What are our objects of study?
An elliptic surface $\pi:X\to C$ captures the notion of a “holomorphically varying family of elliptic curves.” The elements of this family are the fibers $X_c=\inv\pi(c)$ for $c\in C$. While almost all of these are smooth, there will be some finite number of singular fibers, fibers $X_c=\inv\pi(c)$ which are not smooth. Letting $c_1,\dots,c_n\in C$ denote the critical values (i.e. the points above which $X$ is not smooth), letting $\ast C=C\sm\bracks{c_1,\dots,c_n}$, and letting $\ast X=X\sm\bigcup_{i=1}^nX_{c_i}$, the restriction
\[\pi\vert_{\ast X}:\ast X\to\ast C\]is a topological fiber bundle. Thus, via path lifting, is comes equipped with a global monodromy map
\[\pi_1(\ast C, c)\to\Aut(\hom_1(X_c;\Z))\simeq\GL_2(\Z)\]for any fixed $c\in\ast C$. Before saying more about this, we want to look at its local counterpart. Fix any $c_i\in\bracks{c_1,\dots,c_n}$. Since $C$ is a smooth curve, we can find some neighborhood $U\subset C$ around $c_i$ such that $U\cong\Delta=\bracks{z\in\C:\abs z<1}$ with $c_i\in U$ corresponding to the origin $0\in\Delta$. Thus, locally $\pi:X\to C$ looks like an elliptic surface ^{3}
\[f:S\to\Delta\]with a single singular fiber, lying above $0\in\Delta$. Hence, $\ast\Delta=\Delta\sm\bracks0$ is homotopy equivalent to a circle $S^1$, so, fixing a nonzero basepoint $c\in\Delta$, its fundamental group is $\pi_1(\ast\Delta,c)\cong\Z$. As in the global case, the map $S\sm S_0\xto f\ast\Delta$ is a topological fiber bundle, so comes equipped with a local monodromy map
\[\Z\cong\pi_1(\ast\Delta,c)\to\Aut(\hom_1(X_c;\Z))\simeq\GL_2(\Z).\]This map is clearly determined by what it does to the generator $\gamma\in\pi_1(\ast\Delta,c)$ circling the origin once counterclockwise, and so, after choosing a basis for $\hom_1(X_c;\Z)$, is identifiable with a matrix $T\in\GL_2(\Z)$, called the local monodromy matrix. This matrix $T$ is not arbitrary. In fact, up to conjugation (i.e. change of basis), it belongs to one of “finitely many” ^{4} possible monodromy matrices. Furthermore, this local monodromy matrix $T$ actually determines the singular fiber $S_0=\inv f(0)$ up to isomorphism! Thus, elliptic fibrations are locally classified by a single invariant, the monodromy matrix.
With that very brief overview done, let’s get into the meet of things. Consider an elliptic surface $\pi:X\to C$ with critical values $c_1,\dots,c_n\in C$. As before, let $\ast C=C\sm\bracks{c_1,\dots,c_n}$ and let $\ast X=\inv\pi(\ast C)$.
Recall that $\ast X\xto\pi\ast C$ is a topological fiber bundle. The means that $L=\derpush1\pi\Z_X$ is a locally constant sheaf ^{5} on $\ast C$ with stalks/fibers
\[L_c=\hom^1(X_c;\Z)\simeq\Z\oplus\Z.\]As mentioned earlier, this situation gives rise to a global monodromy action $\pi_1(\ast C)\to\GL_2(\Z)$.
The first lecture of Miranda’s Basic Theory of Elliptic Surfaces is a good reference for this. ↩
For this, I am using Compact Complex Surfaces by Barth et al. ↩
We changed notation to make the global and local cases more visually distinguishable. You should think $S=\inv\pi(U)$, $\Delta=U$, $U\subset C$ was taken small enough that $c_j\not\in U$ if $j\neq i$, and $f=\pi\vert_S$. ↩
Two infinite families, and six exceptional matrices ↩
Topologically, $\pi:\ast X\to\ast C$ locally looks like the natural projection $p:T\by U\to U$ where $T\cong(S^1)^2$ is a torus. On these trivial pieces. $\derpush1p\Z_X$ is the sheafification of the presheaf $U\supset V\mapsto\hom^1(\inv p(V);\Z)\simeq\hom^1(T;\Z)$ so $\derpush1p\Z_X$ is the constant sheaf $\hom^1(T;\Z)_ U$. Thus, $\derpush1\pi\Z_X$ is locally constant with stalks isomorphic to $\hom^1(T;\Z)$. ↩
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 ^{1}. 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 ^{2} to make this post be a good one ^{3}. 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 ^{4} 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” ^{5}. $ \newcommand{\CW}{\mathrm{CW}} \newcommand{\Top}{\mathrm{Top}} \newcommand{\Set}{\mathrm{Set}} \newcommand{\Grp}{\mathrm{Grp}} \DeclareMathOperator{\Ho}{Ho} $
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. ^{6} In particular, base spaces will always admit partitions of unity.
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 the post on Brown Representability. 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 $X\to BS^1\simeq\CP^\infty$ ^{7}. 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 $n$ ^{8}.
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.
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.
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.
Our first major goal is to show that
\[\ast\hom(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]\text{ where }c_i\in\hom^{2i}(BU(n);\Z),\]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.
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 ^{9}. 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
\[U_i\cap U_j\xto{\tau_{ij}}U(n+1)\into\GL_{n+1}\C\onto\PGL_n\C.\]This space $\P(E)$, called projectivization of $E$ 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 tautological subbundle
\[L_E:=\bracks{(\l, v)\in\P(E)\by E:v\in\l}\subset\pull\pi E\]whose fiber above a point $\l\in\P(E)$ is the line $\l\subset E_{\pi(\l)}$ it corresponds to. The tautological quotient bundle $Q_E$ on $\P(E)$ is defined by its place in the following exact sequence:
\[0\too L_E\too\pull\pi E\too Q_E\too0.\]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 ^{10}. 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.
The above lemma justifies our earlier claim that $\pull\pi E\simeq L_E\oplus Q_E$.
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
\[0\too L_E\too\pull\pi E\too Q_E\too0\]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 ^{11}, 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.
Returning to our situation of interest, we have a projective bundle $\P(E)\xto\pi B$ with tautological exact sequence
\[0\too L_E\too\pull\pi E\too Q_E\too0.\]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 ^{12}, there must exist some polynomial $p(x)\in\ast\hom(B;\Z)[x]$, say (recall $n+1=\rank E$)
\[p(x)=x^{n+1}+c_1(E)x^n+\dots+c_n(E)x+c_{n+1}(E)\]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 ^{13}. 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.
