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!

  • $\begingroup$ 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. $\endgroup$ – Torsten Schoeneberg Mar 5 at 17:12
  • $\begingroup$ @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? $\endgroup$ – Johnny T. Mar 5 at 18:05
  • $\begingroup$ 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. $\endgroup$ – Torsten Schoeneberg Mar 5 at 21:26
  • $\begingroup$ 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 $0$? $\endgroup$ – Johnny T. Mar 5 at 22:05
  • $\begingroup$ $0 \notin R$. -- $\endgroup$ – Torsten Schoeneberg Mar 5 at 22:42

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