Theorem 1

Assume that V is finite dimensional. The following conditions are equivalent:

(i) Condition (AC) is satisfied and, for all $s\in S,r(s)$ is triangularizable.

(ii) For all $\lambda \in P$, $V^{\lambda}(S)$ is stable under $r(S)$, and $V =\sum_{\lambda\in P}V^{\lambda}(S)$.

If $V =\sum_{\lambda\in P}V^{\lambda}$(S), then $V =\sum_{a\in k}V^a(s)$ for all $s\in S$, and it follows from

Lemma 1 that (ii) implies (i). Assume that condition (i) is satisfied. Lemma 1 and formula (1) imply that each $V^{\lambda}(S)$ is stable under $r(S)$. It remains to prove that $V =\sum_{\lambda\in P}V^{\lambda}(S)$. We argue by induction on dim V. We distinguish two cases.

a) For all $s\in S,r(s)$ has a single eigenvalue $\lambda (s)$. Then $V = V^{\lambda}(S)$.

b) There exists $s\in S$ such that $r(s)$ has at least two distinct eigenvalues. Then V is the direct sum of the $V^a(s)$ for $a\in k$, and dim $V^a(s)<$ dim V for all $a$. Each $V^a(s)$ is stable under $r$(S), and it suffices to apply the induction hypothesis.