# About Levi factor of a standard parabolic subalgebra

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?

• Do you assume now the statement here? And yes, this is only the "reductive Levi subalgebra". Commented May 8, 2019 at 9:39
• 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? Commented May 8, 2019 at 10:21
• Why is what happen? Commented May 8, 2019 at 13:30
• 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. Commented May 8, 2019 at 13:38
• See this post; you should search this site in the future:) Commented May 9, 2019 at 7:53

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.