before getting into things, as usual, a few remarks about this post. First, by a “characteristic class” I mean a functorial way of attaching a cohomology class to a vector bundle <sup id="fnref:3"><a href="#fn:3" class="footnote">1</a></sup>. There are many of these, but because I care more about algebraic/holomorphic things than I do about smooth things, we will basically just look at Chern classes. I want to be quite detailed in this post, so even just focusing on these will require a lot of work. On the upside, the payoff will be a good amount of interesting/neat applications/calculations. Second, I’m going to try <sup id="fnref:1"><a href="#fn:1" class="footnote">2</a></sup> to make this post be a good one <sup id="fnref:30"><a href="#fn:30" class="footnote">3</a></sup>. I feel like many of my recent posts have either been overall meh or have started off good only for me to fumble the ending, and this is starting to really bother me. Hence, for this post, I’m going to try to prioritize quality/detail <sup id="fnref:2"><a href="#fn:2" class="footnote">4</a></sup> over “just producing a finished product” and see how this feels. Third, this post is titled “Characteristic Classes” but, if possible, I’d like to throw in a few things not strictly related to/requiring characteristic classes, but which are both at least somewhat related to them and things I’ve been wanting to improve my understanding of. Finally, characteristic classes are one of those things that can be constructed a dozen different ways, and can be given a million different treatments. Throughout this post, I’ll try to maintain an unapologetically algebro-topological perspective, so don’t expect me to e.g. say the words “connection” or “integration” <sup id="fnref:20"><a href="#fn:20" class="footnote">5</a></sup>.
$
\newcommand{\CW}{\mathrm{CW}}
\newcommand{\Top}{\mathrm{Top}}
\newcommand{\Set}{\mathrm{Set}}
\newcommand{\Grp}{\mathrm{Grp}}
\DeclareMathOperator{\Ho}{Ho}
$</p>
<p>Throughout this post, assume the base space of any fibre bundle is Hausdorff and paracompact, so a CW complex or a manifold or a metric space or something like this. <sup id="fnref:4"><a href="#fn:4" class="footnote">6</a></sup> In particular, base spaces will always admit partitions of unity.</p>
<ol id="markdown-toc">
<li><a href="#principal-g-bundles" id="markdown-toc-principal-g-bundles">Principal $G$-Bundles</a> <ol>
<li><a href="#splitting-principle" id="markdown-toc-splitting-principle">Splitting Principle</a></li>
<li><a href="#cohomology-of-bun" id="markdown-toc-cohomology-of-bun">Cohomology of $BU(n)$</a></li>
</ol>
</li>
<li><a href="#chern-classes" id="markdown-toc-chern-classes">Chern Classes</a></li>
<li><a href="#orientation-and-sphere-bundles" id="markdown-toc-orientation-and-sphere-bundles">Orientation and Sphere Bundles</a></li>
<li><a href="#euler-class" id="markdown-toc-euler-class">Euler Class</a> <ol>
<li><a href="#relation-to-chern-classes" id="markdown-toc-relation-to-chern-classes">Relation to Chern Classes</a></li>
<li><a href="#relation-to-obstruction-theory" id="markdown-toc-relation-to-obstruction-theory">Relation to Obstruction Theory</a></li>
<li><a href="#relation-to-enumerative-geometry" id="markdown-toc-relation-to-enumerative-geometry">Relation to Enumerative Geometry</a></li>
</ol>
</li>
<li><a href="#27-lines-on-a-cubic" id="markdown-toc-27-lines-on-a-cubic">27 Lines on a Cubic</a></li>
</ol>
<h1 id="principal-g-bundles">Principal $G$-Bundles</h1>
<p>The key example of a characteristic class to keep in mind throughout this post is the first Chern class $c_1(E)$ of a complex (topological) line bundle $E\to X$ which was introduced at the end of <a href="../brown-rep">the post on Brown Representability</a>. This class lives in $X$’s second integral cohomology and actually completely characterizes $E$ up to isomorphism (as a topological, complex line bundle over $X$). We constructed this class as the pullback on the canonical generator $\gamma\in\hom^2(\CP^\infty;\Z)\simeq\Z\gamma$ of the cohomology of $\CP^\infty$ along $E$’s classifying map $X\to BS^1\simeq\CP^\infty$ <sup id="fnref:5"><a href="#fn:5" class="footnote">7</a></sup>. Motivated by this, in order to define cohomology classes for higher rank (complex) vector bundles, we will want to understand the cohomology of $BU(n)$ for all $n$ <sup id="fnref:39"><a href="#fn:39" class="footnote">8</a></sup>.</p>
<p>Before calculating $\ast\hom(BU(n);\Z)$, I want to patch up a hole in my introduction to Principal $G$-bundles from the Brown Representability post. In particular, I claimed that such bundles satisfy a homotopy invariance property, but did not proof this. This fact is high-key the starting point to being able to use $G$-bundles productively in homotopy theory, so this is probably actually something worth writing down a proof of. First recall the definition of a principal $G$-bundle.</p>
<div class="definition">
Fix a topological group $G$. A <b>principal $G$-bundle</b> over a space $B$ is a fiber bundle $\pi:P\to B$ with a free and transitive right action by $G$ on the fibers.
</div>
<div class="definition">
Let $\pi_1:P_1\to B_1$ and $\pi_2:P_2\to B_2$ be principal $G$-bundles. A <b>morphism of principal $G$-bundles</b> $f=(\phi,\psi):(P_1,B_1)\to(P_2,B_2)$ is a pair of maps $\phi:P_1\to P_2$ and $\psi:B_1\to B_2$ such that the following commutes
$$\begin{CD}
P_1 @>\phi>> P_2\\
@V\pi_1VV @VV\pi_2V\\
B_1 @>>\psi> B_2
\end{CD}$$
and $\phi(p\cdot g)=\phi(p)\cdot g$ for all $g\in G$. We say $\phi$ <b>lies over</b> $\psi$.
</div>
<p>In order to show that homotopic maps induce isomorphic pullback bundles, the following lemma will be useful. In short, it says that $G$-bundle maps lying above the identity are automatically isomorphisms.</p>
<div class="lemma">
Let $P$ and $P'$ be principal $G$-bundles over $B$, and let $\phi:P\to P'$ be a morphisms lying over the identity $B=B$. Then, $\phi$ is an isomorphism.
</div>
<div class="proof4">
(Injectivity) We first show $\phi$ injective. Pick $p,q\in P$ such that $\phi(p)=\phi(q)$. Since $\phi$ lies over the identity, $p,q$ must belong to the same fiber of $P\to B$. Because $G$ acts transitively on the fibers, there exists some $g\in G$ such that $p\cdot g=q$. Since $\phi$ is a morphism of $G$-bundles, we get that $\phi(p)\cdot g=\phi(q)$, but $\phi(p)=\phi(q)$ and $G$ acts freely on the fibers. This means we must have $g=e$, the identity of $G$, so $q=p\cdot g=p\cdot e=p$.
<br />
(Surjectivity) We now treat surjectivity. Fix some $p'\in P'$ lying over $b\in B$. Let $p\in P$ be any point which also lies over $b$, so $\phi(p)$ and $p'$ belong to the same fiber of $P'$. Hence, by transitivity of the $G$-action, there is some $g\in G$ such that $\phi(p)\cdot g=p'$, and so $p\cdot g\in P$ is a preimage of $p'\in P'$.
<br />
(Continuity of $\inv\phi$) Now that we know $\phi$ is bijective, all that remains is to show that it is an open mapping. For this, we work locally. Pick some small open $U\subset B$ such that we have local trivializations $P\vert_U\simeq U\by G$ and $P'\vert_U\simeq U\by G$. Then, in local coordinates, $\phi\vert_U$ must take the form
$$\phi:(b,g)\mapsto\parens{b,\phi'(b,g)}=\parens{b,\phi'(b,e)g}$$
for some $\phi':U\by G\to G$ satisfying $\phi'(b,gh)=\phi'(b,g)h$. Thus, $\inv\phi$ is of the form
$$(b,g)\mapsto(b,\inv{\phi'(b,e)}g)$$
which is visibly continuous.
</div>
<div class="theorem">
Let $\pi:P\to B'$ be a principal $G$-bundle, and let $f_0\sim f_1:B\rightrightarrows B'$ be two homotopic maps. Then, the bundles $\pull f_0(P)$ and $\pull f_1(P)$ over $B$ are isomorphic.
</div>
<div class="proof4">
Let $F:B\by I\to B'$ be a homotopy from $f_0=F(-,0)$ to $f_1=F(-,1)$. We consider the pullback $\pull FP$, a principal $G$-bundle over $B\by I$. Our goal is essentially to show that $\pull FP$ is of the form $\pull f_0P\by I$; we will in fact show that this is the case for any principal $G$-bundle over $B\by I$. With that said, let $p:Q\to B\by I$ be a principal $G$-bundle. We claim that $Q\simeq Q_0\by I$ where $Q_0:=\inv p(B\by\{0\})\to B$ is the restriction of $Q$ to $0\in I$.
<br />
We seek an isomorphism $Q_0\by I\iso Q$ (lying over the identity on the base). Note that this amounts to extending the canonical map $Q_0\by\{0\}\iso Q_0\into Q$ into one from $Q_0\by I\to Q$. In other words, we are attempting to solve the following lifting problem.
<center>
<img src="https://nivent.github.io/images/blog/char-class/lift.png" width="250" height="100" />
</center>
Now, we're in luck. Since $B\by I$ is paracompact, the map $Q\xto pB\by I$ is a fibration, and so satisfies the homotopy lifting property. This says exactly that we can construct the above dashed map $\phi:Q_0\by I\to Q$ above (so the diagram commutes)! We claim this $\phi$ is an isomorphism. This follows immediately from the lemma.
</div>
<p>With that taken care of, let’s return to our goal of understanding characteristic classes. As was noted earlier, the first step in doing so is determining the cohomology (ring) of $BU(n)$ since this space classifies principal $U(n)$-bundles and so classifies rank $n$ complex vector bundles. We will spend a section going over a technical tool useful for performing this calculation (and also useful for proving things about characteristic classes in general), and then after that we will get our cohomology classes.</p>
<h2 id="splitting-principle">Splitting Principle</h2>
<p>Our first major goal is to show that</p>
<script type="math/tex; mode=display">\ast\hom(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]\text{ where }c_i\in\hom^{2i}(BU(n);\Z),</script>
<p>and so obtain $n$ different characteristic classes for any rank $n$ vector bundle, the Chern classes. In order to do this, we will carry out an inductive argument, relating the cohomology of $BU(n)$ to that of $BU(n-1)\by BU(1)$. This argument will make use of a general construction that allows one to peel of a single (line bundle) summand from an arbitrary vector bundle.</p>
<p>To see what I mean, let $p:E\to B$ be any rank $n+1$ vector bundle. Implicitly cover $B$ by opens $U_i\subset B$ on which $p$ trivializes, and let $\tau_{ij}:U_i\cap U_j\to U(n+1)\subset\GL_{n+1}(\C)$ be $p$’s transition functions <sup id="fnref:6"><a href="#fn:6" class="footnote">9</a></sup>. Then, we can projectivize $E$ in order to form a $\CP^n$-bundle $\pi:\P(E)\to B$ whose fibers are $\inv\pi(b)\simeq\P(\inv p(b))=\P(E_b)$ (I use $E_b$ to denote $E$’s fiber over $b\in B$), and whose transition functions are the compositions</p>
<script type="math/tex; mode=display">U_i\cap U_j\xto{\tau_{ij}}U(n+1)\into\GL_{n+1}\C\onto\PGL_n\C.</script>
<p>This space $\P(E)$, called <b>projectivization of $E$</b> is where we will peel off a summand of $E$. A point $x\in\P(E)$ belongs to the projective space $\P(E)_ {\pi(x)}=\P(E_{\pi(x)})$ and so is identifiable with a line in the fiber $E_{\pi(x)}=\inv p(\pi(x))$. Note that $E$ has a few natural vector bundles. First, there is the pullback bundle $\pull\pi E\to\P(E)$ whose fiber above a point $x\in\P(E)$ is the fiber $E_{\pi(x)}$ of $E\xto pB$ above $\pi(x)\in B$. Inside of $\pull\pi E$, one can find the <b>tautological subbundle</b></p>
<script type="math/tex; mode=display">L_E:=\bracks{(\l, v)\in\P(E)\by E:v\in\l}\subset\pull\pi E</script>
<p>whose fiber above a point $\l\in\P(E)$ is the line $\l\subset E_{\pi(\l)}$ it corresponds to. The <b>tautological quotient bundle</b> $Q_E$ on $\P(E)$ is defined by its place in the following exact sequence:</p>
<script type="math/tex; mode=display">0\too L_E\too\pull\pi E\too Q_E\too0.</script>
<p>Note that if we restrict the above sequence to the fiber $\P(E)_ b=\P(E_b)$ above any point $b\in B$, then we recover the normal sequence of tautological bundles on projective space <sup id="fnref:7"><a href="#fn:7" class="footnote">10</a></sup>. The existence of this sequence allows us to write $\pull\pi E\simeq L_E\oplus Q_E$ as the sum of a line bundle and a rank $n$ vector bundle.</p>
<div class="lemma">
Let $0\too L\too E\too Q\too 0$ be any sequence of complex vector bundles on some base space $B$. Then, $E\simeq L\oplus Q$ as vector bundles.
</div>
<div class="proof4">
Let $m,n,k$ be the ranks of $L,E,Q$, respectively. Since $B$ is paracompact and Hausdorff by assumption, we can cover it by small opens $U_i\subset B$ such that
<ol>
<li> $L,E,Q$ all three trivialize over $U_i$, i.e. $L\vert_{U_i}\simeq U_i\by\C^m$, $E\vert_{U_i}\simeq U_i\by\C^n$, and $Q\vert_{U_i}\simeq U_i\by\C^k$. </li>
<li> There exists a partition of unity $\rho_i:B\to[0,1]$ subordinate to the $U_i$'s. That is, $\supp\rho_i\subset U_i$ for all $i$ (the cover $\{U_i\}$ assumed locally finite) and
$$\sum_i\rho_i(x)=1$$
for all $x\in X$ (all but finitely many terms above are $0$). </li>
</ol>
Locally, the exact sequence $0\too L\too E\too Q\too 0$ of vector bundles looks like the sequence $0\too\C^m\too\C^n\too\C^k\too0$ of vector spaces. This sequence splits (say via $s:\C^k\to\C^n$), and so for every $i$, we can construct a local splitting map
$$\sigma_i:Q\vert_{U_i}\simeq U_i\by\C^k\xto{(\Id,s)}U_i\by\C^n\simeq E\vert_{U_i}$$
realizing $E\vert_{U_i}$ as a sum $E\vert_{U_i}\simeq L\vert_{U_i}\oplus Q\vert_{U_i}$. We want to upgrade these into a global splitting map $\sigma:Q\to E$. To do so, we define
$$\sigma(q)=\sum_i\rho_i(\pi(q))\sigma_i(q)$$
where $q\in Q$ and $\pi:Q\to B$ is the projection map. The formula above makes sense since $\rho_i(\pi(q))=0$ when $q\not\in Q\vert_{U_i}$, and defines a splitting map because $\sigma\vert_{\inv\pi(U_i)}=\sigma_i$ by construction. Thus, $E\simeq L\oplus Q$ as claimed.
</div>
<p>The above lemma justifies our earlier claim that $\pull\pi E\simeq L_E\oplus Q_E$.</p>
<div class="exercise">
Let $EU(n)\to BU(n)$ be the universal rank $n$ vector bundle (really $EU(n)$ should denote the universal principal $U(n)$-bundle, but then you can turn this into a vector bundle by taking the vector bundle associated to the inclusion map $U(n)\into\GL_n\C=\Aut(\C^n)$). By the above discussion, pulling this bundle back to $X:=\P(EU(n))$ splits it into a line bundle and a rank $(n-1)$ vector bundle. Since $X$ splits the universal rank $n$ vector bundle, show that $X\simeq BU(n-1)\by BU(1)$ (they solve the same moduli problem/represent the same functor).
</div>
<p>In order to carry out our induction argument, we still need to be able to relate the cohomology of $\P(E)$ to that of $E$ (e.g. cohomology of $BU(n-1)\by BU(1)$ to that of $BU(n)$ by the exercise above). We do this now. First recall the tautological exact sequence</p>
<script type="math/tex; mode=display">0\too L_E\too\pull\pi E\too Q_E\too0</script>
<p>on $\P(E)$, and let $x=c_1(\dual L_E)\in\hom^2(\P(E);\Z)$ ($\dual L_E$ is the dual line bundle). As earlier remarked, this sequence restricts to the normal tautological exact sequence on the fibers $\P(E)_ b\simeq\P^n$; in particular, $L_E\vert_{\P(E)_ b}$ is the usual tautological line bundle <sup id="fnref:8"><a href="#fn:8" class="footnote">11</a></sup>, and so the canonical generator for $\hom^2(\P(E)_ b;\Z)$ is precisely $x\vert_{\P(E)_ b}$ (this is the reason for taking the dual of $L_E$ before). Taking self cup-products, this means that ${1,x,x^2,\dots,x^{n-1},x^n}\subset\ast\hom(\P(E);\Z)$ are global cohomology classes whose restriction to each fiber $P(E)_ b=\P(E_b)\simeq\P^n$ freely generate the cohomology there (as a module). This allows us to determine the cohomology of $\P(E)$ via the following generalization of the Kunneth formula.</p>
<div class="theorem" name="Leray-Hirsch">
Let $F\xto\iota E\xto\pi B$ be a fiber bundle sequence, and let $R$ be a commutative ring. Assume that there are global cohomology classes $e_1,\dots,e_r\in\ast\hom(E;R)$ on $E$ which, when restricted to each fiber, freely generate the cohomology of said fiber (i.e. $\ast\hom(F;R)$ is a free (graded) $R$-module with basis $\{e_i\vert_F\}_{i=1}^r$). Let $s:\ast\hom(F;R)\to\ast\hom(E;R)$ be the map determined by these $e_i$ (i.e. $s(e_i\vert_F)=e_i$ and we extend linearly). Then, the map
$$\mapdesc f{\ast\hom(B;R)\otimes_R\ast\hom(F;R)}{\ast\hom(E;R)}{\alpha\otimes\beta}{\pull\pi(\alpha)\smile s(\beta)}$$
is an isomorphism of $\ast\hom(B;R)$-modules (not of rings though). Hence, $\ast\hom(E;R)$ is a free $\ast\hom(B;R)$-module with basis $e_1,\dots,e_r$.
</div>
<div class="proof4">
We use the <a href="../spectral-seq">Serre spectral sequence</a>. First note that $\pull\iota:\ast\hom(E;R)\to\ast\hom(F;\R)$ is surjective, and so $\pi_1(B)$ acts trivially on $\ast\hom(F;R)$. Indeed, for any $\gamma\in\pi_1(B)$, it acts via the map $L_\gamma:F\to F$ given on $f\in F$ by lifting $\gamma$ to a path $\wt\gamma:[0,1]\to E$ starting at $f$ and then setting $L_\gamma(f)=\wt\gamma(1)$. Essentially by definition, this means we have a homotopy $h_t:F\to E$ from $h_0=\iota$ to $h_1=L_\gamma$. Hence, $\iota$ and $\iota\circ L_\gamma$ induce the same map on cohomology. Since $\pull\iota=\pull L_\gamma\circ\pull\iota$ and $\pull\iota$ is surjective, we conclude that $\pull L_\gamma=1$, i.e. that $\pi_1(B)$ acts trivially on the cohomology of the fiber.
<br />
This means we are justified in forming the Serre spectral sequence
$$E_2^{p,q}=\hom^p(B;\hom^q(F;R))\implies\hom^{p+q}(E;R).$$
By assumption, $\hom^q(F;R)$ is a free $R$-module (i.e. of the form $R^{\oplus k}$ for some $k$, depending on $q$), so we see immediately that
$$E_2^{p,q}=\hom^p(B;\hom^q(F;R))\simeq\hom^p(B;R)\otimes_R\hom^q(F;R)\simeq E_2^{p,0}\otimes_RE_2^{0,q}.$$
Recall the the differential $d_2$ on the $E_2$-page has bidegree $(2,-1)$, i.e. we have
$$d_2:E_2^{p,q}\simeq E_2^{p,0}\otimes_RE_2^{0,q}\to E_2^{p+2,0}\otimes_RE_2^{0,q-1}\simeq E_2^{p+2,q-1}.$$
By the Leibniz rule, for $a\in E_2^{p,0}$ and $b\in E_2^{0,q}$, we have
$$d_2(a\otimes b)=(a\otimes1)d_2(1\otimes b)+(-1)^p(1\otimes b)d_2(a\otimes 1)=(a\otimes1)d_2(1\otimes b)$$
since $d_2(a\otimes 1)\in E_2^{p+2,-1}=0$. At the same time, $E_\infty^{0,q}=G^0\hom^q(E;R)=\hom^q(E;R)/F^1\hom^q(E;R)$ where $F^1\hom^q(E;R)=\ker(\hom^q(E;R)\to\hom^q(E_0;R))$ where $E_0=F$. In other words,
$$E_\infty^{0,q}\simeq\im(\hom^q(E;R)\to\hom^q(F;R))=\hom^q(F;R)\simeq E_2^{0,q},$$
from which we see that $d_2(b)=0$ for all $b\in E_2^{0,q}$ (so $E_2^{0,q}$ survives to the $E_\infty$-page). Thus, $d_2=0$ in all degrees, so the $E_2$-page is the $E_\infty$-page! This is almost enough for us to conclude that
$$\hom^n(E;R)\simeq\bigoplus_{p+q=n}E_\infty^{p,q}=\bigoplus_{p+q=n}E_2^{p,q}\simeq\bigoplus_{p+q=n}\hom^p(B;R)\otimes_R\hom^q(F;R)$$
which would exactly give the claim. In orer to actually conclude this claim, recall that $E_\infty^{p,q}\simeq G^p\hom^{p+q}(E;R)$ are successive quotients in a filtration
$$0\subset F^{p+q}\hom^{p+q}(E;R)\subset F^{p+q-1}\hom^{p+q}(E;R)\subset\dots\subset F^0\hom^{p+q}(E;R)=\hom^{p+q}(E;R).$$
of $\hom^{p+q}(E;R)$. In particular, we have exact sequences
$$0\too F^{k+1}\hom^n(E;R)\too F^k\hom^n(E;R)\too E_\infty^{k,n-k}\too0.$$
Since $E_{\infty}^{k,n-k}\simeq\hom^k(B;R)\otimes_R\hom^{n-k}(F;R)$, the above sequence splits via the map $\sigma:E_\infty^{k,n-k}\to F^k\hom^n(E;R)$ given by $\sigma(\alpha\otimes\beta)=\pull\pi(\alpha)s(\beta)$ (i.e. $\sigma=f\vert_{E_\infty^{k,n-k}})$, and so
$$F^k\hom^n(E;R)\simeq F^{k+1}\hom^n(E;R)\oplus E_\infty^{k,n-k}.$$
Since this holds for all $k$ and $F^{n+1}\hom^n(E;R)=0$, induction shows that indeed
$$\hom^n(E;R)\simeq\bigoplus_{p+q=n}E_\infty^{p,q}=\bigoplus_{p+q=n}E_2^{p,q}\simeq\bigoplus_{p+q=n}\hom^p(B;R)\otimes_R\hom^q(F;R),$$
so we win.
</div>
<div class="remark">
One can rephrase the hypotheses of the above theorem by saying that $\ast\hom(F;R)$ is a finite free $R$-module, and $\pull\iota:\ast\hom(E;R)\to\ast\hom(F;R)$ is surjective.
</div>
<p>Returning to our situation of interest, we have a projective bundle $\P(E)\xto\pi B$ with tautological exact sequence</p>
<script type="math/tex; mode=display">0\too L_E\too\pull\pi E\too Q_E\too0.</script>
<p>We observed that, letting $x=c_1(\dual L_E)$, the set ${1,x,x^2,\dots,x^{n-1},x^n}\subset\ast\hom(\P(E);\Z)$ freely generates the cohomology of each fiber. Thus, by Leray-Hirsch, $\ast\hom(\P(E);\Z)$ is a free $\ast\hom(B;\Z)$-module with basis $1,x,x^2,\dots,x^{n-1},x^n$. Thus <sup id="fnref:10"><a href="#fn:10" class="footnote">12</a></sup>, there must exist some polynomial $p(x)\in\ast\hom(B;\Z)[x]$, say (recall $n+1=\rank E$)</p>
<script type="math/tex; mode=display">p(x)=x^{n+1}+c_1(E)x^n+\dots+c_n(E)x+c_{n+1}(E)</script>
<p>such that $\ast\hom(\P(E);\Z)\simeq\ast\hom(B;\Z)[x]/(p(x))$. For now, just think of $c_i(E)\in\hom^{2i}(B;\Z)$ above as notation for the coefficient in this polynomial <sup id="fnref:9"><a href="#fn:9" class="footnote">13</a></sup>. In particular, we have an injection $\ast\hom(B;\Z)\into\ast\hom(\P(E);\Z)$. This gives our link between the cohomology of $B$ and that of $\P(E)$. We saw earlier that if we take $B=BU(n)$ and $E=EU(n)$ (modulo the distinction between a vector bundle and a $U(n)$-bundle), then $\P(E)\simeq BU(n-1)\by BU(1)$, so we are ready to carry out our induction argument.</p>
<div class="remark">
In this section, there is no reason to stop after only splitting off a single line bundle. Indeed, if you pull $E$ back to not just $\P(E)$ but to $\P(Q_E)$, then the same arguments show that you can split off two line bundles from $E$. Continuing this pattern let's you show that any for any vector bundle $E\to B$, there's some $f:B'\to B$ such that the pullback bundle $E'=\pull fE\to B'$ splits as a sum of $\rank E$ line bundles. Furthermore, inductively applying the cohomology calculation given here shows that we can even form $B'$ such that $\pull f:\ast\hom(B;\Z)\into\ast\hom(B';\Z)$ is an injection.
<br />
This phenomenon can be described universally. Let $f:(BU(1))^n\to BU(n)$ be the classifying map for the sum of $n$ copies of the universal line bundle (i.e. the classifying map for the universal sum of $n$ line bundles), and consider any rank $n$ vector bundle $E\to B$. This is pulled back along some classifying map $g:B\to BU(n)$, so we can form the pullback $B'=B\by_{BU(n)}BU(1)^n$.
$$\begin{CD}
B' @>h>> B\\
@Vg'VV @VVgV\\
BU(1)^n @>>f> BU(n)
\end{CD}$$
By construction, the pullback bundle $E'=\pull hE\to B'$ has $g\circ h=f\circ g':B'\to BU(n)$ as its classifying map. Since this map factors through $BU(1)^n$, we see that $E'$ splits as the sum of $n$ line bundles, so we can observe this splitting phenomenon without needing any construction. However, this perspective has the downside that it is potentially harder to see that $B$'s cohomology injects into that of $B'$. This is needed, for example, if you want to reduced proofs of properties of Chern classes of general vector bundles to the case of sums of line bundles.
</div>
<h2 id="cohomology-of-bun">Cohomology of $BU(n)$</h2>
<p>Recall that $BU(1)\simeq\CP^\infty$ and so its cohomology ring is $\ast\hom(BU(1);\Z)\simeq\Z[c_1]$ with $c_1\in\hom^2(BU(1);\Z)$ the first Chern class. Also, from the previous section, recall that we have ring isomorphisms</p>
<script type="math/tex; mode=display">\ast\hom(BU(n-1)\by BU(1);\Z)\simeq\ast\hom(BU(n);\Z)[x]/(p_n(x))</script>
<p>where $\deg x=2$ (i.e $x\in\hom^2(BU(n-1)\by BU(1);\Z)$) and</p>
<script type="math/tex; mode=display">p_n(x)=x^n+c_1(EU(n))x^{n-1}+\dots+c_{n-1}(EU(n))x+c_n(E)\in\ast\hom(BU(n);\Z)[x]</script>
<p>is some polynomial <sup id="fnref:11"><a href="#fn:11" class="footnote">14</a></sup>. Our goal this section is to prove the following.</p>
<div class="theorem">
$$\ast\hom(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]$$
where $c_i=c_i(EU(n))\in\hom^{2i}(BU(n);\Z)$ will be called the <b>$i$th Chern class</b>.
</div>
<p>We already know this in the case $n=1$. Let’s see how the argument goes for $n=2$. Here, using Kunneth on the left, we have</p>
<script type="math/tex; mode=display">\Z[c_1,c_1']\simeq\ast\hom(BU(1)\by BU(1);\Z)\simeq\ast\hom(BU(2);\Z)[x]\left./\parens{x^2+c_1(EU(2))x+c_2(EU(2))}\right.</script>
<p>with $\deg c_1=\deg c_1’=2=\deg x$. Now, the fibration $\pi:BU(1)\by BU(1)\to BU(2)$ we are using pulls the universal rank $2$ vector bundle $EU(2)$ into a sum $EU(1)\oplus EU(1)$ of two copies of the universal line bundle <sup id="fnref:12"><a href="#fn:12" class="footnote">15</a></sup>. Thus, recalling how we defined $x$ earlier, we see that we can take $x=-c_1\in\hom^2(BU(1);\Z)$ above. That is, $c_1^2-c_1(EU(2))c_1+c_2(EU(2))=0$. Similarly, by symmetry (e.g. the automorphism $BU(1)\by BU(1)\iso BU(1)\by BU(1)$ permuting the factors), we equally see that $(c_1’)^2-c_1(EU(2))c_1’+c_2(EU(2))=0$. In other words,</p>
<script type="math/tex; mode=display">x^2+c_1(EU(2))x+c_2(EU(2))=(x+c_1)(x+c_1')=x^2+(c_1+c_2')x+c_1c_1'</script>
<p>as polynomials. Matching coefficients, we see that $c_1(EU(2))=c_1+c_1’$ and $c_2(EU(2))=c_1c_1’$ are given by the elementary symmetric polynomials in $c_1,c_1’$. Thus, by Galois theory <sup id="fnref:13"><a href="#fn:13" class="footnote">16</a></sup>, $\Z[c_1(EU(2)),c_2(EU(2))]$ gives a polynomial algebra in $\ast\hom(BU(2);\Z)$. We claim this is the whole thing. This follows from a counting argument. We know, from Leray-Hirsch, that</p>
<script type="math/tex; mode=display">\Z[c_1,c_1']\simeq\ast\hom(BU(2);\Z)\otimes\ast\hom(\P^1;\Z)</script>
<p>additively (i.e. as graded modules). So, letting $S_k$ denote the degree-$k$ part of a graded ring $S$, (cohomology in odd degrees vanishes)</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
k+1
&=\rank_\Z\Z[c_1,c_1']_ {2k}\\
&=\rank_\Z\parens{\ast\hom(BU(2);\Z)\otimes\ast\hom(\P^1;\Z)}_ {2k}\\
&=\rank_\Z\hom^{2k}(BU(2);\Z)+\rank_\Z\hom^{2k-2}(BU(2);\Z)\\
&\ge\rank_\Z\Z[c_1(EU(2)),c_2(EU(2))]_{2k}+\rank_\Z\Z[c_1(EU(2)),c_2(EU(2))]_{2k-2}\\
&=\parens{\floor{\frac k2}+1}+\parens{\floor{\frac{k-1}2}+1}\\
&=k+1.
\end{align*} %]]></script>
<p>This shows that $\Z[c_1(EU(2)),c_2(EU(2))]_ n=\hom^n(BU(2);\Z)$ for all $n$, so</p>
<script type="math/tex; mode=display">\ast\hom(BU(2);\Z)\simeq\Z[c_1,c_2]</script>
<p>with $c_1=c_1(EU(2))$ and $c_2=c_2(EU(2))$ as desired!</p>
<p>We now handle the general case $n\ge2$. Based on how the $n=2$ case played out, let’s update our goal.</p>
<div class="theorem">
Let $f:BU(1)^n\to BU(n)$ be the classifying map for the sum $EU(1)^{\oplus n}$ of $n$ copies of the universal line bundle. Then, $f$ induces an injection
$$\pull f:\ast\hom(BU(n);\Z)\into\ast\hom(BU(1)^n;\Z)\iso\Z[\Ith c1_1,\dots,\Ith cn_1]$$
whose image $\Z[c_1,c_2,\dots,c_n]$ is the polynomial algebra generated by the elementary symmetric polynomials
$$\begin{matrix}
c_1(EU(n)) &=& c_1 &=& \Ith c1_1+\Ith c2_1+\dots+\Ith cn_1\\
c_2(EU(n)) &=& c_2 &=& \Ith c1_1\Ith c2_1+\Ith c1_1\Ith c3_1+\dots+\Ith c{n-1}_1\Ith cn_1\\
&&\vdots\\
c_n(EU(n)) &=& c_n &=& \Ith c1_1\Ith c2_1\cdots\Ith cn_1
\end{matrix}$$
In particular, $\ast\hom(BU(n);\Z)\iso\Z[c_1,c_2,\dots,c_n]$ with $\deg c_i=2i$.
</div>
<div class="proof4">
The isomorphism $\hom(BU(1)^n;\Z)\iso\Z[\Ith c1_1,\dots,\Ith cn_1]$ is simply the Kunneth formula since $B(U(1)^n)\simeq(BU(1))^n$. By the previous section, we know we also have an isomorphism
$$\ast\hom(BU(n-1)\by BU(1);\Z)\simeq\ast\hom(BU(n))[x]\left/\parens{x^n+c_1x^{n-1}+\dots+c_{n-1}x+c_n}\right.$$
where $c_i=c_i(EU(n))\in\hom^{2i}(BU(n);\Z)$. Since we know the theorem already when $n=1$ (and even when $n=2$), we can inductively assume it holds for $n-1$ (and that $n\ge2$) in order to write
$$\ast\hom(BU(n-1)\by BU(1);\Z)\simeq\ast\hom(BU(n-1);\Z)\otimes\ast\hom(BU(1);\Z)\simeq\Z[c_1',c_2',\dots,c_{n-1}',e_1]$$
where the notation $e_1=c_1(EU(1))\in\hom^2(BU(1);\Z)$ is used to avoid a clash with the earlier defined $c_1=c_1(EU(n))\in\hom^2(BU(n);\Z)$. The classifying map $f:BU(1)^n\to BU(n)$ in the theorem statement factors as
$$BU(1)^n\iso BU(1)^{n-1}\by BU(1)\xto{f'\by\Id}BU(n-1)\by BU(1)\to BU(n)$$
where $f':BU(1)^{n-1}\to BU(n-1)$ is the classifying map for $EU(1)^{\oplus(n-1)}$. Thus, on cohomology, $\pull f:\ast\hom(BU(n);\Z)\to\ast\hom(BU(1)^n;\Z)$, is given by the composition (say $e_1\mapsto\Ith cn_1$)
$$\ast\hom(BU(n);\Z)\into\ast\hom(BU(n);\Z)[x]/(p_n(x))\into\Z[\Ith c1_1,\dots,\Ith cn_1],$$
where
$$p_n(x)=x^n+c_1x^{n-1}+\dots+c_{n-1}x+c_n\in\ast\hom(BU(n);\Z)[x].$$
This shows that $\pull f$ is injective, so we only need to show its image is generated by the elementary symmetric polynomials in $\Ith c1_1,\dots,\Ith cn_1$. The polynomial $p_n(x)$ makes sense as a function on $\Z[\Ith c1_1,\dots,\Ith cn_1]$. As in the $n=2$ case handled earlier, from the construction of this polynomial, we easily see that $p_n(-\Ith cn_1)=0$ (since this is the negation of the Chern class of the line bundle being split off). Again, as in the $n=2$ case, there are automorphisms $BU(1)^n\iso BU(1)^n$ given by permuting the factors; this allows us to put any $\ith c_1$ in the role of the split off line bundle, so actually $p_n(-\ith c_1)=0$ for all $i$. Thus, after viewing $p_n$ in the larger ring $\Z[\Ith c1_1,\dots,\Ith cn_1]$ instead of merely $\ast\hom(BU(n);\Z)$, we see that
$$p_n(x)=\prod_{i=1}^n\parens{x+\ith c_n}.$$
By matching coefficients, this says exactly that $c_i:=c_i(EU(n))$ is the $i$th elementary symmetric polynomial in the $\Ith cj_n$'s. This shows that $\Z[c_1,c_2,\dots,c_n]$ is in the image of $\pull f$. In order to finish, we need to show that this is the entire image. We do this by another counting argument. As before, we have
$$\begin{align*}
\Z[c_1',c_2',\dots,c_{n-1}',e_1]
&\simeq\ast\hom(BU(n-1)\by BU(1);\Z)\\
&\simeq\ast\hom(BU(n);\Z)\otimes\ast\hom(\P^{n-1};\Z)
\supset\Z[c_1,c_2,\dots,c_n]\otimes\Z[x]/(x^n)
\end{align*}$$
additively. Keep in mind that $\deg c_i'=2i$, $\deg c_i=2i$, $\deg e_1=2=\deg x$, and cohomology above vanishes in odd degree. Now, observe that (for any $k\ge0$)
$$\begin{align*}
\sum_{i=0}^\infty\rank_\Z\Z[c_1',c_2',\dots,c_{n-1}']_{2k-2i}
&=\rank_\Z\Z[c_1',c_2',\dots,c_{n-1}',e_1]_{2k}\\
&\ge\rank_\Z(\Z[c_1,c_2,\dots,c_{n-1},c_n][x]/(x^n))_{2k}\\
&=\sum_{i=0}^{n-1}\rank_\Z\Z[c_1,c_2,\dots,c_{n-1},c_n]_{2k-2i}\\
&=\sum_{i=0}^{n-1}\sum_{j=0}^\infty\rank_\Z\Z[c_1,c_2,\dots,c_{n-1}]_{2k-2i-2nj}\\
&=\sum_{i=0}^\infty\rank_\Z\Z[c_1,\dots,c_{n-1}]_{2k-2i}
.
\end{align*}$$
where the first equality above comes from conditioning on the exponent of $e_1$ in a monomial, and the second-to-last one comes from condition on the exponent of $c_n$ in a monomial. Since $\deg c_i=\deg c_i'$ for all $i\le n-1$, we see that all expressions above are equal. This can only mean one thing: the theorem statement must be true!
</div>
<h1 id="chern-classes">Chern Classes</h1>
<p>With the conclusion of the previous section, we have obtained our Chern classes, canonical generators of the cohomology ring $\ast\hom(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]$. In this section, we will collect a few basic properties of these classes. First, for visibility’s sake, let’s throw in a definition block.</p>
<div class="definition">
Let $E\to B$ be a vector bundle with classifying map $f:B\to BU(n)$. Then, for $i\in\{0,1,\dots,n\}$, $E$'s <b>$i$th Chern class</b> is the pullback
$$c_i(E):=\pull fc_i\in\hom^{2i}(B;\Z)$$
of the cohomology class $c_i\in\ast\hom(BU(n);\Z)$ constructed in the previous section. Above, $c_0=1\in\hom^0(BU(n);\Z)\simeq\Z$. Putting these together, $E$'s <b>total Chern class</b> is
$$c(E):=\sum_{i=0}^nc_i(E)=1+c_1(E)+c_2(E)+\cdots+c_n(E)\in\ast\hom(B;\Z).$$
Finally, if $i>n$ or $i<0$, we let $c_i(E)=0$.
</div>
<div class="proposition" name="Properties of Chern classes">
The main properties of Chern classes are the following.
<ol>
<li> They are functorial, i.e. $\pull fc_i(E)=c_i(\pull fE)$ always. </li>
<li> They vanish in large degree, i.e. $c_i(E)=0$ if $i>\rank_\C(E)$. </li>
<li> They vanish for trivial bundles, i.e. $c_i(B\otimes\C^n)=0$ unless $i=0$. </li>
<li> They satisfy a product formula, i.e. $c(E\oplus F)=c(E)c(F)$. </li>
<li> If $E$ is the tautological line bundle on $\CP^n$, then $c_1(E)\in\hom^2(\CP^n;\Z)$ is a generator. </li>
</ol>
</div>
<div class="proof4">
The only thing that really warrants a proof is 4., so we'll prove that. This will be an application of the splitting principal. Note that the formula $c(E\oplus F)=c(E)c(F)$ amounts to some finite number of polynomial relations, i.e. it boils down to the claim that $c_i(E),c_j(F),c_k(E\oplus F)$ satisfy some polynomials
$$p_l(x_1,\dots,x_n,y_1,\dots,y_m,z_1,\dots,z_{n+m})$$
with integral coefficients where $n=\rank E$ and $m=\rank F$. Say $B$ is the base space of $E,F$. By the splitting principal, there exists a space $B'\xto fB$ such that $\pull fE,\pull fF$ both split as a sum of line bundles, and $\pull f:\ast\hom(B;\Z)\into\ast\hom(B';\Z)$ is an injection. By functoriality of Chern classes + injectivity of $\pull f$, this means that
$$\begin{align*}
&&&0
&&=p_l(\pull fc_1(E),\dots,\pull fc_n(E),\pull fc_1(F),\dots,\pull fc_m(F),\pull fc_1(E\oplus F),\dots,\pull fc_{n+m}(E\oplus F))\\
&&&&&=\pull fp_l(c_1(E),\dots,c_n(E),c_1(F),\dots,c_m(F),c_1(E\oplus F),\dots,c_{n+m}(E\oplus F))\\
&\implies &&0
&&= p_l(c_1(E),\dots,c_n(E),c_1(F),\dots,c_m(F),c_1(E\oplus F),\dots,c_{n+m}(E\oplus F))
\end{align*}$$
Thus, it suffices to prove the claim in the case that $E$ and $F$ (and hence $E\oplus F$) are both sums of line bundles. By induction, it then suffices to prove this in the case that $E,F$ are themselves both line bundles. Let $f:B\to BU(1)$ and $g:B\to BU(1)$ be the classifying maps for $E,F$, respectively. By functoriality of Chern classes, we can use $(f,g):B\to BU(1)^2$ to reduce from $E,F$ themselves to sum $EU(1)\oplus EU(1)$ of two copies of the universal line bundle. Now, we are done. We showed at the end of the last section that the product formula holds in this case. Indeed, the product formula says that the Chern classes of $EU(1)\oplus EU(1)$ are given by the elementary symmetric functions of the Chern classes of each factor, but this was shown in the last theorem of the previous section.
</div>
<p>We also want to show that $c_1(L\otimes L’)=c_1(L)+c_1(L’)$ when $L,L’$ are both line bundles. For this, <a href="../brown-rep">recall</a> <sup id="fnref:14"><a href="#fn:14" class="footnote">17</a></sup> that $BU(1)\simeq\CP^\infty$ also represents second integral cohomology. That is, we have natural isomorphisms (of functors)</p>
<script type="math/tex; mode=display">\hom^2(-;\Z)\simeq[-,\CP^\infty]\simeq B(-)</script>
<p>where $[X,\CP^\infty]$ is the set of homotopy classes of maps from $X\to\CP^\infty$ and $B(X)$ is the set of isomorphism classes of principal $U(1)$-bundles. Note that $\hom^2(-;\Z)$ is a functor $\Ho(\push\CW)\to\Ab$; that is, it spits about abelian groups, not just sets. As such, we have an “addition natural transformation” $m:\hom^2(-;\Z)\by\hom^2(-;\Z)\to\hom^2(-;\Z)$ such that, for any $X\in\Ho(\push\CW)$ (say $X$ a topological space with the homotopy type of a CW complex <sup id="fnref:15"><a href="#fn:15" class="footnote">18</a></sup>)</p>
<script type="math/tex; mode=display">m_X:\hom^2(X;\Z)\by\hom^2(X;\Z)\to\hom^2(X;\Z)</script>
<p>is the usual addition map on cohomology. There’s similarly an “inversion natural transformation” $i:\hom^2(-;\Z)\to\hom^2(-;\Z)$ which, on any space $X$, simply sends a cohomology class to its negation. We can use the natural isomorphism $\hom^2(-;\Z)\simeq[-,\CP^\infty]$ to translate these into natural transformations</p>
<script type="math/tex; mode=display">m:[-,\CP^\infty]\by[-,\CP^\infty]\to[-,\CP^\infty]\,\text{ and }\, i:[-,\CP^\infty]\to[-,\CP^\infty]</script>
<p>giving a (functorial) group structure to $[X,\CP^\infty]$ for all $X\in\Ho(\push\CW)$ <sup id="fnref:16"><a href="#fn:16" class="footnote">19</a></sup> <sup id="fnref:17"><a href="#fn:17" class="footnote">20</a></sup>. By Yoneda, these must arise from some morphisms (i.e. homotopy classes of continuous maps)</p>
<script type="math/tex; mode=display">m:\CP^\infty\by\CP^\infty\to\CP^\infty\,\text{ and }\, i:\CP^\infty\to\CP^\infty</script>
<p>giving $\CP^\infty$ the structure of a group object in $\Ho(\push\CW)$ (an $H$-space?). Furthermore, we can run this same game with $B(-)$ in place of $\hom^2(-;\Z)$ since $B(X)$, for varying $X$, also has a (functorial) group structure given by taking the tensor product of line bundles. Thus, we arrive at a second pair of maps</p>
<script type="math/tex; mode=display">m':\CP^\infty\by\CP^\infty\to\CP^\infty\,\text{ and }\, i':\CP^\infty\to\CP^\infty</script>
<p>which also give $\CP^\infty$ the structure of a group object in $\Ho(\push\CW)$. We claim that $(m,i)=(m’,i’)$, so the natural isomorphism $\hom^2(-;\Z)\simeq B(-)$ is an isomorphism of functors $\Ho(\push\CW)\to\Ab$ and not just of functors $\Ho(\push\CW)\to\Set$; put more concretely, our claim gives immediately that $c_1(L\otimes L’)=c_1(L)+c_1(L’)$. In order to prove that $(m,i)=(m’,i’)$, we will show that $\CP^\infty$ has a unique structure as a group object in $\Ho(\push\CW)$.</p>
<p>Recall that $\CP^\infty\simeq K(\Z,2)$ <sup id="fnref:18"><a href="#fn:18" class="footnote">21</a></sup>, and so a multiplication map $\mu:\CP^\infty\by\CP^\infty\to\CP^\infty$ corresponds to some element of</p>
<script type="math/tex; mode=display">\hom^2(\CP^\infty\by\CP^\infty;\Z)\simeq\hom^2(\CP^\infty;\Z)\oplus\hom^2(\CP^\infty;\Z)\simeq\Z c_1\oplus\Z c_1'.</script>
<p>Denote this element by $\mu=ac_1+bc_1’$ with $a,b\in\Z$. Let $e:\bracks{* }\to\CP^\infty$ be the identity morphism of its group object structure (i.e. a choice of identity element), so</p>
<script type="math/tex; mode=display">\mu\circ(e\by\Id)=\mu\circ(\Id\by e)=\Id:\CP^\infty\to\CP^\infty.</script>
<p>In terms of cohomology, these equalities say exactly that $b=a=1$ <sup id="fnref:19"><a href="#fn:19" class="footnote">22</a></sup>, which is to say we must have $\mu=c_1+c_1’$. Thus, $\CP^\infty$ has a unique structure as a group object in $\Ho(\push\CW)$, so $(m,i)=(m’,i’)$, so $c_1(L\otimes L’)=c_1(L)+c_1(L’)$ ultimately by abstract nonsense + $\CP^\infty$ having simple cohomology.</p>
<div class="exercise">
Let $E=L_1\oplus L_2\oplus\dots\oplus L_n$ be a sum of $n$ line bundles. Then,
$$\Wedge^kE=\bigoplus_{1\le i_1<\dots< i_k\le n}L_{i_1}\otimes\cdots\otimes L_{i_k},$$
and so
$$c\parens{\Wedge^kE}=\prod_{1\le i_1<\dots< i_k\le n}c(L_{i_1}\otimes\cdots\otimes L_{i_k})=\prod_{1\le i_1<\dots< i_k\le n}(1+x_{i_1}+\dots+x_{i_k})$$
where $x_i=c_1(L_i)$. Notice that the RHS is invariant under the action of the symmetric group of $n$ elements (i.e. permuting the $x_i$ doesn't change the end result), and so gives a polynomial in the Chern classes $c_1(E),\dots,c_n(E)$ of $E$. That is,
$$c\parens{\Wedge^kE}=Q(c_1(E),\dots,c_n(E))$$
for some $Q\in\Z[t_1,\dots,t_n]$. Using the splitting principle to show the above equality holds for all rank $n$ vector bundles $E$. In particular, conclude that $c_1(\det E)=c_1(E)$ for all vector bundles $E$ where $\det E=\Wedge^{\rank E}E$.
<br />
If you feel like, do the same thing for tensor or symmetric powers of line bundles.
</div>
<div class="remark">
Some people don't like filtering vector bundles by rank (i.e. considering $BU(n)$ for some fixed $n$), and would rather just look at all (finite-rank) vector bundles at once. In order to do this, on first observes that $c(\underline\C^n)=1$, where $\underline\C^n$ is the trivial rank $n$ vector bundle on some base space $B$, since $\underline\C^n$ is the pullback of the vector bundle $\C^n\by\{*\}\to\{*\}$ on the one point space (whose cohomology classes obviously vanish); hence,
$$c(E\oplus\underline\C^n)=c(E)c(\underline\C^n)=c(E),$$
which means that characteristic class are a "stable" phenomenon. In order to leverage this, note that there is a map $BU(n)\to BU(n+1)$ which, in the moduli perspective, sends a rank $n$ vector bundle $E$ to the rank $(n+1)$ vector bundle $E\oplus\underline\C$ (i.e. it's the classifying map for the rank $(n+1)$ vector bundle $EU(n)\oplus\underline\C$ on $BU(n)$). By our observation, this map preserves Chern classes, and so Chern classes continue to be well-defined as cohomology classes of the colimit $BU:=\dirlim_{n\to\infty}BU(n)$. One can show that this space characterizes stable isomorphism classes of vector bundles ($E$ and $E'$ are <b>stably isomorphic</b> if $E\oplus\underline\C^n\simeq E'\oplus\underline\C^m$ for some $n,m\ge0$), and that its cohomology group is
$$\ast\hom(BU)\simeq\Z[c_1,c_2,c_3,\dots]\,\,\text{ with }\,\,\deg c_k=2k.$$
In this way, all Chern class come from a single space, and so one is able to consider all (finite rank) vector bundles (up to stable isomorphism) at once.
</div>
<h1 id="orientation-and-sphere-bundles">Orientation and Sphere Bundles</h1>
<p>Let’s now switch gears. We’ve had some fun defining Chern classes, but didn’t really see how they can be applied to concrete problems. To remedy this defect, we won’t take a look at applications of Chern classes; instead, we’ll define a new characteristic class called the Euler class, and then look at some of its applications.</p>
<p>The Euler class is defined for sphere bundles, but, as we will see later, one can easily make sense of the Euler class of an (oriented) vector bundle as well. For this section, we will be working with real vector bundles instead of complex ones. Do not worry though; we will get back to $\C$ soon enough.</p>
<div class="notation">
Taking inspiration from field theory, given a (real or complex) vector bundle $E\to B$, we will let $\units E\subset E$ denote all nonzero elements of $E$ (so $\units E=E\sm\sigma_0(B)$ where $\sigma_0:B\to E$ is the zero section) and similarly let $\units F\subset F$ denote the nonzero elements of a fiber $F$ which is simply $\R^n$ or $\R^{2n}$, up to homeomorphism.
