Orbit-Stabilizer for Finite Group Representations

Niven Achenjang bio photo By Niven Achenjang Comment

One of my professors covered the main result of this post during a class that I missed awhile ago. Using some notes from a friend who attended that class, I want to try to reconstruct the theorem 1. Experience with representation theory will be useful for this post, but I’ll try to cover enough of the basics so that previous exposure isn’t strictly required.

Orbit-Stabalizer

We will begin by continuing our discussion of group actions from last post. Recall the definition

Let $G$ be a group and let $X$ be a set. A (left) group action of $G$ on $X$ is a map $\phi:G\times X\rightarrow X$ satisfying
  • $1\cdot x=x$ for all $x\in X$ where $1\in G$ is the identity
  • $g\cdot(h\cdot x)=(gh)\cdot x$ for all $x\in X$ and $g,h\in G$
where $g\cdot x$ denotes $\phi(g,x)$.

We sometimes write $G\curvearrowright X$ to denote that $G$ acts on $X$. If $X$ has additional structure (e.g. if $X$ is a vector space), then we require our group action to respect $X$’s structure. In general, a group action $G\curvearrowright X$ is a map $G\rightarrow\Aut(X)$ where the automorphisms of $X$ depend on the context 2.

Now, in order to (state and) prove Orbit-Stabalizer, we’ll need to know what those words mean.

Let $G$ be a group acting on a set $X$. Given some $x\in X$, its orbit is $$G\cdot x=\{g\cdot x\mid g\in G\}$$ Furthermore, its stabalizer is $$\Stab(x)=G_x=\{g\in G\mid g\cdot x=x\}$$

Note that the stabalizer of $x\in X$ is a subgroup of $G$ since $g,h\in G_x\implies(g\inv h)\cdot x=g\cdot x=x$. Furthermore, if $G\cdot x=X$ for some $x\in X$, then we say $G$ acts transitively on $X$.

Finally, if $G\curvearrowright X$, then we call $X$ a $G$-set. Naturally, these spaces have their own notion of homomorphisms.

Let $X,Y$ be two $G$-sets. A $G$-map (or $G$-equivariant map or $G$-morphisms) is a map $f:X\rightarrow Y$ s.t. $f(g\cdot x)=g\cdot f(x)$ for all $g\in G$ and $x\in X$. We say $f$ is a $G$-isomorphism if it is bijective.
Show that if $f$ is a $G$-isomorphism, then $\inv f$ is $G$-equivariant.

With our definitions set up, we come to

Let $X$ be a $G$-set. Fix any $Y\subseteq X$ s.t. $g\cdot Y\cap Y\in\{Y,\emptyset\}$ for all $g\in G$ and $G\cdot Y=\{g\cdot y\mid g\in G,y\in Y\}=X$. Finally, let $H=\Stab(Y)=\{g\in G\mid\forall y\in Y:g\cdot y\in Y\}$. Then, $$X\simeq\bigsqcup_{\sigma_i\in G/H}\sigma_iY$$ as $G$-sets where the union is taken over coset representatives of $G/H$ and $\sigma_iY=\{\sigma_i y\mid y\in Y\}$ (note: $\sigma_iy$ is just a formal symbol) and $G$ acts on it via $g\cdot(\sigma_iy)=\sigma_j(h\cdot y)$ for the unique $\sigma_j,h$ s.t. $g\sigma_i=h\sigma_j$.
Let $f:\bigsqcup_{\sigma_i\in G/H}\sigma_iY\to X$ be the map $f(\sigma_iy)=\sigma_i\cdot y$. This map is $G$-equivariant since $$f(g\cdot\sigma_iy)=f(\sigma_j(h\cdot y))=\sigma_j\cdot(h\cdot y)=(\sigma_jh)\cdot y=(g\sigma_i)\cdot y=g\cdot(\sigma_i\cdot y)=g\cdot f(\sigma_iy)$$ where $g\sigma_i=\sigma_jh$. For injectivity, if $\sigma_i\cdot y=\sigma_j\cdot y'$, then $$(\inv\sigma_j\sigma_i)\cdot y=y'\implies\inv\sigma_j\sigma_i\in H\implies\sigma_iH=\sigma_jH\implies\sigma_i=\sigma_j\implies y=y'$$ where the second-to-last implication comes from the fact that we fixed our coset representatives ahead of time. Finally, for surjectivity, fix any $x\in X$. Since $G\cdot Y=X$, there exists $g\in G$ and $y\in Y$ s.t. $g\cdot y=x$. Thus, writing $g=\sigma_jh$, we have that $f(\sigma_j(h\cdot y))=x$.
Let $X$ be a $G$-set, and fix any $x\in X$. Then, $|G\cdot x|=|G:G_x|=|G|/|G_x|$
Apply the above theorem to the $G$-set $G\cdot x$ where $Y=\{x\}$.

It’s worth noting that Orbit-Stabilizer usually only refers to the corollary above, but this stronger version is closer to our main theorem.

