# Wednesday, September 28

## Continued proof

Recall the proposition from last time:

:::{.proposition title="Big!"}
\envlist

a. $\Phi$ spans $H\dual$.
b. If $\alpha\in \Phi$ is a root then $-\alpha\in \Phi$ is also a root.
c. Let \( \alpha\in \Phi, x\in L_{ \alpha}, y\in L_{- \alpha} \), then $[xy] = \kappa(x, y) t_\alpha$.
d. If $\alpha\in \Phi$ then $[L_{ \alpha}, L_{- \alpha}] = \CC t_\alpha$ is 1-dimensional with basis $t_\alpha$.
e. For any \( \alpha \in \Phi \), we have \( \alpha(t_\alpha) = \kappa(t_ \alpha, t_ \alpha) \neq 0 \).
f. (Important) If \( \alpha\in \Phi, x_ \alpha\in L_{ \alpha}\smz \) then there exists some \( y_ \alpha L_{- \alpha} \) in the opposite root space such that \( x_ \alpha, y_ \alpha, h_ \alpha \da [x_ \alpha, y_ \alpha] \) span a 3-dimensional subalgebra $\liesl(\alpha) \leq L$ isomorphic to $\liesl_2(\CC)$.
g. $h_\alpha = {2 t_\alpha \over \kappa(t_ \alpha, t_ \alpha)}$, \( \alpha(h_ \alpha) = 2, h_{ \alpha} = h_{- \alpha} \).

:::

:::{.proof title="continued"}
**Part e**:
We have \( \alpha(t_ \alpha ) = \kappa(t_ \alpha, t_ \alpha) \), so suppose this is zero.
Pick $x\in L_{ \alpha}, y\in L_{ - \alpha}$ such that $\kappa(x, y) = 1$, then

- $[t_ \alpha, x] = 0$,
- $[t_ \alpha, y] = 0$,
- $[x, y] = t_ \alpha$.

Set $S \da \liesl( \alpha) \da \CC\gens{x,y,t_ \alpha}$ and restrict $\ad: L\injects \liegl(L)$ to $S$.
Then $\ad(S) \cong S$ by injectivity, and this is a solvable linear subalgebra of $\liegl(L)$.
Apply Lie's theorem to choose a basis for $L$ such that the matrices for $\ad(S)$ are upper triangular.
Then use that $\ad_L([SS]) = [\ad_L(S) \ad_L(S)]$, which is strictly upper triangular and thus nilpotent.
In particular, $\ad_L (t_ \alpha)$ is nilpotent, but since $t_\alpha\in H$ which is semisimple, so $\ad_L( t_ \alpha)$ is semisimple.
The only things that are semisimple *and* nilpotent are zero, so $\ad_L( t_ \alpha) = 0 \implies t_\alpha = 0$.
This contradicts that $\alpha\in H\dual\smz$.
$\contradiction$

**Part f**:
Given $x_ \alpha\in L_ \alpha\smz$, choose $y_ \alpha \in L_{ - \alpha}$ and rescale it so that
\[
\kappa(x_ \alpha, y _{\alpha} ) = {2\over \kappa(t_ \alpha, t_ \alpha)}
.\]
Set \( h_ \alpha \da {2t_ \alpha\over \kappa(t_ \alpha, t_ \alpha) } \), then by (c), $[x_ \alpha, y_ \alpha] = \kappa( x_ \alpha, y_ \alpha) t_ \alpha = h_ \alpha$.
So 
\[
[ h_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha) }[ t_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha)} \alpha(t_ \alpha) x_ \alpha = 2x_ \alpha
,\]
and similarly $[h_ \alpha, y_ \alpha] = -2 y_ \alpha$.
Now the span $\gens{x_ \alpha, h_ \alpha, y_ \alpha} \leq L$ is a subalgebra with the same multiplication table as $\liesl_2(\CC)$, so $S \cong \liesl_2(\CC)$.

**Part g**:
Note that we would have $h_{ - \alpha} = {2 t_{ - \alpha} \over \kappa( t_ { - \alpha}, t_{- \alpha} ) } = - h_ \alpha$ if $t_{ \alpha} = t_{ - \alpha}$.
This follows from the fact that $H\dual \iso H$ sends $\alpha\mapsto t_\alpha, -\alpha\mapsto t_{- \alpha}$, but by linearity $-\alpha\mapsto -t_{ \alpha}$.
:::

:::{.corollary title="?"}
$L$ is generated as a Lie algebra by the root spaces $\ts{L_ \alpha\st \alpha\in \Phi}$.
:::

