Abstract Nonsense

Decomposing the Group Algebra Into the Direct Sum of Matrix Algebras

Point of post: In this post we show how the group algebra of a finite group can decomposed into the direct sum of group algebras such that the isomorphism between them is unitary when the group algebra is given the usual inner product and the direct sum of matrix algebras the usual direct sum inner product where each summand is given the Hilbert-Schmidt inner product. Moreover, the isomorphism shall in fact be also an associative unital algebra isomorphism.

Motivation

We have seen that the matrix entry functions form an orthonormal basis for the group algebra.  From this we got the important result that $\displaystyle |G|=\sum_{\alpha\in\widehat{G}}d_\alpha^2$. It thus follows from basic linear algebra that as complex vector spaces $\displaystyle \mathcal{A}(G)\cong\bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$ in this post we’ll show much more. Indeed, we’ll show that this isomorphism is also an associative unital algebra isomorphism. Moreover, we’ll even show that if one gives each $\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$ an inner product which is a multiple of the Hilbert-Schmidt inner product and the direct sum the usual inner product on direct sums that the isomorphism is also a unitary map! Thus, in essence the group algebra in all ways important becomes isomorphic to an object we know much, much about.

Decomposing the Group Algebra

For the remainder of the post give every algebra of the form $\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$ the Hilbert-Schmidt inner product with a scaling factor of $\displaystyle \frac{d_\alpha}{|G|^2}$ and every direct sum of matrix algebras the usual inner product on direct sums. With this in mind:

Theorem: Let $G$ be a finite group. Then, the map

$\displaystyle \Phi:\mathcal{A}(G)\to\bigoplus_{\alpha\in\widehat{G}}\text{Mat}_{d_\alpha}\left(\mathbb{C}\right)$

,where when useful we denote $\Phi(f)$ by $\widehat{f}$, given by

$\text{ }$

$\displaystyle \widehat{f}_{\alpha;i,j}=\sum_{g\in G}f(g)\overline{D^{(\alpha)}_{i,j}(g)}$

$\text{ }$

(we’ve implicitly chosen some choice of representative matrix entry functions, but of course as usual it is independent of choice) where $\widehat{f}_{\alpha;i,j}$ denotes the $(i,j)^{\text{th}}$ entry of the $\alpha^{\text{th}}$ coordinate of an element of the codomain is a unitary associative unital algebra isomorphism.

Proof: We first show that $\Phi$ is an associative unital algebra homomorphism. Indeed, note that

$\text{ }$

$\displaystyle \widehat{\delta(e)}_{\alpha;i,j}=\sum_{g\in G}D^{(\alpha)}_{i,j}(g)\delta_e(g)=D^{(\alpha)}_{i,j}(e)=\delta_{i,j}$

where we’ve denoted by $\delta(e)$ what we usually call $\delta_e$. Thus, we have that $\widehat{\delta_e}=\left(I_{d_\alpha}\right)_{\alpha\in\widehat{G}}$ so that the fact that multiplicative identities are mapped to each other is true. Next, to see that $\Phi$ is multiplicative we note that for any $a,b\in\mathcal{A}(G)$ one has

$\text{ }$

\displaystyle \begin{aligned}\widehat{a\ast b}_{\alpha;i,j} &=\sum_{g\in G}\left(a\ast b\right)(g)\overline{D^{(\alpha)}_{i,j}(g)}\\ &=\sum_{g\in G}\sum_{h\in G}a\left(gh^{-1}\right)b(h)\overline{D^{(\alpha)}(gh^{-1}h)}\\ &= \sum_{r=1}^{d_\alpha}\sum_{h\in G}b(h)\overline{D^{(\alpha)}_{r,j}(h)}\sum_{g\in G}a\left(gh^{-1}\right)\overline{D^{(\alpha)}_{i,r}\left(gh^{-1}\right)}\\ &= \sum_{r=1}^{d_\alpha}\left(\sum_{h\in G}b(h)\overline{D^{(\alpha)}_{j,r}(h)}\right)\left(\sum_{g\in g}a(g)\overline{D^{(\alpha)}_{i,r}(g)}\right)\end{aligned}

which is the general entry in the $\alpha^{\text{th}}$ coordinate of $\widehat{a}\widehat{b}$ and so $\widehat{a\ast b}=\widehat{a}\widehat{b}$. Lastly, to verify that $\Phi$ really is an algebra homomorphism we must prove that it is a linear combination. But, this is clear since for every $z,z'\in\mathbb{C}$ and $a,b\in\mathcal{A}(G)$ one has

