# Monday September 30

**Last time:** 
Let $\lieg$ be finite dimensional and $\lieh$ a maximal toral subalgebra.

Then $(\Phi, E)$ is a *root system*, and we obtain a bilinear product 
\begin{align*}
\inner{\wait}{\wait}: E\cross E \to \RR \\
(\alpha, \beta) \mapsto \kappa(t_\alpha, t_\beta)
.\end{align*}

*Examples:*
$\lieg = \liesl(3, \CC)$ and $\lieh = \CC h_1 \oplus \CC h_2$ where

> Todo: Insert clip image h1, h2

\begin{align*}
\alpha_1: \lieh \to \CC \\
h_1 \mapsto 2
h_2 \mapsto -1
.\end{align*}

\begin{align*}
\alpha_2: \lieh \to \CC \\
h_1 \mapsto -1
h_2 \mapsto 2
.\end{align*}

To find $t_{\alpha_i}$, we need to look at $\kappa\mid_\lieh$.

> Todo: Insert phone image

Since we only need the trace, this suffice, and we find

\begin{align*}
\left.\begin{array}{c|cc}{h_{1}} & {h_{2}} \\ {h_{1}} & {12} & {-6} \\ {h_{2}} & {-6} & {12}\end{array}\right]
.\end{align*}

We then get $t_{\alpha_1} = \frac{h_1}{6}$ and $t_{\alpha_2} = \frac{h_2}{6}$. 
Moreover
\begin{align*}
\inner{\alpha_1}{\alpha_1} = \kappa(t_{\alpha_1}, t_{\alpha_1}) = \frac 1 3 \in \QQ \\
\inner{\alpha_1}{\alpha_1} = \frac 1 3 \\ 
\inner{\alpha_1}{\alpha_2} = - \frac 1 6 \\
\inner{\alpha_1}{\alpha_2\dual} = \frac{2\inner{\alpha_1}{\alpha_2}}{\inner{\alpha_2}{\alpha_2}} = -1 \in \ZZ
\inner{\alpha_i}{\alpha_i\dual} = \frac{2\inner{\alpha_i}{\alpha_i}}{\inner{\alpha_i}{\alpha_i}} = 2 \in \ZZ
.\end{align*}

This leads to a nice fact: the matrix $\inner{\alpha_i}{\alpha_j\dual}$ has $\ZZ$ entries, and this is called the *Cartan matrix*.

## Ch.3: Root Systems

### Axiomatics: Reflections

Fix a Euclidean space $E$.

*Definition:*
A *hyperplane* in $E$ is a subspace of codimension 1.
A *reflection* in $E$ is an element $s\in \liegl(E)$ such that

$$
\theset{
E^s \coloneqq \theset{x\in E \suchthat sx = s} \text{ is a hyperplane } H \text{ and } s(x) = -x\quad \forall x\in E \suchthat (x, H) = 0
}$$

For nonzero $\alpha\in E$, its reflection is
\begin{align*}
S_\alpha: E \to E \\
\beta \mapsto \beta - \inner{\beta}{\alpha\dual}\alpha
.\end{align*}

with respect to $H_\alpha = \theset{x\in E \suchthat \inner{x}{\alpha} = 0}$, where $\alpha\dual = \frac{2\alpha}{\inner{\alpha}{\alpha}}$.

**Lemma:**
Let $\Phi \subseteq E$ be finite such that $S_\alpha(\Phi) = \Phi$ for all $\alpha \in \Phi$.

Suppose that $S \in \liegl(E)$ satisfies

1. $S(\Phi) = \Phi$,
2. $S(h) = h$ for all $h\in H$, and
3. $S(\alpha) = -\alpha$ for some $\alpha \in \Phi$,

then $S = S_\alpha$, i.e. this uniquely characterizes $S$

*Proof:*

Let $\tau = S \circ S_\alpha$.
Then $\tau(\Phi) = \Phi$ and $\tau(\alpha) = \alpha$.
This $\tau \actson \RR\alpha$ by 1, and similarly $\tau \actson E/\RR\alpha$ by 1 by picking a representative in $H$.
Moreover, all eigenvalues of $\tau$ are 1.
So the minimal polynomial of $\tau$ divides $(t-1)^{\dim E}$.