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
\[\ast\hom(BU(n-1)\by BU(1);\Z)\simeq\ast\hom(BU(n);\Z)[x]/(p_n(x))\]where $\deg x=2$ (i.e $x\in\hom^2(BU(n-1)\by BU(1);\Z)$) and
\[p_n(x)=x^n+c_1(EU(n))x^{n-1}+\dots+c_{n-1}(EU(n))x+c_n(E)\in\ast\hom(BU(n);\Z)[x]\]is some polynomial ^{14}. Our goal this section is to prove the following.
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
\[\Z[c_1,c_1']\simeq\ast\hom(BU(1)\by BU(1);\Z)\simeq\ast\hom(BU(2);\Z)[x]\left./\parens{x^2+c_1(EU(2))x+c_2(EU(2))}\right.\]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 ^{15}. 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,
\[x^2+c_1(EU(2))x+c_2(EU(2))=(x+c_1)(x+c_1')=x^2+(c_1+c_2')x+c_1c_1'\]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 ^{16}, $\Z[c_1(EU(2)),c_2(EU(2))]$ 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
\[\Z[c_1,c_1']\simeq\ast\hom(BU(2);\Z)\otimes\ast\hom(\P^1;\Z)\]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)
\[\begin{align*} k+1 &=\rank_\Z\Z[c_1,c_1']_ {2k}\\ &=\rank_\Z\parens{\ast\hom(BU(2);\Z)\otimes\ast\hom(\P^1;\Z)}_ {2k}\\ &=\rank_\Z\hom^{2k}(BU(2);\Z)+\rank_\Z\hom^{2k-2}(BU(2);\Z)\\ &\ge\rank_\Z\Z[c_1(EU(2)),c_2(EU(2))]_{2k}+\rank_\Z\Z[c_1(EU(2)),c_2(EU(2))]_{2k-2}\\ &=\parens{\floor{\frac k2}+1}+\parens{\floor{\frac{k-1}2}+1}\\ &=k+1. \end{align*}\]This shows that $\Z[c_1(EU(2)),c_2(EU(2))]_ n=\hom^n(BU(2);\Z)$ for all $n$, so
\[\ast\hom(BU(2);\Z)\simeq\Z[c_1,c_2]\]with $c_1=c_1(EU(2))$ and $c_2=c_2(EU(2))$ as desired!
We now handle the general case $n\ge2$. Based on how the $n=2$ case played out, let’s update our goal.
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.
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, recall ^{17} that $BU(1)\simeq\CP^\infty$ also represents second integral cohomology. That is, we have natural isomorphisms (of functors)
\[\hom^2(-;\Z)\simeq[-,\CP^\infty]\simeq B(-)\]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 ^{18})
\[m_X:\hom^2(X;\Z)\by\hom^2(X;\Z)\to\hom^2(X;\Z)\]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
\[m:[-,\CP^\infty]\by[-,\CP^\infty]\to[-,\CP^\infty]\,\text{ and }\, i:[-,\CP^\infty]\to[-,\CP^\infty]\]giving a (functorial) group structure to $[X,\CP^\infty]$ for all $X\in\Ho(\push\CW)$ ^{19} ^{20}. By Yoneda, these must arise from some morphisms (i.e. homotopy classes of continuous maps)
\[m:\CP^\infty\by\CP^\infty\to\CP^\infty\,\text{ and }\, i:\CP^\infty\to\CP^\infty\]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
\[m':\CP^\infty\by\CP^\infty\to\CP^\infty\,\text{ and }\, i':\CP^\infty\to\CP^\infty\]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 $\Ho(\push\CW)$.
Recall that $\CP^\infty\simeq K(\Z,2)$ ^{21}, and so a multiplication map $\mu:\CP^\infty\by\CP^\infty\to\CP^\infty$ corresponds to some element of
\[\hom^2(\CP^\infty\by\CP^\infty;\Z)\simeq\hom^2(\CP^\infty;\Z)\oplus\hom^2(\CP^\infty;\Z)\simeq\Z c_1\oplus\Z c_1'.\]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
\[\mu\circ(e\by\Id)=\mu\circ(\Id\by e)=\Id:\CP^\infty\to\CP^\infty.\]In terms of cohomology, these equalities say exactly that $b=a=1$ ^{22}, 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.
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.
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.
I claimed the Euler class is naturally defined for sphere bundles, so I should at least say what a sphere bundle is.
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).
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 ^{23}. 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.
This concludes our prelims on sphere bundles, so let’s actually define the Euler class now.
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.
We’ll construct the Euler class via an exercise in my spectral sequence post. 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.
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
\[\begin{CD} S^n @>>> \pull fE @>>> B'\\ @| @VVgV @VVfV \\ S^n @>>> E @>>> B \end{CD}\]of sphere bundles, one obtains a commutative square.
\[\begin{CD} \hom^0(B';\Z) @>\d>> \hom^{n+1}(B';\Z)\\ @A\pull fAA @AA\pull fA \\ \hom^0(B;\Z) @>>\d> \hom^{n+1}(B;\Z) \end{CD}\]The left vertical map sends $1\mapsto1$, and so by commutativity, we see that
\[\pull fe(E)=\pull f\d(1)=\d\pull f(1)=\d(1)=e(\pull fE),\]which says exactly that the Euler class is functorial. ^{24}
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 $u\in\hom^{n+1}(D(E),E;\Z)$ ^{25}.
At this point, we will take some things on faith to avoid repeating ourselves ^{26}. 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 $u=u(E)\in\hom^{n+1}(D(E),E;\Z)$ ^{27}. This Thom class is constructed so that the map ^{28}
\[\mapdesc{\phi}{\hom^p(B;\Z)}{\hom^{p+n+1}(D(E),E;\Z)}{\alpha}{\pull\pi(\alpha)\smile u}\]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.
\[\begin{CD} \cdots @>>> \hom^p(B;\Z) @>\pull\pi>> \hom^p(E;\Z) @>>> \hom^{p-n}(B;\Z) @>\smile e(E)>> \hom^{p+1}(B;\Z) @>>> \cdots\\ @. @V\pull pVV @V\Id VV @VV\phi V @VV\pull pV\\ \cdots @>>> \hom^p(D(E);\Z) @>>> \hom^p(E;\Z) @>>> \hom^{p+1}(D(E),E;\Z) @>>> \hom^{p+1}(D(E);\Z) @>>> \cdots \end{CD}\]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.
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 Euler class of the complex vector bundle $E\to B$ as
\[e(E):=e(S(E_\R))\in\hom^{2n}(B;\Z).\]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
\[e\in\hom^{2n}(BU(n);\Z)\simeq\Z[c_1,c_2,\dots,c_n]_ {2n}\,\,\,\,\text{ where }c_i\in\hom^{2i}(BU(n);\Z).\]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.
