# When is the Lie bracket of two root spaces nonzero

abstract-algebralie-algebrasroot-systemssemisimple-lie-algebras

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.

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).