Exploration 3: Representation theory

• Every representation $G\to\Aut(R^n)$ assigns each $g\in G$ to a specific permutation matrix in $\Aut(R^n)$.
• But this implicitly defines an action of $G$ on $R^n$. The action is given by matrix muliplication in $R^n$, where each $g$ acts by multiplying by its corresponding matrix.
• Conversely, by Cayley’s theorem every group is isomorphic to a permutation group, and therefore can act on $R^{|G|}$ by permuting the $|G|$ basis $R$-vectors. But permutations are automorphisms, so this essentially assigns each element $g\in G$ an automorphism over $R^{|G|}$, i.e. constructing a matrix $R$-representation $G\to\Aut(R^{|G|})$.

Since every element in $R[G]$ is by definition uniquely represented as a linear combination of elements $g\in G$, we know that $G$ spans $R[G]$. Like in polynomial rings, the only linear combination equal to zero is the one where every coefficient is zero, so the elements of $G$ are linearly independent and therefore form a basis of $R[G]$. Having a basis means the elements of $R[G]$ can be represented by $R$-vectors of coefficients, and therefore the group ring $R[G]$ is a free module by construction.

Since $R[G]$ is a free $R$-module, it has a basis whose elements are precisely the elements of the group $G$. It is a property of groups that any element $g\in G$ permutes the elements of $g$ by left-multiplication. Permuting the elements is an automorphism on $G$ and therefore an automorphism on $R^|G|$, thus $R[G]$ is exactly the regular representation $G\to\Aut(R^{|G|})$.

$R[G]$ is a free $R$-module, and $G$ has a natural group action on $R[G]$ by left-multiplication in the ring. Thus it is a $R[G]$-module.

The trick to get $G$-equivariance is to take a sum over all actions of $G$: $$\tilde{\sigma}(m)=\sum_{g\in G}g^{-1}\sigma(gm)$$ which we can do because $G$ is finite. Then we can show that for $h\in G$: $$\begin{aligned} h\tilde{\sigma}(m)&=h\sum_{g\in G}g^{-1}\sigma(gm)\\ &=\sum_{g\in G}hg^{-1}\sigma(gm)\\ &=\sum_{g\in G}h(gh)^{-1}\sigma((gh)m)&\text{ since }g\mapsto gh\text{ simply reorders the sum over }G\\ &=\sum_{g\in G}g^{-1}\sigma(g(hm))\\ &=\tilde{\sigma}(hm) \end{aligned}$$ and thus $\tilde{\sigma}$ is $G$-equivariant.

• Since $M$ is free, it has a basis, and so does the submodule $W$. We can obtain a projection $\pi:M\to W$ by mapping the basis vectors not in $W$ to basis vectors in $W$.
• Recall that endomorphisms like projections can be made $G$-equivariant via an averaging trick. To ensure that the resulting endomorphism $\tilde{\pi}:M\to M$ is still a projection, we need to ensure that it is idempotent and that its image is $W$. It is enough to construct $\tilde{\pi}$ as a map that fixes elements $w\in W$ (thus ensuring idempotence and that the image is at least $W$) and maps other elements in $M$ to an element in $W$ (thus ensuring that the image is at most $W$).
• Given that $|G|$ is a unit in $R$, the map obtained by the averaging trick can be modified to ensure the above properties: $$\tilde{\pi}(m)=\frac{1}{|G|}\sum_{g\in G}g^{-1}\sigma(gm)$$
• This $\tilde{\pi}$ fixes $w\in W$: $$\begin{aligned} \tilde{\pi}(w)&=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi(gw)\\ &=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi(w')&\text{ since }W\text{ is }G\text{-invariant}\\ &=\frac{1}{|G|}\sum_{g\in G}g^{-1}w'&\text{ since }\pi\text{ is a projection onto }W\\ &=\frac{1}{|G|}\sum_{g\in G}w&\text{ since }w'=gw\\ &=\frac{1}{|G|}|G|w\\ &=w \end{aligned}$$ and maps all $m\in M$ to an element in $W$, since $$\begin{aligned} &\pi(gm)\in W&\text{ by definition of }\pi\\ \implies&g^{-1}\pi(gm)\in W&\text{ since $W$ is $G$-invariant}\\ \implies&\frac{1}{|G|}\sum_{g\in G}g^{-1}\sigma(gm)\in W&\text{ by linearity of the group action}\\ \implies&\tilde{\pi}(m)\in W \end{aligned}$$
• Thus $\tilde{\pi}$ as defined is a $G$-equivariant projection onto $W$.

This result comes naturally from the fact that a group ring $R[G]$ encodes all possible linear group actions and therefore its action on a module $M$ models every possible automorphism on the $R$-modules $M$, since automorphisms on $R$-modules are linear by definition.