</div>
<div class="remark">
The cohomological long exact sequence of the pair $(\R^n\sm\{0\},\R^n)$ includes
$$\hom^{k-1}(\R^n;\Z)\too\hom^{k-1}(\R^n\sm\{0\};\Z)\too\hom^k(\R^n,\R^n\sm\{0\};\Z)\too\hom^k(\R^n;\Z)\too\hom^k(\R^n\sm\{0\};\Z).$$
Since $\R^n$ is contractible and $\R^n\sm\{0\}\simeq S^{n-1}$ (homotopy equivalent), we conclude that
$$\hom^k(\R^n,\R^n\sm\{0\};\Z)\simeq\twocases\Z{k=n}0.$$
As a consequence, if $E\to B$ is a rank $n$ real vector bundle, and $F$ is any fiber, then also
$$\hom^k(F,\units F;\Z)\simeq\twocases\Z{k=n}0.$$
</div>
<div class="definition">
Let $\pi:E\to B$ be a real vector bundle of rank $n$. An <b>orientation</b> on $E$ is a "consistent choice of generators of $\hom^n(E_b;\units E_b)$" as $b$ ranges over $B$. That is, it is a choice of generators $u_b\in\hom^n(E_b;\units E_b)$ such that for around each $b\in B$, there is a neighborhood $b\in U\subset B$ and a cohomology class $u_U\in\hom^n(\inv\pi(U),\units{\inv\pi(U)};\Z)$ such that $u_U\vert_{(E_b,\units E_b)}=u_b$ for all $b\in U$. If $E$ admits an orientation, is is called <b>orientable</b>. If it an orientation is chosen, then it is called <b>oriented</b>.
</div>
<div class="remark">
One can equivalently define an orientable real vector bundle as one who can be given transition functions with value in $\SO(n)$. In this perspective, it is easy to see that a real line bundle is orientable iff it is trivial, and that a general real vector bundle $E$ is orientable iff its determinant bundle $\det E:=\Wedge^{\rank E}E$ is orientable (i.e. trivial).
</div>
<p>I claimed the Euler class is naturally defined for sphere bundles, so I should at least say what a sphere bundle is.</p>
<div class="definition">
A <b>sphere bundle</b> $\pi:E\to B$ is a fiber bundle with fiber $F\simeq S^n$ for some $n$. If we want to specify the dimension of the fiber, we may call this an <b>$S^n$-bundle</b> instead. In particular, an $S^0$-bundle is the same thing as a degree 2 covering space.
</div>
<div class="definition">
An $S^n$-bundle $\pi:E\to B$ is called <b>orientable</b> if one can find a compatible choice of local generators for the cohomology of a fiber. That is, it is orientable if there exists cohomology classes $\sigma_b\in\hom^n(E_b;\Z)$ (recall, $E_b:=\inv\pi(B)\simeq S^n$) generating the cohomology of the fiber and satisfying the following condition: for each point $b\in B$, there is some contractible open $U\subset B$ containing $b$ with $E\vert_U\simeq U\by S^n$ which has a generator $\sigma_U\in\hom^n(E\vert_U;\Z)$ such that $\sigma_U\vert_b=\sigma_b$ for all $b\in U$. A choice of compatible local generators is called an <b>orientation</b> and $E$ along with such a choice is said to <b>oriented</b>.
</div>
<div class="remark">
If $E$ has a global cohomology class which restricts to a generator of the cohomology of each fiber, then $E$ is oriented and furthermore, Leray-Hirsch tells us that
$$\ast\hom(E;\Z)\simeq\ast\hom(B;\Z)\otimes\ast\hom(S^n;\Z).$$
</div>
<div class="remark">
The orthogonal group $O(n+1)$ acts on $S^n$, and so some sphere bundles have $O(n+1)$ as their structure group. However, not every sphere bundle is (isomorphic to) one for which this is the case; sometimes the structure group is legitimately bigger than this. However, to define the Euler class, we will need a sphere bundle whose structure group is $SO(n+1)$.
</div>
<p>Before defining Euler classes, let’s tease out the relationship(s) between sphere bundles (the natural homes for Euler classes), real line bundles (the vector bundle analogue of sphere bundles), and complex line bundles (the things I care about).</p>
<div class="definition">
Let $E\to B$ be a rank $(n+1)$ vector bundle. Then, one can form its <b>unit sphere bundle</b> $S(E)$ which is, up to homotopy, just $E$ minus its zero section (i.e. $\units E$). This is an $S^n$-bundle with structure group $O(n+1)$. This is called the unit sphere bundle because, if you give $E$ a Riemannian metric, it is isomorphic to the sphere bundle obtained by restricting each fiber of $E$ to its unit sphere.
</div>
<div class="proposition">
A vector bundle $E\to B$ is orientable if and only if $S(E)$ is orientable.
</div>
<div class="proof4">
This is really just an earlier remark that, by looking at the long exact sequence of the pair $(F,\units F)$, we have a natural isomorphism
$$\hom^{k-1}(\units F;\Z)\iso\hom^k(F,\units F;\Z).$$
This let's you carry orientation back and forth between $E$ and $S(E)$.
</div>
<div class="corollary">
A vector bundle $E\to B$ is orientable if and only if $S(\det E)$ is a disconnected double cover of $B$.
</div>
<div class="corollary">
Every vector bundle over a simply connected base is orientable.
</div>
<p>What about complex vector bundles? Well, given a complex vector bundle $E\to B$ of rank $n$, let $E_\R\to B$ be its underlying real vector bundle (of rank $2n$). This bundle is always orientable (and even oriented). Let $e_1,\dots,e_n$ be a complex basis of $E_b$, so $e_1,ie_i,e_2,ie_2,\dots,e_n,ie_n$ gives a real basis of $(E_\R)_ b$. This defines a canonical orientation of this fiber <sup id="fnref:21"><a href="#fn:21" class="footnote">23</a></sup>. One can turn this into a cohomology class by considered an (orientation-preserving) map $\Delta^n\to((E_\R)_ b,\punits{E_\R}_ b)$ which represents a homology class $\alpha$, and then considering the unique cohomology class $u_b\in\hom^n((E_\R)_ b,\punits{E_\R}_ b)$ which is a generator and satisfies $\angles{u_b,\alpha}=1$. From naturality of this construction (i.e. it not depending on the complex basis you choose in the beginning), one can show that it gives an orientation of $E_\R\to B$. Hence, for any complex vector bundle $E$, we can obtain an oriented sphere bundle $S(E_\R)$, and so we will be able to make sense of the Euler class of a complex vector bundle.</p>
<p>This concludes our prelims on sphere bundles, so let’s actually define the Euler class now.</p>
<h1 id="euler-class">Euler Class</h1>
<p>We’ll construct the Euler class in a bit of an unusual way. I think one usually first proves a “Thom isomorphism” relating the cohomology of the total space of an oriented real vector bundle $E\to B$ with that of the pair ($E,\units E$) (shifted up $\rank E$ places) via cupping with an “orientation class” $u\in\hom^n(E,\units E;\Z)$. Then, one uses the natural maps $\hom^n(E,\units E;\Z)\to\hom^n(E;\Z)\iso\hom^n(B;\Z)$ to define the Euler class as the image of $u$ in $\hom^n(B;\Z)$. However, I looked up a proof of the existence of this orientation class, and the one I saw seemed long and annoying to the read… so um, we’re gonna do something else and hope no complications arise when we try to prove things about it.</p>
<p>We’ll construct the Euler class via an exercise in my <a href="../spectral-seq">spectral sequence post</a>. Namely, we’ll show that oriented sphere bundles come equipped with a natural long exact sequence in cohomology, and then the Euler class will arise as the image of $1\in\hom^0(B;\Z)$ under a map appearing in this exact sequence.</p>
<div class="theorem" name="Gysin sequence">
Let $S^n\to E\to B$ be an oriented $S^n$ bundle. Then, there exists an exact sequence
$$\cdots\too\hom^p(B;\Z)\xtoo\alpha\hom^p(E;\Z)\xtoo\beta\hom^{p-n}(B;\Z)\xtoo\d\hom^{p+1}(B;\Z)\too\cdots$$
</div>
<div class="proof4">
We'll construct this using the Serre spectral sequence, so we first need to know that $\pi_1(B)$ acts trivially on $\ast\hom(S^n;\Z)$. This follows from $E$ being oriented. Locally, $E$ looks like $S^n\by U$ (for $U\subset B$ open) and so loops on $B$ locally do nothing to $\ast\hom(S^n)$. Since we have a consistent choice of generator for $\hom^n(S^n;\Z)$ (the only nonzero (reduced) cohomology group), any $\gamma\in\pi_1(B)$ will take the preferred generator on $S^n\by U$ to the preferred generator on $S^n\by V$ when in the overlap $U\cap V$, and so we see that $\gamma$ acts by the identity on cohomology.
<br />
We can know form the Serre spectral sequence $E_2^{p,q}=\hom^p(B;\hom^q(S^n;\Z))\implies\hom^{p+q}(E)$. We have $E_2^{p,q}=0$ when $q\not\in\{0,n\}$ and so the only page with non-trivial differentials (remember the $E_k$ page has differentials with bidegree $(k,1-k)$) is the $E_{n+1}$ page. Its differentials look like
$$\begin{CD}
0 @>>> E_{n+1}^{p,n} @>\d_{n+1}>> E_{n+1}^{p+n+1,0} @>>> 0\\
@. @| @|\\
0 @>>> \hom^p(B;\Z) @>>> \hom^{p+n+1}(B;\Z) @>>> 0
\end{CD}$$
Thus, the objects of the $E_{n+2}=E_\infty$ page are
$$E_\infty^{p,q}=\begin{cases}
\hom^p(B;\Z)/\im\d_{n+1} &\text{if }q=0\\
\ker\d_{n+1} &\text{if }q=n\\
0&\text{otherwise}
\end{cases}$$
Thus, the only nonzero objects of the $p$-diagonal of the $E_\infty$-page are
$$E_\infty^{p,0}=\hom^p(B;\Z)/\im\d_{n+1}\,\,\text{ and }\,\,E_\infty^{p-n,n}=\ker\d_{n+1}.$$
The elements of the $p$-diagonal of the $E_\infty$-page are successive quotients in a filtration
$$0\subset F^p\hom^p(E;\Z)\subset F^{p-1}\hom^p(E;\Z)\subset\cdots\subset F^0\hom^p(E;\Z)=\hom^p(E;\Z),$$
so but most of quotients are $0$. In particular, $E_\infty^{k,p-k}=F^k\hom^p(E)/F^{k+1}\hom^p(E)=0$ for all $k\not\in\{p,p-n\}$, so the above filtration really looks like
$$0\subset F^p\hom^p(E;\Z)=\cdots=F^{p-n+1}\hom^p(E;\Z)\subset F^{p-n}\hom^p(E;\Z)=\cdots=\hom^p(E;\Z).$$
Since $F^p\hom^p(E;\Z)\simeq E_\infty^{p,0}$, we have the following exact sequence
$$\begin{CD}
0 @>>> F^{p-n+1}\hom^p(E;\Z) @>>> F^{p-n}\hom^p(E;\Z) @>>> E_\infty^{p-n,n} @>>> 0\\
@. @| @| @| \\
0 @>>> \hom^p(B;\Z)/\im\d_{n+1} @>\alpha'>> \hom^p(E;\Z) @>\beta'>> \ker\d_{n+1} @>>> 0
\end{CD}$$
At this point, we are basically done. Consider the compositions
$$\alpha:\hom^p(B;\Z)\to\hom^p(B;\Z)/\im d_{n+1}\xto{\alpha'}\hom^p(E;\Z)$$
and
$$\beta:\hom^p(E;\Z)\xto{\beta'}\ker d_{n+1}\into\hom^{p-n}(B;\Z).$$
These fit into a sequence
$$\cdots\too\hom^{p-n-1}(B;\Z)\xtoo{d_{n+1}}\hom^p(B;\Z)\xtoo{\alpha}\hom^p(E;\Z)\xtoo{\beta}\hom^{p-n}(B;\Z)\xtoo{d_{n+1}}\hom^{p+1}(B;\Z)\too\cdots$$
which we claim is exact. First, $\alpha'$ is injective, so we have $\ker\alpha=\im\d_{n+1}$ which gives exactness at $\hom^p(B;\Z)$. Second, $\ker\beta=\ker\beta'=\im\alpha'=\im\alpha$ so we get exactness as $\hom^p(E;\Z)$ as well. Finally, $\beta'$ is surjective so $\im\beta=\ker\d_{n+1}$ which gives exactness at $\hom^{p-n}(B;\Z)$. Thus the above long sequence is exact everywhere, and we win.
</div>
<div class="definition">
Let $E\to B$ be an oriented $S^n$-bundle. Then its <b>Euler class</b> $e(E)\in\hom^{n+1}(B;\Z)$ is defined as the image of $1\in\hom^0(B;\Z)$ under the map $\d:\hom^0(B;\Z)\to\hom^{n+1}(B;\Z)$ constructed above (this is the differential on the $E_{n+1}$-page of the Serre spectral sequence).
</div>
<div class="remark">
The reason the above definition is well-defined is that the $\hom^0(B;\Z)$ in this long exact sequence is really $\hom^0(B;\hom^n(S^n;\Z))$ which has a canonical generator (i.e. canonical choice of $1$) exactly because $S^n\to E\to B$ is oriented. When I say that $e(E)=\d(1)$ (the image of $1\in\hom^0(B;\Z)$), this is really saying that $e(E)=\d(u)$ where $u\in\hom^n(S^n;\Z)$ (recall $\hom^0(B;\hom^n(S^n;\Z))\simeq\hom^n(S^n;\Z)$ with the isomorphism once again canonical) is the preferred generator.
</div>
<div class="remark">
Note that if $e(E)=0$, then by exactness, there's some $a\in\hom^n(E;\Z)$ such that $\beta(a)=1\in\hom^0(B;\Z)$, but as we mentioned above $\hom^0(B;\Z)\simeq\hom^n(S^n;\Z)$ with $1$ if the former corresponding to the preferred generator $u$ of the latter. Thus, if $e(E)=0$, we get a global cohomology class $a\in\hom^n(E;\Z)$ restricting to the preferred generator on each fiber. In this way, the Euler class is an "obstruction to the existence of such a global class" or a "measure of how twisted $E$ is" or "something like that." Note that, by Leray-Hirsch, this shows that $e(E)=0$ entails that $\ast\hom(E;R)$ is a free $\ast\hom(B;R)$-module with basis $u\in\hom^n(S^n;\Z)$, i.e. there are isomorphisms
$$\hom^k(B;\Z)\iso\hom^{n+k}(E;\Z)$$
sending $\hom^k(B;\Z)\ni\alpha\longmapsto\pull\pi(\alpha)\smile a\in\hom^{n+k}(E;\Z)$ when $e(E)=0$.
</div>
<div class="remark">
Let $E\to B$ be an oriented $S^n$-bundle, and let $\bar E\to B$ denote $E$ with the reverse orientation. This amounts to negating the choice of preferred generator of $\hom^n(S^n;\Z)$, so we see that $e(\bar E)=-e(E)$.
</div>
<div class="exercise">
I don't know how hard this is because I've never tried doing it myself, but show that the map $\d:\hom^k(B;\Z)\to\hom^{k+n+1}(B;\Z)$ above is given by taking the cup product with the Euler class.<br />Also show that $\alpha:\hom^p(B;\Z)\to\hom^p(E;\Z)$ is just the pullback $\pull\pi$ along the projection $\pi:E\to B$.<br />When $p=n$, the sequence includes the map $\beta:\hom^n(E;\Z)\to\hom^0(B;\Z)$. Show that composing this with the natural map $\hom^0(B;\Z)\iso\hom^n(S^n;\Z)$ gives the usual restriction map $\hom^n(E;\Z)\to\hom^n(S^n;\Z)$.
</div>
<p>The only property of the Euler class that I think we will need to know for now is that it is functorial. The point here is that the Serre spectral sequence is itself functorial (this is clear from its construction), and so given a pullback</p>
<script type="math/tex; mode=display">\begin{CD}
S^n @>>> \pull fE @>>> B'\\
@| @VVgV @VVfV \\
S^n @>>> E @>>> B
\end{CD}</script>
<p>of sphere bundles, one obtains a commutative square.</p>
<script type="math/tex; mode=display">\begin{CD}
\hom^0(B';\Z) @>\d>> \hom^{n+1}(B';\Z)\\
@A\pull fAA @AA\pull fA \\
\hom^0(B;\Z) @>>\d> \hom^{n+1}(B;\Z)
\end{CD}</script>
<p>The left vertical map sends $1\mapsto1$, and so by commutativity, we see that</p>
<script type="math/tex; mode=display">\pull fe(E)=\pull f\d(1)=\d\pull f(1)=\d(1)=e(\pull fE),</script>
<p>which says exactly that the Euler class is functorial. <sup id="fnref:23"><a href="#fn:23" class="footnote">24</a></sup></p>
<p>Now that we have seen a direct construction of the Euler class, let’s return to this whole “Thom isomorphism” thing I alluded to earlier. The main point of this thing was/is to construct a Thom/orientation class $u\in\hom^n(E,\units E;\Z)$ for an oriented vector bundle $E\to B$. I prefer to think in terms of sphere bundles, so we are really after a certain cohomology class $u\in\hom^n(D(E), S(E);\Z)$ where, given a rank $n$ oriented vector bundle $E\to B$, $S(E)$ as usual denotes its associated unit sphere bundle and $D(E)$ denotes its analogously defined unit disk bundle $D(E)\to B$ (with fibers homeomorphic to the $n$-disk, $D^n$). Let’s take things one step further. Say, as has been the case in this section, we start with an oriented $S^n$ bundle $E\to B$ instead. How should we obtain a Thom class now? The first thing we would like to do is “fill in” the fibers of this bundle in order to obtain a $D^{n+1}$-bundle $D(E)\to B$ with $E$ as its “fiberwise boundary.” Then, the Thom class will be a certain cohomology class $u\in\hom^{n+1}(D(E),E;\Z)$ <sup id="fnref:33"><a href="#fn:33" class="footnote">25</a></sup>.</p>
<div class="proposition">
Let $\pi:E\to B$ be an $S^n$-bundle. Then, there exists a $D^{n+1}$-bundle $p:D(E)\to B$ with an inclusion $E\into D(E)$ realizing $\inv\pi(b)$ as the boundary of $\inv p(b)$ for all $b\in B$.
</div>
<div class="proof4">
Let's begin with a bit of intuition; the actual proof will be rather short. We want to "fill in" the fibers of $\pi$, and topologically, this corresponds to forming the cones $C\inv\pi(b)\cong CS^n\cong D^{n+1}$ where, in general for a top. space $X$,
$$CX:=(X\by I)/(X\by\{0\}).$$
Hence, one way to form $D(E)$ would be simply take the cones on each fiber (over a local trivialization) and then glue these together by lifting $\pi$'s transition functions from the fibers to their cones. However, it would be preferable to perform some global construction directly to $E$ which replaces all fibers with their cones. The key observation here is as follows: to form the cone of a fiber, you first form its cylinder and then collapse one end of the cylinder (i.e. one copy of the fiber) to a point. Since the map $\pi:E\to B$ already collapses each fiber to a (different) point, you can form the "global" cylinder $E\by I$ and then simply identity points by their $\pi$-image in $B$ in order to form the cones of all the fibers at once.
<br />
With that said, let $D(E):=M_\pi$ be the mapping cylinder
$$M_\pi:=(E\by I)\cup_{(E\by\{0\})}B=\parens{(E\by I)\sqcup B}/((e,0)\sim\pi(e)),$$
and let $p:D(E)\to B$ be the natural "projection to the base" map (i.e. $p(e,t)=\pi(e)$ for $(e,t)\in E\by I$ and $p(b)=b$ when $b\in B$). Then, by construction, for any $b\in B$, we have
$$\inv p(b)=M_{\pi\vert_{\inv\pi(b)}}=(\inv\pi(b)\by I)\cup_{(\inv\pi(b)\by\bracks0)}\bracks b=(\inv\pi(b)\by I)/(\inv\pi(b)\by\bracks0)=C\inv\pi(b)\cong D^{n+1}.$$
Furthermore, by restricting the locally trivial neighborhoods for $\pi$, we see that $p$ is still a fiber bundle. Finally, the map $E\into D(E)$ is simply $e\mapsto(e,1)$.
</div>
<p>At this point, we will take some things on faith to avoid repeating ourselves <sup id="fnref:34"><a href="#fn:34" class="footnote">26</a></sup>. Let $\pi:E\to B$ be an oriented $S^n$-bundle as usual. By the proposition above, we have a “fiber sequence pair” $(D^{n+1}, S^n)\to(D(E), E)\to B$. Given this, just as we used the Serre spectral sequence for the fibration $S^n\to E\to B$ in order to construct the Euler class, one can use the Serre spectral for the fiber sequence pair $(D^{n+1}, S^n)\to(D(E), E)\to B$ (i.e. a spectral sequence $E_2^{p,q}=\hom^p(B;\hom^q(D^{n+1},S^n;\Z))\implies\hom^{p+q}(D(E),E;\Z)$) to construct a Thom class $u=u(E)\in\hom^{n+1}(D(E),E;\Z)$ <sup id="fnref:35"><a href="#fn:35" class="footnote">27</a></sup>. This Thom class is constructed so that the map <sup id="fnref:37"><a href="#fn:37" class="footnote">28</a></sup></p>
<script type="math/tex; mode=display">\mapdesc{\phi}{\hom^p(B;\Z)}{\hom^{p+n+1}(D(E),E;\Z)}{\alpha}{\pull\pi(\alpha)\smile u}</script>
<p>is an isomorphism for all $p$. Furthermore, this map gives the below isomorphism of exact sequences between the Gysin sequence and the long exact sequence for the pair $(D(E),E)$. Recall that $(D^{n+1},S^n)\to(D(E),E)\xto{(p,\pi)}B$ is our pair of fiber bundles over $B$, and that the fibration $p:D(E)\to B$ is a homotopy equivalence since the fiber is contractible.</p>
<script type="math/tex; mode=display">\begin{CD}
\cdots @>>> \hom^p(B;\Z) @>\pull\pi>> \hom^p(E;\Z) @>>> \hom^{p-n}(B;\Z) @>\smile e(E)>> \hom^{p+1}(B;\Z) @>>> \cdots\\
@. @V\pull pVV @V\Id VV @VV\phi V @VV\pull pV\\
\cdots @>>> \hom^p(D(E);\Z) @>>> \hom^p(E;\Z) @>>> \hom^{p+1}(D(E),E;\Z) @>>> \hom^{p+1}(D(E);\Z) @>>> \cdots
\end{CD}</script>
<p>In particular, the Euler class $e(E)\in\hom^{n+1}(B;\Z)$ is the preimage of the restriction $u(E)\vert_{D(E)}\in\hom^{n+1}(D(E);\Z)$ of the Thom class to the total space.</p>
<h2 id="relation-to-chern-classes">Relation to Chern Classes</h2>
<p>We haven’t thought about classifying spaces in a while; let’s change that. Let $E\to B$ be a rank $n$ complex vector bundle. This gives rise to an oriented $S^{2n-1}$-bundle $S(E_\R)\to B$, and so we can define the <b>Euler class</b> of the complex vector bundle $E\to B$ as</p>
<script type="math/tex; mode=display">e(E):=e(S(E_\R))\in\hom^{2n}(B;\Z).</script>
<p>We saw at the end of the last section that the Euler class is functorial as a cohomology class of sphere bundles, and this extends to its functoriality as a cohomology class of complex vector bundles. Thus, just as with all characteristic classes of complex vector bundles, the Euler class (on complex vector bundles) must really correspond to some universal cohomology class</p>
<script type="math/tex; mode=display">e\in\hom^{2n}(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]_ {2n}\,\,\,\,\text{ where }c_i\in\hom^{2i}(BU(n);\Z).</script>
<p>We aim to figure out which class it is. Necessarily, $e=p(c_1,c_2,\dots,c_n)$ is some polynomial in the Chern classes, and so we just need to determine which one it is. Determining a polynomial in Chern classes is exactly the type of situation one uses the splitting principle for, and so we are instantly reduced to determining the Euler class of a sum $L_1\oplus\dots\oplus L_n$ of line bundles. This will have two parts; what’s the Euler class of a line bundle, and what’s the Euler class of a sum of 2 line bundles? Once we know these, we just induct and have our answer in general.</p>
<p>We will start with computing the Euler class of a sum of line bundles, so let $L_1\xto{\pi_1}B$ and $L_2\xto{\pi_2}B$ be complex line bundles. Note that their “internal direct sum” $L_1\oplus L_2\to B$ is the pullback of their “external direct sum” $L_1\by L_2\to B\by B$ along the diagonal map $\Delta:B\to B\by B,b\mapsto(b,b)$. That is, we have a pullback diagram.</p>
<script type="math/tex; mode=display">\begin{CD}
L_1\oplus L_2 @>>> L_1\by L_2\\
@V\pi_1\oplus\pi_2VV @VV\pi_1\by\pi_2V\\
B @>>\Delta> B\by B
\end{CD}</script>
<p>Now, we can compute $e(L_1\oplus L_2)$ using functoriality + knowledge of the cohomology of a product. We have $\ast\hom(B\by B;\Z)\simeq\ast\hom(B;\Z)\otimes\ast\hom(B;\Z)$ and, by considering the two projections $B\by B\rightrightarrows B$, one sees that $e(L_1\by L_2)=e(L_1)\otimes e(L_2)\in\hom^4(B\by B;\Z)$. Pulling back along the diagonal maps corresponds to taking cup products, so</p>
<script type="math/tex; mode=display">e(L_1\oplus L_2)=\pull\Delta e(L_1\by L_2)=e(L_1)\smile e(L_2)\in\hom^4(B;\Z).</script>
<p>Thus, in general, the Euler class $e(L_1\oplus L_2\oplus\cdots\oplus L_n)$ of a sum is the product $e(L_1)e(L_2)\dots e(L_n)$ of the Euler classes.</p>
<p>We still need to determine the Euler class of a line bundle in terms of its first Chern class. That is, it is clear that $e(L)=kc_1(L)$ for some $k\in\Z$ independent of $L$, but we still don’t know $k$ <sup id="fnref:24"><a href="#fn:24" class="footnote">29</a></sup>. Luckily, we can determine $k$ by looking at a simple universal case. Recall that $BU(1)\simeq\CP^\infty$, and let $\iota:\CP^1\into\CP^\infty\simeq BU(1)$ be the natural inclusion map, so $\pull\iota:\hom^2(BU(1);\Z)\iso\hom^2(\CP^1;\Z)$ is an isomorphism. Thus, we are reduced to determining the Euler class of the tautological line bundle $E\to\CP^1$ on $\CP^1$. This is the line bundle whose fiber above a point $\l\in\CP^1$ is the line $\l\subset\C^2$ represented by that point. Hence, by definition, we see that the sphere bundle $S(E_\R)$ associated to it is the Hopf fibration</p>
<script type="math/tex; mode=display">S^1\to S^3\to S^2.</script>
<p>The Euler class of this bundle comes from the differential on the $E_2$-page of its Serre spectral sequence. That page looks like</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{array}{c | c c c c}
\tbf q & \\
1 & \hom^0(S^2;\hom^1(S^1))\simeq\Z a & 0 & \hom^2(S^2;\hom^1(S^1))\simeq\Z c_1a\\
0 & \hom^0(S^2;\hom^0(S^1))\simeq\Z & 0 & \hom^2(S^2;\hom^0(S^1))\simeq\Z c_1 \\\hline
& 0 & 1 & 2 & \tbf p
\end{array} %]]></script>
<p>where $a\in\hom^2(S^1)$ is the preferred generator and $c_1\in\hom^2(S^2)$ is the first Chern class of the tautological line bundle $E\to\CP^1\simeq S^2$. The only nontrivial differential on this page (or any page thereafter) is $\d_2:\Z a\to\Z c_1$. Since neither $\Z a$ nor $\Z c_1$ survive to the $E_\infty$-page (the $1$- and $2$-diagonals of the $E_\infty$-page are all $0$ since $\hom^2(S^3)=\hom^1(S^3)=0$), we see that this map must be an isomorphism, so $d_2(a)=\pm c_1$. Since, by definition, $e(E)=d_2(a)$, we see that $e(E)=\pm c_1(E)$.</p>
<p>Thus, for a general complex line bundle $L\to B$, we have $e(L)=\pm c_1(L)$. In particular, if $L$ is given the “correct” orientation, then $e(L)=c_1(L)$. Since $L$, being a complex line bundle, comes equipped with a canonical orientation, one suspects that this one (its complex orientation) is the correct one, and so that we can safely write $e(L)=c_1(L)$ (implicitly endowing $L$ with its complex orientation). This is indeed the case, but I’m not sure if we will be able to show it <sup id="fnref:27"><a href="#fn:27" class="footnote">30</a></sup>.</p>
<p>In any case, we can cheat. Let’s just adopt the convention that when we write $e(L)$ for $L$ a complex line bundle, we’re implicitly taking its Euler class with respect to the “correct” orientation (i.e. always $L$’s complex orientation or never $L$’s complex orientation), and, having adopted this convention, we can now safely write $e(L)=c_1(L)$. Combining this with the fact that $e(L_1\oplus L_2)=e(L_1)e(L_2)$ and with the splitting principal, one sees that for a general complex vector bundle $E\to B$, we have <sup id="fnref:28"><a href="#fn:28" class="footnote">31</a></sup></p>
<script type="math/tex; mode=display">e(E)=c_n(E).</script>
<h2 id="relation-to-obstruction-theory">Relation to Obstruction Theory</h2>
<p>This sign issue we ran into will be resolved in the next section. In the current one, we’ll look at what might be our first actual application of characteristic classes in this post: the Euler class gives an obstruction to the existence of (non-vanishing) sections. In other words, if your Euler class is nonzero, then your sphere bundle has no sections.</p>
<div class="theorem">
Let $S^n\to E\xto\pi B$ be an oriented sphere bundle with a section $\sigma:B\to E$. Then, $e(E)=0\in\hom^{n+1}(B;\Z)$.
</div>
<div class="proof4">
Recall that the Euler class originates from the Gysin sequence
$$\cdots\too\hom^0(B;\Z)\xtoo\d\hom^{n+1}(B;\Z)\xtoo\alpha\hom^{n+1}(E;\Z)\too\cdots$$
where, as per an exercise, $\alpha=\pull\pi:\ast\hom(B;\Z)\to\ast\hom(E;\Z)$. Since $\pi\circ\sigma=\Id_B$, we see that $\pull\sigma\circ\pull\pi=\Id_{\ast\hom(B;\Z)}$ so $\pull\pi$ is injective (in all degrees). By exactness of the above sequence, this shows that $\d:\hom^0(B;\Z)\to\hom^{n+1}(B;\Z)$ is the zero map, so $e(E)=\d(1)=0$.
</div>
<div class="corollary">
Let $E\to B$ be an oriented real vector bundle with a non-vanishing section. Then, $e(E)=0$.
</div>
<div class="corollary">
Let $E\to B$ be a complex vector bundle with a non-vanishing section. Then, $c_n(E)=0$.
</div>
<p>One can naturally wonder if the converse is true. That is, if $e(E)=0$, then must there necessarily be a section of $\pi$? The answer to this turns out to be no, and the issue is essentially that $S^n$ has higher homotopy groups <sup id="fnref:29"><a href="#fn:29" class="footnote">32</a></sup> beyond the $\pi_n(S^n)\simeq\Z$ from which the Euler class ultimately originates. In general, when faced with lifting problems like this (can you lift a map against a fibration, possibly extending an initial lift on some subspace), one obtains a sequence $\omega_k\in\hom^{k+1}(\text{blah};\pi_k(F))$ ($F$ the fiber) of cohomology classes such that a lift exists iff all of these cohomology classes vanish. In this framework, the Euler class $e(E)\in\hom^{n+1}(B;\Z)=\hom^{n+1}(B;\pi_n(S^n))$ is what’s called a <b>primary obstruction class</b> since it is the first one which can be nonzero in the situation of constructing a section of a sphere bundle <sup id="fnref:32"><a href="#fn:32" class="footnote">33</a></sup>.</p>
<h2 id="relation-to-enumerative-geometry">Relation to Enumerative Geometry</h2>
<p>This is the most exciting part of this whole post. The Euler class. can. be. used. to. count.</p>
<p>We will make sense of this in this section, and then end the post with an example. By using the Euler class to count, I mean that the Euler class, in nice situations, encodes the number of zeros of a generic section of its vector bundle. Essentially, we will refine the result that a bundle with a section with $0$ zeros has Euler class equal to $0$.</p>
<p>For this application, we will need not just the Euler class itself, but also the Thom class. Before, proving things, let’s recall Poincare duality.</p>
<div class="theorem" name="Poincare duality">
Let $M$ be a compact, oriented $n$-manifold with (possibly empty) boundary. Then, there is a fundamental class $[M]\in\hom_n(M,\del M)$ such that
$$\mapdesc{P}{\hom^k(M,\del M;\Z)}{\hom_{n-k}(M;\Z)}{\alpha}{[M]\frown\alpha}$$
is an isomorphism for all $k$. Above $\frown:\hom_m(M,\del M)\by\hom^k(M,\del M)\to\hom_{m-k}(M;\Z)$ denotes the cap product.
</div>
<p>As suggested by the fact that we recalled the above theorem, for the results of this section, we will need to assume our base space in a compact manifold. Given this, we will show that the Euler class of an oriented real vector bundle is Poincare dual to the zero set of a generic section of the bundle. When the dimension of the base space equals the rank of the bundle, the a generic section will have a zero-dimensional zero set (i.e. a finite, discrete set of zeros), and so in that case, the Euler class will simply count the number of zeros.</p>
<div class="lemma">
Let $S^n\to E\to M$ be an oriented $S^n$-bundle over a compact $k$-manifold. Let $\sigma_0:M\to D(E)$ denote the zero section of the associate disk bundle. Then,
$$\sigma_{0,*}[M]=\pm[D(E)]\frown u(E)\in\hom_k(D(E);\Z),$$
i.e. the Thom class is Poincare dual to the zero section of $p:D(E)\to M$.
</div>
<div class="proof4">
Non-connected spaces don't exist, so assume $B$ connected. Note that $D(E)$ is $(k+n+1)$-dimensional (oriented) compact manifold with boundary $E$. Hence, we have isomorphisms
$$\Z=\hom^0(M;\Z)\xto{\pull\pi(-)\smile u}\hom^{n+1}(D(E),E;\Z)\xto{[D(E)]\frown}\hom_k(D(E);\Z)\xto{\push p}\hom_k(M;\Z)=\Z.$$
The generator $1\in\hom^0(M;\Z)$ maps to $[D(E)]\frown u\in\hom_k(D(E);\Z)$ which must be a generator of $\hom_k(D(E);\Z)$. At the same time, $\sigma_0:M\to D(E)$ is a homotopy equivalence and so sends the generator $[M]\in\hom_k(M;\Z)$ to a generator $\sigma_{0,*}[M]\in\hom_k(D(E);\Z)$. The claim follows.
</div>
<p>The above lemma is our first inclination that the Thom class (and hence the Euler class) has anything to do with sections of its bundle. Next, we will show how one can use Thom classes to construct the cohomology class dual to a submanifold $N\subset M$. For notational convenience, given a compact, oriented $k$-dimensional submanifold $\iota:N\into M$ (here, $\dim M=n$), let $\ast{[N]}\in\hom^{n-k}(M;\Z)$ denote the Poincare dual to $\push\iota[N]\in\hom_k(M;\Z)$.</p>
<p>Now, let’s take a very brief detour into the theory of vector bundles on smooth manifolds. Let $N\into M$ be compact, oriented smooth manifolds of dimensions $k$ and $n$, respectively. Then, there exists vector bundles $TN\to N$ and $TM\to M$ of ranks $k$ and $n$, respectively, called the tangent bundles of $N$ and $M$. In particular, $TN$ is a subbundle of $TM\vert_N$, the tangent bundle of $M$ restricted to $N$, and so fits into an exact sequence (of vector bundles on $N$)</p>
<script type="math/tex; mode=display">0\too TN\too TM\vert_N\too N_{N/M}\too0,</script>
<p>where the rank $(n-k)$ vector bundle $N_{N/M}$ is by definition the <b>normal bundle of $N$ in $M$</b>. Intuitively, the tangent bundle $TM$ contains all the directions one can move along in $M$ (and similarly for $TN$), so the normal bundle $N_{N/M}$ contains all the directions in $M$ which are perpendicular to $N$. The amazing fact is that there exists a “tubular neighborhood” $U\subset M$ of $N$ with a smooth embedding $U\into N_{N/M}$ sending $N\subset U\subset M$ to the zero section in $N_{N/M}$; in other words, the (unit disk bundle of) the normal bundle embeds back into the manifold. Accepting this, attached to $N\into M$ is a Thom class</p>
<script type="math/tex; mode=display">u_N:=u(N_{N/M})\in\hom^{n-k}(D(N_{N/M},S(N_{N/M});\Z)\simeq\hom^{n-k}(U,U\sm N;\Z).</script>
<p>By excision, we have an isomorphism $\hom^{n-k}(M,M\sm N;\Z)\to\hom^{n-k}(U,U\sm N;\Z)$, and this former space naturally restricts to $\hom^{n-k}(M;\Z)$. Let $u_N^M$ denote the image of $u(N_{N/M})$ under the composite map</p>
<script type="math/tex; mode=display">\hom^{n-k}(D(N_{N/M}),S(N_{N/M});\Z)\iso\hom^{n-k}(U,U\sm N;\Z)\iso\hom^{n-k}(M,M\sm N;\Z)\to\hom^{n-k}(M;\Z).</script>
<p>This $u_N^M$ is the Poincare dual of $N$.</p>
<div class="lemma">
Let $N\into M$ be the inclusion of a closed, oriented $k$-dimensional manifold into a compact, oriented $n$-dimensional manifold. Then,
$$\ast{[N]}=\pm u_N^M\in\hom^{n-k}(M;\Z).$$
</div>
<div class="proof4">
Let $\iota:N\into M$ be the inclusion map. It is equivalent to show that
$$\push\iota[N]=[X]\frown u_N^M\in\hom_k(M;\Z).$$
For this, we consider the following commutative diagram (commutativity from naturality of the cap product)
$$\begin{CD}
\hom_n(M;\Z)\otimes\hom^{n-k}(M,M\sm N;\Z) @>>> \hom_n(M,M\sm N;\Z)\otimes\hom^{n-k}(M,M\sm N;\Z) @>>> \hom_n(U, U\sm N;\Z)\otimes\hom^{n-k}(U, U\sm N;\Z)\\
@VVV @V\frown VV @VV\frown V\\
\hom_n(M;\Z)\otimes\hom^{n-k}(M;\Z) @>\frown >>\hom_k(M;\Z) @<<< \hom_k(U)
\end{CD}$$
where $U\subset M$ is a tubular neighborhood of $N$. Start with $[M]\otimes u_N^M$ in the bottom left. By construction of $u_N^M$, we can lift this to the top left and then pass to the top right, ending up with $[U]\otimes u_N$. By previous lemma, $[U]\frown u_N=\pm\push\iota[N]\in\hom_k(U)$ (since $\iota:N\to U$ is identified with the zero section of $N_{N/M}\to N$). Thus, by commutativity of the diagram, we must have
$$[M]\frown u_N^M=[U]\frown u_N=\pm\push\iota[N]\in\hom_k(M)$$
as claimed.
</div>
<p>If one is more careful above (actually, more careful in the first lemma of this section), then they can remove the $\pm$, and just conclude that $\ast{[N]}=\pm u_N^M\in\hom^{n-k}(M;\Z).$ However, what we have above is good enough for our purposes; in the end, we’ll want to count zeros of a section, and so we’ll know that the correct answer will be a positive number. We’ll obtain a result saying that the Euler class computes this count up to sign, so we can always get the correct result by computing the Euler class and then taking the absolute value of what we get.</p>
<div class="theorem">
Let $E\to M$ be a smooth, oriented rank $n$ real vector bundle over a compact, oriented manifold $M$. Let $\psi$ be a section whose graph is transverse to the zero section (i.e. a "generic section"), and let $Z=\inv\psi(0)\subset B$. Then,
$$e(E)=\ast{[Z]}\in\hom^n(B;\Z).$$
</div>
<div class="proof4">
Let $u\in\hom^n(E,\units E;\Z)$ denote the Thom class of $E$, and let $u\vert_E$ be its image under the map $\hom^n(E,\units E;\Z)\to\hom^n(E;\Z)$. Let $U\subset M$ be a tubular neighborhood of $Z$, and let $u_Z\in\hom^n(U,U\sm Z;\Z)$ be the Thom class of $N_{Z/M}$. Because everything in sight has secretly been given compatible orientations (and because I'm secretly neglecting to show that $E\vert_Z$ is isomorphic to the normal bundle $N_{Z/M}$), $\ast\psi\vert_Uu\in\hom^n(U,U\sm Z;\Z)$ restricts to the preferred generator of each fiber of the normal bundle, so $\ast\psi\vert_Uu=u_Z$. Given this, one applies excision to $(U,U\sm Z)\into(M,M\sm Z)$ followed by natural map $\hom^n(M,M\sm Z;\Z)\to\hom^n(M;\Z)$ in order to obtain the identity
$$\ast\psi(u\vert_E)=u_Z^M\in\hom^n(B;\Z).$$
Above, the LHS is the Euler class $e(E)$ while the RHS is $\ast{[Z]}$ by the most recent lemma.
</div>
<div class="corollary">
Let $E\to M$ be a smooth, oriented rank $n$ real vector bundle over a compact, oriented $n$-dimensional manifold $M$. Then,
$$[M]\frown e(E)\in\hom_0(M;\Z)=\Z$$
is the number of zeros of a generic section, up to sign.
</div>
<div class="corollary">
Let $E\to B$ be a holomorphic rank $n$ vector bundle over an $n$-dimensional complex manifold (so $2n$-dimensional as a smooth manifold). Then,
$$[B]\frown c_n(E)\in\hom_0(B;\Z)=\Z$$
is the number of zeros of a generic section, up to sign.
