# Wednesday, October 19

:::{.proof title="of lemma A, continued"}
We're proving $\ell( \sigma) = n(\sigma) \da \size( \Phi^- \intersect \sigma(\Phi^-))$
by induction on $\ell(\sigma)$, where we already checked the zero case.
Assume the result for all $\tau$ with $\ell( \tau) \leq \ell( \sigma)$ for $\tau \in W$.
Write \( \sigma = s_1\cdots s_t \) with $s_i \da s_{ \alpha_i}, \alpha_i\in \Delta$ reduced.
Set $\tau \da \sigma s_t = s_1\cdots s_{t-1}$ which is again reduced with $\ell(\tau) = \ell( \sigma) - 1$.
By the deletion condition, $s_1 \cdots s_{t-1}( \alpha_t) > 0$, so $s_1\cdots s_{t-1}s_t (\alpha_t) = s_1 \cdots s_{t-1}(- \alpha_t) < 0$.
Thus $n(\tau) = n( \sigma) - 1$, since $s_t$ permutes $\Phi^+\smts{\alpha_t}$, so
\[
\ell( \sigma) - 1 = \ell( \tau) = n( \tau) = n( \sigma) -1 \implies \ell( \sigma) = n( \sigma)
.\]
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:::{.remark}
This is useful for finding reduced expressions, or at least their length: just compute how many positive roots change sign under $\sigma$.
Using the deletion condition and lemma A, it's clear that any expression for $\sigma$ as a product of simple reflections can be converted into a *reduced* expression by deleting pairs of simple reflections, and this terminates after finitely many steps.
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:::{.lemma title="B"}
Recall that the open Weyl chambers are the complements of hyperplanes.
The closure of any Weyl chamber is a fundamental domain for the action $W\actson \EE$.
:::

## 10.4: Irreducible root systems

:::{.definition title="Irreducible root systems"}
A root system $\Phi \subseteq \EE$ is **irreducible** if it cannot be partitioned into mutually orthogonal nonempty subsets.
Otherwise, $\Phi$ is **reducible**.
:::

:::{.proposition title="?"}
Let $\Delta \subseteq \Phi$ be a simple system.
Then $\Phi$ is irreducible iff $\Delta$ is irreducible, i.e. $\Delta$ cannot be partitioned into nonempty orthogonal subsets.
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:::{.proof title="?"}
$\Phi$ reducible implies $\Delta$ reducible: write $\Phi = \Phi_1 \disjoint \Phi_2$ where $( \Phi_1, \Phi_2) = 0$; this induces a similar partition of $\Delta$.
Then $(\Delta, \Phi_2) = 0 \implies (\EE, \Phi_2) = 0 \implies \EE = \emptyset$ using nondegeneracy of the bilinear form. $\contradiction$