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.
\[\begin{CD} L_1\oplus L_2 @>>> L_1\by L_2\\ @V\pi_1\oplus\pi_2VV @VV\pi_1\by\pi_2V\\ B @>>\Delta> B\by B \end{CD}\]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
\[e(L_1\oplus L_2)=\pull\Delta e(L_1\by L_2)=e(L_1)\smile e(L_2)\in\hom^4(B;\Z).\]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.
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 $k$ ^{29}. 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
\[S^1\to S^3\to S^2.\]The Euler class of this bundle comes from the differential on the $E_2$-page of its Serre spectral sequence. That page looks like
\[\begin{array}{c | c c c c} \tbf q & \\ 1 & \hom^0(S^2;\hom^1(S^1))\simeq\Z a & 0 & \hom^2(S^2;\hom^1(S^1))\simeq\Z c_1a\\ 0 & \hom^0(S^2;\hom^0(S^1))\simeq\Z & 0 & \hom^2(S^2;\hom^0(S^1))\simeq\Z c_1 \\\hline & 0 & 1 & 2 & \tbf p \end{array}\]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 $e(E)=\pm c_1(E)$.
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 ^{30}.
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 ^{31}
\[e(E)=c_n(E).\]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.
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 ^{32} 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 primary obstruction class since it is the first one which can be nonzero in the situation of constructing a section of a sphere bundle ^{33}.
This is the most exciting part of this whole post. The Euler class. can. be. used. to. count.
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 $0$.
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.
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.
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 $\push\iota[N]\in\hom_k(M;\Z)$.
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 $N$)
\[0\too TN\too TM\vert_N\too N_{N/M}\too0,\]where the rank $(n-k)$ vector bundle $N_{N/M}$ is by definition the normal bundle of $N$ in $M$. 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
\[u_N:=u(N_{N/M})\in\hom^{n-k}(D(N_{N/M},S(N_{N/M});\Z)\simeq\hom^{n-k}(U,U\sm N;\Z).\]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
\[\hom^{n-k}(D(N_{N/M}),S(N_{N/M});\Z)\iso\hom^{n-k}(U,U\sm N;\Z)\iso\hom^{n-k}(M,M\sm N;\Z)\to\hom^{n-k}(M;\Z).\]This $u_N^M$ is the Poincare dual of $N$.
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.
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!
Let’s see some of this in action.
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 $X$?
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
\[S=\bracks{(v,p):v\in p}\subset\C^4\by G\to G\]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 $X$!
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 $X$!
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
\[\Sym^3\dual E=\Sym^3(\dual L_1\oplus\dual L_2)=(\dual L_1)^{\otimes 3}\oplus((\dual L_1)^{\otimes2}\otimes\dual L_2)\oplus(\dual L_1\otimes(\dual L_2)^{\otimes2})\oplus(\dual L_2)^{\otimes3}.\]Thus, taking the total Chern class of both sides, we see that
\[\begin{align*} c(\Sym^3\dual E) &=(1-3x_1)(1-2x_1-x_2)(1-x_1-2x_2)(1-3x_2)\\ &=1 - 6(x_1+x_2) + (11x_1^2+32x_1x_2+11x_2^2) - 6(x_1^3 + 8x_1^2x_2 + 8x_1x_2^2 + x_2^3) + x_1x_2(18x^2 + 45x_1x_2 + 18x_2^2)\\ &= 1 - 6c_1(E) + (11c_1(E)^2 + 10c_2(E)) - 6(c_1(E)^3 + 5c_1(E)c_2(E)) + c_2(E)(18c_1(E)^2 + 9c_2(E)) \end{align*}\]Thus, for any rank $2$ complex vector bundle $E\to B$, we have
\[c_4(\Sym^3\dual E)=c_2(E)(18c_1(E)^2 + 9c_2(E)).\]Um, now we switch gears and cheat. I actually don’t know a quick and easy way to calculate the above cup products ^{34}, 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 any of its sections in order to determine the number of lines on $X$.
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
\[X:x^3+y^3+z^3+w^3=0.\]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
\[(az+bw)^3+(cz+dw)^3+z^3+w^3=0\]as polynomials in $\C[z,w]$. Equating coefficients, this means that we need
\[\begin{align*} a^3 + c^3 &= -1\\ b^3 + d^3 &= -1\\ a^2b &= -c^2d\\ ab^2 &= -cd^2 \end{align*}\]If $a,b,c,d$ are all non-zero, then
\[a^3=(a^2b)^2/(ab^2)=-(c^2d)^2/(cd^2)=-c^3\implies0=a^3+c^3=-1,\]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
\[x=\zeta_3^iw\,\,\,\,\text{ and }\,\,\,\,y=\zeta_3^jz\]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.
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. ↩
No promises ↩
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. ↩
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. ↩
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 ↩
Some things I say may be false without this. For example, I think you need this for fiber bundles to be fibrations. ↩
“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 $\hom^2(\CP^\infty;\Z)$. ↩
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 ↩
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 $(x,\tau_{ij}(x)v)\in\inv p(U_j)\simeq U_j\by\C^{n+1}$. ↩
$\pull\pi E\vert_{\P(E)_ b}\simeq\C^{n+1}\by\P(E)_ b$ 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 $\pull\pi E\vert_{E_b}\simeq\pull qE_b\simeq\C^{n+1}\by\P(E)_ b$. ↩
It is $\ints{\P^n}(-1)$ if you are familiar with this notation. ↩
I really should have started with a rank $n$ vector bundle. ↩
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. ↩
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 $c_1(EU(n))$. ↩
$c_1$ corresponds to the first Chern class on the first factor, and $c_1’$ corresponds to the first Chern class on the second factor. ↩
I think that’s how people usually prove algebraic independence of symmetric polynomials ↩
This is not the simplest way to do this, but more practice with Yoneda-type reasoning is never bad ↩
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. ↩
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. ↩
Also, I should have also include an “identity natural transformation” $e:\bracks{0}\to\hom^2(-;\Z)$, but oh well. ↩
$\CP^\infty$ is like unironically the best topological space; fight me ↩
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 $e\by\Id:\CP^\infty\to\CP^\infty\by\CP^\infty$. ↩
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. ↩
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 $E’\simeq\pull fE$. ↩
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 ↩
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 ↩
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. ↩
You may want a relative version on Leray-Hirsch to prove this ↩
At this point, I don’t even think we know that $k\neq 0$. ↩
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. ↩
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 ↩
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 ↩
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. ↩
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). ↩
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 \spec and sheaf 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 ^{3}, and then focus on curves ^{4}.
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 ^{5}.
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.
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
\[\smooth_{\R^n}(U)=\bracks{\text{smooth functions }f:U\to\R}\]on $\R^n$ is indeed a sheaf, and hence the pair $(\R^n,\smooth_{\R^n})$ form a so-called
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.
