# Friday September 27

**Last time:**
We saw $\Phi \subseteq \lieh^* = \theset{\alpha: \lieh \to \CC}$.

Suppose $\lieg$ is semisimple and $\lieh$ is a maximal toral subalgebra and take $F = \CC$.

**Propositions:**

a. $\dim \lieg_\alpha = 1 ~\forall \alpha \in \Phi$
b. $\CC \alpha \intersect \Phi = \theset{\pm \alpha}$, and $2\alpha \not\in\Phi$ where $c\alpha: \lieh \to \CC,~ h \mapsto c\actson \alpha(h)$. Moreover, $M = \lieh \oplus \left( \bigoplus_{c\neq0} \lieg_{c\alpha} \right)$
c. If $\alpha, \beta \in \Phi$ and $\beta \neq -\alpha$ Let $p,q \in \ZZ$ be the largest such that $\beta -p\alpha$ and $b + q\alpha$ are in $\Phi$. Moreover, $\beta(h_\alpha) = \kappa(t_\beta, t_\alpha) = p-q \in \ZZ$.

*Proof of (c):*

Set $M = \sum_{i\in \ZZ} \lieg_{\beta + i\alpha}$, which is an $\liesl(2, \alpha)$ module.
By (a), we have $\dim \lieg_{\beta + i\alpha} = 1 \iff \beta + i\alpha \in \Phi$.
But for all $x\in \lieg_{\beta + i\alpha}$, we have $[h, x] = (\beta + i\alpha)(h)x$ for all $h\in\lieh$.
But then $[h_\alpha, x] = (\beta(h_\alpha) + i \alpha(h_\alpha))x = (\beta(h_\alpha) + 2i)x$

Then $\lieg_{\beta + i\alpha} \subseteq M_{\beta(h_\alpha) + 2i}$, so $\beta(h_\alpha) \in \ZZ$.

Moreover, $\mathrm{Wt}(M) = 2\ZZ$ or $2\ZZ + 1$, and in particular $\dim M_0 + \dim M_1 = 1$.

Thus $M$ is irreducible, and $M \cong L(m)$ for some $m \in \ZZ_{\geq 0}$.
Moreover, $\mathrm{Wt}(M) = \theset{m, m-2, \cdots -m}$, and$simg \lieg_{\beta + i\alpha} = 1$ for all $i \in [-p, q]$.
Thus $\beta + i\alpha \in \Phi$.

*Proof of 8.3(e):* 
$\alpha(t_\alpha) \neq 0$.
The claim is that for all $\beta \in Phi$, there exists an $r\in \QQ$ such that $\beta(h) = r\alpha(h)$ for all $h \in [\lieg_\alpha, \lieg_{-\alpha}]$.

There are two cases: if $\beta = -\alpha$, then we're done by the previous argument.

Otherwise, $\beta \neq -\alpha$. 
Take $M = \bigoplus_{i \in\ZZ} \lieg_{\beta + i\alpha}$.

Then,
\begin{align*}
\Tr_M(\ad h) 
&= \sum_i \Tr_M( (\ad e_i \circ \ad f_i) -(\ad f_i \circ \ad e_i) ) \\
&= \sum_i \Tr_{\lieg_{\beta + i\alpha}}(\ad h) \\
&= \sum (\beta + i\alpha)(h) \dim \lieg_{\beta + i\alpha} \\
&= \sum_i \dim \lieg_{\beta + i\alpha} \beta(h) + \sum_i i\dim(\lieg_{\beta + i\alpha}) \\
\implies \beta(h) &= 
\frac{
-\sum_i \dim \lieg_{\beta + i\alpha}
}{
\sum_i \dim \lieg_{\beta + i\alpha}
}\alpha(h)
.\end{align*}

Now consider the killing form $\kappa(t_\beta, t_\alpha) = \beta(t_\alpha) = r \alpha(t_\alpha)$, where the last equality is what we are claiming.