A faithful representation is injective, i.e. its domain $G$ is isomorphic to its image $\Aut(M)$. This means that only one element $g\in G$ maps to the identity automorphism, so only one element $g\in G$ acts as identity on $M$.

If two elements $x,y\in R[G]$ have the same action on $M$, then $x-y$ must be the zero action (sending all elements of $M$ to zero). But $x-y=0$ implies $x=y$.

• Recall that if a projection exists on $M$, then $M$ is isomorphic to $\ker\sigma\oplus\im\sigma$. So it is enough to define a projection $\sigma:M\to M$ where $\ker\sigma\iso M/W$ and $\im\sigma=W$. Let’s see what those conditions imply.
  • To ensure that $\ker\sigma\iso M/W$, we need to ensure that every element in $\ker\sigma$ differs by some element in $W$.
  • The requirement $\im\sigma=W$ implies that $\sigma$ maps $M$ to the $G$-invariant submodule $W$ of $M$. Thus we need $\sigma$ to be $G$-equivariant ($g\sigma(m)=\sigma(gm)$) so that it preserves $G$-invariance.
  • Finally, to be a projection, $\sigma$ must be idempotent.
• We’ll start with the requirement that $\sigma$ is $G$-equivariant. In order to get $G$-equivariance, one trick is to take the sum of all products with $g$: $$\tilde{\sigma}(m)=\sum_{g\in G}gm$$ Then we can show that for $h\in G$: $$\begin{aligned} h\tilde{\sigma}(m)&=h\sum_{g\in G}gm\\ &=\sum_{g\in G}hgm&\text{ by linearity in }R[G]\text{-modules}\\ &=\sum_{g\in G}gm&\text{ since }g\mapsto hg\text{ is an automorphism on }G\\ &=\tilde{\sigma}(hm) \end{aligned}$$ where $\sum_{g\in G}hgm=\sum_{g\in G}gm$ because the act of multiplying every element of the sum by $h$ just permutes the order of the sum. Therefore, $\tilde{\sigma}$ commutes with the action of $G$ on $M$, and is therefore $G$-equivariant.
• Next up is ensuring that the elements of $\ker\sigma$ differ by elements of $W$. One way is to send every $gm$ to $W$ via some projection $\pi:M\to W$. We redefine $\tilde{\sigma}$: $$\tilde{\sigma}(m)=\sum_{g\in G}\pi(gm)$$ Then $\tilde{\sigma}$ is a linear combination of elements of $W$. This means that $\im\tilde{\sigma}$ is a subset of $W$, and we can show that two elements of $\ker\tilde{\sigma}$ differ by $\pi(gm)-\pi(gn)\in W$: $$\begin{aligned} \tilde{\sigma}(m)-\tilde{\sigma}(n)&=\sum_{g\in G}\pi(gm)-\sum_{g\in G}\pi(gn)\\ &=\sum_{g\in G}\pi(gm)-\pi(gn) \end{aligned}$$
• Finally, $\sigma$ must be a projection. Therefore, it must be idempotent, and every element of $W$ must be mapped to. Without losing the previous properties, we can have $\sigma(w)=w$ for every $w\in W$ by taking the inverse action of $G$, and dividing by $|G|$ (which works since $\char R\nmid |G|$): $$\sigma(m)=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi(gm)$$ because $$\begin{aligned} \sigma(w)&=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi(gw)\\ &=\frac{1}{|G|}\sum_{g\in G}g^{-1}\pi(w')&\text{ since }W\text{ is }G\text{-invariant}\\ &=\frac{1}{|G|}\sum_{g\in G}g^{-1}w'&\text{ since }\pi\text{ is a projection onto }W\\ &=\frac{1}{|G|}\sum_{g\in G}w&\text{ since }w'=gw\\ &=\frac{1}{|G|}|G|w\\ &=w \end{aligned}$$ Therefore $\sigma(w)=w$ for all $w\in W$, meaning $W\subseteq\im\sigma$. By definition, $\sigma(m)$ remains a linear combination of elements of $W$, and therefore $\im\sigma\subseteq W$. Therefore $\im\sigma=W$.
• Finally, since $\sigma$ is an idempotent endomorphism, we know that $M$ decomposes into $\ker\sigma\oplus\im\sigma$. Since $\ker\sigma\iso M/W$ and $\im\sigma=W$ by construction, we have $M\iso M/W\oplus W$.

• If the $R[G]$-module $M$ corresponding to the representation $\rho$ doesn’t reduce into irreps, there is a proper $G$-invariant submodule $W$ of $M$.
• By the above theorem, which we can use since $\char R\nmid |G|$, there is some complementary $G$-invariant submodule $W'$ such that $M\iso W\oplus W'$.
• By recursively decomposing proper $G$-invariant submodules $W$ of each of the factors, you get smaller and smaller submodules. Since $G$ is finite, this process eventually stops when you are left with a direct sum of irreps that is isomorphic to the given representation $\rho$.