# How to show this $L$-module is simple? (related to the root space decomposition of semisimle Lie algebras)

Given a semisimple Lie algebra (finite dimensional over a field $$K$$ characteristic $$0$$ and algebraically closed), there exists a root space decomposition $$L = H \oplus \oplus_{\alpha \in R} L_{\alpha},$$ where $$H$$ is a maximal toral subalgebra, $$R = \{\alpha \in H^* : L_{\alpha} \not = 0 , \alpha \not = 0 \}$$ and $$L_{\alpha} = \{ x \in L : ad h(x) = \alpha(h) x \ \forall h \in H \}.$$

I want to prove that $$(L_{\alpha} + L_{-\alpha} + [L_{\alpha},L_{-\alpha}])$$-module $$\sum_{j \in \mathbb{Z}} L_{\beta+ j \alpha}$$ is simple when $$\alpha$$ and $$\beta$$ are linearly independent roots.
I know that each $$L_{\alpha}$$ is one dimensional for $$\alpha \in R$$.

Any comments would be appreciated. Thank you!

• Do you know that for any roots $\gamma, \delta$ whose sum is $\in R$, we have $[L_\gamma, L_\delta] =L_{\gamma+\delta}$? That should be quite helpful. – Torsten Schoeneberg Mar 5 at 17:12
• @TorstenSchoeneberg Yes, but I couldn't really make use of it because $0 \not = x \in L_{\gamma}, 0 \not = y \in L_{\delta}$ implies $[x,y] \in L_{\gamma + \delta}$ but $[x,y]$ could be $0$ when $L_{\gamma + \delta} \not = 0$. Can I rule this situation out somehow? – Johnny T. Mar 5 at 18:05
• So you do not (yet) have the equality I stated, just the easy inclusion "$\subseteq$". The proof for the other inclusion which I know relies on basic representation theory of $\mathfrak{sl}_2$ (which taken a bit further would prove your claim as well). Maybe there are simpler arguments. – Torsten Schoeneberg Mar 5 at 21:26
• oops I missed that, but I thought it can't be true for all cases: we know that $[L_{\alpha}, L_{-\alpha}]$ is one dimensional when $\alpha$ is a root but $L_0 = H$ which is not necessarily $1$ dimensional. I guess it holds when the sum is not $0$? – Johnny T. Mar 5 at 22:05
• $0 \notin R$. -- – Torsten Schoeneberg Mar 5 at 22:42