Hence, we have a choice in our definition of a morphism $(X,\ints X)\to(Y,\ints Y)$ of ringed space. It could either ^{8} 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.
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.
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.
First, to get this out of the way, a
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 $\mfm_{f(x)}\subset\ints{Y,f(x)}$).
Now, schemes are basically algebraic manifolds, except the phrase “algebraic manifold” already refers to something else. 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 $\ints A(\spec A)=A$ ^{9}. 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,
One can give other, possibly more concrete, constructions of $\ints A$.
Any locally ringed space isomorphic to the pair $(\spec A,\ints A)$ is called an affine scheme. In general, a scheme 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.
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 $I\subset A$.
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 $S$-points of $X$. For example, if $X=\spec\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)$ and $S=\spec A$, then
\[X(A)=\bracks{\spec A\to\spec\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)}=\bracks{\Z[x_1,\dots,x_n]/(f_1,\dots,f_m)\to A}=\bracks{a_1,\dots,a_n\in A:f_i(a_1,\dots,a_n)=0\,\forall i}\]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).
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.
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.
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.
It will be useful to prove a universal property for maps to affine space, so let’s do that.
Our main application of this will be in the case of the following space (possibly only when $n=1$).
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 kernel,cokernel,image to be the sheafifications of the following presheaves
\[(\pker f)(U)=\ker(\msF(U)\to\msG(U)).\] \[(\pim f)(U)=\im(\msF(U)\to\msG(U)).\] \[(\pcoker f)(U)=\coker(\msF(U)\to\msG(U)).\]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 exact 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 $U\subset X$.
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 global sections functor $\Gamma(\msF)=\msF(X)$.
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 ^{10}. 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 ^{11}. We will see later ^{12} 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
\[0\too\ms A\too\ms B\too\ms C\too0\]of sheaves on $X$, we get a long exact sequence
\[0\too\hom^0(\ms A)\too\hom^0(\ms B)\too\hom^0(\ms C)\too\hom^1(\ms A)\too\hom^1(\ms B)\too\hom^1(\ms C)\too\dots\]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,
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.
Before moving one, we’ll make one quick definition.
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
\[\msE_p(U)=\bracks{s:U\to E:p\circ s=1_U}\]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 $E$.
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.
Given a vector bundle $\ms E$ on $X$ and a point $x\in X$, the fiber of $\ms E$ above $x$ is the vector space
\[\ms E(x):=\ms E_x\otimes\kappa(x)=\ms E_x\otimes\ints{X,x}/\mfm_x=\ms E_x/\mfm_x\ms E_x\]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
\[\ms E(X)\too\msE_x\too\ms E(x).\]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.
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
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.
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 $\DeclareMathOperator{\Set}{Set}P_n:\Sch_k\to\Set$
\[P_n(S)=\bracks{\parens{\msL,(s_0,\dots,s_n)}:s_0,\dots,s_n\in\Gamma(\msL)\text{ have no common zeros}}\]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
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 $\P^n$.
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,
\[D_+(x_i)\simeq\spec k\sqbracks{\frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i}}\simeq\A^n_k\]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 $x\leftrightarrow\inv y$.
We know can move away from the abstract generalities a bit and focus on something a little more down-to-earth.
One of the nice things about smooth curves is that their local rings are about as nice as you could ask for.
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 function field of $C$.
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.
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 effective if $D\ge0$.
Fix a smooth curve $C$. Let $k(C)_ C$ denote the constant sheaf with stalks equal to $k(C)$. ^{13} 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
\[\ints C(D)(U)=\bracks{f\in k(C):v_p(f)+v_p(D)\ge0\,\forall p\in U}.\]This is a line bundle (exercise ^{14}) 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 Divisor class group of $C$. We’ve shown that $\Cl C\into\Pic C$, and we’ve claimed this is actually an isomorphism.
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.
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 residue field at $x$ is the field
\[\kappa(x):=\ints{X,x}/\mfm_x\]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
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 $C\to\P^1$.
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).
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
\[\begin{matrix} \push f:& \Div X &\too& \Div Y && \pull f:&\Div Y &\too& \Div X\\ &[p] &\longmapsto& [\kappa(p):\kappa(f(p))]\cdot[f(p)] &&& [q] &\longmapsto& \sum_{p\to q}e(p/q)\cdot[p] \end{matrix}\]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
\[\inv f(\eta_Y)=\inv f\parens{\bigcap_{W\ni\eta_Y}W}=\bigcap_{W\ni\eta_Y}\inv f(W)\ni\eta_X\]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.
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 $k$.
For an affine open $\spec A\subset C$, let $\Omega_{A/k}$ denote the module of (Kahler) differentials which is the $A$-module generated by the symbols $\d a$ for all $a\in A$ subject to the relations
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 ^{15}, we can glue together the various sheaves $\wt{\Omega_{A/k}}$ to get a global sheaf $\omega_C=\Omega_{C/k}$ of differentials on $C$.
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 genus of $C$ to be $h^0(\omega_C):=\dim_k\hom^0(\omega_C)$ the dimension of global sections of $\omega_C$.
Now, one of the main utilities of differentials in algebraic geometry is their appearance in the following surprising and very useful theorem.
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.
Now, given Serre duality ^{16}, 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 $\msF$ ^{17}. The second part is a formula for $\chi(\ints X)$. We will begin, perhaps unsurprisingly, with the first part.
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 ^{18}, 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 complete if the image of any morphism $C\to\P_k^n$ is closed.
The correct notions of “algebraically compact” and “algebraically Hausdorff” are being proper and separated, but for out purposes, completeness as just defined suffices ^{19}. From now on ^{20}, assume all curves are complete. With this assumption made
Thus, $\chi(\ints C)=1-g$. In summary, the takeaways are
\[\begin{align*} \chi(\ints C(D))=\deg D+1-g && h^1(\ints C(D))=h^0(\omega_C(-D)) \end{align*}\]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.
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).
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 $k$.
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.
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
\[h^0(\ints E(D_n))=h^0(\ints E(-D_n))-h^0(\ints E(D_n))=\deg D_n+\chi(\ints E)=n\]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
\[v_p(c_0+c_2X+c_3Y+c_4X^2+c_5XY)=\min\bracks{i:c_i\neq0}\le5\implies c_0+c_2X+c_3Y+c_4X^2+c_5XY\neq0.\]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 ^{21}
\[aY^2+a_1XY+a_3Y=bX^3+a_2X^2+a_4X+a_6.\]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
\[a^3Y^2+a_1a^2XY+a_3aY=a^3X^3+a_2a^2X^2+a_4aX+a_6.\]Finally, dividing by $a^3$, we see that our elliptic curve $E$ satisfies a relation of the form
\[\label{wform}\begin{align}Y^2+a_1XY+a_3Y=X^3+a_2X^2+a_4X+a_6.\end{align}\]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 $(Y-(a_2X+a_3)/2)/216$ ^{22} to get a relation of the form $Y^2=X^3+aX+b$.
