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Let $\mathfrak{g}$ be a complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$. Fix a Borel subalgebra $\mathfrak{b}$ containing $\mathfrak{h}$ and a parabolic subalgebra $\mathfrak{p}$ containing $\mathfrak{b}$. Let $I \subseteq\Delta$ be the subset of simple roots corresponding to $\mathfrak{p}$. Denote by $\Phi_I$ the subsystem generated by $I$, i.e., $\Phi_I:=\Phi\cap\sum_{\alpha\in I}\mathbb{Z}\alpha$.

Levi decomposition of $\mathfrak{p}$ gives $\mathfrak{p}=\mathfrak{l}\oplus \mathfrak{u}$, where $\mathfrak{l} := \mathfrak{h}\oplus\sum_{\alpha\in \Phi_I}\mathfrak{g}_\alpha$ is the Levi subalgebra of $\mathfrak{p}$ and $\mathfrak{u}:=\sum_{\alpha\in\Phi^+\backslash\Phi_I^+}\mathfrak{g}_\alpha$ is the nilpotent radical of $\mathfrak{p}$.

My question: I remember the Levi subalgebra of a Lie algebra must be semisimple. However, $\mathfrak{l}$ is only reductive. For example, when $I=\emptyset$, $\mathfrak{l}$ becomes $\mathfrak{h}$, which is definitely not semisimple.
Why does this happen? What did I miss?

James Cheung
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  • Do you assume now the statement here? And yes, this is only the "reductive Levi subalgebra". – Dietrich Burde May 08 '19 at 09:39
  • Yes. I assume the statement there. According to Wikipedia, any finite-dimensional real Lie algebra g is the semidirect product of a solvable ideal and a semisimple subalgebra. One is its radical, a maximal solvable ideal, and the other is a semisimple subalgebra, called a Levi subalgebra. The Levi decomposition implies that any finite-dimensional Lie algebra is a semidirect product of a solvable Lie algebra and a semisimple Lie algebra. Why is that happen? – James Cheung May 08 '19 at 10:21
  • Why is what happen? – Dietrich Burde May 08 '19 at 13:30
  • Why the Levi subalgebra of $\mathfrak{p}$ is not semisimple? Is it because of the field? By the way, do you know the Levi decomposition for complex Lie algebra? – James Cheung May 08 '19 at 13:32
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    Because it is not a "Levi subalgebra" but just a reductive Levi subalgebra, which is something different. Concerning your second question, yes, I know the Levi decomposition. – Dietrich Burde May 08 '19 at 13:38
  • If you know the Levi decomposition for complex Lie algebra, would you telling me where to find the exact statement? Since I can only find out the statement for finite dimensional real Lie algebra. – James Cheung May 09 '19 at 06:41
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    See this post; you should search this site in the future:) – Dietrich Burde May 09 '19 at 07:53
  • Now I know that Levi decomposition also works for complex finite dimensional Lie algebra. How come the decomposition $\mathfrak{p}=\mathfrak{l}+\mathfrak{u}$ gives a reductive Levi subalgebra but not a semisimple one? Does $\mathfrak{p}=\mathfrak{l}+\mathfrak{u}$ not even a Levi decomposition from the very beginning? – James Cheung May 09 '19 at 09:08
  • I know the Levi subalgebra in the sense of the Levi decomposition is defined to be the semisimple part of the Levi decomposition. What is the precise definition for reductive Levi subalgebra? – James Cheung May 09 '19 at 09:24

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A Levi subalgebra is semisimple. The problem is in the statement. The component $\mathfrak u$ in the decomposition should be the radical of the Lie algebra $\mathfrak p$, that is defined as the maximal solvable ideal of $\mathfrak p$.

In the case $I=\emptyset$, we have that $$\mathfrak p = \mathfrak h \oplus \sum_{\alpha \in \Phi^+} \mathfrak g_\alpha$$ that is a solvable algebra (in fact, it's a Borel subalgebra for $\mathfrak g$). So $\mathfrak p = \mathfrak u$ and the Levi factor is trivial.

In general, for the parabolic subalgebra $\mathfrak p$ associated with the subset $I$ of the simple roots, the radical $\mathfrak u$ is of the form $$\mathfrak u = \bigcap_{\alpha \in I} \ker(\alpha) \oplus \sum_{\beta \in \Phi^+\backslash \Phi^+_I} \mathfrak g_{\beta}$$ The proof that this is an ideal is done by computing brackets with each component of the root space decomposition of $\mathfrak p$. Also, $\mathfrak u$ is solvable because it's a subalgebra of $\mathfrak b$. The maximality comes by checking that we can find an $\mathfrak{sl}_2$-triple in any bigger ideal, and then we would lose solvability.

Teodora
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