A Quick Intro to Representations of Finite Groups

Now that we’ve seen Orbit-Stabilizer, we’ll need to introduce some definitions from representation theory.

Fix a group $G$ and a vector space $V$ over a field $\F$. A (linear) representation of $G$ is a map $\rho:G\rightarrow\GL_{\F}(V)$. Given such a map, we call $V$ a $G$-rep, and morphisms of $G$-reps are $G$-equivariant linear maps. Finally $\theta:G\to\GL_{\F}(U)$ is a subrepresentation if $U\subseteq V$ and $\theta(g)=\rho(g)\mid_U$ for all $g\in G$.

When studying linear representations of groups, there are two main perspectives one can take. Everything can be done in terms of an explicit representation (i.e. the map $\rho$ above) or in terms of modules over the group ring. Since I haven’t talked about modules on this blog before 3, I’ll stick to the explicit representation approach and leave exercises to translate things into statements about modules for the interested reader.

Prove that a linear representation of $G$ is the same thing as an $\F[G]$-module. 4

Thankfully, we don’t need a lot of representation theory for the main result of this post. We only need to know a few different types of linear representations. Also, in case I ever forget to mention this, for the rest of this post, assume all vector spaces are finite-dimensional and assume that all groups are finite.

A permutation representation of $G$ on a finite-dimensional $\F$-vector space $V$ is a linear representation $\rho:G\rightarrow\GL(V)$ in which the elements of $G$ act by permuting some basis $B=\{b_1,\dots,b_n\}$ for $V$.
Consider the symmetric group $S_n$ acting on $\C^n=\bigoplus_{i=1}^n\C e_i$ via $\sigma\cdot e_i=e_{\sigma(i)}$.
Let $G$ be any finite group, and consider $\C[G]\simeq\bigoplus_{g\in G}\C g$ as vector spaces. This is the regular representation when $G$ acts via $h\cdot g=hg$ on the basis.

Finally, we need the notion of induced representations. This let’s you take a representation of a group $H$ and canoncially construct a representation of a larger group $G\supseteq H$. The construction is very reminiscent of the Orbit-Stabilizer theorem.

Let $H\le G$ be a subgroup of $G$, and let $V$ be an $H$-rep. Fix a complete set of coset representatives $\sigma_1=e,\dots,\sigma_n\in G$ s.t. $G/H=\{\sigma_iH:0\le i\le n\}$ and $n=|G/H|$. Then, as a vector space, the induced representation from $H$ to $G$ is $$\Ind_H^GV=\bigoplus_{i=1}^n\sigma_iV$$ where $\sigma_iV=\{\sigma_iv\mid v\in V\}$ is a space of formal symbols. This is given a $G$-action as follows: given some $\sigma_iv\in\Ind_H^GV$, there's a unique $\sigma_j$ and $h\in H$ s.t. $g\sigma_i=\sigma_jh$. We define $g\cdot\sigma_iv=\sigma_j(h\cdot v)$.
Prove that, as $\F[G]$-modules, we have $$\Ind_H^GV\simeq\F[G]\otimes_{\F[H]}V$$ so induction is really just extension of scalars.
The regular representation is $\Ind_1^G\F$ where $1$ denotes the trivial group and $G$ acts trivially (i.e. by the identity) on $\F$.

Orbit-Stabilizer v2

This is where we’ll prove the main result, which roughly says that (almost-)permutation representations are induced representations.

Let $V$ be a $G$-rep with a decomposition $V\simeq\bigoplus_{i=0}^nV_i$ as a vector space s.t. for all $i,j\in\{0,\dots,n\}$, there exists a $g\in G$ s.t. $g\cdot V_i=V_j$, and let $H=\Stab(V_0)$. Then, $$V\simeq\Ind_H^GV_0$$
We will show this by constructing an explicit isomorphism. Let $f:\Ind_H^GV_0\rightarrow V$ be the map given by $$f(\sigma_iv_0)=\sigma_i\cdot v_0$$ This is easily seen to be $G$-equivariant, and it is linear by construction. For surjectivity, it suffices to find preimages for elements of the form $v_i\in V_i$. Given such an element, there exists some $g_i\in G$ and $w_i\in V_0$ s.t. $g_i\cdot w_i=v_i$. Now, we can write $g_i=h_i\ith\sigma_j$ for a unique $h_i\in H$ and coset representative $\ith\sigma_j$. Doing so gives us that $f(\ith\sigma_j(h_i\cdot w_i))=v_i$ so $f$ is surjective as claimed. Finally, we need to show that $f$ is injective, so fix some $w=\sum_{\sigma_i\in G/H}\sigma_i\ith v_0\in\ker f$. This means that $\sum_{\sigma_i\in G/H}\sigma_i\cdot\ith v_0=0$, but we claim that $\sigma_i\cdot\ith v_0$ and $\sigma_j\cdot\Ith vj_0$ belong to different summands (i.e. different $V_i$'s) which forces $\sigma_i\cdot\ith v_0=0\implies\ith v_0=0$ for all $i$ so $w=0$. To prove the claim, suppose that $\sigma_i\cdot\ith v_0,\sigma_j\cdot\Ith vj_0\in V_k$ for some $k$. Then, $$\inv\sigma_j\sigma_i\cdot\ith v_0\in V_0\implies\inv\sigma_j\sigma_i\in H\implies\sigma_j=\sigma_i$$ and we win.