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
\[0\too\ints E(-p)\too\ints E\too\ints p\too0.\]Twist by $D_3$ (i.e. tensor with $\ints E(D_3)$) and look in cohomology to get
\[0\too\hom^0(\ints E(D_3-p))\too\hom^0(\ints E(D_3))\too\kappa(p)\too\hom^1(\ints E(D_3-p))=0,\]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)
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.
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
\[\mapdesc f{E(k)}{\Pic^0E}{p}{[p]-[\infty]}.\]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
\[0\too\ints E(p-q)\too\ints E(p)\too\ints q\too0\]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 $p\neq q\in E$.
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.
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 ^{23}. We will use without proof the existence of the morphism $m$.
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 $k(E)/\pull\phi k(E’)$.
Unless otherwise stated, assume that basically all isogenies below are nonzero.
One nice properties of isogenies is that they are automatically group homomorphisms.
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.
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
\[(\wh\phi\circ\phi)\circ\wh\phi=\wh\phi\circ(\deg\phi)=(\deg\phi)\circ\wh\phi\]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
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.
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 algebraically closed 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.
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 ^{24}
\[\# E(k)[m]=\#\inv m(\infty)=m^2\]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 $G\simeq\zmod m\oplus\zmod m$.
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 ^{25}. This map visibly has degree $p$ and separable degree $1$, so we see that
\[\# E[p^e]=\deg_s[p^e]=\deg_s(\wh\phi\circ\phi)^e=(\deg_s\wh\phi)^e\in\bracks{1,p^e}\]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 $E[p^e]\simeq\zmod{p^e}$.
All in all, we have shown the following.
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 $G_\Q:=\Gal(\Qbar/\Q)$.
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 $\bar E/\Qbar$ ^{26}, so the results of the previous (sub)section apply to $E(\Qbar)$. In particular, fixing a prime $\l$, for all $n\ge1$, we have
\[E(\Qbar)[\l^n]=\Zmod{\l^n}\oplus\Zmod{\l^n}.\]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
\[\spec\Qbar\xto{\spec\sigma}\spec\Qbar\xto xE.\]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
\[G_\Q\too\Aut(E(\Qbar)[\l^n])\simeq\Aut\parens{\Zmod{\l^n}\oplus\Zmod{\l^n}}\]for all $n\ge1$. Letting, $T_\l(E):=\invlim_{n\ge1}E(\Qbar)[\l^n]$ be the ($\l$-adic) Tate module and taking an inverse limit of the above maps, we get our desired representation
\[\rho_{E,\l}:G_\Q\too\Aut(T_\l(E))\simeq\Aut(\Z_\l\oplus\Z_\l)\simeq\GL_2(\Z_\l)\into\GL_2(\Q_\l)\]of $G_\Q$.
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.
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) ↩
I’ll include exercise. I encourage you to do some (but probably not all) of them just to have practice thinking about this stuff. ↩
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 ↩
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 ↩
collection of charts ↩
All of this data is still there, but now neatly packaged into the structure sheaf $\ints X$ ↩
In either case, the morphism of sheaves is not required to have much of anything to do with the map on underlying spaces ↩
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 $A=k[x_1,\dots,x_n]/\sqrt I$. ↩
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. ↩
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 ↩
Really, I’ll claim without proof ↩
I know this notation is trash, but it’s also short-lived ↩
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))) ↩
You may also want to appeal to Nike’s trick: 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 $b\in B$) ↩
Plus all the other stuff I’ve asked you to take for granted ↩
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) ↩
This boils down to the fact that the unit ideal $(1)=A$ of a ring $A$ has a finite set of generators. ↩
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. ↩
and also retroactively for some of the earlier stuff (e.g. Serre duality) ↩
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 ↩
I think these are the right substitutions. I haven’t actually checked ↩
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 $E\to\spec k\from S$.
Actually, I think there’s a slicker way to do this. Maybe I’ll come back and type something up at some point… ↩
The middle expression here really shows that I made some poor choices of notation ↩
I did not touch on this before, but morphisms between smooth curves are in bijection field maps between their function fields. ↩
$\bar E$ is the “base change of $E$ to $\Qbar$”. Intuitively, it results from taking $E/\Q$ and then extending scalars from $\Q$ to $\Qbar$. ↩
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 ^{2} 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 ^{3}. 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 ^{4}. Hence, unless stated otherwise, assume all spaces below are based, connected CW complexes. ^{5}
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
\[\mapdesc T{[X,S^1]}{\hom^1(X;\Z)}{[f]}{\pull f\alpha}\]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.
We first show that $T$ is injective.
To show that $T$ is surjective, we’ll first need a simple lemma.
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 $X$.
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
\[\Hom(G,G)=\Hom(\pi_n(K(G,n)),G)\simeq\Hom(\hom_n(K(G,n)),G)\simeq\hom^n(K(G,n);G)\]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 $n=1$).
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 $h_X$?
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…
That’s all.
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
\[\mapdesc {T(A)}{[A,X]}{F(A)}f{\pull fx}\]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.
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 $\pi_n(X)=[S^n,X]=F(S^n)$.
Fix a contravariant functor $F:\push\CW\to\Set$ satisfying the wedge and Mayer-Vietoris axioms.
This may seem like a strong condition, but we’ll see that universal pairs are relatively easy to come by.
Whew. That was the brunt of the argument. We now show that maps to universal pairs can always be extended.
Now, we prove Brown Representability.
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.
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 CW Approximation of $X$. There are a couple things worth noting.
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
\[\pi_n(Z)=[S^n,Z]\simeq[S^n,X]=\pi_n(X)\]with middle isomorphism coming from compositing with $f$.
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
\[[-,Z(X)]\simeq[-,X]\xto{\push g}[-,Y]\simeq[-,Z(Y)].\]Hence, by Yoneda (or by looking at the image of the identity in $[Z(X),Z(X)]$), $g$ must correspond to some map $Z(g):Z(X)\to Z(Y)$.
At last, our earlier claim that the homotopy category of (based, connected) topological spaces was equivalent to that of CW complexes has some justification ^{6}.
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
\[\hom^n(X;G)\to\hom^n(U;G)\oplus\hom^n(V;G)\to\hom^n(U\cap V;G)\]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 $K$.
To see that $K$ is an Eilenberg-Maclane space, we simply observe that
\[\pi_k(K)=[S^k,K]=\hom^n(S^k;G)=\twocases G{n=k}0.\]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 $\Omega$-spectrum $K_n$ - 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.
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
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.
\[h(p)=(u,g)\implies h(p\cdot g')=(u,gg').\]The main motivation for studying these (at least, as far as I’m concerned) comes from the following theorem which I will not prove.
