Peter–Weyl decomposition of a group representation rather than group algebra

Question

Consider a finite or compact group $G$. The Peter–Weyl decomposition is usually formulated for the group algebra $\mathbb{C}[G]\simeq\bigoplus_i \operatorname{End}(V_i)$, where $V_i$ are the spaces of irreducible representations of $G$. But any representation $T$ defines an object analogous to $\mathbb{C}[G]$, the span of $T(g)$, which I will denote $\mathbb{C}[T]$ by analogy. In fact, $\mathbb{C}[G]\simeq\mathbb{C}[L]$, where $L$ is the left regular representation. If $T=\sum_in_iT_i$ is the isotypic decomposition of $T$ into irreducibles then it seems to be true that $$ \mathbb{C}[T]\simeq\bigoplus_i\mathbb{C}[T_i]\simeq\bigoplus_i\operatorname{End}(V_i), $$ with only irreducibles from $T$ included. The Peter–Weyl theorem is a special case with $T=L$, while the Burnside's theorem is a special case with an irreducible $T$. I think, such a decomposition is used implicitly in the answer to Linear relations among permutation matrices, for example. This might be a case of the Artin–Wedderburn theorem, but it uses a more abstract algebraic language and when I saw it in the context of group representations it was also usually applied to the group algebra.

On the other hand, standard proofs of the Peter–Weyl theorem (Serre's, etc.) do not seem to generalize straightforwardly. We do get a fragment of ‘Fourier transform’ $\mathbb{C}[T]\to\bigoplus_i\mathbb{C}[n_iT_i]$ by restricting to the isotypic components, and it is not hard to show that $\mathbb{C}[nT]\simeq\mathbb{C}[T]$ for any $T$. The ‘Fourier transform’ is easily injective, but for surjectivity the proofs resort to special properties of $L$ or $\mathbb{C}[G]$ or $L^2(G)$, like dimension count for finite groups. On the right, it is $\sum_i(\dim T_i)^2$, which for $T=L$ can be matched to $|G|$ known in advance. But for general $T$ one must show directly that isotypic restrictions can be specified independently (which would not be true if some $T_i$ were isomorphic, say) without using special properties. Is there a nice direct proof of this in the spirit of linear algebra/harmonic analysis and/or reference where it is covered?

This might be related to completeness of matrix elements, but to use that we would have to lift this into $L^2(G)$ and prove something like $\mathbb{C}[T]\simeq M(T)$ for the subspace spanned by matrix elements $T_{ij}(g)$ taken in some basis. The problem is then to show that the natural map $M(T)\to\mathbb{C}[T]$ is surjective.

Answer 1

You don't say what kind of a group $G$ is but I'm going to assume for simplicity that $G$ is finite. Then, yes, it follows from Artin-Wedderburn. The point is that once we know that $\mathbb{C}[G] \cong \prod \text{End}(V_i)$ as algebras we also know what modules look like in terms of the RHS: modules over a finite direct product canonically break up into a finite direct sum of modules over each factor, and modules over $\text{End}(V_i)$ are direct sums of copies of $V_i$.

So if we consider the action of $\mathbb{C}[G]$ on a representation $V \cong \bigoplus_i n_i V_i$ then it follows immediately that the factors $\text{End}(V_j)$ where $n_j = 0$ act trivially, while the factors $\text{End}(V_i)$ where $n_i > 0$ act on each individual isotypic component and hence linear independently.

It may be more conceptually satisfying to instead use here the Jacobson density theorem or the double commutant theorem.

Edit: Here are some details about a different way to organize the argument. We really just need to get clear on what the two-sided ideals of $\mathbb{C}[G]$ are. We have that

1. The matrix algebra $M_n(\mathbb{C})$ has no nontrivial two-sided ideals. This is true more generally of $M_n(D)$ for $D$ a simple ring, and we have more generally that the two-sided ideals of $M_n(R)$ are naturally in bijection with the two-sided ideals of $R$.

2. The two-sided ideals of a finite product $\prod R_i$ of rings are the tuples of two-sided ideals $\prod I_i$ of each factor.

3. Combining 1 and 2, the two-sided ideals of a finite product $\prod M_{n_i}(\mathbb{C})$ of matrix algebras are products $\prod I_i$ where each $I_i$ is either the entire matrix algebra $M_{n_i}(\mathbb{C})$ or zero. So the quotients $\prod M_{n_i}(\mathbb{C})$ are exactly given by deleting some of the factors.

Now let $V$ be a complex representation of a finite group $G$ (not necessarily finite-dimensional). We get an induced homomorphism $\mathbb{C}[G] \to \text{End}(V)$ whose image, by 3, is obtained by deleting some of the factors in the Artin-Wedderburn decomposition $\mathbb{C}[G] \cong \prod_i M_{n_i}(\mathbb{C})$. A factor acts nontrivially on $V$ iff the corresponding irreducible appears in $V$ (this follows from thinking about the isotypic decomposition, I can give more details here if desired), so we conclude that the image of $\mathbb{C}[G]$ is the product of the $M_{n_i}(\mathbb{C})$ corresponding to all the irreducibles appearing in $V$ as desired.