Abstract Nonsense

Crushing one theorem at a time

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


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 22, 2011 - Posted by | Algebra, Representation Theory | , , ,

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