# Monday, September 26

## 8.1 Continued

:::{.remark}
A Lie algebra is **toral** iff every element is semisimple -- this exists because any semisimple Lie algebra contains at least one semisimple element and you can take its span.
Let $H$ be a fixed maximal toral subalgebra, then we have a root space decomposition
\[
L = C_L(H) \oplus \bigoplus _{\alpha\in \Phi \subseteq H\dual} L_\alpha, \qquad L_\alpha \da \ts{ x\in L \st [hx] = \alpha(h) x \,\forall h\in H}
.\]
Let $L$ be semisimple and finite dimensional over $\CC$ from now on.
:::

:::{.proposition title="?"}
1.
\[
[L_\alpha, L_\beta] \subseteq L_{ \alpha + \beta} \qquad \forall \alpha, \beta\in H\dual
.\]
2.
\[
x\in L_{\alpha}, \alpha\neq 0 \implies \ad_x \text{ is nilpotent}
.\]
3. If \( \alpha, \beta\in H\dual \) and \( \alpha + \beta\neq 0 \) then $L_ \alpha \perp L_ \beta$ relative to $\kappa_L$, the Killing form for $L$.
:::

:::{.proof title="?"}
\envlist

1. Follows from the Jacobi identity.

2. Follows from (1), that $\dim L < \infty$, and the root space decomposition.
  This is because $L_{\alpha}\neq L_{\alpha + \beta}\neq L_{\alpha + 2\beta} \neq \cdots$ and there are only finitely many $\beta$ to consider.

3. If \( \alpha + \beta\neq 0 \) then $\exists h\in H$ such that $(\alpha + \beta)(h) \neq 0$.
  For $x\in L_\alpha, y\in L_ \beta$,
  \[
  \alpha(h) \kappa(x,y) 
  &= \kappa([hx], y) \\
  &= -\kappa([xh], y) \\
  &= -\kappa(x, [hy]) \\
  &= -\beta(h)\kappa(x,y) \\
  &\implies ( \alpha + \beta)(h) \kappa(x,y) = 0 \\
  &\implies \kappa(x,y)=0 \text{ since } (\alpha + \beta)(h) \neq 0
  .\]

:::

:::{.corollary title="?"}
$\ro{\kappa}{L_0}$ is nondegenerate, since $L_0 \perp L_{\alpha}$ for all $\alpha\in \Phi$, but $\kappa$ is nondegenerate.
Moreover, if $L_{ \alpha} \neq 0$ then $L_{- \alpha}\neq 0$ by (3) and nondegeneracy.
:::

## 8.2: $C_L(H)$

:::{.proposition title="?"}
Let $H \leq L$ be maximal toral, then $H = C_L(H)$.
:::

:::{.proof title="?"}
Skipped, about 1 page of dense text broken into 7 steps.
Uses the last corollary along with Engel's theorem.
:::

:::{.observation}
If $L$ is a classical Lie algebra over $\CC$, we choose $H$ to be diagonal matrices in $L$, and $x\in L\sm H$ is non-diagonal, then there exists an $h\in H$ such that $[hx]\neq 0$.
Note that toral implies abelian and nonabelian implies nontoral, thus there is no abelian subalgebra of $L$ properly containing $H$ -- adding any nontoral element at all to $H$ makes it nonabelian.
This same argument shows $C_L(H) = H$ since nothing else commutes with $H$.
This implies that $L = H \oplus_{ \alpha \in \Phi} L_\alpha$.
:::

:::{.corollary title="?"}
$\ro{\kappa}{H}$ is nondegenerate.
:::

:::{.remark}
As a result, $\kappa$ induces an isomorphism $H \iso H\dual$ by $h\mapsto \kappa(h, \wait)$ and $H\dual \iso H$ by $\phi\mapsto t_\phi$, the unique element such that $\kappa(t_\phi, \wait) = \phi(\wait)$.
In particular, given \( \alpha\in \Phi \subset H\dual \) there is some $t_\alpha\in H$.
The next 3 sections are about properties of $\Phi$:

- Orthogonality,
- Integrality,
- Rationality.

