# Abstract Nonsense

## Schur’s Lemma (First and Second Forms)

Point of post: In this post we discuss Schur’s lemma, and the necessary prerequisite concepts, such as intertwinors.

Motivation

Schur’s lemma will function as one of the main tools in proving theorems. In essence, Schur’s lemma says that the kinds of maps between two representation spaces of a finite group $G$ which “commute” with irreps (in a sense made precise below) are very limited. In essence, any such map is either the zero map or invertible. We shall use this basic concept to prove other “versions” of Schur’s lemma which will prove useful in different situations.

Intertwinors

Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(V\right)$ and $\rho':G\to\mathcal{U}\left(W\right)$  two representations. Then a map $T:V\to W$ is called an interwinor for $\rho$ and $\rho'$ if for each $g\in G$ the following diagram commutes:



Our first theorem regarding intertwinors will play a pivotal role in our proof of the first form of Schur’s lemma.

Theorem: Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and $\rho':G\to\mathcal{U}\left(\mathscr{W}\right)$ be two representations of $G$. Then, if $T$ is an intertwinor for $\rho$ and $\rho'$ then $\ker T$ is invariant under $\rho$ and $\text{im } T$ is invariant under $\rho'$.

Proof: Suppose that $v\in\ker T$ then for each $g\in G$ one has that $T\rho_g(v)=\rho'_gT(v)=\rho'_g(\bold{0})=\bold{0}$ and thus $\rho_g(v)\in ker T$. Since $v\in\ker T$ was arbitrary it follows that $\ker T$ is invariant under $\rho$.

Secondly, assume that $T(v)\in\text{im }T$ then $\rho'_g(T(v))=T(\rho_g(v))\in\text{im }T$. The conclusion follows. $\blacksquare$

Schur’s Lemma

With just this simple fact about intertwinors we can prove the first form of Schur’s lemma:

Theorem(Schur’s Lemma First Form): Let $G$ be a finite group and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and $\rho':G\to\mathcal{U}\left(\mathscr{W}\right)$  be irreps with $\deg\rho,\deg\rho'>0$. If $T$ is an intertwinor for $\rho$ and $\rho'$ then either $T\equiv\bold{0}$ or $T$ is invertible.

Proof: Suppose that $T\not\equiv \bold{0}$ then $\ker T$ is a proper $\rho$-invariant subspace of $\mathscr{V}$ and since $\rho$ is an irrep it follows that $\ker T=\{\bold{0}\}$ and thus $T$ is injective. Suppose then that $T$ isn’t surjective and $\text{im }T\subsetneq\mathscr{W}$, then $\text{im }T$ is a proper $\rho'$-invariant subspace of $\mathscr{W}$ and thus by assumption that $\rho'$ is an irrep it follows that $\text{im }T=\{\bold{0}\}$ which contradicts that $\ker T=\{\bold{0}\}$. It follows that $\text{im }T=\mathscr{W}$ and thus $T$ is surjective. Combining the two gives the desired result. $\blacksquare$

Corollary: If $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and $\rho':G\to\mathcal{U}\left(\mathscr{W}\right)$ are irreps and there exists a non-zero intertwinor $T:\mathscr{V}\to\mathscr{W}$ then $\rho\simeq \rho'$.

Proof: Indeed, by Schur’s lemma we know that $T$ must be an isomorphism such that $T\rho_gT^{-1}=\rho'_g$, and thus by earlier theorem it follows that $\rho\simeq\rho'$ as desired. $\blacksquare$

We now prove a somewhat stronger version of Schur’s lemma when we are only interested in the case when $\rho=\rho'$ and $\mathscr{V}=\mathscr{W}$. Put more directly:

Theorem(Schur’s Lemma Second Form): Let $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ and $T:\mathscr{V}\to\mathscr{V}$ be an intertwinor for $\rho$ and itself. Then, $T=\lambda\mathbf{1}$ for some $\lambda$.

Proof: Since $\mathscr{V}$ is a complex vector space we know that $T$ admits at least one eigenvalue $\lambda$. Let $T'=T-\lambda\mathbf{1}$. We note then that $\ker T'$ is non-trivial since if $u$ is an eigenvalue corresponding to $\lambda$ then $\text{span}\{u\}\subseteq\ker T'$. This said we see that

$\rho_gT'=\rho_g\left(T-\lambda\mathbf{1}\right)=\rho_gT-\rho_g(\lambda\mathbf{1})=T\rho_g-\rho_g(\lambda\mathbf{1})=(T-\lambda\mathbf{1})\rho_g=T'\rho_g$

and thus $T'$ is an intertwinor, but since $T'$ can’t be an isomorphism since it’s kernel is non-trivial it follows from Schur’s lemma that $T'=\bold{0}$ from where it follows that $T'=\lambda\mathbf{1}$ as desired. $\blacksquare$

References:

1.Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

2. Serre, Jean Pierre. Linear Representations of Finite Groups. New York: Springer-Verlag, 1977. Print

January 27, 2011 -

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