We want to show that $\tau \divides (t-1)^N$ for some large $N$, which forces $\tau \divides \gcd((t-1)^{\dim E}, t^N - 1) = 1$.
For any $\beta \in \Phi$ and $k > \abs{\Phi}$, not all vectors $\beta, \tau(\beta), \cdots \tau^k(\beta)$.
So $\beta = \tau^{k_\beta}(\beta)$ for some $k_\beta$ depending on $\beta$ (noting that $\tau$ is invertible.)

Multiplying all of these $k_\beta$s together, we can get some $k_\Phi$ that is larger than $\abs \Phi$, and so $\beta = \tau^{k_\Phi}$ for *all* $\beta \in \Phi$.
But then $\tau^{k_\Phi} = 1$ in $\liegl(E)$.

### Root Systems

**Definition:**
A subset $\Phi$ of $E$ a Euclidean space is called a *root system* iff

1. $\abs \Phi < \infty, 0\not\in\Phi$, and $E = \bigoplus_{\alpha\in\Phi} \RR \alpha$
2. $\alpha \in \phi \implies \CC \alpha \intersect \Phi = \theset{\pm \alpha}$
3. $\alpha \in \Phi \implies S_\alpha(\Phi) = \Phi$
4. $\alpha, \beta \in \Phi \implies \inner{\beta}{\alpha\dual} \in \ZZ$.

**Definition:**
The *rank* of a root system is the dimension on $E$.

**Definition:**
The *Weyl Group* of $\Phi$ is defined as 
$$
W = \generators{S_\alpha \mid \alpha \in \Phi} \subseteq \liegl(E)
$$

Note that $W \injects \Sigma_{\abs \Phi}$, a permutation group of size $\abs \Phi$.

**Lemma:**
If $g \in \liegl(E)$ and $g(\Phi) = \Phi$, then for all $\alpha, \beta \in \Phi$, we have
\begin{align*}
g s_\alpha g\inv = s_{g(\alpha)}, \\
\inner{\beta}{\alpha\dual} = \inner{g(\beta)}{g(\alpha)\dual},\\
\inner{\beta}{\alpha\dual} = \inner{w(\beta)}{w(\alpha)\dual}\quad \forall w\in W
.\end{align*}

*Proof:* 
Check 1-3 in Lemma 9.1.

*Proof of 1:*
We have
\begin{align*}
g s_\alpha g\inv(g(\beta)) = g s_\alpha (\beta) \in g(\Phi) = \Phi \quad \forall \beta \in \Phi, \\
.\end{align*}

*Proof of 2:*
We have
\begin{align*}
\theset{g(\beta) \suchthat \beta \in \Phi} = \Phi \implies g s_\alpha g\inv (\Phi) = \Phi \quad \forall h\in gH_\alpha \\
.\end{align*}

and so $gs_\alpha g\inv(h) = gg\inv(h) = h$, so $h$ is a fixed point of this map.

*Proof of 3:*
We have
$gs_\alpha g\inv(g(\alpha) = gs_\alpha(\alpha) = -g(\alpha)$, and so $gs_\alpha g\inv = s_{g(\alpha)})$ by Lemma 9.1.

Finally, we have 
\begin{align*}
g s_\alpha g\inv (g(\beta)) = g(s_\alpha(\beta)) = g(\beta - \inner{\beta}{\alpha\dual}\alpha) = g(\beta) - \inner{\beta}{\alpha\dual}g(\alpha) \\
= \\
s_{g(\alpha)} = g(\beta) - \inner{g(\beta)}{g(\alpha)\dual}g(\alpha)
.\end{align*}