When is the Lie bracket of two root spaces nonzero? Let $L$ be a semisimple Lie algebra with root space decomposition $L = L_0 \oplus \bigoplus_{\alpha \in \Phi} L_\alpha$. For roots $\alpha, \beta \in \Phi$, we always have $[L_\alpha, L_\beta] \subseteq L_{\alpha + \beta}$.
I know how to show that $[L_{\alpha}, L_{-\alpha}]$ is nonzero. In other cases, when can we guarantee that $[L_\alpha, L_\beta]$ is nonzero assuming that $\alpha + \beta$ is also a root? This feels like it should be really important in recovering a semisimple Lie algebra from its root system, but I have been unable to find a simple answer.
 A: This is always true; in fact, $[L_\alpha, L_\beta] = L_{\alpha+\beta}$ for any two roots $\alpha, \beta$ such that $\alpha+\beta$ is a root (which by definition entails that $L_{\alpha+\beta} \neq 0$).
It holds in particular if your semisimple Lie algebra is split, i.e. you defined your roots via some Cartan subalgebra consisting of ad-diagonalisable matrices (as is automatically the case e.g. over $\mathbb C$), and where it turns out all root spaces are one-dimensional. But it even holds in the more general case that you defined your (possibly non-reduced) root system via some maximal split toral subalgebra (which might not be a maximal toral = Cartan subalgebra), where root spaces might have dimension $\ge 2$.
In any case it's a good question, because the proof I know uses, even in the split case, a little more "well-known" theory than one might expect.
Hint 1: In the representation theory of $\mathfrak{sl}_2$, it is well-known that the eigenvalues of $h := \pmatrix{1&0\\0&-1}$ on a finite-dimensional representation $V$ are all integers $\lambda$, and of course we have a decomposition $V \simeq \bigoplus V_\lambda$ into corresponding eigenspaces. It is further well-known, but the crucial fact for what you want, that if both $\lambda$ and $\lambda +2$ are such eigenvalues, then (EDIT: thanks @Joppy for pointing out there was a mistake here earlier, I hope it's good now)

*

*for $\lambda \ge -1$, the element $x :=\pmatrix{0&1\\0&0}$ maps $V_\lambda$ surjectively onto $V_{\lambda+2}$

*for $\lambda \le -1$, the element $x :=\pmatrix{0&1\\0&0}$ maps $V_\lambda$ injectively into $V_{\lambda+2}$
Hint 2: $\mathfrak{sl}_2$-triples. More precisely, you want an $h_\alpha$ in your maximal split toral subalgebra (= Cartan subalgebra if we're in the split case) with $\beta(h_\alpha)= \check{\alpha}(\beta)$, and a good $x_\alpha \in L_\alpha$, and consider $V:=\sum_{i \in \mathbb Z} L_{\beta + i \alpha}$ as an $\mathfrak{sl}_2$-representation.
Now hint 2 together with just the "surjective" part of hint 1 gives the assertion for $\check{\alpha}(\beta) \ge -1$. However, if that is not the case, then we can just flip the roles of $\alpha$ and $\beta$ (because of $\alpha \neq -\beta$, at least one of $\check{\alpha}(\beta)$ and $\check{\beta}(\alpha)$ is always $\ge -1$, even in the case of non-reduced root systems), and still get $[L_{\beta}, L_{\alpha}]=[L_{\alpha}, L_{\beta}]=L_{\alpha+\beta}$.
If we are in the split case, i.e. our maximal split toral subalgebra equals its own centralizer $L_0$, then the same idea proves that all root spaces $L_\alpha$ are one-dimensional: because $ad(x_\alpha):L_0 \rightarrow L_\alpha$ is surjective, but by definition has image spanned by $x_\alpha$. (Then the "injective" part of hint 1 together with hint 2 again suffices to conclude in general, even without flipping $\alpha$ and $\beta$.)
For the split case, compare also the answers to Why are root spaces of root decomposition of semisimple Lie algebra 1 dimensional? and How to show this $L$-module is simple? (related to the root space decomposition of semisimle Lie algebras).
