# Monday September 23

**Last time:**
$\mathfrak{h}$ is a *toral* subalgebra if it contains only semisimple elements, and implies that there is a *root space decomposition*
$$
\lieg = \mathfrak g_0 \oplus \bigoplus_{\alpha \in \Phi} \lieg_\alpha
$$

where $\lieg_\alpha = \theset{x\in\lieg \suchthat [h, x] = \alpha(h)x ~\forall h\in\mathfrak h}$ and $\Phi = \theset{\alpha: \mathfrak h \to \CC \suchthat \lieg_\alpha\neq 0, \alpha \neq 0}$ and $\lieg_0 = C_\lieg(\mathfrak h)$.

Take larger $\mathfrak h$ yields finer decompositions, and a maximal $\mathfrak h$ gives $\dim \lieg_\alpha  = 1 \forall \alpha \in Phi$. 

**Corollary:** 
$\restrictionof{\kappa}{\lieg_0}$ is nondegenerate.

## The Centralizer of $\mathfrak h$

If $x,y \in \mathrm{End}(V)$ where $V$ is finite dimensional, $xy=yx$, and $y$ is nilpotent, then $xy$ is nilpotent and $\Tr(xy)=0$.

**Proposition:**
If $\mathfrak h\subseteq \lieg$ is a maximal toral subalgebra, then $\mathfrak h = \lieg_0$.

*Proof:*

**Step 1:**
If $x\in \lieg_0$, then $x_s, x_n \in \lieg_0$.

If $x\in\lieg_0$, then $\ad x(\mathfrak h) \subseteq 0$.
By proposition 4.2, $\ad x_s(\mathfrak h) \subseteq 0, \ad x_n(\mathfrak h)$, and so $x_s, x_n \in \lieg_0$.

**Step 2:**
$\theset{x_s \suchthat x\in\lieg_0} \subseteq \mathfrak h$.

If $x\in \lieg_0$, then by step 1 we have $x_s \in \lieg_0$ and so $\mathfrak h + \CC x_s$ is toral, and thus $x_s \in \mathfrak h$.

**Step 3:**
$\restrictionof{\kappa}{\mathfrak h}$ is non-degenerate.

We want to show that $\kappa(h, x) = 0 ~\forall x\in\lieg \implies h = 0$.
By the corollary, it suffices to show that $\kappa(h, x) = 0 ~\forall x\in \lieg_0$.
By step 2, it suffices to check this only for $x\in \lieg_0$ such that $x=x_n$.

If $x=x_n$, then $\ad x_n$ is nilpotent and $\ad h$ commutes with $\ad x$ because $[h, x] = 0$ (since $x\in \lieg_0$).
By the lemma, $\Tr(\ad h \circ \ad x) = 0$, since $\ad h = \kappa(h, x)$.

**Step 4:**
$\lieg_0$ is nilpotent.

Pick $x\in \lieg_0$. 
Then by step 2, $x_s \in \mathfrak h$, so $\ad x_s: \lieg_0 \selfmap$ is a zero map and thus nilpotent.

So $\ad x_n$ is nilpotent, meaning that $\ad x$ is nilpotent.
By Engel's theorem, this implies that $\lieg_0$ itself is nilpotent.

**Step 5:**
$\lieg_0$ is abelian.

Suppose that $I \coloneqq [\lieg_0, \lieg_0] \neq 0$. 
We have $I \normal \lieg_0$, and $I$ is not nilpotent whereas $\lieg_0$ is.

By Lemma 3.3, we have $I \intersect Z(\lieg_0) \neq 0$, so pick $x$ in the intersection.
Note that $\kappa(\mathfrak h, I) = \kappa(\mathfrak h, [\lieg_0, \lieg_0])$, which by associativity equals $\kappa([\mathfrak h, \lieg_0], \lieg_0) = 0$.

