Generalized weight spaces of a nilpotent subalgebra

Let $H\in \mathfrak{h}$ and $N=\dim \mathfrak{g}$. By [L3] applied to ${\mathrm{ad}}_H$, the spaces $V_{\lambda ,H}=\ker ({\mathrm{ad}}_H-\lambda {)}^N$ are ${\mathrm{ad}}_H$-invariant and $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda ,H}$. Moreover $\mathfrak{h}\subseteq {\mathfrak{g}}_0$: for $H^{\mathrm{\prime }}\in \mathfrak{h}$, [L1] gives $({\mathrm{ad}}_H{)}^n{H}^{\mathrm{\prime }}=0$ for some $n$, so $H^{\mathrm{\prime }}\in {\mathfrak{g}}_0$ because the condition defining ${\mathfrak{g}}_0$ is vacuous at $\alpha =0$. This is (iii).

We claim each $V_{\lambda ,H}$ is stable under ${\mathrm{ad}}_Y$ for every $Y\in \mathfrak{h}$. By [L1] the operator ${\mathrm{ad}}_H$ is nilpotent on $\mathfrak{h}$, so there is $m\ge 0$ with $({\mathrm{ad}}_H{)}^{m}Y=0$; put $N=\dim \mathfrak{g}$, so that $({\mathrm{ad}}_H-\lambda {)}^{N}X=0$ for $X\in V_{\lambda ,H}$ by the definition of $V_{\lambda ,H}$ and [L3]. The operator identity $({\mathrm{ad}}_H-\lambda {)}^n[Y,X]={\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)[({\mathrm{ad}}_H{)}^{n-k}Y,({\mathrm{ad}}_H-\lambda {)}^{k}X]$ holds for every $n\ge 0$ by induction on $n$, because ${\mathrm{ad}}_H$ is a derivation: the case $n=0$ is trivial and the induction step applies ${\mathrm{ad}}_H-\lambda$ to both sides and uses Pascal's rule. Taking $n=m+N$, every summand vanishes, since either $n-k\ge m$, so $({\mathrm{ad}}_H{)}^{n-k}Y=0$, or $k\ge N$, so $({\mathrm{ad}}_H-\lambda {)}^{k}X=0$. Hence $({\mathrm{ad}}_H-\lambda {)}^{m+N}[Y,X]=0$ and $[Y,X]\in V_{\lambda ,H}$. As $Y\in \mathfrak{h}$ was arbitrary, every $V_{\lambda ,H}$ is stable under ${\mathrm{ad}}_{\mathfrak{h}}$.

Fix a basis $H_1,\dots ,H_r$ of $\mathfrak{h}$; iterating step 1.2 over the pairwise compatible decompositions $\mathfrak{g}={⨁}_{\lambda }{V}_{\lambda ,H_j}$ refines the direct sum decomposition to $\mathfrak{g}={⨁}_{({\lambda }_1,\dots ,{\lambda }_r)}(V_{{\lambda }_1,H_1}\cap \cdots \cap V_{{\lambda }_r,H_r})$, and each summand is stable under ${\mathrm{ad}}_H$ for every $H\in \mathfrak{h}$. For a tuple $({\lambda }_1,\dots ,{\lambda }_r)$ with nonzero summand $W$ let $\alpha \in {\mathfrak{h}}^{\ast }$ be the linear functional with $\alpha (H_j)={\lambda }_j$. By [L2], the solvable algebra $\mathfrak{h}$ acts triangularly on $W$ in a suitable basis $v_1,\dots ,v_s$; the diagonal entries of such a triangular form are eigenvalues of ${\mathrm{ad}}_{H_j}$ on $W$ for each $j$, and since $W\subseteq V_{{\lambda }_j,H_j}$ the only eigenvalue of ${\mathrm{ad}}_{H_j}$ there is ${\lambda }_j$, so every diagonal entry equals $\alpha$. Hence each $v_i$ satisfies $({\mathrm{ad}}_H-\alpha (H){)}^i{v}_i=0$ for all $H\in \mathfrak{h}$, and therefore $W\subseteq {\mathfrak{g}}_{\alpha }$. In particular only finitely many ${\mathfrak{g}}_{\alpha }$ are nonzero.

Since each ${\mathfrak{g}}_{\alpha }$ is a linear subspace by definition and stable under every ${\mathrm{ad}}_H$ by step 1.2, and since an element of ${\mathfrak{g}}_{\alpha }$ satisfies the generalized eigenvalue condition for each $H_j$ with value $\alpha (H_j)$, we have ${\mathfrak{g}}_{\alpha }\subseteq V_{\alpha (H_1),H_1}\cap \cdots \cap V_{\alpha (H_r),H_r}$; combined with step 2.1 and the injectivity of the map $({\lambda }_1,\dots ,{\lambda }_r)\mapsto {\sum }_j{\lambda }_j{e}_j$ on the dual basis, this gives ${\mathfrak{g}}_{\alpha }=V_{\alpha (H_1),H_1}\cap \cdots \cap V_{\alpha (H_r),H_r}$ and the direct sum decomposition (ii), with only finitely many nonzero terms. This proves (i) and (ii).

For $X\in {\mathfrak{g}}_{\alpha }$, $Y\in {\mathfrak{g}}_{\beta }$ and $H\in \mathfrak{h}$, the binomial expansion gives $({\mathrm{ad}}_H-(\alpha +\beta )(H){)}^n[X,Y]={\sum }_{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right)[({\mathrm{ad}}_H-\alpha (H){)}^{k}X,({\mathrm{ad}}_H-\beta (H){)}^{n-k}Y]$; choosing $n\ge 2P$ where $P$ bounds the two vanishing exponents for $X$ and $Y$, so that for every $k$ either $k\ge P$ or $n-k\ge P$, every summand is zero, hence $[X,Y]\in {\mathfrak{g}}_{\alpha +\beta }$, which is (iv). If $\mathfrak{g}=0$ all spaces are zero and every assertion is vacuous.