$\text{ }$

\displaystyle \begin{aligned}\widehat{z a+z' b}_{\alpha;i,j} &=\sum_{g\in G}\left(za+z'b\right)(g)\overline{D^{(\alpha)}_{i,j}(g)}\\ &=z\sum_{g\in G}a(g)\overline{D^{(\alpha)}_{i,j}(g)}+z'\sum_{g\in G}b(g)\overline{D^{(\alpha)}_{i,j}(g)}\\ &=z\widehat{a}+z'\widehat{b}\end{aligned}

Thus, if we can prove that $\Phi$ is unitary then it will be injective, but we know that the domain and codomain are of finite and equal dimension and so $\Phi$ is an injective if and only if its an isomorphism and so we’ll be done. So, to see that $\Phi$ is unitary we note that we only have to prove that $\langle a,a\langle=\langle \widetilde{a},\widetilde{a}\rangle$ for every $a\in\mathcal{A}(G)$. Indeed,

\displaystyle \begin{aligned}\left\langle \widehat{D^{(\alpha_0)}_{i_0,j_0}},\widehat{D^{(\beta_0)}_{k_0,\ell_0}}\right\rangle &= \sum_{\alpha\in\widehat{G}}\frac{d_\alpha}{|G|^2}\left\langle \left[\sum_{g\in G}D^{(\alpha_0)}_{i_0,j_0}(g)\overline{D^{(\alpha)}_{i,j}(g)}\right],\left[\sum_{h\in G}D^{(\beta_0)}_{k_0,\ell_0}(h)\overline{D^{(\alpha)}_{k,\ell}(h)}\right]\right\rangle_{\text{HS}}\\ &= \sum_{\alpha\in\widehat{G}}\frac{d_\alpha}{|G|^2}\left\langle \left[\frac{|G|}{d_{\alpha_0}}\delta_{\alpha_0,\alpha}\delta_{i_0,i}\delta_{j_0,j}\right],\left[\frac{|G|}{d_{\beta_0}}\delta_{\alpha,\beta_0}\delta_{k,k_0},\delta_{\ell,\ell_0}\right]\right\rangle_{\text{HS}}\\ &= \sum_{\alpha\in\widehat{G}}d_\alpha\sum_{i,j=1}^{d_{\alpha}}\left(\frac{1}{d_{\alpha_0}}\delta_{\alpha,\alpha_0}\delta_{i_0,i}\delta_{j,j_0}\right)\overline{\left(\frac{1}{d_{\beta_0}}\delta_{\alpha,\beta_0}\delta_{k_0,i}\delta_{\ell_0,j}\right)}\\ &= \frac{1}{d_{\alpha_0}d_{\beta_0}}\delta_{\alpha_0,\beta_0}\delta_{i_0,k_0}\delta_{j_0,\ell_0}\sum_{\alpha\in\widehat{G}}d_\alpha\sum_{i,j=1}^{d_\alpha}\delta_{\alpha,\alpha_0}\delta_{i,i_0}\delta_{j,j_0}\\ &= \frac{1}{d_{\alpha_0}d_{\beta_0}}\delta_{\alpha_0,\beta_0}\delta_{i_0,k_0}\delta_{j_0,\ell_0}d_{\alpha_0}\sum_{i,j=1}^{d_{\alpha_0}}\delta_{i,i_0}\delta_{j,j_0}\\ &= \frac{1}{\sqrt{d_{\alpha_0}}\sqrt{d_{\beta_0}}}\delta_{i_0,k_0}\delta_{j_0,\ell_0}\end{aligned}

Thus, $\Phi$ carries the orthonormal basis $\displaystyle \left\{\sqrt{d_\alpha}D^{(\alpha)}_{i,j}\right\}_{\alpha\in\widehat{G};i,j\in[d_\alpha]}$ to an orthonormal basis, and thus $\Phi$ is unitary as claimed. The conclusion follows from previous discussion. $\blacksquare$

References:

Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print.