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Let $L$ be a semisimple Lie algebra and $\Phi$ be the root set of $L.$ Let $L_{\alpha}$ be the root subspaces of $L$ corresponding to the root $\alpha \in \Phi.$ Show that if $\alpha, \beta, \alpha + \beta \in \Phi$ then $[L_{\alpha}, L_{\beta}] = L_{\alpha + \beta}.$

I can see why $[L_{\alpha}, L_{\beta}] \subseteq L_{\alpha + \beta}.$ But I can't prove that they are equal. For this we just need to show that $[L_{\alpha}, L_{\beta}] \neq (0)$ because the root subspaces are one dimensional. But how can I conclude that the elements of $L_{\alpha}$ acts non-trivially on the elements of $L_{\beta}\ $? I am looking for a simple straightforward answer.

Thanks for your time.

ACB
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    As far as I remember, one can use the fact that $\bigoplus_{k\in\mathbb{Z}}L_{\beta+k\alpha}$ is a representation of $\mathfrak{sl}2$ (two of its generators are from $L{\alpha}$ and $L_{-\alpha}$). After that one can apply the repesetation theory of $\mathfrak{sl}_2$. – richrow Mar 15 '23 at 20:38
  • @richrow$:$ There should be some positive integers $r,q$ such that all the roots of the $\alpha$-string through $\beta$ are lying between $\beta-r \alpha$ and $\beta +q \alpha.$ In fact, one can show that for any integer $k$ with $-r\lt k\lt q,$ the corresponding element $\beta+k \alpha$ of this string is also a root of $L.$ Since any finite dimensional $\mathfrak{sl}2$-module has symmetric weight spaces, that allows one to conclude that $\beta (h{\alpha})=r-q,$ where $h_{\alpha}$ spans the one dimensional Cartan subalgebra of the copy of $\mathfrak{sl}_2$ corresponding to the root $\alpha.$ – ACB Mar 15 '23 at 20:58
  • @richrow$:$ This is what I know. But I can't see how it implies that $[L_{\alpha}, L_{\beta}] \neq (0)$ if $\alpha, \beta, \alpha + \beta \in \Phi.$ Any help in this regard would be greatly appreciated. Thanks. – ACB Mar 15 '23 at 21:03
  • Does this answer your question? When is the Lie bracket of two root spaces nonzero? – Torsten Schoeneberg Mar 16 '23 at 01:24
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    As seen in that duplicate, I think even the most "simple straightforward" answer will need some representation theory of $\mathfrak{sl}_2$. I have never seen a proof that would avoid that. – Torsten Schoeneberg Mar 16 '23 at 01:25
  • @TorstenSchoeneberg$:$ Sorry, it may sound a bit rude to you but I personally don't like advertising. I just went through your answer and I personally feel that your answer is not addressing the issue that I am looking for. You have said if you take any finite dimensional $\mathfrak {sl}2$ module then $V{\lambda}$ maps surjectively onto $V_{\lambda + 2}$ if $\lambda \geq -1.$ If you translate this into my language as in my first comment it turns out that $V = L$ and $\lambda = r - q.$ But the problem is that $L_{\beta} \subseteq L_{r-q}.$ ...continued – ACB Mar 16 '23 at 04:58
  • So even if the adjoint action of $x_{\alpha} \in L_{\alpha} \setminus {0}$ from $L_{r-q} \longrightarrow L_{r-q+2}$ is surjective (obviously that will happen only if $r-q \geq -1$ according to your answer) the corresponding map from $L_{\beta} \longrightarrow L_{\alpha + \beta}$ is not necessarily surjective. Unless you clarify these facts clearly in your answer please don't mark my question as a duplicate to the one you have answered. It's a humble request. Thanks. – ACB Mar 16 '23 at 05:03
