Corollary (Frobenius reciprocity)

Suppose that A and B are semisimple algebras that are finite-dimensional over a commutative field K and that $f$ is K-linear. Then the K-vector spaces S, T, D, E, and H are finite-dimensional, and we have the equalities

In particular, when K is algebraically closed, we have D = E = K and

Since A-module S is simple, it is monogenous, and D is a linear subspace of Hom$_K(S,S)$; hence S and D are finite-dimensional over K. For an analogous reason, T and E are finite-dimensional over K. Finally, H is a linear subspace of Hom$_K(S,T)$; it is therefore also finite-dimensional. Formula (21) then follows from Proposition 11 because the dimension of H over K is equal to [H : D][D : K] and to [H : E][E : K] (II, §1, No. 13, p. 222, Proposition 25). By Theorem 1 of VIII, p. 47, if K is algebraically closed, we have D = E = K. The second part of the corollary then follows from the first.

Let A and B be semisimple algebras that are finite-dimensional over a commutative field K, and let $f$ be a homomorphism of K-algebras from A to B.

Let $\mathscr{S}(A)$ be the set of classes of simple A-modules; for every $\lambda \in \mathscr{S}$ (A), let $S_{\lambda}$ be a module of class $\lambda$ and $D_{\lambda}$ its centralizer. Then $D_{\lambda}$ is an algebra of finite degree over K; we denote this degree by $d_{\lambda}$. We define $\mathscr{S}$(B), $T_\mu$, $E_\mu$, and $e_\mu$ for $\mu$ in $\mathscr{S}(B)$ analogously. The Grothendieck group $K_0(A)$ has the family $([S_{\lambda}])_{\lambda\in\mathscr{S}(A)}$ as a basis, and $K_0(B)$ has $([T_\mu])_{\mu\in\mathscr{S}(B)}$ as a basis. Let $(a_{\mu\lambda})$ be the matrix of $f^*: K_0(A)\rightarrow K_0(B)$ and $(b_{\lambda \mu})$ the matrix of $f_*: K_0(B)\rightarrow K_0(A)$ with respect to these bases. By definition, we have

for $\lambda$ in $\mathscr{S}(A)$ and $\mu$ in $\mathscr{S}(B)$. We denote the dimension of the vector space Hom$_A(S_{\lambda}, f_*(T_\mu))$ over the field K by $h_{\lambda \mu}$. By the corollary above, we have

When the field K is algebraically closed, we have $d_{\lambda}=e_\mu= 1$; consequently, we have

In other words, the matrices of $f_*$ and $f^*$ with respect to the given bases of $K_0(A)$ and $K_0(B)$ are each other’s transposes.