Multiplicities in the Regular Representation of a Semisimple Algebra

Question

Let $R$ be a finite dimensional associative algebra over a field $k$ and suppose that $R$ is semisimple, i.e., that we can express $R$ as a direct sum of left $R$-modules $$R\cong \oplus_i S_i^{\oplus n_i},$$ where the $S_i$ are non-isomorphic simple modules.

Fact: If $R=kG$ is the group algebra of a finite group $G$ over an algebraically closed field $k$ then the multiplicity $n_i$ is equal to the dimension of the simple module $S_i$: $$n_i=\dim_k S_i.$$

Question: At what level of generality does this fact remain true?

Answer 1

The multiplicity $n_i$ can be calculated as $$ n_i = \frac{\dim_k S_i}{\dim_k \operatorname{End}_R(S_i)} \cdotp $$ This can be seen in (at least) two ways:

First: If $M$ is any finite-dimensional $R$-module and $M \cong \bigoplus_i S_i^{\oplus n_i}$ is a decomposition into pairwise non-isomorphic simple $A$-modules $S_i$ then \begin{align*} \operatorname{Hom}_R(M, S_i) \cong \operatorname{Hom}_R\left( \bigoplus_j S_j^{\oplus n_j}, S_i \right) &\cong \prod_j \operatorname{Hom}_R(S_j, S_i)^{n_j} \\ &\cong \operatorname{Hom}_R(S_i, S_i)^{n_i} = \operatorname{End}_R(S_i)^{n_i} \end{align*} as $k$-vector spaces, where we used for the third isomorphim that $\operatorname{Hom}_R(S_j, S_i) = 0$ for all $j \neq i$ by Schur’s lemma. It follows that $$ \dim_k \operatorname{Hom}_R(M,S_i) = n_i \dim_k \operatorname{End}_R(S_i) $$ and therefore that $$ n_i = \frac{\dim_k \operatorname{Hom}_R(M,S_i)}{\dim_k \operatorname{End}_R(S_i)} \cdotp $$ For $M = R$ we have that $\operatorname{Hom}_R(R, S_i) \cong S_i$ as $k$-vector spaces and therefore $$ n_i = \frac{\dim_k \operatorname{Hom}_R(R,S_i)}{\dim_k \operatorname{End}_R(S_i)} = \frac{\dim_k S_i}{\dim_k \operatorname{End}_R(S_i)} \cdotp $$

Second: By the theorem of Artin-Wedderburn we have that $$ R \cong \operatorname{Mat}_{n_1}(D_1) \times \dotsb \times \operatorname{Mat}_{n_r}(D_r). $$ for skew fields $D_1, \dotsc, D_r$ over $k$. Then $D_1^{n_1}, \dotsc, D_r^{n_r}$ is a set of representatives for the isomorphism classes of simple $R$-modules, $D_i^{n_i}$ appears with multiplicity $n_i$ in $R$ and $\operatorname{End}_R(D_i^{n_i}) \cong D_i^\mathrm{op}$ for all $i = 1, \dotsc, r$. We may assume that $S_i = D_i^{n_i}$ for all $i$. It then follows that $$ \dim_k S_i = \dim_k D_i^{n_i} = n_i \dim_k D_i = n_i \dim_k D_i^{\mathrm{op}} = n_i \dim_k \operatorname{End}_R(S_i). $$

Answer 2

The general description of $R$ is given by the Artin-Wedderburn Theorem: $R$ is a product of matrix algebras over finite-dimensional division algebras over $k$.

If $k$ is algebraically closed, then the only finite-dimensional division algebra over $k$ is itself, and your theorem follows from the representation theory of the matrix algebras $M_{n \times n}(k)$.