# Abstract Nonsense

## Consequence of the Decomposition of the Group Algebra Into Matrix Algebras

Point of post: In this post we shall use our previous results concerning the decomposition of the group algebra to prove that if one writes an element of $\mathcal{A}(G)$ in a particular way then that form is conducive to computing polynomials.

Motivation

We saw in our last post that the group algebra is isomorphic to a direct sum of matrix algebras. We shall use this fact to derive an interesting fact about the group algebra. Namely, we know that for every choice of matrix entry functions one has that the group algebra is a direct sum of the subalgebras of the form $\Lambda^{(\alpha)}$ where $\Lambda^{(\alpha)}=\text{span}_{\mathbb{C}}\left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\text{ and }i,j\in[d_\alpha]\right\}$. Thus, every element $f\in\mathcal{A}(G)$ has a decomposition of the form $\displaystyle \sum_{\alpha\in\widehat{G}}f^{(\alpha)}=f$ where $f^{(\alpha)}\in\Lambda^{(\alpha)}$. We shall show that with this decomposition it is much simpler to calculate $p(f)$ for some polynomial $p\in\mathbb{C}[x]$.  Namely, we’ll show the awesome result that $\displaystyle p(f)$ is actually equal to $\displaystyle \sum_{\alpha\in\widehat{G}}p\left(f^{(\alpha)}\right)$. In fact, this isn’t surprising as we shall see that each $\Lambda^{(\alpha)}$ shall act analogously to $\widetilde{\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)}$ sitting inside $\displaystyle \bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$.

Ramifications of the Group Algebra Decomposition

We recall that the map

$\displaystyle \Phi:\mathcal{A}\left(G\right)\to\bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$

where $\displaystyle \left(\Phi(f)\right)_{\alpha;i,j}=\sum_{g\in G}f(g)\overline{D^{(\alpha)}_{i,j}(g)}$ is a unitary algebra isomorphism where we’ve fixed some set of representative matrix entry functions. We will use this to derive a very, very useful result concerning the deomposition of the group algebra. Namely, if we let

$\displaystyle \Lambda^{(\alpha)}=\text{span}_\mathbb{C}\left\{D^{(\alpha)}_{i,j}:i,j\in[d_\alpha]\right\}$

(which we know are subalgebras) then we have that $\displaystyle \mathcal{A}(G)=\bigoplus_{\alpha\in\mathcal{A}}\Lambda^{(\alpha)}$. Thus, for every $f\in\mathcal{A}(G)$ we may write $f$ uniquely as $\displaystyle \sum_{\alpha\in\widehat{G}}f^{(\alpha)}$ where $f^{(\alpha)}\in\Lambda^{(\alpha)}$. But, since $\Phi$ is an isomorphism of algebras we know that this means that

$\displaystyle f=\Phi^{-1}\left(\sum_{\alpha\in\widehat{G}}\Phi\left(f^{(\alpha)}\right)\right)$

Note though that $\displaystyle \Phi\left(f^{(\alpha)}\right)=\widetilde{\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)}$ (where the $\widetilde{\text{ }}$ notation is defined as before for general algebras). Indeed, for each coordinate $\beta\ne\alpha$ we have that

$\displaystyle \Phi\left(f^{(\alpha)}\right)_{\beta;i,j}=\sum_{g\in G}f^{(\alpha)}\overline{D^{(\beta)}_{i,j}(g)}=0$

where the last equality is clear since for some suitable constants $\displaystyle f^{(\alpha)}=\sum_{k,\ell}c_{k,\ell}D^{(\alpha)}_{k,\ell}$ and using theorthogonality relation between the matrix entry functions. It clearly follows then that for any polynomial $p\in\mathbb{C}[x]$ with no constant term that

$\displaystyle p\left(\sum_{\alpha\in\widehat{G}}\Phi\left(f^{(\alpha)}\right)\right)=\sum_{\alpha\in\widehat{G}}p\left(\Phi\left(f^{(\alpha)}\right)\right)$

and since $\Phi$ is an algebra homomorphism this can be rewritten

$\displaystyle \sum_{\alpha\in\widehat{G}}\Phi\left(p\left(f^{(\alpha)}\right)\right)$

Putting all this together we get the following interesting theorem:

Theorem: Let $f\in\mathcal{A}(G)$ and $\displaystyle \sum_{\alpha\in\widehat{G}}f^{(\alpha)}$ the unique representation of $f$ in terms of $\displaystyle \bigoplus_{\alpha\in\widehat{G}}\Lambda^{(\alpha)}$. Then, for any $p\in\mathbb{C}[x]$ with no constant term one has

$\text{ }$

$\displaystyle p(f)=\sum_{\alpha\in\widehat{G}}p\left(f^{(\alpha)}\right)$

$\text{ }$

Proof: From the above and the fact that $\Phi^{-1}$ is clearly an algebra homomorphism we have that

\displaystyle \begin{aligned}p(f) &=p\left(\Phi^{-1}\left(\Phi\left(\sum_{\alpha\in\widehat{G}}f^{(\alpha)}\right)\right)\right)\\ &=\Phi^{-1}\left(p\left(\sum_{\alpha\in\widehat{G}}\Phi\left(f^{(\alpha)}\right)\right)\right)\\ &=\Phi^{-1}\left(\Phi\left(\sum_{\alpha\in\widehat{G}}p\left(f^{(\alpha)}\right)\right)\right)\\ &=\sum_{\alpha\in\widehat{G}}p\left(f^{(\alpha)}\right)\end{aligned}

The conclusion follows. $\blacksquare$

References:

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