  • @TorstenSchoeneberg$:$ We should look for injectivity not surjectivity. Injectivity of $\text {ad} (x_{\alpha})$ guarantees that when it's restricted to $L_{\beta}$ the restricted map from $L_{\beta} \longrightarrow L_{\alpha + \beta}$ is also injective and hence in particular non-zero. Now since every root subspace of a semisimple Lie algebra is one-dimensional we are through. This should be the argument. Now I think it's clear things up. Fine! Thanks. – ACB Mar 16 '23 at 05:30
  • @TorstenSchoeneberg$:$ But the problem is that I can't guarantee that one of $\alpha (h_{\beta})$ or $\beta(h_{\alpha})$ is less than $-1$ which would ensure injectivity of $\text {ad} (x_{\beta})$ (resp. $\text {ad} (x_{\alpha})$). – ACB Mar 16 '23 at 10:09
  • @TorstenSchoeneberg$:$ I am clueless as to what makes us conclude that the surjection from $L_{\lambda} \longrightarrow L_{\lambda + 2}$ induces a surjection from $L_{\beta} \longrightarrow L_{\alpha + \beta}.$ – ACB Mar 16 '23 at 10:15
  • We do not apply the theory to $V=L$, but to $V = \bigoplus_{k\in \mathbb Z} L_{\beta+ k \alpha}$ (in case $\lambda := \check{\alpha}(\beta) \ge -1$), or to $V = \bigoplus_{k\in \mathbb Z} L_{\alpha + k \beta}$ (in case $\lambda := \check{\beta}(\alpha) \ge -1$). Then by definition, $V_\lambda = L_\beta$ in the first case, and $V_\lambda = L_\alpha$ in the second case, and in both cases $V_{\lambda +2} = L_{\alpha+\beta}$. – Torsten Schoeneberg Mar 16 '23 at 19:07
  • I read my other answer again and I find those things stated there. --- Your idea of using injectivity instead might also work, there we would need that one of $\check{\alpha}(\beta)$ or $\check{\beta}(\alpha)$ is $\le -1$. For the case at hand, that's good enough, but note that the other answer handles even more general cases, in particular the non-reduced one (occurring over fields like $\mathbb R$ instead of $\mathbb C$), where $\alpha$ and $2\alpha$ are both roots, and we still have surjective but not injective $[L_\alpha, L_\alpha] \rightarrow L_{2\alpha}$. – Torsten Schoeneberg Mar 16 '23 at 19:28
  • @TorstenSchoeneberg$:$ Why is $L_{\lambda} = L_{\beta}\ $? According to the notation in Humphreys $L_{\lambda} = {x \in L\ |\ [h_{\alpha},x] = \lambda x }$ which clearly contains $L_{\beta}$ if $\lambda = \beta (h_{\alpha})$ but how can they be equal? I am still confused. – ACB Mar 16 '23 at 19:57
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    Again, $V$ is not all of $L$ but just $V:=\bigoplus_{k \in \mathbb Z} L_{\beta +k\alpha}$. Write an element $v\in V$ as $(v_k){k \in \mathbb Z}$ according to this decomposition. Choose your $\mathfrak{sl}_2$-triple including semisimple element $h\alpha$ being the coroot to $\alpha$, which by definition means $\beta(h_\alpha)=\check{\alpha}(\beta) =: \lambda$, and of course $\alpha(h_\alpha)=2$. – Torsten Schoeneberg Mar 17 '23 at 01:03
  • For any $k$, $v \in L_{\beta +k\alpha} \implies [h_\alpha,v] = ((\beta +k\alpha)(h_\alpha)) v = (\beta(h_\alpha) + k \alpha(h_\alpha))v = (\lambda+2k)v$; so $L_{\beta+k\alpha} \subseteq V_{\lambda+2k}$ for all $k$, in particular $k=0$ (you seem to accept this). Now take $v\in V \setminus L_\beta$, i.e. it has some nonzero component $v_k$ in some $L_{\beta+k\alpha}$ for $k \neq 0$. Then the $k$-th component of $[h_\alpha, v]$ is $(\lambda+2k)v_k$ which is not equal to $\lambda v_k$, the $k$-th component of $\lambda v$. So $v \notin V_\lambda$. – Torsten Schoeneberg Mar 17 '23 at 01:03
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    Got it. Many many thanks for your kind help @TorstenSchoeneberg. – ACB Mar 17 '23 at 02:16

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