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$\DeclareMathOperator{\ad}{ad}$ Let $L$ be a (finite dimensional) semisimple Lie algebra. Let $H$ be a maximal toral subalgebra of $L$. Consider a representation $\pi: L \to \mathfrak{gl}(V)$. It is then a basic fact that we have a submodule $$ V' = \bigoplus_{\lambda \in H^*} V_\lambda, $$ where $V_\lambda = \{v \in V \mid \forall h \in H: \pi(h)(v) = \lambda(h)v \}$.

In the case that $\pi = \ad_L$ I do understand that $V' = V$ since it then follows that $\ad_L(H)$ is a family of pairwise commuting endomorphisms, which is therefore (and the fact that all $h \in H$ are $\ad$-semisimple) simultaneously diagonalizable. If $x_1,x_2,\cdots,x_l$ is a basis for which all $\ad_L(h), \ (h \in H)$ are diagonalizable, then each $x_i$ belongs to some weight space, whence we conclude that $V' =V$.

In the general case, I cannot see why $V=V'$ if $\dim V < \infty$.

For a same argument to hold, it would suffice to know that $\pi(H)$ is simultaneously diagonalizable. Is this true in the f.d. case ?

This is by the way lemma 20.1 in Humphrey's book.

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Let $W=\sum_{\lambda\in H^{\ast}}V_{\lambda}$. The sum is direct because eigenvectors in different eigenspaces are linearly independent. Hence we have $W=V'$. Now $V$ is semisimple by Weyl's theorem, because $L$ is semisimple. So we may assume that $V$ is already simple. Since $V$ is finite dimensional, it contains a non-zero vector $v$ which is an eigenvector for the action of $H$, by Lie's Theorem, since $H$ is abelian, hence solvable. But then $V=V'$, since $V$ has no proper submodule.

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  • $\begingroup$ Thank you. This is a quite nice interplay of non-trivial theorems, I think. $\endgroup$
    – user42761
    Jun 1, 2015 at 9:47

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