By step 3, we have $\mathfrak h \intersect I = 0$.
By step 2, $x\neq x_s$, and thus $x_n \neq 0$.
But we also have $x\in Z(\lieg_0)$, so $[x, \lieg_0] = 0$ and $\ad x(\lieg_0) \subseteq 0$.
By Proposition 4.2, this holds for $x_s, x_n$ as well, which are both in the center.
So for all $y\in \lieg_0$, $\ad y$ commutes with $\ad x_n$, which is nilpotent.

By the lemma, this implies that $0 = \Tr(\ad y \circ \ad x_n) = \kappa(x_n, y)$ for all $y\in \lieg_0$.
So $x_n = 0$.

**Step 6:**
Suppose $\lieg_0 \not\subset \mathfrak h$.
By step 2, there exists an $x\in \lieg_0$ such that $x\not\in\mathfrak h$, where $x_n \neq 0$.
By step 5, $[x_n, y] = 0$ for all $y\in\lieg_0$.
Then $\ad x$ (which is nilpotent) commutes with $\ad y$.
By the lemma, $0 = \kappa(x_n, y)$ for all $y\in\lieg_0$, and thus $x_n = 0$. $\qed$

> Main idea: Choose a maximal toral subalgebra to get a nice root space decomposition, and so it coincides with $\lieg_0$.

**Corollary:**
$\restrictionof{\kappa}{\lieg}$ is nondegenerate.

Thus for all $\alpha \in \mathfrak h^*$, there exists a unique $t_\alpha \in \mathfrak h$ such that $\alpha = \kappa(t_\alpha, \wait): \mathfrak h \to \CC$.

In other words, there is an identification
\begin{align*}
\mathfrak h &\mapsvia{1-1} \mathfrak h^* \\
h &\mapsto \kappa(h, \wait) \\
t_\alpha &\leftarrow \alpha
.\end{align*}

**Definition:**
A subalgebra $\mathfrak h \subseteq \lieg$ is a *Cartan subalgebra* if $\mathfrak h$ is nilpotent and 
$$
\mathfrak h = N_\lieg(\mathfrak h) = \theset{x\in \lieg \suchthat [x, h] \subseteq \mathfrak h}.
$$

Note that $N_\lieg(\mathfrak h)$ is the largest subalgebra of $\lieg$ in which $\mathfrak h$ is an ideal.

**Remark:** 
If $\lieg$ is semisimple and finite dimensional with $\mathrm{char}(F) = 0$, we will have a correspondence:
\begin{align*}
\theset{\text{CSAs of $\lieg$}} \iff \theset{\text{maximal toral subalgebras of $\lieg$}}
.\end{align*}

Maximal toral subalgebras advantages over Cartan subalgebra definition:

- Yields the finest root space decomposition
- $\mathfrak h^* = \mathfrak h$, Weyl group?
- Existence is easy compared to CSAs

On the other hand, CSA advantages:

- All CSAs are conjugate under $G$ (some group to be defined)
- The dimensions of all CSAs are the same, giving a well-defined notion of dimension ($\mathrm{rank} \lieg = \dim \mathfrak h$).

## 8.3: Orthogonality Properties

> From now on, $\mathfrak h$ will be a maximal toral subalgebra.

**Proposition:**
Let $\alpha \in \Phi$. Then

a. $\Phi$ spans $\lieh^*$
b. $-\alpha \in \Phi$
c. $\forall x\in \lieg_\alpha, y\in\lieg_{-\alpha}$, we have $[x,y] = \kappa(x, y)t_\alpha$
d. $[\lieg_\alpha, \lieg_{-\alpha}] = \CC t_\alpha$ (let the nonzero scalar be $\lambda$)
e. $\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) \neq 0$.
f. For any nonzero $e_\alpha \in \lieg_\alpha$, there exists a unique $f_\alpha \in \lieg_{-\alpha}$ such that $[e_\alpha, f_\alpha] = h_\alpha \coloneqq \frac{\lambda}{\kappa(t_\alpha, t_\alpha)} t_\alpha$. Moreover, $\generators{e_\alpha, f_\alpha, h_\alpha} = \liesl(2, \CC)$.