</div>
<p>At this point one may reasonably wonder why this is such a big deal. The point (at least my point) is the following: say you want to calculate the number of some type of geometric object. It often happens that the objects you want to count are given by zero set of a well-chosen section on a suitable line bundle. Once you realize this in your specific case, the above theorem tells you that computing this count amounts to calculating a characteristic class. Even if calculating characteristic classes isn’t your thing, the theorem still tells you that your geometric count is (largely) independent of the section you choose to count it! That means, even if you problem naturally presents you with one section of your line bundle, the above results says you can resolve it by counting the zeros of the section which is easiest to work with!</p>
<p>Let’s see some of this in action.</p>
<h1 id="27-lines-on-a-cubic">27 Lines on a Cubic</h1>
<p>As is probably unsurprising by this point, we will end this post by determining the number of lines (copies of $\CP^1$) on a (complex) cubic surface. Let $F=F(x,y,z,w)$ be a degree 3 homogeneous polynomial, and let $X=\bracks{F=0}\subset\CP^3$ be the cubic surface it determines. What can one count the number of lines on $X$?</p>
<p>Well, consider $G=\mrm{Gr}(2,4)$ (or $\mrm{Gr}(1,3)$ depending on who you ask), the Grassmannian manifold consisting of $2$-dimensional subspaces of $\C^4$ (equivalently, of lines in $\CP^3$). Note that, as a complex manifold, the dimension of $G$ is $2(4-2)=4$. On $G$, one has the tautological subbundle</p>
<script type="math/tex; mode=display">S=\bracks{(v,p):v\in p}\subset\C^4\by G\to G</script>
<p>which is a rank $2$ holomorphic vector bundle $S\to G$ whose fiber $S_p$ above a point $p\in G$ is the plane represented by that point. Consider the dual bundle $\dual S=\Hom(S,\C)\to G$ whose fiber $\dual S_p$ over a point $p\in G$ is the space of linear functions $S_p\to\C$. Since $S_p\subset\C^4$, we see that every linear functional $S_p\to\C$ is the restriction of some linear functional $\C^4\to\C$ on $\C^4$ (i.e. $\dual{(\C^4)}\to\dual S_p$ is surjective for all $p$). Letting $x,y,z,w$ suggestively denote a fixed basis for $\C^4$, we get that every linear functional on $\C^4$ (and so every linear functional on $S_p$) is given by some homogeneous linear polynomial in $x,y,z$, and $w$. That is, sections of $\dual S\to G$ correspond to homogeneous linear polynomials in the variables $x,y,z,w$. Thus, forming the symmetric bundle $\Sym^3\dual S\to G$, we get that sections of it correspond to homogeneous degree $3$ polynomials in $x,y,z,w$. In particular, there exists a section $\sigma_F:G\to\Sym^3\dual S$ corresponding to the polynomial $F$ used to define our cubic surface $X$!</p>
<p>Now, what does it mean for some $p\in G$ to be a zero of $\sigma_F$, i.e. when is $\sigma_F(p)=0$? Well, $\sigma_F(p)\in\Sym^3\dual S_p$ is the linear functional $F\in\Sym^3\dual{(\C^4)}$ restricted to the plane $S_p\subset\C^4$, so $\sigma_F(p)=0$ if and only if $F(c_1,c_2,c_3,c_4)=0$ for all $c_1x+c_2y+x_3z+c_4w\in S_p\subset\C^4$ (here, $c_i\in\C$). That is, $\sigma_F(p)=0$ iff $F$ vanishes along the plane $S_p\subset\C^4$. Now, we observe that a $2$-plane in $\C^4$ in precisely a line in $\CP^3$, so points of $G$ can be viewed a parameterizing all the lines on $\CP^3$, given some $p\in G$, we have $\sigma_F(p)=0$ if and only if $F$ vanishes along the line (in $\CP^3$) represented by $p$. In other words, $\sigma_F(p)=0$ iff the line $p\subset\CP^3$ lives in the set $\bracks{F=0}=:X$; the zeros of $\sigma_F$ are precisely the lines in $X$!</p>
<p>This brings us to the home stretch. Since $\rank S=2$, we easily see that $\rank\Sym^3\dual S=4=\dim_\C G$, so the number of lines on $X$ is (Poincare dual to) $c_4(\Sym^3\dual S)$. Let’s compute this. By the splitting principle, to determine $c_4(\Sym^3\dual E)$ for a general (rank 2) vector bundle $E$, we can assume that $E=L_1\oplus L_2$ is a sum of line bundles. Let $x_1=c_1(L_1)$ and $x_2=c_1(L_2)$, so $c(E)=(1+x_1)(1+x_2)=1+(x_1+x_2)+x_1x_2$. Note that</p>
<script type="math/tex; mode=display">\Sym^3\dual E=\Sym^3(\dual L_1\oplus\dual L_2)=(\dual L_1)^{\otimes 3}\oplus((\dual L_1)^{\otimes2}\otimes\dual L_2)\oplus(\dual L_1\otimes(\dual L_2)^{\otimes2})\oplus(\dual L_2)^{\otimes3}.</script>
<p>Thus, taking the total Chern class of both sides, we see that</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
c(\Sym^3\dual E)
&=(1-3x_1)(1-2x_1-x_2)(1-x_1-2x_2)(1-3x_2)\\
&=1 - 6(x_1+x_2) + (11x_1^2+32x_1x_2+11x_2^2) - 6(x_1^3 + 8x_1^2x_2 + 8x_1x_2^2 + x_2^3) + x_1x_2(18x^2 + 45x_1x_2 + 18x_2^2)\\
&= 1 - 6c_1(E) + (11c_1(E)^2 + 10c_2(E)) - 6(c_1(E)^3 + 5c_1(E)c_2(E)) + c_2(E)(18c_1(E)^2 + 9c_2(E))
\end{align*} %]]></script>
<p>Thus, for any rank $2$ complex vector bundle $E\to B$, we have</p>
<script type="math/tex; mode=display">c_4(\Sym^3\dual E)=c_2(E)(18c_1(E)^2 + 9c_2(E)).</script>
<p>Um, now we switch gears and cheat. I actually don’t know a quick and easy way to calculate the above cup products <sup id="fnref:38"><a href="#fn:38" class="footnote">34</a></sup>, so we won’t compute them. Instead, we’ll use the observation that $c_4(\Sym^3\dual S)$ can be computed using the zeros of <b>any</b> of its sections in order to determine the number of lines on $X$.</p>
<p>That is, at this point, we know that the number of lines on $X$ is given by $c_4(\Sym^3\dual S)$ for $S\to\mrm{Gr}(2,4)$ the tautological subbundle. This has absolutely no dependence on $X$, so we know already that every cubic surface (over $\C$) has the same number of lines! Thus, it suffices to just pick one and count how many lines it has. For this, we introduce the Fermat cubic surface</p>
<script type="math/tex; mode=display">X:x^3+y^3+z^3+w^3=0.</script>
<p>Up to a permutation of coordinates, every line in $\CP^3$ is given by two linear equations of the form $x=az+bw$ and $y=cz+dw$ for some $a,b,c,d\in\C$. This will lie on $X$ if and only if</p>
<script type="math/tex; mode=display">(az+bw)^3+(cz+dw)^3+z^3+w^3=0</script>
<p>as polynomials in $\C[z,w]$. Equating coefficients, this means that we need</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
a^3 + c^3 &= -1\\
b^3 + d^3 &= -1\\
a^2b &= -c^2d\\
ab^2 &= -cd^2
\end{align*} %]]></script>
<p>If $a,b,c,d$ are all non-zero, then</p>
<script type="math/tex; mode=display">a^3=(a^2b)^2/(ab^2)=-(c^2d)^2/(cd^2)=-c^3\implies0=a^3+c^3=-1,</script>
<p>which is nonsense. Hence, possibly after renaming, we may assume $a=0$. Then, $c^3=-1$, so $d=0$ and $b^3=-1$ as well. This gives $9$ (since $3$ cube roots of $-1$) possible choices of $(a,b,c,d)$ giving rise to $9$ distinct lines on $X$. These lines are</p>
<script type="math/tex; mode=display">x=\zeta_3^iw\,\,\,\,\text{ and }\,\,\,\,y=\zeta_3^jz</script>
<p>for some $i,j\in\bracks{0,1,2}$. The rest of the lines are given by permuting the coordinates. Base on the form these lines take, we see that we get a set of $9$ lines for every partition of the set $\bracks{x,y,z,w}$ into subsets of size $2$. Thus, there are $9\cdot3=27$ lines on $X$, and so $27$ lines on any complex cubic surface.</p>
<div class="footnotes">
<ol>
<li id="fn:3">
<p>This may make more sense by the end of the post if it doesn’t right now, but because of the existence of classifying spaces, characteristic classes can equivalently be thought of as cohomology classes in $\ast\hom(BU(n);A)$ for some $n$ (the rank of the vector bundle) and abelian group $A$, assuming you only care about complex vector bundles. <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:1">
<p>No promises <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:30">
<p>I didn’t quite meet this goal. Started strong but ran into issues by the end (lacking details once I start talking about Euler classes) as seems to be the norm. <a href="#fnref:30" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>Except in footnotes where I’ll reserve the power to be vague, to be handwavy, to give potentially bad intuition, and to say potentially incorrect things. <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:20">
<p>I will admit that thinking in terms of e.g. integrating differential forms is more intuitive than in terms of e.g. cupping cohomology classes. However, tough luck; I like working abstractly even when intuiting actually geometry <a href="#fnref:20" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>Some things I say may be false without this. For example, I think you need this for fiber bundles to be fibrations. <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>“Isn’t $-\gamma$ also a generator?” you ask. Yes, but it’s actually a different one. The point is that $\CP^2$ has a complex structure, so a canonical orientation, so a canonical choice of generator $\gamma\in\hom^2(\CP^2;\Z)$, and this is “the same” generator we use for $\hom^2(\CP^\infty;\Z)$. <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:39">
<p>Ostensibly, $BU(n)$ characterizes principal $U(n)$-bundles, not complex rank-$n$ vector bundles, so what gives? Well, two things: (1) every complex rank $n$ vector bundle arises from some principal $U(n)$-bundle and (2) the universal rank $n$ vector bundle is the vector bundle $(EU(n)\by_{U(n)}\C^n)\to BU(n)$ over $BU(n)$ where $U(n)$ acts on $\C^n$ by matrix-vector multiplication as one would expect <a href="#fnref:39" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>Fix any $x\in U_i\cap U_j$. The point $(x,v)\in\inv p(U_i)\simeq U_i\by\C^{n+1}$ is identified with the point $(x,\tau_{ij}(x)v)\in\inv p(U_j)\simeq U_j\by\C^{n+1}$. <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>$\pull\pi E\vert_{\P(E)_ b}\simeq\C^{n+1}\by\P(E)_ b$ is trivial since it is pullback from a vector bundle over a point. That is, letting $q=\pi\vert_{\P(E)_ b}:\P(E)_ b\to{b}$, we have $\pull\pi E\vert_{E_b}\simeq\pull qE_b\simeq\C^{n+1}\by\P(E)_ b$. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>It is $\ints{\P^n}(-1)$ if you are familiar with this notation. <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:10">
<p>I really should have started with a rank $n$ vector bundle. <a href="#fnref:10" class="reversefootnote">↩</a></p>
</li>
<li id="fn:9">
<p>As the notation suggests, these coefficients are exactly the Chern classes of $E$. I think I’ve heard that this way of obtaining Chern classes was discovered by Grothendieck, but don’t quote me on that. <a href="#fnref:9" class="reversefootnote">↩</a></p>
</li>
<li id="fn:11">
<p>Above, one should really write $\pull\pi c_1(EU(n))$ for $\pi:BU(n-1)\by BU(1)\to BU(n)$ instead of just $c_1(EU(n))$. <a href="#fnref:11" class="reversefootnote">↩</a></p>
</li>
<li id="fn:12">
<p>$c_1$ corresponds to the first Chern class on the first factor, and $c_1’$ corresponds to the first Chern class on the second factor. <a href="#fnref:12" class="reversefootnote">↩</a></p>
</li>
<li id="fn:13">
<p>I think that’s how people usually prove algebraic independence of symmetric polynomials <a href="#fnref:13" class="reversefootnote">↩</a></p>
</li>
<li id="fn:14">
<p>This is not the simplest way to do this, but more practice with Yoneda-type reasoning is never bad <a href="#fnref:14" class="reversefootnote">↩</a></p>
</li>
<li id="fn:15">
<p>Strictly speaking, $X$ does not need to have the homotopy type of a CW-complex, but if it doesn’t you need to be extra careful. In particular, $\hom^2(X;\Z)$ would not represent singular cohomology, but instead singular cohomology of a CW approximation. <a href="#fnref:15" class="reversefootnote">↩</a></p>
</li>
<li id="fn:16">
<p>We use the same letters to denote these technically different maps in order to emphasize how closely related they are, and totally not because I was too lazy to come up with new names. <a href="#fnref:16" class="reversefootnote">↩</a></p>
</li>
<li id="fn:17">
<p>Also, I should have also include an “identity natural transformation” $e:\bracks{0}\to\hom^2(-;\Z)$, but oh well. <a href="#fnref:17" class="reversefootnote">↩</a></p>
</li>
<li id="fn:18">
<p>$\CP^\infty$ is like unironically the best topological space; fight me <a href="#fnref:18" class="reversefootnote">↩</a></p>
</li>
<li id="fn:19">
<p>e.g. the cohomology class corresponding to $\mu\circ(e\by\Id)$ is $ac_1\in\Z c_1’=\hom^2(\CP^\infty;\Z)$ (cohomology of right factor of $\CP^\infty\by\CP^\infty$). One sees by tracing identifications that forming the cohomology class corresponding to $\mu\circ(e\by\Id)$ amounts to pulling back $\mu$’s cohomology class along the map $e\by\Id:\CP^\infty\to\CP^\infty\by\CP^\infty$. <a href="#fnref:19" class="reversefootnote">↩</a></p>
</li>
<li id="fn:21">
<p>There are many equivalent definitions of orientations of various objects. One way to define the orientation of a real vector space $V$ is by saying that it is a connected component of $\Wedge^{\dim V}V\sm{0}$. In other words, it is an equivalence class of ordered bases where two bases are considered equivalent if they differ by a matrix with positive determinant. <a href="#fnref:21" class="reversefootnote">↩</a></p>
</li>
<li id="fn:23">
<p>In case you’re wondering where we used that we were considering the pullback bundle $\pull fE\to B’$ of $E$ and not just an arbitrary vector bundle $E’\to B’$ over $B’$ with a map $E’\to E$ to $E$ (lying over the given map $B’\xto fB$), the answer is no where. In general, if $E’\to E$ is a bundle map over $B’\xto fB$, then $E’\simeq\pull fE$. <a href="#fnref:23" class="reversefootnote">↩</a></p>
</li>
<li id="fn:33">
<p>I think one of my worse decisions this post has been to use the same symbol $E$ as the default for all my fiber bundles, whether they be (oriented) real (or complex) vector bundles, sphere bundles, principal $G$-bundles, or what have you <a href="#fnref:33" class="reversefootnote">↩</a></p>
</li>
<li id="fn:34">
<p>In case you are wondering, yes, I could just restructure this post by showing the existence of the Thom class first and then using its existence to define the Euler class in order to avoid this omission <a href="#fnref:34" class="reversefootnote">↩</a></p>
</li>
<li id="fn:35">
<p>Going through this yourself might be good practice. You should end up getting that $\hom^{p+n+1}(D(E),E;\Z)\simeq\hom^p(B;\Z)$ and that the Thom class $u\in\hom^{n+1}(D(E),D;\Z)$ is the unique cohomology class restricting to the preferred generator of $\hom^{n+1}(D^{n+1},S^n;\Z)\simeq\hom^n(S^n;\Z)$ on each fiber. <a href="#fnref:35" class="reversefootnote">↩</a></p>
</li>
<li id="fn:37">
<p>You may want a relative version on Leray-Hirsch to prove this <a href="#fnref:37" class="reversefootnote">↩</a></p>
</li>
<li id="fn:24">
<p>At this point, I don’t even think we know that $k\neq 0$. <a href="#fnref:24" class="reversefootnote">↩</a></p>
</li>
<li id="fn:27">
<p>Computing differentials in spectral sequences is hard… I spent ~3 hours trying to carefully work this out because I couldn’t find a source that does it (all the ones I saw either defined $c_1(L)=e(L)$ or showed $c_1(L)=\pm e(L)$ and left it at that), and in the end, I concluded that if I didn’t stop thinking about it, I’d lose my mind. <a href="#fnref:27" class="reversefootnote">↩</a></p>
</li>
<li id="fn:28">
<p>I don’t think I remarked this earlier, but from the multiplicativity $c(E\oplus F)=c(E)c(F)$ of the total Chern class, one sees that $c_{\mrm{top}}(E\oplus F)=c_{\mrm{top}}(E)c_{\mrm{top}}(F)$. In particular, $c_n(L_1\oplus\cdots\oplus L_n)=c_1(L_1)\dots c_1(L_n)$ when $L_1,\dots,L_n$ are line bundles <a href="#fnref:28" class="reversefootnote">↩</a></p>
</li>
<li id="fn:29">
<p>If $B$ does not have higher cohomology groups (i.e. if $\hom^k(B)=0$ for all $k>n+1$), then it is true that existence of a section is equivalent to vanishing of the Euler class <a href="#fnref:29" class="reversefootnote">↩</a></p>
</li>
<li id="fn:32">
<p>Technically speaking, the primary obstruction class has its own definition and we haven’t shown that it equals the Euler class yet, but it does. <a href="#fnref:32" class="reversefootnote">↩</a></p>
</li>
<li id="fn:38">
<p>Note that the coefficients, $18$ and $9$, add up to $27$. Also note that, even without being able to determine these cup products, we can conclude that the number of lines on a cubic surface must be some multiple of $9$ (independent of the chosen surface). <a href="#fnref:38" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>For a long time I’ve been telling myself that I should try and get a solid understanding of characteristic classes, so why not do that now?Classification of Elliptic Surfaces without Multiple Fibers2020-06-09T00:00:00+00:002020-06-09T00:00:00+00:00https://nivent.github.io/blog/elliptic-surface-globalA Tour of Some Number Theory: Part I: Elliptic Curves2020-03-31T00:00:00+00:002020-03-31T00:00:00+00:00https://nivent.github.io/blog/tour-nt-i<p>I have a vague vision in my head of a series of posts designed to help me, and maybe also you, understand (parts of) the relationship between a few objects of interests in algebraic number theory: elliptic curves, selmer groups, binary quartic forms, and modular forms <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup>. I don’t know the whole story encompassing these objects, but allegedly, they are all linked in one way or another, and I have soft plans to understand these links over the next $n$ days. To aid in this, I hope to write (medium length to long <sup id="fnref:8"><a href="#fn:8" class="footnote">2</a></sup>) posts on each of these objects, drawing attention to their relationships to each other, and also including some facts/results I find interesting even if they don’t necessarily lie at the intersection of the study of e.g. elliptic curves and quartic forms. It seems most appropriate to begin this series with elliptic curves since they are the main reason I care about any of the other stuff.</p>
<p>While trying to come up with a rough idea for what I was going to say in this post, I ran into the issue of what my definition of an elliptic curve should be. On the one hand, I don’t think I will need anything too fancy for the main things I want to talk about in these posts, so I could probably get away with saying an elliptic curve $E$ is the zero set of a polynomial of the form $E:y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$ and not worry about having to prove anything annoying. On the other hand, having access to things like Riemann-Roch and the Picard group is really useful for understanding the group structure on elliptic curves, but being able to use these things requires much more setup. In the end, I decided that since I have already said words like <a href="../comm-alg">\spec</a> and <a href="../cover-fundgrp-sheaf">sheaf</a> on this blog before, the added work needed to be able to say Riemman-Roch is worth it, so I’ll start with a quick and probably mostly unhelpful introduction to schemes <sup id="fnref:7"><a href="#fn:7" class="footnote">3</a></sup>, and then focus on curves <sup id="fnref:2"><a href="#fn:2" class="footnote">4</a></sup>.</p>
<p>Finally, just to make things clear, for the main purpose of this post (setting the basics of elliptic curves, so we can later see how they connect to some other things), you can probably just skip to the section titled “Elliptic Curves,” read from there, and take some things for granted. The parts before that section are mostly here for technical completeness and for me to see if I actually know anything about geometry <sup id="fnref:23"><a href="#fn:23" class="footnote">5</a></sup>.</p>
<ol id="markdown-toc">
<li><a href="#a-quickish-introduction-to-schemes" id="markdown-toc-a-quickish-introduction-to-schemes">A Quick(ish) Introduction to Schemes</a> <ol>
<li><a href="#schemes" id="markdown-toc-schemes">Schemes</a></li>
<li><a href="#maps-to-affine-schemes" id="markdown-toc-maps-to-affine-schemes">Maps to Affine Schemes</a></li>
<li><a href="#sheaf-cohomology" id="markdown-toc-sheaf-cohomology">Sheaf Cohomology</a></li>
<li><a href="#line-bundles" id="markdown-toc-line-bundles">Line Bundles</a></li>
<li><a href="#projective-space" id="markdown-toc-projective-space">Projective Space</a></li>
</ol>
</li>
<li><a href="#generalities-on-algebraic-curves" id="markdown-toc-generalities-on-algebraic-curves">Generalities on Algebraic Curves</a> <ol>
<li><a href="#divisors" id="markdown-toc-divisors">Divisors</a></li>
<li><a href="#differentials" id="markdown-toc-differentials">Differentials</a></li>
<li><a href="#riemann-roch" id="markdown-toc-riemann-roch">Riemann-Roch</a></li>
</ol>
</li>
<li><a href="#elliptic-curves" id="markdown-toc-elliptic-curves">Elliptic Curves</a> <ol>
<li><a href="#group-law" id="markdown-toc-group-law">Group Law</a></li>
<li><a href="#isogenies" id="markdown-toc-isogenies">Isogenies</a></li>
<li><a href="#torsion-points" id="markdown-toc-torsion-points">Torsion Points</a></li>
</ol>
</li>
<li><a href="#l-adic-representations" id="markdown-toc-l-adic-representations">$\l$-adic Representations</a></li>
</ol>
<h1 id="a-quickish-introduction-to-schemes">A Quick(ish) Introduction to Schemes</h1>
<p>We will begin with some motivation for the eventual definition of a scheme. This will motivation will take the form of a definition of a $\smooth$ manifold. Usually, one thinks of a $\smooth$ manifold $X$ as a topological space satisfying some technical conditions (e.g. must be Hausdorff and paracompact) which moreover, and most essentially, “locally looks like $\R^n$”. The standard way to make sense of this last condition, is to say that $X$ is covered by opens (i.e. charts) $\bracks{U_\alpha}_ {\alpha\in A}$ each coming with a homeomorphism $\phi_\alpha:U_\alpha\to\R^n$. However, this is not enough, because we want $X$ to be $\smooth$, so we require furthermore that the “transition functions” $\phi_{\alpha\beta}:\phi_\alpha(U_{\alpha\beta})\to\phi_\beta(U_{\beta\alpha})$ (here, $U_{\alpha\beta}:=U_\alpha\cap U_\beta$) defined by $\phi_{\alpha\beta}=\phi_\beta\circ\inv\phi_\alpha$ are $\smooth$ in the normal calculus sense (since they are maps between open subsets of $\R^n$). The point of having all these charts and smooth overlap conditions, is that they allow you to make sense of the notion of a smooth function $X\to\R$ because smoothness is a local condition and we already know what it means for a function $\R^n\to\R$ to be smooth.</p>
<p>This might make you wonder, if we really care about smooth functions $X\to\R$, why not start with these instead of starting with charts? One can do this, and this is closer to what one does in the algebraic setting. Still on the topic of manifolds, note that the presheaf</p>
<script type="math/tex; mode=display">\smooth_{\R^n}(U)=\bracks{\text{smooth functions }f:U\to\R}</script>
<p>on $\R^n$ is indeed a sheaf, and hence the pair $(\R^n,\smooth_{\R^n})$ form a so-called <bf>ringed space</bf> (i.e. a space with a choice of sheaf of rings). With this as a prototypical example of a ringed space (in the setting of $\smooth$ manifolds), one can now define a $\smooth$ manifold to be a (Hausdorff, paracompact) topological space $X$ with a sheaf $\ints X$ such that $(X,\ints X)$ is locally isomorphic to $(\R^n,\smooth_{\R^n})$, i.e. there exists an open cover $\bracks{U_\alpha}$ of $X$ such that $(U_\alpha,\ints{U_\alpha})\simeq(\R^n,\smooth_{\R^n})$ for all $\alpha$. Here, $\ints{U_\alpha}$ is the sheaf $\ints{U_\alpha}(V)=\ints X(U_\alpha\cap V)$ for any open $V\subset U_\alpha$. This is the same definition of smooth manifold as the standard one, but now we don’t worry about transition functions or about including a choice of atlas <sup id="fnref:3"><a href="#fn:3" class="footnote">6</a></sup> in the datum of a smooth manifold <sup id="fnref:4"><a href="#fn:4" class="footnote">7</a></sup>.</p>
<p>There are a few things I should clarify. First, I never said what a morphism of ringed spaces is, and so the requirement that $(U,\ints{U_\alpha})$ be isomorphic to $(\R^n,\smooth_{\R^n})$ is not yet well-defined. Intuitively, a morphism $f:(X,\ints X)\to(Y,\ints Y)$ between ringed spaces should be a continuous map $X\to Y$ and a map of sheaves $\ints X\to\ints Y$, but these sheaves live on different spaces, so we need a way to transfer sheaves from one space to another. There are two natural ways to do this.</p>
<div class="definition">
Let $f:X\to Y$ be a continuous map, and let $\msF$ be a presheaf on $X$. Then, the <b>direct image presheaf</b> $\push f\msF$ on $Y$ is given by
$$(\push f\msF)(V)=\msF(\inv f(V)).$$
This is a sheaf when $\msF$ is in which case we call it the direct image sheaf (as opposed to presheaf).
</div>
<div class="definition">
Let $f:X\to Y$ be a continuous map, and let $\msG$ be a presheaf on $Y$. Then, the <b>inverse image sheaf</b> $\inv f\msG$ on $X$ is the sheafification of the presheaf
$$U\mapsto\msG(f(U)):=\dirlim_{U'\supset f(U)}\msG(U')$$
where the direct limit is taken over opens containing $f(U)$ (think of this presheaf as spitting out germs of sections above $f(U)$). We do not denote this by $\pull f$ because that is a different operation we will see later.
</div>
<p>Hence, we have a choice in our definition of a morphism $(X,\ints X)\to(Y,\ints Y)$ of ringed space. It could either <sup id="fnref:5"><a href="#fn:5" class="footnote">8</a></sup> be a pair $(f’,\sharp f)$ of a continuous function $f’:X\to Y$ and a morphism $\sharp f:\ints Y\to\push f\ints X$ or a pair $(f’,f^\flat)$ of a continuous function $f’:X\to Y$ and a morphism $f^\flat:\inv f\ints Y\to\ints X$. Thankfully for us, these two definitions are equivalent because of the following.</p>
<div class="proposition">
Let $f:X\to Y$ be continuous, and let $\Ab(X)$ denote the category of abelian sheaves on $X$. Then, the functors $\push f:\Ab(X)\rightleftarrows\Ab(Y):\inv f$ are adjoint. That is, we have functorial isomorphisms
$$\Hom_{\Ab(X)}(\inv f\msG,\msF)\iso\Hom_{\Ab(Y)}(\msG,\push f\msF).$$
</div>
<div class="proof4">
Take my word for it or go through the trouble of proving this yourself. If you choose the latter, it helps to first construct natural maps $\msG\to\push f\inv f\msG$ and $\inv f\push f\msF\to\msF$ for $\msG\in\Ab(Y)$ and $\msF\in\Ab(X)$. Then the above isomorphism is given by applying $\push f$ or $\inv f$ followed by composing with one of these maps, which you can show by computing what these compositions do at stalks.
</div>
<p>Now that we have a notion of morphism of ringed spaces, our earlier definition of a smooth manifold as a ringed space that looks locally like $(\R^n,\smooth_{\R^n})$ makes sense.</p>
<div class="exercise">
Show that the ringed space definition of a $\smooth$ manifold is the same as the standard definition, including showing that sections $s\in\ints X(U)$ of the structure sheaf can be identified with smooth functions $U\to\R$.
</div>
<div class="exercise">
Show that if $f=(f',\sharp f):(X,\ints X)\to(Y,\ints Y)$ is a morphism of ringed spaces, then it induces local maps $\sharp f_x:\ints{Y,f(x)}\to\ints{X,x}$ on stalks.
</div>
<p>Now that we know what ringed spaces are and have seen how we can use sheaves to nicely describe spaces formed by gluing together ones we care about, let’s define schemes.</p>
<h2 id="schemes">Schemes</h2>
<p>First, to get this out of the way, a <bf>locally ringed space</bf> $(X,\ints X)$ is a ringed space such that the stalks $\ints{X,x}$ are local rings. For a locally ringed space $(X,\ints X)$ and a point $x\in X$, we define the <b>residue field at $x$</b> to be</p>
<script type="math/tex; mode=display">\kappa(x):=\ints{X,x}/\mfm_x\text{ where }\mfm_x\text{ is the (unique) maximal ideal of }\ints{X,x}.</script>
<p>A morphism of locally ringed spaces is a morphism $f:(X,\ints X)\to(Y,\ints Y)$ of ringed spaces such that the induced maps $\ints{Y,f(x)}\to\ints{X,x}$ are local (i.e. $\inv f(\mfm_x)=\mfm_{f(x)}$ where $\mfm_x\subset\ints{X,x}$ is the maximal ideal and similarly for $\mfm_{f(x)}\subset\ints{Y,f(x)}$).</p>
<p>Now, schemes are basically algebraic manifolds, except the phrase “algebraic manifold” already <a href="https://www.wikiwand.com/en/Algebraic_manifold">refers to something else</a>. Our model space/protypical example will be the affine schemes $\spec A$ where $A$ is any (commutative) ring (with unity). Recall that $\spec A=\bracks{\text{prime ideals }\mfp\subset A}$ and we topologize it by giving it the Zariski topology whose closed sets are the ones of the form $V(I)=\bracks{\mfp\supset I}$ for $I\subset A$ an ideal. We need to give this space a structure sheaf $\ints A=\ints{\spec A}$ which we want to think of as the “sheaf of (regular) functions”. It is clear that we should have $\ints A(\spec A)=A$ <sup id="fnref:6"><a href="#fn:6" class="footnote">9</a></sup>. Similarly, given $a\in A$, the basic open $D(a)=\bracks{\mfp\in\spec A:a\not\in\mfp}$ (“points where $a$ does not vanish”) satisfies $D(a)\simeq\spec A_a$ (at least topologically), so we should have $\ints A(D(a))=A_a$. Now, because sheaves are defined locally, because these $D(a)$ for a base for the topology on $\spec A$, and because $D(a)\cap D(b)=D(ab)$, this actually uniquely characterizes a sheaf on $\spec A$, which we unsurprisingly call $\ints A$. That is,</p>
<div class="proposition">
Let $A$ be a ring. Then, there is a unique sheaf $\ints A$ on $\spec A$ such that $\ints A(D(a))=A_a$ for any basic/distinguished open $D(a)\subset\spec A$. Slightly more concretely, for any open $U\subset\spec A$, we have
$$\ints A(U)=\invlim_{D(a)\subset U}A_a.$$
</div>
<div class="proof4">
Uniqueness is easy because any sheaf is determined by what it does on a base. We have a presheaf defined (the transition maps are what they have to be), so we'd only need to show that this is a sheaf. To do this, you can reduce to coverings (of a given open $U$) back basic opens $D(a_i)$ ($i\in I$), and then the ability to uniquely glue (compatible) sections overs of these basic opens will ultimately boil down to the universal property of localizations (and in particular, uniqueness of the localization map $A_{a_i}\to A_{a_ia_j}$).
</div>
<p>One can give other, possibly more concrete, constructions of $\ints A$.</p>
<div class="exercise">
Show that $\ints A$ is the sheafification of the presheaf $U\mapsto\inv S_UA$ where $$S_U=\bracks{a\in A:\mfp\in U\implies a\not\in\mfp}=\bigcap_{\mfp\in U}(A\sm\mfp).$$
</div>
<div class="exercise">
Fix an $A$-module $M$. Given a closed set $Z\subset\spec A$, let $I(Z)=\bigcap_{\mfp\in Z}\mfp$. Given an open $U\subset\spec A$, show that
$$\wt M(U)=\dirlim_n\Hom_A(I(X\sm U)^n,M)$$
defines a sheaf $\wt M$ on $\spec A$, and in particular, $\wt A\simeq\ints A$. These sheaves are called <b>quasi-coherent</b>. In general, if $X$ is a (not necessarily affine) scheme, any sheaf which locally looks like $\wt M$ is called quasi-coherent.
</div>
<div class="exercise">
Show that $\ints{A,\mfp}=A_\mfp$, i.e. that the stalk of the structure sheaf is the localization at the prime.
</div>
<p>Any locally ringed space isomorphic to the pair $(\spec A,\ints A)$ is called an <b>affine scheme</b>. In general, a <b>scheme</b> is a (locally) ringed space $(X,\ints X)$ which is locally isomorphic to affine schemes. A morphism of schemes is just a morphism of the underlying locally ringed spaces.</p>
<p>It is generally useful to know when a scheme $S$ is affine because affine schemes are the easiest to work with. We will not say much about figuring this out in general except to claim without proof that if $S\into\spec A$ is a closed immersion (to be defined below), then $S$ is affine, and furthermore, $S\simeq\spec A/I$ for some (not necessarily radical) ideal $I\subset A$.</p>
<p>I should also mention that if $X$ is a fixed scheme, and $S$ is another scheme, then we let $X(S)$ denote the set of morphisms $S\to X$. If $S=\spec A$ is affine, then we also write $X(A)=X(\spec A)$ for this set. We call the set $X(S)$ the set of <b>$S$-points of $X$</b>. For example, if $X=\spec\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)$ and $S=\spec A$, then</p>
<script type="math/tex; mode=display">X(A)=\bracks{\spec A\to\spec\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)}=\bracks{\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)\to A}=\bracks{a_1,\dots,a_n\in A:f_i(a_1,\dots,a_n)=0\,\forall i}</script>
<p>really should be thought of as the set of points of the vanishing set $V(f_1,\dots,f_m)$ with coordinates in $A$. If we are working with $Y$-schemes $X\to Y$ and $S\to Y$, then $X(S)$ only consists of the maps $S\to X$ which respect the given maps $X\to Y$ and $S\to Y$ (In this case, we really should denote the set $X_Y(S)$, but it’s usually clear from context what we mean).</p>
<p>Now, schemes are too general to successfully study all at once, so there’s a whole host of adjectives one can put in front of a scheme or morphism of schemes in order to make things more tractable. Usually, one spends weeks getting a feel for all these various adjectives, but we don’t have time for that, so I’ll just go ahead a define a few with little motivation/intuition.</p>
<div class="definition">
A scheme $X$ is called <b>reduced</b> if $\ints X(U)$ is a reduced ring (i.e. has no nilpotents) for all open $U\subset X$. It is called <b>integral</b> if $\ints X(U)$ is a domain for all open $U\subset X$ (equivalently, $X$ is reduced and irreducible). It is called <b>locally noetherian</b> if $\ints X(U)$ is noetherian for all open $U\subset X$. It is called <b>noetherian</b> if it is locally noetherian and compact (or quasi-compact as algebraic geometers like to call it) as a topological space.
</div>
<div class="definition">
A morphism $f:X\to Y$ of schemes is called an <b>open immersion</b> if $f$ is topologically an embedding onto an open subset $U\subset Y$, and the map $f:X\to U$ is an isomorphism (where the structure sheaf on $U$ is $\ints U=\ints Y\vert_U$). It is called a <b>closed immersion</b> if it is topologically an embedding onto a closed subset of $Y$, and the map $\ints Y\to\push f\ints X$ is surjective.
</div>
<p>The definition of a closed immersion is different from what one might expect because there is not a unique scheme structure on a closed subset of a scheme. For example, the closed set ${(2)}\subset\spec\Z$ is the underlying topological space for the closed subschemes $\spec\F_2\into\spec\Z$ and $\spec\Z/(4)\into\spec\Z$, but $\spec\F_2\neq\spec\Z/(4)$. The requirement that $\ints Y\to\push f\ints X$ be surjective mirrors the fact that, in the affine case, closed subschemes correspond to quotients of the ring of global functions.</p>
<div class="definition">
A morphism $f:X\to Y$ of schemes is said to be <b>locally of finite type</b> if for any open affine $\spec A\subset Y$ and open affine $\spec B\subset\inv f(\spec A)$, the restricted map $f:\spec B\to\spec A$ (really, it's action on global sections) turns $B$ into a finitely generated $A$-algebra. We say it is <b>finite</b> if, for any affine open $\spec A\subset Y$, its preimage $\inv f(\spec A)=\spec B$ is affine an $B$ is a finitely generate $A$-module. We say $f$ is <b>quasi-compact</b> if for any quasi-compact $U\subset Y$, its preimage $\inv f(U)\subset X$ is quasi-compact too. We say $f$ is <b>of finite type</b> if it is both quasi-compact and locally of finite type.
</div>
<p>There are other adjectives, like separated and proper, that I may throw around every now and then. Don’t worry about them too much. Maybe if I end up putting (a version of) this post online, I’ll come back here later and actually define them. I don’t feel like doing it yet.</p>
<h2 id="maps-to-affine-schemes">Maps to Affine Schemes</h2>
<p>It will be useful to prove a universal property for maps to affine space, so let’s do that.</p>
<div class="proposition">
Let $A$ be a ring, and let $(X,\ints X)$ be a scheme. The natural map
$$\alpha:\Hom(X,\spec A)\to\Hom(A,\ints X(X))$$
taking a morphism $f$ to its action on global sections $\alpha(f)=\sharp f_{\spec A}:\ints A(A)\to\ints X(X)$ is bijective.
</div>
<div class="proof4">
(Injectivity) Pick $f,g:X\rightrightarrows\spec A$ with $\alpha(f)=\alpha(g)$. We first claim that they agree on topological spaces. To see this, note that, for any $x\in X$, we have a commutative diagram
$$\begin{CD}
A @>\alpha(f)>> \ints X(X)\\
@VVV @VVV\\
A_{f(x)} @>\sharp f_x>> \ints{X,x}
\end{CD}$$
so taking the preimage of the maximal ideal $\mfm_x\subset\ints{X,x}$ under the composition $A\to\ints{X,x}$ along both routes, we see that
$$f(x)=\inv{\alpha(f)}("\mfm_x")$$
where "\mfm_x" with quotes is the preimage of $\mfm_x$ under the map $\ints X(X)\to\ints{X,x}$. Now, we have an analogous diagram with $g$ in place of $f$, so using that $\alpha(f)=\alpha(g)$, we see that
$$f(x)=\inv{\alpha(f)}("\mfm_x")=\inv{\alpha(g)}("\mfm_x")=g(x),$$
and hence $f,g$ agree on topological spaces. To see that they agree on sheaves, let $\mfp=f(x)=g(x)\in\spec A$, note that we have a commutative diagram
<center>
<img src="https://nivent.github.io/images/blog/tour-nt-i/toaff.png" width="200" height="200" />
</center>
By the universal property of $A_\mfp$, there is a unique map $A_\mfp\to\ints{X,x}$ factoring through the given map $A\to\ints{X,x}$, so $\sharp f_x=\sharp g_x$. Hence, $\sharp f=\sharp g$ as they agree on stalks, so $f=g$.
<br />
(Surjectivity) Pick some $\phi:A\to\ints X(X)$, and let $\{U_i\}_{i\in I}$ be an affine cover of $X$. Let $\phi_i:A\to\ints X(U_i)$ be the composition $A\xto\phi\ints X(X)\to\ints X(U_i)$, and let $f_i=\spec\phi_i:U_i=\spec\ints X(U_i)\to\spec A$, so, in particular, $\alpha(f_i)=\phi_i$. Note that $f_i\vert_{U_i\cap U_j}=f_j\vert_{U_i\cap U_j}$ for all $i,j\in I$ by injectivity of $\alpha$, so these $f_i$'s glue to give a global map $f:X\to\spec A$ and evidently, $\alpha(f)=\phi$.
</div>
<p>Our main application of this will be in the case of the following space (possibly only when $n=1$).</p>
<div class="definition">
Fix a ring $A$. <b>Affine $n$-space over $A$</b> is the scheme $\A^n_A:=\spec A[x_1,\dots,x_n]$.
</div>
<div class="corollary">
Fix a ring $A$, and let $X$ be an $A$-scheme (i.e. a scheme with a given morphism $X\to\spec A$). Then, a morphism (as $A$-schemes) $X\to\A^n_A$ is the same thing as the choice of $n$ global sections $s_1,\dots,s_n\in\ints X(X)$.
</div>
<h2 id="sheaf-cohomology">Sheaf Cohomology</h2>
<p>It is impossible to study geometry without making use of cohomology, so I guess I should take some time to define a cohomology theory on sheaves. Fix a topological space $X$. Then, the category $\Ab(X)$ of abelian sheaves is an abelian category (i.e. you can talk about kernels, cokernels, exactness, and all that good stuff). In particular, given a morphism $f:\msF\to\msG$ of sheaves on $X$, we define its <b>kernel,cokernel,image</b> to be the sheafifications of the following presheaves</p>
<script type="math/tex; mode=display">(\pker f)(U)=\ker(\msF(U)\to\msG(U)).</script>
<script type="math/tex; mode=display">(\pim f)(U)=\im(\msF(U)\to\msG(U)).</script>
<script type="math/tex; mode=display">(\pcoker f)(U)=\coker(\msF(U)\to\msG(U)).</script>
<p>As it turns out, one does not need to sheafify in the case of kernels, so $\ker f=\pker f,\pim f=(\pim f)^+,$ and $\pcoker f=(\pcoker f)^+$ where $^+$ is my notation for sheafification. With these defined, a sequence $\ms A\xto f\ms B\xto g\ms C$ of sheaves is called <b>exact</b> if $\ker g=\im f$. Note that this is weaker than requiring $\ker(\ms B(U)\to\ms C(U))=\im(\ms A(U)\to\ms B(U))$ for all open $U\subset X$.</p>
<div class="exercise">
Show that $\ms A\xto f\ms B\xto g\ms C$ is exact iff the sequence of maps $\ms A_x\xto{f_x}\ms B_x\xto{g_x}\ms C_x$ on stalks is exact for all $x\in X$. In particular, a map of sheaves is surjective, injective, or bijective iff it is so on all stalks.
</div>
<p>Now that we have a notion of exactness, we can talk about a functor being left/right exact and then form derived functors. Of note, let $\Gamma:\Ab(X)\to X$ be the <b>global sections functor</b> $\Gamma(\msF)=\msF(X)$.</p>
<div class="proposition">
$\Gamma$ is left exact.
</div>
<div class="proof4">
There are a few ways to see this.
<ul>
<li> You could show directly that given a short exact sequence $0\to\ms A\to\ms B\to\ms C\to0$ of sheaves, the sequence $0\to\ms A(X)\to\ms B(X)\to\ms C(X)$ remains exact. The main point here is that if you have a section $s\in\ker(\ms B(X)\to\ms C(X))$, then exactness at the sheaf level gives you local sections of $\ms A$ which map to (restrictions of) $s$ (i.e. by picking representatives of elements at the stalks), and these then glue by injectivity of $\ms A\to\ms B$. </li>
<li> Alternatively, you could use the general fact that a functor is left exact iff it preserves kernels. Here, we clearly have that for $f:\ms A\to\ms B$ a morphism of sheaves, $\Gamma(\ker f)=\ker\parens{\Gamma(f):\ms A(X)\to\ms B(X)}$. </li>
<li> A third possibility would be to show/observe that we have a natural isomorphism of functors
$$\Gamma(\msF)=\Hom_{\Ab(X)}(\Z_X,\msF)$$
where $\Z_X$ is the constant sheaf on $X$ with stalks equal to $\Z$, and then use that $\Hom$-functors are usually left exact. </li>
</ul>
In any case, the conclusion is the same.
</div>
<p>This is (almost) all we need to define the sheaf cohomology groups $\hom^i(X,\msF)$ as the right-derived functors of $\Gamma$. As a technical condition, in order for this construction to exist, we need to know that $\Ab(X)$ has enough injective, i.e. that every abelian sheaf on $X$ embeds in an injective sheaf. This is true and can be deduced from the fact that $\Ab$, the category of abelian groups, has enough injectives + a clever construction <sup id="fnref:9"><a href="#fn:9" class="footnote">10</a></sup>. We won’t do this in detail here partly because I’m lazy, and partly because we care mainly about curves, and so only care about cohomology in degrees 0,1 <sup id="fnref:11"><a href="#fn:11" class="footnote">11</a></sup>. We will see later <sup id="fnref:15"><a href="#fn:15" class="footnote">12</a></sup> that $\hom^1$ for the sheaves we care about (i.e. line bundles on curves) has a rather concrete description that will make it useful for computations. For now, the main things to know about cohomology are (1) that given a short exact sequence</p>
<script type="math/tex; mode=display">0\too\ms A\too\ms B\too\ms C\too0</script>
<p>of sheaves on $X$, we get a long exact sequence</p>
<script type="math/tex; mode=display">0\too\hom^0(\ms A)\too\hom^0(\ms B)\too\hom^0(\ms C)\too\hom^1(\ms A)\too\hom^1(\ms B)\too\hom^1(\ms C)\too\dots</script>
<p>in cohomology, and (2) cohomology for sheaves supported on a closed set $Z\subset X$ can be computed either on $Z$ or on $X$. Specifically,</p>
<div class="theorem">
Let $j:Z\into X$ be a closed immersion, and let $\msF$ be a sheaf of abelian groups on $Z$. Then, $\hom^i(Z,\msF)=\hom^i(X,\push j\msF)$ for all $i$.
</div>
<p>The idea here is that there a special acyclic (i.e. higher cohomology vanishes) sheaves called “flasque sheaves,” and the pushforward of a flasque resolution of $\msF$ is a flasque resolution of $\push j\msF$ (with the same global sections), so their cohomologies are literally computed by the same complex. This is an example of the intuition/slogan that “sheaves on a closed subset $Z\subset X$ are the same thing as sheaves on $X$ which vanish outside of $Z$.” Because of this, we will often be lazy and omit the pushforward when considering sheaves on $X$ and sheaves on a closed subset in the same breath.</p>
<p>Before moving one, we’ll make one quick definition.</p>
<div class="definition">
Given a sheaf $\msF$ on a space $X$, its <b>Euler characteristic</b> is
$$\chi(\msF):=\sum_{n\ge0}(-1)^n\dim\hom^n(\msF)$$
whenever this is well-defined.
</div>
<h2 id="line-bundles">Line Bundles</h2>
<p>Returning to our manifold motivation, (differential) geometers really seem to like vector bundles. If you’re studying a manifold, then its common to also try to understand its tangent bundle, cotangent bundle, exterior powers of these, etc. With this in mind, it might make sense to define algebraic vector bundles. To motivate the definition more, consider a topological vector bundle $p:E\to B$ of rank $n$. From $p$, one can construct the sheaf</p>
<script type="math/tex; mode=display">\msE_p(U)=\bracks{s:U\to E:p\circ s=1_U}</script>
<p>of local sections of $p$. Because $p$ is a vector bundle, it locally looks like $\R^n\by U\to U$, and so $\msE_p$ locally looks like $\ints U^{\oplus n}$ where $\ints U$ is the sheaf of (continuous or smooth or whatever) functions on $U$. In the algebraic context, we will take this sheaf $\msE_p$ as the definition of a vector bundle instead of the topological space $E$.</p>
<p>Let $(X,\ints X)$ be a ringed space. An $\ints X$-module $\msF$ is a sheaf on $X$ such that $\msF(U)$ is an $\ints X(U)$-module for all $U\subset X$, and such that the restriction maps $\msF(U)\to\msF(V)$ (when $V\subset U$) are compatible with restriction maps $\ints X(U)\to\ints X(V)$ in the evident sense. These form a category $\DeclareMathOperator{\Mod}{Mod}\Mod(X)=\Mod(X,\ints X)$ whose morphisms are exactly what you would expect.</p>
<div class="definition">
An $\ints X$-module $\msF$ is said to be <b>locally free of rank $n$</b> if there exists an open cover $\bracks{U_i}$ of $X$ such that $\msF\vert_{U_i}\simeq\ints{U_i}^{\oplus n}$ for all $i\in I$. In this case, we also call $\msF$ a <b>vector bundle of rank $n$</b>, or a <b>line bundle</b> if $n=1$.
</div>
<p>Given a vector bundle $\ms E$ on $X$ and a point $x\in X$, the <b>fiber of $\ms E$ above $x$</b> is the vector space</p>
<script type="math/tex; mode=display">\ms E(x):=\ms E_x\otimes\kappa(x)=\ms E_x\otimes\ints{X,x}/\mfm_x=\ms E_x/\mfm_x\ms E_x</script>
<p>whose dimension (over the residue field $\kappa(x)$) is equal to the rank of $E$ (since $\ms E_x\simeq\ints{X,x}^{\oplus\rank E}$). Given a section $s\in\msE(X)$ and a point $x\in X$, the value of this section at the point is the image $s(x)\in\ms E(x)$ of $s$ under the natural map</p>
<script type="math/tex; mode=display">\ms E(X)\too\msE_x\too\ms E(x).</script>
<p>I guess the main thing we need to know about line bundles is how they play with direct/inverse images and cohomology. I’ll just quote some results.</p>
<div class="proposition">
Let $f:X\to Y$ be a morphism of ringed spaces, and let $\msF$ be an $\ints X$-module. Then, $\push f\msF$ is an $\ints Y$-module.
</div>
<div class="proof4">
Easy.
</div>
<p>It is not the case that $\push f\msF$ is a line bundle when $\msF$ is. It is not even necessarily the case that $\push f\ints X$ is a line bundle (e.g. let $f$ be constant). Furthermore, the inverse image $\inv f\msG$ of an $\ints Y$-module does not even have to be an $\ints X$-module. To remedy this situation, we introduce</p>
<div class="definition">
Let $f:X\to Y$ be a morphism of ringed spaces, and let $\msG$ on an $\ints Y$-module. Then, the <b>inverse image</b> or <b>pullback module</b> is
$$\pull f\msG:=\inv f\msG\otimes_{\inv f\ints Y}\ints X$$
where the tensor operation on modules is the sheafification of the naive tensor presheaf.
</div>
<div class="proposition">
If $\msG$ is a vector bundle, then so is $\pull f\msG$ (and they have the same rank).
</div>
<div class="proof4">
Formation of the inverse image commutes with restriction, so we reduce to the case that $\msG=\ints Y^{\oplus n}$ where we get $\pull f\msG=\ints X^{\oplus n}$.
</div>
<h2 id="projective-space">Projective Space</h2>
<p>At this point, we know what schemes are, we know what cohomology is, and we even know what line bundles are. All we have left before moving on is to (sort of) see an example of a non-affine scheme. I won’t actually construct projective space where because that would be annoying, but I’ll at least tell you about it.</p>
<p>Fix a field $k$, and let $\DeclareMathOperator{\Sch}{Sch}\Sch_k$ denote the category of $k$-schemes, that is schemes $S$ equipped with a morphism $S\to\spec k$. It is clear that for any affine open $\spec A\subset S$ inside a $k$-scheme, $A$ is a $k$-algebra. Furthermore, given an $\ints S$-module $\msF$ on a $k$-scheme, its cohomology groups $\hom^i(\msF)$ are actually $k$-vector spaces. To study the geometry of curves, we want to understand their line bundles, so we introduce the following function $\DeclareMathOperator{\Set}{Set}P_n:\Sch_k\to\Set$</p>
<script type="math/tex; mode=display">P_n(S)=\bracks{\parens{\msL,(s_0,\dots,s_n)}:s_0,\dots,s_n\in\Gamma(\msL)\text{ have no common zeros}}</script>
<p>which spits out the collection of all sets of $n+1$ linearly independent sections of a line bundle on $P_n$. The main thing one needs to know about this functor is</p>
<div class="theorem">
The functor $P_n$ is representable. That is, there exists a $k$-scheme $\P^n=\P_k^n$ such that, for all $k$-schemes $S$, we have
$$P_n(S)=\P^n(S):=\Hom_{\Sch_k}(S,\P^n).$$
We call this scheme <b>$n$-dimensional projective space over $k$.</b> It is integral, of finite type, and proper.
</div>
<p>One takeaway from the above theorem is that, whatever projective space is, we know that we can give a map into it by specifying a line bundle along with some linearly independent global sections of it. In particular, the identity morphism $\P^n\to\P^n$ corresponds to some line bundle $\ints{\P^n}(1)$ on $\P^n$ with $n+1$ linearly independent global sections (which we think of as “homogeneous coordinates” on $\P^n$). Via Yoneda-type reasoning, given any morphism $f:S\to\P^n$ of $k$-schemes, the data on $S$ determining this morphism is the line bundle $\msL:=\pull f\ints{\P^n}(1)$ with sections the pull-backs of the homogeneous coordinates of $\P^n$.</p>
<p>I do not think this is apparent from the above characterization, but it is a fact that $\P^n$ can be covered by $n+1$ affine opens, commonly denoted $D_+(x_i)$ for $i=0,\dots,n$. In fact, mirroring the classical construction of projective space,</p>
<script type="math/tex; mode=display">D_+(x_i)\simeq\spec k\sqbracks{\frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i}}\simeq\A^n_k</script>
<p>for all $i$, and these affines are glued how you would expect. For example, to form $\P^1$, we glue $D_+(x_0)=\spec A\sqbracks{\frac{x_1}{x_0}}$ to $D_+(x_1)=\spec A\sqbracks{\frac{x_0}{x_1}}$ via the map $f:D_+(x_0)\to D_+(x_1)$ sending $x_0/x_1\mapsto x_1/x_0$. Put, perhaps more clearly, $\P^1$ is formed by gluing $\spec k[x]$ to $\spec k[y]$ (along the open subsets formed by removing the origin) via $x\leftrightarrow\inv y$.</p>
<h1 id="generalities-on-algebraic-curves">Generalities on Algebraic Curves</h1>
<p>We know can move away from the abstract generalities a bit and focus on something a little more down-to-earth.</p>
<div class="definition">
Let $R$ be a local ring with maximal ideal $\mfm$ and residue field $k$. We say $R$ is a <b>regular local ring</b> if $\dim R=\dim_k\mfm/\mfm^2$ where $\dim R$ is its Krull dimension.
</div>
<div class="definition">
Given a field $k$, a <b>curve</b> $C$ over $k$ is an integral, separated scheme of finite type over $k$ whose underlying topological space is 1-dimensional. If all of $C$'s local rings are regular, then we say $X$ is <b>nonsingular (or smooth)</b>. We will sometimes write $C/k$ to denote that $C$ is a curve over $k$.
</div>
<p>One of the nice things about smooth curves is that their local rings are about as nice as you could ask for.</p>
<div class="proposition">
If $C$ is a smooth curve over a field $k$, then its local rings $\ints{C,p}$ at the closed points $p\in C$ are Dedekind domains.
</div>
<div class="proof4">
Fix a closed point $p\in C$. Recall that
$$\ints{C,p}=\dirlim_{U\ni p}\ints C(U).$$
First, since $C$ is integral, $\ints C(U)$ is a domain for all open $U$, so $\ints{C,p}$ is a domain too. Second, let $\spec A\ni p$ be an affine neighborhood around $p$, and let $\mfp\subset A$ be the prime ideal corresponding to $p$. Since $C\to\spec k$ is finite type, $A$ is a finitely generated $k$-algebra and so noetherian by Hilbert basis. Hence, $\ints{C,p}=\ints{A,p}=A_\mfp$ is noetherian as well. Finally, $\spec A\subset C$ is an open subset of an irreducible space, so $\dim A=\dim\spec A=\dim C=1$. Since $\mfp\subset A$ is a nonzero prime (since $p$ a closed point), we see that $\dim A_\mfp=\dim A=1$, so $\ints{C,p}$ is a 1-dimensional noetherian domain. To show that it is a Dedekind domain (in fact, a dvr since it is local), we only need to show that it is integrally closed. We will in fact do one better by showing that it is a UFD. By Kaplansky's Criterion, it suffices to show that every nonzero prime ideal of $\ints{C,p}$ contains a prime element. Since $\ints{C,p}$ is 1-dimensional, local this is the same as requiring that its maximal ideal $\mfm\subset\ints{C,p}$ be principal, but this is an immediate consequence of Nakyama's lemma + the fact that $\dim_k\mfm/\mfm^2=\dim\ints{C,p}=1$.
</div>
<p>The above theorem tell us that the local rings $\ints{C,p}$ of a smooth curve are discrete valuation rings (when $p$ is a closed point), so for a closed point $p\in C$, we let $v_p:\units{k(C)}\to\Z$ denote the corresponding discrete valuation where $k(C)=\Frac\ints{C,q}$ (for any possibly non-closed $q\in C$) is the <b>function field</b> of $C$.</p>
<h2 id="divisors">Divisors</h2>
<p>Now, the main purpose of this section is to gain some understanding of the structure of line bundles on a smooth curve. To that end, we make the following definition.</p>
<div class="definition">
Let $C$ be a smooth curve over $k$. The group $\Div C$ of <b>divisors on $C$</b> is the free abelian group on the closed points of $C$. If $D=\sum n_p\cdot[p]\in\Div C$, then we let $v_p(D)=n_p$ be its coefficient corresponding to the (closed) point $p$.
</div>
<p>Given two divisors $D,E\in\Div C$, we write $D\ge E$ if $v_p(D)\ge v_p(E)$ for all $p\in C$. We say a divisor $D$ is <b>effective</b> if $D\ge0$.</p>
<div class="example">
Let $f\in k(C)$ be a nonzero meromorphic function. Then,
$$(f):=\sum_{p\in C}v_p(f)\cdot[p]$$
is a so-called <b>principal divisor</b>.
</div>
<p>Fix a smooth curve $C$.
Let $k(C)_ C$ denote the constant sheaf with stalks equal to $k(C)$. <sup id="fnref:12"><a href="#fn:12" class="footnote">13</a></sup> Note that $k(C)_ C(U)=k(C)$ for all $U\subset C$ since $C$ is irreducible. Given a divisor $D\in\Div C$, let $\ints X(D)\subset k(C)_ C$ denote the subsheaf</p>
<script type="math/tex; mode=display">\ints C(D)(U)=\bracks{f\in k(C):v_p(f)+v_p(D)\ge0\,\forall p\in U}.</script>
<p>This is a line bundle (exercise <sup id="fnref:13"><a href="#fn:13" class="footnote">14</a></sup>) and the map $D\mapsto\ints C(D)$ gives a group homomorphism $\Div C\to\Pic C$, where $\Pic C$ is the group of line bundles (group operation given by tensoring). This map is surjective (we’ll see a dumb proof of this later) and its kernel is given exactly by the subgroup of principal divisors. To see this second part, note that if $f\in k(C)$, then multiplication by $f$ gives an isomorphism $\ints C(f)\iso\ints C$. Conversely, if $\ints C\iso\ints C(D)$, then $D=(1/f)$ where $f$ is the image of $1\in\Gamma(\ints C)$. With that said, let $\Cl C=\Div C/\units{k(C)}$ denote the <b>Divisor class group</b> of $C$. We’ve shown that $\Cl C\into\Pic C$, and we’ve claimed this is actually an isomorphism.</p>
<div class="exercise">
Let $p\in C$ be a closed point in a smooth curve. Show that the line bundle $\ints C(-p)$ can be naturally identified with the "ideal sheaf" $\msI_p$ given on an open $U\subset C$ by
$$\msI_p(U)=\bracks{s\in\ints C(U):s(p)=0\text{, i.e. }s_p\in\mfm_p\subset\ints{C,p}}.$$
</div>
<p>Hence, the study of line bundles on $C$ is tied up in the study of its divisors. We’re interested in understanding the sizes $h^0(D)=\dim_k\hom^0(\ints C(D)),h^1(D)=\dim_k\hom^1(\ints C(D))$ of the cohomology groups of $C$’s divisors. We’ll gain this understanding in the form of the Riemann-Roch formula which will appear in a few (sub)sections.</p>
<p>An important notion when studying divisors on curves, is the notion of a divisor’s degree. First, a bit of notation. Given a (possibly non-closed) point $x$ is a locally ringed space $(X,\ints X)$, the <b>residue field at $x$</b> is the field</p>
<script type="math/tex; mode=display">\kappa(x):=\ints{X,x}/\mfm_x</script>
<p>where $\mfm_x\subset\ints{X,x}$ is the (unique) maximal ideal, i.e. $\kappa(x)$ is the residue field of the local ring at $x$. With that said</p>
<div class="definition">
Let $D\in\Div C$ be a divisor on a smooth curve $C$ over a field $k$. Then, the degree of $D$ is
$$\deg D=\sum_{p\in C}[\kappa(p):k]v_p(D).$$
</div>
<p>We wish to show that this descends to a map on class groups, i.e. that principal divisors have degree $0$. To do this, we will first show that nonzero elements of the function field $k(C)$ of $C$ are basically just morphisms $C\to\P^1$.</p>
<p>Let $f\in k(C)$ be nonzero. Let $Z,P\subset C$ be its set of zeros/poles, respectively. Then, $f\in\Gamma(C\sm P,\ints C$ and so determines a map $C\sm P\xto f\A^1_k$. Similarly, $1/f$ determines a map $C\sm Z\xto{1/f}\A^1_k$. We can glue these two maps together in other to form a map $C\xto f\P^1$. Using that $k(\P^1)=k(t)=\Frac k[t]$, we can recover $f\in k(C)$ from the morphism $C\to\P^1$ it defines as the image of $t$ under the map $k(\P^1)\to k(C)$ arising from this morphism (i.e. the induced map on stalks at the generic points).</p>
<p>Now, we’d like to be able to use morphisms between curves to relate divisors on them. To this end, let $f:X\to Y$ be a morphism (non constant) between smooth curves. We use $f$ to define two maps on divisors</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{matrix}
\push f:& \Div X &\too& \Div Y && \pull f:&\Div Y &\too& \Div X\\
&[p] &\longmapsto& [\kappa(p):\kappa(f(p))]\cdot[f(p)] &&& [q] &\longmapsto& \sum_{p\to q}e(p/q)\cdot[p]
\end{matrix} %]]></script>
<p>where $e(p/q)=v_p(t_q)$ where $t_q\in\ints{Y,q}$ is a uniformizer. Furthermore, since $f$ is non-constant (and all curves are assumed irreducible), it must be surjective (its image is an irreducible subset of a 1-dimensional space and bigger than 1 point). From this, we can conclude that $f$ maps $\eta_X$, the generic point of $X$, to $\eta_Y$, the generic point of $Y$. This is because the generic point of a space (which always exists for irreducible schemes) is the unique point contained in all of its open sets and so</p>
<script type="math/tex; mode=display">\inv f(\eta_Y)=\inv f\parens{\bigcap_{W\ni\eta_Y}W}=\bigcap_{W\ni\eta_Y}\inv f(W)\ni\eta_X</script>
<p>where we used surjectivity of $f$ to know that $\inv f(W)\neq\emptyset$ for every open set $W$ (containing $\eta_Y$). The upshot of this is that $f$ induces a map $k(Y)=\ints{Y,\eta_Y}\to\ints{X,\eta_X}=k(X)$ on function fields, and so we define $\deg f=[k(X):k(Y)]$. Now, we get the following theorem.</p>
<div class="theorem">
Let $f:X\to Y$ be a finite morphism between curves. Then, (1) $\deg\pull fD=(\deg f)\deg D$, (2) $\deg\push fD=\deg D$, and (3) $\push f\pull fD=(\deg f)D$ for any divisor $D$.
