Reference or solution needed: Proof of $[\mathfrak{g}_\alpha,\mathfrak{g}_\beta]\subseteq\mathfrak{g}_{\alpha+\beta}$

Question

Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{h}\leq\mathfrak{g}$ be a sub Lie algebra of $\mathfrak{g}$. Define $\Delta(\mathfrak{g},\mathfrak{h})\subseteq\mathfrak{h}^*$ to be the set of roots of $\mathfrak{g}$. I would like to prove that $$[\mathfrak{g}_\alpha,\mathfrak{g}_\beta]\subseteq\mathfrak{g}_{\alpha+\beta}\forall \alpha,\beta\in\Delta\cup\{0\}$$ Where $$\mathfrak{g_\alpha}:=\{g\in\mathfrak{g}|\exists n\in\mathbb{N}~s.t (ad(h)-\alpha(h))^ng=0\forall h\in\mathfrak{h}\}$$ I got stuck in trying to find an expression for $$(ad(h)-\alpha(h)-\beta(h))^{n+m-1}[X,Y]$$for $X\in\mathfrak{g}_\alpha$ and $Y\in\mathfrak{g}_\beta$ and $(ad(h)-\alpha(h))^nX=0$, $(ad(h)-\beta(h))^mY=0$

Answer 1

As mentioned in a comment, Bourbaki's book on Lie Groups and Algebras contains the following far more general statement (ch. VII paragraph 1 no. 1 prop. 2(ii)).

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Ridiculously general setup:

Let $S$ be any set, $V, V', W$ be vector spaces over some field $k$, and $r: S \rightarrow End(V)$, $r':S\rightarrow End(V')$, $q: S \rightarrow End(W)$ be any maps. For any map $\lambda :S \rightarrow k$, call $$V^\lambda(S) :=\{v\in V: \text{ for all } s\in S \text{ there is } n \in \mathbb{N}: (r(s)-\lambda(s))^n \,v = 0\}$$

and analogously for the other spaces.

Claim: Let $B: V\times V' \rightarrow W$ be a bilinear map such that

$$ q(s) B(v,v') = B(r(s)v, v') + B(v, r'(s)v') \qquad \qquad (\ast)$$

for all $s \in S, v\in V, v'\in V'$. Then the image (w.r.t. the map $B$) of $V^\lambda(S) \times V'^\mu(S)$ lies in $W^{\lambda+\mu}(S)$.

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(Yours is the case $S=\mathfrak h$, $V=V'=W=\mathfrak g$, $r=r'=q= (ad: h \mapsto ad(h)=[h, \cdot])$, $B = [ \cdot, \cdot]$, and condition $(\ast)$ is the Jacobi identity.)

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As for a proof, bilinearity of $B$ gives

$$ (q(s)-\lambda(s)-\mu(s)) B(v,v') = B((r(s)-\lambda(s))v, v') + B(v, (r'(s)-\mu(s))v')$$

(which settles your case $n=m=1$), and now the one crucial step is to show via induction that more generally for all $k \ge 1$,

$$ (q(s)-\lambda(s)-\mu(s))^k B(v,v') = \sum_{i=0}^k \binom{k}{i}B((r(s)-\lambda(s))^i v, (r'(s)-\mu(s))^{k-i}v')$$

which gives the expression you are after in your case $k = m+n-1$.