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 $\mathfrak{g}$?

  • $\begingroup$ Rarely semi-simple, because $\mathfrak h$ is too big, but invariably reductive. $\endgroup$ – paul garrett Mar 23 at 1:21
  • $\begingroup$ 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? $\endgroup$ – James Cheung Mar 23 at 1:34
  • 1
    $\begingroup$ 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 $I$. $\endgroup$ – paul garrett Mar 23 at 1:39
  • $\begingroup$ One thing that helps would be to figure out the centre of $\mathfrak{l}_I$. $\endgroup$ – Torsten Schoeneberg Mar 23 at 8:57

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