Proposition 3

The sum $\sum_{\lambda\in P}V^{\lambda}(S)$ is direct. The sum $\sum_{\lambda\in P}V_{\lambda}(S)$ is direct.

The second assertion is a consequence of the first; hence it suffices to prove that. We distinguish several cases.

$a) S$ is empty. The assertion is trivial.

$b) S$ is reduced to a single element $s$. Let $\lambda_0, \lambda_1, . . . , \lambda_n$ be distinct elements of $k$. For $i= 0,1, . . . , n$, let $v_i\in V^{\lambda_i}(s)$ and assume that $v_0=v_1+\cdots +v_n$. It suffices to prove that $v_0$ = 0. For $i= 0, . . . , n$, there exists an integer $q_i>$ 0 such that $(r(s)-\lambda_i)^{q_i}v_i$ = 0. Consider the polynomials $P(X) =\prod_{i\geq 1}(X-\lambda_i)^{q_i}$ and $Q(X) = (X-\lambda_0)^{q_0}$. We have $Q(r(s))v_0= 0$, and

$P(r(s))v_0=\sum_{i=1}^nP(r(s))v_i= 0$. Since P and Q are relatively prime, the Bezout identity proves that $v_0= 0$.

$c) S$ is finite and non-empty. We argue by induction on the cardinal of S. Let $s\in S$ and $S'= S$-- $\{s\}$. Let $(v_{\lambda})_{\lambda\in P}$ be a finitely-supported family of elements of V such that $\sum_{\lambda\in P}v_{\lambda}= 0$ and $v_{\lambda}\in V^{\lambda}(S)$. Let $\lambda_0\in P$. Let $P'$ be the set of $\lambda \in P$ such that $\lambda |S'=\lambda_0|S'$. By the induction hypothesis applied to $S'$, we have $\sum_{\lambda\in P'}v_{\lambda}= 0$. If $\lambda , \mu$ are distinct elements of $P',\lambda (s)\not=\mu(s)$.

Since the sum $\sum_{\alpha\in k}V^{\alpha}(s)$ is direct by b), and since $v_{\lambda}\in V^{\lambda(s)}(s),v_{\lambda}= 0$ for all $\lambda \in P'$, and in particular $v_{\lambda_0}= 0$, which we had to prove.

d) General case. Let $(v_{\lambda})_{\lambda\in P}$ be a finitely-supported family of elements of V such that $\sum_{\lambda\in P}v_{\lambda}= 0$ and $v_{\lambda}\in V^{\lambda}(S)$. Let $P'$ be the finite set of $\lambda \in P$ such that $v_{\lambda}\not= 0$, and let $S'$ be a finite subset of S such that the conditions $\lambda \in P'$, $\mu\in P',\lambda |S'=\mu|S'$ imply that $\lambda =\mu$. We have $v_{\lambda}\in V^{\lambda|S'}(S')$; applying c), we see that $v_{\lambda}= 0$ for $\lambda \in P'$, which completes the proof.

Recall that, if $x\in$ End(V), we denote by ad $x$ the map $y \rightarrow xy-yx= [x, y]$ from End(V) to itself.