# Wednesday October 30

Last time:

\begin{align*}
W \actson \lieh^*, \lambda \mapsto w(\lambda),
W \actson \lieh, \lieh \mapsto w \cdot \lieh
\end{align*}

such that $\lambda(w\cdot h) = (w\inv \lambda ) (h) \forall \lambda \in \lieh^*$.

We then get compatible squares:

\begin{center}
\begin{tikzcd}
\lieh^* \arrow[dd, "w"] \arrow[rr] &  & \lieh \arrow[dd, "w\cdot"] &  & \alpha \arrow[dd] \arrow[rr] &  & t_\alpha \arrow[dd]            \\
                                   &  &                            &  &                              &  &                                \\
\lieh^* \arrow[rr]                 &  & \lieh                      &  & w\alpha \arrow[rr]           &  & w \cdot t_\alpha = t_{w\alpha}
\end{tikzcd}
\end{center}

**Proposition:**

a. $\Theta_i \coloneqq \exp(\ad e_i) \circ \exp(\ad (-f_i)) \circ \exp(\ad e_i)$,
b. $\Theta_i(\lieh) = \lieh$, so it fixes Cartan subalgebra.
c. $\Theta_i\mid_\lieh = s_i$ where $s_i$ is the Weyl group action

*Proof of (a):*

We want to show that $\exp(\ad e_i)$ is well-defined as an automorphism of $\lieg$. 
It suffices to check that $\ad e_i$ is *locally nilpotent*, i.e. for all $x\in \lieg$, there exists some $n_x > 0$ such that $\ad(e_i)^n = 0$.
We will also need to check that $\exp \ad e_i$ is a derivation.

To see the local nilpotency, we can check

\begin{align*}
(\ad e_i)^n([x, y]) = \sum_{t=0}^n {n \choose t} \left[ (\ad e_i)^t, (\ad e_i)^{n-t} \right]
\end{align*}

for all $x, y \in \lieg$. 

If $x,y$ are locally nilpotent, then $[x, y]$ is as well.

It thus suffices to check that $\ad e_i$ acts on generators in a nilpotent way.

A direct computation shows $\ad e_i = [e_i, e_i] = 0$, and $(\ad e_i)^{1 - a_{ji}}(e_j) = 0$ by the Serre relations.

We also find that $\ad e_i (h_j) = [e_i, h_j] = -[h_j, e_i] = -a_{ij} e_i$, and applying it again yields $(\ad e_i)^2(h_j) = -a_{ij}[e_i, e_i] = 0$.

We have $\ad e_i (h_j) = 0$, and applying $\ad e_i h_i$ multiple times yields $h_i, [e_i, h_i], 0$, so $\ad^{3} e_i(h_i) = 0$.

*Proof of (b):*

By a direct computation, we have $\Theta_i(h_j) = h_j - a_{ij} h_i \in \lieh$. (See CJ's notes for full computation.)

*Proof of (c):*

Consider computing $s_i \cdot h_j$. 
This is the unique element satisfying $\lambda(s_i \cdot h_j) = (s_i\inv \lambda)(h_j)$, but we can compute

\begin{align*}
(s_i\inv \lambda) (h_j) &= h_j - a_{ij} h_i = \Theta_i(h_j)
.\end{align*}

$\qed$

**Theorem (Serre):**
Fix $\Phi \supseteq \Pi = \theset{\alpha_1, \cdots, \alpha_\ell}$ and $I = \theset{1, \cdots, \ell}$.
Define $A$ by $a_{ij} = (\alpha_j, \alpha_i\dual)$.
Let $\lieg = \lieg(A)$ be the algebra generated by these elements.

