# Abstract Nonsense

## Matrix Entry Functions (Pt. II)

Point of post: This post is a continuation of this one.

Matrix Entry Functions

Suppose that for each $\alpha\in\widehat{G}$ we had created some matrix representation $D^{(\alpha)}:G\to\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$. We can then produce $d_\alpha^2$ elements of the group algebra $\mathcal{A}(G)$. Indeed, for a general $k\times k$ matrix $A$ let $A_{i,j}$ denote the $(i,j)^{\text{th}}$ entry of $A$. Then, for each $i,j\in[d_\alpha]$ define $D^{(\alpha)}_{i,j}:G\to\mathbb{C}$ by $\left(D^{(\alpha)}_{i,j}\right)(g)=D^{(\alpha)}_{i,j}(g)$. We call these the matrix entry functions induced by $\alpha$. Then, considering that $\left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\text{ and }i,j\in[d_\alpha]\right\}\subseteq\mathcal{A}\left(G\right)$ it makes sense to speak of the span of this set, in particular denote $\text{span}\left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\text{ and }i,j\in[d_\alpha]\right\}$ by $\Lambda$. We now derive some fundamental facts about $\Lambda$.

Theorem: $\Lambda$ separates points. (By separates points we mean that for any distinct $x,y\in G$ there exists some $f\in\Lambda$ such that $f(x)=1$ and $f(y)=0$).

Proof: Let $x\in G$. Define $f:G\to\mathbb{C}$ by $f(z)=\left\langle L_z(\delta_e),\delta_x\right\rangle$ where $L$ denotes the usual left representation of $G$ into $\mathcal{U}\left(\mathcal{A}(G)\right)$ and $\langle\cdot,\cdot\rangle$ is the usual inner product on the group algebra. Now, from previous theorem we know that $L_z(\delta_e)=\delta_z$ and thus $f(z)=\left\langle \delta_z,\delta_x\right\rangle$. But, recall that $\left\{\delta_g\right\}_{g\in G}$ is orthonormal with respect to this inner product and thus clearly $f(x)=1$ and $f(z)=0$ for $z\ne x$. Thus, the theorem will be proven if we can show that $f\in\Lambda$. To do this we merely note that since $L$ is a representation of $G$ we know that

$\displaystyle L\cong\bigoplus_{\alpha\in\Omega}\rho^{(\alpha)}$

for some finite $\Omega\subseteq \widehat{G}$ (and the $\rho^{(\alpha)}$ were the one’s fixed at the outset of this post). In particular, there exists some fixed unitary map $W$ such that for each $z\in G$ one has that

$\displaystyle L_z=W\bigoplus_{\alpha\in\Omega}\rho^{(\alpha)}(z)W^{-1}\quad\mathbf{(1)}$

Thus, if $\mathcal{B}$ denotes some ordering of the basis $\left\{\delta_g\right\}_{g\in G}$ and $\mathcal{B}'$ denotes the canonical ordered basis for $\displaystyle \bigoplus_{\alpha\in\Omega}\mathscr{V}^{(\alpha)}$ (where these are the representation spaces for the fixed $\rho^{(\alpha)}$) then $\mathbf{(1)}$ implies that for some fixed matrix $M$ one has that

$\displaystyle \left[L_z\right]_{\mathcal{B}}=M\left[\bigoplus_{\alpha\in\Omega}\rho^{(\alpha)}(z)\right]_{\mathcal{B}'}M^{-1}$

but $\mathcal{B}'$ is such that this can be rewritten as

$\displaystyle \left[L_z\right]_{\mathcal{B}}=M\bigoplus_{\alpha\in\Omega}D^{(\alpha)}(z)M^{-1}\quad\mathbf{(2)}$

Note though that by definition and using the bilinearity of the inner product, and the fact that $\left\{\delta_g\right\rangle_{g\in G}$ is orthonormal we see that

\displaystyle \begin{aligned}f(z) &= \left\langle L_z(\delta_e),\delta_x\right\rangle\\ &= \left\langle \sum_{g\in G}\left(L_z\right)_{g,e}\delta_g,\delta_x\right\rangle\\ &= \left(L_z\right)_{x,e}\end{aligned}

note though that upon comparison of $\left(L_z\right)_{x,e}$ with the $(x,e)^{\text{th}}$ entry of

$\displaystyle M\bigoplus_{\alpha\in\Omega}D^{(\alpha)}(z)M^{-1}$

we may conclude that $\left(L_z\right)_{x,e}\in\Lambda$ from where the conclusion follows. $\blacksquare$

We now show that $\Lambda$ has the property that the  pointwise product of two elements of $\Lambda$ is an element of $\Lambda$.

Theorem: Let $D^{(\alpha)}_{i,j},D^{(\beta)}_{k,\ell}\in\Lambda$, then $D^{(\alpha)}_{i,j}D^{(\beta)}_{k,\ell}\in\Lambda$ where $\left(D^{(\alpha)}_{i,j}D^{(\beta)}_{k,\ell}\right)(g)=D^{(\alpha)}_{i,j}(g)D^{(\beta)}_{k,\ell}(g)$.

Proof: Let $\alpha,\beta\in \widehat{G}$ be arbitrary and let $\rho^{(\alpha)},\rho^{(\beta)}$ be the prefixed representatives of $\alpha,\beta$ respectively. Recall then $\rho^{(\alpha)}\otimes\rho^{(\beta)}:G\to\mathscr{V}^{(\alpha)}\otimes\mathscr{V}^{(\beta)}$ is a representation. This then implies that

$\displaystyle \rho^{(\alpha)}\otimes\rho^{(\beta)}\cong\bigoplus_{\gamma\in\Omega}\rho^{(\gamma)}$

for some finite $\Omega\subseteq\widehat{G}$. In particular, if $\mathcal{B}$ denotes the usual ordered tensor basis for $\mathscr{V}^{(\alpha)}\otimes\mathscr{V}^{(\beta)}$ and $\mathcal{B}'$ the usual ordered direct sum basis for $\displaystyle \bigoplus_{\gamma\in\Omega}\mathscr{V}^{(\gamma)}$ we have that for some fixed matrix $M$

$\displaystyle \left[\left(\rho^{(\alpha)}\otimes\rho^{(\beta)}\right)(g)\right]_{\mathcal{B}}=M\left[\bigoplus_{\gamma\in\Omega}\rho^{(\gamma)}(g)\right]_{\mathcal{B}'}M^{-1}$

for every $g\in G$. But because of how $\mathcal{B},\mathcal{B}'$ were chosen we may rewrite this as

$\displaystyle D^{(\alpha)}(g)\otimes D^{(\beta)}(g)=M\bigoplus_{\gamma\in\Omega}D^{(\gamma)}(g)M^{-1}$

where the left side denotes the Kronecker product. The result then follows by comparing the entries of the left and right and sides of this equation. $\blacksquare$

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.