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?
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.