# Wednesday September 25

**Today:** 
Properties of the root space when the toral subalgebra is maximal.

**Last time:**
We have $\lieg = \lieg \oplus \left( \bigoplus_{\alpha \in \Phi} \lieg_\alpha \right)$ where $\kappa\mid_{\lieh}$ is nondegenerate.
We also have a correspondence
\begin{align*}
\lieh \iff \lieh^*
h \mapsto \kappa(\lieh, \wait) \\
t_\alpha \leftarrow \alpha \coloneqq \kappa(t_\alpha, \wait)
.\end{align*}

## Orthogonality Properties

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

a. $\Phi$ spans $\lieh^*$
b. $-\alpha \in \Phi$
c. $[x, y] = \kappa(x,y)t_\alpha$ for all $x\in \lieg_\alpha$ and $y\in \lieg_{-\alpha}$
d. $[\lieg_\alpha, \lieg_{-\alpha}] = \CC t_\alpha$
e. $\alpha(t_\alpha) = \kappa(t_\alpha, t_\alpha) \neq 0$
f. For each nonzero $e_\alpha \in \lieg_\alpha$, there exists a unique $f_\alpha \in \lieg_{-\alpha}$ such that $[e_\alpha, f_\alpha] = h_\alpha \coloneqq \frac{2}{\kappa(t_\alpha, t_\alpha)}t_\alpha$. Moreover, $\generators{e_\alpha, t_\alpha, h_\alpha} \cong \liesl(2, \CC)$.

*Proof of (a):*
We want to show that $h \in \lieh$ implies that if $\alpha(h) = 0$ for all $\alpha \in \Phi$, then $h = 0$.

Take $x\in \lieg_\alpha$. 
Then $[h, x] = \alpha(h)x = 0$. 
So $[\lieh, \lieg] = 0$ because $\lieh$ is abelian.
But then $[h, \lieg] = 0$, or $h \in Z(\lieg) = 0$ since $\lieg$ is semisimple.

*Proof of (b):*
By Proposition 8.1c, we have $\kappa(\lieg_\alpha, \lieg_\beta) = 0$ for all $\beta \neq -\alpha$.

If $-\alpha \not\in\Phi$, then $\lieg_{-\alpha} = 0$.
So $\kappa(\lieg_\alpha, \lieg) = 0$ by the non-degeneracy of $\kappa$.

*Proof of (c):*
For all $h\in\lieh$, we have
\begin{align*}
\kappa(h, [x, y]) = \kappa([h, x], y) \\
= \kappa(\alpha(h)x, y) \\
= \kappa(t_\alpha, h)\kappa(x, y) \\
= \kappa(\kappa(x,y)t_\alpha, h) \\
= \kappa(h, \kappa(x,y)t_\alpha)
.\end{align*}

which implies that $\kappa(h, [x, y] - \kappa(x,y)t_\alpha) = 0$, which forces the second argument to be zero by non-degeneracy.

*Proof of (d):*
We will show that (d) implies (c), i.e. $[\lieg_\alpha, \lieg_{-\alpha}] \subseteq \CC t_\alpha$.

We want to show $\kappa(x,y)$ is not always zero.

Pick any nonzero $x\in\lieg_\alpha$.
Then $\kappa(x, \lieg_\beta) = 0$ for all $\beta\neq -\alpha$.
If $\kappa(x, \lieg_{-\alpha}) = 0$, then $\kappa(x, \lieg) = 0$. 
By non-degeneracy, this forces $x=0$.

*Proof of (e):*
We will skip this for now, and revisit with methods from later sections that make this proof simpler.

*Proof of (f):*
Let $e_\alpha \neq 0$ in $\lieg_\alpha$. 
Then there exists a $y\in\lieg_{-\alpha}$ such that $\kappa(e_\alpha, y) \neq 0$.
Set $f_\alpha \in \lieg_{-\alpha}$ such that $\kappa(e_\alpha, f_\alpha) = \frac{2}{\kappa(t_\alpha, t_\alpha)}$.

