# Is this subalgebra semisimple?

Denote by $$\mathfrak{g}$$ a complex semisimple Lie algebra and let $$\mathfrak{h}$$ be a Cartan subalgebra of $$\mathfrak{g}$$. Let $$\Phi$$ be the root system of $$(\mathfrak{g},\mathfrak{h})$$ and denote by $$\mathfrak{g}_\alpha$$ the root subspace of $$\mathfrak{g}$$ corresponding to a root $$\alpha$$. We fix a choice of positive roots $$\Phi^+$$, and let $$\Delta$$ be the corresponding subset of simple roots in $$\Phi^+$$. Note that each subset $$I\subseteq\Delta$$ generates a root system $$\Phi_I\subseteq\Phi$$, with positive roots $$\Phi_I^+:=\Phi_I\cap \Phi^+$$.

Let $$\mathfrak{l}_I:=\mathfrak{h}\oplus\sum_{\alpha\in\Phi_I}\mathfrak{g}_\alpha$$.

My questions:

1. Is $$\mathfrak{l}_I$$ complex semsimple? If so, why?

2. Is $$\mathfrak{l}_I$$ a Levi subalgebra of $$\mathfrak{g}$$?

• Rarely semi-simple, because $\mathfrak h$ is too big, but invariably reductive. – paul garrett Mar 23 at 1:21
• However, when $I=\Delta$, we get $\mathfrak{l}_I=\mathfrak{g}$ is complex semisimple while the $\mathfrak{h}$ is still big. How to explain this? – James Cheung Mar 23 at 1:34
• Well, in the semi-simple case all of $\mathfrak h$ is hit by brackets from the root spaces, while this is not so for proper $I$. – paul garrett Mar 23 at 1:39
• One thing that helps would be to figure out the centre of $\mathfrak{l}_I$. – Torsten Schoeneberg Mar 23 at 8:57