Abstract Nonsense

Crushing one theorem at a time

*-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

 


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January 21, 2011 - Posted by | Algebra, Representation Theory | , , ,

4 Comments »

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