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 ^{7} which assigns to a space $X$, the set $B(X)$ of its principal $G$-bundles up to their natural notion of equivalence ^{8}.
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.
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 $\bigvee B_\alpha$ ^{9}. 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.
As a consequence of the paragraph above the exercise, we now know that our classifying spaces $BG$ 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?
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 delooping of $G$.
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
\[G\to EG\to BG\]which induces a long exact sequence in homotopy
\[\cdots\too\pi_{n+1}(BG)\to\pi_n(G)\to\pi_n(EG)\to\pi_n(BG)\to\pi_{n-1}(G)\to\cdots.\]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 $\Omega BG$.
We now show a converse to some of what we’ve seen.
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.
Given a complex rank $n$ vector bundle $E\to B$, one would initially expect its structure group to be $\GL_n(\C)$ ^{10}, 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 ^{11}. 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
\[S^1\to S^{2n+1}\to\CP^n\]for all $n$. These sequences are compatible in the relevant sense, so we can take limit as $n\to\infty$ ^{12} to get a bundle
\[S^1\to S^\infty\to\CP^\infty.\]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 $f:B\to\CP^\infty$ ^{13}. However, we can say even more. Not only is $\CP^\infty$ a $BS^1$, it is also $K(\Z,2)$ ^{14} so we have isomorphisms
\[B(X)\simeq[X,\CP^\infty]\simeq\hom^2(X;\Z).\]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 (first) Chern class.
Technically, it doesn’t quite concern functors on the category Top of topological spaces, but I’ll say more about this later. ↩
maps inducing an isomorphism on all homotopy groups ↩
Of course, this latter point about CW approximation is not logically necessary to start and prove Brown representability for CW complexes ↩
Really, I should call this something like $\Ho(\push\CW)$, but who’s got the time to type all of that? ↩
Certainly assume all spaces are based no matter what ↩
Knowing this was not logically necessary for anything we’ve done ↩
This is a functor because the (categorical) pullback of a bundle is a bundle ↩
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. ↩
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 ↩
I think I even claimed this before ↩
The point is that Gram-Schmidt gives a deformation retraction $\GL_n(\C)\to U(n)$ which then induces a homotopy equivalence $B\GL_n(\C)\simeq BU(n)$. ↩
Form the direct limits $S^\infty=\dirlim S^n=\dirlim S^{2n+1}$ and $\CP^\infty=\dirlim\CP^n$ ↩
One can show that the line bundle associated to the bundle $S^\infty\to\CP^\infty$ is the so-called tautological line bundle 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. ↩
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). ↩
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.
Specifically, I’ll speak abstractly about two common sources of spectral sequence ^{2}. Because I really wanna talk about this material ^{3}, 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$\dots$ ^{4}
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.
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.
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
For the next step, we need to know how to define kernels and cokernels.
Cokernels are define similarly as being initial with respect to maps $\nu:Y\to N$ such that $\nu\circ f=0$. Now,
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 $a=b$ ^{5}. We would like every monomorphism to arise in this way. Similarly, cokernels are epimorphisms, and we would like them to be the only epimorphisms ^{6}.
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.
Fix an additive category $\msA$. A chain complex ^{7} $A_\bullet$ is a sequence
\[A_\bullet:\cdots\too A_1\xtoo{d_1} A_0\xtoo{d_0} A_{-1}\too\cdots\]of morphisms in $\msA$ such that $d_n\circ d_{n+1}=0$ for all $n$. A chain map $f:A_\bullet\to B_\bullet$ is a commutative diagram
\[\begin{CD} \cdots @>>> A_1 @>d_1>> A_0 @>d_0>> A_{-1} @>>> \cdots\\ &@Vf_{-1}VV @Vf_0VV @Vf_1VV\\ \cdots @>>> B_1 @>d_1>> B_0 @>d_0>> B_{-1} @>>> \cdots \end{CD}\]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.
Before going on, I should maybe also say what homology is.
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 $\msA$. ^{8}
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.
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
\[F^p\hom^n(A^\bullet)=\bracks{\alpha\in\hom^n(A^\bullet)\mid\exists x\in F^pA^n:\alpha=[x]},\]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
\[0\too F^{p+1}\pull\hom(A^\bullet)\too F^p\pull\hom(A^\bullet)\too G^p\pull\hom(A^\bullet)\too0.\]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.”
We start by denoting the associated graded complex by
\[E_0^{p,q}:=G^pA^{p+q}\text{ }\text{ with differential }\text{ }d_0^{p,q}:E^{p,q}_ 0\to E_0^{p,q+1}.\]Denote the cohomology of this complex by
\[E_1^{p,q}:=\hom^{p+q}(G^pA^\bullet)=\frac{\ker\big({d_0^{p,q}:E_0^{p,q}\to E_0^{p,q+1}}\big)}{\im\parens{d_0^{p,q-1}:E_0^{p,q-1}\to E_0^{p,q}}}=\frac{\ker\big({d_0^{p,q}:G^pA^{p+q}\to G^pA^{p+q+1}}\big)}{\im\parens{d_0^{p,q-1}:G^pA^{p+q-1}\to G^pA^{p+q}}},\]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
\[\mapdesc{d_1^{p,q}}{E_1^{p,q}}{E_1^{p+1,q}}{\alpha}{[dx]}\]One easily sees that $d_1^{p,q}\circ d_1^{p+1,q}=0$ ^{9}, and so we are justified in defining
\[E_2^{p,q}:=\hom^p(E_1^{\bullet, q})=\frac{\ker\big({d_1^{p,q}:E_1^{p,q}\to E_1^{p+1,q}\big)}}{\im\parens{d_1^{p-1,q}:E_1^{p-1,q}\to E_1^{p,q}}}.\]As a sanity check to make sure things make sense, try doing the following.
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. $dx\in F^{p+2}A^{p+q+1}$ ^{10}. This suggest we can get a map $d_2^{p,q}:E_2^{p,q}\to E_2^{p+2,q-1}$.
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
\[\begin{align*} Z_r^{p,q} &=\frac{F^pA^{p+q}\cap\inv d\parens{F^{p+r}A^{p+q+1}}+F^{p+1}A^{p+q}}{F^{p+1}A^{p+q}}\\ B_r^{p,q} &=\frac{F^pA^{p+q}\cap d\parens{F^{p-r+1}A^{p+q-1}}+F^{p+1}A^{p+q}}{F^{p+1}A^{p+q}} \end{align*}\]and let $E_r^{p,q}=Z_r^{p,q}/B_r^{p,q}$. On these objects, we can define a differential
\[\mapdesc{d_r^{p,q}}{E_r^{p,q}}{E_r^{p+r,q-r+1}}{[z]}{[dz]}\]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 ^{11}.