By (c), we have
\begin{align*}
[e_\alpha, f_\alpha] = \kappa(e_\alpha, t_\alpha)t_\alpha \\
= \frac{2}{\kappa(t_\alpha, t_\alpha)}t_\alpha \\
= h_\alpha
.\end{align*}

and
\begin{align*}
[h_\alpha, e_\alpha] = \frac{2}{\kappa(t_\alpha, t_\alpha)}[t_\alpha, e_\alpha] \\
= \frac{2}{\kappa(t_\alpha, t_\alpha)} \alpha(t_\alpha) e_\alpha \\
= 2 e_\alpha
.\end{align*}

and similarly $[h_\alpha, f_\alpha] = -2 f_\alpha$.

**Definition:**
Let $\liesl(2, \alpha) = \generators{e_\alpha, f_\alpha, h_\alpha}$ as in (f).
A priori, this depends on a choice of $e_\alpha \neq 0$.
We will show that this only depends on $\alpha$.

## Orthogonality/Integrality Properties

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

a. $\dim \lieg_\alpha = 1$. (Note that in general, $\dim \lieg_0 = \dim \lieh \geq 1$)
b. $\CC\alpha \intersect \Phi = \theset{\pm \alpha}$
c. If $\beta \in \Phi$ such that $\alpha + \beta \in \Phi$, then $[\lieg_\alpha, \lieg_\beta] = \lieg_{\alpha + \beta}$.
d. If $\beta \neq -\alpha \in \Phi$, then let $p,q \in \ZZ$ be the largest such that $\beta - p\alpha$ and $\beta + q\alpha$ are both in $\Phi$. Then $\beta + i\alpha \in \Phi$ for every $-p \leq i \leq q$, and
\begin{align*}
\beta(h_\alpha) = \kappa(t_\beta, h_\alpha) = \frac{2\kappa(t_\beta, t_\alpha)}{\kappa(t_\alpha, t_\alpha)} = p-q \in \ZZ
.\end{align*}

*Proof of (a) and (b):*

Let $M = \lieh \oplus \left( \bigoplus_{c\neq 0} \lieg_{c\alpha} \right) \leq \lieg$ as a subspace.
By a routine check, $M$ is an $\liesl(2, \alpha)$ submodule of $\lieg$.
Recall that $M = \bigoplus_{\lambda \in \ZZ}$ as a direct sum of vector spaces.
Applying Weyl's theorem, we also have $M = \bigoplus_{m \in \ZZ_{\geq 0}} L(m)^{\oplus \mu_m}$ as a direct sum of (irreducible?) modules.

For $\lieh$, if we have $[h_\alpha, h] = 0$ for all $h\in \lieh$, then $h \in M_0$.
For $\lieg_{c\alpha}$, $[h_\alpha, x] = c\alpha(h_\alpha) x$ for all $x\in \lieg_{c\alpha}$.
But this equals $2cx$.
So this implies that $\lieg_{c\alpha} \subseteq M_{2c}$.

Thus $2c \in \ZZ$, and thus $c \in \frac 1 2 \ZZ$, and $M_0 = \lieh$.

We then have $\dim M_0 = \sum_{m\in 2\ZZ} \mu_m$.
So write $h = \CC t_\alpha \oplus \ker \alpha$ as vector spaces.
Consider the action $\liesl(2, \alpha) \actson \ker \alpha$, which is trivial since $h \in \ker \alpha$.
We $[h_\alpha, h] = 0$, $[e_\alpha, h] = -\alpha(h)e_\alpha = 0$ since $h\in\ker\alpha$, and similarly $[f_\alpha, h] = 0$.

Thus $\ker \alpha = L(0)^{\oplus \dim \lieh - 1}$. 
Moreover, $\liesl(2, \alpha) = L(2) = \mathrm{span}(e_\alpha, t_\alpha, f_\alpha)^T$.
But this forces the case that there is no other summand of the form $L(k)$ for $k$ even in $M$.

Then $\lieg_{2\alpha} \subseteq M_4$, which must be zero.
So $2\alpha \not\in \Phi$, so $2\alpha$ is not a root.
("Twice a root is never a root")

So $\frac 1 2 \alpha \not\in \Phi$, otherwise we could apply this argument to conclude that $\alpha$ is not a root and reach a contradiction.
Thus $M_1 = 0$, since $c\neq \frac 1 2$ implies that there is not summand of the form $L(k)$ for $k$ odd in $M$.
But this forces $M = \lieh \oplus \liesl(2, \alpha)$.

> Motto: reduce the complexity by using the $\liesl(2)$ module structure and its representation theory!