Suppose that $\alpha(t_\alpha) = 0$.
Then $\kappa(t_\beta, t_\alpha) = 0$ for all $\beta \in Phi$.
By the non-degeneracy of $\kappa$, we have $t_\alpha = 0$ and thus $\alpha = 0$.

## Summary 

We have $\lieg$ semisimple, finite dimensional, and $\lieh$ a maximal toral subalgebra (i.e. the Cartan subalgebra).
This implies that $\kappa$ is nondegenerate, and we have a correspondence
\begin{align*}
\lieh \iff \lieh\dual \\
h \mapsto \kappa(h, \wait) \\
t_\alpha \leftarrow \alpha
.\end{align*}

This gives a symmetric bilinear form $(\wait, \wait): \lieh\dual \to \lieh\dual$.

For $\alpha \in Phi$, define its *coroot* $\alpha\dual = \frac{2}{(\alpha, \alpha)}\alpha$.

Note that $(\wait)\dual$ is not linear: note that
\begin{align*}
(2\alpha)\dual = 
\frac{2}{(2\alpha, 2\alpha)} 2\alpha 
= \frac{\alpha}{(\alpha, \alpha)} 
= \frac{\alpha\dual}{2}
.\end{align*}

Assume that $\Phi = \theset{\alpha_i}$.
Define $E_\QQ = \bigoplus_{i=1}^\ell \QQ_{\alpha_i}$, and $E = \RR \tensor_\QQ E_\QQ$.

**Lemma:**
If $\alpha, \beta \in \Phi$, then

a. $(\beta, \alpha) \in \QQ$,
b. $(\wait, \wait)$ on $E_\QQ$ is positive definite, i.e. $x\neq 0 \implies (x, x) > 0$.

An immediate consequence of (b) is that $(\wait, wait)$ on $E$ is an inner product.

*Proof:*
For all $\lambda, \mu \in \lieh\dual$, we have 
\begin{align*}
(\lambda, \mu) 
&= \kappa(t_\lambda, t_\mu) \\
&= \Tr_\lieg(\ad t_\lambda \circ \ad t_\mu) \\
&= \Tr_\lieg(...) + \sum_{\alpha \in \Phi} \Tr_{\lieg_\alpha}(...) \\
&= 0 + \sum_{\alpha \in \Phi} \alpha(t_\lambda) \alpha(t_\mu) \\
&= \sum_[\alpha \in \Phi] = \kappa(t_\alpha, t_\lambda) \kappa(t_\alpha, t_\mu) \\
&= \sum_{\alpha \in \Phi} (\alpha, \lambda)(\alpha, \mu) \\
.\end{align*}

So pick $\lambda = \mu = \alpha \in \Phi$.
Then $(\alpha, \alpha) = \sum_{\beta \in \Phi} (\beta, \alpha)^2$.

Then

\begin{align*}
\frac{1}{(\alpha, \alpha)} = \sum_{\beta \in \Phi} \left(\frac{(\beta, \alpha\dual)}{2} \right)^2
.\end{align*}

where $(\beta, \alpha\dual) = \cdots = \beta(h_\alpha)\in \ZZ$.

This means that $(\alpha, \alpha) \in \QQ_{> 0}$.

**Summary of Properties Proved:**

Let $\alpha, \beta \in \Phi$. Then

1. $0 \not\in \Phi$ and $\Phi$ spans $E$
2. $\CC\alpha \intersect \Phi = \theset{\pm \alpha}$
3. $\beta - (\beta, \alpha\dual)\alpha \in \Phi$
4. $(\beta, \alpha\dual) \in \ZZ$

Thus the assignment $(\lieg, \lieh) \mapsto (\Phi, E)$ defines a **root system**. 
This only works when $\lieg$ is semisimple and $\lieh$ is maximal toral.

*Proof of (3):*

We computed $(\beta, \alpha\dual) = p-q$.
Then $-p \leq -(\beta, \alpha\dual) = q-p \leq q$.
So this must be something on the root stream.