# If $\mathfrak{h}$ is a maximal ideal, then there exists a subalgebra $\mathfrak{l}$, such that $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{l}$

Definition: Let $$\mathfrak{g}$$ be a Lie algebra, a ideal $$\mathfrak h \subset \mathfrak g$$ is called a maximal ideal, if for all ideal $$\mathfrak{h}_1 \subset \mathfrak{g}$$ such that $$\mathfrak{h}\subset \mathfrak{h}_1$$, then $$\mathfrak{h}_1 = \mathfrak{h}$$ or $$\mathfrak{h}_1 = \mathfrak{g}$$.

I'm trying to solve the following exercise

Question: Let $$\mathfrak g$$ be a Lie algebra and $$\mathfrak{h}\subset \mathfrak{g}$$ a maximal ideal, then there exists a subalgebra $$\mathfrak{l}$$ such that $$\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{l}$$.

Can anyone help me?

## Some ideas

Consider the projection $$\pi:\mathfrak{g}\to\mathfrak{g}/\mathfrak{h},$$ once $$\mathfrak{h}$$ is a maximal ideal, then $$\mathfrak{g}/\mathfrak{h}$$ is simple. If $$\mathfrak{i}\subset \mathfrak{g}/\mathfrak{h}$$ is a ideal of $$\mathfrak{g}/\mathfrak{h}$$, then $$\pi^{-1}(\mathfrak{i})$$ is a ideal of $$\mathfrak{g}$$ satisfying $$\mathfrak{h}\subset \pi^{-1}(\mathfrak{i})$$, using the maximality of $$\mathfrak{h}$$, $$\mathfrak{i}=\{0\}$$ or $$\mathfrak{i}=\mathfrak{g}/ \mathfrak{h}$$. How shoud I proceed? I tryied to use Levi decomposition but I did not get many results.

N.B: $$\oplus$$ denotes the direct sum of vector space, $$\mathfrak{g}$$ is a finite dimensional Lie algebra.

• Are you working assuming finite-dimensional? – user10354138 May 25 at 12:16
• Yes, $\mathfrak{g}$ is a finite dimensial Lie algebra, I added this information on the question. – Matheus Manzatto May 25 at 13:34

Maybe I've figured out how to do the question. Consider $$\pi: \mathfrak{g}\to \mathfrak{g}/\mathfrak{h},$$ the usual projection.
By the Levi decomposition theorem $$\mathfrak g = \mathfrak s\oplus \mathfrak r(\mathfrak g)$$, where $$\mathfrak r(\mathfrak g)$$ is the solvable ideal of $$\mathfrak{g}$$ and $$\mathfrak s$$ is a semisimple subalgebra. Since $$\mathfrak g / \mathfrak h$$ is simple and $$\pi(\mathfrak r(\mathfrak g))$$ is solvable, then $$r(\mathfrak g) \subset \mathfrak h$$. Moreover $$\pi(\mathfrak{s})$$ is surjective, implying $$\frac{\mathfrak s}{\mathfrak s \cap \mathfrak h} \cong \mathfrak{g}/\mathfrak{h}.$$
Once $$\mathfrak{s}$$ is semisimple there exist simple ideals $$\mathfrak{i}_0,\mathfrak{i}_1,...,\mathfrak{i}_n$$, such that $$\mathfrak{s} =\mathfrak{i}_0\oplus \mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n,$$ since $$\mathfrak{g}/\mathfrak{h}$$ is simple, we can suppose without loss of generality $$\mathfrak{s}\cap \mathfrak{h} = \mathfrak{i}_1\oplus\cdots\oplus\mathfrak{i}_n,$$ because every ideal of $$\mathfrak{s}$$ has the form $$\mathfrak{i}_{k_1}\oplus \dots \oplus \mathfrak{i}_{k_m}$$.
Then \begin{align} \mathfrak{g} &= \mathfrak{s}\oplus \mathfrak r(\mathfrak g)\\ & = (\mathfrak{i}_0\oplus \mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n)\oplus r(\mathfrak g)\\ &= \mathfrak{i}_0\oplus( \mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n\oplus r(\mathfrak g)), \end{align} once $$\pi( \mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n\oplus r(\mathfrak g)) = \{0\}$$, we conclude $$\mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n\oplus r(\mathfrak g))\subset \mathfrak{h},$$ finally, comparing dimensions $$\mathfrak{i}_1\oplus \dots\oplus\mathfrak{i}_n\oplus r(\mathfrak g)= \mathfrak h$$.
Implying $$\mathfrak g = \mathfrak i_0 \oplus \mathfrak h,$$ and finishing the demonstration.