Corollary 2

Assume that V is finite dimensional and that condition (AC) is satisfied. Denote by $V^+(S)$ the vector subspace $\sum_{s\in S}((\bigcap_{i\geq 1}r(s)^iV)$. Then:

(i) $V^0(S)$ and $V^+(S)$ are stable under $r(S)$;

(ii) $V = V^0(S)\oplus V^+(S)$;

(iii) every vector subspace W of V, stable under $r(S)$ and such that $W^0(S) = 0$, is contained in $V^+(S)$;

(iv) $\sum_{s\in S}r(s)V^+(S) = V^+(S)$.

Moreover, $V^+(S)$ is the only vector subspace of V with properties (i) and (ii). For any extension $k'$ of $k,(V\otimes_kk')^+(S) = V^+(S)\otimes_kk'$.

The last assertion is immediate. Thus, taking Prop. 1 into account, in proving the others we can assume that $k$ is algebraically closed. By Th. 1, V = $\sum_{\lambda\in P}V^{\lambda}$(S), and the $V^{\lambda}(S)$ are stable under $r(S)$. If $s\in S$, the characteristic

polynomial of $r(s)|V^{\lambda}(S)$ is $(X-\lambda (s))^{dim V^{\lambda}(S)}$; it follows that $\bigcap_{i\geq 1}r(s)^iV^{\lambda}(s)$

is zero if $\lambda (s) = 0$ and is equal to $V^{\lambda}(S)$ if $\lambda (s)\not= 0$; hence,

which proves (i), (ii) and (iv). If W is a vector subspace of V stable under $r$(S), then $W =\sum_{\lambda\in P}W^{\lambda}(S)$ and $W^{\lambda}(S) = W\cap V^{\lambda}(S)$. If $W^0(S) = 0$, we see that $W\subset V^+$(S), which proves (iii).

Let $V'$ be a vector subspace of V stable under $r(S)$ and such that $V'\cap V^0(S) = 0$. Then ${V'}^0(S) = 0$, so $V'\subset V^+(S)$ by (iii). If, in addition, $V = V^0(S) + V'$, we see that $V'= V^+(S)$. Q.E.D.

We sometimes call $(V^0(S),V^+(S))$ the Fitting decomposition of V, or of the map $r: S\rightarrow$ End(V). If S reduces to a single element $s$, we write $V^+(s)$ or $V^+(r(s))$ instead of $V^+(\{s\})$. We have that $V = V^0(s)\oplus V^+(s), V^0(s)$ and $V^+(s)$ are stable under $r(s),r(s)|V^0(s)$ is nilpotent and $r(s)|V^+(s)$ is bijective.