Proposition 9

Let $\mathfrak{h}$ be a nilpotent Lie algebra and M a finite dimensional $\mathfrak{h}$-module. Denote by P the set of maps from $\mathfrak{h}$ to $k$.

(i) Each $M^{\lambda}(\mathfrak{h})$ is an $\mathfrak{h}$-submodule of M. If $x_M$ is triangularizable for all $x\in \mathfrak{h}$, then $M =\sum_{\lambda\in P}M^{\lambda}(\mathfrak{h})$.

(ii) If $k$ is infinite, there exists $x\in \mathfrak{h}$ such that $M^0(x) = M^0(\mathfrak{h})$.

(iii) If $k$ is of characteristic 0, and if $\lambda \in P$ is such that $M^{\lambda}(\mathfrak{h})\not= 0$, then $\lambda$ is a linear form on $\mathfrak{h}$ vanishing on $[\mathfrak{h},\mathfrak{h}]$, and $M_{\lambda}(\mathfrak{h})\not= 0$.

(iv) If $f: M\rightarrow N$ is a surjective homomorphism of finite dimensional $\mathfrak{h}$-modules, then $f(M^{\lambda}(\mathfrak{h})) = N^{\lambda}(\mathfrak{h})$ for all $\lambda \in P$.

(v) If N is a finite dimensional $\mathfrak{h}$-module, and B a bilinear form on $M\times N$ invariant under $\mathfrak{h}$, then $M^{\lambda}(\mathfrak{h})$ and $N^\mu(\mathfrak{h})$ are orthogonal relative to B if $\lambda +\mu\not= 0$. Moreover, if B is non-degenerate then so is its restriction to $M^{\lambda}(\mathfrak{h})\times N^{-\lambda}(\mathfrak{h})$ for all $\lambda \in P$.

Part (i) follows from no. 1, Lemma 1 and Th. 1. Part (ii) follows from no. 2, Prop. 7. Part (iv) follows from no. 1, Cor. 3 of Th. 1. We prove (iii). We can assume that $M = M^{\lambda}(\mathfrak{h})$. Then, for all $x\in \mathfrak{h},\lambda (x) =$ (dim $M$)$^{-1}$Tr($x_M$); this proves that $\lambda$ is linear (which also follows from Prop. 5) and that $\lambda$ vanishes on $[\mathfrak{h},\mathfrak{h}]$. Consider the map $\rho :\mathfrak{h}\rightarrow$ End$_k(M)$ defined by

from the above, this is a representation of $\mathfrak{h}$ on M, and $\rho (x)$ is nilpotent for all $x\in \mathfrak{h}$. By Engel’s theorem (Chap. I, §4, no. 2, Th. 1), there exists $m\not= 0$ in M such that $\rho (x)m= 0$ for all $x\in \mathfrak{h}$, so $m\in M_{\lambda}(\mathfrak{h})$.

The first assertion of (v) follows from no. 1, Prop. 2 (ii). To prove the second, we can assume that $k$ is algebraically closed in view of Prop. 1 of no. 1; it then follows from the first and the fact that $M =\sum_{\lambda}M^{\lambda}(\mathfrak{h}), N =\sum_{\lambda}N^{\lambda}(\mathfrak{h})$, cf. (i).