:::

## 8.3: Orthogonality properties of $\Phi$.

:::{.proposition title="Big!"}
\envlist

a. $\Phi$ spans $H\dual$.
b. If $\alpha\in \Phi$ is a root then $-\alpha\in \Phi$ is also a root.
c. Let \( \alpha\in \Phi, x\in L_{ \alpha}, y\in L_{- \alpha} \), then $[xy] = \kappa(x, y) t_\alpha$.
d. If $\alpha\in \Phi$ then $[L_{ \alpha}, L_{- \alpha}] = \CC t_\alpha$ is 1-dimensional with basis $t_\alpha$.
e. For any \( \alpha \in \Phi \), we have \( \alpha(t_\alpha) = \kappa(t_ \alpha, t_ \alpha) \neq 0 \).
f. (Important) If \( \alpha\in \Phi, x_ \alpha\in L_{ \alpha}\smz \) then there exists some \( y_ \alpha L_{- \alpha} \) in the opposite root space such that \( x_ \alpha, y_ \alpha, h_ \alpha \da [x_ \alpha, y_ \alpha] \) span a 3-dimensional subalgebra $\liesl(\alpha) \leq L$ isomorphic to $\liesl_2(\CC)$.
g. $h_\alpha = {2 t_\alpha \over \kappa(t_ \alpha, t_ \alpha)}$, \( \alpha(h_ \alpha) = 2, h_{ \alpha} = h_{- \alpha} \).

:::

:::{.proof title="?"}
\envlist

a. If it does not span, choose $h \in H\smz$ with \( \alpha(h) = 0 \) for all \( \alpha\in \Phi \).
  Then $[h, L_ \alpha] = 0$ for all $\alpha$, but $[h H] = 0$ since $H$ is abelian.
  Using the root space decomposition, $[h L] =0$ and so $h\in Z(L) = 0$ since $L$ is semisimple. $\contradiction$

b. Follows from proposition 8.2 and $\kappa(L_ \alpha, L_ \beta) = 0$ when \( \beta\neq \alpha \).

c. Let $h\in H$, then
\[
\kappa(h, [xy]) 
&= \kappa([hx], y) \\
&= \alpha(h) \kappa(x, y)\\
&= \kappa(t_\alpha, h) \kappa(x, y) \\
&= \kappa( \kappa(x, y) t_ \alpha, h) \\
&= \kappa( h, \kappa(x, y) t_ \alpha) \\
&\implies \kappa(h, [xy] - \kappa(x,y)t_\alpha) = 0 \\
&\implies [xy] = \kappa (x,y) t_ \alpha
,\]
where we've used that $[xy]\in H$ and $\kappa$ is nondegenerate on $H$ and $[L_{ \alpha}, L_{ - \alpha}] \subseteq L_0 = H$.

d. (c) shows $[L_{ \alpha}, L_{ - \alpha}]$ is spanned by $t_ \alpha$ if it is nonzero.
  Let $x\in L_ \alpha\smz$, then if $\kappa(x, L_{ - \alpha}) = 0$ then $\kappa$ would have to be degenerate, a contradiction.
  So there is some $y\in L_{ - \alpha}$ with \( \kappa(x, y) \neq 0 \).
  Moreover $t_\alpha\neq 0$ since $\alpha\neq 0$ and $\alpha\mapsto t_\alpha$ is an isomorphism.
  Thus $[xy] = \kappa(x,y) t_ \alpha$.

e. Suppose \( \alpha(t_\alpha) = \kappa(t_{ \alpha}, t_{ \alpha}) = 0\), then for $x\in L_{\alpha}, y\in L_{ - \alpha}$, we have $[t_ \alpha, x] = \alpha(t_ \alpha)x = 0$ and similarly $[t_ \alpha, y] = 0$.
  As before, find $x\in L_{ \alpha}, y\in L_{ - \alpha}$ with $\kappa(x,y)\neq 0$ and scale one so that $\kappa(x, y) = 1$.
  Then by (c), $[x, y] = t_ \alpha$, so combining this with the previous formula yields that $S \da \gens{x, y, t_ \alpha}$ is a 3-dimensional solvable subalgebra.[^dne_omg]
  Taking $\ad: L\injects \liegl(L)$, which is injective by semisimplicity, and similarly $\ro{\ad}{S}: S\injects \ad(S) \leq \liegl(L)$.
  We'll use Lie's theorem to show everything here is a commutator of upper triangular, thus strictly upper triangular, thus nilpotent and reach a contradiction.

[^dne_omg]: 
Note that this don't actually exist! We're in the middle of a contradiction.

:::