# Monday, May 03

:::{.remark}
Recall that if $\lieg \in \liealg^\fd$ and $\ch(k) = 0$, there exists a semisimple subalgebra \( \mathcal{L}\leq \lieg  \) such that \( \lieg \cong \rad(\lieg) \semidirect \mathcal{L}  \).
Note that this is a *direct* product on the level of vector spaces.
:::

:::{.proof title="?"}
Since $\lieg / \sqrt \lieg$ is semisimple, it suffices to show that the following SES splits:
\[
0 \to \sqrt \lieg \to \lieg \to \lieg/\sqrt \lieg \to 0
.\]

Case 1;
If $\sqrt \lieg$ is abelian, then these extensions are classified by $H^2(\lieg /\sqrt \lieg; \sqrt \lieg) = 0$ by Whitehead's second lemma.
So there is only one extension, the split extension.


Case 2:
If $\sqrt \lieg$ is not abelian, we use induction on the *derived length* of $\sqrt \lieg$, which is the length of the derived series.
Recall that $\sqrt \lieg$ is solvable, so its derived series eventually terminates and this is well-defined.

So set \( \mathfrak{r}\da [ \sqrt{\lieg} , \sqrt{\lieg}]  \normal \lieg \supsetneq \sqrt \lieg \).
Now $\sqrt{\lieg} / \mathfrak{r}$ is abelian, so the following extension splits:
\[
0 \to {\sqrt \lieg \over \mathfrak{r}  } \to {\lieg \over \mathfrak{r} } \to { \lieg \over \sqrt{\mathfrak{g} }} \to 0
.\]
There is a subalgebra (not necessarily an ideal) \( \lieg \leq \lieg \) with \( \mathfrak{r}\leq \mathfrak{h} \) such that
\[
\lieg / \mathfrak{r} 
&\cong \sqrt{\lieg} / \mathfrak{r} \semidirect \lieh/ \mathfrak{r} \\
\lieh/\lier 
&\cong {\lieg/\lier \over \sqrt \lieg / \lier } \cong \lieg / \sqrt \lier
,\]
and the latter is semisimple.
We have obvious inclusions
\[
\lier \da [ \sqrt{\lieg}, \sqrt{\lieg} ] \subseteq \lieh \intersect \sqrt{\lieg} \subseteq \sqrt \lieh
,\]
where \( \lieh \intersect \sqrt{\lieg} \) is a solvable ideal.
But then $\lieg/\lier$ is semisimple, so $\lier = \sqrt \lieh$ -- otherwise, $\sqrt \lieh / \lier$ is a nonzero solvable ideal in $\lieh / \lier$, which would be a contradiction.
So these inclusions are equalities, and we have a SES of Lie algebras
\[
0 \to \lier \to \lieh \to \lieh/\lier \to 0
.\]
Since $\lier$ has smaller derived length than $\sqrt \lieg$, inductively this sequence splits.
Thus there exists a semisimple subalgebra \( \mathcal{L}\leq \lieh  \) such that \( \lieh \cong \lier \semidirect \mathcal{L}  \), and so
\[
\mathcal{L}\cong \lieh / \lier \cong \lieg /\sqrt \lieg 
.\]
Using that $\lieg/\lier \cong \sqrt \lieg/\lier \semidirect \lieh/\lier$, it just remains to check that things intersect properly.
We have \( \lieh \intersect \sqrt \lieg = \lier \) and \( \mathcal{L} \intersect \lier = 0  \) by cancelling denominators, and so we can conclude
\( \lieg \cong \sqrt \lieg \semidirect \mathcal{L}  \).
:::