Now $\Delta$ reducible implies $\Phi$ reducible:
write $\Delta =\Delta_1 \disjoint \Delta_2$ with $(\Delta_1, \Delta_2) = 0$.
Let $\Phi_i$ be the roots which are $W\dash$conjugate to an element of $\Delta_i$.
Then elements in $\Phi_i$are obtained from $\Delta_i$ by adding and subtracting only elements of $\Delta_i$, so $(\Phi_1, \Phi_2) = 0$ and $\Phi = \Phi_1 \union \Phi_2$ by a previous lemma that every $\beta\in \Phi$ is conjugate to some $\alpha\in \Delta$.
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:::{.lemma title="A"}
Let $\Phi \contains \Delta$ be irreducible.
Relative to the partial order $\leq$ on roots, there is a unique maximal root $\tilde \alpha$.
In particular, if $\beta\in \Phi$ and $\beta\neq \tilde \alpha$, then $\height( \beta) < \height( \tilde \alpha)$ and $(\tilde \alpha, \alpha) \geq 0$ for all $\alpha\in \Delta$.
Moreover, one can write $\tilde \alpha = \sum _{\alpha\in \Delta}$ with $k_\alpha > 0$, i.e. it is a sum where every simple root appears.
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:::{.proof title="?"}
**Existence**:
Let $\tilde \alpha$ be any maximal root in the ordering.
Given \( \alpha \in \Delta, (\tilde \alpha, \alpha) \geq 0 \) -- otherwise $s_ \alpha(\tilde \alpha)= \tilde \alpha-\inp{ \tilde \alpha}{\alpha} \alpha > \alpha$, a contradiction. $\contradiction$
Write $\tilde \alpha = \sum_{\alpha\in \Delta} k_ \alpha \alpha$ with $k_ \alpha \in \ZZ_{\geq 0}$, where it's easy to see these are all non-negative.
Suppose some $k_\gamma = 0$, then $(\tilde \alpha, \gamma)\leq 0$ -- otherwise $s_\gamma( \tilde \alpha) = \tilde \alpha - \inp{\tilde \alpha}{\gamma} \gamma$ has both positive and negative coefficients, which is not possible.
Since $(\tilde \alpha, \alpha) \geq 0$, we must have $( \tilde \alpha, \gamma) = 0$.
So write
\[
0 = (\tilde \alpha, \gamma) = \sum_{ \alpha\in \Delta} k_ \alpha( \alpha, \gamma) \leq 0
,\]
so $( \alpha, \gamma)= 0$ whenever $k_\alpha \neq 0$, otherwise this expression would be strictly $< 0$.
Thus \( \Delta = \Delta_1 \disjoint \Delta_2 \) where \( \Delta_1 = \ts{\alpha\in \Delta \st K_ \alpha\neq 0} \) and \( \Delta_2 = \ts{ \alpha\in \Delta\st k_ \alpha = 0 } \). This is an orthogonal decomposition of $\Delta$, since any $\gamma \in \Delta_2$ is orthogonal to any $\alpha\in \Delta_1$.
Note that $\Delta_1\neq \empty$ since $\tilde\alpha \neq 0$, and if $\Delta_2\neq \empty$ then this is a contradiction, so $\Delta_2$ must be empty.
So no such $\gamma$ exists.

**Uniqueness**: let $\tilde \alpha$ be any maximal root in the ordering and let $\tilde \alpha'$ be another such root.
Then $(\tilde \alpha, \tilde \alpha') = \sum_{\alpha\in \Delta} k_ \alpha (\alpha, \tilde \alpha')$ with $k_ \alpha > 0$ and $(\alpha, \tilde \alpha') \geq 0$.
So $(\tilde \alpha, \tilde \alpha ') > 0$ since $\Delta$ is a basis for $\EE$ and anything orthogonal to a basis is zero by nondegeneracy of the form.
Since $\tilde \alpha \neq 0$, it is not orthogonal to everything.
By Lemma 9.4, either $\tilde \alpha, \tilde \alpha '$ are proportional (which was excluded in the lemma), in which case they are equal since they're both positive, or otherwise \( a \da \tilde \alpha - \tilde \alpha' \in \Phi \) is a root.
In the latter case, $a > 0 \implies \tilde \alpha > \tilde \alpha'$ or $a< 0 \implies \tilde \alpha < \tilde \alpha'$, both contradicting maximality.
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:::{.remark}
If $\beta = \sum_{\alpha\in \Delta} m_ \alpha \alpha \in \Phi^+$, then $m_ \alpha \leq k_ \alpha$ for all $\alpha$ since $\beta \leq \alpha$.
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:::{.lemma title="B"}
If $\Phi$ is irreducible then $W$ acts irreducibly on $\EE$ (so there are no $W\dash$invariant subspaces).
In particular, the $W\dash$orbit of a root spans $\EE$.
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:::{.proof title="?"}
Omitted.
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:::{.lemma title="C"}
If $\Phi$ is irreducible, then at most two root lengths occur, denoted **long** and **short** roots.
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:::{.proof title="?"}
Omitted.
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:::{.example title="?"}
$B_2$ has 4 long roots and 4 short roots, since they fit in a square:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-19_09-57.pdf_tex} };
\end{tikzpicture}


Similarly $G_2$ has long and short roots, fitting into a star of David.
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:::{.lemma title="D"}
If $\Phi$ is irreducible then the maximal root $\tilde \alpha$ is a long root.
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:::{.remark}
There is also a unique maximal short root.
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:::{.proof title="?"}
Omitted.
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