This wasn’t the proof I had in mind. I imagined (and still do) that it was possible to directly apply the original orbit-stabilizer by letting $X$ be a (well-chosen) basis for $V$ and $Y$ be a (well-chosen) basis for $V_0$. However, in trying to make this work, I ran in to issues getting a well-defined action of $G$ on $B$. Basically, $H=\Stab(V_0)$ can act nontrivially so it’s possible that $h\cdot B_0\not\subseteq B$ which is troublesome. I still hold out hope that this idea can be salvaged in general5, so

See if you can come up with a proof of the above that applies the original Orbit-Stabilizer theorem (e.g. apply it to a basis of V and then extend linearly). If you can, let me know.

Even though the proof is a little unsatisfying, we have proven what we set out to prove, so let’s end with a couple examples. $\newcommand{\trv}{\underline{\text{Trv}}}\newcommand{\alt}{\underline{\text{Alt}}}$

Consider $S_n\curvearrowright\Sym^2\C^n$ where $\C^n=\bigoplus\C e_i$ and $S_n$ acts by permuting the $e_i$. Restricting this action to the basis $B=\{e_ie_j:i,j\in\{1,\dots,n\}\}$, we see there are two $S_n$-orbits $$\begin{matrix} B_0 &=& \brackets{e_ie_j:i\neq j} && \Stab(\C e_1e_2) &=& S_2\times S_{n-2}\\ B_1 &=& \brackets{e_1^2,\dots,e_n^2} && \Stab(\C e_1^2) &=& S_{n-1} \end{matrix}$$ Thus we can write $\Sym^2\C^n=V\oplus W$ where $V=\C B_0=\bigoplus_{i\neq j}\C e_ie_j$ and $W=\C B_1=\bigoplus_{i=1}^n\C e_i^2$. Furthermore, $S^n$ acts transitively on these decompositions of $V,W$ so applying our theorem ($V_0=\C e_1e_2$ and $W_0=\C e_1^2$) yields $$\Sym^2\C^n\simeq\parens{\Ind_{S_2\times S_{n-2}}^{S_n}\trv\otimes\trv}\oplus\parens{\Ind_{S_{n-1}}^{S_n}\trv}$$ where $\trv$ is the trivial 1-dimensional $S_k$ representation sending each element to the number 1.
This time, let's look at $S_n\curvearrowright\parens{\Wedge^2\C^n}\otimes\C^n$ where $\C^n=\bigoplus e_i$ and $S_n$ again acts by permuting the $e_i$. We have a basis $B=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i< j\}$ but it's not fixed by $S_n$ (e.g. $(12)\cdot(e_1\wedge e_2)\otimes e_3=(e_2\wedge e_1)\otimes e_3\not\in B$), so we'll look instead at the spanning set $B'=\{(e_i\wedge e_j)\otimes e_k:i,j,k\in\{1,\dots,n\},i\neq j\}$ which is fixed by $S_n$. This has the following orbits $$\begin{matrix} B_0 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j\neq k} && \Stab(\C (e_1\wedge e_2\otimes e_3)) &=& S_2\times S_{n-3}\\ B_1 &=& \brackets{(e_i\wedge e_j)\otimes e_k:i\neq j,k\in\{i,j\}} && \Stab(\C(e_1\wedge e_2\otimes e_1)) &=& S_{n-2} \end{matrix}$$ It's worth noting that $(12)\cdot(e_1\wedge e_2)\otimes e_1=-(e_1\wedge e_2)\otimes e_2$ so we can switch whether $k=i$ or $k=j$ in $B_1$ above. Applying our theorem to (the span of) each orbit and summing them up, we get that $$\Wedge^2\C^n\otimes\C^n\simeq\parens{\Ind_{S_2\times S_{n-3}}^{S_n}\alt\otimes\trv}\oplus\parens{\Ind_{S_{n-2}}^{S_n}\trv}$$ where $\alt$ is the alternating 1-dimensional $S_k$ representation sending each element to its sign.
  1. which, unsurprisingly, is a version of Orbit-stabilizer for representations of finite groups 

  2. for X a set, they are (self) bijections 

  3. but really should at some point 

  4. This includes proving that $\F[G]$-linear maps are $G$-equivariant and that submodules correspond to subrepresentations 

  5. It certainly can be in the case that H does indeed act trivially (or at least stabalizes the basis)… Question: is there always a basis B_0 s.t. Stab(V_0) is contained in Stab(B_0)? 

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