# Monday, October 17

## 10.3: The Weyl group

:::{.theorem title="?"}
Fix a base for $\Phi$.

a. ($W$ acts transitively on the set of Weyl chambers)
  If $\gamma\in \EE$ is regular (not on a root hyperplane), there exists $\sigma\in W$ such that $( \sigma(\delta), \alpha)> 0$ for all $\alpha\in \Delta$, i.e. $\sigma( \gamma) \in \mcc(\Delta)$, the dominant Weyl chamber relative to $\Delta$.

b. ($W$ acts transitively on bases)
  If $\Delta'$ is another base for $\Phi$, then there exists $\sigma\in W$ such that $\sigma( \Delta') = \Delta$, so $W$ acts transitively on bases.

c. (Every orbit of $W\Phi$ contains a simple root)
  If $\beta\in \Phi$ then there exists a $\sigma\in W$ such that $\sigma( \beta)\in \Delta$.

d. ($W$ is generated by simple roots) 
  $W = \gens{s_ \alpha \st \alpha\in \Delta}$ is generated by *simple* roots.

e. (Stabilizers are trivial) 
  If $\sigma( \Delta) = \Delta$ for some $\sigma\in W$, then $\sigma = 1$.

:::

:::{.proof title="?"}
**Part c**:
Set $W' \da \gens{s_ \alpha\st \alpha\in \Delta}$, we'll prove (c) with $W$ replaced $W'$, which is larger.
First suppose \( \beta\in \Phi^+ \) and consider $W' \beta \intersect \Phi^+$.
This is nonempty since it includes $\beta$ and is a finite set, so choose $\gamma$ in it of minimal height.
Claim: $\height(\gamma) = 1$, making $\gamma$ simple.
If not, supposing $\height( \gamma) > 1$, write $\gamma = \sum_{ \alpha\in \Delta} k_ \alpha \alpha$ with $k_ \alpha > 0$.
Since $\gamma\neq 0$, we have $(\gamma, \gamma) > 0$, so substitute to yield
\[
0 < (\gamma, \gamma) = (\gamma, \sum_{\alpha \in \Delta} k_ \alpha \alpha) = \sum_{\alpha\in \Delta} k_ \alpha (\gamma, \alpha)
,\]
so $(\gamma, \alpha)>0$ for some \( \alpha \in \Delta \), and $s_{ \alpha} \gamma = \gamma - \inp \gamma \alpha \alpha\in \Phi^+$ is positive where $\inp \gamma \alpha > 0$.
This is a contradiction, since it has a smaller height.
Note that if $\beta\in \Phi^-$ then $-\beta\in \Phi^+$ and there exists a $\sigma\in W'$ such that $\sigma( - \beta) = \alpha\in \Delta$.
So $\sigma( \beta) = - \alpha$, and $s_ \alpha \sigma( \beta) = s_ \alpha( - \alpha) = \alpha \in \Delta$.

**Part d**:
Given $\beta$, pick $\sigma\in W'$ such that $\sigma\inv( \beta) = \alpha\in \Delta$.
Then 
\[
s_ \beta = s_{ \sigma( \alpha)} = \sigma s_{ \alpha} \sigma\inv \in W'
,\]
so $W \leq W' \leq W$, making $W = W'$.

**Parts a and b**:
Recall $\rho = {1\over 2}\sum _{ \beta\in \Phi^+}$ and choose $\sigma\in W$ such that $( \sigma(\delta), \rho)$ is maximal (picking from a finite set).
Given $\alpha\in \Delta$, we have $s_\alpha \sigma\in W$, and so
\[
( \sigma(\delta), \rho) 
&\geq ( \sigma(\delta), \rho) \\
&= (\sigma(\delta), s_ \alpha \rho) \\
&= (\sigma(\delta), \rho - \alpha) \\
&= (\sigma(\delta), \rho ) - ( \sigma( \delta), \alpha)
,\]
and so $( \sigma( \delta), \alpha)\geq 0$ for all $\alpha\in \Delta$.
Importantly, $\gamma$ is regular, so this inequality is structure for all $\alpha\in \Delta$.
So $W$ acts transitively on the Weyl chambers, and consequently on simple systems (i.e. bases for $\Phi$) by the discussion at the end of $\S 10.1$.2

**Part e**:
Suppose $\sigma( \Delta) = \Delta$ and $\sigma \neq 1$, and write $\sigma = \prod_{1\leq i \leq t} s_i$ with $s_i \da s_{\alpha_i}$ for \( \alpha_i \in \Delta\) with $t \geq 1$ minimal.
Note $\sigma( \Delta) = \Delta$ and $\alpha_t \in \Delta$, we have \( \sigma( \alpha_t) > 0 \) and $\prod_{1\leq i\leq t}(\alpha_t) = \prod_{1\leq i \leq t-1}s_i (-\alpha_t)$ so $\prod_{1\leq i\leq t-1} s_i(\alpha_t) < 0$.
This fulfills the deletion condition, so $\prod_{1\leq i \leq t} = s_1\cdots \hat{s_u}\cdots \hat{s_t}$ which is of smaller length.
:::

:::{.remark}
In type $A_n$, $\size W(A_n) \approx n!$, and since bases biject with $W$ there are many choices of bases.
:::

:::{.definition title="?"}
Let $\Delta \subseteq \Phi$ be a base and write \( \sigma\in W \) as \( \sigma = \prod_{1\leq i \leq t} s_{\alpha_i} \) with $\alpha_i\in \Delta$ and $t$ minimal.
We say this is a **reduced expression** for $\sigma$ and say $t$ is the **length** of $\sigma$, denoted $\ell( \sigma)$.
By definition, $\ell(1) = 0$.
:::

:::{.remark}
Since $W \leq \GL(\EE)$, there is a map $\det: W\to \GL_1(\RR) = \RR\units$.
The determinant of a reflection is $-1$ by writing it in a basis about the fixed hyperplane, and so $\det\sigma = (-1)^{\ell( \sigma)}$ and in fact $\det: W\to \ts{\pm 1}$.
Thus $\ell( \sigma \sigma') \equiv \ell( \sigma) + \ell( \sigma)\mod 2$.

Note also that if $\sigma' = s_ \alpha$ for $\alpha$ simple, then $\ell( \sigma s_{ \alpha}) = \ell( \sigma) \pm 1$.
The proof: $\ell( \sigma s_ \alpha)\leq \ell( \sigma) + 1$, similarly for $\sigma s_ \alpha$, and use $\det( \sigma s_ \alpha) = - \det \sigma$.
:::

:::{.warnings}
Reduced expressions are not unique: for $A_2$, one has $s_ \alpha s_ \beta s_ \alpha = s_ \beta s_ \alpha s_ \beta$, and these two reflections do not commute.
:::

:::{.remark}
Some temporary notation for this section: for $\sigma\in W$, set
\[
n( \sigma) \da \size( \Phi^- \intersect \sigma(\Phi^+))
,\]
the number of positive roots that $\sigma$ sends to negative roots.
:::

:::{.lemma title="A"}
For all $\sigma\in W$,
\[
n( \sigma) = \ell( \sigma)
.\]
:::

:::{.proof title="?"}
Induct on $\ell(\sigma)$: if zero, then $\sigma = 1$ and $n(1) = 0$ since it fixes all positive roots.
If $\ell( \sigma ) = 1$ then \( \sigma = s_{ \alpha} \) for some simple $\alpha$, and we know from the last section that $\sigma$ permutes $\Phi^+\smts{ \alpha}$ and \( \sigma( \alpha) = - \alpha \), so $n( \sigma) = 1$.
:::