# Friday, September 09

## 5.1: The Killing form

:::{.remark}
For the rest of the course: $k = \kbar$ and $\characteristic k = 0$.
Theorem from last time: $L$ is semisimple iff its Killing form $\kappa(x, y) \da \trace(\ad_x \ad_y)$ is nondegenerate.
:::

:::{.proof title="?"}
Let $S = \Rad(\kappa) \normal L$, which is easy to check using "invariance" (associativity) of the form.
Given $s,s'\in S$, the restricted form $\kappa_S(x, y) = \trace(\ad_{S, s} \ad_{S, s'}) = \trace(\ad_{L, s} \ad_{L, s'})$, which was proved in a previous lemma. 
But this is equal to $\kappa(s, s') = 0$.
In particular, we can take $s\in [SS]$, so by (the corollary of) Cartan's criterion for solvable Lie algebras, $S$ is solvable as a Lie algebra and thus solvable as an ideal in $L$.

$\implies$:
Since $\Rad(L)$ is the sum of all solvable ideals, we have $S \subseteq \Rad(L)$, but since $L$ is semisimple $\Rad(L) = 0$ and thus $S=0$.

$\impliedby$:
Assume $S=0$.
If $I \normal L$ is a solvable ideal so $I^{(n)} = 0$ for some $n\geq 0$.
If $I^{(n-1)} \neq 0$, it is a nonzero abelian ideal -- since we want to show $\Rad(L) = 0$, we don't want this to happen!
Thus it STS every abelian ideal is contained in $S$.

So let $I \normal L$ be an abelian ideal, $x\in I$, $y\in L$.
Define an operator
\[
A_{xy}^2 \da (\ad_x \ad_y)^2: L \mapsvia{\ad_y} L \mapsvia{\ad_x} I \mapsvia{\ad_y} I \mapsvia{\ad_x} 0
,\]
which is zero since $[II] =0$.
Thus $A_{xy}$ is a nilpotent endomorphism, which are always traceless, so $0 = \trace(\ad_x \ad_y) = \kappa(x, y)$ for all $y\in L$, and so $x\in S$.
Thus $I \subseteq S$.
:::

:::{.warnings}
$\Rad(\kappa) \subseteq \Rad(L)$ always, but the reverse containment is not always true -- see exercise 5.4.
:::

## 5.2: Simple Ideals of $L$

:::{.definition title="Direct sums of Lie algebras"}
Let $L_i\in \Lie\Alg\slice k$, then their **direct sum** is the product $L_1 \times L_2$ with bracket
\[
[x_1 \oplus x_2, y_1 \oplus y_2] \da [x_1 y_1] \oplus [x_2 y_2]
.\]
:::

:::{.remark}
In particular, $[L_1, L_2] = 0$, and thus any ideal $I_1 \normal L_1$ yields an ideal $I_1 \oplus 0 \normal L_1 \oplus L_2$.
Moreover, if $L = \bigoplus I_i$ is a vector space direct sum of ideals of $L$, this is automatically a Lie algebra direct sum since $[I_i, I_j] = I_i \intersect I_j = 0$ for all $i\neq j$.
:::

:::{.warnings}
This is *not* true for subalgebras!
Also, in this theory, one should be careful about whether direct sums are as vector spaces or (in the stronger sense) as Lie algebras.
:::

:::{.theorem title="?"}
Let $L$ be a finite-dimensional semisimple Lie algebra.
Then there exist ideals $I_n$ of $L$ such that $L = \bigoplus I_i$ with each $I_j$ simple as a Lie algebra.
Moreover every simple ideal if one of the $I_j$.
:::

:::{.proof title="?"}
Let $I \normal L$ and define
\[
I^\perp \da \ts{x\in L \st \kappa(x, I) = 0 }
,\]
the orthogonal complement of $I$ with respect to $\kappa$.
This is an ideal by the associativity of $\kappa$.
Set $J\da I \intersect I^\perp \normal L$, then $\kappa([JJ], J) = 0$ and by Cartan's criterion $J$ is a solvable ideal and thus $J = 0$, making $L$ semisimple.

From the Endman-Wildon lemma in the appendix (posted on ELC, lemma 16.11), $\dim L = \dim I + \dim I^\perp$ and $L = I \oplus I^\perp$, so now induct on $\dim L$ to get the decomposition when $L$ is not simple.
These are semisimple ideals since solvable ideals in the $I, I^\perp$ remain solvable in $L$.
Finally let $I\normal L$ be simple, then $[I, L] \subseteq L$ is a ideal (in both $L$ and $I$), which is nonzero since $Z(L) = 0$.
Since $I$ is simple, this forces $[I, L] = I$.
Writing $L = \bigoplus I_i$ as a sum of simple ideals, we have
\[
I = [I, L] = [I, \bigoplus I_i] = \bigoplus [I, I_i] 
,\]
and by simplicity only one term can be nonzero, so $I = [I, I_j]$ for some $j$.
Since $I_j$ is an ideal, $[I, I_j] \subseteq I_j$, and by simplicity of $I_j$ we have $I = I_j$.
:::

:::{.corollary title="?"}
Let $L$ be semisimple, then $L = [LL]$ and all ideals and homomorphic images (but not subalgebras) are again semisimple.
Moreover, every ideal of $L$ is a sum of simple ideals $I_j$ of $L$.
:::

:::{.proof title="?"}
Take the canonical decomposition $L = \bigoplus I_i$ and check
\[
[L, L] = [\bigoplus I_i, \bigoplus I_j] = \bigoplus [I_i, I_i] = \bigoplus I_i
,\]
where in the last step we've used that the $I_i$ are (not?) abelian.
Let $J \normal L$ to write $L = J \bigoplus J^\perp$, both of which are semisimple as Lie algebras.
In particular, if $\phi: L\to L'$, set $J \da \ker \phi \normal L$.
Then $\im \phi = L/J \cong J^\perp$ as Lie algebras, using the orthogonal decomposition, so $\im \phi$ is semisimple.
Finally if $J \normal L$ then $L = J \oplus J^\perp$ with $J$ semisimple, so by the previous theorem $J$ decomposes as  $J = \oplus K_i$ with $K_i$ (simple) ideals in $J$ -- but these are (simple) ideals in $L$ as well since it's a direct sum.
Thus the $K_i$ are a subset of the $I_j$, since these are the only simple ideals of $L$.
:::