</div>
<div class="proof4">
Maybe I'll come back and fill this in when I'm feeling less lazy.
</div>
<div class="corollary">
Let $C$ be a smooth curve, and pick $f\in\units{k(C)}$. Then, $\deg(f)=0$.
</div>
<div class="proof4">
From the discussion on $f$ as a map $C\to\P^1$, it is clear that $(f)=\pull f[0]-\pull f[\infty]$, so $\deg(f)=(\deg f)\deg([0]-[\infty])=0$.
</div>
<h2 id="differentials">Differentials</h2>
<p>Our next aim is to construct the so-called canonical bundle which will be a non-arbitrary line bundle we can write down for any smooth curve $C$. Fix a smooth curve $C$ over a field $k$.</p>
<p>For an affine open $\spec A\subset C$, let $\Omega_{A/k}$ denote the <b>module of (Kahler) differentials</b> which is the $A$-module generated by the symbols $\d a$ for all $a\in A$ subject to the relations</p>
<ul>
<li> $\d c=0$ if $c\in k$ </li>
<li> $\d(a+b)=\d a+\d b$ </li>
<li> $\d(ab)=a\d b+b\d a$ </li>
</ul>
<div class="exercise">
A <b>$k$-linear derivation</b> on $A$ is a $k$-linear map $d:A\to M$ to an $A$-module $M$ such that $d(ab)=adb+bda$ for all $a,b\in A$ and $d(c)=0$ if $c\in k$. Show that the map $A\ni a\mapsto da\in\Omega_{A/k}$ is the universal $k$-linear derivation in the sense that for any $A$-module $M$, we have
$$\DeclareMathOperator{\Der}{Der}\Hom_A(\Omega_{A/k},M)\iso\Der_k(A,M)$$
with isomorphism given by composition with this map.
</div>
<div class="exercise">
For a multiplicative set $S\subset A$, we have $\sinv\Omega_{A/k}\simeq\Omega{\sinv A/k}$.
</div>
<p>From a previous exercise, the $A$-module $\Omega_{A/k}$ gives rise to an $\ints A$-module $\wt{\Omega_{A/k}}$ on $\spec A$. Because formation of the module of differentials plays well with localizations <sup id="fnref:14"><a href="#fn:14" class="footnote">15</a></sup>, we can glue together the various sheaves $\wt{\Omega_{A/k}}$ to get a global sheaf $\omega_C=\Omega_{C/k}$ of differentials on $C$.</p>
<div class="theorem">
The sheaf $\omega_C$ is a line bundle.
</div>
<div class="proof4">
Omitted. See e.g. Hartshorne.
</div>
<p>Differentials are useful in differential topology/geometry, so this sheaf is probably rather important. In case this quantity pops up again later, let’s go ahead and define the <b>genus</b> of $C$ to be $h^0(\omega_C):=\dim_k\hom^0(\omega_C)$ the dimension of global sections of $\omega_C$.</p>
<p>Now, one of the main utilities of differentials in algebraic geometry is their appearance in the following surprising and very useful theorem.</p>
<div class="theorem">
Let $\msL$ be a line bundle on a curve $C$. Then,
$$\hom^1(\msL)\simeq\dual{\hom^0(\omega_C\otimes\dual\msL)}.$$
In particular,
$$h^1(\msL)=h^0(\omega_C\otimes\dual\msL)$$
where $\dual\msL=\Hom(\msL,\ints C)$ is $\msL$'s dual bundle.
</div>
<p>Originally, I planned on giving a proof of this theorem, but sadly, I think doing so would send us too far afield. Usually one proves a vast generalization of the above applying to higher dimensional schemes and far more sheaves than just bundles, but carrying this out would require more than a subsection of a blog post. There is a simpler proof just in the case of curves, but even it is too involved for me to justify reproducing here, so uh, just believe me on this one.</p>
<h2 id="riemann-roch">Riemann-Roch</h2>
<p>Now, given Serre duality <sup id="fnref:16"><a href="#fn:16" class="footnote">16</a></sup>, proving Riemann-Roch will be child’s play. In general, “Riemann-Roch” type theorems consist of two parts. The first part is a formula, usually of the form $\chi(\msF)=f(\msF)+\chi(\ints X)$, for computing the Euler characteristic of a vector bundle in terms of that of the structure sheaf. In a sense that is hard to make precise, the function $f$ usually only depends coarsely on $\msF$ <sup id="fnref:17"><a href="#fn:17" class="footnote">17</a></sup>. The second part is a formula for $\chi(\ints X)$. We will begin, perhaps unsurprisingly, with the first part.</p>
<div class="theorem">
Let $D$ be a divisor on a smooth curve $C$, and let $\msL=\ints C(D)$. Then,
$$\chi(\msL)=\deg D+\chi(\ints C).$$
</div>
<div class="proof4">
We induct on $\abs{\deg D}$. Fix a divisor $D$ for which the claimed equality is known to hold (e.g. $D=0$), and let $p\in C$ be a closed point. We'll show that it holds for $D+p$ as well (there's also a $D-p$ case, but I'll leave that to you). Note that $\ints C(-p)$ is naturally a subsheaf of $\ints C$, and the quotient $\ints p$ is a sheaf on the closed set $\{p\}\subset C$ (i.e. the stalks $\ints{p,x}\subset\ints{C,x}$ vanish at all $x\in C$ except $x=p$). Hence, we have an exact sequence
$$0\too\ints C(-p)\too\ints C\too\ints p\too0.$$
Now, tensoring with vector bundles is exact (because exactness can be checked at stalks and bundles are free modules at stalks) so we can tensor the above with $\ints C(D+p)$ to get an exact sequence
$$0\too\ints C(D)\too\ints C(D+p)\too\ints p\too0.$$
The cokernel does not change e.g. because it is a line bundle on a point and there's only one of those. Looking in cohomology, we get the sequence
$$0\too\hom^0(\ints C(D))\too\hom^0(\ints C(D+p))\too\kappa(p)\too\hom^1(\ints C(D))\too\hom^1(\ints C(D+p))\too0$$
where $\hom^1(\ints p)=0$ since this sheaf is supported on a 0-dimensional space. Now, since the alternating sum of dimensions in any exact sequence starting at ending with $0$ is equal to $0$, we get
$$h^0(\ints C(D))-h^0(\ints C(D+p))+[\kappa(p):k]-h^1(\ints C(D))+h^1(\ints C(D+p))=0$$
from which we conclude that
$$\chi(\ints C(D+p))=\chi(\ints C(D))+[\kappa(p):k]=\deg D+\chi(\ints C)+[\kappa(p):k]=\deg(D+p)+\chi(\ints C)$$
and so win.
</div>
<p>For the second part, we need to calculation $\chi(\ints C)=h^0(\ints C)-h^1(\ints C)$. We know from Serre duality that $h^1(\ints C)=h^0(\omega_C)=g$, so we really just need to calculate $h^0(\ints C)$. To do this, I’ll need to make explicit an assumption that I have been making in my head this whole time. The stated definition of curve allows for schemes like the affine line $\A^1_k$, but this is space, for example, is somehow incomplete (think “non-compact”). Now, topological compactness is not as useful a notion for schemes as it is for manifolds; for example, all affine schemes are compact <sup id="fnref:19"><a href="#fn:19" class="footnote">18</a></sup>, but $\A_k^n$ should not be considered complete since it is the analogue of $\R^n$. Hence, the right notion of algebraic compactness/completeness will be something else. For our purposes, it suffices to reason as follows: projective space $\P_k^n$ should be “algebraically compact” (& also “algebraically Hausdorff”), so if $C$ if complete, then the image of any map $C\to\P_k^n$ should be “algebraically compact” as well (since $C$ is), but this is just saying that the image should be closed (since $\P_k^n$ is “algebraically compact and Hausdorff”), so we say a curve $C$ is <b>complete</b> if the image of any morphism $C\to\P_k^n$ is closed.</p>
<p>The correct notions of “algebraically compact” and “algebraically Hausdorff” are being proper and separated, but for out purposes, completeness as just defined suffices <sup id="fnref:20"><a href="#fn:20" class="footnote">19</a></sup>. From now on <sup id="fnref:18"><a href="#fn:18" class="footnote">20</a></sup>, assume all curves are complete. With this assumption made</p>
<div class="proposition">
Let $C$ be a (complete) curve over $k$. Then, $\hom^0(\ints C)=k$.
</div>
<div class="proof4">
Pick any global section $s\in\hom^0(\ints C)$. This corresponds to some morphism $s:C\to\A^1_k$. We can embed $\A_k^1\into\P_k^1$ to view $s$ as a morphism $C\to\P_k^1$ whose image is necessarily closed since $C$ is complete. This shows that $s(C)\neq\A_k^1$ (since $\A_k^1$ is not closed in $\P_k^1$) and that $s(C)$. Since $s(C)$ is closed in $\A_k^1$, it must then be a finite union of points, but $C$ is irreducible, so $s(C)$ is as well which means that $s(C)$ must be a single point, hence $s$ is constant.
</div>
<p>Thus, $\chi(\ints C)=1-g$. In summary, the takeaways are</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
\chi(\ints C(D))=\deg D+1-g && h^1(\ints C(D))=h^0(\omega_C(-D))
\end{align*} %]]></script>
<p>where, given a line bundle $\msL$, $\msL(D)$ is shorthand for $\msL\otimes\ints C(D)$. Here are two consequences of the work that we have done.</p>
<div class="proposition">
Given a smooth curve $C$, $\deg\omega_C=2g-2$.
</div>
<div class="proof4">
By Riemann-Roch + Serre duality, $g-1=\deg\omega_C+1-g$.
</div>
<div class="proposition">
Every line bundle on a smooth curve $C$ comes from a divisor.
</div>
<div class="proof4">
Fix a line bundle $\msL\in\Pic C$. Mirroring the proof of Riemann-Roch, we can show that for any $D\in\Div C$, we have $\chi(\msL(D))=\deg D=\chi(\msL)$. Hence, for $D$ of large degree, we have $\chi(\msL(D))>0\implies h^0(\msL(D))>0$, so $\msL(D)$ has some nonzero global section $s$. Let $D'$ be the effective (all nonzero coefficients) divisor
$$D'=\sum_{p\in C}v_p(s)\cdot[p].$$
Then, multiplication by $s$ given an isomorphism $\ints C(D')\iso\msL(D)$, so $\msL\simeq\ints C(D-D')$.
</div>
<p>Worth noting: contained in the above proof is an argument for the fact that divisors with negative degrees have no nonzero global sections (any global section gives an effective divisor and you can’t have an effective divisor of negative degree).</p>
<h1 id="elliptic-curves">Elliptic Curves</h1>
<p>That was quite a bit longer than I originally intended, but we made it. We can now turn our attention to the actual intended focus of this post: elliptic curves. Fix a field $k$.</p>
<div class="definition">
An <b>elliptic curve</b> over $k$ is a pair $E=(C,\infty)$ such that $C/k$ is a smooth (complete) curve of genus $1$ over $k$, and $\infty\in C(k)$ is a chosen rational point (in particular, $\kappa(\infty)=k$).
</div>
<p>This definition may be different from one you have seen before, so we will first show that it is actually the same as the classical one corresponding to the vanishing set of a cubic polynomial.</p>
<p>Fix an elliptic curve $E/k$. For any $n\in\Z$, let $D_n=n[\infty]\in\Div E$. Note that the canonical bundle $\omega_E$ is of degree $\deg\omega_E=2g(E)-2=0$ and has a nonzero global section $s\in\hom^0(\omega_E)$. Hence, $\omega_E=\ints E(s)$ where $(s)\in\Div E$ is an effective divisor of degree $0$, i.e. $\omega_E\simeq\ints E$ is trivial. Now, applying Riemann-Roch to the divisor $D_n$ defined before, we have</p>
<script type="math/tex; mode=display">h^0(\ints E(D_n))=h^0(\ints E(-D_n))-h^0(\ints E(D_n))=\deg D_n+\chi(\ints E)=n</script>
<p>for all $n\ge1$ (since divisors with negative degree have no global sections). The constant $1$ function lives in $\hom^0(\ints E(D_n))$ for all $n\ge0$ (and generates it when $n=0$), so we can complete it to basis $\bracks{1,X}$ of $\hom^0(\ints E(D_2))$ which we then in turn complete to a basis $\bracks{1,X,Y}$ of $\hom^0(\ints E(D_3))$. Necessarily, $X$ has a double pole at $\infty$ (since $h^0(\ints E(D_n))=1$ for $n\in\bits$) and similarly $Y$ has a triple pole at $\infty$. Now, note that $\bracks{1,X,Y,X^2,XY}\subset\hom^0(\ints E(D_5))$ and we claim that these form a basis. This is because they all have different orders of vanishing at $p$, so for any $c_0,c_2,\dots,c_5\in k$, not all zero, we have</p>
<script type="math/tex; mode=display">v_p(c_0+c_2X+c_3Y+c_4X^2+c_5XY)=\min\bracks{i:c_i\neq0}\le5\implies c_0+c_2X+c_3Y+c_4X^2+c_5XY\neq0.</script>
<p>Now, $\hom^0(\ints E(D_6))$ contains the 7 elements $1,X,Y,X^2,XY,Y^2,X^3$ which therefore must satisfy some nontrivial linear relation of the form <sup id="fnref:21"><a href="#fn:21" class="footnote">21</a></sup></p>
<script type="math/tex; mode=display">aY^2+a_1XY+a_3Y=bX^3+a_2X^2+a_4X+a_6.</script>
<p>Necessarily, two of the involved functions (those with nonzero coefficient) must have the same valuation at $P$, so $a,b\neq0$. We can divide through by $b$ to assume that $b=1$, and then replace $X,Y$ with $aX,aY$ to obtain</p>
<script type="math/tex; mode=display">a^3Y^2+a_1a^2XY+a_3aY=a^3X^3+a_2a^2X^2+a_4aX+a_6.</script>
<p>Finally, dividing by $a^3$, we see that our elliptic curve $E$ satisfies a relation of the form</p>
<script type="math/tex; mode=display">\label{wform}\begin{align}Y^2+a_1XY+a_3Y=X^3+a_2X^2+a_4X+a_6.\end{align}</script>
<p>I just said finally, but you potentially expected a simpler looking end result. This is the best we can do over an arbitrary field $k$, but if we further suppose that $\Char k\neq2,3$ then we can (simultaneously) replace $X$ by $(X-3a_1^2-12a_2)/36$ and $Y$ by $(Y-(a_2X+a_3)/2)/216$ <sup id="fnref:22"><a href="#fn:22" class="footnote">22</a></sup> to get a relation of the form $Y^2=X^3+aX+b$.</p>
<p>Before talking about the more arithmetic side of elliptic curves, I would like to explain to what extent these relations define the curve $E$, so fix $a_1,a_2,a_3,a_4,a_6\in k$ such that (\ref{wform}) holds. Then, we claim that $\parens{\ints E(D_3),(1,X,Y)}$ determines a morphism $f:E\to\P_k^2$. We need to check that $1,X,Y$ have no common zeros. Since these generate $\hom^0(\ints E(D_3))$, it suffices to check that there’s no (closed) $p\in E$ s.t. $s(p)=0\in\ints E(D_3)(p)\simeq\kappa(p)$ for all $s\in\hom^0(\ints E(D_3))$. To see this, consider the exact sequence</p>
<script type="math/tex; mode=display">0\too\ints E(-p)\too\ints E\too\ints p\too0.</script>
<p>Twist by $D_3$ (i.e. tensor with $\ints E(D_3)$) and look in cohomology to get</p>
<script type="math/tex; mode=display">0\too\hom^0(\ints E(D_3-p))\too\hom^0(\ints E(D_3))\too\kappa(p)\too\hom^1(\ints E(D_3-p))=0,</script>
<p>where the last equality comes from Serre duality. The map $\hom^0(\ints E(D_3))\too\kappa(p)$ above is the evaluation map, so we see that it is surjective, i.e. that some section has nonzero evaluation at $p$. Hence, we do get a morphism $f:E\too\P_k^2$, and further analysis can be used to show that it is a closed embedding. Once you know this, one can show that the section $X,Y$ of $\ints E(D_3)$ extend to global sections of the line bundle $\ints{\P_k^2}(1)$ on $\P^2$ (i.e. that $X,Y$ are really linear functions on $\P^2$). Hence, the relation (\ref{wform}) is really prescribing a way to write $E$ as a hypersurface in $\P^2$ (i.e. $E$ is the vanishing set of the homogenization of that polynomial)</p>
<h2 id="group-law">Group Law</h2>
<p>Now, the main source of interest in elliptic curves ultimately stems from the fact that their rational points form a group (actually, their $S$-points for any $k$-scheme $S$ do). We will describe how in this section.</p>
<p>Let $E/k$ be an elliptic curve with chosen basepoint $\infty\in E(k)$. Let $\Pic^0(E)\subset\Pic E=\ker\deg$ be the subgroup of (linear equivalence classes of) degree $0$ divisors. Consider the map</p>
<script type="math/tex; mode=display">\mapdesc f{E(k)}{\Pic^0E}{p}{[p]-[\infty]}.</script>
<p>Let $X,Y\in\hom^0(\ints E(3[\infty]))$ be as in the previous section. We aim to show that the above map is a bijection. This will allows us to pullback the group structure on $\Pic^0E$ to one on $E(k)$. We first show injectivity. Suppose that we have $p,q\in E(k)$ such that $[p]-[\infty]=[q]-[\infty]$ so $[p]=[q]$. Consider the exact sequence</p>
<script type="math/tex; mode=display">0\too\ints E(p-q)\too\ints E(p)\too\ints q\too0</script>
<p>This gives an injection $\hom^0(\ints E(p-q))\into\hom^0(\ints E(p))$. If $p\neq q$, then every section of $\ints E(p-q)$ must vanish at $q$, but $\hom^0(\ints E(p))$ is generated by the nowhere vanishing section $1$, so $\hom^0(\ints E(p-q))=0$ in this case which shows that $[p]\neq[q]\in\Cl E$ if $p\neq q\in E$.</p>
<p>Now, we show that $f$ is surjectivity. Let $D\in\Div E$ be a degree 0 divisor. Then, $D+[\infty]$ is degree 1, so there’s some nonzero global section $s\in\hom^0(\ints E(D+[\infty]))$. As usual, this means that $D+[\infty]$ is equivalent to some degree 1, effective divisor $E$ which must be of the form $E=[p]$ for some $p\in E(k)$. Thus, $D=[p]-[\infty]=f(p)$, so $f$ is bijective.</p>
<p>This allow us to define a group law on $E(k)$ via $p+q=\inv f(f(p)+f(q))$. As constructed, this group law is potentially arbitrary. However, one can show that it is actually induced from a morphism $m:E\by E\to E$ of $k$-schemes. In fancier words, this group law is a manifestation of the fact that $E$ is a $k$-group scheme <sup id="fnref:24"><a href="#fn:24" class="footnote">23</a></sup>. We will use without proof the existence of the morphism $m$.</p>
<div class="exercise">
Show that the group law on $E(k)$ is characterized by the following 2 properties
<ol>
<li> The basepoint $\infty\in E(k)$ is the identity element. </li>
<li> For any 3 points $p,q,r\in E(k)$ "lying on a line" (i.e. $p+q+r=(s)$ for some $s\in\hom^0(\ints E(D_3))$), we have
$$p+q+r=0.$$
</li>
</ol>
</div>
<h2 id="isogenies">Isogenies</h2>
<div class="definition">
Let $E,E'$ be elliptic curves over $k$. An <b>isogeny</b> $\phi:E\to E'$ is a morphism respecting base points, i.e. one such that $\phi(\infty)=\infty'$.
</div>
<p>Since $E,E’$ are 1-dimensional, irreducible, we either have $\phi(E)=\infty’$ or $\phi(E)=E’$ for any isogeny $\phi$. Like before, given a non-constant isogeny $\phi:E\to E’$, we let $\deg\phi$ denote the degree of the field extension $k(E)/\pull\phi k(E’)$.</p>
<p>Unless otherwise stated, assume that basically all isogenies below are nonzero.</p>
<div class="example">
For an elliptic curve $E$ and an integer $m\ge0$, there is a multiplication by $m$ isogeny $E\xto mE$. We will see later that $\deg m=m^2$.
</div>
<p>One nice properties of isogenies is that they are automatically group homomorphisms.</p>
<div class="proposition">
Let $\phi:E_1\to E_2$ be an isogeny. Then, the induced map $E_1(k)\to E_2(k)$ is a group homomorphism.
</div>
<div class="proof4">
Let $f_i:E_i(k)\to\Pic^0E_i$ ($i=1,2$) be the map you expect. Then, $\phi$ induces a group homomorphism $\push\phi:\Pic E_1\to\Pic E_2$ given on prime divisors (i.e. closed points) by $[p]\mapsto[\phi(p)]$. Hence, for $p,q\in E_1(k)$, we have
$$\begin{align*}
\phi(p+q)
&=\phi(\inv f_1(f_1(p)+f_1(q)))\\
&=\inv f_2(\push\phi(f_1(p)+f_1(q)))\\
&=\inv f_2(\push\phi(f_1(p))+\push\phi(f_1(q)))\\
&=\inv f_2(f_2(\phi(p))+f_2(\phi(q)))\\
&=\phi(p)+\phi(q)
\end{align*}$$
Above we used the fact, which follows directly form the definitions, that $\push\phi(f_1(p))=f_2(\phi(p))$.
</div>
<p>The above shows that isogenies basically correspond to homomorphisms between Picard groups. Under this correspondence, an isogeny $\phi:E_1\to E_2$ is paired with the homomorphism $\push\phi:\Pic^0E_1\to\Pic^0E_2$. However, there is another homomorphism of Picard groups naturally associated to $\phi$: the pullback $\pull\phi:\Pic E_2\to\Pic E_1$, so this too should correspond to some isogeny.</p>
<div class="definition">
Let $\phi:E_1\to E_2$ be an isogeny. Its <b>dual isogeny</b> $\wh\phi:E_2\to E_1$ is given by the composition
$$E_2\iso\Pic^0E_2\xto{\pull\phi}\Pic^0E_1\iso E_1.$$
</div>
<div class="exercise">
Show that formation of the dual isogeny satisfies
$$\wh{\phi+\psi}=\wh\phi+\wh\psi$$
and
$$\wh{\phi\circ\psi}=\wh\psi\circ\wh\phi$$
when each expression is defined.
</div>
<div class="exercise">
Let $\phi_m:E\to E$ denote the multiplication by $m$ map on some elliptic curve $E$. Then, show that its dual isogeny $\wh\phi_m:E\to E$ is still multiplication by $m$ (hint: use that $\wh\phi+\wh\psi=\wh{\phi+\psi}$ in general)
</div>
<p>Now, let $\phi:E_1\to E_2$ be an isogeny. The composition $\phi\circ\wh\phi:E_2\to E_2$ corresponds to the map $\push\phi\pull\phi:\Pic^0E_2\to\Pic^0E_2$ which is just multiplication by $(\deg\phi)$. Hence, $\phi\circ\wh\phi:E_2\to E_2$ is also multiplication by $\deg\phi$. We claim the same is true for the other composition $\wh\phi\circ\phi:E_1\to E_1$, i.e. that this is multiplication by $(\deg\phi)$. To see this, note that</p>
<script type="math/tex; mode=display">(\wh\phi\circ\phi)\circ\wh\phi=\wh\phi\circ(\deg\phi)=(\deg\phi)\circ\wh\phi</script>
<p>where the last equality comes from $\wh\phi$ being a homomorphism. Since $\wh\phi$ is surjective, we conclude that $\wh\phi\circ\phi=\deg\phi$ (as functions). As a consequence of the last exercise, we get</p>
<div class="corollary">
The multiplication by $m$ map has degree $m^2$.
</div>
<p>We can also show that $\deg\phi=\deg\wh{\phi}$ in general. From our above calculations on compositing an isogeny with its dual, this will follow from the following.</p>
<div class="proposition">
Let $\phi:E_1\to E_2$ be an isogeny, and let $\psi:E_2\to E_1$ be a map such that $\psi\circ\phi$ is multiplication by $m=\deg\phi$. Furthermore, $\psi=\wh\phi$. In particular, $\wh{\wh\phi}=\phi$.
</div>
<div class="proof4">
We know that $\wh\phi\circ\phi=m$, so we have $(\psi-\wh\phi)\circ\phi=0$, the constant map. Since $\phi$ is surjective, this shows that $\psi-\wh\phi=0$, so we win.
<br />
For the claim that $\wh{\wh\phi}=\phi$, this follows from the fact that
$$\phi\circ\wh\phi=m=\wh m=\wh{\phi\circ\wh\phi}=\wh{\wh\phi}\circ\wh\phi=\deg\wh\phi.$$
</div>
<h2 id="torsion-points">Torsion Points</h2>
<p>We now turn to understanding the structure of torsion points of an elliptic curve. In order to have a chance of getting nice results, fix an <b>algebraically closed</b> field $k$, and let $E/k$ be an elliptic curve. Let $E(k)[m]=\ker(m:E(k)\to E(k))$ denote the set of $m$-torsion points of $E(k)$ (note: since $k$ is algebraically closed, $E(k)$ is the set of all closed points) where $m\ge1$ is an integer. Our understanding of $E(k)[m]$ will rest on the following lemma relating the algebra of a morphism to its geometry.</p>
<div class="lemma">
Let $\phi:C_1\to C_2$ be a finite morphism of smooth curves over $k$. Then, there exists a dense open $U\subset C_2$ such that $\#\inv\phi(q)=\deg_s\phi$ for all closed $q\in U$, where $\deg_s=[k(C_1):k(C_2)]_s$ is the <b>separable degree</b> of $\phi$.
</div>
<div class="proof4">
Start with an affine open $V=\spec A$ around the generic point $\eta_2\in C_2$ of $C_2$. Since $\phi$ is finite, we can write $U:=\inv\phi(V)=\spec B$ with $\phi:\spec B\to\spec A$ giving $B$ the structure of a finite $A$-module. Since $C_1,C_2$ are smooth curves, we claim that the rings $A,B$ are Dedekind domains. It is clear that they are 1-dimensional (e.g. since open subsets of irreducible spaces are irreducible), noetherian domains, so we only need to show that they are integrally closed. Take $A$ for example. We have
$$A=\bigcap_{\mfp\in\spec A}A_\mfp\subset\Frac A=k(C_2).$$
By an earlier proposition, $A_\mfp$ is integrally closed for all $\mfp$, so $A$ is an intersection of integrally closed rings, and hence integrally closed. The same reasoning applies to $B$. Thus, $A\into B$ is an integral (because finite) extension of Dedekind domains.
<br />
Hence, given some $q\in V$ (corresponding to a prime $\mfp\in\spec A$), determining $\inv\phi(q)$ is the same as determining the primes $\mfP\in\spec B$ lying above the given $\mfp\in\spec A$. Since every prime in a purely inseparable extension of Dedekind domains completely ramifies, as far as determining the set $\inv\phi(q)$ is considered, we lose no information if we assume that $\Frac B=k(C_1)$ is separable over $\Frac A=k(C_2)$. It is now a standard result from the theory of Dedekind domains that in any such situation, a (nonzero) prime $\mfp\in\spec A$ factors in $B$ as a product
$$\mfp B=\prod_{i=1}^g\mfP_i^{e_i}$$
of primes $\mfP_i\in\spec B$ with $n=[k(C_1):k(C_2)]=\sum_{i=1}^ge_if_i$ where $f_i=[B/\mfP_i/A/\mfp]$. Furthermore, the ramification indices $e_i$ are all equal to $1$ except for the (finite set of) primes dividing a certain "discriminant ideal" $\Delta_{B/A}\subset A$. Thus, we see that $\#\inv\phi(q)=\deg_s\phi$ for any (nonzero) $q$ in the open set $\spec A\sm V(\Delta_{B/A})$, and so we win.
</div>
<div class="corollary">
Let $\phi:E_1\to E_2$ be an isogeny of elliptic curves over $k$. Then, $\#\inv\phi(q)=\deg_s\phi$ for all $q\in E_2(k)$.
</div>
<div class="proof4">
Apply lemma to get this for a dense open, and then take any point and translate it into that dense open.
</div>
<p>In the previous section we calculated that $\deg m=m^2$ (where $m$ is denoting both an integer and the multiplication by that integer map). With this in mind, fix any nonzero $m\in\Z$ such that $p:=\Char k\nmid m$. In this case, multiplication by $m$ is separable, so we get that <sup id="fnref:25"><a href="#fn:25" class="footnote">24</a></sup></p>
<script type="math/tex; mode=display">\# E(k)[m]=\#\inv m(\infty)=m^2</script>
<p>Let $G=E(k)[m]$. For any $d\mid m$, let $G[d]$ denote its $d$-torsion subgroup, so $G[d]=E(k)[d]$. Thus, $G$ is an abelian group of size $m^2$ whose $d$-torsion subgroup has size $d^2$ for all $d\mid m$. An argument inducting on the number of primes dividing $m$ then shows that this implies that $G\simeq\zmod m\oplus\zmod m$.</p>
<p>Now, what if $m=p^e$ for some $e\ge1$ (Here, we’re assuming $p=\Char k\neq0$)? In this case, consider the $p$th power Frobenius map $\phi:E\to E$ which is the morphism $E\to E$ corresponding to the $p$ power map $k(E)\to k(E)$ on $E$’s function field <sup id="fnref:26"><a href="#fn:26" class="footnote">25</a></sup>. This map visibly has degree $p$ and separable degree $1$, so we see that</p>
<script type="math/tex; mode=display">\# E[p^e]=\deg_s[p^e]=\deg_s(\wh\phi\circ\phi)^e=(\deg_s\wh\phi)^e\in\bracks{1,p^e}</script>
<p>where the ambiguity at the ends rests upon whether $\wh\phi$ is separable or inseparable. In the case that it is separable, $E[p^e]$ is a group of order $p^e$ whose $p^n$-torsion subgroup is of order $p^n$ for all $n\le e$. Another easy induction argument shows that this implies that $E[p^e]\simeq\zmod{p^e}$.</p>
<p>All in all, we have shown the following.</p>
<div class="theorem">
Let $m\in\Z$ be a nonzero integer, and let $p=\Char k\ge0$. Then,
<ul>
<li> If $m\neq0\in k$, i.e. if $p\nmid m$, we have
$$E[m]=\Zmod m\oplus\Zmod m.$$
</li>
<li> If $\Char k=p>0$, then one of the following is true. Either
$$E[p^e]=\{\infty\}$$
for all $e\ge1$, or
$$E[p^e]=\Zmod{p^e}$$
for all $e\ge1$.
</li>
</ul>
</div>
<h1 id="l-adic-representations">$\l$-adic Representations</h1>
<p>To end this monster of a blog post, we will show how to use an Elliptic curve $E/\Q$ to construct an $\l$-adic representation of the absolute Galois group $G_\Q:=\Gal(\Qbar/\Q)$.</p>
<p>Fix an elliptic curve $E/\Q$. Note that the set (really, group) $E(\Qbar)$ (morphisms as $\Q$-schemes) is naturally isomorphic to $\bar E(\Qbar)$ (mophisms as $\Qbar$-schemes) for a uniquely determined elliptic curve $\bar E/\Qbar$ <sup id="fnref:27"><a href="#fn:27" class="footnote">26</a></sup>, so the results of the previous (sub)section apply to $E(\Qbar)$. In particular, fixing a prime $\l$, for all $n\ge1$, we have</p>
<script type="math/tex; mode=display">E(\Qbar)[\l^n]=\Zmod{\l^n}\oplus\Zmod{\l^n}.</script>
<p>Now, note that $G_\Q$ has a natural (left) action on $E(\Qbar)$. Spelled out, given $\Qbar$-point $x:\spec\Qbar\to E$ and an automorphism $\sigma\in G_\Q$, we let $\sigma\cdot x\in E(\Qbar)$ be the composition</p>
<script type="math/tex; mode=display">\spec\Qbar\xto{\spec\sigma}\spec\Qbar\xto xE.</script>
<p>Since multiplication by $m$ is defined over $\Q$, this action commutes with the multiplication by $m$ map, and so restrictions to a linear action $G_\Q\actson E(\Qbar)[m]$ for all $m\ge1$. Thus, we have compatible maps</p>
<script type="math/tex; mode=display">G_\Q\too\Aut(E(\Qbar)[\l^n])\simeq\Aut\parens{\Zmod{\l^n}\oplus\Zmod{\l^n}}</script>
<p>for all $n\ge1$. Letting, $T_\l(E):=\invlim_{n\ge1}E(\Qbar)[\l^n]$ be the <b>($\l$-adic) Tate module</b> and taking an inverse limit of the above maps, we get our desired representation</p>
<script type="math/tex; mode=display">\rho_{E,\l}:G_\Q\too\Aut(T_\l(E))\simeq\Aut(\Z_\l\oplus\Z_\l)\simeq\GL_2(\Z_\l)\into\GL_2(\Q_\l)</script>
<p>of $G_\Q$.</p>
<p>Assuming I continue this series and things work out roughly the way I hope they do, we will in a later post show that this representation is irreducible most of the time.</p>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>Not necessarily in this order <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>I feel like I keep semi-accidentally writing super long posts, and I’m not a fan of this (especially this one. It probably did not need to be anywhere near this long) <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>I’ll include exercise. I encourage you to do some (but probably not all) of them just to have practice thinking about this stuff. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>On the off chance I succeed in writing this post and putting it online, I should mention that you can probably go straight to the part about curves, pretend the word scheme does not exist, and still manage to follow. I hope to make things fairly concrete because then there is less theory I need to remember how to set up <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:23">
<p>The part before elliptic curves is kinda messy right now because of a long series of rewriting it as I realized setting things up would be a more involved process than i anticipated. If you read it, watch out for typos/mistakes (In particular, watch out for places where I’m implicitly assuming a field is algebraically closed when I shouldn’t be). If you find mistakes (or if things are unclear), leave a comment <a href="#fnref:23" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>collection of charts <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>All of this data is still there, but now neatly packaged into the structure sheaf $\ints X$ <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>In either case, the morphism of sheaves is not required to have much of anything to do with the map on underlying spaces <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>In the classical setting, $A=k[x_1,\dots,x_n]/\sqrt I$ is a f.g. $k$-algebra and $\spec A$ corresponds to the set $V(I)=\bracks{(a_1,\dots,a_n)\in k^{\oplus n}:f(a_1,\dots,a_n)=0\,\forall f\in I}\subset k^{\oplus n}$, so the global functions on $V(I)=\spec A$ are given by $A=k[x_1,\dots,x_n]/\sqrt I$. <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:9">
<p>Given $\msF\in\Ab(X)$, embed the stalk $\msF_x\into I_x$ into an injective group. For each $x\in X$, let $j_x:{x}\into X$ denote the inclusion, and consider the sheaf $\msI=\prod_{x\in X}j_{x,* }(I_x)$ where $I_x$ is viewed as a sheaf on ${x}$. This is an injective sheaf into which $\msF$ embeds. <a href="#fnref:9" class="reversefootnote">↩</a></p>
</li>
<li id="fn:11">
<p>It’s an (annoying to prove) theorem that cohomology (for the sheaves we care about) vanish in degrees above the dimension of the underlying space <a href="#fnref:11" class="reversefootnote">↩</a></p>
</li>
<li id="fn:15">
<p>Really, I’ll claim without proof <a href="#fnref:15" class="reversefootnote">↩</a></p>
</li>
<li id="fn:12">
<p>I know this notation is trash, but it’s also short-lived <a href="#fnref:12" class="reversefootnote">↩</a></p>
</li>
<li id="fn:13">
<p>Hints: (1) A uniformizer in $\ints{C,p}\subset k(C)$ is like a local coordinate centered at $p$ and (2) the function field $k(C)$ is the stalk at the generic point $\eta\in C$ (i.e. the point contained in all open sets (i.e. the point whose closure is all of $C$ (i.e. the point corresponding to the zero ideal in any affine open))) <a href="#fnref:13" class="reversefootnote">↩</a></p>
</li>
<li id="fn:14">
<p>You may also want to appeal to <b>Nike’s trick</b>: for $\spec A,\spec B\subset X$ affines in a general scheme $X$, we can cover the intersection $\spec A\cap\spec B$ with affines $U$ that are distinguished in both $\spec A,\spec B$ (i.e. $U=\spec A_a\subset\spec A$ and $U=\spec B_b\subset\spec B$ for some $a\in A$ and $b\in B$) <a href="#fnref:14" class="reversefootnote">↩</a></p>
</li>
<li id="fn:16">
<p>Plus all the other stuff I’ve asked you to take for granted <a href="#fnref:16" class="reversefootnote">↩</a></p>
</li>
<li id="fn:17">
<p>For example, if you are working over $\C$ where you have access to topological methods, then $f$ is generally (and I think always) a function of the chern classes of $\msF$ which only see the bundle’s underlying topology (and not its complex/holomorphic structure) <a href="#fnref:17" class="reversefootnote">↩</a></p>
</li>
<li id="fn:19">
<p>This boils down to the fact that the unit ideal $(1)=A$ of a ring $A$ has a finite set of generators. <a href="#fnref:19" class="reversefootnote">↩</a></p>
</li>
<li id="fn:20">
<p>I haven’t thought this through, so I could be wrong, but for curves, the given definition of completeness is the same as being proper. Briefly, Riemann-Roch (+ some work) shows that a complete curve $C$ can be embedded in some projective space $\P_k^N$ (as a closed subset), but projective space is proper and so its closed subsets are too. <a href="#fnref:20" class="reversefootnote">↩</a></p>
</li>
<li id="fn:18">
<p>and also retroactively for some of the earlier stuff (e.g. Serre duality) <a href="#fnref:18" class="reversefootnote">↩</a></p>
</li>
<li id="fn:21">
<p>This it the standard way to label the indices for some reason. I guess note that the index + the valuation of the corresponding monomial always equals 6 <a href="#fnref:21" class="reversefootnote">↩</a></p>
</li>
<li id="fn:22">
<p>I think these are the right substitutions. I haven’t actually checked <a href="#fnref:22" class="reversefootnote">↩</a></p>
</li>
<li id="fn:24">
<p>By Yoneda, to show this, it suffices to give $E(S)$ a (functorial in $S$) group law for all $k$-schemes $S$. I believe, but have not checked, that one can do this by showing that $E(S)$ is in natural bijection with $\Pic^0(E\by_kS)$ where $E\by_kS$ is the (categorical) pullback of $E\to\spec k\from S$.<br /> Actually, I think there’s a slicker way to do this. Maybe I’ll come back and type something up at some point… <a href="#fnref:24" class="reversefootnote">↩</a></p>
</li>
<li id="fn:25">
<p>The middle expression here really shows that I made some poor choices of notation <a href="#fnref:25" class="reversefootnote">↩</a></p>
</li>
<li id="fn:26">
<p>I did not touch on this before, but morphisms between smooth curves are in bijection field maps between their function fields. <a href="#fnref:26" class="reversefootnote">↩</a></p>
</li>
<li id="fn:27">
<p>$\bar E$ is the “base change of $E$ to $\Qbar$”. Intuitively, it results from taking $E/\Q$ and then extending scalars from $\Q$ to $\Qbar$. <a href="#fnref:27" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>I have a vague vision in my head of a series of posts designed to help me, and maybe also you, understand (parts of) the relationship between a few objects of interests in algebraic number theory: elliptic curves, selmer groups, binary quartic forms, and modular forms 1. I don’t know the whole story encompassing these objects, but allegedly, they are all linked in one way or another, and I have soft plans to understand these links over the next $n$ days. To aid in this, I hope to write (medium length to long 2) posts on each of these objects, drawing attention to their relationships to each other, and also including some facts/results I find interesting even if they don’t necessarily lie at the intersection of the study of e.g. elliptic curves and quartic forms. It seems most appropriate to begin this series with elliptic curves since they are the main reason I care about any of the other stuff. Not necessarily in this order ↩ I feel like I keep semi-accidentally writing super long posts, and I’m not a fan of this (especially this one. It probably did not need to be anywhere near this long) ↩Brown Representability2019-12-26T00:00:00+00:002019-12-26T00:00:00+00:00https://nivent.github.io/blog/brown-rep<p>Let’s prove and discuss one of my favorite theorems from algebraic topology. It gives a characterization of when a contravariant “homotopy functor” from topological spaces <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup> to sets is representable, i.e. is (isomorphic to one) of the form $F(X)=[X,S]$ where $S$ is $F$’s representing object, and the notation $[X,S]$ denotes the set of homotopy classes of maps $X\to S$. The nice part about this is that many nice/important spaces one encounters when studying algebraic topology are secretly just representing objects for some functor that people care about. Instead of constructing these spaces directly, one can infer their existence from Brown Representability, and then use properties of the functor they represent to understand them as objects. Vice versa, knowing that a functor is representable gives you additional control over its behavior/properties. This will make more sense later when I discuss some example applications.
$
\newcommand{\CW}{\mathrm{CW}}
\newcommand{\Top}{\mathrm{Top}}
\newcommand{\Set}{\mathrm{Set}}
\DeclareMathOperator{\Ho}{Ho}
$</p>
<p>Before starting, I should probably make clear in what category we’re actually working. Based on the fact that I used the phrase “homotopy functor” and that I’m called a functor representable when it spits out homotopy classes of maps, possibly the first category that comes to mind is something like: objects are topological spaces and morphisms are homotopy classes of maps. I’ll call this the naive homotopy category of Top because it is not quite what we want; in addition to considering maps only up to homotopy, you also want to artificially invert weak equivalences <sup id="fnref:2"><a href="#fn:2" class="footnote">2</a></sup> so that they become isomorphisms. In order to not have to worry about annoyances coming from artificially adding morphisms, we’ll just work with CW complexes instead of arbitrary spaces. The point is Whitehead’s theorem says that weak equivalences already are honest-to-God homotopy equivalences for CW complexes, and CW approximation tells us that every topological space is weakly equivalent to a CW complex, so these two together tells us that the (non-naive) homotopy category is equivalent to the naive homotopy category restricted to CW complexes <sup id="fnref:3"><a href="#fn:3" class="footnote">3</a></sup>. Lastly, because we’re doing algebraic topology and because we’re not savages, we want to work in a category of based spaces/maps so throughout we’ll assume that every space $X$ we encounter has some chosen basepoint, which we’ll denote by $x\in X$ or $*\in X$ depending on what’s convenient, and every map $f:X\to Y$ respects basepoints (i.e. $F(*)=*$). Finally, we need to assume all our spaces are connected. With all that said, let $\push\CW$ denote the category whose objects are based connected CW-complexes, and whose morphisms are (based) homotopy classes of continuous maps <sup id="fnref:4"><a href="#fn:4" class="footnote">4</a></sup>. Hence, unless stated otherwise, assume all spaces below are based, connected CW complexes. <sup id="fnref:5"><a href="#fn:5" class="footnote">5</a></sup></p>
<h1 id="a-warmup">A Warmup</h1>
<p>Before tackling the main theorem, let’s work through a case that is easy enough to do “by hand.” Namely, we’ll prove that first singular cohomology is represented by the circle, i.e. $\hom^1(X;\Z)\simeq[X,S^1]$. We know that $\hom^1(S^1;\Z)\simeq\Z$, so let $\alpha\in\hom^1(S^1;\Z)$ be a generator, and consider the map</p>
<script type="math/tex; mode=display">\mapdesc T{[X,S^1]}{\hom^1(X;\Z)}{[f]}{\pull f\alpha}</script>
<p>where $[f]$ denote the equivalence class of a map $f:X\to S^1$. We aim to show that $T$ is a bijection. In fact, in this case, $\hom^1(-;\Z)$ is not just a functor to Set, but actually a functor to $\Ab$, so one may wonder if $T$ could end up being a group isomorphism and not just a bijection. This is in fact the case. Give $[X,S^1]$ the group structure it gets from $S^1\subset\C$ being a (topological) group; I’ll leave it as an exercise to show that this map $T$ is a homomorphism (this should just be definition chasing). Because the proofs of injectivity/surjectivity of $T$ that we will see are of different flavors, we will give them separately.</p>
<p>We first show that $T$ is injective.</p>
<div class="proposition">
$T$ is injective.
</div>
<div class="proof4">
Fix $f:X\to S^1$ such that $\pull f\alpha=0$. Then, for any $[g]\in\pi_1(X)$ (i.e. $g:S^1\to X$), we have that $\push f(g)=f\circ g:S^1\to S^1$ is nullhomotpic since it's a degree $0$ map (as $\pull{(f\circ g)}=\pull g(\pull f\alpha)=0$ in cohomology and $\pi_1(S^1)=[S^1,S^1]\simeq\Z$ with the isomorphism given by the degree map). Thus, $f$ induces the zero map on fundamental groups (and so does so on all homotopy groups since $\pi_n(S^1)=0$ for $n\neq1$). Since $\push f\pi_1(X)=0$, we can lift $f$ to a map $\wt f:X\to\R$. In other words, $f$ factors as $X\xto{\wt f}\R\to S^1$. Since $\R$ is contractible, $f$ is nullhomotpic.
</div>
<p>To show that $T$ is surjective, we’ll first need a simple lemma.</p>
<div class="lemma">
Let $f:X\to Y$ be a map, and let $CX=(X\by I)/(X\by\{1\})$ be the cone on $X$. Then, $f$ extends to a map $CX\to Y$ ($X\into CX$ via $x\mapsto(x,0)$) iff it is nullhomotopic.
</div>
<div class="proof4">
$(\to)$ Say $f$ extends to a map $CX\to Y$. Then, $f$ factors as $X\to CX\to Y$ but $CX$ is contractible, so $f$ is nullhomotopic.<br />
($\from$) Say $h_t:X\to Y$ is a homotopy with $h_0=f$ and $h_1=*$ (constant map to basepoint). Then, $h_t$ is a map $X\by I\to Y$ which is constant on $X\by\{1\}$, but this is exactly a map $CX\to Y$, so we win.
</div>
<div class="proposition">
$T$ is surjective.
</div>
<div class="proof4">
This argument will have quite a bit more setup than the last. We start with a few observations. Note that Hurewicz + Universal Coefficients shows that
$$\hom^1(X;\Z)\simeq\Hom(\hom_1(X;\Z),\Z)\simeq\Hom(\pi_1(X),\Z)$$
since $\Z$ is abelian. Under these isomorphisms, one sees that for $f:X\to S^1$, we have $T(f)(g)=\alpha(f\circ g)$ where $\alpha\in\Hom(\pi_1(S^1),\Z)$ is still the generator, and $T(f)\in\Hom(\pi_1(X),\Z)$ and $g:S^1\to X$. This perspective let's us rephrase surjectivity of $T$ as the ability to extend certain maps. Specifically, let $A=\bigvee_{\gamma\in\pi_1(X)}S^1_\gamma$ where each $S^1_\gamma$ is a copy of the circle $S^1$, and let $h:A\to X$ be a map s.t. the restriction $h_\gamma:=h\vert_{S^1_\gamma}:S^1_\gamma\to X$ satisfies $[h_\gamma]=\gamma$ for all $\gamma\in\pi_1(X)$. Now, let $\beta\in\hom^1(X;\Z)$ be an arbitrary cohomology class, and let $g_\beta:A\to S^1$ be a map such that $g_\beta\vert_{S^1_\gamma}:S^1_\gamma\to S^1$ is a degree $\beta(\gamma)$ map for all $\gamma$. Then, $\beta$ is in the image of $T$ iff there exists some $f_\beta:X\to S^1$ such that the following diagram commutes
<center>
<img src="https://nivent.github.io/images/blog/brown-rep/extend.png" width="150" height="100" />
</center>
i.e. if we can extend $g_\beta$ to a map on $X$. In order to make this language of extension more natural, we replace $X$ with the mapping cylinder $M_h$ (this is legal since the projection $M_h\to X$ is a homotopy equivalence), and so assume that $h:A\to X$ is inclusion of a subcomplex. Finally, observe that $S^1$ is a $K(\Z,1)$, and so $S^1\simeq\Omega K(\Z,2)$ is (homotopically equivalent to) a loop space. With all that said, we can now actually prove this thing. Thinking of $S^1$ as a loop space, the map $A\xto{g_\beta}S^1\simeq\Omega K$ (with $K=K(\Z,2)$) gives rise to a homotopy $H_t:A\to K$ given by $H_t(a)=g_\beta(a)(t)$ (recall, a loop is a map $I\to K$, equal on the endpoints). Note that $H_1(a)=g_\beta(a)(1)=*$ since all loops are based, so $H_t$ is a nullhomotopy. By the previous lemma, this means $H_t$ is equivalently a map $CA\to K$. Similarly, $H_0(a)=*$ so $H_0$ extends to a constant map $X\to K$. Combining this with previous map gives us a map $G_0:X\cup CA\to K$ (constant on $X$, and $H_t$ on $CA$). This map is nullhomotpic via $G_t(x)=*$ for $x\in X$ and $G_t(a,s)=H_{s+(1-s)t}(a)$ for $(a,s)\in A\by I$. Now, $X\cup CA$ is a subcomplex of $CX$, and so the inclusion $X\cup CA\into CX$ satisfies the homotopy extension property. The constant map $\overline G_1:CX\to K$ extends the starting endpoint of the homotopy $G_{1-t}:X\cup CA\to K$, and so we can continue $\overline G_1$ to an entire homotopy $\overline G_{1-t}:CX\to K$ which extends $G_{1-t}$. In particular, $\overline G_0$ is a map $CX\to K$ extending the map $G_0:X\cup CA\to K$ which was constant on $X$; i.e., $\overline G_0$ is a homotopy $X\by I\to K$ which is constant at both endpoints; i.e. $\overline G_0$ is a map $X\to\Omega K\simeq S^1$. This is our desired extension $f_\beta$.
