# Friday, October 07

:::{.lemma title="?"}
Let $\Phi \subseteq \EE$ be a root system with Weyl group $W$.
If $\sigma\in \GL(\EE)$ leaves $\Phi$ invariant then 
\[
\sigma s_{\alpha} \sigma\inv = s_{ \sigma( \alpha)} \qquad\forall \alpha\in \Phi
\]
and 
\[
( \beta, \alpha\dual) = ( \sigma(\beta), \sigma(\alpha)\dual ) \qquad \forall \alpha, \beta \in \Phi
.\]
:::

:::{.warnings}
\[
( \sigma( \beta), \sigma( \alpha) ) \neq (\beta, \alpha)
,\]
i.e. the $(\wait)\dual$ is important here since it involves a ratio.
Without the ratio, one can easily scale to make these not equal.
:::

:::{.definition title="?"}
Two root systems \( \Phi \subseteq \EE, \Phi' \subseteq \EE' \) are **isomorphic** iff there exists $\phi: \EE\to \EE'$ of vector spaces such that $\phi(\Phi) = \Phi'$ such that 
\[
(\varphi( \beta), \varphi(\alpha)\dual) = (\beta, \alpha) \da {2 (\beta, \alpha) \over (\alpha, \alpha)} \qquad\forall \alpha, \beta \in \Phi
.\]
:::

:::{.example title="?"}
One can scale a root system to get an isomorphism:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-19.pdf_tex} };
\end{tikzpicture}

:::

:::{.remark}
Note that if $\phi: \Phi \iso \Phi'$ is an isomorphism, then
\[
\varphi(s_{ \alpha}( \beta)) = s_{ \varphi( \alpha)}( \varphi(\beta)) \qquad \forall \alpha, \beta\in \Phi \implies \varphi \circ s_{ \alpha} \varphi\inv = s_{ \varphi( \alpha)}
.\]
So $\phi$ induces an isomorphism of Weyl groups 
\[
W &\iso W' \\
s_{\alpha} &\mapsto s_{ \varphi( \alpha)}
.\]

By the lemma, an automorphism of $\Phi$ is the same as an automorphism of $\EE$ leaving $\Phi$ invariant.
In particular, $W\injects \Aut( \Phi)$.
:::

:::{.definition title="Dual root systems"}
If $\Phi \subseteq \EE$ is a root system then the **dual root system** is
\[
\Phi\dual \da \ts{ \alpha\dual \st \alpha\in \Phi}, \qquad \alpha\dual \da {2\alpha\over (\alpha, \alpha)}
.\]
:::

:::{.exercise title="?"}
Show that $\Phi\dual$ is again a root system in $\EE$.
:::

:::{.remark}
One can show $W( \Phi) = W( \Phi\dual)$ and $\inp {\lambda}{ \alpha\dual} \alpha\dual = \inp{ \lambda}{ \alpha} \alpha = (\lambda, \alpha\dual)$ for all \( \alpha\in \Phi, \lambda\in \EE \), so \( s_{\alpha\dual} = s_{\alpha} \) as linear maps on $\EE$.
:::

## 9.3: Example(s)

:::{.definition title="Ranks of root systems"}
Let \( \Phi \subseteq \EE \) be a root system, then $\ell \da \dim_\RR \EE$ is the **rank** of $\Phi$.
:::

:::{.remark}
Rank 1 root systems are given by choice of $\alpha\in \RR$:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-31.pdf_tex} };
\end{tikzpicture}

:::

:::{.remark}
Recall ${2( \beta, \alpha) \over (\alpha, \alpha)} \in \ZZ$, and from linear algebra, $\inp{v}{w} = \norm v \cdot \norm w \cos( \theta)$ and $\norm{\alpha}^2 = ( \alpha, \alpha)$.
We can thus write
\[
\inp{ \beta}{ \alpha} = {2( \beta, \alpha) \over (\alpha, \alpha)} = 2{\norm \beta\over \norm \alpha} \cos (\theta), \qquad \inp \alpha \beta= 2{\norm \alpha\over \norm \beta} \cos( \theta)
,\]
and so
\[
\inp \alpha \beta\inp \beta \alpha = 4\cos^2( \theta)
,\]
noting that $L_{ \alpha, \beta} \da \inp \alpha \beta, \inp \beta \alpha$ are integers of the same sign.
If positive, this is in QI, and if negative QII.
This massively restricts what the angles can be, since $0 \leq \cos^2( \theta) \leq 1$.

