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

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

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