:::{.proof title="?"}
It STS $H \subseteq \gens{\ts{L_\alpha}_{\alpha\in \Phi}}$.
Given $\alpha\in \Phi$,
\[
\exists x_\alpha\in L_ \alpha, y_ \alpha\in L_{- \alpha} \quad\text{such that}\quad
\gens{x_ \alpha, y_ \alpha, h_ \alpha\da [x_ \alpha, y_ \alpha] } \cong \liesl_2(\CC)
.\]
Note any $h_\alpha\in \CC\units t_ \alpha$ corresponds via $\kappa$ to some $\alpha\in H\dual$.
By (a), $\Phi$ spans $H\dual$, so $\ts{t_\alpha}_{\alpha\in \Phi}$ spans $H$.
:::

## 8.4: Integrality properties of $\Phi$

:::{.remark}
Any $\alpha\in \Phi$ yields $\liesl(\alpha) \cong \liesl(\alpha)$, and in fact that the generators entirely determined by the choice of $x_\alpha$.
View $L\in\mods{\liesl(\alpha)}$ via $\ad$.
:::

:::{.lemma title="?"}
If $M \leq L \in \mods{\liesl(\alpha)}$ then all eigenvalues of $h_\alpha\actson M$ are integers.
:::

:::{.proof title="?"}
Apply Weyl's theorem to decompose $M$ into a finite direct sum of irreducibles in $\mods{ \liesl_2(\CC) }$.
The weights of $h_\alpha$ are of the form $m, m-2,\cdots, -m+2, -m\in \ZZ$.[^hwcomplex]


[^hwcomplex]: 
This fails for infinite dimensional modules, e.g. Verma modules. The highest weight can be any complex number.

:::

:::{.example title="?"}
Let $M = H + \liesl(\alpha) \leq L \in \mods{\liesl( \alpha)}$, which one can check is actually a submodule since bracketing either lands in $\liesl(\alpha)$ or kills elements.
What does Weyl's theorem say about this submodule?
There is some intersection.
Set $K \da \ker\alpha \subseteq H$, so $\codim_H K = 1$ by rank-nullity.
Note that $h_\alpha \not\in K$, so $M = K \oplus \liesl(\alpha)$.
Moreover $\liesl(\alpha)\actson K$ by zero, since bracketing acts by $\alpha$ which vanishes on $K$.
So $K\cong \CC\sumpower{n+1}$ decomposes into trivial modules.
:::

:::{.example title="?"}
Let $\beta\in \Phi\union\ts{0}$ and define $M \da \bigoplus _{c\in \CC} L_{\beta + c\alpha}$, then $L \leq L \in \mods{\liesl( \alpha)}$.
It will turn out that $L_{\beta+ c \alpha} \neq 0 \iff c\in [-r, q] \subseteq \ZZ$ with $r, q\in \ZZ_{\geq 0}$.
:::

:::{.proposition title="A"}
Let $\alpha\in \Phi$, then the root spaces $\dim L_{\pm \alpha} = 1$, and the only multiples of $\alpha$ which are in $\Phi$ are $\pm \alpha$.
:::

:::{.proof title="?"}
Note $L_\alpha$ can only pair with $L_{- \alpha}$ to give a nondegenerate Killing form.
Set 
\[
M \da \bigoplus _{c\in \CC} L_{c \alpha} = H \oplus \bigoplus _{c \alpha\in \Phi} L_{c \alpha}
.\]
By Weyl's theorem, this decomposes into irreducibles.
This allows us to take a complement of the decomposition from before to write $M = H \bigoplus \liesl(\alpha) \oplus W$, and we WTS $W = 0$ since this contains all $L_{c \alpha}$ where $c\neq \pm 1$.
Since $H \subseteq K \oplus \liesl( \alpha)$, we have $W \intersect H = 0$.
If $c_ \alpha$ is a root of $L$, then \( h_ \alpha \) has $(c \alpha)(h_ \alpha) = 2c$ as an eigenvalue, which must be an integer by a previous lemma.
So $c\in \ZZ$ or $c\in \ZZ + {1\over 2}$.

Suppose $W\neq 0$, and let $V(s)$ (or $L(s)$ in modern notation) be an irreducible $\liesl( \alpha)\dash$submodule of $W$ for $s\in \ZZ_{\geq 0}$.
If $s$ is even, $V(s)$ contains an eigenvector $w$ for $h_ \alpha$ of eigenvalue zero by applying $y_\alpha$ $s/2$ times.
We can then write $w = \sum_{c\in \CC} v_{c \alpha}$ with $v_{ ca } \in L_{ c \alpha}$, and by finiteness of direct sums we have $v_{c \alpha} = 0$ for almost every $c\in \CC$.
Then
\[
0
&= [h_ \alpha, w] \\
&= \sum_{c\in \CC} [h_ \alpha, v_{ c \alpha} ] \\
&= \sum_{c\in \CC} (c\alpha)( h _{\alpha} )v_{c \alpha} \\
&= \sum 2c v_{c \alpha} \\
&\implies v_{c \alpha} = 0 \text{ when } c\neq 0
,\]
forcing $w\in H$, the zero eigenspace.
But $w\in W$, so $w\in W \intersect H = 0$. $\contradiction$
:::