# Abstract Nonsense

## Schur’s Lemma (*-representation Form)

Point of post: In this post we discuss the notion of Schur’s lemma as it applies to irreducible $\ast$-representations, and with this show that every irrep of an abelian group must have degree one.

Motivation

In our last post we saw how the existence of a non-zero intertwinor for an irrep guarantees the equivalence of the two irreps, since any such intertwinor must be an isomorphism. We also saw how if the two irreps and representation spaces coincided that the intertwinor must, in fact, be a multiple of the identity map. In this post we describe the notion of an irreducible $\ast$-representation and then give a version of Schur’s second lemma for such representations. This will allow us to conclude, after some work, that every irrep of an abelian group must have degree one.

Irreducible $\ast$-representations and the $\ast$-representation Form of Schur’s Lemma

Let $G$ be a finite group and $\mathcal{A}\left(G\right)$ the group algebra. Then, if $\mathscr{V}$ is a pre-Hilbert space and $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$  a  $\ast$-representation we call $\mathscr{W}\leqslant\mathscr{V}$ $\varrho$-invariant (or invariant when $\varrho$ is evident) if $\mathscr{W}$ is invariant under $\varrho_a$ for each $a\in\mathcal{A}\left(G\right)$. We then, unsurprisingly, define a $\ast$-representation $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$ to be irreducible if the only $\varrho$-invariant subspaces of $\mathscr{V}$ are $\{\bold{0}\}$ and $\mathscr{V}$.

Our first result result about irreducible $\ast$-representations is a fundamental one. Namely:

Theorem: Let $G$ be a finite group, $\mathcal{A}\left(G\right)$ the group algebra, and $\mathscr{V}$ a pre-Hilbert space. Then if $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$ is an irreducible $\ast$-representation then the induced representation $\rho:G\to\mathcal{U}\left(G\right):g\mapsto \varrho(\delta_g)$ is irreducible. Conversely, if $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ is an irrep then the induced $\ast$-representation

$\displaystyle\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right):a\mapsto \sum_{g\in G}a(g)\rho_g$

Proof: Suppose first that $\varrho$ is an irrep and $\mathscr{W}\leqslant\mathscr{V}$ is invariant under $\rho_g$ for each $g\in G$. We note then that for every $a\in\mathcal{A}\left(G\right)$ one has

$\displaystyle \varrho_a\left(\mathscr{W}\right)=\sum_{g\in G}a(g)\rho_g\left(\mathscr{W}\right)\subseteq\sum_{g\in G}a(g)\mathscr{W}\subseteq\mathscr{W}$

and thus $\mathscr{W}$ is $\varrho$-invariant, but by assumption this implies that $\mathscr{W}$ is either $\mathscr{V}$ or $\{\bold{0}\}$. From where it follows that $\mathscr{W}$ is $\rho$-invariant implies that $\mathscr{W}$ is either $\{\bold{0}\}$ or $\mathscr{V}$ and thus $\rho$ is an irrep.

Conversely, suppose that $\rho:G\to\mathcal{U}\left(G\right)$ is an irrep and let $\varrho$ be the induced irrep. Assume then that $\mathscr{W}$ is invariant under $\varrho$ then we have in particular that $\mathscr{W}$ is invariant under $\varrho(\delta_g)=\rho_g$ for each $g\in G$ and thus $\mathscr{W}$ is $\rho$-invariant and thus by assumption $\mathscr{W}$ is either $\mathscr{V}$ or $\{\bold{0}\}$, and by applying similar logic as last time we may conclude that $\mathscr{W}$ is $\varrho$-invariant implies that $\mathscr{W}$ is $\{\bold{0}\}$ or $\mathscr{V}$ and thus $\varrho$ is irreducible. $\blacksquare$

With this we now have that third form of Schur’s lemma, the $\ast$-representation form:

Theorem(Schur’s Lemma Third Form-$\ast$-representation Form): Let $G$ be a finite group, $\mathcal{A}\left(G\right)$ the group algebra, and $\mathscr{V},\mathscr{W}$ a pre-Hilbert spaces. Then if $\varrho':\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$ and $\varrho':G\to\text{End}\left(\mathscr{W}\right)$ are two irreducible $\ast$-representations and $T:\mathscr{V}\to\mathscr{W}$ such that $T\varrho_g=\varrho'_gT$ for every $g\in G$ then either $T=\bold{0}$ or $T$ is an isomorphism. Moreover, if $\varrho=\varrho'$ and $\mathscr{V}=\mathscr{W}$ then $T=\lambda\mathbf{1}$ for some $\lambda\in\mathbb{C}$.

Proof: We note first that since $\varrho$ is irreducible the induced representations $\rho:G\to\to\mathcal{U}\left(\mathscr{V}\right)$ and $\rho':G\to\mathcal{U}\left(\mathscr{W}\right)$ \$ must be irreps, but by definition $T\rho_g=T\varrho(\delta_g)=\varrho'(\delta_g)T=\rho'_gT$ from where it follows by the first form of Schur’s lemma that $T$ is either zero or an isomorphism.

Assume now that $\varrho=\varrho'$ and $\mathscr{V}=\mathscr{W}$ then using the fact that the induced representation $\rho$ is irreducible and $T\rho_g=\rho_gT$ for each $g\in G$ implies by the second form of Schur’s lemma that $T=\lambda\mathbf{1}$ for some $\lambda\in\mathbb{C}$. $\blacksquare$

As a quick application of Schur’s Lemmas we prove the extremely interesting following theorem:

Theorem: Let $G$ be a finite abelian group, $\mathscr{V}$ a pre-Hilbert space, and $\rho:G\to\mathcal{U}\left(\mathscr{V}\right)$ an irreducible representation. Then, $\deg\rho=1$.

Proof: Consider the induced $\ast$-representation $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$. Since $\rho$ is an irrep we know that $\varrho$ is an irreducible $\ast$ representation. Note though that for any $g\in G$ and any $a\in\mathcal{A}\left(G\right)$ one has that

\displaystyle \begin{aligned}\rho_g\varrho(a) &=\rho_g\sum_{h\in G}a(h)\rho_h\\ &=\sum_{h\in G}a(h)\rho_g\rho_h\\ &=\sum_{g\in G}a(h)\rho_{gh}\\ &=\sum_{h\in G}a(h)\rho_{hg}\\ &=\sum_{h\in G}a(h)\rho_{h}\rho_g\\ &=\varrho(a)\rho_g\end{aligned}

and thus by the third form of Schur’s lemma we may conclude that there exists $\{\lambda_g\}_{g\in G}\subseteq\mathbb{C}$ such that $\rho_g=\lambda_g\mathbf{1}$ for each $g\in G$. But, this clearly implies that every subspace of $\mathscr{V}$ is invariant under every $\rho_g$ and thus $\rho$-invariant. Thus, if $\dim\mathscr{V}>1$ we could choose $\mathscr{W}\leqslant\mathscr{V}$ with $0<\dim\mathscr{W}<\dim\mathscr{V}$ and by earlier remark this would be $\rho$-invariant contradicting the irreducibility of $\rho$. It follows that $\dim\mathscr{V}=\deg\rho=1$.

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