</div>
<p>At last, we have shown that $[X,S^1]\simeq\hom^1(X;\Z)$ for connected $\CW$-complexes $X$. Now, in light of our discussion that the homotopy category of (based, connected) CW complexes is equivalent to that of (based, connected) topological spaces, it may be tempting to conclude from this that in fact $[X,S^1]\simeq\hom^1(X;\Z)$ for all based, connected topological spaces $X$. However, this is false. The point is that, in the homotopy category, weak equivalences are meant to be isomorphisms, but singular cohomology is not invariant under weak equivalences, so $\hom^1(-;\Z)$ isn’t a functor on $\Ho(\push\Top)$, the homotopy category on topological spaces. The correct functor on $\Ho(\push\Top)$ corresponding to singular cohomology on $\push\CW$ is $\hom^1(Z(-);\Z)$ where $Z(X)$ is a (functorial) $CW$ approximation of $X$.</p>
<p>Before moving onto Brown representability in full generality, it is worth mentioning an extension of the argument here. In this section, we showed that $[X,K(\Z,1)]\simeq\hom^1(X;\Z)$; later in this post, we’ll use Brown representability to say furthermore that, for any abelian group $G$ and $n\ge1$, we have $[X,K(G,n)]\simeq\hom^n(X;G)$. The isomorphism here is again given by pulling back a “universal cohomology class on $K(G,n)$” which is defined as the image of the identity map under the isomorphism</p>
<script type="math/tex; mode=display">\Hom(G,G)=\Hom(\pi_n(K(G,n)),G)\simeq\Hom(\hom_n(K(G,n)),G)\simeq\hom^n(K(G,n);G)</script>
<p>coming from Hurewicz + Universal Coefficients. If you read over the proof of surjectivity given here, it converts essentially immediately into a proof of surjective for general $K(G,n)$. The injectivity, however, made more usage of specific features of $S^1$ (e.g. it has a contractible universal cover) so one needs a different strategy to prove injectivity for arbitrary Eilenberg-Maclane spaces (although not when $n=1$).</p>
<h1 id="the-theorem">The Theorem</h1>
<p>Now, that we’re all warmed up, let’s get into the meat of things. We seek a characterization for when a (contravariant) functor $F:\push\CW\to\Set$ is representable. Let’s start by finding some necessary conditions, so fix a space $X$ and consider the functor $h_X=[-,X]:\push\CW\to\Set$; what are some general properties of $h_X$?</p>
<p>Before answering this, let’s introduce a bit of notation. Given a contravariant function $F:\push\CW\to\Set$ and an open embedding $i:U\into V$, we get a map $F(i):F(V)\to F(U)$. Given any $x\in F(V)$, denote its image $F(i)(x)\in F(U)$ as $x\vert_U$, taking inspiration from the notation for restricting functions. Now, returning to our functor $h_X$, it satisfies…</p>
<ol>
<li> (<b>Wedge Axiom</b>)
First, $h_X$ turns wedges (i.e. coproducts in $\push\CW$) into products. Symbolically,
$$h_X(\bigvee_\alpha A_\alpha)=\sqbracks{\bigvee_\alpha A_\alpha,X}\simeq\prod_\alpha[A_\alpha,X]=\prod_\alpha h_X(A_\alpha),$$
since a map out of a coproduct is the same thing as a map out of each factor.
</li>
<li> (<b>Mayer-Vietoris Axiom</b>)
Secondly, we can glue together maps agreeing on an overlap. In other words, if $A=U\cup V$ and $u\in h_X(U),v\in h_X(V)$ are maps agreeing on $U\cap V$ (i.e. they have the same image under the maps $h_X(U)\to h_X(U\cap V)\from h_X(V)$, i.e. $u\vert_{U\cap V}=v\vert_{U\cap V}$), then they glue to some map $a\in h_X(A)$ (i.e. $a\vert_U=u$ and $a\vert_V=v$).
</li>
</ol>
<p>That’s all.</p>
<div class="theorem" name="Brown Representability">
Let $F:\push\CW\to\Set$ be a contravariant functor which satisfies the wedge and Mayer-Vietoris axioms. Then, there exists some $X\in\push\CW$ such that $F\simeq h_X$, i.e. there exists isomorphisms (i.e. bijections) $F(A)\simeq h_X(A)=[A,X]$ for all $A\in\push\CW$ which are functorial in $A$.
</div>
<p>Note that the $X$ guaranteed by this theorem is necessarily unique up to unique isomorphism (e.g. by Yoneda). Similarly, the natural isomorphism $F\simeq h_X$ (which really goes the other way) is necessarily given by pulling back the “universal structure” $x\in F(X)=h_X(X)=[X,X]$ which is the element corresponding to the identity map $X\to X$. That is, it must be of the form</p>
<script type="math/tex; mode=display">\mapdesc {T(A)}{[A,X]}{F(A)}f{\pull fx}</script>
<p>Finally, note that $F(*)={*}$ is a one-element set. This is because the wedge axiom tells us that $F(X\vee*)=F(X)\by F(*)$ but $X\vee*\simeq X$, so $F(X\vee*)=F(X)$ and the bijection corresponds to the projection map $F(X)\by F(*)\to F(X)$. This is only possible if $F(*)$ is a singleton.</p>
<p>With that said, how do we prove this thing? Well, we’re working with CW complexes which are basically determined by their homotopy groups, so maybe we should find a space $X$ with the correct homotopy groups. Note that, if $F$ is representable with representing object $X$, we have $\pi_n(X)=[S^n,X]=F(S^n)$.</p>
<p>Fix a contravariant functor $F:\push\CW\to\Set$ satisfying the wedge and Mayer-Vietoris axioms.</p>
<div class="definition">
A pair $(X,u)$ with $X\in\push\CW$ and $u\in F(X)$ is called <b>universal</b> if the homomorphism $T_u=T_{X,u,n}:\pi_n(X)\to F(S^n)$ given by $T_u:[f]\mapsto\pull fu$ is an isomorphism for all $n$.
</div>
<p>This may seem like a strong condition, but we’ll see that universal pairs are relatively easy to come by.</p>
<div class="lemma">
Given any pair $(Z,z)$ with $Z\in\push\CW$ and $z\in F(z)$, there exists a universal pair $(X,u)$ with $Z$ a subcomplex of $X$ and $u\vert_Z=z$.
</div>
<div class="proof4">
We construct $X$ from $Z$ by inductively attaching cells. Specifically, we seek pairs $(X_n,u_n)$ such that $Z\subset X_n$, the map $T_n=T_{u_n}:\pi_k(X_n)\to F(S^k)$ is an isomorphism for $0 < k < n$, and $T_n$ is surjective for $k=n$. We start with $n=1$ (We can't start with $n=0$ since $S^0$ is disconnected). Let $X_1=Z\bigvee_\alpha S^1_\alpha$ where $\alpha$ ranges over the elements of $F(S^1)$. By the Wedge axiom, there exists some $u_1\in F(X_1)$ with $u_1\vert_Z=z$ and $u_1\vert_{S^1_\alpha}=\alpha$ for all $\alpha\in F(S^1)$, so this gives $(X_1,u_1)$.
<br />
Now, suppose we have $(X_n,u_n)$ constructed, and we seek to construct $(X_{n+1},u_{n+1})$. Let $X_n'=X_n\bigvee_\alpha S^{n+1}_\alpha$ where $\alpha$ ranges over all $F(S^{n+1})$. Use the Wedge axiom to find some $u_n'\in F(X_n')$ such that $u_n'\vert_{X_n}=u_n$ and $u_n'\vert_{S^{n+1}_\alpha}=\alpha$. Then, $\pi_{n+1}(X_n')$ surjects onto $F(S^{n+1})$, but we don't yet have that $\pi_n(X_n')$ maps isomorphically onto $F(S^n)$. For each $\beta\in\ker\parens{\pi_n(X_n')\to F(S^n)}$, consider an $(n+1)$-cell $e^{n+1}_\beta$ and glue it to $X$ via an attaching map $\phi_\beta:\del e^{n+1}_\beta=S^n\to X_n'$ representing $\beta$ (i.e. $[\phi_\beta]=\beta\in\pi_n(X_n')$). Call the result of gluing all these cells $X_{n+1}$, i.e. $X_{n+1}$ is the pushout
$$\begin{CD}
\bigvee_\beta S^n_\beta @>\bigvee\phi_\beta>> X_n'\\
@VVV @VVV\\
\bigvee_\beta e^{n+1}_\beta @>>> X_{n+1}
\end{CD}$$
By construction, $u_n'\vert_{S_\beta^n}=\pull\beta(u_n')=0$ for all $\beta$. Similarly, $F(\bigvee_\beta e^{n+1}_\beta)=0$ since this wedge is contractible, so Mayer-Vietoris let's us extend $u_n'$ to some $u_{n+1}\in F(X_{n+1})$. Now, because $X_{n+1}$ is formed by attaching cells of dimension $n+1$ to $X_n'$, we have that $\pi_k(X_n')\to\pi_k(X_{n+1})$ is an isomorphism for $k\le n-1$ and a surjection for $k=n$, so the same is true of $T_{u_{n+1}}:\pi_k(X_{n+1})\to F(S^k)$ by staring at the below commutative diagram.
<center>
<img src="https://nivent.github.io/images/blog/brown-rep/triag.png" width="350" height="200" />
</center>
It is also clear that $T_{u_{n+1}}$ is still surjective from $\pi_{n+1}(X_{n+1})\to F(S^{n+1})$ essentially since the $S^{n+1}_\alpha$ from before have been left unchanged (gluing cells commutes with forming wedge products). Thus, we only need to show that $\pi_n(X_{n+1})\to F(S^n)$ is injective in order to be finished with the induction. Pick some $\gamma\in\ker\parens{\pi_n(X_{n+1})\to F(S^n)}$. Since the top map in the above diagram is surjective, this $\gamma$ comes from some $\beta\in\ker(\pi_n(X_n')\to F(S^n))$. However, by construction, any such $\beta$'s image in $\pi_n(X_{n+1})$ is trivial (it's represented by a map factoring through the contractible $e_\beta^{n+1}$), so $\gamma=0$ and $T_{u_{n+1}}$ is indeed injective.
<br />
With the induction done, let $X=\bigcup_{n\ge1}X_n$. We first seek to find some $u\in F(X)$ such that $u\vert_{X_n}=u_n$ for all $n$. Consider the <b>mapping telescope</b> $T$ of the inclusions $X_1\into X_2\into X_3\into\dots$ which is just the subcomplex $T=\bigcup_iX_i\by[i,i+1]$ of $X\by[1,\infty)$. We have that $X\by[1,\infty)$ deformation retracts onto $T$ (essentially since $X_i\by[1,\infty)$ deformation retracts onto $X_i\by[i,i+1]$) and so the natural projection $T\to X$ is a homotopy equivalence. Now, let $A\subset T$ be the union of the subcomplexes $\parens{X_i\by[i,i+1]}$ with $i$ odd along with the line $\{*\}\by[1,\infty)$. Let $B\subset T$ be the same with $i$ even. Then,
$$A\simeq\bigvee_{i\ge1}X_{2i-1},\, B\simeq\bigvee_{i\ge1}X_{2i},\, A\cap B\simeq\bigvee_{i\ge1}X_i,\,\text{ and }A\cup B=T.$$
The wedge axiom gives some $a\in F(A)$ and $b\in F(B)$ such that $a\vert_{X_{2i-1}}=u_{2i-1}$ and $b\vert_{X_{2i}}=u_{2i}$ for all $i$. Furthermore, the fact that $u_i\vert_{X_j}=u_j$ for all $1\le j\le i$ shows us that $a\vert_{A\cap B}=b\vert_{A\cap B}$, so the Mayer-Vietoris axiom gives some $u\in F(T)$ s.t. $u\vert_{X_i\by\{i+1\}}=u_i$ for all $i$. Since $T\simeq X$, we can view this $u$ as an element of $F(X)$ s.t. $u\vert_{X_i}=u_i$ for all $i$.
<br />
Finally, to see that $(X,u)$ is indeed universal, use that the isomorphism $T_{u_n}:\pi_k(X_n)\to F(S^n)$ (say, $k< n$) factors as
$$\pi_k(X_n)\to\pi_k(X)\xto{T_u}F(S^k).$$
The map $\pi_k(X_n)\to\pi_k(X)$ is an isomorphism for all $k < n-1$ so we must also have that $T_u$ is an isomorphism for $k < n-1$. Since $n$ is arbitrary, we win.
</div>
<p>Whew. That was the brunt of the argument. We now show that maps to universal pairs can always be extended.</p>
<div class="lemma">
Let $(X,u)$ be a universal pair, and let $(A,B)$ be a CW-complex $A$ with a subcomplex $B\subset A$. Then, for each $a\in F(A)$ and each map $f:B\to X$ with $\pull f(u)=a\vert_B$, there exists some $g:A\to X$ extending $f$ with $\pull g(u)=a$.
</div>
<div class="proof4">
We first replace $X$ with the mapping cylinder of $f$ in order to assume that $f$ is inclusion of a subcomplex. Let $Z=A\cup_BX$, their union with the two copies of $B$ identified. By Mayer-Vietoris, there is some $z\in F(Z)$ such that $z\vert_A=a$ and $Z\vert_X=u$. Use the previous lemma to embed $(Z,z)$ into a universal pair $(X',u')$. The inclusion $(X,u)\into(X',u')$ induces an isomorphism on homotopy groups (since both universal) so is a homotopy equivalence and hence $X'$ deformation retracts onto $X$. This deformation retraction gives a homotopy (rel $B$) of the inclusion $A\into X'$ to a map $g:A\to X$. We have $\pull g(u)=a$ since $u'\vert_X=u$ and $u'\vert_A=a$.
</div>
<p>Now, we prove Brown Representability.</p>
<div class="theorem">
Brown Representability is true.
</div>
<div class="proof4">
Let $(X,u)$ be a universal pair. We claim that it is "well-named" in the sense that the natural transformation $T_u:h_X\to F$ where the map $T_u(A):[A,X]\to F(A)$ is $T_u(A)(f)=\pull fu$ is actually an isomorphism (i.e. $T_u(A)$ is an isomorphism for all $A$). Fix some $A\in\push\CW$. Applying the previous lemma to the pair $(A,*)$ shows that $T_u(A)$ is surjective. To see that $T_u(A)$ is also injective, suppose that $T_u(f_0)=T_u(f_1)$, i.e. $\pull f_0(u)=\pull f_1(u)$. Let $B=A\by\del I\cup\{*\}\by I$. We seek to apply the previous lemma to the pair $(A,B)$. Take $a\in F(A\by I)$ to be $a=\pull p\pull f_0(u)=\pull p\pull f_1(u)$ (here, $p:A\by I\to A$ is the projection map), and let $f:B\to X$ be the map given by $f_0$ on $A\by\{0\}$, $f_1$ on $A\by\{1\}$, and constant to basepoint on $\{*\}\by I$. One easily sees that $\pull f(u)$ is $(\pull f_0(u),\pull f_1(u))$ under the identification $F(B)\simeq F(A)\by F(A)$ (coming from the fact that $B\simeq A\vee A$) which does indeed say that $\pull f(u)=a\vert_B$. Thus, the previous lemma let's us extend $f$ to $A\by I$, i.e. gives us a homotopy between $f_0$ and $f_1$, so $T_u$ is injective, and we win.
</div>
<p>A natural question at this point is, “What is this useful for?” and so we’ll spend the rest of this post giving a partial answer to that.</p>
<h1 id="an-application-cw-approximation">An Application: CW Approximation</h1>
<p>We’ll start with something simple, but kinda cute. Let $X$ be an arbitrary based, connected topological space. Then, we evidently get a contravariant functor $F:\push\CW\to\Set$ given by $F(A)=[A,X]$. It is clear that this satisfies the Wedge and Mayer-Vietoris axioms, so we get from this a CW complex $Z$ such that $[-,Z]\simeq[-,X]$. This $Z$ is called a <b>CW Approximation</b> of $X$. There are a couple things worth noting.</p>
<p>First, we get a map $f:Z\to X$ which is a weak equivalence. This map is the universal structure alluded to before, the image of the identity map under the isomorphism $[Z,Z]\simeq[Z,X]$. This being a weak equivalence comes from the fact that</p>
<script type="math/tex; mode=display">\pi_n(Z)=[S^n,Z]\simeq[S^n,X]=\pi_n(X)</script>
<p>with middle isomorphism coming from compositing with $f$.</p>
<p>Second, the formation of CW approximations is functorial. Indeed, if we a map $g:X\to Y$ between arbitrary based, connected topological spaces, then $g$ induces a natural transformation $\push g:[-,X]\to[-,Y]$. Letting $Z(X),Z(Y)$ denote their CW approximations, we get another natural transformation</p>
<script type="math/tex; mode=display">[-,Z(X)]\simeq[-,X]\xto{\push g}[-,Y]\simeq[-,Z(Y)].</script>
<p>Hence, by Yoneda (or by looking at the image of the identity in $[Z(X),Z(X)]$), $g$ must correspond to some map $Z(g):Z(X)\to Z(Y)$.</p>
<p>At last, our earlier claim that the homotopy category of (based, connected) topological spaces was equivalent to that of CW complexes has some justification <sup id="fnref:6"><a href="#fn:6" class="footnote">6</a></sup>.</p>
<h1 id="an-application-singular-cohomology">An Application: Singular Cohomology</h1>
<p>Next, we deliver on an earlier promise to show that $\hom^n(X;G)\simeq[X,K(G,n)]$ when $X$ is a (based, connected) CW complex. For this, we need to show that Singular cohomology satisfies the wedge and Mayer-Vietoris axioms. Showing that is satisfies the wedge axiom is left as an exercise if you haven’t seen it before; for Mayer-Vietoris, we apply the Mayer-Vietoris sequence. In particular, if $X=U\cup V$, then we get an exact sequence</p>
<script type="math/tex; mode=display">\hom^n(X;G)\to\hom^n(U;G)\oplus\hom^n(V;G)\to\hom^n(U\cap V;G)</script>
<p>where the second map gives the difference of the pullbacks of cohomology classes along the inclusion maps $U\cap V\rightrightarrows U,V$. Exactness of this sequence says exactly that if we have $u\in\hom^n(U;G)$ and $v\in\hom^n(V;G)$ restricting to the same element of $\hom^n(U\cap V;G)$, then they come from some $x\in\hom^n(X;G)$. Thus, Brown representability applies and we know that $\hom^n(X;G)\simeq[X,K]$ for some space $K$.</p>
<p>To see that $K$ is an Eilenberg-Maclane space, we simply observe that</p>
<script type="math/tex; mode=display">\pi_k(K)=[S^k,K]=\hom^n(S^k;G)=\twocases G{n=k}0.</script>
<p>Furthermore, any reduced cohomology theory $h^n$ will satisfy these axioms and so be of the form $h^n(X)=[X,K_n]$ for some sequence $K_n$ of spaces. One can verify that we must have $K_n\simeq\Omega K_{n+1}$ coming from the fact that $h^n(X)\simeq h^{n+1}(\Sigma X)$. One can also show that, conversely, given a so-called <b>$\Omega$-spectrum</b> $K_n$ - a sequence of spaces with (weak) homotopy equivalences $K_n\to\Omega K_{n+1}$ - the functors $h^n=[-,K_n]$ form a reduced cohomology theory.</p>
<h1 id="an-application-classifying-spaces">An Application: Classifying Spaces</h1>
<p>This application will be a bit more involved than the last two. In it, we’ll use Brown representability to define spaces which classify certain types of fiber bundles (e.g. real or complex vector bundles). We’ll essentially partition the class of fiber bundles by their structure groups (e.g. a rank $n$ real vector bundle has structure group $\GL_n(\R)$ since this what controls how we glue trivial bundles to get non-trivial ones). To this end, we define</p>
<div class="definition">
Fix a topological group $G$. A <b>principal $G$-bundle</b> over a space $B$ is a fiber bundle $\pi:P\to B$ with a free and transitive right action by $G$ on the fibers.
</div>
<p>The point of this free, transitive right action is that it causes the fiber space of $P$ to be homeomorphic to $G$ except they have “forgotten which point was the identity.” I should also mention that for $\pi:P\to B$ to be a principal $G$-bundle, we require that any local trivialization $h:\inv\pi(U)\iso U\by G$ intertwines the right action with right translation, i.e.</p>
<script type="math/tex; mode=display">h(p)=(u,g)\implies h(p\cdot g')=(u,gg').</script>
<p>The main motivation for studying these (at least, as far as I’m concerned) comes from the following theorem which I will not prove.</p>
<div class="theorem">
Given any fiber bundle $\pi:E\to B$ with fiber $F$ and structure group $\Aut(F)$, there exists a principal $\Aut(F)$-bundle $P$ such that $E=P\by_GF:=(P\by F)/G$ where $G=\Aut(F)$ acts on $(P\by F)$ via $(p,f)\cdot g=(p\cdot g,\inv g\cdot f)$. This principal bundle $P$ is called the <b>frame bundle</b> of $E$, and $E$ is called a <b>fiber bundle associated to $P$</b>.
</div>
<p>So, for example, every complex vector bundle comes from some principal $\GL_n(\C)$-bundle. This theorem turns the problem of understanding various kinds of bundles into one of just understanding these principal $G$-bundles. With that said, let $B:\push\CW\to\Set$ be the functor <sup id="fnref:10"><a href="#fn:10" class="footnote">7</a></sup> which assigns to a space $X$, the set $B(X)$ of its principal $G$-bundles up to their natural notion of equivalence <sup id="fnref:7"><a href="#fn:7" class="footnote">8</a></sup>.</p>
<p>Now, we claim that $B$ satisfies the wedge and mayer-vietoris axioms which will give the existence of a space $BG$ such that the set of principal $G$-bundles on $X$ (up to isomorphism) is in canonical bijection with the set $[X,BG]$ of (homotopy classes of) maps from $X$ into $BG$. This is a big deal. It is very nonobvious that such a space should exist.</p>
<p>It is easy to see that $B$ satisfies the wedge axiom. If I have bundles $\pi_\alpha:P_\alpha\to B_\alpha$ over a collection of spaces, then I can glue them all to get a bundle over their wedge $\bigvee B_\alpha$ <sup id="fnref:8"><a href="#fn:8" class="footnote">9</a></sup>. Conversely, any bundle over a wedge can be restricted to one over each factor. The Mayer-Vietoris axiom takes a little more work out, but is still not hard. It basically says that defining a bundle over $U\cup V$ is the same as giving bundles over $U$ and $V$ that agree on $U\cap V$. This is true. The main point in this is that the conditions of being a bundle are all local and so bundles are amenable to gluing constructions/arguments.</p>
<div class="exercise">
Homotopy invariance says that any principal $G$-bundle over a contractible space is trivial (i.e. of the form $X\by G$). Use this and Mayer-Vietoris to show that $B(S^n)\simeq\pi_{n-1}(G)$.
</div>
<p>As a consequence of the paragraph above the exercise, we now know that our <b>classifying spaces</b> $BG$ exist for all topological groups $G$. As usual, the set $B(BG)=[BG,BG]$ has a universal structure corresponding to the identity map, and we call this the universal bundle $\pi_G:EG\to BG$. What properties do these spaces have?</p>
<p>The exercise shows that $\pi_n(BG)\simeq\pi_{n-1}(G)$ (i.e. $\pi_n(\Omega BG)\simeq\pi_n(G)$). In light of this, we will first show that indeed $G\simeq\Omega BG$ (already, this computation of homotopy groups shows that $BG\simeq K(G,1)$ when $G$ is a discrete group). Because of this. $BG$ is sometimes called a <b>delooping</b> of $G$.</p>
<p>Actually, we’ll first show that $EG$ is contractible because this will suffice to show that $G\simeq\Omega BG$. We have a fiber sequence</p>
<script type="math/tex; mode=display">G\to EG\to BG</script>
<p>which induces a long exact sequence in homotopy</p>
<script type="math/tex; mode=display">\cdots\too\pi_{n+1}(BG)\to\pi_n(G)\to\pi_n(EG)\to\pi_n(BG)\to\pi_{n-1}(G)\to\cdots.</script>
<p>One can trace identifications to see that the maps $\pi_n(BG)\to\pi_{n-1}(G)$ above are exactly the isomorphisms $B(S^n)\iso\pi_{n-1}(G)$ from the exercise, so $\pi_n(EG)=0$ for all $n$. Since $EG$ is a CW-complex, this shows that it is contractible. Thus, $G$ is homotopy equivalent to the homotopy fiber of the constant map $*\to BG$, but this is just $\Omega BG$.</p>
<p>We now show a converse to some of what we’ve seen.</p>
<div class="proposition">
Let $\pi:E\to B$ be a principal $G$-bundle with contractible total space $E$. Then, $B\simeq BG$.
</div>
<div class="proof4">
Essentially by definition, we have that $\pi$ is induced by some map $f:B\to BG$. That is we have a commutative diagram like below where the right square is a pullback square.
$$\begin{CD}
G @>>> E @>\pi>> B\\
@| @VVV @VVfV\\
G @>>> EG @>\pi_G>> BG
\end{CD}$$
This diagram induces a homomorphism of long exact sequences
$$\begin{CD}
\pi_n(G) @>>> \pi_n(E) @>>> \pi_n(B) @>>> \pi_{n-1}(G) @>>> \pi_{n-1}(E)\\
@| @| @V\push fVV @| @|\\
\pi_n(G) @>>> \pi_n(EG) @>>> \pi_n(BG) @>>> \pi_{n-1}(G) @>>> \pi_{n-1}(EG)
\end{CD}$$
the maps $\pi_n(E)\to\pi_n(EG)$ are isomorphism because both groups are $0$. Now, the 5 lemma (or a diagram chase) show that the maps $\pi_n(B)\to\pi_n(BG)$ are isomorphisms as well. Since these are all CW complexes, we win by Whitehead.
</div>
<p>This proposition gives us a way of actually constructing these spaces. To end this post, I’ll sketch the construction of a space classifying complex line bundles.</p>
<p>Given a complex rank $n$ vector bundle $E\to B$, one would initially expect its structure group to be $\GL_n(\C)$ <sup id="fnref:11"><a href="#fn:11" class="footnote">10</a></sup>, but we can take it to be something smaller. In particular, using Graham-Schmidt (or whatever it’s called), you can give all locally trivial pieces orthonormal bases, and then your transition function lands in $U(n)$, the group of $n\by n$ unitary matrices <sup id="fnref:15"><a href="#fn:15" class="footnote">11</a></sup>. In particular, the structure group of a complex line bundle can be taken to be $U(1)=S^1$. Thus, we seek a bundle $S^1\to E\to B$ with $E$ contractible. Well, if we let $B=\CP^n$, then we can take $E=S^{2n+1}$ and the natural quotient map $E\to B$ makes $E$ an $S^1$-bundle over $B$. That is, we have a bundle</p>
<script type="math/tex; mode=display">S^1\to S^{2n+1}\to\CP^n</script>
<p>for all $n$. These sequences are compatible in the relevant sense, so we can take limit as $n\to\infty$ <sup id="fnref:12"><a href="#fn:12" class="footnote">12</a></sup> to get a bundle</p>
<script type="math/tex; mode=display">S^1\to S^\infty\to\CP^\infty.</script>
<p>Now, $S^\infty$ is contractible (essentially because $\pi_k(S^n)=0$ when $n$ big and $k$ fixed), so this shows that $\CP^\infty\simeq BS^1$. In light of the first theorem in this section, this means that any complex line bundle $\pi:E\to B$ ultimately arises from some map $f:B\to\CP^\infty$ <sup id="fnref:13"><a href="#fn:13" class="footnote">13</a></sup>. However, we can say even more. Not only is $\CP^\infty$ a $BS^1$, it is also $K(\Z,2)$ <sup id="fnref:14"><a href="#fn:14" class="footnote">14</a></sup> so we have isomorphisms</p>
<script type="math/tex; mode=display">B(X)\simeq[X,\CP^\infty]\simeq\hom^2(X;\Z).</script>
<p>That is, the set (really group because of tensor products of line bundles) of isomorphism classes of (topological) complex line bundles on a space $X$ is isomorphic to the second integral cohomology group $\hom^2(X;\Z)$. Given a line bundle $\pi:E\to X$, its corresponding cohomology class $c_1(E)\in\hom^2(X;\Z)$ is called its <b>(first) Chern class</b>.</p>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>Technically, it doesn’t quite concern functors on the category Top of topological spaces, but I’ll say more about this later. <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>maps inducing an isomorphism on all homotopy groups <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>Of course, this latter point about CW approximation is not logically necessary to start and prove Brown representability for CW complexes <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>Really, I should call this something like $\Ho(\push\CW)$, but who’s got the time to type all of that? <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>Certainly assume all spaces are based no matter what <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>Knowing this was not logically necessary for anything we’ve done <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:10">
<p>This is a functor because the (categorical) pullback of a bundle is a bundle <a href="#fnref:10" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>It is not obvious that this functor is well-defined (i.e. that homotopically equivalent spaces have “the same” sets of principal $G$-bundles), but it is. I won’t prove this either. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>The total space of this glued bundle will not be the wedge of the $P_\alpha$’s since you need to glue the total spaces together along fibers <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:11">
<p>I think I even claimed this before <a href="#fnref:11" class="reversefootnote">↩</a></p>
</li>
<li id="fn:15">
<p>The point is that Gram-Schmidt gives a deformation retraction $\GL_n(\C)\to U(n)$ which then induces a homotopy equivalence $B\GL_n(\C)\simeq BU(n)$. <a href="#fnref:15" class="reversefootnote">↩</a></p>
</li>
<li id="fn:12">
<p>Form the direct limits $S^\infty=\dirlim S^n=\dirlim S^{2n+1}$ and $\CP^\infty=\dirlim\CP^n$ <a href="#fnref:12" class="reversefootnote">↩</a></p>
</li>
<li id="fn:13">
<p>One can show that the line bundle associated to the bundle $S^\infty\to\CP^\infty$ is the so-called <b>tautological line bundle</b> on $\CP^\infty$ which is the map $R^\infty\to\CP^\infty$ where the fiber above a point $\l\in\CP^\infty$ (i.e. a line) is the line that point corresponds to. This means all complex line bundles are pullbacks of this universal, tautological line bundle. <a href="#fnref:13" class="reversefootnote">↩</a></p>
</li>
<li id="fn:14">
<p>You may note furthermore that $S^1\simeq K(\Z,1)$ so $BK(\Z,1)\simeq K(\Z,2)$. In general, we have that $BK(G,n)\simeq K(G,n+1)$ coming from our earlier computation of homotopy groups of $BG$. Of course, this only makes sense when $K(G,n)$ is (homotopy equivalent to) a topological group … which is, I think, maybe always the case (you can see this by showing that $B(G\by G)\simeq BG\by BG$ and then inductively writing $K(G,n+1)=BK(G,n)$. Start with $K(G,0)=G$ with discrete topology and you have an abelian group still at each step). <a href="#fnref:14" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>Let’s prove and discuss one of my favorite theorems from algebraic topology. It gives a characterization of when a contravariant “homotopy functor” from topological spaces 1 to sets is representable, i.e. is (isomorphic to one) of the form $F(X)=[X,S]$ where $S$ is $F$’s representing object, and the notation $[X,S]$ denotes the set of homotopy classes of maps $X\to S$. The nice part about this is that many nice/important spaces one encounters when studying algebraic topology are secretly just representing objects for some functor that people care about. Instead of constructing these spaces directly, one can infer their existence from Brown Representability, and then use properties of the functor they represent to understand them as objects. Vice versa, knowing that a functor is representable gives you additional control over its behavior/properties. This will make more sense later when I discuss some example applications. $ \newcommand{\CW}{\mathrm{CW}} \newcommand{\Top}{\mathrm{Top}} \newcommand{\Set}{\mathrm{Set}} \DeclareMathOperator{\Ho}{Ho} $ Technically, it doesn’t quite concern functors on the category Top of topological spaces, but I’ll say more about this later. ↩Spectral Sequences2019-08-19T01:00:00+00:002019-08-19T01:00:00+00:00https://nivent.github.io/blog/spectral-seq<p>Back when I was in high-school, I became really interested in this thing called “machine learning.” The main idea is that you bombard some algorithm with a ton of examples (of a task being performed or of objects being classified), and then you cross your fingers and hope it has managed to reliably learn how to do what the examples showcased. One big draw of this approach is that there are many tasks where it is not clear how to accomplish them, but where it feels like there has to be enough information present that accomplishing them is possible. For example, imagine writing a program that takes in an image and tells you if there’s a bird in it; this is hard to do algorithmically, but certainly the pixel values of an image contain enough information to decide whether or not there’s a bird there <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup>. The moral of this detour is that sometimes we find ourselves in situations where we have lots of information available to tackle some problem, but figuring out how to utilize all that information is quite tricky. Imagine, for example, you have some topological space $X$ with a filtration
$
\newcommand{\msC}{\ms C}
\newcommand{\msA}{\ms A}
\DeclareMathOperator{\Ch}{Ch}
\DeclareMathOperator{\CoCh}{CoCh}
\DeclareMathOperator{\Fil}{Fil}
\DeclareMathOperator{\Tot}{Tot}
$</p>
<script type="math/tex; mode=display">X^0\subseteq X^1\subseteq X^2\subseteq\cdots\subseteq X</script>
<p>such that $X=\dirlim X^k$. The example to keep in mind is $X$ a CW-complex, and $X^k$ its $k$-skeleton. Intuitively, the cohomology groups $\hom^n(X^k)$ should approximate $\hom^n(X)$, so if you know all of them, then you should have enough information to say something about $\hom^n(X)$. Figuring out exactly what you can say in this situation (and others) is the aim of spectral sequences, which, if you haven’t guessed yet, are the stars of this post.</p>
<p>Specifically, I’ll speak abstractly about two common sources of spectral sequence <sup id="fnref:18"><a href="#fn:18" class="footnote">2</a></sup>. Because I really wanna talk about this material <sup id="fnref:2"><a href="#fn:2" class="footnote">3</a></sup>, I won’t do my usual thing of trying to write as if everything I’ve written previously on this blog forms a dense subset of what I’m assume the reader knows. Instead, I’ll assume you’re comfortable with the words like cohomology, category, and other things that start with a c, then go from there. With that said, I’ll briefly define abelian categories, say what a spectral sequence is, given an example, and depending on how I’m feeling at the end, either say the word hypercohomology or save that for a future post. Let’s get started$\dots$</p>
<h1 id="abelian-categories-briefly">Abelian Categories Briefly</h1>
<p>An abelian category is one that behaves like the category of abelian groups. These form the main setting for much of homological algebra, so it’s probably worthwhile to see how they’re defined at least once. I’ll build up the definition piece by piece.</p>
<div class="definition">
A category $\msC$ is called <b>preadditive</b> if every hom set $\Hom_{\msC}(A,B)$ (where $A,B\in\ob\msC$) is an abelian group such that composition of morphisms is bilinear.
</div>
<p>To get to the next step, we need the notion of a biproduct, which is just an object that is both a product (think direct product) and coproduct (think direct sum). I guess I’ll define what a coproduct (of two objects) is, and leave defining product to you.</p>
<div class="definition">
Let $\msC$ be a category, and fix objects $X_1,X_2\in\ob\ms C$. A <b>coproduct</b> of $X_1$ and $X_2$ is an object $X$ (often denoted $X_1\amalg X_2$) together with a pair of morphisms $\iota_1:X_1\to X,\iota_2:X_2\to X$ satisfying the below universal property:
<ul>
<li>
for every object $Y\in\msC$ with maps $f_1:X_1\to Y$ and $f_2:X_2\to Y$, there exists a unique morphism $f:X\to Y$ such that the below diagram commutes
<center>
<img src="https://nivent.github.io/images/blog/spectral-seq/coprod.png" width="350" height="100" />
</center>
</li>
</ul>
In short, it's an object such that "maps out of the coproduct are maps out of each factor." In medium, it is an object that is initial in the category of objects of $\msC$ with maps from $X_1$ and $X_2$.
</div>
<div class="remark">
The above defines a binary coproduct. You can easily extend this definition to that of a coproduct of objects $X_i$ where $i\in I$ ranges through an arbitrary index set.
</div>
<div class="example">
Coproducts in $\textrm{Set}$ are disjoint unions, and coproducts in $\textrm{Ab}$ are direct sums.
</div>
<p>Now, define products in a dual way, i.e. “maps into products are maps into each factor,” and then say a biproduct is an object that is both a product and coproduct (of the same set of objects). This brings us to</p>
<div class="definition">
A preadditive category $\msC$ is called <b>additive</b> if every finite set of objects has a biproduct, and a zero object $0\in\msC$ (i.e. one that is both initial and final) exists.
</div>
<div class="notation">
Given $A,B\in\msC$, an additive category, let $0_{AB}\in\Hom_{\msC}(A,B)$ denote the morphism given composed from $A\to0\to B$.
</div>
<p>For the next step, we need to know how to define kernels and cokernels.</p>
<div class="definition">
Let $\msC$ be an additive category (or even just a category with zero morphisms), and fix any morphism $f:X\to Y$ in $\msC$. Then, an object $M$ is called a <b>kernel</b> of $f$, usually denoted $\ker f$, if it is final in the category of maps to $A$ "killed by $f$". I mean to say, there exists a morphism $\mu:M\to X$ such that for any morphism $\alpha:A\to X$ such that $f\circ\alpha=0_{AY}$, there exists a unique $a:A\to M$ making the following diagram commutes
<center>
<img src="https://nivent.github.io/images/blog/spectral-seq/ker.png" width="250" height="100" />
</center>
</div>
<p>Cokernels are define similarly as being initial with respect to maps $\nu:Y\to N$ such that $\nu\circ f=0$. Now,</p>
<div class="definition">
An additive category is <b>preabelian</b> if every morphism has both a kernel and a cokernel.
</div>
<p>Almost there. Note that if $\mu:\ker f\to X$ is the kernel of a map $f:X\to Y$, then $\mu$ is a monomorphism, i.e. if we have $a:A\to\ker f$ and $b:A\to\ker f$ such that $\mu\circ a=\mu\circ b$, then $a=b$ <sup id="fnref:3"><a href="#fn:3" class="footnote">4</a></sup>. We would like every monomorphism to arise in this way. Similarly, cokernels are epimorphisms, and we would like them to be the only epimorphisms <sup id="fnref:4"><a href="#fn:4" class="footnote">5</a></sup>.</p>
<div class="definition">
Let $\msC$ be a preabelian category. A monomorphism is called <b>normal</b> if it is also a kernel, and an epimorphism is called <b>conormal</b> (or still just <b>normal</b>) if it is also a cokernel.
</div>
<div class="definition">
A preabelian category is an <b>abelian category</b> if every monomorphism is normal and every epimorphism is conormal.
</div>
<p>To sum it all up, an abelian category is one with a zero object, all binary biproducts, all kernels and cokernels, and where all monomorphisms/epimorphisms are normal. Lot’s of the language used in the theory of abelian groups (quotients, subobjects, exactness, etc.) carries over to arbitrary abelian categories in a fairly straightforward manner, so you can often get away with imagining elements of abstract abelian categories as just being abelian groups or $R$-modules, or what have you. Before moving on, here’s one important example of an abelian category.</p>
<p>Fix an additive category $\msA$. A <b>chain complex</b> <sup id="fnref:5"><a href="#fn:5" class="footnote">6</a></sup> $A_\bullet$ is a sequence</p>
<script type="math/tex; mode=display">A_\bullet:\cdots\too A_1\xtoo{d_1} A_0\xtoo{d_0} A_{-1}\too\cdots</script>
<p>of morphisms in $\msA$ such that $d_n\circ d_{n+1}=0$ for all $n$. A <b>chain map</b> $f:A_\bullet\to B_\bullet$ is a commutative diagram</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{CD}
\cdots @>>> A_1 @>d_1>> A_0 @>d_0>> A_{-1} @>>> \cdots\\
&@Vf_{-1}VV @Vf_0VV @Vf_1VV\\
\cdots @>>> B_1 @>d_1>> B_0 @>d_0>> B_{-1} @>>> \cdots
\end{CD} %]]></script>
<p>whose rows are chain complexes. Let $\Ch_\bullet(\msA)$ denote the category whose objects are chain complexes in $\msA$, and whose morphisms are chain maps.</p>
<div class="exercise">
Show that $\Ch_\bullet(\msA)$ is an abelian category if $\msA$ is (hint: do everything componentwise).
</div>
<p>Before going on, I should maybe also say what homology is.</p>
<div class="definition">
Given a category $\msC$ and a morphism $f:X\to Y$ in $\msC$, the <b>image</b> of $f$ is an object $I$, usually denoted $\im f$, with a monomorphism $\iota:I\to Y$ that is initial among morphisms that $f$ factors through. That is, there exists a morphism $e:X\to I$ such that for any morphisms $e':X\to I'$ and $m':I'\to Y$ where $f=m'\circ e'$, there is a unique morphism $v:I\to I'$ making the below diagram commute
<center>
<img src="https://nivent.github.io/images/blog/spectral-seq/im.png" width="250" height="100" />
</center>
</div>
<div class="remark">
When $\msC$ is an abelian category, we have $\im f\simeq\ker(\coker f)$.
</div>
<div class="definition">
Let $A_\bullet=\bracks{d_n:A_n\to A_{n-1}}$ be a chain complex. Its <b>$n$th homology object</b> should intuitively be $\hom_n(A_\bullet)=\ker d_n/\im d_{n+1}$. I guess the way we make this formal/general/whatever is by setting
$$\hom_n(A_\bullet):=\coker\parens{\im d_{n+1}\to\ker d_n}$$
where the map $\im d_{n+1}\to\ker d_n$ exists (and is unique) because the composition $\im d_{n+1}\to A_n\xto{d_n}A_{n+1}$ is the zero map.
</div>
<h1 id="spectral-sequence-of-a-filtered-complex">Spectral Sequence of a Filtered Complex</h1>
<p>Now we start developing the good stuff. I should probably start by saying that I will only be considering spectral sequences in cohomology in this post, so, for example, we’ll we be working in the (abelian) category $\CoCh(\msA)$ of cochain complexes of an abelian category $\msA$. <sup id="fnref:6"><a href="#fn:6" class="footnote">7</a></sup></p>
<p>For the remainder of this section, fix some abelian category $\msA$. Probably a good place to start here is with the definition of a filtered complex.</p>
<div class="definition">
A <b>decreasing filtration</b> $F$ on an object $A\in\msA$ is a famility $(F^nA)_{n\in\Z}$ of subobjects of $A$ such that
$$A\supseteq\dots\supseteq F^nA\supseteq F^{n+1}A\supseteq\dots\supseteq0.$$
A <b>filtered object</b> of $\msA$ is a pair $(A,F)$ consisting of an object $A\in\msA$ and a decreasing filtration $F$ on $A$. A <b>morphism</b> $(A_1,F_1)\to(A_2,F_2)$ of filtered objects is a morphism $\phi:A_1\to A_2$ such that $\phi(F_1^kA_1)\subset F^k_2A_2$ for all $k\in\Z$. We thus arrive at the category $\Fil(\msA)$ of filtered objects.
</div>
<div class="definition">
Given a filtered object $(A,F)$, we get <b>associated graded objects</b> $G^kA=F^kA/F^{k+1}A$.
</div>
<div class="definition">
A filtration $F$ on an object $A$ is called <b>bounded</b> (or <b>finite</b>) if there exists $n,m$ such that $F^nA=A$ and $F^mA=0$.
</div>
<div class="definition">
A <b>filtered cochain complex</b> is an object in the category $\CoCh(\Fil(\msA))$ (equivalently, one in $\Fil(\CoCh(\msA))$).
</div>
<p>Let $A^\bullet=\bracks{d_n:A^n\to A^{n+1}}$ be a filtered cochain complex. Its differential $d$ induces a well-defined differential $G^pA^n\to G^pA^{n+1}$, so we get an associated graded chain complex $G^pA^\bullet$. Furthermore, we get a natural filtration on cohomology given by</p>
<script type="math/tex; mode=display">F^p\hom^n(A^\bullet)=\bracks{\alpha\in\hom^n(A^\bullet)\mid\exists x\in F^pA^n:\alpha=[x]},</script>
<p>which has assocated graded pieces $G^p\hom^n(A^\bullet)$. If we’re lucky this grading will actually determine the cohomology of $A^\bullet$ via the short exact sequences</p>
<script type="math/tex; mode=display">0\too F^{p+1}\pull\hom(A^\bullet)\too F^p\pull\hom(A^\bullet)\too G^p\pull\hom(A^\bullet)\too0.</script>
<p>It’s sometimes easier to compute $\pull\hom(G^pA^\bullet)$ than $G^p\pull\hom(A^\bullet)$, so we may wonder about comparing the two in order to eventual get a handle on $\pull\hom(A^\bullet)$. It turns out that we can get from one to the other via a series of “successive approximations.”</p>
<p>We start by denoting the associated graded complex by</p>
<script type="math/tex; mode=display">E_0^{p,q}:=G^pA^{p+q}\text{ }\text{ with differential }\text{ }d_0^{p,q}:E^{p,q}_ 0\to E_0^{p,q+1}.</script>
<p>Denote the cohomology of this complex by</p>
<script type="math/tex; mode=display">E_1^{p,q}:=\hom^{p+q}(G^pA^\bullet)=\frac{\ker\big({d_0^{p,q}:E_0^{p,q}\to E_0^{p,q+1}}\big)}{\im\parens{d_0^{p,q-1}:E_0^{p,q-1}\to E_0^{p,q}}}=\frac{\ker\big({d_0^{p,q}:G^pA^{p+q}\to G^pA^{p+q+1}}\big)}{\im\parens{d_0^{p,q-1}:G^pA^{p+q-1}\to G^pA^{p+q}}},</script>
<p>which we think of as a “first-order approximation” to $\pull\hom(A^\bullet)$. Let’s explicitly construct a second-order approximation. Note that a cohomology class $\alpha\in E_1^{p,q}$ can be represented by a chain $x\in F^pA^{p+q}$ with differential $dx\in F^{p+1}A^{p+q+1}$. With this in mind, we define</p>
<script type="math/tex; mode=display">\mapdesc{d_1^{p,q}}{E_1^{p,q}}{E_1^{p+1,q}}{\alpha}{[dx]}</script>
<p>One easily sees that $d_1^{p,q}\circ d_1^{p+1,q}=0$ <sup id="fnref:7"><a href="#fn:7" class="footnote">8</a></sup>, and so we are justified in defining</p>
<script type="math/tex; mode=display">E_2^{p,q}:=\hom^p(E_1^{\bullet, q})=\frac{\ker\big({d_1^{p,q}:E_1^{p,q}\to E_1^{p+1,q}\big)}}{\im\parens{d_1^{p-1,q}:E_1^{p-1,q}\to E_1^{p,q}}}.</script>
<p>As a sanity check to make sure things make sense, try doing the following.</p>
<div class="exercise">
Suppose that $F^{-1}A^\bullet=0$ and $F^1A^\bullet=A^\bullet$ (so $F^0A^\bullet$ is just some subcomplex). Show that in this case, we have $E_2^{p,q}=G^p\hom^{p+q}(A^\bullet)$.
</div>
<p>Returning to general filtrations, we can continue to construct higher order approximations. Doing one more step before handling general approximations, note that an $\alpha\in E_2^{p,q}$ can be represented by some $\st\alpha\in E_1^{p,q}$ with differential $d_1\st\alpha=0\in E_1^{p+1,q}$. Since $d_1\st\alpha=[dx]$ where $x\in F^pA^{p+q}$ is any chain representing $\alpha$, we can take $dx$ to be the zero element of $\ker(d_0^{p+1,q})$, which is to say we can take $x$ s.t. $dx\in F^{p+2}A^{p+q+1}$ <sup id="fnref:9"><a href="#fn:9" class="footnote">9</a></sup>. This suggest we can get a map $d_2^{p,q}:E_2^{p,q}\to E_2^{p+2,q-1}$.</p>
<p>Based on what we’ve seen so far, it seems as though elements of an $r$th order approximation $E_r^{p,q}$ should be ultimately represented by cycles $x\in F^pA^{p+q}$ such that $dx\in F^{p+r}A^{p+q+1}$. This turns out to be exactly the case. For $r\ge0$, define</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
Z_r^{p,q} &=\frac{F^pA^{p+q}\cap\inv d\parens{F^{p+r}A^{p+q+1}}+F^{p+1}A^{p+q}}{F^{p+1}A^{p+q}}\\
B_r^{p,q} &=\frac{F^pA^{p+q}\cap d\parens{F^{p-r+1}A^{p+q-1}}+F^{p+1}A^{p+q}}{F^{p+1}A^{p+q}}
\end{align*} %]]></script>
<p>and let $E_r^{p,q}=Z_r^{p,q}/B_r^{p,q}$. On these objects, we can define a differential</p>
<script type="math/tex; mode=display">\mapdesc{d_r^{p,q}}{E_r^{p,q}}{E_r^{p+r,q-r+1}}{[z]}{[dz]}</script>
<p>where $z\in F^pA^{p+q}\cap\inv d(F^{p+r}A^{p+q+1})$. Note that $d_r$ has bidegree $(r,1-r)$. With these definitions set up, we have the following <sup id="fnref:10"><a href="#fn:10" class="footnote">10</a></sup>.</p>
<div class="theorem">
<ol>
<li> The map $d_r^{p,q}$ is well-defined with $d_r^{p,q}\circ d_r^{p-r,q+r-1}=0$. </li>
<li> $E_{r+1}$ is given by taking the cohomology of $E_r$, i.e.
$$E_{r+1}^{p,q}\simeq\frac{\ker\parens{d_r^{p,q}:E_r^{p,q}\to E_r^{p+r,q-r+1}}}{\im\parens{d_r^{p-r,q+r-1}:E_r^{p-r,q+r-1}\to E_r^{p,q}}}$$
</li>
<li> $E_1^{p,q}=\hom^{p+q}(G^pA^\bullet)$. </li>
<li> If the filtration of $A^\bullet$ is bounded, then for every $p,q$, for $r$ sufficiently large, we have $E_r^{p,q}=G^p\hom^{p+q}(A^\bullet)$. In this case, we say $E_1$ <b>converges</b> (or <b>abuts</b>) to $\hom^{p+q}(A^\bullet)$, and denote this $E_1^{p,q}\implies\hom^{p+q}(A^\bullet)$. </li>
</ol>
</div>
<div class="proof4">
For convenience, let $K_r^{p,q}=F^pA^{p+q}\cap\inv d(F^{p+r}A^{p+q+1})$ and $I_r^{p,q}=F^pA^{p+q}\cap d(F^{p-r+1}A^{p+q-1})$.
<ol>
<li>
It's clear by definitions that every element of $E_r^{p,q}$ can be represented by some $z\in K_r^{p,q}$ just by definition. Given such a $z$, we have $dz\in K_r^{p+r,q-r+1}$ because $dz\in F^{p+r}A^{p+q+1}$ and $d^2z=0$. Hence, it remains to show that if $z\in I_r^{p,q}$, then $dz\in I_r^{p+r,q-r+1}$. Well, if $z\in I_r^{p,q}$, then $z=dw$ with $w\in F^{p-r+1}A^{p+q-1}$, so $dz=d^2w=0\in F^{p+r}A^{p+q+1}$. Hence, $dz\in I_r^{p+r,q-r+1}$ as desired. It's clear that $d_r^2=0$ since $d^2=0$.