First, an easy case: suppose $L_{ \alpha, \beta} = 4$, so $\cos^2( \theta) = 1\implies \cos( \theta) = \pm 1\implies \theta= 0, \pi$.

- If $0$, then $\alpha,\beta$ are in the same 1-dimensional subspace and thus $\beta = \alpha$.
  In this case, \( \inp \beta \alpha = 2 = \inp \alpha \beta \).
- If $\pi$, then $\alpha = - \beta$.
  Here $\inp \beta \alpha = -2$.

So assume \( \beta\neq \pm \alpha \), and without loss of generality $\norm \beta\geq \norm \alpha$, or equivalently $\inp \alpha \beta \leq \inp \beta \alpha$.
Note that if $\inp \alpha \beta\neq 0$ then
\[
{ \inp \beta \alpha\over \inp \alpha \beta} = {\norm{ \beta}^2 \over \norm{ \alpha}^2}
.\]

The other possibilities are as follows:

| $\inp \alpha\beta$ 	| $\inp \beta\alpha$ 	| $\theta$ 	| $\norm{\beta}^2/\norm{\alpha}^2$ 	|
|--------------------	|--------------------	|----------	|----------------------------------	|
| 0                  	| 0                  	| $\pi/2$  	| Undetermined                     	|
| 1                  	| 1                  	| $\pi/3$  	| 1                                	|
| -1                 	| -1                 	| $2\pi/3$ 	| 1                                	|
| 1                  	| 2                  	| $\pi/4$  	| 2                                	|
| -1                 	| -2                 	| $3\pi/4$ 	| 2                                	|
| 1                  	| 3                  	| $\pi/6$  	| 3                                	|
| -1                 	| -3                 	| $5\pi/6$ 	| 3                                	|

Cases for the norm ratios:

- $1: A_2$
- $2: B_2 = C_2$
- $3: G_2$

These are the only three irreducible rank 2 root systems.
:::

:::{.lemma title="?"}
Let \( \alpha, \beta\in\Phi \) lie in distinct linear subspaces of $\EE$.
Then

1. If $(\alpha, \beta) > 0$, i.e. their angle is strictly acute, then $\alpha - \beta$ is a root 
2. If $(\alpha, \beta) < 0$ then \( \alpha + \beta \) is a root.

:::

:::{.proof title="?"}
Note that (2) follows from (1) by replacing $\beta$ with $-\beta$.
Assume $(\alpha, \beta) > 0$, then by the chart \( \inp \alpha \beta =1 \) or \( \inp \beta \alpha = 1 \).
In the former case, 
\[
\Phi\ni s_{ \beta}( \alpha) = \alpha - \inp \alpha \beta \beta = \alpha- \beta
.\]
In the latter,
\[
s_{ \alpha}(\beta) = \beta- \alpha \in \Phi\implies - (\beta- \alpha) = \alpha- \beta\in \Phi
.\]
:::

:::{.remark}
Suppose $\rank( \Phi) = 2$.
Letting \( \alpha\in \Phi \) be a root of shortest length, since $\RR\Phi = \EE$ there is some $\beta \in \EE$ not equal to $\pm \alpha$.
Without loss of generality assume $\angle_{\alpha, \beta}$ is obtuse by replacing $\beta$ with $-\beta$ if necessary:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-57.pdf_tex} };
\end{tikzpicture}

Also choose $\beta$ such that $\angle_{ \alpha, \beta}$ is maximal.

**Case 0**:
If $\theta = \pi/2$, one gets $\AA_1\times \AA_1$:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-59.pdf_tex} };
\end{tikzpicture}

We'll continue this next time.
:::