Theorem 1

a) The set $\Theta^{ss}(A)$ consists of the coefficients of the finite-dimensional semisimple representations of A. It is an $(A,A)$-sub-bimodule of $A^*$.

b) For every $S\in \mathscr{S}_K$, the isotypical component of Θ(A) of type S is equal to $\Theta_S(A)$. The left A-module $\Theta^{ss}(A)$ is the direct sum of the submodules $\Theta_S(A)$, where S runs through $\mathscr{S}_K$.

c) For every S in $\mathscr{S}_K$, the right A-module $S^*$ is simple, and $\Theta_S(A)$ is the isotypical component of type $S^*$ of the right A-module Θ(A). The mapping that sends S to cl($S^*$) is a bijection from $\mathscr{S}_K$ to the set of classes of simple $A^o$-modules of finite dimension over K.

d) Viewed as a right A-module, Θ(A) has socle $\Theta^{ss}(A)$.

Let E be a semisimple A-module of finite dimension over K. Every coefficient of E belongs to the image of an A-linear mapping from E to $A^*$ (VIII, p. 378, Proposition 2) and therefore belongs to $\Theta^{ss}(A)$. Conversely, let $f\in \Theta^{ss}(A)$. Then the A-module $Af$ is finite-dimensional over K and is semisimple. By the remark in VIII, p. 367$,f$ is a coefficient of $Af$.

For every S in $\mathscr{S}_K$, the isotypical component of Θ(A) of type S is generated by the images of the A-linear mappings from S to $A^*$; it is therefore equal to $\Theta_S(A)$ by Proposition 2, a) of VIII, p. 378. On the other hand, if S is a simple A-module that is not finite-dimensional over K, then the isotypical component of Θ(A) of type S is zero: indeed, every simple submodule of Θ(A) is monogenous and therefore finite-dimensional over K. Since the socle of Θ(A) is the direct sum of its isotypical components (VIII, p. 65, Proposition 4), this proves b).

Let S be a simple A-module of finite dimension over K. Since the K-vector space S is not reduced to 0, the same holds for $S^*$. Let E be a submodule of the right A-module $S^*$; its orthogonal $E'$ in S is an A-submodule of S. Since S is simple, we have either $E'= 0$, in which case $E = S^*$, or $E'= S$, in which case E = 0. Hence $S^*$ is a simple right A-module.

We have $\Theta (A^o) = \Theta (A)$ (VIII, p. 376); we identify right A-modules with left $A^o$-modules. Since every vector space of finite dimension over K is isomorphic to its bidual, the above proves that the mapping $S\mapsto$ cl($S^*$) is a bijection from $\mathscr{S}_K$ to the set of classes of simple $A^o$-modules of finite dimension over K. Now, for S in $\mathscr{S}_K$, the isotypical component of $\Theta (A^o)$ of type $S^*$ is equal to $\Theta_{S^*}(A^o)$ by assertion b) applied to the algebra $A^o$, and we have $\Theta_{S^*}(A^o) = \Theta_S(A)$. Assertions c) and d) follow immediately.