</li>
<li>
Note that $K_{r+1}^{p,q}\subset K_r^{p,q}$, so there's a natual induced injection $Z_{r+1}^{p,q}\into Z_r^{p,q}$. Composing this with the quotient map gives a map $\phi:Z_{r+1}^{p,q}\to E_r^{p,q}$. Now, for $z\in K_{r+1}^{p,q}$, we have $\phi([z])\in\ker d_r^{p,q}$ since $d_r(\phi([z]))=[dz]\in E_r^{p+r,q-r+1}$ and $dz\in F^{p+r+1}A^{p+q+1}=F^{(p+r)+1}A^{(p+r)+(q-r+1)}$. In fact, we have that $\phi$ surjects onto $\ker d_r^{p,q}$ because its image, by definition, contains every $z$ s.t. $dz\in F^{p+r+1}A^{p+q+1}$. Now, we claim that $\ker\phi=B_r^{p,q}$ which suffices to prove 2. It's clear that $I_{r+1}^{p,q}\subset I_r^{p,q}$, so $B_r^{p,q}\subset\ker\phi$. Conversely, suppose that $z\in\ker\phi\subset Z_{r+1}^{p,q}$. Well, $\phi(z)=[z]$, so this means that $z\in B_r^{p,q}$. Hence, $z\in B_r^{p,q}\cap Z_{r+1}^{p,q}$, so $z\in B_{r+1}^{p,q}$.
</li>
<li>
It's clear from definitions that $Z_1^{p,q}=\ker\parens{G^pA^{p+q}\to G^pA^{p+q+1}}$ and that $B_1^{p,q}=\im\parens{G^pA^{p+q-1}\to G^pA^{p,q}}$. Part 3 follows.
</li>
<li>
Suppose now that there exists $n,m\in\Z$ such that $F^nA^\bullet=A^\bullet$ and $F^mA^\bullet=0$ (so $m>n$). Fix any $p,q\in\Z$, and choose any $r>\max\{m-p,p-n+1,0\}$. Then,
$$F^{p+r}A^{p+q+1}\subseteq F^mA^{p+q+1}=0\text{ and }F^{p-r+1}A^{p+q-1}\supseteq F^nA^{p+q-1}=A^{p+q-1},$$
so $Z_r^{p,q}=\parens{F^pA^{p+q}\cap\ker d+F^{p+1}A^{p+q}}/F^{p+1}A^{p+q}$, and $B_r^{p,q}=\parens{F^pA^{p+q}\cap\im d+F^{p+1}A^{p+q}}/F^{p+1}A^{p+q}$. With these descriptions stated, we obviously have a surjective map $F^p\hom^{p+q}(A^\bullet)\onto E_r^{p,q}$. The kernel of this map will be the cohomology classes $\alpha\in F^p\hom^{p+q}(A^\bullet)$ represented by a cycle $x\in F^{p+1}A^{p+q}$; that is, the kernel is exacly $F^{p+1}A^{p+q}$, so we get our desired isomorphism $G^p\hom^{p+q}\iso E_r^{p,q}$.
</li>
</ol>
</div>
<p>Before ending this section, I should maybe mention some standard terminology. The data summarized in the above theorem (i.e. objects $E_r^{p,q}$ with differentials of bidgree $(r,1-r)$ such that $E_{r+1}$ is the cohomology of $E_r$) is collectively known as a <b>(cohomological) spectral sequence</b>. For fixed $r$, the objects $E_r^{p,q}$ form the <b>$r$th page</b> (or <b>$E_r$-page</b>) of the sequence. In general, if $E_r^{p,q}$ only depends on $p,q$ for $r$ sufficiently large, then we denote this object by $E_\infty^{p,q}$. Hence, for the spectral sequence of a bounded filtered complex $A^\bullet$, we have $E_\infty^{p,q}=G^p\hom^{p+q}(A^\bullet)$. Spectral sequences are usually drawn as a 2d grid with $p$ increasing to the right, and $q$ increasing as you move vertically upwards.</p>
<div class="exercise">
Show that this spectral sequence is compatible with cup products in the sense that, on each page, we get an induced map
$$\mapdesc{\smile}{E_r^{p,q}\by E_r^{s,t}}{E_r^{p+s,q+t}}{([x],[y])}{[x\smile y]}$$
such that $d_r(\alpha\smile\beta)=d_r(\alpha)\smile\beta+(-1)^{p+q}\alpha\smile d_r(\beta)$.
<br />
Actually, hold off on this exercise for now, and then do it specifically for the Serre spectral sequence construction later.
</div>
<h1 id="a-neat-application">A Neat Application</h1>
<p>Now that we’ve seen a way of constructing spectral sequences, let’s spend some time looking at their applications on topology. In this section, we’ll (re)prove one result one usually gets without spectral sequences, and then in the next section we’ll look into something more substantive. Namely, we’ll first show that singular and cellular cohomology agree.</p>
<p>Fix a CW-complex $X$, and let $X^k$ denote its $k$-skeleton. Let $C^\bullet(X)$ denote its singular cochain complex, so $C^n(X)=\Hom_{\Z}(C_n(X),\Z)$ where $C_n(X)$ is the free abelian group generated by maps $\Delta^n\to X$. We filter this by setting <sup id="fnref:16"><a href="#fn:16" class="footnote">11</a></sup></p>
<script type="math/tex; mode=display">F^pC^n(X)=\bracks{\phi\in C^n(X):\phi\vert_{C_n(X^p)}=0}=\ker\parens{C^n(X)\to C^n(X^p)},</script>
<p>where the map $C^n(X)\to C^n(X^p)$ is the natural restriction map. This is indeed a decreasing filtration, so we get a spectral sequence $E_{p,q}^0=G^pC^{p+q}(X)\implies\hom^{p+q}(X)$. We claim that $E_{p,q}^0\simeq C^{p+q}(X^{p+1},X^p)$, the group of relative cochains. Note that we have a homomorphism of short exact sequences</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{CD}
0 @>>> F^{p+1}C^{p+q}(X) @>>> C^{p+q}(X) @>>> C^{p+q}(X^{p+1}) @>>> 0\\
&@VVV @VVV @VVV\\
0 @>>> F^pC^{p+q}(X) @>>> C^{p+q}(X) @>>> C^{p+q}(X^p) @>>> 0
\end{CD} %]]></script>
<p>where the middle map is the identity, the right map is the natural restriction map, and left map is the unique one making the diagram commute. Now, since the middle map is an isomorphism, the snake lemma tells us that the left map is injective, and that its cokernel is isomorphic to $\ker(C^p(X^{p+1})\to C^p(X^p))$, so</p>
<script type="math/tex; mode=display">E_0^{p,q}=G^pC^{p+q}(X)=F^pC^{p+q}(X)/F^{p+1}C^{p+q}(X)\simeq\ker\parens{C^{p+q}(X^{p+1})\to C^{p+q}(X^p)}=C^{p+q}(X^{p+1},X^p)</script>
<p>as claimed. Now, by definition, cohomology of this page gives relative cohomology, so $E_1^{p,q}=\hom^{p+q}(X^{p+1},X^p)$ <sup id="fnref:15"><a href="#fn:15" class="footnote">12</a></sup>. Recall that $\hom^{p+q}(X^{p+1},X^p)=0$ if $q\neq1$, so the only nontrivial differentials on the $E_1$ page are $d_1^{p,1}:\hom^{p+1}(X^{p+1},X^p)\to\hom^{p+2}(X^{p+2},X^{p+1})$. One easily checks that these agree with the differentials defining cellular cohomology, so the $E_2$ is given by</p>
<script type="math/tex; mode=display">E^2_{p,q}=\twocases{\hom_{\text{cell}}^{p+1}(X)}{q=1}0.</script>
<p>There are no more nontrivial differentials past this point, so $E_2^{p,q}=E_\infty^{p,q}$. Finally, since each diagonal <sup id="fnref:12"><a href="#fn:12" class="footnote">13</a></sup> only contains one nonzero object, we conclude that $\hom^p(X)\simeq E_\infty^{p-1,1}\simeq\hom^p_{\mrm{cell}}(X)$ <sup id="fnref:13"><a href="#fn:13" class="footnote">14</a></sup>, so singular and cellular cohomology agree.</p>
<h1 id="serre-spectral-sequence">Serre Spectral Sequence</h1>
<p>In this section, I’ll need to assume more topology background that in the previous sections; in particular, you should know about fibrations and their long exact sequences in homotopy. Here, we’ll construct the Serre spectral sequence which is used to relate the (co)homologies of the base and fiber spaces of a fibration, to that of its total space. With that said, let’s just do it <sup id="fnref:17"><a href="#fn:17" class="footnote">15</a></sup> <sup id="fnref:19"><a href="#fn:19" class="footnote">16</a></sup>.</p>
<div class="theorem" name="Serre Spectral Sequence">
Let
$$F\too E\xtoo\pi B$$
be a (weak/Serre) fibration with $B$ simply-connected. Then, there exists a spectral sequence $E_r^{p,q}$ with $E_2$-page
$$E_2^{p,q}=\hom^p(B;\hom^q(F))\implies\hom^{p+q}(E).$$
</div>
<div class="proof4">
For simplicities sake, assume that $B$ is a finite-dimensional CW-complex. To get the theorem for all CW-complexes, you can make a limit argument at the end. To then get the full theorem, you can use CW approximation. We won't bother writing either of those arguments out in detail.
<br />
Let $B^p$ denote $B$'s $p$-skeleton, and let $X^p=\inv\pi(B^p)$, so $X=\dirlim X^p$. Then, like before, we can filter $C^\bullet(X)$ by setting $F^pC^\bullet(X)=C^\bullet(X,X^{p-1})$ and so obtain a spectral sequence $E_0^{p,q}\implies\hom^{p+q}(X)$ with $E_1$-page $E_1^{p,q}=\hom^{p+q}(X^p,X^{p-1})$. Our goal is to calculate the $E_2$-page of this sequence. Namely, we will show that the differential $E_1^{p,q}\xto{d_1}E_1^{p+1,q}$ is "the same as"
$$\Hom(\hom_p(B^p,B^{p-1}),\hom^q(F))\to\Hom(\hom_{p+1}(B^{p+1},B^p),\hom^q(F)).$$
Note that cohomology of the above is just $\hom^p(B;\hom^q(F))$, (cellular) cohomology with coefficients in $\hom^q(F)$. We will first show that the groups match up, i.e. that $\hom^{p+q}(X^p,X^{p-1})\simeq\Hom(\hom_p(B^p,B^{p-1}),\hom^q(F))$. Let $\bracks{D_\alpha^p}_{\alpha\in A}$ be the collection of $p$-cells of $B$, let $S_\alpha^{p-1}=\del D_\alpha^p$, and let $\phi_\alpha:D_\alpha^p\to B^p$ be the characteristic map of the $p$-cell $D_\alpha^p$ for each $\alpha\in A$. Then, $\hom_p(B^p,B^{p-1})$ is free abelian with basis $\{D_\alpha^p\}$, so
$$\Hom(\hom_p(B^p,B^{p-1}),\hom^q(F))\simeq\Hom(\Z^{\oplus A},\hom^q(F))\simeq\prod_{\alpha\in A}\hom^q(F).$$
Now, define the spaces $(\wt D_\alpha^p,\wt S_\alpha^p)$ via the pullback squares
$$\begin{CD}
(\wt D_\alpha^p,\wt S_\alpha^{p-1}) @>\wt\phi_\alpha>> (X^p, X^{p-1})\\
@VVV @VVV\\
(D_\alpha^p, S_\alpha^{p-1}) @>\phi_\alpha>> (B^p, B^{p-1})
\end{CD}$$
Note that $\wt D_\alpha^p$ and $\wt S_\alpha^{p-1}$ are not literal disks/spheres, but instead look for like "disk x fiber" or "sphere x fiber." We claim that
$$\prod_{\alpha\in A}\hom^{p+q}(\wt D_\alpha^p,\wt S_\alpha^{p-1})\simeq\hom^{p+q}(X^p,X^{p-1}).$$
Since $B$ is a CW-complex, there exists some open neighborhood $N\subset B^p$ containing $B^{p-1}$ that deformation retracts onto $B^{p-1}$. That is, there exists some homotopy $h_t:N\to N$ with $h_0$ the identity and $h_1$ a retraction onto $B^{p-1}$. Since $\pi$ is a fibration, letting $\wt N=\inv\pi(N)$, we can lift $h_t$ to a homotopy $\wt h_t:\wt N\to\wt N$ wtih $\st h_0$ the identity and $\st h_1$ a map whose image lies in $X^{p-1}$. From this, we conclude that the inclusion $X^{p-1}\into\wt N$ is a homotopy equivalence. This means that $\pull\hom(X^p,X^{p-1})\iso\pull\hom(X^p,\wt N)$. At the same time, excision shows us that $\pull\hom(X^p,\wt N)\simeq\pull\hom(X^p\sm X^{p-1},\wt N\sm X^{p-1})$, but the RHS is now
$$\pull\hom\parens{\bigsqcup_{\alpha\in A}(\wt D_\alpha^p,\wt S_\alpha^{p-1})}\simeq\prod_{\alpha\in A}\pull\hom(\wt D_\alpha^p,\wt S_\alpha^{p-q}),$$
so we're happy. The next step is to now construct isomorphisms
$$\eps_\alpha:\hom^{p+q}(\wt D_\alpha^p,\wt S_\alpha^{p-1})\iso\hom^q(F).$$
Let $\wt D^p\onto D^p$ be any fibration over a $p$-dimensional disk. Write $\wt S^{p-1}:=\partial D^p=D^{p-1}_-\cup_{S^{p-2}}D^{p-1}_+$ as a union of a northern and southern hemisphere. Then, cohomology of the triple $(\wt D^p,\wt S^{p-1},\wt D^{p-1}_+)$ gives an isomorphism $\hom^{p+q-1}(\wt S^{p-1},\wt D^{p-1}_+)\iso\hom^{p+q}(\wt D^p,\wt S^{p-1})$, and excision (remove the complement of $\wt D_-^{p-1}$) gives an isomorphism $\hom^{p+q-1}(\wt S^{p-1},\wt D_+^{p-1})\iso\hom^{p+q-1}(\wt D_-^{p-1},\wt S^{p-2})$. Combining these two gives,
$$\hom^{p+q}(\wt D^p,\wt S^{p-1})\simeq\hom^{p+q-1}(\wt D_-^{p-1},\wt S^{p-2}).$$
Applying this to $\wt D_\alpha^p$ + a little bit of induction gives a map
$$\eps_\alpha:\hom^{p+q}(\wt D_\alpha^p,\wt S_\alpha^{p-1})\iso\hom^q(\wt D_\alpha^0).$$
Now, $\wt D_\alpha^0$ is the fiber over some $0$-cell $D_\alpha$, which is not necessarily $F$ (the fiber over our chosen basepoint $*$). However, lifting a (contractible) path from $D_\alpha^0$ to $F$ gives a (canonical) isomorphism $\hom^q(\wt D_\alpha^0)\simeq\hom^q(F)$, so we're done comparing groups. It's now left to show that the below square (whose vertical maps are isomorphisms) commutes
$$\begin{CD}
\hom^{p+q}(X^p,X^{p-1}) @>d_1>> \hom^{p+q+1}(X^{p+1},X^p)\\
@VVV @VVV\\
\Hom(\hom_p(B^p,B^{p-1}),\hom^q(F)) @>>> \Hom(\hom_{p-1}(B^p,B^{p-1}),\hom^q(F))
\end{CD}$$
We won't bother doing this here because I'm lazy, so just trust me when I say this is the case. QED.
</div>
<div class="remark">
The assumption that the base is simply connected is sufficient for our purposes in this post, but not strictly necessary. More generally, whenever $F\to E\xto\pi B$ is a fiber sequence with $\pi_1(B)$ acting on $\ast\hom(F)$ trivially, we get the same spectral sequence. Even more generally, when $\pi_1(B)$ acts on $\ast\hom(F)$ nontrivially, one can still get a spectral sequence abutting to the cohomology of the base space, but the $\hom^p(B;\hom^q(F))$ in the $E_2$ page can no longer be interpreted as singular cohomology; one now has to consider "cohomology with local coefficients" or "cohomology of a local system (locally constant sheaf)".
</div>
<div class="remark">
You do note need to use integral cohomology. The same argument(s) give a spectral sequence with $E_2^{p,q}=\hom^p(B;\hom^q(F;A))implies\hom^{p+q}(E;A)$ for any abelian group $A$.
</div>
<p>This is our first serious, readily applicable spectral sequence. Perhaps unsurprisingly, it turns out to be really useful, so let’s see it in action.</p>
<h3 id="hurewicz-isomorphism">Hurewicz isomorphism</h3>
<p>We’ll first use Serre’s spectral sequence to prove Hurewicz’s theorem that a spaces first nontrivial homotopy group and first nontrivial homology group coincide <sup id="fnref:8"><a href="#fn:8" class="footnote">17</a></sup>. Before proving this, we’ll first need to prove a lemma (also named Hurewicz) elucidating the connection between $\pi_1$ and $H_1$ <sup id="fnref:11"><a href="#fn:11" class="footnote">18</a></sup>.</p>
<div class="lemma">
$\hom_1(X)\simeq\ab{\pi_1(X)}$.
</div>
<div class="proof4">
For any $x\in X$, fix some path $\lambda_x:I\to X$ from our basepoint $*$ to $x$. Let $\Delta^n$ denote the standard $n$-simplex, and note that $\Delta^1\simeq I=[0,1]$, so paths and singular 1-simplices are the same thing. Hence, it seems fruitful to consider the map
$$\mapdesc \phi{\pi_1(X)}{\hom_1(X)}{[f]}{[f]}$$
send the homotopy class of a loop $f:I\to X$ (based at $*$) to its homology class as a singular 1-simplex $f:\Delta\to X$. We should probably start by showing that this map is well-defined. First note that if $f,g$ are loops in $X$ with concatenation $fg$, then, as singular 1-chains, $fg$ is homologous to $f+g$. This is because $fg-f-g$ is the boundary of the 2-simplex $\Delta^2\to\Delta^1\xto{fg}X$ given by first squashing two faces of the triangle onto the third and then applying the loop $fg$. Hence, to show that $h$ is well-defined, it suffices to show that nullhomotopic loops have trivial homology classes. Let $h_t:I\to X$ be a homotopy from the constant loop $h_0=*$ to $h_1$. View this as a map $h:I\by I\to X,(t,s)\mapsto h_t(s)$. Note that $I\by I$ is a square, and so cutting diagonally, we can view it as the union of two simplices $\sigma_1,\sigma_2$. When taking the boundary of $(\sigma_1-\sigma_2)$, the diagonal of the square will cancel out, and we'll be left with $h_1$ plus a bunch of constant simplices. Each constant 1-simplex is the boundary of the constant 2-simplex with the same image, so those are all homologically trivial. Hence, $h_1$ is homologically trivial as well, so $\phi$ is well-defined. $\phi$ is also clearly surjective since a 1-simplex is a cycle iff it's a loop. Since $\hom_1(X)$ is abelian, $\phi$ factors through a map $\ab\phi:\ab{\pi_1(X)}\to\hom_1(X)$, which we claim is an isomorphism (i.e. injective). Note that we have a map $\psi:C_1(X)\to\ab\pi_1(X)$ given on a simplex $\sigma:I\to X$ by $\psi(\sigma)=[\lambda_{\sigma(1)}\sigma\overline{\lambda_{\sigma(0)}}]$. This map vanishes on the boundary of 2-simplices, and so induces a homomorphism $\push\psi:\hom_1(X)\to\ab\pi_1(X)$. Finally, if $f:I\to X$ is a loop, then $\psi(h(f))=[\lambda_*f\overline{\lambda_*}]=[f]\in\ab\pi_1(X)$, so $h$ is injective and we win.
</div>
<p>Alright, got that out of the way. Before proving Hurewicz, I should note that while I’ve only talked about spectral sequences in cohomology, spectral sequences in homology exist as well. In particular, there’s one associated to an increasing filtration of a chain complex, and this gives rise to a Serre spectral sequence in homology which, for a fibration $F\to E\to B$, looks like $\hom_p(B;\hom_q(F))\implies\hom_{p+q}(E)$ <sup id="fnref:20"><a href="#fn:20" class="footnote">19</a></sup>. This is what we’ll use in the below proof.</p>
<div class="theorem" name="Hurewicz">
Let $X$ be a simply connected space. Then the first nontrivial homology and homotopy groups agree, i.e. given any $n\ge1$, if $\pi_k(X)=0$ for all $1\le k< n$, then $\hom_k(X)=0$ for all $1\le k< n$ and $\hom_n(X)\simeq\pi_n(X)$.
</div>
<div class="proof4">
We induct on $n$. The base case $n=1$ holds since $X$ is assumed simply connected and $\hom_1(X)=\ab{\pi_1(X)}$. Now, suppose $n>1$. Note that $\pi_k(\Omega X)\simeq\pi_{k+1}(X)$ for all $k$, so $\pi_k(\Omega X)=0$ for $1\le k< n-1$. This means that $\hom_k(\Omega X)=0$ in the same range, and $\hom_{n-1}(\Omega X)=\pi_{n-1}(\Omega X)=\pi_n(X)$. With this information in mind, consider the Serre spectral sequence induced by the path fibration $\Omega X\into PX\to X$. Induction tells us that $\redhom_k(X)=0$ for all $k< n$, so the $E^n$-page of this spectral sequence looks like (all the relevant differentials before this page were trivial)
$$\begin{array}{c | c c c c}
\tbf q & \\
n-1 & \hom_{n-1}(\Omega X) & 0 & \dots & 0 & ? \\
n-2 & 0 & 0 & \dots & 0 & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 0 & 0 & \dots & 0 & 0\\
0 & \Z & 0 & \dots & 0 & \hom_n(X) \\\hline
& 0 & 1 & \dots & n-1 & n & \tbf p
\end{array}$$
and there's a differential $d^n_{n,0}:\hom_n(X)\to\hom_{n-1}(\Omega X)\simeq\pi_n(X)$. Since this sequence is concentrated in the first quadrant, this is the last nontrivial differential that the $(n,0)$ and $(0,n-1)$ spots participate in, so $E_{n,0}^\infty=E_{n,0}^{n+1}$ and $E_{0,n-1}^\infty=E_{0,n-1}^{n+1}$. At the same time, $PX$ is contractible, so $E^\infty_{p,q}=0$ for all $p,q$. Thus, $d^n_{n,0}$ must be an isomorphism in order to kill both the $(0,n-1)$ and $(n,0)$ slots. Hence, $\hom_n(X)\simeq\pi_n(X)$ as claimed.
</div>
<div class="corollary">
Fix any $n>1$. We have $\pi_k(S^n)=0$ if $k< n$, and $\pi_n(S^n)=\Z$.
</div>
<h3 id="pi_4s3zmod2">$\pi_4(S^3)=\zmod2$</h3>
<p>For our final application of spectral sequences this post, we’ll compute a nontrivial homotopy group of a sphere. First note that the Hopf fibration $S^1\to S^3\to S^2$ gives $\pi_k(S^3)\simeq\pi_k(S^2)$ for all $k>2$, so last corollary already showed that $\pi_3(S^2)\simeq\Z$. This section, we’ll see that $\pi_4(S^2)\simeq\pi_4(S^3)\simeq\zmod2$. <sup id="fnref:25"><a href="#fn:25" class="footnote">20</a></sup></p>
<p>First let $S^3\to K(\Z,3)$ be a map inducing an isomorphism on $\pi_3$. Let $X$ be the homotopy fiber of this map, and let $Y$ be the homotopy fiber of the map $X\to S^3$. Then, $Y\simeq\Omega K(\Z,3)=K(\Z,2)\simeq\CP^\infty$. Furthermore, the long exact sequence of the original fibration $X\to S^3\to K(\Z,3)$ shows that $\pi_k(X)=0$ for $k\le3$ (use that $\pi_3(S^3)\to\pi_3(K(\Z,3))$ is an isomorphism) and that $\pi_4(X)\simeq\pi_4(S^3)$. By Hurewicz, this means that $\hom_4(X)\simeq\pi_4(S^3)$ and we’ll compute this by looking at the Serre spectral sequence (in cohomology, where we have cup products) of the left fibration $\CP^\infty\to X\to S^3$. The $E_3$-page of this sequence looks like</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{array}{c | c c c c}
6 & \Z a^3 & & & \Z a^3x\\
4 & \Z a^2 & & & \Z a^2x\\
2 & \Z a & & & \Z ax\\
0 & \Z1 & & & \Z x \\\hline
& 0 & 1 & 2 & 3
\end{array} %]]></script>
<p>Above, $x\in\hom^3(S^3)$ is a generator as is $a\in\hom^2(\CP^\infty)$. The cup product in cohomology let’s us use these to write down generators for the cohomology groups $\hom^p(S^3;\hom^q(\CP^\infty))$. All the groups in odd numbered rows are 0 as are all groups in columns $p\not\in{0,3}$. The nontrivial differentials are $d_3^{0,2q}:\Z a^q\to\Z a^{q-1}x$ where $q\ge1$. Since $X$ is 3-connected, we have $\hom^k(X)=0$ for all $k\le3$, and so $d_3^{0,2}$ must be an isomorphism, i.e. (we can choose $x$ s.t.) $da=x$ where by $da$ we really mean $d_3^{0,2}(a)$, but don’t want to write that every time. Now, using the fact that $d$ is a derivation, we have <sup id="fnref:21"><a href="#fn:21" class="footnote">21</a></sup></p>
<script type="math/tex; mode=display">d(a^2)=(da)a+a(da)=2ada=2ax\text{ and in general }d(a^q)=na^{q-1}da=qa^{q-1}x.</script>
<p>Thus, the differential $d_3^{0,2q}$ is really just multiplication by $q$. Since this is the last page with nontrivial differentials, we get that, on the $E_\infty=E_4$-page, the only nonzero object on the $n=5$ diagonal is $E_\infty^{3,2}\simeq\zmod2$ and that the $n=4$ diagonal is 0 everywhere. Thus, $\hom^4(X)=0$ and $\hom^5(X)=\zmod2$. In general, we get $\hom^{2k}(X)=0$ and $\hom^{2k+1}(X)=\zmod k$. Because these groups all have rank $0$, universal coefficients tells us that</p>
<script type="math/tex; mode=display">\hom^k(X)=\Ext^1(\hom_{k-1}(X),\Z),</script>
<p>and so <sup id="fnref:22"><a href="#fn:22" class="footnote">22</a></sup> we get that</p>
<script type="math/tex; mode=display">\hom_n(X)\simeq\twocases{\zmod k}{n=2k\text{ is even}}0.</script>
<p>In particular, $\pi_4(S^2)\simeq\pi_4(S^3)\simeq\hom_4(X)\simeq\zmod2$.</p>
<h3 id="exercises-">Exercises <sup id="fnref:14"><a href="#fn:14" class="footnote">23</a></sup></h3>
<div class="exercise" name="Thom-Gysin sequence">
Let $S^n\to E\to B$ be a fibration with $B$ simply connected. Construct a long exact sequence
$$\cdots\too\hom^{p+n}(B;\Z)\too\hom^{p+n}(E;\Z)\too\hom^p(B;\Z)\too\hom^{p+n+1}(B;\Z)\too\hom^{p+n+1}(E;\Z)\too\cdots$$
Use this sequence to show that if $S^n\to S^{n+k}\to S^k$ is a sphere fibration, then $n=k-1$.
</div>
<div class="exercise">
Let $F\to E\to S^n$ be a fibration with $n>1$. Construct a long exact sequence relating the cohomology of $F$ to that of $E$. Use this to determine the cohomology of the loop space $\Omega S^n$.
</div>
<h1 id="sepctral-sequence-of-a-double-complex">Sepctral Sequence of a Double Complex</h1>
<p>At this point, things start to slow down. We move away from topology and back into pure homological algebra in order to construct another type of spectral sequence. We won’t see examples of this one in this post, but we might in future posts.</p>
<p>Fix once again some ambient abelian category $\msA$, and let $A^{\bullet,\bullet}\in\CoCh(\CoCh(A))$ be some double complex. That is to say, we have a collection $A^{p,q}\in\msA$ of objects of $\msA$ with commuting horizontal $d^{p,q}:A^{p,q}\to A^{p+1,q}$ and vertical $\del^{p,q}:A^{p,q}\to A^{p,q+1}$ differentials.</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{matrix}
d^2=0 && \del^2=0 && d\circ\del=\del\circ d
\end{matrix} %]]></script>
<p>Complexes are good for defining (co)homology, and the usual way we get (co)homology from a double complex is by passing to its total complex <sup id="fnref:23"><a href="#fn:23" class="footnote">24</a></sup>.</p>
<div class="definition">
The <b>total complex</b> $\Tot(A^{\bullet,\bullet})\in\CoCh(\msA)$ has objects
$$\Tot(A^{\bullet,\bullet})^n=\bigoplus_{n=p+q}A^{p,q}$$
with differential
$$d^n=\sum_{n=p+q}\parens{d^{p,q}+(-1)^p\del^{p,q}}.$$
</div>
<p>Directly computing the cohomology of the total complex can be tricky, but luckily it comes with a natural filtration, and hence a natural spectral sequence <sup id="fnref:24"><a href="#fn:24" class="footnote">25</a></sup>. This filtration is</p>
<script type="math/tex; mode=display">F^p\Tot(A^{\bullet,\bullet})^n=\bigoplus_{\substack{n=i+j\\i\ge p}}A^{i,j}.</script>
<p>From this, we get a spectral sequence $E_r^{p,q}$ which, under mild assumptions (e.g. for each $n\in\Z$ there are only finitely many nonzero $A^{p,q}$ with $p+q=n$), converges to $\hom^{p+q}(\Tot(A^{\bullet,\bullet}))$. We claim that the $E_2$-page of this sequence is given by the “naive double cohomology”</p>
<script type="math/tex; mode=display">E_2^{p,q}\simeq\hom^p_d(\hom^q_\del(A^{\bullet,\bullet}))</script>
<p>given by taking vertical cohomology followed by horizontal cohomology. One we show this, we’ll call it a day. The $0$th page is given by</p>
<script type="math/tex; mode=display">E_0^{p,q}=F^p\Tot(A^{\bullet,\bullet})^{p+q}/F^{p+1}\Tot(A^{\bullet,\bullet})^{p+q}=A^{p,q},</script>
<p>with differential $d_0:E_0^{p,q}\to E_0^{p,q+1}$ equal to the vertical differential $\del$. Hence,</p>
<script type="math/tex; mode=display">E_1^{p,q}=\hom^q_\del(A^{p,\bullet}).</script>
<p>Any $[a]\in E_1^{p,q}$ is represented by some $a\in A^{p,q}$ with $\del a=0$, so the differential $d_1:E_1^{p,q}\to E_1^{p+1,q}$ on the $E_1$-page acts on these representatives just like $d$ does. Thus, the $E_2$-page is indeed</p>
<script type="math/tex; mode=display">E_2^{p,q}=\hom^p_d\parens{\hom^q_\del\parens{A^{\bullet,\bullet}}}</script>
<p>as claimed.</p>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>Replace “bird” with something less ill-defined if you want <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:18">
<p>While writing this, I repeatedly felt the need to add examples/exercises, so things quickly became more hands on than I first planned <a href="#fnref:18" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>And because singular (co)homology is one of the few things I know I never want to bother developing on this blog (along with the basics of linear algebra, most of point-set topology, and maybe some other stuff) <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>Just apply the universal property to $\alpha:=\mu\circ a=\mu\circ b$. <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>For intuition, in $\mathrm{Ab}$, monomorphisms are injective maps, so this is saying every subgroup should be a kernel (i.e. all quotients exist). Similarly, epimorphisms are surjective maps, so the image of every map should be the domain modulo some subgroup (i.e. first isomorphism) <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>If you’re reading this post, you should already know what these are. I’m just defining them here because I want this first section to stand alone as an introduction to abelian categories regardless of the fact that it fits into the broader context of this whole post. <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>I’ll likely say chain when I mean to say cochain many times below <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>From now on, I’ll start being a little more sloppy with my $d$’s, not always explitily giving them their upper indexing. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:9">
<p>There’s some dubious reasoning here, but I’m only trying to build intuition so that’s fine <a href="#fnref:9" class="reversefootnote">↩</a></p>
</li>
<li id="fn:10">
<p>Don’t read the proof of this; do it for yourself. The only reason I have it typed up here is that I’ve never worked through it on my own before this. <a href="#fnref:10" class="reversefootnote">↩</a></p>
</li>
<li id="fn:16">
<p>This filtration always works anytime you can write $X$ as a direct limit of spaces $X=\dirlim X^p$, and is the one I alluded to way back in the beginning of this post. <a href="#fnref:16" class="reversefootnote">↩</a></p>
</li>
<li id="fn:15">
<p>If I were smart, I wouldv’e subtracted 1 when defining the filtration, so that this would read $\hom^{p+q}(X^p,X^{p-1})$ instead. <a href="#fnref:15" class="reversefootnote">↩</a></p>
</li>
<li id="fn:12">
<p>i.e. $\bracks{E_2^{p,q}:p+q=n}$ for some fixed $n$. <a href="#fnref:12" class="reversefootnote">↩</a></p>
</li>
<li id="fn:13">
<p>Secretely, we need that $X$ is finite (i.e. $X=X^n$ for some $n$) so the filtration is bounded. However, we can always write $X\approx\dirlim X^p$ as a direct limit of finite CW-complexes, and cohomology commutes with direct limits (i.e. $\pull\hom(\dirlim X^p)=\invlim\pull\hom(X^p)$), so we win by taking limits. <a href="#fnref:13" class="reversefootnote">↩</a></p>
</li>
<li id="fn:17">
<p>I probably should have mentioned this earlier, but as far as I’m concerned, all topological spaces $X$ are path-connected and based with basepoint denoted by $* \in X$. <a href="#fnref:17" class="reversefootnote">↩</a></p>
</li>
<li id="fn:19">
<p>If any of you know a nice way to get footnotes working inside of html blocks, please tell me <a href="#fnref:19" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>Secretly, non-simply connected spaces don’t exist. <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:11">
<p>Probably you know this result already, but I’ve never bothered to pay attention to a proof of it before, so it feels worth writing up. <a href="#fnref:11" class="reversefootnote">↩</a></p>
</li>
<li id="fn:20">
<p>Also worth mentioning that the differential $d^r$ on the $r$th page of a homological spectral sequence has bidegree $(-r, r-1)$. <a href="#fnref:20" class="reversefootnote">↩</a></p>
</li>
<li id="fn:25">
<p>By Freudenthal suspension, we actually get the stronger result that $\pi_{n+1}(S^n)\cong\zmod2$ for all $n\ge3$. <a href="#fnref:25" class="reversefootnote">↩</a></p>
</li>
<li id="fn:21">
<p>$a$ and $da$ commute because $a$ lives in even degree. Remember that $ab=(-1)^{\deg(a)\deg(b)}ba$ in general when taking cup products <a href="#fnref:21" class="reversefootnote">↩</a></p>
</li>
<li id="fn:22">
<p>I think you may need to use that $\hom_{k-1}(X)$ is finitely generated and so looks like $\Z^r\oplus\zmod{p_1^{k_1}}\oplus\dots\oplus\zmod{p_g^{k_g}}$ and then use that Ext splits over direct sums in the first factor <a href="#fnref:22" class="reversefootnote">↩</a></p>
</li>
<li id="fn:14">
<p>It feels weird having an “exercises” (sub)section, but I didn’t want to give the impression that these were under the Hurewicz (sub)heading <a href="#fnref:14" class="reversefootnote">↩</a></p>
</li>
<li id="fn:23">
<p>Unimportant technical detail: in general, abelian categories are only required to have finite direct sums (“biproducts”). However, if you have a really big double complex (i.e. $A^{p,q}\neq0$ even when $p<0$ and/or $q<0$), then formation of the total complex can involve infinite direct sums, and so may not always be possible. <a href="#fnref:23" class="reversefootnote">↩</a></p>
</li>
<li id="fn:24">
<p>Actually, two natural filtrations/spectral sequences, but I’ll only mention one of them <a href="#fnref:24" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>Back when I was in high-school, I became really interested in this thing called “machine learning.” The main idea is that you bombard some algorithm with a ton of examples (of a task being performed or of objects being classified), and then you cross your fingers and hope it has managed to reliably learn how to do what the examples showcased. One big draw of this approach is that there are many tasks where it is not clear how to accomplish them, but where it feels like there has to be enough information present that accomplishing them is possible. For example, imagine writing a program that takes in an image and tells you if there’s a bird in it; this is hard to do algorithmically, but certainly the pixel values of an image contain enough information to decide whether or not there’s a bird there 1. The moral of this detour is that sometimes we find ourselves in situations where we have lots of information available to tackle some problem, but figuring out how to utilize all that information is quite tricky. Imagine, for example, you have some topological space $X$ with a filtration $ \newcommand{\msC}{\ms C} \newcommand{\msA}{\ms A} \DeclareMathOperator{\Ch}{Ch} \DeclareMathOperator{\CoCh}{CoCh} \DeclareMathOperator{\Fil}{Fil} \DeclareMathOperator{\Tot}{Tot} $ Replace “bird” with something less ill-defined if you want ↩$\ell$-adic Representations of Elliptic Curves2019-08-12T00:00:00+00:002019-08-12T00:00:00+00:00https://nivent.github.io/blog/elliptic-irrep<p>The title of this post is a bit of a misnomer. We won’t be talking about representations of elliptic curves, but about representations attached to (associated with?) elliptic curves. In particular, the goal of this post is to prove that given an elliptic curve $E/\Q$ defined over the rationals, and a prime $\l$, the $\l$-adic representation</p>
<script type="math/tex; mode=display">G_{\Q}\to\GL(T_\l(E))\iso\GL_2(\Z_\l)\subset\GL_2(\Q_\l)</script>
<p>is irreducible where $G_{\Q}=\Gal(\Qbar/\Q)$ is the absolute Galois group of $\Q$ and $T_\l(E)=\invlim E[\l^n]$ is $E$’s $\l$-adic Tate module. If you don’t know what some of these words mean, don’t worry; I’ll explain. <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup></p>
<p>It’ll be quite a while before we get into defining and proving irreducibility of these representations since doing so requires a lot of ideas I have not introduced on this blog before. To begin, we’ll introduce some of the basics of the general theory of algebraic curves before focussing specifically on elliptic curves. Since I’ve said the word $\spec$ here before, I could set things up in those terms <sup id="fnref:7"><a href="#fn:7" class="footnote">2</a></sup>, but I won’t; instead, I’ll take an approach that’s more concrete. <sup id="fnref:2"><a href="#fn:2" class="footnote">3</a></sup> With that said, let’s get started$\dots$</p>
<p>Note: I’m writing this post like a madman (in a rushed manner manner split over several days without much of a game plan ahead of time), so it likely contains more mistakes than usual (e.g. there may be some circular arguments here). Some time after finishing it and putting it online, I’ll take another look at it and try to resolve all these issues. Until then, recovering a coherent post from what’s below is left as an exercise to the reader <sup id="fnref:8"><a href="#fn:8" class="footnote">4</a></sup>.</p>
<h1 id="algebraic-varieties">Algebraic Varieties</h1>
<p>This section is mostly definitions <sup id="fnref:3"><a href="#fn:3" class="footnote">5</a></sup>, and so can be skipped and referred back to whenever you see something but don’t know what it means.</p>
<p>Fix a field $k$ with algebraic closure $\bar k$. Let $\bar k[X]=\bar k[x_1,\dots,x_n]$ be a polynomial ring in $n$ variables. Let $\A^n=\A^n(\bar k)=\bracks{P=(p_1,\dots,p_n):p_i\in\bar k}$ denote <b>affine $n$-space</b> (over $\bar k$).</p>
<div clas="definition">
Let $I\subset\bar k[X]$ be an ideal. Its <b>vanishing set</b> is
$$V(I)=\bracks{P\in\A^n:f(P)=0\text{ for all }f\in I}.$$
Any set $C$ of the form $C=V(I)$ for some $I$ is called an <b>affine algebraic set</b>.
</div>
<div class="definition">
Given any set $C\subset\A^n$, we associate to it the ideal
$$I(C)=\bracks{f\in\bar k[X]:f(P)=0\text{ for all }P\in C}.$$
When $C$ is an algebraic set, we see that $C$ is <b>defined over $k$</b>, denoted $C/k$, exactly when $I(C)$ can be generated by polynomials in $k[X]$; equivalently, exactly when $I(C)=I(C/K)\bar k[X]$ where $I(C/K)=I(C)\cap k[X]$.
</div>
<div class="remark">
Let $G=G_{\bar k/k}=\Gal(\bar k/k)$ be the absolute Galois group of $k$. Then, $G$ naturally acts on $\A^n$, and because $f(P^\sigma)=f(P)^\sigma$ for any $f\in\bar k[X]$ and $\sigma\in G$, we see that $G$ even acts on algebraic sets defined over $k$.
</div>
<p>Note that Hilbert’s basis theorem tells us that $\bar k[X]$ is noetherian, so any algebraic set is given by the vanishing of only finitely many polynomials.</p>
<div class="definition">
When $C$ is defined over $k$, we write
$$C(K)=C\cap\A^n(K)=\bracks{P\in C:P^\sigma=P\text{ for all }\sigma\in G_{\bar k/k}}$$
for the set of <b>$k$-rational points</b> on $C$.
</div>
<div class="definition">
An affine algebraic set $C/k$ is called an <b>affine variety</b> if $I(C)$ is a prime ideal in $\bar k[X]$. In this case, its <b>affine coordinate ring</b> is $k[C]=k[X]/I(C/k)$, and its <b>function field</b> is $k(C)=\Frac k[C]$.
</div>
<div class="remark">
Note that an element $f\in\bar k[C]$ induces a well-defined function $f:C\to\bar K$ since we only killed functions vanishing along $C$. Furthermore, the action of $G_{\bar k/k}$ on $k[X]$ descends to one on $\bar k[V]$ (and $\bar k(V)$). As with points, we denote the action of $\sigma\in G_{\bar k/k}$ on $f$ by $f\mapsto f^\sigma$, so we see that
$$(f(P))^\sigma=f^\sigma(P^\sigma)$$
always.
</div>
<p>That was a lot of definitions, and there are only more to come. If you haven’t seen material like this before, this may be a good place to mention that my <a href="../comm-alg">previous post on algebraic geometry</a> might be helpful for understanding why some of the definitions here are reasonable. For example,</p>
<div class="definition">
Let $V$ be a variety. Then, its <b>dimension</b> is $\dim(V)=\trdeg_{\bar k}\bar k(V)$.
</div>
<div class="example">
If $V=\A^n$, then $I(V)=(0)$, $\bar k[V]=\bar k[X]$, and $\bar k(V)=\bar k(X)=\bar k(x_1,\dots,x_n)$, so $\dim V=n$.
</div>
<div class="example">
If $f\in\bar k[X]$ and $V=V(f)$ is the vanishing set of a single (irreducible) nonconstant polynomial, then $\dim(V)=n-1$.
</div>
<div class="definition">
Let $V$ be a variety, and fix some $P\in V$. Let $f_1,\dots,f_m\in\bar k[X]$ generate $I(V)$. We say that $V$ is <b>nonsingular</b> (or <b>smooth</b>) <b> at $P$</b> if the $m\by n$ matrix
$$\parens{\pderiv{f_i}{X_j}(P)}_{1\le i\le m\atop1\le j\le n}$$
has rank $n-\dim(V)$$. We say that $V$ is <b>nonsingular</b> (or <b>smooth</b>) if it's nonsingular at every point.
</div>
<div class="exercise">
Let $V/k$ be a $1$-dimensional variety. Prove that $k[V]$ is a Dedekind domain when $V$ is smooth.
</div>
<div class="definition">
Let $V$ be a variety, and fix a point $P\in V$. Consider the evaluation at $P$ homomorphism $\bar k[V]\to\bar k,f\mapsto f(P)$. Its kernel is the maximal ideal $\mfm_P$ of functions vanishing at $P$. The localization $\bar k[V]_P:=\bar k[V]_{\mfm_P}$ is call the <b>local ring of $V$ at $P$</b>. That is, elements of $\bar k[V]_P$ look like $F=f/g$ where $f,g\in\bar k[V]$ and $g(P)\neq0$. Given such an element, $F(P)=f(P)/g(P)\in\bar k$ is well-defined. Functions in $\bar k[V]_P$ are said to be <b>regular</b> (or <b>defined</b>) at $P$.
</div>
<p>We now turn to projective algebraic sets and projective varities. <b>Projective $n$-space (over $k$)</b>, denoted $\P^n=\P^n(\bar k)$, is the set of lines through the origin in $\A^{n+1}$. Succienctly, $\P^n=(\A^{n+1}\sm{0})/\units{\bar k}$. More explicitly,</p>
<script type="math/tex; mode=display">\P^n=\bracks{\sqbracks{p_0:p_1:\dots:p_n}:p_i\in\bar k}\left/\parens{\sqbracks{p_0:p_1:\dots:p_n}\sim\sqbracks{\lambda p_0:\lambda p_1:\dots:\lambda p_n}\text{ for all }\lambda\in\units{\bar k}}\right..</script>
<p>Note that $G_{\bar k/k}$ acts on $\P^n$ be acting on each coordinate individually. Since individual coordinates in projective space are not well-defined, evaluating a polynomial at a projective point doesn’t make sense. However, for homogeneous polynomials, we can still check if one vanishes at a point.</p>
<div class="definition">
A polynomial $f\in\bar k[X]=\bar k[X_0,\dots,X_n]$ is said to be <b>homogeneous of degree $d$</b> if
$$f(\lambda X_0,\dots,\lambda X_n)=\lambda^df(X_0,\dots,X_n)$$
for all $\lambda\in\bar K$. Similarly, an ideal $I\subset\bar k[X]$ is <b>homogeneous</b> if it is generated by homogeneous polynomials.
</div>
<p>Given a homogenous polynomial $f$, the answer to the question, “does $f(P)=0$?” does not depend on how we write down $P$, so we can define a vanishing set $V(f)\subset\P^n$. Similarly, we get some vanishing set $V(I)\subset\P^n$ for any homogenous ideal $I$.</p>
<div class="definition">
A <b>(projective) algebraic set</b> is any set of the form $C=V(I)$ for some homogeneous ideal $I\subset\bar k[X]$. If $C$ is a projective algebraic set, then its homogeneous ideal $I(C)$ is the one generated by
$$\bracks{f\in\bar k[X]:f\text{ is homogeneous and }f(P)=0\text{ for all }P\in C}.$$
Such a $C$ is <b>defined over $k$</b>, denoted $C/k$, if $I(C)$ can be generated by homogeneous polynomials in $k[X]$. If $C$ is defined over $k$, then its <b>$k$-rational points</b> are
$$C(k)=C\cap\P^n(k)=\bracks{P\in C:P^\sigma=P\text{ for all }\sigma\in G_{\bar k/k}}=\bracks{[p_0:\dots:p_n]\in C:p_i\in k}.$$
</div>
<div class="example">
A <b>line</b> in $\P^2$ is the vanishing set of a linear equation
$$aX+bY+cZ=0$$
where $a,b,c\in\bar k$ not all zero.
</div>
<div clas="definition">
A projective algebraic set $C$ is called a <b>(projective) variety</b> if its homogeneous ideal $I(C)$ is prime in $\bar k[X]$.
</div>
<p>Note that there are many ways of embedding $\A^n\subset\P^n$. For example, each set</p>
<script type="math/tex; mode=display">U_i=\bracks{\sqbracks{p_0:\dots:p_n}\in\P^n:p_i\neq0}</script>
<p>is easily seen to be a copy of $\A^n$, e.g. via the natural bijection $\phi_i:\A^n\to U_i$ given by</p>
<script type="math/tex; mode=display">(p_1,\dots,p_n)\mapsto\sqbracks{p_1,\dots,p_{i-1},1,p_i,\dots,p_n}.</script>
<p>Hence, given any projective algebraic set $C$ with homogeneous ideal $I(C)$ (and a choice of $i$), $\inv\phi_i(V\cap U_i)$, which we call $C\cap\A^n$, is an affine algebraic set with ideal</p>
<script type="math/tex; mode=display">I(V\cap\A^n)=\bracks{f(Y_1,\dots,Y_{i-1},1,Y_{i+1},\dots,Y_n):f(X_0,\dots,X_n)\in I(C)}.</script>
<p>This shows that any projective algebraic set (resp. variety) is covered by a bunch of affine algebraic sets (resp. varieties) $C\cap U_0,\dots,C\cap U_n$. The process of replacing a homogeneous $f(X_0,\dots,X_n)$ with $f(Y_1,\dots,Y_{i-1},1,Y_i,\dots,Y_n)$ is called <b>dehomogenization with respect to $X_i$</b>, and can be reversed by taking $f(Y)\in\bar k[Y]=\bar k[Y_1,\dots,Y_n]$ to
<script type="math/tex">\ast f(X_0,\dots,X_n)=X_i^{\deg f}f\parens{\frac{X_0}{X_i},\frac{X_1}{X_i},\dots,\frac{X_{i-1}}{X_i},\frac{X_{i+1}}{X_i},\dots,\frac{X_n}{X_i}},</script>
the <b>homogenization of $f$ with respect to $X_i$</b>.</p>
<p>This let’s us take an affine algebraic set $V\subset\A^n$ to its <b>projective closure</b> which is the projective algebraic set whose homogenous ideal is generated by $\bracks{\ast f(X):f\in I(V)}$. When we do this to an affine variety $V$, we get out a projective variety. Because we can move back and forth between the affine and projective worlds like this, we often make definitions on projective varieties by referring to the analagous thing on one of their affine covers. For example,</p>
<div class="definition">
Let $V/k$ be a projective variety, and choose $\A^n\subset\P^n$ such that $V\cap\A^n\neq\emptyset$. The <b>dimension</b> of $V$ is $\dim(V)=\dim(V\cap\A^n)$. We make similar definitions for the function field $k(V)$ of $V$, the local ring $\bar k[V]_P$ of $V$ at $P$, and the questions of whether $V$ is smooth at a point $P$ or whether a function $F\in\bar k(V)$ is regular at some $P\in V$. As an example of one of these, the <b>local ring of $V$ at $P$</b>, denoted $\bar k[V]_P$, is the local ring of $V\cap\A^n$ at $P$.
</div>
<p>At this point, we’ve more-or-less laid the groundwork for what a (projective) variety is. So, the next thing would be to describe maps between them. One quirky thing about varieties is that their maps are not required to be defined everywhere. <sup id="fnref:4"><a href="#fn:4" class="footnote">6</a></sup></p>
<div class="definition">
Let $V_1,V_2\subset\P^n$ be projective varieties. A <b>rational map</b> from $V_1$ to $V_2$ is a map of the form
$$\phi:V_1\too V_2,\text{ }\text{ }\text{ }\phi=[f_0:\dots:f_n],$$
where $f_i\in\bar k(V_1)$ such that for every $P\in V_1$ for which all $f_i$ are defined, we have
$$\phi(P)=[f_0(P):\dots:f_n(P)]\in V_2.$$
We say that $\phi$ is defined over $k$ if there exists some $\lambda\in\units{\bar k}$ s.t. $\lambda f_0,\dots,\lambda f_n\in k(V_1)$.
</div>
<div class="remark">
We have $G_{\bar k/k}$ acting on $\phi$ in the obvious way, e.g.
$$\phi^\sigma(P)=\sqbracks{f^\sigma_0(P):\dots:f_n^\sigma(P)}.$$
Note that $\phi$ is defined over $k$ iff $\phi=\phi^\sigma$ for all $\sigma\in G_{\bar k/k}$.
</div>
<div class="definition">
A rational map $\phi=[f_0:\dots:f_n]:V_1\to V_2$ is <b>regular</b> (or <b>defined</b>) at $P\in V_1$ if there is a function $g\in\bar k(V_1)$ such that
<ol>
<li> each $gf_i$ is regular at $P$, and </li>
<li> there is some $i$ for which $(gf_i)(P)\neq0$. </li>
</ol>
When this is the case, we set
$$\phi(P)=\sqbracks{(gf_0)(P):\dots:(gf_n)(P)}.$$
A rational map that is regular at every point is called a <b>morphism</b>.
