# Abstract Nonsense

## Using Orthogonality Relations to Compute Convolutions of Characters and Matrix Entry Functions

Point of post: In this post we will use our past results about the pairwise inner product of characters and matrix entry functions to compute their convolutions.

Motivation

In past posts we obainted certain relations between the pairwise inner product (as elements of the group algebra) of matrix entry functions and irreudcible characters. We shall use these relations to compute the convolution of matrix entry functions with eachother and likewise for characters.

Pairwise Convolution of Matrix Entry Functions and Characters

Theorem: For any $\alpha,\beta\in\widehat{G}$, $i,j\in[d_\alpha]$, and $k,\ell\in[d_\beta]$ one has that $\displaystyle D^{(\alpha)}_{i,j}\ast D^{(\beta)}_{k,\ell}=\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{j,k}D^{(\alpha)}_{i,\ell}$.

Proof: We merely note that for any $x\in G$ one has that

\displaystyle \begin{aligned}\left(D^{(\alpha)}_{i,j}\ast D^{(\beta)}_{k,\ell}\right)(x) &= \sum_{g\in G}D^{(\alpha)}_{i,j}\left(xg^{-1}\right)D^{(\beta)}_{k,\ell}(g)\\ &= \sum_{g\in G}\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{i,p}(x)\overline{D_{j,p}^{(\alpha)}(g)}D^{(\beta)}_{k,\ell}(g)\\ &=\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{i,p}(x)\sum_{g\in G}\overline{D^{(\alpha)}_{j,p}(g)}D^{(\beta)}_{k,\ell}(g)\\ &=\sum_{p=1}^{d_\alpha} \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{p,\ell}\delta_{j,k}\delta_{\alpha,\beta}\\ &= \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{j,k}D^{(\alpha)}_{i,\ell}(x)\end{aligned}

Since $x\in G$ was arbitrary the conclusion follows. $\blacksquare$

We can make a similar statement about the convolution of two irreducible characters. Namely:

Theorem: Let $G$ be a finite group. Then, for every $\alpha,\beta\in\widehat{G}$ one has that $\displaystyle \chi^{(\alpha)}\ast\chi^{(\beta)}=\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\chi^{(\alpha)}$.

Proof: It’s the same old computation. Namely, let $x\in G$ be arbitrary. Then

\displaystyle \begin{aligned}\left(\chi^{(\alpha)}\ast\chi^{(\beta)}\right)(x) &= \sum_{g\in G}\chi^{(\alpha)}\left(xg^{-1}\right)\chi^{(\beta)}(g)\\ &=\sum_{g\in G} \sum_{p=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,p}\left(xg^{-1}\right)D^{(\beta)}_{q,q}(g)\\ &=\sum_{p,s=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,s}(x)\sum_{g\in G}\overline{D^{(\alpha)}_{p,s}(g)}D^{(\beta)}_{q,q}(g)\\ &= \sum_{p,s=1}^{d_\alpha}\sum_{q=1}^{d_\beta}D^{(\alpha)}_{p,s}(x)\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\delta_{p,q}\delta_{q,s}\\ &=\frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\sum_{p,s=1}^{d_\alpha}D^{(\alpha)}_{p,s}(x)\delta_{p,q}\delta_{p,s}^2\\ &= \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\sum_{p=1}^{d_\alpha}D^{(\alpha)}_{p,p}(x)\\ &= \frac{|G|}{d_\alpha}\delta_{\alpha,\beta}\chi^{(\alpha)}(x)\end{aligned}

Since $x$ was arbitrary the conclusion follows. $\blacksquare$

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

March 2, 2011 -