# Abstract Nonsense

## Matrix Entry Functions Form an (almost) Orthonormal Basis

Point of post: In this post we derive the result that the matrix entry functions form an orthonormal basis for the group algebra, thus deriving the fundamental result that $\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|$.

Motivation

In our last post we showed how fixing, for each $\alpha\in\widehat{G}$, some representative $\rho^{(\alpha)}:G\to\mathscr{V}^{(\alpha)}$ and some orthonormal ordered basis $\mathcal{B}^{(\alpha)}$ enabled us to form $d_\alpha^2$ ‘matrix entry functions’ which were elements of the group algebra $\mathcal{A}(G)$. We further derived some important properties about the span $\Lambda$ of these matrix entry functions (that it is closed under pointwise product and that it separates points). In this post we take this further and show, using a simple case of the Stone-Weierstrass theorem, that $\Lambda=\mathcal{A}(G)$. Moreover, we show that in the usual inner product on $\mathcal{A}(G)$ we have that the set of matrix entry functions is  almost (up to a scalar factor) orthonormal. Thus, it will follow that the set of matrix entry functions is an orthonormal basis for $\mathcal{A}(G)$ and thus we will derive the fundamental result that $\#\left(\widehat{G}\right)<\infty$ and much sharper that $\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|$.

Orthogonality of the Matrix Entry Functions

Let $G$ be a finite group and $\mathcal{D}=\left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\text{ and }i,j\in[d_\alpha]\right\}$ be as before . We claim that in the usual inner product on the group algebra  $\mathcal{D}$ is (up to a scalar factor) orthonormal. More formally:

Theorem: For any $D^{(\alpha)}_{i,j}D^{(\beta)}_{k,\ell}\in\mathcal{D}$ one has the relation

$\displaystyle \left\langle D^{(\alpha)}_{i,j},D^{(\beta)}_{k,\ell}\right\rangle=\frac{1}{d_\alpha}\delta_{\alpha,\beta}\delta_{i,k}\delta_{j,\ell}$

Proof: Fix some $\alpha,\beta\in\widehat{G}$.  We then let $B:\mathbb{C}^{d_\alpha}\to\mathbb{C}^{d_\beta}$ be any linear map. We define then

$\displaystyle \widetilde{B}=\frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}BD^{(\beta)}(g)^{-1}$

We note then that $\widetilde{B}$ is an intertwinor for $D^{(\alpha)}$ and $D^{(\beta)}$. Indeed, for any $g_0\in G$

\displaystyle \begin{aligned}D^{(\alpha)}(g_0)\widetilde{B} &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}(g_0)D^{(\alpha)}(g)BD^{(\beta)}(g)^{-1}\\ &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}(gg_0)BD^{(\beta)}\left(gg_0g_0^{-1}\right)^{-1}\\ &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}(g)BD^{(\beta)}\left(gg_0\right)^{-1}\\ &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}(g)BD^{(\beta)}(g)^{-1}D^{(\beta)}(g_0)\\ &= \widetilde{B}D^{(\beta)}(g_0)\end{aligned}

It then follows from Schur’s lemma that $\widetilde{B}=0$ if $D^{(\alpha)}\not\cong D^{(\beta)}$ (i.e. if $\alpha\ne\beta$) and $\widetilde{B}=cI_{d_\alpha}$ otherwise. If $\alpha=\beta$ then we note then that for the $c$ one has that

\displaystyle \begin{aligned}c &= \frac{1}{d_\alpha}\text{tr}\left(\widetilde{B}\right)\\ &= \frac{1}{d_\alpha}\text{tr}\left(\frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}BD^{(\alpha)}(g)^{-1}\right)\\ &= \frac{1}{d_\alpha}\frac{1}{|G|}\sum_{g\in G}\text{tr}\left(D^{(\alpha)}(g)BD^{(\alpha)}(g)^{-1}\right)\\ &= \frac{1}{d_\alpha}\text{tr}\left(B\right)\end{aligned}

and thus either way $\displaystyle \widetilde{B}=\frac{1}{d_\alpha}\delta_{\alpha,\beta}\text{tr}\left(B\right)$. Note that this was true for any linear mapping $B:\mathbb{C}^{d_\alpha}\to\mathbb{C}^{d_\beta}$. In particular, fix $i\in[d_\alpha]$ and $k\in[d_\beta]$ and let the $(r,s)^{\text{th}}$ entry of $B$ (thought of as a matrix) be given by $\delta_{j,r}\delta_{\ell,s}$. We then compute (by just brute force) that the $(i,k)^{\text{th}}$ entry of $\widetilde{B}$ is

\displaystyle \begin{aligned}\frac{1}{|G|}\sum_{g\in G}\sum_{r=1}^{d_\alpha}\sum_{s=1}^{d_\beta}D^{(\alpha)}_{i,r}(g)B_{r,s}D^{(\beta)}_{s,k}(g^{-1}) &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}_{i,j}D^{(\beta)}_{\ell,k}\left(g^{-1}\right)\\ &= \frac{1}{|G|}\sum_{g\in G}D^{(\alpha)}_{i,j}(g)\overline{D^{(\beta)}_{k,\ell}(g)}\\ &= \left\langle D^{(\alpha)}_{i,j},D^{(\beta)}_{k,\ell}\right\rangle\end{aligned}

But, noting that $\text{tr}(B)=\delta{j,\ell}$ we see that $\displaystyle \widetilde{B}=\frac{1}{d_\alpha}\delta_{\alpha,\beta}\delta_{j,\ell}$ and thus the the $(i,k)^{\text{th}}$ entry of $\widetilde{B}$ is $\displaystyle \frac{1}{d_\alpha}\delta_{\alpha,\beta}\delta_{j,\ell}\delta_{i,k}$. Thus, comparing these two forms for the $(i,k)^{\text{th}}$ entry of $\widetilde{B}$ gives that $\displaystyle \left\langle D^{(\alpha)}_{i,j},D^{(\beta)}_{k,\ell}\right\rangle=\frac{1}{d_\alpha}\delta_{\alpha,\beta}\delta_{j,\ell}\delta_{i,k}$ and since $i,j,k,\ell,\alpha,\beta$ were arbirary the conclusion follows. $\blacksquare$

Corollary: $\mathcal{D}$ is a linearly independent subset of $\mathcal{A}(G)$.

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

February 23, 2011 -

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