</div>
<div class="exercise">
Show that we could equivalently define a rational map $\phi:V_1\to V_2$ as one of the form
$$\phi=[\phi_0(X):\dots:\phi_n(X)]$$
where the $\phi_i(X)\in\bar k[X]=\bar k[x_0,\dots,x_n]$ are all homogeneous of the same degree, but not all in $I(V_1)$, and
$$f(\phi_0(X),\dots,\phi_n(X))\in I(V_1)$$
for every $f\in I(V_2)$.
</div>
<div class="definition">
A morphism $\phi:V_1\to V_2$ is called an <b>isomorphism</b> if there exists a morphism $\psi:V_2\to V_1$ such that $\phi\circ\psi$ and $\psi\circ\phi$ are both identity maps. We say that $V_1/k$ and $V_2/k$ are <b>isomorphic over $k$</b> if $\phi,\psi$ are both defined over $k$.
</div>
<p>Phew. Alright, I think we can move on to the next section now. I hope you have all of this <a href="https://www.youtube.com/watch?v=xseGkDSt8ec">memorized</a>.</p>
<h1 id="algebraic-curves">Algebraic Curves</h1>
<p>An <b>algebraic curve</b> is a projective variety of dimension 1. For curves, $\bar k[C]$ is nicer than just an arbitrary integral domain.</p>
<div class="proposition">
Let $C$ be a curve smooth at $P\in C$. Then, $\bar k[C]_P$ is a dvr.
</div>
<div class="proof4">
You first accept the fact that $\dim_{\bar k}(\mfm_P/\mfm_P^2)=\dim C=1$ exactly when $C$ is smooth at $P$. Then, you use Nakayama's lemma to conclude that $\mfm_P$ is a principal ideal of the local ring $\bar k[C]_P$. Once you know the maximal ideal is principal, show that we have a dvr is done a la <a href="../dedekind-domain">the proof here</a> that local Dedekind domains are dvrs.
</div>
<div class="corollary">
$\bar k[C]$ is a Dedekind domain when $C$ is a smooth curve.
</div>
<div class="proposition">
Let $C$ be a curve with smooth point $P\in C$. The <b>valuation on $\bar k[C]_P$</b> is given by
$$\mapdesc{\ord_P}{\bar k[C]_P}{\Z_{\ge0}\cup\{\infty\}}f{\sup\bracks{d\in\Z:f\in\mfm_P^d}}.$$
This extends naturally to a valuation, still denoted $\ord_P$, on $\bar k(C)=\Frac\bar k[C]_P$. A <b>uniformizer</b> for $C$ at $P$ is any function $t\in\bar k(C)$ with $\ord_P(t)=1$.
</div>
<div class="definition">
With $C,P$ as above and $f\in\bar k(C)$, we say that $f$ has a <b>zero</b> at $P$ if $\ord_P(f)>0$, $f$ has a <b>pole</b> at $P$ if $\ord_P(f)<0$, $f$ is <b>regular</b> at $P$ if $\ord_P(f)\ge0$, and $f(P)=\infty$ if $f$ has a pole at $P$.
</div>
<div class="remark">
Let $C/k$ be a smooth curve. Note that given some $f\in\bar k(C)$, we get an associated rational map, also denoted $f$, which is
$$f:C\too\P^1,\text{ }\text{ }\text{ }P\longmapsto[f(P):1].$$
This map turns out to actually be a morphism (where $[f(P):1]$ is interpreted as $[1:0]$ if $f$ has a pole at $P$), and all rational maps $C\to\P^1$ arise in this fashion (except the constant map to $[1:0]$).
</div>
<p>Now, here’s some nice information about regular functions that we’ll just take for granted.</p>
<div class="proposition">
Let $C$ be a smooth curve and fix $f\in\bar k(C)$ with $f\neq0$. Then, there are only finitely many points of $C$ at which $f$ has a pole or zero. Furthermore, if $f$ has no poles, then $f\in\bar K$ is constant.
</div>
<div class="remark">
Let $\phi:C_1\to C_2$ be a morphism of curves. Then, we get an induced map $\pull\phi:K(C_2)\to K(C_1)$ on function fields, $\pull\phi(f)=f\circ\phi$.
</div>
<div class="definition">
Let $\phi:C_1\to C_2$ be a morphism of curves. We will define the <b>degree</b> $\deg\phi$ of $\phi$. If $\phi$ is constant, we set $\deg\phi=0$. Else, we set
$$\deg\phi=\sqbracks{K(C_1):\pull\phi K(C_2)}.$$
Furthermore, still assuming $\phi$ nonconstant, for $P\in C_1$, we define the <b>ramification index</b> (or <b>degree</b>) <b>of $\phi$ at $P$</b> to be
$$e_\phi(P)=\ord_P(\pull\phi t_{\phi(P)}),$$
where $t_{\phi(P)}\in K(C_2)$ is a uniformizer at $\phi(P)$. We say that $\phi$ is <b>unramified</b> at $P$ if $e_\phi(P)=1$.
</div>
<div class="proposition">
Let $\phi:C_1\to C_2$ be a morphism of curves. Then, $\phi$ is either surjective or constant. Furthermore, when it's surjective, we have
$$\sum_{P\in\inv\phi(Q)}e_\phi(P)=\deg\phi$$
for every $Q\in C_2$.
</div>
<p>WIth this, let’s move onto differentials.</p>
<h3 id="differentials">Differentials</h3>
<p>Because calculus is quite useful for doing analysis/geometry over $\R$ and $\C$, we’d like something similar for studying curves over any field $k$.</p>
<div class="definition">
Let $C$ be a curve. The space of <b>meromorphic differential forms</b> on $C$, denoted $\Omega_C$ or $\omega_C$, is the $\bar k(C)$-vector space generated by symbols of the form $\dx$ with $x\in\bar k(C)$, subject to the rules
<ul>
<li> $\d(x+y)=\dx+\dy$ </li>
<li> $\d(xy)=x\dy+y\dx$ </li>
<li> $\d a=0$ </li>
</ul>
for all $x,y\in\bar k(C)$ and $a\in\bar k$.
</div>
<div class="remark">
Let $\phi:C_1\to C_2$ be a nonconstant map of curves. This gives rise to an associated map of functions fields $\pull\phi:\bar k(C_2)\to\bar k(C_1),f\mapsto f\circ\phi$ as well as an induced map on differentials $\pull\phi:\Omega_{C_2}\to\Omega_{C_1}$,
$$\pull\phi\parens{\sum f_i\dx_i}=\sum\parens{\pull\phi f_i}\d\parens{\pull\phi x_i}.$$
</div>
<p>Now, it’ll be useful to know that $\Omega_C$ is a $1$-dimensional $\bar k(C)$-vector space for any curve $C$; I will not prove this, but it’s good to know. As a consequence of this (+ something else I didn’t bother mentioning), any $\omega\in\Omega_C$ can be written as $\omega=g\dt$ for a unique $g\in\bar k(C)$ where $t\in\bar k(C)$ is some fixed uniformizer. We’ll denote this $g$ by $\omega/\dt$. Note that the quantity $\ord_P(\omega/\dt)$ does not depend on the choice of uniformizer $t$, and so we simply denote it by $\ord_P(\omega)$. Furthermore, $\ord_P(\omega)\neq0$ for only finitely many $P\in C$ (when $\omega\neq0$). Call $\omega$ <b>holomorphic</b> (I don’t know what the standard term is) is $\ord_P(\omega)\ge0$ for all $P\in C$.</p>
<h3 id="divisors--riemann-roch">Divisors & Riemann-Roch</h3>
<p>This is the point in this blog post where I may actually start bothering to prove things I claim. <sup id="fnref:5"><a href="#fn:5" class="footnote">7</a></sup> We’ll state and prove one of the main tools in the study of algebraic curves: the Riemann-Roch theorem. Before we can state it though, we first need to introduce some terminology:</p>
<div class="definition">
Fix a curve $C$. A <b>divisor</b> $D$ on $C$ is an element of the <b>divisor group</b> $\Div(C)=\Z^{\oplus C}$, the free abelian group generated by points of $C$. That is, $D$ is a (finite) formal sum
$$D=\sum_{P\in C}n_P[P],$$
where $n_P\in\Z$. We denote $n_P$ by $\ord_P(D)$. The <b>degree</b> of $D$ is $\deg D=\sum_{P\in C}n_P$. Note that, if $C$ is defined over $k$, then $G_{\bar k/k}$ acts on $\Div(C)$ and we can say what it means for a divisor to be defined over $k$ (it means that $D$ is fixed by $G_{\bar k/k}$).
</div>
<div class="remark">
Given a map $\phi:C_1\to C_2$ of curves, and a divisor $D\in\Div(C_2)$, we get a divisor $\pull\phi D\in\Div(C_1)$ defined by
$$\pull\phi D=\sum_{Q\in C_2}\sum_{P\in\inv\phi(Q)}\ord_Q(D)e_\phi(P)[P].$$
It is clear that $\deg(\pull\phi D)=(\deg\phi)(\deg D)$.
</div>
<div class="remark">
When $C$ is a smooth curve, to each $f\in\units{\bar k(C)}$, we associate the divisor
$$(f)=\div(f)=\sum_{P\in C}\ord_P(f)[P],$$
and similarly for differentials $\omega\in\Omega_C$. In the case that $f\in\units{\bar k(C)}$, view $f$ as a map $C\to\P^1$, and note that $\div(f)=\pull fD$ where $D=[0:1]-[1:0]\in\Div(\P^1)$, so $\deg\div (f)=\deg(f)\deg D=0$.
</div>
<div class="definition">
Divisors of the form $(f)$ are called <b>principal</b>, and form a subgroup of $\Div(C)$ since $(f)-(g)=(f/g)$. Two divisors $D_1,D_2$ are called <b>linearly equivalent</b>, denoted $D_1\sim D_2$ if $D_1-D_2$ is principal. The <b>Picard group</b> $\Pic(C)$ of $C$ is the quotient of $\Div(C)$ by its subgroup of principal divisors.
</div>
<div class="remark">
Because $\dim_{\bar k(C)}\Omega_C=1$, we have $\omega_1=f\omega_2$ with $f\in\units{\bar k(C)}$ for any two nonzero deifferentials $\omega_1,\omega_2\in\Omega_C$. Hence, $\div(\omega_1)=\div(f)+\dim(\omega_2)$ and so we can define the <b>canonical divisor (class)</b> on $C$ to be the image of $\dim(\omega)$ in $\Pic(C)$ for any nonzero differential $\omega\in\Omega_C$.
</div>
<p>Part of the utility of divisors is that they are a convenient way of describing the location and number of zeros/poles of functions. To make this more precise, given two divisors $D_1,D_2$, we say that $D_1\ge D_2$ if $\ord_P(D_1)\ge\ord_P(D_2)$ for all $P\in C$. If $D\ge0$, then we say that $D$ is <b>effective</b>. Note that for a function $f\in\units{\bar k(C)}$, point $P\in C$, and number $n\in\Z_{>0}$, we have $\div(f) + n[P]\ge0$ iff $f$ has a pole of order at most $n$ at $P$, and we similarly have $\div(f) - n[P]\ge0$ iff $f$ has a zero or order at least $n$ at $P$.</p>
<div class="definition">
Let $D\in\Div(C)$. We associate to $D$ the $\bar k$-vector space
$$\mc L(D)=\bracks{f\in\units{\bar k(C)}:\div(f)+D\ge0}\cup\{0\},$$
and let $\l(D)=\dim_{\bar k}(D)$ denote its dimension.
</div>
<div class="proposition">
Let $D\in\Div(C)$. Then, (1) $\deg D<0\implies\l(D)=0$, and (2) $\mc L(D)$ only depends (up to isomorphism) on $D$'s class in $\Pic(C)$.
</div>
<div class="proof4">
(1) Suppose $f\in\mc L(D)$ and $f\neq0$. Then, $\div(f)+D\ge0$ so
$$0\le\deg(\div(f)+D)=\deg\div(f)+\deg(D)=\deg(D).$$
(2) Pick $D'\in\Div(C)$ with $D-D'=\div(g)$ ($g\in\units{\bar k(C)}$). Then the map $\mc L(D)\too\mc L(D'),f\longmapsto fg$ is an isomorphism.
</div>
<p>A natural question to ask, and one mostly answered by Riemann-Roch, is, “Can we calculate $\l(D)$ for an arbitrary divisor?” Our main tool for getting a handle of $\l(D)$ <sup id="fnref:6"><a href="#fn:6" class="footnote">8</a></sup> is a certain 7-term exact sequence.</p>
<div class="definition">
Given a divisor $D\in\Div(C)$, define the $\bar k$-vector space
$$\Omega(D)=\Omega_C(D)=\bracks{\omega\in\Omega_C:\dim(\omega)+D\ge0}\cup\{0\}.$$
</div>
<div class="notation">
Given a $\bar k$-vector space $V$, let $\ast V=\dual V=\Hom_{\bar k}(V,\bar k)$ denote its dual vector space.
</div>
<div class="lemma">
Let $C$ be a smooth curve. Fix a divisor $D\in\Div(C)$ and a point $P\in C$. Then, there is an exact sequence
$$0\to\mc L(D-P)\to\mc L(D)\to\bar k\to\dual{\Omega(P-D)}\to\dual{\Omega(-D)}\to0.$$
</div>
<p>Before we can prove this lemma, we have to be able to say what all these maps are. While most of them are fairly straightforward (at least when $D=0$), the map $\bar k\to\dual{\Omega(P-D)}$ is more involved. To define it, first fix a uniformizer $t\in\bar k(C)$ at $P$ (i.e. $\ord_P(t)=1$). Now, given any differential form $\omega\in\Omega_C$, we can write $\omega=g\dt$ for some unique $g\in\bar k(C)$. Forming the completion $\wh{\bar k[C]_ P}=\invlim_n\bar k[C]_ P/\mfm_P^n\bar k[C]_ P$, and letting $\bar k(C)_ P:=\Frac\parens{\wh{\bar k[C]_ P}}$, one notes that $\bar k(C)_ P$ is isomorphic to the ring of formal power series in $t$ with coefficients in $\bar k$ <sup id="fnref:9"><a href="#fn:9" class="footnote">9</a></sup>. This let’s us write a <b>Taylor</b> (or <b>Laurent</b>) <b>expansion</b> for $g$:</p>
<script type="math/tex; mode=display">g=\sum_{n\in\Z}a_n(g)t^n\text{ }\text{ where }\text{ }a_n(g)\in\bar k</script>
<p>and there exists some $m>0$ s.t. $n<-m\implies a_n(g)=0$. With this in mind, define</p>
<script type="math/tex; mode=display">\res_P(\omega)=a_{-1}(g)</script>
<p>modelled after residues in complex analysis <sup id="fnref:10"><a href="#fn:10" class="footnote">10</a></sup>. Note that $\ord_P(g)$ is the smallest $n$ such that $a_n(g)\neq0$, so $\res_P(\omega)=0$ for any holomorphic form $\omega$. In our to be able to effectively employ residues, one needs to know the following. <sup id="fnref:11"><a href="#fn:11" class="footnote">11</a></sup></p>
<div class="lemma">
Let $C$ be a smooth curve, and let $\omega\in\Omega_C$ be any differential form. Then,
$$\sum_{P\in C}\res_P(\omega)=0.$$
</div>
<div class="proof4">
First note that $\res_P(\omega)=0$ for any point $P\in C$ where $\omega$ does not have a pole. Hence, the sum under consideration is actually finite. To do show this, we'll start by showing it in the special case of $C=\P^1=\bar k\cup\{\infty\}$. In this case, $\bar k(C)=\bar k(t)$, and $t_c=t-c$ is a uniformizer at $c\in\bar k\subset\P^1$ while $t_\infty=\frac1t$ is a uniformizer at $\infty$. Now, write $\omega=g\dt$ for some $g\in\bar k(t)$. For any $c\in\bar k$, we have
$$\omega=g\dt=g\d(t-c)=g\dt_c=g(t_c+c)\dt_c\implies\res_c(\omega)=$$
$$\omega=g\dt=g\inv{\parens{\dt_\infty/\dt}}\dt_\infty=-gt^ 2\dt_\infty=-g(1/t_\infty)t_\infty^{-2}\implies\res_\infty(\omega)=$$
</div>
<p>Now, we’ll prove a special case of the previous lemma, and leave the construction of this exact sequence for an arbitrary divisor $D$ as an exercise.</p>
<div class="lemma">
Let $C$ be a smooth curve, and fix a point $P\in C$. Then, there is an exact sequence
$$0\too\mc L(D-P)\too\mc L(D)\too\bar k\too\dual{\Omega(P-D)}\too\dual{\Omega(-D)}\too0.$$
</div>
<div class="proof4">
We should probably start by saying what the maps are.
<ol>
<li> The first map $\mc L(D-P)\to\mc L(D)$ is the natural inclusion map. </li>
<li>
The second map $\mc L(D)\to\bar k$ is "evaluation" at $P$, i.e. $f\mapsto (t^{\ord_P(D)}f)(P)$
</li>
<li>
The third map $\bar k\to\Omega(P-D)^\vee$ sends $c\in\bar k$ to
$$\omega\longmapsto c\cdot\res_P(t^{-\ord_P(D)}\omega)$$
</li>
<li>
The fourth map $\dual{\Omega(P-D)}\to\dual{\Omega(-D)}$ is the natural restriction map.
</li>
</ol>
We now check exactness.
<ul>
<li> Exactness at $\mc L(D-P)$ is obvious because this is literally an inclusion map. </li>
<li> Exactness at $\mc L(D)$ follows from the fact that $\mc L(D-P)$ is the space of maps $f\in\mc L(D)$ s.t. $\ord_P(f)+\ord_P(D)\ge1$. </li>
<li>
Exactness at $\bar k$ is the statement that there exists $\omega\in\Omega(P-D)$ such that $\res_P(t^{-\ord_P(D)}\omega)\neq0\iff(t^{\ord_P(D)}f)(P)=0$ for all $f\in\mc L(D)$.
</li>
</ul>
</div>
<h1 id="elliptic-curves">Elliptic Curves</h1>
<h3 id="group-law">Group Law</h3>
<h3 id="isogenies">Isogenies</h3>
<h3 id="torsion-points">Torsion Points</h3>
<h1 id="l-adic-representations">$\l$-adic Representations</h1>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>There is a lot to explain, so I anticipate this becoming one of my longer (and also more fun) posts to date…. In the end, this became my <a href="../solving-pell">second post</a> that I feel is overly ambitious and so hard to follow. If you’ve read some of this, please tell me how hard it is to understand so I can know for sure if I really am trying to do too much all at once. <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>I’ve also said the word sheaf before, so I could really try to be fancy <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>My main reference for this post is Silverman who doesn’t use $\spec$, and rather not have to do the extra work associated with translating terminology and making sure I don’t say anything false. <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>I’m sure things aren’t as bad as I make them sound hear, but gotta set the bar low so it’s easier to exceed expectations. <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>This section is basically just the first chapter of Silverman because I’m unoriginal. <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>Think of meromorphic maps from complex analysis <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>Could you imagine how long this post would be if I did this the whole time? <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>Including proving Riemann-Roch <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:9">
<p>At the very least, showing that the underlying additive group of this ring is formall power series can be done relatively easily in much the same manner as in my <a href="../abs-val-p-adic">p-adic post</a> <a href="#fnref:9" class="reversefootnote">↩</a></p>
</li>
<li id="fn:10">
<p>This value does not depend on the choice of uniformizer $t$, but to avoid having to show this, we fixed a canonical choice in the beginning (if you’re bothered by this, show independence as an exercise). <a href="#fnref:10" class="reversefootnote">↩</a></p>
</li>
<li id="fn:11">
<p>Working over $\C$, one can show this easily using Stoke’s theorem. However, we’re doing algebraic geometry, not complex geometry, so we need to be more creative <a href="#fnref:11" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>The title of this post is a bit of a misnomer. We won’t be talking about representations of elliptic curves, but about representations attached to (associated with?) elliptic curves. In particular, the goal of this post is to prove that given an elliptic curve $E/\Q$ defined over the rationals, and a prime $\l$, the $\l$-adic representationA Nice Lemma about Dedekind Domains2019-08-12T00:00:00+00:002019-08-12T00:00:00+00:00https://nivent.github.io/blog/dedekind-lemma<p>Often times in these posts, the main focus is some big/nice theorem/result; however, this time there’s <a href="https://www.youtube.com/watch?v=F8mYLi3PGOc">a nice lemma about Dedekind domains</a> that I think merits its own post. This is partially because I’m still shocked that it’s true, and partially because I don’t know where else it is written down. After proving it, I will (maybe briefly <sup id="fnref:9"><a href="#fn:9" class="footnote">1</a></sup>) mention one of its uses <sup id="fnref:1"><a href="#fn:1" class="footnote">2</a></sup>. However, I think this use is less exciting than the lemma itself. The gist of the lemma is that lattices over Dedekind domains can be modified at a single prime.</p>
<h1 id="the-lemma">The Lemma</h1>
<p>I guess I might as well start with stating this thing. Before I can do that though, it’s probably worthwhile to define what I mean by a lattice.</p>
<div class="definition">
Let $A$ be an integral domain with fraction field $F$. Let $V$ be an $F$-vector space. Then, an <b>$A$-lattice</b> (or <b>lattice over $A$</b>) $L\subset V$ is a torsion-free, finitely generated $A$-module. Its <b>rank</b>
is $\dim_F(F\otimes_AL)$.
</div>
<div class="lemma" name="This Post's Main Lemma">
Let $R$ be a Dedekind domain with fraction field $F$, $\mfp$ a maximal ideal of $R$, and $R_\mfp$ the localization of $R$ at $\mfp$. For an $R$-lattice $L$ and $R_\mfp$-lattice $L_0$, both in $F^n$, there exists a unique $R$-lattice $L'$ in $F^n$ such that $L'_\mfp=L_0$ and $L_\mfq=L'_\mfq$ for all maximal ideals $\mfq\neq\mfp$.
</div>
<p>Notice above that we have hard equalities. These aren’t abstract isomorphisms; all of these lattices are literal subsets of our fixed vector space $F^n$, and these are literal equalties of sets.</p>
<p>What freaks me out about this lemma is that it (almost) tells us that we can consider some global $R$-lattice as a collection of local $R_\mfp$-lattices, make arbitrary changes to the local lattices, and then stitch all these changes together to get some global transformation of the lattice we started with. I feel like you generally don’t have this much global control at the local level.</p>
<p>Anyways, let’s prove this thing. The first thing we’ll want to do is reduce the statement of the lemma to something more manageable. Right now, we have infinitely many local conditions we need our fabled lattice $L’$ to satisfy (one for each prime of $R$), and infinity is a pretty big number. It would be nice if we only had finitely many constraints instead. This brings us to our first step. <sup id="fnref:2"><a href="#fn:2" class="footnote">3</a></sup></p>
<div class="claim">
Same setup as the main lemma. There exists some $r\in R$ such that, for proving the main lemma, it suffices to satisfy some local conditions just for the (finitely many) primes $\mfq\mid(r)$ as well as one at $r$ itself.
</div>
<div class="proof4">
Let's start by letting $M$ be the $R$-span of an $R_\mfp$-basis for $L_0$. Then, $M$ is a (global) lattice and will be our surrogate for $L_0$. Fix a uniformizer $\pi$ for $\mfp$ (i.e. a generator of $\mfp R_\mfp$). Note that $R_\mfp[1/\pi]=F$, and
$$M_\mfp[1/\pi]=M_F=F^n=L_F=L_\mfp[1/\pi]$$
where the notation $M_F$ refers to the $F$-span of $M$. Hence, $L[1/\pi]$ and $M[1/\pi]$ are $R[1/\pi]$-lattices which become the same after localizing at $\mfp'=\mfp R[1/\pi]$. We claim that there exists some $s\in R[1/\pi]\sm\mfp'$ such that $L[1/\pi]_s=M[1/\pi]_s$. To find such an $s$, (implicitly fix bases for $M[1/\pi]$ and $R[1/\pi$], then) let $B\in\GL_n(F)$ be a change of basis matrix between $L[1/\pi]_{\mfp'}$ and $M[1/\pi]_{\mfp'}$. Since $F=R[1/\pi]_{\mfp'}$ and $B$ has only finitely many entries, we can write $B=s'B'$ with $B'\in\GL_n(R[1/\pi])$ and $s'\in R[1/\pi]\sm\mfp'$. Hence, we see that we can take $s=s'\det(B')$ since then inverting $s$ inverts both $s'$ and $\det(B')$.
<br />
Now, it's harmless to replace $s\in R[1/\pi]$ with its numerator in some fractional expression, so we can form $r=s\pi\in R\sm\{0\}$. We have $M_r=L_r$ as $R_r$-lattices inside $F^n$. Now, let $M'=L\cap M$, an $R$-lattice in $F^n$, and observe that $M'_r=L_r=M_r$. We claim that, for proving the main lemma, it suffices to find an $R$-lattice $L'$ containing $M'$ such that (i) $L'_r=M_r'$ and (ii) for each of the finitely many primes $\mfq\mid(r)$, we have $L'_\mfq=L_\mfq$ if $\mfq\neq\mfp$ and $L'_\mfp=L_0$ (i.e. $M_\mfp$) otherwise. This is because for $\mfq\nmid(r)$, we also would have $L'_\mfq=L_\mfq$ since $L'_r=M_r'=L'_r$ (i.e. $L'_\mfq=(L'_r)_{\mfq'}$ where $\mfq'=\mfq L_r$).
</div>
<p>Hence, we’ve reduced proving the main lemma to constructing this $R$-lattice $L’$ mentioned at the end of the above argument. Writing $V=F^n$ for ease of notation, we can view $L’$ as a finitely generated $R$-submodule of the (huge) $R$-module $V/M’$. While $V/M’$ may seem largely unwildly, we actually have some hope of being able to understand/work with it since it is torsion (e.g. think of $\qz$). To see that it’s torsion, note that $M_F’=V$, so every element of $V$ looks like $\frac mr$ with $m\in M’$ and $r\in R$, and so is killed (in $V/M’$) when multiplied by its denominator. Now, to make working with $V/M’$ easy, we’ll prove a general fact that torsion $R$-modules decompose into sums of (some of) their localizations.</p>
<div class="claim">
Let $N$ be a torsion $R$-module, and fix any $a\in R$. Then, the natural map
$$N\too N\sqbracks{\frac1a}\oplus\parens{\bigoplus_{\mfq\mid(a)}N_\mfq}$$
is an isomorphism.
</div>
<div class="proof4">
It suffices to check this after localizing at each prime $\mfP$ of $R$. If $\mfP\mid(a)$, then, because $N$ is torsion, localizing kills all $N_\mfq$ on the RHS except for $N_\mfP$. Furthermore, since $N_\mfP$ is a dvr (in particular, a PID) and $a\in N_\mfP$ is a non-unit, we see that
$$N\sqbracks{\frac1a}_\mfP=N_\mfP\sqbracks{\frac1a}=0$$
as well, so this map is an isomorphism at all primes dividing $(a)$. Similarly, if $\mfP\nmid(a)$, then, since $N$ is torsion, all $N_\mfq$ are killed after localizing at $\mfP$ while, since $a\not\in\mfP$,
$$N\sqbracks{\frac1a}_\mfP=N_\mfP,$$
so this map is also an isomorphism at all primes not dividing $(a)$. Thus, it is an isomorphism locally which implies that it is one globally.
</div>
<p>In our particular case, the above says that</p>
<script type="math/tex; mode=display">\frac V{M'}\iso\parens{\frac V{M'}}\sqbracks{\frac1r}\oplus\parens{\bigoplus_{\mfq\mid(r)}\parens{\frac V{M'}}_ \mfq}</script>
<p>Now, we’re practically done. Using the description on the RHS, take $L’$ to be the $R$-submodule of $V/M’$ given by inserting the chosen $R_\mfq$-submodule of each $(V/M’)_ \mfq$ for $\mfq\mid(r)$ (and $0$ for $(V/M’)[1/r]$). One needs to check that this module is finitely generated over $R$. However, this is clear since a finitely generated torsion $R_\mfq$-module is just a finitely generated module over $R_\mfq/(\mfq R_\mfq)^N=R/\mfq^N$ for some large $N$ (and hence also finitely generated over $R$). There are only finitely many primes dividing $(r)$, so we win.</p>
<p>There we have it. Lattices over Dedekind domains can be modified one prime at a time.</p>
<h1 id="its-use">Its Use</h1>
<p>For this part of the post, you’ll probably want to have some familiarity with completions of rings (e.g. be comfortable with inverse limits) and adeles (e.g. know what a restricted direct product is).</p>
<p>We’ll switch up notation in this section. Previously, given a Dedeking domain $R$ with prime ideal $\mfp$, we used $R_\mfp$ to denote its localization at $\mfp$. From now on, we’ll instead let $R_\mfp$ denote $R$’s completion at $\mfp$. That is,</p>
<script type="math/tex; mode=display">R_\mfp:=\invlim_nR/\mfp^n</script>
<p>is the valuation ring (i.e. elements with norm $\le1$) of the (analytic) completion $F_\mfp$ of $F$ with respect to the absolute value $\abs x_\mfp=\eps^{-v_\mfp(x)}$ where $0<\eps<1$ and $v_\mfp(x)$ is the largest power of $\mfp$ dividing $(x)$.</p>
<p>The point of this section will be to use the main lemma to prove that (isomorphism classes of) finitely generated projective $R$-modules of rank $n$ are classified via a certain double coset space. <sup id="fnref:3"><a href="#fn:3" class="footnote">4</a></sup> Before stating this, it will be useful to define/prove a few things.</p>
<div class="proposition">
Let $\wh R=\invlim R/\mfa$ be the "profinite" completion of $R$; this inverse limit is taken along all nonzero ideals of $R$, with each $R/\mfa$ discrete. Then,
$$\wh R\simeq\prod_\mfp R_\mfp$$
as topological rings.
</div>
<div class="proof4">
We'll first construct the left-to-right map. A map to a direct product is a map to each factor, so fix some prime $\mfp$; we'll construct a map $\wh R\to R_\mfp$. Now, a map into an inverse limit is a (consistent choice of a) map into each factor, so we really just need to map $\wh R\to R/\mfp^n$. However, we have a natural choice of such a map since $\mfp^n$ is an ideal of $R$ and $\wh R=\invlim R/\mfa$ is formed along all ideals (for the same reason, these maps are consistent and so do in fact give a map $\wh R\to R_\mfp$). We claim that the map $\wh R\to\prod_\mfp R_\mfp$ that we've constructed is continuous. This is the case iff all projections $\wh R\to R/\mfp^n$ are continuous, and this is the case since given $a\in R/\mfp^n$, its preimage is the open set
$$\wh R\cap\sqbracks{\prod_{\mfa\neq\mfp^n}R/\mfa\by\{a\}}.$$
Now, since $\wh R$ is compact and $\prod_\mfp R_\mfp$ is Hausdorff, to show that this map is an isomorphism of topological rings, it suffices to show that it is a bijection. For injectivity, fix some $a=\parens{a_\mfa}_\mfa\in\ker\parens{\wh R\to\prod_\mfp R_\mfp}$. The statement that this is in the kernel is exactly the statement that $a_{\mfp^n}=0$ for all primes $\mfp$ and natural $n\ge0$. Because $R$ is a Dedekind domain, we can write
$$\mfa=\prod_{\mfp\mid\mfa}\mfp^{v_\mfp(\mfa)}$$
as a finite product of prime powers, so
$$R/\mfa\simeq\prod_{\mfp\mid\mfa}R/\mfp^{v_\mfp(\mfa)}.$$
Hence, $a_\mfa=0$ for all $\mfa$ as its projection to each prime power factor is trivial. Thus, $a=0$. By the same kind of Chinese Remainder Theorem argument, we see that this map is surjective, and hence an isomorphism of topological rings.
</div>
<div class="corollary">
$$\wh\Z\simeq\prod_p\Z_p$$
</div>
<p>Our next definition will be that of the <b>adele ring</b> $\A_R=F\otimes_R\wh R$, where $F=\Frac(R)$. Note that when $R=\Z$ (or, more generally $R=\ints K$ for $K$ a number field), this is just the ring of finite adeles instead of the usual adele ring over $\Q$ (or $K$) which also contains information about $R$’s infinite places. Now, if you’ve seen adeles before, you’re probably used to seeing them defined via some kind of restricted direct product instead of this weird tensor. The two definitions are equivalent.</p>
<div class="proposition">
$$\A_R\simeq\prodp_{\mfp}\parens{F_\mfp,R_\mfp}$$
where we remember that $F_\mfp$ denotes the completion of $F$ at $\mfp$ and $R_\mfp$ denotes the completion of $R$ at $\mfp$ (the valuation ring of $F_\mfp$). Similarly,
$$\GL_n(\A_R)\simeq\prodp_{\mfp}\parens{\GL_n(F_\mfp),\GL_n(R_\mfp)}.$$
</div>
<div class="proof4">
We'll only prove the first part here. Note that we have a map
$$\mapdesc\phi{\A_R}{\prodp_\mfp\parens{F_\mfp,R_\mfp}}{x\otimes\parens{r_\mfp}_\mfp}{\parens{xr_\mfp}_\mfp}$$
using the description $\A_R\simeq F\otimes_R\wh R\simeq F\otimes_R\prod_\mfp R_\mfp$. This map is well-defined because $xr_\mfp\not\in R_\mfp\implies v_\mfp(x)\neq0$ and this latter part can only be true for finitely many primes. To see that this is an isomorphism, first note that ($a,b,c,d\in R$)
$$\frac ab\otimes(r_\mfp)_\mfp+\frac cd\otimes(s_\mfp)_\mfp=\frac{ad}{bd}\otimes(r_\mfp)_\mfp+\frac{bc}{bd}\otimes(s_\mfp)_\mfp=\frac1{bd}\otimes(adr_\mfp)_\mfp+\frac1{bd}\otimes(bcs_\mfp)_\mfp=\frac1{bd}\otimes\parens{adr_\mfp+bcs_\mfp}_\mfp,$$
so all tensors here are simple. Now, injectivity is clear because every ring in sight is an integral domain. Surjectivity is also easy to see essentially because collecting the denominators of the (finitely many) non-integral components of the restricted direct product gives you an element of $F$.
<br />
Showing that $\GL_n(\A_R)$ is also a restricted direct product is left as an exercise.
</div>
<p>Cool. Now that we know something about adeles, we can prove the thing we alluded to at the beginning of this section. Recall that an $R$-module $P$ is called <b>projective</b> if any of the following hold</p>
<ul>
<li> Every short exact sequence $0\too K\too E\too P\too 0$ splits </li>
<li> There exists an $R$-module $Q$ such that $P\oplus Q$ is $R$-free </li>
<li> The (covariant) function $\Hom_R(P,-)$ is exact </li>
</ul>
<p>When $R$ is a Dedekind domain (and $P$ is finitely generated), this is just a fancy way to say that $P$ is torsion-free <sup id="fnref:4"><a href="#fn:4" class="footnote">5</a></sup>. Given a projective $R$-module $P$, its <b>rank</b> is $\dim_F(P\otimes_RF)$ where $F=\Frac(R)$ <sup id="fnref:5"><a href="#fn:5" class="footnote">6</a></sup>.</p>
<p>One last thing: for proving the below theorem, it will be helpful to know that (finitely presented) projective modules are “locally free” in the below sense. <sup id="fnref:7"><a href="#fn:7" class="footnote">7</a></sup></p>
<div class="proposition">
Let $M$ be a finitely presented $R$-module. Then, TFAE
<ul>
<li> $M$ is projective. </li>
<li> There exists $r_1,\dots,r_k\in R$ s.t. $(r_1,\dots,r_k)=R$ and $M[\frac1{r_i}]$ is a free $R[\frac1{r_i}]$-module for all $i$. </li>
</ul>
</div>
<div class="proof4">
Somewhere in <a href="http://math.stanford.edu/~church/teaching/210A-F17/math210A-F17-hw3-sols.pdf">here</a>.
</div>
<p>Finally, the one use of our main lemma I know is proving the following:</p>
<div class="theorem">
Let $R$ be a Dedekind domain with fraction field $F$. Then, the set of isomorphism classes of finitely generated projective $R$-modules of rank $n$ can be identified with the double coset space $\GL_n(\wh R)\sm\GL_n(\A_R)\,/\,\GL_n(F)$.
</div>
<div class="proof4">
First convince yourself that $\GL_n(\wh R)\simeq\prod_\mfp\GL_n(R_\mfp)$. Now we'll describe the map
$$\bracks{\text{isomorphism classes of finitely generated projective $R$-modules of rank $n$}}\too\GL_n(F)\sm\GL_n(\A_R)\,/\,\GL_n(\wh R).$$
Let $P$ be such an $R$-module, fix an $F$-linear isomorphism $f:F\otimes_RM\iso F^n$, and choose any prime $\mfp\subset R$. Now, fix $r_1,\dots,r_m$ (I'm pretty sure, secretly, $m=n$ but that's not needed for this proof) as in the previous proposition, and let $\bracks{\ith e_j}_{j=1}^n$ be an $R[\frac1{r_i}]$-basis for $M[\frac1{r_i}]$. Choosing a basis like this gives rise to a map $\ith g:F\otimes_{R[\frac1{r_i}]}M[\frac1{r_i}]\iso F^n$. Noting that
$$F\otimes_{R[\frac1{r_i}]}M\sqbracks{\frac1{r_i}}\simeq F\otimes_{R[\frac1{r_i}]}R\sqbracks{\frac1{r_i}}\otimes_RM\simeq F\otimes_RM,$$
and comparing $\ith g$ with $f$ gives us some $\ith T\in\GL_n(F)$.
$$\begin{CD}
F\otimes_RM @>f>> F^n\\
@VVV @VV\ith TV\\
F\otimes_{R\sqbracks{\frac1{r_i}}}M\sqbracks{\frac1{r_i}} @>\ith g>> F^n
\end{CD}$$
Now we make use of the prime $\mfp$ we fixed earlier. Since $(r_1,\dots,r_m)=R$, there must exist some $k=k_\mfp\in\{1,\dots,m\}$ such that $r_k\not\in\mfp$, so $\bracks{\Ith ek_j}_{j=1}^n$ descends to an $R_\mfp$-basis of $M_\mfp$, and $\Ith Tk$ descends to a map $\Ith Tk_\mfp\in\GL_n(F_\mfp)$. Now, we investigate when we further have that $\Ith Tk_\mfp\in\GL_n(R_\mfp)$. By construction, $\det(\Ith Tk_\mfp)=\det(\Ith Tk)$, so since there are only finitely many $\ith T$, we claim that we will have $\Ith Tk_\mfp\in\GL_n(R_\mfp)$ for all but finitely many $\mfp$. In more detail, we can first fix some $a\in R$ s.t. the matrix for $a\ith T$ has $R$-entries for all $i$. Hence, if both $a$ and $\det(a\Ith Tk)$ are invertible in $R_\mfp$, then $\Ith Tk_\mfp$ will be an element of $\GL_n(R_\mfp)$. Thus, any prime $\mfq$ with $\Ith T{k_\mfq}_\mfq\not\in\GL_n(R_\mfq)$ must divide
$$a\prod_{i=1}^m\det\parens{a\ith T}\in R.$$
There are only finitely many such primes, so (we claim that) our desired map is
$$\mapdesc\psi{\bracks{f:F\otimes_RM\iso F^n}}{\GL_n(\wh R)\sm\GL_n(\A_R)\,/\,\GL_n(F)}f{\parens{\Ith T{k_\mfp}_\mfp}_\mfp}$$
where we really should have chosen notation that made it clear that $\Ith T{k_\mfp}_\mfp$ depends on $f$. To show that this map is well-defined, we will show that changing the isomorphism $f$ affects our result by an element of $\GL_n(F)$ on the left, while changing the local bases affects the result by an element of $\GL_n(\wh R)$ acting on the right.<br />
Let $\phi:M\to N$ be an isomorphism of projective $R$-modules. Then, we get a commutative diagram
$$\begin{CD}
F\otimes_RM @>f>> F^n\\
@V{1\otimes\phi}VV @VVTV\\
F\otimes_RN @>g>> F^n
\end{CD}$$
where $T\in\GL_n(F)$. Now, using the composition $g\circ(1\otimes\phi)$ in place of $F$ has the effect of turning $\ith T$ into $\ith T\inv T$ as one sees by staring at the below commutative diagram
$$\begin{CD}
F\otimes_RM @>g>> F^n\\
@A{1\otimes\phi}AA @AATA\\
F\otimes_RM @>f_\mfp>> F^n\\
@VVV @VV\ith TV\\
F\otimes_{R\sqbracks{\frac1{r_i}}}M\sqbracks{\frac1{r_i}} @>\ith g>> F^n
\end{CD}$$
Hence, changing our isomorphism really does correspond to an action of $\GL_n(F)$ on the right.<br />
Now, fix a prime $\mfp\subset R$ and an $R_\mfp$-basis $\bracks{f_j}_{j=1}^n$ for $M_\mfp$. What would happen to $\Ith Tk_\mfp$ if we used this basis instead of $\bracks{\Ith ek_j}_{j=1}^n$? Well, this alternate choice of basis gives rise to a map $g_\mfp:F_\mfp\otimes_{R_\mfp}M_\mfp\iso F_\mfp^n$, and then again, you stare at a diagram and see that this would have the effect of replacing $\Ith Tk_\mfp$ with $T_\mfp\Ith Tk_\mfp$ where $T_\mfp$ is what makes the below commute
$$\begin{CD}
F_\mfp\otimes_{R_\mfp}M_\mfp @>f>> F^n_\mfp\\
@VVV @VV{\Ith Tk_\mfp}V\\
F_\mfp\otimes_{R_\mfp}M_\mfp @>\Ith gk_\mfp>> F^n_\mfp\\
@VVV @VV{T_\mfp}V\\
F_\mfp\otimes_{R_\mfp}M_\mfp @>g_\mfp>> F^n_\mfp
\end{CD}$$
Clearly, $T_\mfp\in\GL_n(\mfp)$ since it corresponds to changing $R_\mfp$-bases from our original choice to our new choice, so changing all local bases at once corresponds to an action of $\GL_n(\wh R)$ on the left as desired.<br />
So far, all we've done is defined this map; we still need to show that it's a bijection. We start with surjectivity. Fix some representative $(T_\mfp)_\mfp$ of an element of $\GL_n(\wh R)\sm\GL_n(\A_R)\,/\,\GL_n(F)$ (i.e. $T_\mfp\in\GL_n(F_\mfp)$). Since $T_\mfp\in\GL_n(R_\mfp)$ almost always, we can act by $\GL_n(\wh R)$ on the left to assume that $T_\mfp=I$, the identity for all but finitely many $\mfp$. The point of doing this is that taking $L=R^n$ and letting $f:F\otimes_RL\iso F^n$ be the natural isomorphism, we get that $\psi(f)$ is represented by the identity in every slot. Hence, $\psi(f)_\mfp=T_\mfp$ for all but finitely many $\mfp$, and we use our main lemma to change $L$, one prime at a time, in the finiely many differing slots to arrive at a new lattice $L'$ for which $\psi(L')=(T_\mfp)_\mfp$, proving surjectivity. <br />
I'll leave injectivity for the reader because I mainly just cared about applying the main lemma, and I've done that now.
</div>
<p>I don’t actually know why this theorem is useful, but it probably is for something. I guess, geometrically, rank $n$ projective $R$-modules correspond to rank $n$ vector bundles over $\spec R$, so this gives some characterization of vector bundles in terms of this double coset space. <sup id="fnref:8"><a href="#fn:8" class="footnote">8</a></sup> When $n=1$, we see that the Picard group $\Pic(R)$ is $\units{\wh R}\sm\units{\A_R}\,/\,\units F$. Maybe one could use this to prove finiteness of class groups when $R=\ints K$ is the ring of integers of some number field or something; I really don’t know.</p>
<h1 id="an-exercise">An Exercise</h1>
<p>The simplest Dedekind domains are PIDs. Try proving the results of this post just for PIDs to see if you can make the proofs much simpler.</p>
<div class="footnotes">
<ol>
<li id="fn:9">
<p>Spoiler: I accidentally ended up spending most of my time on this one use instead of keeping focus on the lemma throughout <a href="#fnref:9" class="reversefootnote">↩</a></p>
</li>
<li id="fn:1">
<p>If you know other uses of this lemma, please let me know. <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>This is like a lemma for a lemma. What do you call that? <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>Assuming I’m not remembering things incorrectly, when $n=1$, this gives the Picard group of $R$. <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>Exercise: prove this (hint: use the fact that $R$ is locally a PID and (finitely generated) torsion-free modules over PIDs are free) <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>Secretly, the rank of a projective $R$-module $P$ is supposed to be a function $r:\spec(R)\to\N_{\ge0}$ given by $r(\mfp)=\dim_{R/\mfp}(P\otimes_RR/\mfp)$, but this function is constant (and equal to what I wrote outside this footnote) when $R$ is not stupid (e.g. when $R$ is an integral domain) <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>We say an $R$-module $M$ is finitely presented if there exists a short exact sequence $0\to A\to F\to M\to 0$ with $F$ a finitely generated free $R$-module and $A$ finitely generated. When $R$ is Noetherian, this is equivalent to being finitely generated. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>Something something the geometry of Dedekind domains is controlled by adeles something something? <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>Often times in these posts, the main focus is some big/nice theorem/result; however, this time there’s a nice lemma about Dedekind domains that I think merits its own post. This is partially because I’m still shocked that it’s true, and partially because I don’t know where else it is written down. After proving it, I will (maybe briefly 1) mention one of its uses 2. However, I think this use is less exciting than the lemma itself. The gist of the lemma is that lattices over Dedekind domains can be modified at a single prime. Spoiler: I accidentally ended up spending most of my time on this one use instead of keeping focus on the lemma throughout ↩ If you know other uses of this lemma, please let me know. ↩The Duality Between Algebra and Geometry2019-07-28T00:00:00+00:002019-07-28T00:00:00+00:00https://nivent.github.io/blog/dual-alg-geo<p>I talked a little bit about the topic of this post’s title in a <a href="../comm-alg">recent post</a>, but I want to stress that this <a href="https://www.youtube.com/watch?v=F8mYLi3PGOc">duality between algebra and geometry</a> goes beyond <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup> this $\spec$ business. In particular, I’m going to discuss a more “topological” setting where we see nice interplay between algebra and geometry <sup id="fnref:2"><a href="#fn:2" class="footnote">2</a></sup>: relating a (compact, Hausdorff) topological space to its ring of (real-valued) continuous functions.</p>
<h1 id="prelim-on-separation-axioms">Prelim on Separation Axioms</h1>
<p>This might just be because there’s no point-set topology class at my school <sup id="fnref:3"><a href="#fn:3" class="footnote">3</a></sup>, but I get the sense that too many people don’t know about the theory of separation axioms for topological spaces. Sadly, I do not think I have enough space in this post to develop this theory, but I can state some needed highlights.</p>
<h1 id="x--c0x">$X ``=” C^0(X)$</h1>
<h1 id="swans-theorem">Swan’s Theorem</h1>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>I don’t actually know that much about this area, so maybe what I’m gonna talk about in this post is somehow the same as this $\spec$ stuff. That would surprise me though. <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>It really bothers me that topology and geometry are very similar in general character, but the only word I know for capturing both of them at once is “geometry”. Like, sometimes I say “geometry” and mean “topology/geometry” but other times I say it and mean just “geometry”. How’s anyone supposed to understand what I’m saying? <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>Which is really a shame <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>I talked a little bit about the topic of this post’s title in a recent post, but I want to stress that this duality between algebra and geometry goes beyond 1 this $\spec$ business. In particular, I’m going to discuss a more “topological” setting where we see nice interplay between algebra and geometry 2: relating a (compact, Hausdorff) topological space to its ring of (real-valued) continuous functions. I don’t actually know that much about this area, so maybe what I’m gonna talk about in this post is somehow the same as this $\spec$ stuff. That would surprise me though. ↩ It really bothers me that topology and geometry are very similar in general character, but the only word I know for capturing both of them at once is “geometry”. Like, sometimes I say “geometry” and mean “topology/geometry” but other times I say it and mean just “geometry”. How’s anyone supposed to understand what I’m saying? ↩Dedekind Domains Done Right2019-03-26T00:00:00+00:002019-03-26T00:00:00+00:00https://nivent.github.io/blog/dedekind-domain<p>I’ve wanted to write this post for a long time now. Dedekind domain’s are, objectively, the best rings in existence <sup id="fnref:2"><a href="#fn:2" class="footnote">1</a></sup>, and their greatness stems from one fact: ideals in a Dedekind domain factor uniquely into (finite) products of prime ideals. However, I’ve never seen a proof of this fact that I liked (i.e. one that’s straightforward enough for me to actually remember) <sup id="fnref:1"><a href="#fn:1" class="footnote">2</a></sup>, and so this post is my attempt to remedy this situation. Like (almost) always, I’ll start by introducing some background I’ll need in the proof, and then I’ll actually get into the good stuff in a separate section.</p>
<h1 id="background">Background</h1>
<p>I’ll start with a fact about localizations that I think I’ve used before on this blog, but never actually stated/proven.</p>
<p>First, observe that given a ring $R$, a multiplicative set $S\subset R$, and an $R$-module $M$, we can form an $\sinv R$-module $\sinv M$ which we call the localization of $M$ at $S$ (away from $S$? I can never remember what preposition to use). The construction is exactly what you expect: elements of $\sinv M$ are formal fractions $\frac ms$ with $m\in M$ and $s\in S$, and we say that</p>
<script type="math/tex; mode=display">\frac ms=\frac nt\iff\exists u\in S:u\cdot(t\cdot m-s\cdot n)=0,</script>
<p>where we’ve explicitly used $\cdot$ to emphasize that $m,n$ are module elements while $u,s,t$ are ring elements. If you know about tensor products, then you can show that we also have</p>
<script type="math/tex; mode=display">\sinv M\simeq M\otimes_R\sinv R,</script>
<p>so localization is really just extension of scalars. Now, onto the fact.</p>
<div class="proposition">
Let $M$ be an $R$-module. Then, $M=0\iff M_\mfm=0$ for all maximal ideals $\mfm\subset M$.
</div>
<div class="proof4">
On of these directions is easy, so we'll prove the other one. Assume that $M_\mfm=0$ for all maximal ideals $\mfm\subset R$, and suppose that $M\neq0$. Fix some nonzero $x\in M$, and let $\mfm$ a maximal ideal containing
$$\ann(x)=\bracks{r\in R:r\cdot x=0}\subsetneq R.$$
Then, $\frac x1=\frac01\in M_\mfm$ so there exists some $u\in R\sm\mfm\subset R\sm\ann(x)$ such that $u\cdot x=0$. This means that $u\in(R\sm\ann(x))\cap\ann(x)=\emptyset$, which is just wonderful.
</div>
<p>The real utility of this proposition comes from the fact that given an $R$-linear map $f:M\to N$, and a multiplicative set $S\subset M$, we can always form the $\sinv R$-linear map $\sinv f:\sinv M\to\sinv N$ given by $\sinv f(m/s)=f(m)/s$.</p>
<div class="exercise">
Show that we always have $\im(\sinv f)=\sinv\im(f)$ and $\ker(\sinv f)=\sinv\ker(f)$, so localization commutes with taking images and kernels. Also show localization commutes with forming quotients if you feel like it.
</div>
<div class="corollary">
Let $f:M\to N$ be a map between $R$-modules. Then, $f$ is an injection (or surjection or bijection) iff $f_\mfm$ is an injection (or surjection or bijection) for all maximal $\mfm\subset R$.
