# Monday September 2

Recall the killing form:
\begin{align*}
\kappa: \lieg^2 \to \FF \\
(x,y) \mapsto \mathrm{tr}(\ad_x \circ \ad_y)
.\end{align*}

and Cartan's criteria:

1. $\lieg$ is solvable  $\iff \kappa(x, y) = 0 ~\forall x\in [\lieg, \lieg], ~y\in\lieg$.
2. $\lieg$ is semisimple $\iff$ $\kappa$ is non-degenerate.

**Theorem:**
If $\lieg$ is semisimple, then

a. $\lieg = \bigoplus_{i=1}^n I_i$ for some $I_i \normal \lieg$ which are all simple.
b. Every simple ideal $I \normal \lieg$ is one of the $I_i$.
c. $\kappa_{I_i} = {\kappa_\lieg}\mid_{I_i \cross I_i}$.

![Image](figures/2019-09-09-09:40.png)\

*Proof of (a):*
Use induction on $\dim \lieg$. If $\lieg$ has no nonzero proper ideals, then $\lieg$ is simple and we're done.

Otherwise, let $I_1$ be a minimal nonzero ideal of $\lieg$.
Then $I_1^\perp \normal \lieg$ is also an ideal, and thus $I \coloneqq I_1 \intersect I_1^\perp \normal \lieg$ is as well.
Then for all $x\in [I, I]$, we must have $\kappa(x, y)= 0$ for any $y\in I \subseteq I_1^\perp$.
So $I$ is solvable, and thus $I= 0$. So $\lieg = I_1 \oplus I_1^\perp$.

Note that any ideal of $I_1^\perp$ is also an ideal of $\lieg$, which implies that $\mathrm{rad}(I_1^\perp) \subseteq \mathrm{rad}(\lieg)$, which is zero since $\lieg$ is semisimple, and thus $I_1^\perp$ is semisimple as well.

By the inductive hypothesis, $I_1^\perp = I_2 \oplus \cdots \oplus I_n$ where each $I_j \normal I_i^\perp$ is simple.
Then $I_j \normal \lieg \implies [I_1, I_j] \subset I_1 \intersect I_j$, since $I_1$ has no contribution.
But this is a subset of $I_1 \intersect I_1^\perp = 0$. $\qed$

*Proof of (b):*
If $I \normal \lieg$, then $[I, \lieg] \normal I$ because $[[I, \lieg], I] \subseteq [I, I] \subseteq [I, \lieg]$.

Since $\lieg$ is semisimple, $0 = \mathrm{rad}(\lieg) \supseteq Z(\lieg)$. So $[I, \lieg] \neq 0$, and thus $[I, \lieg] = I$ since $I$ is simple.
But then $[I, \lieg] = \bigoplus [I, I_i]$ is simple as well.
So only one direct summand can survive, since otherwise this would produce at least 2 nontrivial ideals, and $[I, \lieg] = [I, I_i]$ for some $i$.

So for all $j\neq i$, we must have $I_j \intersect I = I_j \intersect [I, I_i] = 0$, and so $I \subseteq I_i$.
But then $I = I_i$ since $I_i$ itself is simple, and we're done.

*Proof of (c):*

(Without using the simplicity of $I_i$)

For $x,y\in I_i$, we have

![Image](figures/2019-09-09-09:45.png)\

## Inner Derivations

Recall that $\ad \lieg \subseteq \mathrm{Der} \lieg$, and in fact (lemma) this is an ideal.

**Theorem:**
If $\lieg$ is semisimple, then $\ad \lieg = \mathrm{Der} \lieg$.

*Proof of lemma:*

For all $\delta \in \mathrm{Der} \lieg$ and all $x,y \in \lieg$, we have

\begin{align*}
[\delta, \ad_x](y) &= \delta([x, y]) - [x, \delta(y)] \\
&= [\delta(x), y] \\
&= [\ad_{\delta(x)}](y)
,\end{align*}

and so $[\delta, \ad x] \subseteq \ad \lieg$. $\qed$

*Proof of theorem:*

If $\lieg$ is semisimple, then $0 = \mathrm{rad} \lieg \supseteq Z(\lieg) = \ker \ad$. Thus $\ad \lieg \cong g/\ker \ad \cong \lieg$ is also semisimple.

This means that $\kappa_{\ad \lieg}$ is non-degenerate, and thus $\ad \lieg \intersect (\ad \lieg)^\perp = 0$, where $(\ad \lieg)^\perp \normal \mathrm{Der}(\lieg)$.

> Note that the non-degeneracy of $\kappa$ already forces $(\ad \lieg)^\perp = 0$.

Then $[(\ad \lieg)^\perp, \ad \lieg] = 0$, and so for all $\delta \in (\ad \lieg)^\perp$, we have $\delta(x) = [\delta, \ad x]$ by the lemma, but we've shown that this is zero.

But then $\delta$ must be zero because $\ad$ is an isomorphism, and in particular it is injective.
This means that $(\ad \lieg)^\perp = 0$, and thus $\ad \lieg = \lieg$. $\qed$

We can use this to define an abstract Jordan decomposition by pulling back decompositions on adjoints:

![Image](figures/2019-09-09-10:01.png)\