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 (cohomological) spectral sequence. For fixed $r$, the objects $E_r^{p,q}$ form the $r$th page (or $E_r$-page) 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.
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.
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 ^{12}
\[F^pC^n(X)=\bracks{\phi\in C^n(X):\phi\vert_{C_n(X^p)}=0}=\ker\parens{C^n(X)\to C^n(X^p)},\]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
\[\begin{CD} 0 @>>> F^{p+1}C^{p+q}(X) @>>> C^{p+q}(X) @>>> C^{p+q}(X^{p+1}) @>>> 0\\ &@VVV @VVV @VVV\\ 0 @>>> F^pC^{p+q}(X) @>>> C^{p+q}(X) @>>> C^{p+q}(X^p) @>>> 0 \end{CD}\]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
\[E_0^{p,q}=G^pC^{p+q}(X)=F^pC^{p+q}(X)/F^{p+1}C^{p+q}(X)\simeq\ker\parens{C^{p+q}(X^{p+1})\to C^{p+q}(X^p)}=C^{p+q}(X^{p+1},X^p)\]as claimed. Now, by definition, cohomology of this page gives relative cohomology, so $E_1^{p,q}=\hom^{p+q}(X^{p+1},X^p)$ ^{13}. 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
\[E^2_{p,q}=\twocases{\hom_{\text{cell}}^{p+1}(X)}{q=1}0.\]There are no more nontrivial differentials past this point, so $E_2^{p,q}=E_\infty^{p,q}$. Finally, since each diagonal ^{14} only contains one nonzero object, we conclude that $\hom^p(X)\simeq E_\infty^{p-1,1}\simeq\hom^p_{\mrm{cell}}(X)$ ^{15}, so singular and cellular cohomology agree.
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 ^{16} ^{17}.
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.
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 ^{18}. Before proving this, we’ll first need to prove a lemma (also named Hurewicz) elucidating the connection between $\pi_1$ and $H_1$ ^{19}.
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 $\hom_p(B;\hom_q(F))\implies\hom_{p+q}(E)$ ^{20}. This is what we’ll use in the below proof.
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 $\pi_4(S^2)\simeq\pi_4(S^3)\simeq\zmod2$. ^{21}
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
\[\begin{array}{c | c c c c} 6 & \Z a^3 & & & \Z a^3x\\ 4 & \Z a^2 & & & \Z a^2x\\ 2 & \Z a & & & \Z ax\\ 0 & \Z1 & & & \Z x \\\hline & 0 & 1 & 2 & 3 \end{array}\]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 ^{22}
\[d(a^2)=(da)a+a(da)=2ada=2ax\text{ and in general }d(a^q)=na^{q-1}da=qa^{q-1}x.\]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
\[\hom^k(X)=\Ext^1(\hom_{k-1}(X),\Z),\]and so ^{23} we get that
\[\hom_n(X)\simeq\twocases{\zmod k}{n=2k\text{ is even}}0.\]In particular, $\pi_4(S^2)\simeq\pi_4(S^3)\simeq\hom_4(X)\simeq\zmod2$.
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.
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.
\[\begin{matrix} d^2=0 && \del^2=0 && d\circ\del=\del\circ d \end{matrix}\]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 ^{25}.
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 ^{26}. This filtration is
\[F^p\Tot(A^{\bullet,\bullet})^n=\bigoplus_{\substack{n=i+j\\i\ge p}}A^{i,j}.\]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”
\[E_2^{p,q}\simeq\hom^p_d(\hom^q_\del(A^{\bullet,\bullet}))\]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
\[E_0^{p,q}=F^p\Tot(A^{\bullet,\bullet})^{p+q}/F^{p+1}\Tot(A^{\bullet,\bullet})^{p+q}=A^{p,q},\]with differential $d_0:E_0^{p,q}\to E_0^{p,q+1}$ equal to the vertical differential $\del$. Hence,
\[E_1^{p,q}=\hom^q_\del(A^{p,\bullet}).\]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
\[E_2^{p,q}=\hom^p_d\parens{\hom^q_\del\parens{A^{\bullet,\bullet}}}\]as claimed.
Replace “bird” with something less ill-defined if you want ↩
While writing this, I repeatedly felt the need to add examples/exercises, so things quickly became more hands on than I first planned ↩
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) ↩
Secretly, you can skip the first part on abelian categories. Immediately after writing it, I forgot that I wanted to work with an abstract category instead of just $R$-modules, and so had $R$-modules in mind as I wrote everything else (e.g. I talk about elements of objects of the category which is kind of a no-no in general). So um, there are a few option. You can just pretend we’re working with $R$-modules throughout (and everything will be fine especially since all the example computations are done with abelian groups), or if you want things to still work in general, you can use the (non-trivial) fact that every abelian category embeds in a category of $R$-modules (or maybe using Yoneda/functor-of-points type reasoning? I haven’t thought about this approach) ↩
Just apply the universal property to $\alpha:=\mu\circ a=\mu\circ b$. ↩
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) ↩
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. ↩
I’ll likely say chain when I mean to say cochain many times below ↩
From now on, I’ll start being a little more sloppy with my $d$’s, not always explitily giving them their upper indexing. ↩
There’s some dubious reasoning here, but I’m only trying to build intuition so that’s fine ↩
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. ↩
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. ↩
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. ↩
i.e. $\bracks{E_2^{p,q}:p+q=n}$ for some fixed $n$. ↩
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. ↩
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 $* \in X$. ↩
If any of you know a nice way to get footnotes working inside of html blocks, please tell me ↩
Secretly, non-simply connected spaces don’t exist. ↩
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. ↩
Also worth mentioning that the differential $d^r$ on the $r$th page of a homological spectral sequence has bidegree $(-r, r-1)$. ↩
By Freudenthal suspension, we actually get the stronger result that $\pi_{n+1}(S^n)\cong\zmod2$ for all $n\ge3$. ↩
$a$ and $da$ commute because $a$ lives in even degree. Remember that $ab=(-1)^{\deg(a)\deg(b)}ba$ in general when taking cup products ↩
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 ↩
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 ↩
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. ↩
Actually, two natural filtrations/spectral sequences, but I’ll only mention one of them ↩
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. ^{1}
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 ^{2}, but I won’t; instead, I’ll take an approach that’s more concrete. ^{3} With that said, let’s get started$\dots$
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 ^{4}.
This section is mostly definitions ^{5}, and so can be skipped and referred back to whenever you see something but don’t know what it means.
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 affine $n$-space (over $\bar k$).
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.