</div>
<div class="proof4">
Use exercise to apply the proposition to $\ker(f)$ and $\coker(f)=N/\im(f)$.
</div>
<p>That’s one thing down. The second thing we’ll need is generalized Cayley-Hamilton.</p>
<div class="theorem" name="Cayley-Hamilton">
Let $M$ be an $R$-module generated by $n$ elements, and let $T:M\to M$ be some $R$-linear map. Then, there exists a monic polynomial $f\in R[x]$ of degree $n$ such that $f(T)=0\in\End_R(M)$.
</div>
<div class="proof4">
First, fix a surjection $\pi:R^n\to M$, and note that we can form a commutative square
<center>
<img src="https://nivent.github.io/images/blog/dedekind-domain/ch.png" width="150" height="100" />
</center>
and so lift $T$ to a map $\alpha:R^n\to R^n$. The top map comes from the fact that $R^n$ is free so we can construct it by choosing any lifts of $T(\pi(e_i))$ for $e_1,\dots,e_n$ a basis for $R^n$. Furthermore, this lift is nice in that $p(\alpha)\in\End_R(R^n)$ lifts $p(T)\in\End_R(M)$ for any $p\in R[x]$, so to prove the theorem, it suffices to find some $f\in R[x]$ s.t. $f(\alpha)=0$. Since $R^n$ is free, we have a matrix $(a_{ij})_{i,j=1}^n$ defined by writing $\alpha(e_j)=\sum_{i=1}^na_{ij} e_i$. With this in mind, let $A=\Z\sqbracks{\bracks{x_{ij}}_{i,j=1}^n}$, and consider the universal matrix
$$S=\Mat{x_{11}}\cdots{x_{1n}}\vdots\ddots\vdots{x_{n1}}\cdots{x_{nn}}\in\End_A(A^n).$$
Now, let $\pi_T:A\to R$ be the homomorphism determined by $\pi_T(x_{ij})=a_{ij}$ for all $i,j$. With this map, we are reduced to finding some $g\in A[x]$ s.t. $g(S)=0$ since then applying $\pi_T$ gives some $f\in R[x]$ s.t. $f(\alpha)=0$. Luckily for us, $A$ is a domain, so we can take $g(x)=\det(xI-S)$ where $I$ is the identity matrix. To see why this works note that we can embed $\Frac(S)\into\C$ (since $\trdeg_{\Q}\Frac(S)<\trdeg_{\Q}\C$ and $\C$ algebraically closed) and so we are reduced to Cayley-Hamilton over $\C$ where this choice of $g$ definitely works.
</div>
<h1 id="the-good-stuff">The Good Stuff</h1>
<p>Now we’re here. Before getting in Dedekind domains, let’s briefly discuss dvrs (which we’ll see are just local Dedekind domains).</p>
<div class="definition">
Let $K$ be a field and $\Gamma$ be a totally ordered abelian group. A <b>valuation</b> $v:\units K\to\Gamma$ is a map such that
<ol>
<li> $v(ab)=v(a)+v(b)$ for all $a,b\in\units K$. </li>
<li> $v(a+b)\ge\min(v(a),v(b))$ with equality if $v(a)\neq v(b)$ for all $a,b\in\units K$.</li>
</ol>
We sometimes also say $v(0)=\infty$. A valuation is <b>discrete</b> if $\Gamma=\Z$.
</div>
<div class="definition">
Given a discrete valuation $v:\units K\to\Z$, its associated <b>discrete valuation ring (dvr)</b> is
$$R=\bracks{\alpha\in\units K:v(\alpha)\ge0}\cup\{0\}.$$
</div>
<p>Dvrs are very nice as detailed in the following theorem.</p>
<div class="theorem">
Let $R$ be a dvr with valuation $v:\units{\Frac(R)}\to\Z$. Then,
<ol>
<li> $R$ is a domain. </li>
<li> $\alpha\in\units R\iff v(\alpha)=0$. </li>
<li> $\mfm=\bracks{\alpha\in R:v(\alpha)>0}$ is maximal (this and 2. imply that $R$ is local) </li>
<li> $\mfm=(t)$ for any $t\in R$ with $v(t)=1$. </li>
<li> $R$ is a PID </li>
<li> Any $x\in R$ is of the form $x=ut^n$ with $u\in\units R$ and $v(t)=1$. (this and 5. imply that every ideal of $R$ is of the form $(t^n)$ for some $n$) </li>
</ol>
</div>
<div class="proof4">
<ol>
<li>
If $a,b\in R$ are nonzero, then $v(ab)=v(a)+v(b)$ is finite, so $ab\neq0$.
</li>
<li>
$ab=1$ implies that $v(a)+v(b)=v(1)=0$. Since $v(a),v(b)\ge0$, we must have $v(a)=v(b)=0$. Conversely, assume $v(a)=0$, and pick $b\in K$ s.t. $ab=1$. Then, $v(b)=v(a)+v(b)=v(1)=0$, so $b\in R$ and hence $a\in\units R$.
</li>
<li>
If $a,b\in\mfm$ and $r\in R$, then $v(rb)=v(r)+v(b)>0$ and $v(a+b)\ge\min(v(a),v(b))>0$ so $\mfm$ is an ideal. Since it's literally all the nonunits by 2., it's maximal and the only maximal ideal.
</li>
<li>
Fix $t\in R$ with $v(t)=1$, and fix any nonzero $r\in\mfm$. There exists $\alpha\in\Frac(R)$ such that $\alpha t=r$, so $v(\alpha)+1=v(\alpha)+v(t)=v(r)>0$. Thus, $v(\alpha)\ge0$, so $\alpha\in R$ and $r\in(t)$.
</li>
<li>
Let $I\subset R$ be an ideal and fix $t\in I$ with minimal valuation. Let $r\in I$ be any other nonzero element, and again fix $\alpha\in\Frac(R)$ s.t. $\alpha t=r$. Then, $v(\alpha)+v(t)=v(r)\ge v(t)$, so $v(\alpha)\ge 0$ which menas $\alpha\in R$ and $r\in(t)$, so $I=(t)$.
</li>
<li>
Fix some $x\in R$, and let $n=v(x)$. There's some $u\in\Frac(R)$ such that $ut^n=x$, so $v(u)+n=v(u)+v(t^n)=v(x)=n$. Hence, $v(u)=0$, so $u\in\units R$.
</li>
</ol>
</div>
<p>And finally, what is a Dedekind domain?</p>
<div class="definition">
An integral domain $R$ is a <b>Dedekind domain</b> if it is 1-dimensional, integrally closed, and noetherian.
</div>
<div class="example">
Any PID is a Dedekind domain.
</div>
<div class="example">
Let $K/\Q$ be a finite extension, and let $\ints K\subset K$ be the integral closure of $\Z$ in $K$. Then, $\ints K$ is Dedekind (More generally, the integral closure of a Dedekind domain in a field is Dedekind).
</div>
<div class="example">
Let $k$ be algebraically closed, and let $f\in k[x_1,\dots,x_n]$ be irreducible and "smooth," i.e. for all $c\in k^n$, (at least) one of $f(c),\pderiv f{x_1}(c),\dots,\pderiv f{x_n}(c)$ is nonzero. Then, $k[x_1,\dots,x_n]/(f)$ is a Dedekind domain.
</div>
<p>I’ll leave verifying these examples up to you. Our first lemma is that local Dedekind domains are dvrs.</p>
<div class="lemma">
Let $A$ be a local Dedekind domain. Then, $A$ is a dvr.
</div>
<div class="proof4">
Let $\mfm\subset A$ be its maximal ideal, and fix any $t\in\mfm\sm\mfm^2$. We claim that $\mfm=(t)$. First note that $\bar\mfm\subset A/(t)$ is the unique maximal ideal of the local 0-dimensional noetherian (hence local artinian) ring $A/(t)$ and so is nilpotent. Hence, $\mfm^n\subset(t)$ for some $n\ge1$. Suppose that $n>1$ and that $\mfm^{n-1}\not\subset(t)$; choose any $r\in\mfm^{n-1}\sm(t)$. Note that, given any $m\in\mfm$, we have $rm\in(t)\cap\mfm^n\subset(t)\cap\mfm^2$, so the multiplication map
$$\frac rt:\mfm\to\mfm$$
is $A$-linear. Hence, Cayley-Hamilton implies that $\frac rt\in\Frac(A)$ is integral over $A$. Since $A$ is integrally closed, we have $r/t\in A$ which contradicts $r\in\mfm^{n-1}\sm(t)$. Hence, $\mfm^{n-1}\subset(t)$, and induction then shows that in fact $\mfm\subset(t)\subsetneq A$, so $\mfm=(t)$. Now, given any $x\in A\sm\{0\}$, let $v(x)\in\Z_{\ge0}$ be the highest power of $t$ dividing $x$. This is well-defined because if $t^n\mid x\neq0$ for all $n$, then we could form the infinite chain
$$\parens{x}\subsetneq\parens{\frac xt}\subsetneq\parens{\frac x{t^2}}\subsetneq\parens{\frac x{t^3}}\subsetneq\cdots,$$
which contradicts $A$ being noetherian. Furthermore, it is clear that $v(ab)=v(a)+v(b)$ always and that $v(a+b)\ge\min(v(a),v(b))$ with equality if $v(a)\neq v(b)$ because that's how division works. Now, note that $v$ extends to a map $\units{\Frac(A)}\to\Z$ via $v(x/y)=v(x)-v(y)$. To finish, we need to show that
$$A=\bracks{\alpha\in\units{\Frac(A)}:v(\alpha)\ge0}\cup\{0\}.$$
This is because if $v(\alpha)\ge0$, then we can write $\alpha=x/y$ for some $x,y\in A$ with $v(y)=0$. However, $v(y)=0\implies y\in A\sm(t)=A\sm\mfm=\units A$ so $\alpha=x/y\in A$.
</div>
<div class="corollary">
A ring $A$ is a dvr iff $A$ is a local Dedekind domain iff $A$ is a local PID.
</div>
<p>The above was most of the work in proving this theorem about factoring ideals into primes. The rest of the proof is essentially the observation that localizing a Dedekind domain at a prime gives you a local Dedekind domain.</p>
<div class="theorem" name="Structure Theorem">
Let $A$ be a Dedekind domain, and let $(0)\neq I\subset A$ be an ideal. Then, $I$ factors into a unique finite product of prime ideals.
</div>
<div class="proof4">
Fix any nonzero prime (i.e. maximal) $\mfp\subset A$. Since localizations preserve noetherianness and integral closednesss and since the only prime inside of $\mfp$ is $(0)$ (i.e. $\mfp$ has height 1), we see that $A_\mfp$ is a local Dedekind domain, and so a dvr. Hence, $IA_\mfp=(\mfp A_\mfp)^{v_{\mfp}(I)}$ for some $v_{\mfp}(I)\in\Z_{\ge0}$. Let
$$J=\prod_{(0)\neq\mfp\in\spec A}\mfp^{v_\mfp(I)}=\bigcap_{(0)\neq\mfp\in\spec A}\mfp^{v_\mfp(I)},$$
and note that $I\subseteq J$ since $I\subseteq IA_\mfp\cap A=\mfp^{v_\mfp(I)}$ for all $\mfp$. Since the inclusion $I\into J$ is (by construction) locally an isomorphism (i.e. $I_\mfp=J_\mfp$), we conclude that in fact $I=J$, so $I$ is a product of primes. This product is unique since the exponents are recoverable via localization, so we only need to show that this product is finite (i.e. that $v_\mfp(I)=0$ almost always). This is equivalent to the claim that $I$ is contained in only finitely many maximal ideals, but that's true since $A/I$ is Artinian and Artinian rings only have finitely many maximal ideals. Thus, we win.
</div>
<p>Well, that wasn’t so bad, was it? I feel like this is a nice proof because it doesn’t require many technical lemmas and you can even get away with not mentioning fraction ideals. To end this post, you can try proving a converse to the structure theorem.</p>
<div class="exercise">
Let $A$ be an integral domain such that every nonzero ideal $I\subset A$ factors uniquely into a finite product of prime ideals. Show that $A$ is a Dedekind domain.
</div>
<div class="footnotes">
<ol>
<li id="fn:2">
<p>Fight me <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:1">
<p>Although Serre comes close in his “Local Fields” book (maybe I should also admit that I’ve only seen two approaches aside from the one I’ll show here). I should also add that when I say “never seen,” I mean excluding the approach in this post; this proof was outlined in one of my classes, but I don’t know where else it’s written down. <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>I’ve wanted to write this post for a long time now. Dedekind domain’s are, objectively, the best rings in existence 1, and their greatness stems from one fact: ideals in a Dedekind domain factor uniquely into (finite) products of prime ideals. However, I’ve never seen a proof of this fact that I liked (i.e. one that’s straightforward enough for me to actually remember) 2, and so this post is my attempt to remedy this situation. Like (almost) always, I’ll start by introducing some background I’ll need in the proof, and then I’ll actually get into the good stuff in a separate section. Fight me ↩ Although Serre comes close in his “Local Fields” book (maybe I should also admit that I’ve only seen two approaches aside from the one I’ll show here). I should also add that when I say “never seen,” I mean excluding the approach in this post; this proof was outlined in one of my classes, but I don’t know where else it’s written down. ↩Some Classic Affine Algebraic Geometry2019-03-24T00:00:00+00:002019-03-24T00:00:00+00:00https://nivent.github.io/blog/comm-alg<p>This will serve as background for a post I wanna write on Dedekind domains. The idea here is to blaze through <sup id="fnref:1"><a href="#fn:1" class="footnote">1</a></sup> some geometry so I can use words like “Artinian ring” and “Krull dimension” when discussing Dedekind domains <sup id="fnref:8"><a href="#fn:8" class="footnote">2</a></sup>. The plan is to talk about integeral extensions of rings (+ other stuff), spectra of rings, dimensions of rings, and then Artinian rings. In a sense, this might be more commutative algebra than (classic) algebraic geometry, but what’s the difference? Note that all rings in this post are commutative with unity (and all ring maps preserve the unit).</p>
<h1 id="a-mess-of-words-i-dont-think-ive-defined-on-this-blog-before">A Mess of Words I Don’t Think I’ve Defined on this Blog Before</h1>
<p>For completeness, before getting into integral extensions I want to really quickly fill a couple gaps in this blog by saying what a module and an algebra are.</p>
<div class="definition">
Let $R$ be a ring. A <b>(left) $R$-module</b> $M$ is an abelian group with an $R$-action $R\by M\to M$ satisfying the following
<ol>
<li> $r(m_1+m_2)=rm_1+rm_2$ always.</li>
<li> $(r_2r_1)m=r_2(r_1m)$ always.</li>
<li> $(r_1+r_2)m=r_1m+r_2m$ always.</li>
<li> $1m=m$ where $1\in R$ is the multiplicative unit</li>
</ol>
</div>
<div class="remark">
An $R$-module structure of a group $M$ is the same thing as a ring homomorphism $R\to\End(M)$ into $M$'s endomorphism group.
</div>
<div class="remark">
If $R$ is a field, then an $R$-module is an $R$-vector space. If $R=\Z$, then an $R$-module is an abelian group.
</div>
<div class="definition">
Let $R$ be a ring. An <b>$R$-algebra $A$</b> is a ring with an $R$-module structure such that
$$r(a_1a_2)=(ra_1)a_2=a_1(ra_2).$$
</div>
<div class="remark">
An $R$-algebra is the same thing as a ring $A$ with a ring map $R\to Z(A)$ into the center of $A$.
</div>
<p>Modules, and in particular algebras, will feature heavily in this post. Modules in general can be rather poorly behaved, but we can maintain our sanity if we restrict ourselves to modules over a commutative ring that satisfies a nice finiteness property.</p>
<div class="definition">
An $R$-module $M$ is called <b>Noetherian</b> if it satisfies the <b>ascending chain condition</b> on submodules; that is, any chain
$$M_1\subseteq M_2\subseteq M_3\subseteq\cdots\subseteq M$$
on submodules eventually stabilizes (i.e. $M_n=M_{n+1}=\cdots$ for some $n\ge1$). If $R$ is Noetherian as an $R$-module, then we call it a <b>Noetherian ring</b>.
</div>
<div class="remark">
I never defined what a submodule is, but I think it's not that hard to figure out. However, while I'm on the subject of submodules, two things: (1) quotient modules always exist (i.e. given any submodule $N\subseteq M$, the natural quotient $M/N$ can be given an $R$-module structure) and (2) an $R$-submodule of $R$ is the same thing as an ideal.
</div>
<p>The Noetherian property is a kind of natural generalization being a PID. This is maybe not immediately obvious from the given definition, but this next theorem will help.</p>
<div class="proposition">
Let $M$ be an $R$-module. Then $M$ is Noetherian iff every submodule of $M$ is finitely generated.
</div>
<div class="proof4">
$(\to)$ Assume $M$ is Noetherian, and let $N\subset M$ be a submodule. Fix some $x_1\in N$, and let $N_1=Rx_1$ be the module it generates. Inductively choose $x_n\in M$ ($n>1$) such that $x_n\in N\sm N_{n-1}$ if $N_{n-1}\neq N$, and $x_n=x_{n-1}$ otherwise; set $N_n=\sum_{i=1}^nRx_i$. This gives an ascending chain
$$N_1\subseteq N_2\subseteq N_3\subseteq\cdots\subseteq M$$
of submodules of $M$ (submodules of $N$ even), and so must stabalize at the $m$th step for some $m$. Then, $N_m=N$ (otherwise, $N_{m+1}$ would be bigger by construction), so $N$ is generated by $m<\infty$ elements.
<br />
$(\from)$ Exercise.
</div>
<div class="corollary">
A ring is Noetherian iff every ideal is finitely generated.
</div>
<div class="remark">
This proposition makes it clear that subrings/submodules of a Neotherian ring/module are also noetherian (e.g. since submodules of $N\subset M$ are still submodules of $M$ and submodules of $M/N$ are submodules of $M$ containing $N$).
</div>
<p>Another nice thing to know about the Noetherian property is that almost every ring/module you will ever care about is Noetherian.</p>
<div class="theorem">
Let $R$ be a noetherian ring. Then every finitely generated (f.g.) $R$-module is noetherian.
</div>
<div class="proof4">
The main part of the proof is to show that noetherianess is preserved under extensions (i.e. if $N\subset M$ and $M/N$ are noetherian, then so is $M$), so we only prove this (to finish, show f.g. free modules $R^{\oplus m}$ are noetherian, and every f.g. module is a quotient of a f.g. free module). Consider a short exact sequence of $R$-modules
$$0\too N\too M\too M/N\too 0$$
where $N$ and $M/N$ are Noetherian. Let $M'\subset M$ be any $R$-submodule, so we get another short exact sequence
$$0\too N\cap M'\xtoo fM'\xtoo gM'/(N\cap M')\too0$$
where $N\cap M'\subseteq N$ and $M'/(N\cap M')\simeq (N+M')/N\subseteq M/N$. Hence, these modules are finitely generated, say by $\{e_1,\dots,e_n\}\subseteq N\cap M'$ and $\{\bar f_1,\dots,\bar f_m\}\subseteq M/(N\cap M')$, respectively. Now, pick any $m\in M'$, and write (non-uniquely) $g(m)=r_1\bar f_1+\dots+r_m\bar f_m$ with $r_i\in R$. Because $g$ is surjective, we can pick (non-unique) lifts $f_1,\dots,f_m\in M'$ of $\bar f_1,\dots,\bar f_m$, so $m-(r_1f_1+\dots+r_mf_m)\in\ker g=\im f$. Hence,
$$m=(r_1f_1+\dots+r_mf_m)+(s_1e_1+\dots+s_ne_n)$$
for some $s_j\in R$ (where we've identified $e_i\in N\cap M'$ with $f(e_i)\in M'$ since $f$ is injective). Thus, $M'$ is generated by $\{e_1,\dots,e_n\}\cup\{f_1,\dots,f_m\}$, so every submodule of $M$ is finitely generated.
</div>
<div class="theorem" name="Hilbert Basis">
Let $R$ be a Noetherian ring. Then, $R[x]$ is Noetherian as well.
</div>
<div class="proof4">
Exercise (hint: given an ideal $I\subseteq R[x]$, first consider the ideal of leading coefficients of elements of $I$)
</div>
<div class="corollary">
Let $R$ be a Noetherian ring. Then every finitely generated $R$-algebra is Noetherian.
</div>
<p>I think that’s all the highlights about Noetherianess. Let’s get into integral extensions.</p>
<div class="definition">
Let $\phi:R\to S$ be a ring map (e.g. an injection of rings). Then, $s\in S$ is <b>integral over $R$</b> if it is the root of a monic polynomial in $R[x]$. If all $s\in S$ are integral over $R$, then $S$ is <b>integral over $R$</b> or <b>an integral extension of $\phi(R)$</b>.
</div>
<div class="definition">
Let $\phi:R\to S$ be a ring map. The <b>integral closure of $R$ in $S$</b> is the set
$$\bar R=\{s\in S:s\text{ is integral over }R\}.$$
We say that $R$ is <b>integrally closed in $S$</b> if $\bar R=R$, and we say that a domain $R$ is <b>integrally closed</b> if it is integrally closed in $\Frac(R)$.
</div>
<div class="proposition">
Integral closures are rings.
</div>
<div class="proof4">
Let $\phi:R\to S$ be a ring map, and let $a,b\in S$ be integral over $R$. We need to show that $ab$ and $a+b$ are integral over $R$ as well (I guess we also need that $-a$ is integral over $R$). This is a consequence of the following lemma (consider the ring $R[a,b]$).
</div>
<div class="lemma">
Let $\phi:R\to S$ be a ring map. Then, $s\in S$ is integral over $R$ iff there exists a a subring $S'\subseteq S$ containing $s$ which is finitely generated as an $R$-module.
</div>
<div class="proof4">
$(\to)$ Assume that $s\in S$ is integral over $R$, and let $d$ be the degree of a monic polynomial in $R[x]$ vanishing at $s$. Then, $R[s]\subseteq S$ is a ring generated as an $R$-module by $\{s^k\}_{k=0}^{d-1}$.
<br />
$(\from)$ Let $S'\subseteq S$ be a subring containing $s$ which is finitely generated as an $R$-module. Write $S'=\sum_{i=1}^nRe_i$ with $e_i\in S'$. Note that the multiplication by $s$ map $m_s:S'\to S', x\mapsto sx$ is $R$-linear, and so can be represented (non-uniquely) by some matrix. That is, $m_s(e_j)=\sum_ia_{ij}e_i$ for some $a_{ij}\in R$, so
$$\Mat s\dots0\vdots\ddots\vdots0\dots s\vVec{e_1}\vdots{e_n}=\Mat{a_{11}}\dots{a_{1n}}\vdots\ddots\vdots{a_{n1}}\dots{a_{nn}}\vVec{e_1}\vdots{e_n}.$$
Subtracting now gives
$$\Mat {s-a_{11}}\dots{-a_{1n}}\vdots\ddots\vdots{-a_{n1}}\dots{s-a_{nn}}\vVec{e_1}\vdots{e_n}=\vVec0\vdots0.$$
Call the matrix on the left $M$. By Cramer's rule, we can form its adjugate $M^{\mrm{adj}}$ so that $M^{\mrm{adj}}M=\det M\cdot I$ where $I$ is the identity matrix. Multiplying the above equation by the adjugate gives $(\det M)e_i=0$ for all $i$. Since the $e_i$'s generate $S'$, we conclude $(\det M)x=0$ for all $x\in S'$. Taking $x=1$, gives $\det M=0$. Finally, $\det M$ literally is a monic polynomial in with $R$-coefficients vanishing at $s$, so we win.
</div>
<div class="remark">
A finitely generated $R$-algebra $S$ is integral over $R$ iff $S$ is finitely generated as an $R$-module.
</div>
<p>Integral extensions are the ring-theoretic analogue of algebraic extensions from field theory. Indeed, if $R$ is a field, then $x$ is integral over $R$ iff it is algebraic over $R$. Integral extensions have many nice properties, some of which are collected below.</p>
<div class="proposition">
Integral closures are integrally closed.
</div>
<div class="proof4">
Exercise.
</div>
<div class="proposition">
UFDs are integrally closed.
</div>
<div class="proof4">
Exercise (Hint: this is basically the rational root theorem).
</div>
<div class="proposition">
Let $\phi:A\to B$ be a ring map, and let $S\subset A$ be a multiplicative set. If $B$ is integral over $A$, then $\phi(S)^{-1}B$ is integral over $\sinv A$.
</div>
<div class="proof4">
Exercise.
</div>
<div class="corollary">
Let $S\subset R$ be a multiplicative set, and suppose that $R$ is integrally closed. Then, $\sinv R$ is integrally closed as well.
</div>
<div class="proposition">
Let $A\subset B$ be an integral extension of rings. Then, $A$ is a field iff $B$ is a field.
</div>
<div class="proof4">
$(\to)$ Suppose that $A$ is a field, and fix any $b\in B$. Then, $b$ is the root of some
$$f(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\in A[x].$$
This means that
$$-a_0=b(b^{n-1}+a_{n-1}b^{n-2}+\cdots+a_1),$$
and hence $\inv b=\frac{-1}{a_0}(b^{n-1}+a_{n-1}b^{n-2}+\dots+a_1)\in B$.
<br />
($\from$) This direction is proved similarly using that $\inv a\in B$ for any nonzero $a\in A$.
</div>
<div class="proposition">
Let $A\subset B$ be an integral extension of rings. Let $\mfa\subsetneq B$ be an ideal, and let $\mfa'=\mfa\cap A$. Then, $B/\mfa$ is integral over $A/\mfa'$.
</div>
<div class="proof4">
Just take a monic polynomial and reduce it mod $\mfa'$.
</div>
<div class="proposition">
Let $A\subset B$ be an integral ring extension, and let $\mfp\subset B$ be a prime ideal. If $\mfp\cap A$ is maximal, then so is $\mfp$.
</div>
<div class="proof4">
$B/\mfp$ is integral over the field $A/(\mfp\cap A)$.
</div>
<div class="theorem" name="Going up">
Let $A\subset B$ be an integral ring extension, and let $\mfp\subset A$ be prime. Then, there's some prime ideal $\mfp'\subset B$ with $\mfp'\cap A=\mfp$.
</div>
<div class="proof4">
Let $S=A\sm\mfp$, so $\sinv B$ is integral over the local ring $\sinv A=A_{\mfp}$ with unique maximal ideal $\sinv\mfp$. Let $\mfm\subset\sinv B$ be any maximal ideal. Then, $\mfm'=\mfm\cap\sinv A$ is prime, so $\sinv B/\mfm$ is a field integral over $\sinv A/\mfm'$; hence, $\mfm'$ is maximal and so must be $\sinv\mfp$. Thus, $\mfp'=\mfm\cap B$ is a prime of $B$ laying over $\mfp$ since
$$(\mfm\cap B)\cap A=\mfm\cap A=(\mfm\cap\sinv A)\cap A=\sinv\mfp\cap A=\mfp.$$
</div>
<p>Alright. I think that’s more than enough about integral extensions for now. Bottom line: they’re preserved by reasonal operations and they play nicely with primes.</p>
<h1 id="finally-some-geometry">Finally, Some Geometry</h1>
<p>Let’s actually justify the use of the word geometry in the title of this post by talking of $\spec$. Hilbert’s Nullstellensatz (which we’ll prove later) shows that, for an algebrically closed field $k$, points in $k^n$ are in bijection with maximal ideals of $k[x_1,\dots,x_n]$. Since we’d like to do geometry in purely algebraic settings (i.e. over an arbitrary ring $R$), we may think that a good replacement for $k^n$ is the set of maximal ideals of $R$. However, there’s a better choice; the set of prime ideals of $R$.</p>
<div class="definition">
Let $R$ be a ring. Its <b>spectrum</b> is
$$\spec R=\bracks{\mfp\subset R:\mfp\text{ is prime}}.$$
</div>
<p>Now, these spectres or whatever are supposed to be our replacements for things like $\C^n$, so they better be geometric in some sense. At the very least, they better have a topology.</p>
<div class="definition">
Given an ideal $I\subseteq R$, its "vanishing set" is $V(I)=\bracks{\mfp\in\spec R:\mfp\supseteq I}$. The <b>Zariski topology</b> on $\spec R$ is given by having closed sets be $V(I)$ for $I$ an ideal.
</div>
<div class="exercise">
Prove that this is an actual topology. Also prove that $V(I)$ is homeomorphic to $\spec R/I$. If you haven't had enough of this exercise, show that the open set $\spec R\sm V(r)$ (where $r\in R$) is homeomorphic to $\spec R_r=\spec R[1/r]$.
</div>
<div class="remark">
Given a ring map $f:R\to S$, we get an induced map on spectra $\spec S\to\spec R$ given by sending $\mfp\subset S$ to $\inv f(\mfp)\subset R$. This map is easily seen to be continuous, and so $\spec$ is a contravariant functor from Rng to Top.
</div>
<p>The idea behind the Zariski topology is that “zero sets of polynomials” should be closed <sup id="fnref:2"><a href="#fn:2" class="footnote">3</a></sup>, so how do we get from that to this? Well, think of $\spec R$ as some variety (e.g. if $R=\C[x,y]/(y^2-x^3-x)$, then $\spec R$ is the curve given by $y^2=x^3-x$) and we want elements of $R$ to be functions on $\spec R$ (i.e. intuitively, elements of $R$ should give well-defined polynomials on $\spec R$ (see previous parenthetical)). The natural way of realizing this is to say that $r(\mfp)= r\pmod\mfp$ where $r\in R$ and $\mfp\in\spec R$ <sup id="fnref:3"><a href="#fn:3" class="footnote">4</a></sup>, so $r(\mfp)=0$ precisely when $r\in\mfp$. With this in mind, the points of $\spec R$ vanishing on each function in an ideal $I$ <sup id="fnref:4"><a href="#fn:4" class="footnote">5</a></sup> are exactly those that contain $I$.</p>
<p>We now have a topological space associated to an arbitrary ring $R$. How should we think about it geometrically? Well, motivated by Hilbert’s Nullstellensatz, we should think of the maximal ideals of $R$ as the (closed) points of $\spec R$ (point here used geometrically as a 0-dimensional thing. Any element of $\spec R$ could be reasonably called a point since it’s a space). Motivated by the exercise, the non-maximal prime ideals should correspond to higher-dimensional subvarieties: curves and hyperplanes and whatnot. This does beg the question though: what do I mean by dimension here? To answer that, I first need to define irreducible sets. <sup id="fnref:5"><a href="#fn:5" class="footnote">6</a></sup></p>
<div class="definition">
A subvariety $V(I)\subset\spec R$ (note: $I$ arbitrary) is called <b>irreducible</b> if writing $V(I)=V(J_1)\cup V(J_2)$ requires that $I=J_1$ or $I=J_2$.
</div>
<div class="exercise">
Show that $V(I)$ is irreducible iff $V(I)=V(\mfp)$ with $\mfp\subset R$ prime.
</div>
<p>Irreducible varieties are the ones that we care about; these are things like 1 point, 1 curve, 1 plane, etc. Intuitively, an irreducible variety cannot contain two things of the same dimension (otherwise you could write it as the union of those two things), so if $V(I)\subset V(J)$ with both irreducible, you would expect that $\dim V(I)<\dim V(J)$.</p>
<div class="definition">
Let $X=\spec R$. Then, $\dim X$ is defined to be the length of the longest chain $X_1\subsetneq X_2\subsetneq\dots\subsetneq X_n=X$ of irreducible subvarieties of $X$. Similarly, $\dim R$ is defined to be $\dim(\spec R)$.
</div>
<div class="remark">
By the previous exercise, the dimension of a ring is the length of its longest chain of prime ideals.
</div>
<div class="remark">
It is fairly easy to see that $\dim A\ge\dim A_\mfp+\dim A/\mfp$ for all $\mfp\in\spec A$. It turns out that this is always an equality (at least when $A$ is a finitely generated algebra over an algebraically closed field $k$). I won't prove this in this post, but after finishing this section, you should be able to do this by inducting on $\dim A$ and making use of Noether normalization.
</div>
<div class="example">
Let $\A_k^n=k[x_1,\dots,x_n]$ where $k$ is a field. Then, $\dim\A_k^n=n$. In general, $\dim R[x]=1+\dim R$ when $R$ is noetherian.
</div>
<div class="exmaple">
$\dim\Z=1$. In general, $\dim R\le1$ if $R$ is a PID.
</div>
<div class="exmaple">
$\dim k=0$ if $k$ is a field (this is not an iff!).
</div>
<div class="remark">
$\dim R=0$ iff all prime ideals are maximal. For $R$ a domain, $\dim R=1$ iff all nonzero prime ideals are maximal.
</div>
<p>For the rest of this section, I’ll discuss some alternative definitions of dimension when your ring $R$ is niceish. Fix a field $k$. Unless otherwise stated, assume that $R$ is a finitely generated $k$-algebra for the rest of the section. This means that we can write $R=k[x_1,\dots,x_n]/I$.</p>
<div class="definition">
An <b>affine irreducible variety</b> is an irreducible set $\spec k[x_1,\dots,x_n]/I=V(I)\subset\A_k^n$.
</div>
<div class="definition">
Let $\phi:\spec B\to\spec A$ be a morphism induced by a $k$-algebra homomorphism $f:A\to B$. We say that $\phi$ is <b>finite</b> if $B$ is integral over $f(A)$.
</div>
<div class="theorem">
If $\phi:\spec B\to\spec A$ is finite, then
<ol>
<li> $\phi$ is a closed map </li>
<li> for any $\mfp\in\spec A$, $\inv\phi(\mfp)$ is finite. </li>
<li> $\phi$ is injective iff it was induced by a surjective map of rings </li>
</ol>
</div>
<div class="proof4">
I'll only prove 2. Assume $\phi$ is finite and induced by $f:A\to B$. Fix any $\mfp\in\spec A$, and let $B_\mfp=f(A\sm\mfp)^{-1}B$. Then, $B_\mfp$ is an integral extension of $A_\mfp=(A\sm\mfp)^{-1}A$, and so a finitely generated $A_\mfp$-module. Note that primes of $B_\mfp$ are in bijection with primes of $B$ away (i.e. disjoint) from $f(A\sm\mfp)$, so in particular, any $\mfq\in\inv\phi(\mfp)$ corresponds to a unique prime of $B_\mfp$. Now, let $\mfm=\mfp A_\mfp$ be the unique maximal ideal of $A_\mfp$, and consider the $A_\mfp/\mfm$-algebra $B_\mfp/f(\mfm)$. Note that $B_\mfp/f(\mfm)$ is a finitely generated $A_\mfm/\mfm$-module (i.e. a finite-dimensional vector space), and so has only finitely many prime ideals (we'll see this when we talk about Artinian rings). However, prime ideals of $B_\mfp/f(\mfm)$ are prime ideals of $B_\mfp$ containing $f(\mfm)$, and so are a superset of $\inv\phi(\mfp)$; thus, we win.
</div>
<div class="lemma" name="Noether normalization">
Let $X\subset\A_k^n$ be an irreducible subvariety. Then, there's a map $\pi:\A_k^n\onto\A_k^d$ such that the composition $X\to\A_k^n\onto\A_k^d$ is finite and surjective.
</div>
<div class="proof4">
We'll prove this assuming $k$ infinite (the general case is similar but with a non-linear transformation). Note that, by induction, it suffices to prove this in the case that $X=\spec k[x_1,\dots,x_n]/(f)$ is an irreducible hypersurface, so that's what we'll do. For an aribtrary nonzero $c=(c_1,c_2,\dots,c_{n-1})\in k^{n-1}$, consider the projection $\A_k^n\to\A_k^{n-1}$ induced by $y_i=c_ix_n+x_i$ for $i=1,\dots,n-1$. Write $f=f_d+f_{d-1}+\dots+f_0$ where $d=\deg f$ and $f_i$ is homogeneous of degree $i$ for all $i$. Then, $x_n$ satisfies the polynomial
$$g(T)=f(y_1-c_1T,y_2-c_2T,\dots,y_{n-1}-c_{n-1}T,T)\in k[y_1,\dots,y_{n-1}][T].$$
Furthermore, the leading term of $g(T)$ is $f_d(-c_1T,-c_2T,\dots,-c_{n-1}T,T)=(-T)^df_d(c_1,c_2,\dots,c_{n-1},1)$. Since $f_d\neq0\in k[x_1,\dots,x_n]$ and $k$ is infinite, we can fix a choice of $r=(r_1,\dots,r_n)\in k^n$ s.t. $f_d(r)\neq0$. Possibly after reordering, we may assume $r_n\neq0$, so $f_d(r/r_n)=f_d(r)/r_n^d\neq0$. Thus, choosing $c=(r_1/r_n,\dots,r_{n-1}/r_n)\in k^{n-1}$ causes the composite $\pi_c:X\to\A_k^n\to\A_k^{n-1}$ induced by the map
$$\mapdesc\phi{k[y_1,\dots,y_{n-1}]}{k[x_1,\dots,x_n]/(f)}{y_i}{c_ix_n+x_i+(f)}$$
to be finite; this choice of $c$ makes $x_n$ integral over the image, and the other $x_i$ are also integral over the image since $x_i=\phi(y_i)-c_ix_n$ and being integral is closed under ring operations. Hence, we only need to show that $\pi_c$ is surjective (i.e. that $\phi$ is injective). Fix some $p(y_1,\dots,y_{n-1})\in\ker\phi$ so
$$p(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1})=f(x_1,\dots,x_n)q(x_1,\dots,x_n).$$
Now, divide $p(y_1,\dots,y_{n-1})$ by $f(y_1-c_1y_n,\dots,y_{n-1}-c_{n-1}y_n,y_n)$ in $k(y_1,\dots,y_{n-1})[y_n]$ to get
$$p(y_1,\dots,y_{n-1})=\st q(y_1,\dots,y_n)f(y_1-c_1y_n,\dots,y_{n-1}-c_{n-1}y_n,y_n)+r(y_1,\dots,y_n).$$
Suppose that $p\neq0$; we can apply $\phi$ to $y_i$ ($i < n$) and send $y_n\mapsto x_n$, resulting in
$$p(c_1x_n+x+1,\dots,c_{n-1}x_n+x_{n-1})=r(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1},x_n),$$
so the total degree of $p$ is the total degree of $r$ which is less than the total degree of $f$. This contradicts $p(c_1x_n+x_1,\dots,c_{n-1}x_n+x_{n-1})=f(x_1,\dots,x_n)q(x_1,\dots,x_n)$, so we must have $p=0$. Hence, $\phi$ is injective, and $\pi_c$ is surjective.
</div>
<p>This is a nice lemma. For example, we can use it to give another definition of dimension.</p>
<div class="proposition">
Let $R=k[x_1,\dots,x_n]/\mfp$ where $\mfp$ is prime, and let $\spec R\onto\A_k^d$ be the map from Noether normalization. Then, $d=\dim R$.
</div>
<div class="proof4">
We have that $R$ is integral over $k[x_1,\dots,x_d]$. Hence, any chain of primes in $R$ restricts to a chain of primes in $k[x_1,\dots,x_d]$, and any chain of primes in $k[x_1,\dots,x_d]$ lifts to a chain of primes in $R$ (Going up), so $\dim R=\dim k[x_1,\dots,x_d]=d$.
</div>
<div class="corollary">
Let $R=k[x_1,\dots,x_n]/\mfp$ where $\mfp$ is prime. Then, $\trdeg_k\Frac R=\dim R$.
</div>
<div class="proof4">
Noether normalization gives the existence of $d=\dim R$ algebraically independent elements $y_1,\dots,y_d\in R$ such that $R$ is integral of $S=k[y_1,\dots,y_d]$. Let $F=\Frac(R)$ and $K=\Frac(S)$. Since $R$ is integeral over $S$, $F/K$ is algebraic so $\trdeg_kF=\trdeg_kS=d$.
</div>
<p>We also (finally) get a nice proof of the Nullstellensatz.</p>
<div class="lemma" name="Zariski's">
Let $k$ be a field, $A$ a finitely generated $k$-algebra, and $\mfm$ a maximal ideal of $A$. Then, $A/\mfm$ is a finite degree field extension of $k$.
</div>
<div class="proof4">
Note that $\dim(A/\mfm)=0$ since it's a field, so Noether normalization gives the existence of a surjective, finite map $\spec(A/\mfm)\to\spec k$. This means that $A/\mfm$ is integral over $k$ and hence an algebraic extension. Since $A$ was finitely generated as a $k$-algebra, we conclude that $A/\mfm$ is a finitely generated algebraic extension of $k$ (i.e. a finite extension of $k$).
</div>
<div class="definition">
Let $k$ be a field. For $S\subset k^n$, let
$$I(S)=\bracks{f\in k[x_1,\dots,x_n]:\forall x\in S,f(x)=0}.$$
Conversely, for $J\subset k[x_1,\dots,x_n]$, let
$$V(J)=\bracks{x\in k^n:\forall f\in J,f(x)=0}.$$
</div>
<div class="theorem" name="Hilbert's Nullstellensatz">
Let $k$ be an algebraically closed field. Then, maximal ideals of $k[x_1,\dots,x_n]$ are in bijection with points of $k^n$: $(a_1,\dots,a_n)\in k^n\mapsto\mfm_a=(x_1-a_1,\dots,x_n-a_n)$.
</div>
<div class="proof4">
Let $\mfm$ be a maximal ideal of $k[x_1,\dots,x_n]$. Then, $k[x_1,\dots,x_n]/\mfm$ is a finite degree extension of the algebrically closed field $k$, and so must itself be $k$. Pick a $k$-isomorphism $\phi:k[x_1,\dots,x_n]/\mfm\to k$, and let $a_i=\phi(x_i)$. Then, $\mfm\supseteq\mfm_a$ where $a=(a_1,\dots,a_n)$, so $\mfm=\mfm_a$ since $\mfm_a$ is maximal and $\mfm$ is proper.
</div>
<h1 id="artinian-rings">Artinian Rings</h1>
<p>I think at this point, we’ve had a nice little introduction to algebraic geometry. I mentioned at the beginning that this post was motivated primarily by my desire to give a nice treatment of Dedekind domains. For that, I’ll need to make use of some facts about Artinian rings, so this is what we’ll end on. Artinian rings are dual to Noetherian ones, but as we’ll see, they are much more constrained.</p>
<div class="definition">
A ring $A$ is called <b>Artinian</b> if it satisfies the <b>descending chain condition</b> on ideals; that is, any chain
$$A\supseteq I_0\supseteq I_1\supseteq\dots$$
of ideals stabilizes.
</div>
<div class="example">
If $k$ is a field, then $k[x]/(x^n)$ is Artinian.
</div>
<p>We want to understand the structure of Artininan rings. We first observe that they only have finitely many maximal ideals.</p>
<div class="proposition">
Let $A$ be an Artinian ring. Then, $A$ has finitely many maximal ideals.
</div>
<div class="proof4">
Suppose otherwise and let $\mfm_1,\mfm_2,\dots$ be an infinite list of distinct maximal ideals. Then,
$$\mfm_1\supseteq\mfm_1\cap\mfm_2\supseteq\mfm_1\cap\mfm_2\cap\mfm_3\supseteq\cdots$$
is a descending chain of ideals, so $\mfm_1\cap\cdots\cap\mfm_n=\mfm_1\cap\cdots\cap\mfm_n\cap\mfm_{n+1}$ for some $n$. Then, Chinese remainder theorem gives
$$\prod_{i=1}^n\frac A{\mfm_i}\simeq\prod_{i=1}^{n+1}\frac A{\mfm_i},$$
which is absurd.
</div>
<p>This proposition is actually stronger than it may seem at first, since in fact all prime ideals of $A$ are maximal.</p>
<div class="proposition">
Let $A$ be an Artinian ring. Then, $\dim A=0$.
</div>
<div class="proof4">
Let $\mfp\subset A$ be prime. Then, $A/\mfp$ is an Artinian domain. Pick any nonzero $x\in A/\mfp$ and consider the chain $(x)\supseteq(x^2)\supseteq(x^3)\supseteq\dots$. This stabilizers, so $x^n=ux^{n+1}$ for some $n$ and some unit $u$. Thus,
$$0=x^n(ux-1).$$
Since $A/\mfp$ is a domain and $x\neq0$, we must have $ux=1$, so $x$ is a unit.
</div>
<div class="corollary">
$A$ Artinian $\implies\spec A$ finite and discrete.
</div>
<p>We’ll next show one of the stranger properties of Artinian rings: they’re isomorphic to the product of their localizations. To do this, we’ll need to make use of Nakayama’s lemma which I’ll state without proof <sup id="fnref:6"><a href="#fn:6" class="footnote">7</a></sup>. We only need the first version, but the second is also nice.</p>
<div class="lemma" name="Nakayama's">
Let $R$ be a commutative ring, and let $\mf M=\cap\mfm$ where $\mfm$ ranges over maximal ideals of $R$. Let $M$ be a finitely generated $R$-module. Then, $\mf M\cdot M=M\implies M=0$.
</div>
<div class="corollary">
Let $R$ be a local ring with maximal ideal $\mfm$, and let $M$ be a finitely generated $R$-module. Them, $M/\mfm M$ is a finite dimensional $R/\mfm$-vector space. A subset $X\subset M$ generates $M$ iff its image $\bar X\subset M/\mfm M$ generates $M/\mfm M$.
</div>
<div class="proposition">
Let $A$ be a local Artinian ring. Then, its maximal ideal is nilpotent.
</div>
<div clas="proof4">
Let $\mfm\subset A$ be maximal. The chain $\mfm\supseteq\mfm^2\supseteq\cdots$ tells us that $\mfm^n=\mfm^{n+1}$ for some $n$. Since $\mfm$ is the only maximal ideal, Nakayama says that $\mfm^n=0$.
</div>
<div class="proposition">
Let $A$ be Artinian. Then,
$$A\simeq\prod_{\mfm\in\spec A}A_\mfm.$$
</div>
<div class="proof4">
Since $\spec A$ is finite, we can pick some $N$ large enough that $(\mfm A_\mfm)^N=\mfm^NA_\mfm=0$ for all $\mfm\in\spec A$. Hence, $(\mfm^NA_\mfm)\cap A=\mfm^N=0$ for all $\mfm\in\spec A$. Fix a prime $\mfm\in\spec A$, and consider the natural map $f:A\to A_\mfm$. It's clear that $\ker f\supseteq\mfm^N$, but we claim that this is an equality. Pick any $a\in\ker f$. Then, $a/1=0/1$ so there's some $s\not\in\mfm$ s.t. $sa=0\in\mfm^N$. As $s\not\in\mfm$, an easy induction argument shows that $a\in\mfm^N$, so $\ker f=\mfm^N$. We also claim that $f$ is surjective. Pick some $s\not\in\mfm$. The image of $s$ in $A/\mfm^N$ is invertible (one stupid way to see this is that $(s)+\mfm^N=A$ by checking locally), so there's some $a\in A$ s.t. $sa-1\in\mfm^N$. Hence,
$$sa-1\in\mfm^NA_\mfm\implies sa-1=0\in A_\mfm\implies\frac a1=\frac1s,$$
so $f$ is surjective (for elements with non-unit numerator, use that $f$ is multiplicative). Thus, $A/\mfm^N\simeq A_\mfm$. To finish, let $J=\prod_{\mfm\in\spec A}\mfm$ be the nilradical of $A$. Then, $J^N=0$ since locally, $J^N_\mfm=\mfm^NA_\mfm=0$, and so Chinese remainder theorem gives
$$A\simeq A/J^N\simeq\prod_{\mfm\in\spec A}\frac A{\mfm^N}\simeq\prod_{\mfm\in\spec A}A_\mfm$$
as desired.
</div>
<div class="corollary">
Let $A$ be Artinian. Then, $A$ is Noetherian.
</div>
<div class="proof4">
By the proposition (and the easy fact that finite products of noetherian rings are noetherian), it suffices to prove this when $A$ is local. Let $\mfm\subset A$ be its unique maximal ideal, and write $\mfm^N=0$. Then,
$$0=\mfm^N\subseteq\mfm^{N-1}\subseteq\cdots\subseteq\mfm\subseteq A$$
is a finite filtration of $A$ giving rise to the short exact sequences ($0\le n < N$)
$$0\too\mfm^{n+1}\too\mfm^n\too\mfm^n/\mfm^{n+1}\too0.$$
Since $0$ is Noetherian and extensions of noetherian rings are noetherian (I don't think I proved this either but it's also easy), it suffices to show that $\mfm^n/\mfm^{n+1}$ is noetherian for all $n$. However, $\mfm^n/\mfm^{n+1}$ is a vector space over $A/\mfm$, and is finite-dimensional because otherwise you could get an infinite descending chain (which you can then pull back to $A$). Since it's a finite dimensional vector space, it's also noetherian (e.g. because it's a free $A/\mfm$-module and so noetherien as an $A/\mfm$-module, but every ideal is an $A/\mfm$-submodule so an ascending chain of ideals is an ascending chain of $A/\mfm$-submodules which then must stabilize). Thus, we win.
</div>
<p>In the end, we’ve shown that every Artinian ring is a $0$-dimensional, Noetherian ring. One can in fact show the converese, so a ring is Artinian precisely when it’s 0-dimensional and Noetherian. I’ll leave the reverse direction as an exercise <sup id="fnref:7"><a href="#fn:7" class="footnote">8</a></sup>.</p>
<div class="footnotes">
<ol>
<li id="fn:1">
<p>This means there will be omitted definitions/examples and probably a lot of things left as exercises <a href="#fnref:1" class="reversefootnote">↩</a></p>
</li>
<li id="fn:8">
<p>I should admit that most of my intuition comes from the rings under consideration are (finitely generated) algebras over a field, and so some of the things I claim to be true may only be true in this case <a href="#fnref:8" class="reversefootnote">↩</a></p>
</li>
<li id="fn:2">
<p>The connection to polynomials comes from the fact that clasically people tend to work with finitely generated $k$-algebras, so your ring $R$ looks like $R=k[x_1,\dots,x_n]/I$. <a href="#fnref:2" class="reversefootnote">↩</a></p>
</li>
<li id="fn:3">
<p>Note that this is weird because $r$ doesn’t have a (nice) well-defined domain. The image of every $\mfp\in\spec R$ lies in a different ring. <a href="#fnref:3" class="reversefootnote">↩</a></p>
</li>
<li id="fn:4">
<p>Equivalently, on each of the generators of $I$ <a href="#fnref:4" class="reversefootnote">↩</a></p>
</li>
<li id="fn:5">
<p>I should mention that I think the term variety is usally reserved for irreducible sets (and general $V(I)$ may instead be called algebraic sets), but oh well. <a href="#fnref:5" class="reversefootnote">↩</a></p>
</li>
<li id="fn:6">
<p>Exercise: prove it <a href="#fnref:6" class="reversefootnote">↩</a></p>
</li>
<li id="fn:7">
<p>It should be possible to show that the nilradical is nilpotent, and then $A$ is the product of its localizations (each with a nilpotent maximal ideal). From this, it’s not hard to conclude that $A$ is artinian. <a href="#fnref:7" class="reversefootnote">↩</a></p>
</li>
</ol>
</div>This will serve as background for a post I wanna write on Dedekind domains. The idea here is to blaze through 1 some geometry so I can use words like “Artinian ring” and “Krull dimension” when discussing Dedekind domains 2. The plan is to talk about integeral extensions of rings (+ other stuff), spectra of rings, dimensions of rings, and then Artinian rings. In a sense, this might be more commutative algebra than (classic) algebraic geometry, but what’s the difference? Note that all rings in this post are commutative with unity (and all ring maps preserve the unit). This means there will be omitted definitions/examples and probably a lot of things left as exercises ↩ I should admit that most of my intuition comes from the rings under consideration are (finitely generated) algebras over a field, and so some of the things I claim to be true may only be true in this case ↩