# Abstract Nonsense

## *-representations of the Group Algebra

Point of post: This is a continuation of this post

We now show that not only is every mapping from $\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{A}\left(G\right)\right)$ of that form a $\ast$-representation but that every $\ast$-representation is of this form for some representation $G\to\mathcal{A}\left(G\right)$. More formally:

Theorem: Let $G$ be a finite group, $\mathcal{A}\left(G\right)$  the group algebra on $G$, and $\mathscr{V}$ a pre-Hilbert space. Then, if $\varrho:\mathcal{A}\left(G\right)\to\text{End}\left(\mathscr{V}\right)$ is a $\ast$-representation then there exists some representation $\rho:G\to\text{End}\left(\mathscr{V}\right)$ such that for each $a\in \mathcal{A}\left(G\right)$ one has

$\displaystyle \varrho(a)=\sum_{g\in G}a(g)\rho_g$.

Proof: Define $\rho:G\to\text{End}\left(\mathscr{V}\right)$ by $g\mapsto \varrho(\delta_g)$. We first claim that in fact $\rho\left(G\right)\subseteq\mathcal{U}\left(G\right)$. Indeed, we recall that for finite-dimensional spaces unitarity  of $T$ is equivalent to $T^\ast=T^{-1}$.  Note though that for each $g\in G$ one has

$\displaystyle \rho_g^\ast\rho_g=\varrho(\delta_g)^\ast\varrho(\delta_g)=\varrho\left(\delta_g^\ast\ast \delta_g\right)$

but a  quick computation shows that:

$\delta_g ^{\ast}(h)=\overline{\delta_g\left(h^{-1}\right)}=\begin{cases}1 & \mbox{if}\quad h=g^{-1}\\ 0 & \mbox{if}\quad g\ne h^{-1}\end{cases}=\delta_{g^{-1}}(h)$

so that $\delta_g ^\ast=\delta_{g^{-1}}$ and so

$\displaystyle \rho_g ^\ast \rho_g=\varrho\left(\delta_{g^{-1}}\ast\delta_g\right)=\varrho\left(\delta_e\right)=\mathbf{1}$

and similarly $\rho_g \rho_g^\ast=\mathbf{1}$ from where $\rho_g^\ast=\rho_g ^{-1}$ and thus unitarity of $\rho_g$ follows. Thus, to prove that $\rho$ is a representation it suffices to show that it’s a homomorphism, but this follows immediately from the fact that for any $g,h\in G$

$\displaystyle \rho_g\rho_h=\varrho\left(\delta_g\right)\varrho\left(\delta_h\right)=\varrho\left(\delta_g\ast\delta_h\right)=\varrho\left(\delta_{gh}\right)=\rho_{gh}$

To finish the argument it suffices to note that for any $a\in\mathcal{A}\left(G\right)$ we have that

$\displaystyle \varrho\left(a\right)=\varrho\left(\sum_{g\in G}a(g)\delta_g\right)=\sum_{g\in G}a(g)\varrho\left(\delta_g\right)=\sum_{g\in G}a(g)\rho_g$

as desired. $\blacksquare$

Remark: The representation $\rho_g=\varrho\left(\delta_g\right)$ is called the representation induced by $\varrho$.

This shows that essentially there is a one-to-one correspondence between $\ast$-representations of $G$ on $\mathscr{V}$ and representations of $G$ on $\mathscr{V}$.

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 21, 2011 -

1. […] saw in our last post that -representations of the group algebra on a pre-Hilbert space are, in a sense, one-to-one. In […]

Pingback by Representation Theory: Left Regular Representation « Abstract Nonsense | January 23, 2011 | Reply

2. […] have just seen that the map given by is a -representation on . But, by previous theorem we know that this induces a representation given by . At first this may seem like quite an […]

Pingback by Representation Theory: Left Regular Representation (Pt. II) « Abstract Nonsense | January 23, 2011 | Reply

3. […] the group algebra, and a pre-Hilbert space. Then if is an irreducible -representation then the induced representation is irreducible. Conversely, if is an irrep then the induced […]

Pingback by Representation Theory: Schur’s Lemma (*-representation Form) « Abstract Nonsense | January 27, 2011 | Reply

4. […] let denote the circle group, then we know that . That said, by our lemma that . But, using the fact that for an abelian group every equivalence class is of degree one (and the fact that is […]

Pingback by Clearer Proof for the Number of Degree One Irrep Classes « Abstract Nonsense | May 6, 2011 | Reply