Then

a. $\lieg = n^- \oplus \lieh \oplus n$ as vector spaces, where $n^- \cong \tilde n^- / s^-$, $\lieh \cong \tilde\lieh$, and $n \cong \tilde n / s^+$. 
b. $\lieg = \oplus_{\mu \in \lieh^*} \lieg_\mu$ as vector spaces, where $\lieg_\mu = \theset{x\in\lieg \suchthat [h, x] = \mu(h) x \forall h\in \lieh}$
c. $\dim \lieg_\lambda = \dim \lieg_\mu$ if $\lambda \in W_\mu$,
d. $\dim \lieg = \ell + \abs \Phi$,
e. $\lieg$ is semisimple,
f. $\lieh$ is a Cartan subalgebra with root system $\Phi$.

*Proofs:*

a. Follows from Theorem 18.2b and Lemma b.
b. Similar to Theorem 18.2a.

*Proof of (c):*

We may assume that $\lambda = s_i \mu$.
Pick $x\in \lieg_\lambda$.
Then for all $h\in \lieh$, we have

\begin{align*}
[\Theta_i(h), \Theta_i(x)] &= \Theta_i(h, x) \\
&= \lambda(h)\Theta_i(x) \\
&= \lambda(\Theta_i\inv (h)) \Theta_i(x) \\
&= \lambda(s_i\inv \cdot h) \Theta_i(x) \\
&= (s_i\lambda) \Theta_i(x)
,\end{align*}

so $\Theta_i(x) \in \lieg_{s_i\lambda}$, and thus $\Theta_i(\lieg_\lambda) \subseteq \lieg_{s_i\lambda}$.

Replacing $\Theta_i$ with $\Theta_i\inv$ and $\lambda$ by $s_i \lambda$, we find $\Theta_i\inv (\lieg_{s_i \lambda}) \subseteq \lieg_{s_i s_i \lambda} = \lieg_\lambda$,
and so $\lieg_\lambda \cong \lieg_{s_i \lambda}$, i.e. $\lieg_{s_i \lambda} \subseteq \Theta_i(\lieg)$.

*Proof of (d):*

By Corollary 18.2b, we have

\begin{align*}
\dim \lieg_{k \alpha_{ii}} =
\begin{cases}
1, & k= \pm 1 \\
0, & k\not \in \theset{0, \pm 1} \\
\ell, & k = 0
\end{cases}
.\end{align*}

Thus $\tilde \lieg_0 = \tilde \lieh$.

Since $s_{ij}^+$ is of height $1+a_{ji} \geq 2$, we have $\dim \lieg_{\alpha_i} = \dim \tilde \lieg_{\alpha_i} = 1$ for all $i\in I$.
Thus for any $\alpha \in \Phi$, we have $\alpha = w \alpha_i$ for some element of the Weyl group $w\in W$.

By parts (a) and (c), we have $\dim \lieg_{\alpha} = 1$, so $\dim \lieg_{k\alpha}$ satisfies the same cases as $\dim \lieg_{k\alpha_{ii}}$ above.

It remains to show that there are no other root spaces, i.e. $\lieg_\mu = 0$ if $\mu \not\in \ZZ\alpha$ for all $\alpha \in \Phi$.

We can show this by considering reflections about hyperplanes again, i.e. that $\alpha \in \Phi \implies H_\mu \neq H_\alpha$.

If this is the case, it implies that there exists an $h \in \lieh$ such that $h \in H_\mu \setminus H_\alpha$ for all $\alpha \in \Phi$.
But then $\mu(h) = 0$ when $h \not\in H_\alpha$ for all $\alpha \in \Phi$, so pick $w\in W$ such that $w\inv a_i(h) \in C(\Pi)$, the fundamental chamber.
Thus $w\inv \alpha_i (h) > 0$ for all $i$, and is equal to $\alpha_i(w \cdot h)$, and
$$
0 = \mu(h) = \kappa(t_\mu, h) = \cdots = (w\mu)(w \cdot h)
$$

Writing $w_\mu = \sum_{i=1}^\ell m_i \alpha_i$, we have $0 = \sum_{i=1}^\ell m_i \alpha_i(w\cdot h)$, we find that note all $m_i$ have the same sign, which is a contradiction. $\qed$