Abstract Nonsense

Crushing one theorem at a time

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

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February 23, 2011 - Posted by | Algebra, Representation Theory | , ,

6 Comments »

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