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 previous post on algebraic geometry might be helpful for understanding why some of the definitions here are reasonable. For example,
We now turn to projective algebraic sets and projective varities. Projective $n$-space (over $k$), 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,
\[\P^n=\bracks{\sqbracks{p_0:p_1:\dots:p_n}:p_i\in\bar k}\left/\parens{\sqbracks{p_0:p_1:\dots:p_n}\sim\sqbracks{\lambda p_0:\lambda p_1:\dots:\lambda p_n}\text{ for all }\lambda\in\units{\bar k}}\right..\]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.
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 $I$.
Note that there are many ways of embedding $\A^n\subset\P^n$. For example, each set
\[U_i=\bracks{\sqbracks{p_0:\dots:p_n}\in\P^n:p_i\neq0}\]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
\[(p_1,\dots,p_n)\mapsto\sqbracks{p_1,\dots,p_{i-1},1,p_i,\dots,p_n}.\]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
\[I(V\cap\A^n)=\bracks{f(Y_1,\dots,Y_{i-1},1,Y_{i+1},\dots,Y_n):f(X_0,\dots,X_n)\in I(C)}.\]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 dehomogenization with respect to $X_i$, and can be reversed by taking $f(Y)\in\bar k[Y]=\bar k[Y_1,\dots,Y_n]$ to \(\ast f(X_0,\dots,X_n)=X_i^{\deg f}f\parens{\frac{X_0}{X_i},\frac{X_1}{X_i},\dots,\frac{X_{i-1}}{X_i},\frac{X_{i+1}}{X_i},\dots,\frac{X_n}{X_i}},\) the homogenization of $f$ with respect to $X_i$.
This let’s us take an affine algebraic set $V\subset\A^n$ to its projective closure 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,
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. ^{6}
Phew. Alright, I think we can move on to the next section now. I hope you have all of this memorized.
An algebraic curve is a projective variety of dimension 1. For curves, $\bar k[C]$ is nicer than just an arbitrary integral domain.
Now, here’s some nice information about regular functions that we’ll just take for granted.
WIth this, let’s move onto differentials.
Because calculus is quite useful for doing analysis/geometry over $\R$ and $\C$, we’d like something similar for studying curves over any field $k$.
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 $\omega$ holomorphic (I don’t know what the standard term is) is $\ord_P(\omega)\ge0$ for all $P\in C$.
This is the point in this blog post where I may actually start bothering to prove things I claim. ^{7} 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:
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 effective. 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 $P$.
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 $\l(D)$ ^{8} is a certain 7-term exact sequence.
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 $\bar k$ ^{9}. This let’s us write a Taylor (or Laurent) expansion for $g$:
\[g=\sum_{n\in\Z}a_n(g)t^n\text{ }\text{ where }\text{ }a_n(g)\in\bar k\]and there exists some $m>0$ s.t. $n<-m\implies a_n(g)=0$. With this in mind, define
\[\res_P(\omega)=a_{-1}(g)\]modelled after residues in complex analysis ^{10}. 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. ^{11}
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.
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 second post 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. ↩
I’ve also said the word sheaf before, so I could really try to be fancy ↩
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. ↩
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. ↩
This section is basically just the first chapter of Silverman because I’m unoriginal. ↩
Think of meromorphic maps from complex analysis ↩
Could you imagine how long this post would be if I did this the whole time? ↩
Including proving Riemann-Roch ↩
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 p-adic post ↩
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). ↩
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 ↩
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.
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.
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.
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. ^{3}
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.
In our particular case, the above says that
\[\frac V{M'}\iso\parens{\frac V{M'}}\sqbracks{\frac1r}\oplus\parens{\bigoplus_{\mfq\mid(r)}\parens{\frac V{M'}}_ \mfq}\]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.
There we have it. Lattices over Dedekind domains can be modified one prime at a time.
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).
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,
\[R_\mfp:=\invlim_nR/\mfp^n\]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 $(x)$.
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. ^{4} Before stating this, it will be useful to define/prove a few things.
Our next definition will be that of the adele ring $\A_R=F\otimes_R\wh R$, 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.
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 projective if any of the following hold
When $R$ is a Dedekind domain (and $P$ is finitely generated), this is just a fancy way to say that $P$ is torsion-free ^{5}. Given a projective $R$-module $P$, its rank is $\dim_F(P\otimes_RF)$ where $F=\Frac(R)$ ^{6}.
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. ^{7}
Finally, the one use of our main lemma I know is proving the following:
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. ^{8} 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.
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.
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. ↩
This is like a lemma for a lemma. What do you call that? ↩
Assuming I’m not remembering things incorrectly, when $n=1$, this gives the Picard group of $R$. ↩
Exercise: prove this (hint: use the fact that $R$ is locally a PID and (finitely generated) torsion-free modules over PIDs are free) ↩
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) ↩
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. ↩
Something something the geometry of Dedekind domains is controlled by adeles something something? ↩
This might just be because there’s no point-set topology class at my school ^{3}, 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.
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? ↩
Which is really a shame ↩
I’ll start with a fact about localizations that I think I’ve used before on this blog, but never actually stated/proven.
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
\[\frac ms=\frac nt\iff\exists u\in S:u\cdot(t\cdot m-s\cdot n)=0,\]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
\[\sinv M\simeq M\otimes_R\sinv R,\]so localization is really just extension of scalars. Now, onto the fact.
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 $\sinv f(m/s)=f(m)/s$.
That’s one thing down. The second thing we’ll need is generalized Cayley-Hamilton.
Now we’re here. Before getting in Dedekind domains, let’s briefly discuss dvrs (which we’ll see are just local Dedekind domains).
Dvrs are very nice as detailed in the following theorem.
And finally, what is a Dedekind domain?
I’ll leave verifying these examples up to you. Our first lemma is that local Dedekind domains are dvrs.
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.
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.
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. ↩
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.
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.
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.
Another nice thing to know about the Noetherian property is that almost every ring/module you will ever care about is Noetherian.
I think that’s all the highlights about Noetherianess. Let’s get into integral extensions.
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.
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.
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 $R$.
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.
The idea behind the Zariski topology is that “zero sets of polynomials” should be closed ^{3}, 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 $\mfp\in\spec R$ ^{4}, 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 $I$ ^{5} are exactly those that contain $I$.
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. ^{6}
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 $\dim V(I)<\dim V(J)$.
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 $R=k[x_1,\dots,x_n]/I$.
This is a nice lemma. For example, we can use it to give another definition of dimension.
We also (finally) get a nice proof of the Nullstellensatz.
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.
We want to understand the structure of Artininan rings. We first observe that they only have finitely many maximal ideals.
This proposition is actually stronger than it may seem at first, since in fact all prime ideals of $A$ are maximal.
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 ^{7}. We only need the first version, but the second is also nice.
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 ^{8}.
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 ↩
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 $R=k[x_1,\dots,x_n]/I$. ↩
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. ↩
Equivalently, on each of the generators of $I$ ↩
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. ↩
Exercise: prove it ↩
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. ↩