# Serre's Theorem, Continued (Friday, November 11)

:::{.proof title="of Serre's theorem, continued"}

Recall that we have
\[
L = N^- \oplus H \oplus N^+ \da Y/\gens{(S_{ij}^-)} \oplus \CC\gens{h_1,\cdots, h_\ell} \oplus X/\gens{ (S_{ij}^+) }
.\]

:::{.remark title="Step 6"}
For $1\leq i\leq \ell$, note $\ad_{L, x_i}$ (and similarly $\ad_{L, y_i}$) is locally nilpotent on $L$.
Let $M \subseteq L$ be the subspace of elements on which $\ad_{x_i}$ acts nilpotently.
By the Leibniz rule, $(\ad_{x_i})^{m+n}([uv])$ when $(\ad_{x_i})^m(v) = 0$ and $(\ad_{x_y})^n(u) = 0$, so $M \leq L$ is a Lie subalgebra.
By the Serre relations, $(\ad_{x_i})^2(h_j) = 0$ and $(\ad_{x_i})^3(y_j) = 0$, so the generators of $L$ are in $M$ and thus $L = M$.
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:::{.remark title="Step 7"}
Defining $\tau_i \da \exp(\ad_{x_i}) \circ \exp(\ad_{-y_i}) \circ \exp(\ad_{x_i}) \in \Aut(L)$, by lemma (B) we have $\tau_i(L_ \lambda) = L_{s_i \lambda}$ where $s_i \da s_{\alpha_i}$ and \( \lambda\in H\dual \).
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:::{.remark title="Step 8"}
Let \( \lambda, \mu \in H\dual \) and suppose \( w \lambda = \mu \); we want to show $\dim L_ \lambda = \dim L_{ \mu}$.
Note $W$ is generated by simple reflections $s_i$, so it STS this when $w=s_i$, whence it follows from lemma (B).
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:::{.remark title="Step 9"}
Clearly $\dim (L_0)_{\alpha_i} = 1$ since it's spanned by $x_i$.
Then $\dim (L_0)_{k \alpha_i} = 0$ for $k\neq 0, \pm 1$, so $\dim L_{\alpha_i} \leq 1$ and $\dim L_{k \alpha} = 0$ for $k\neq 0, \pm 1$.
Since $x_i\in L_{\alpha_i}$ has a nonzero image in $L$, $\dim L_{\alpha_i} = 1$.
:::

:::{.remark title="Step 10"}
If \( \beta\in \Phi \), conjugate it to a simple root using \( \beta = w \alpha_i \) with $w\in W, \alpha_i \in \Delta$.
By step 8, $\dim L_{ \beta} = 1$ and $L_{k \beta} = 0$ for $k\neq 0, \pm 1$.
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:::{.remark title="Step 11"}
Suppose \( L_ \lambda \neq 0 \) where \( \lambda\neq 0 \).
Then \( \lambda\in \ZZ_{\geq 0} \Delta \) or \( \ZZ_{\leq 0} \Delta \), i.e. all coefficients are the same sign.
Suppose \( \lambda\not\in \Phi \), then \( \lambda\in \ZZ \Phi \) by (10).
Exercise 10.10 yields $\exists w\in W$ such that $w \lambda\ZZ \Delta$ with both positive and negative coefficients.
Thus $w \lambda$ can not be a weight, and by step 8, $0 = \dim L_{w \lambda} = \dim L_ \lambda$.
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:::{.remark title="Step 12"}
Writing $L = N^- \oplus H \oplus N^+$ with $H = L_0, N^+ = \sum_{ \lambda > 0} L_{ \lambda} = \sum_{\beta\in \Phi^+} L_{ \lambda}$ and $N^- = \sum_{ \lambda < 0} L_{ \lambda} = \sum_{ \beta\in \Phi^-} L_ \beta$, by step 10 we can conclude $\dim L = \ell + \size\phi < \infty$.
This shows that $H$ is toral, i.e. its elements are ad-semisimple.
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:::{.remark title="Step 13"}
We have that $L$ is a finite-dimensional Lie algebra.
To show semisimplicity, we need to now $L$ has no nonzero solvable ideals, and as before it's ETS $L$ has no nonzero *abelian* ideals.
Suppose \( A \subseteq L \) is an abelian ideal; we WTS $A = 0$.
Since $[H,A] \subseteq A$ and $H\actson L$ diagonally, $H\actson A$ diagonally as well and thus
\[
A = (A \intersect H) \oplus \bigoplus _{\alpha\in \Phi} (A \intersect L_{ \alpha})
.\]
If \( A \intersect L_{\alpha} \neq 0 \) then \( A \intersect L_{ \alpha} = L_{\alpha} \), which is 1-dimensional.

> Note: the argument in Humphreys here may not be quite right, so we have to do something different.

Now $\exists w\in W$ such that $w \alpha = \alpha_i \in \Delta$ as in step 8, so write $w = s_{i_1}\cdots s_{i_t}$.
Then $\tau_{i_1}\cdots \tau_{i_t} (L_ \alpha) = L_ \alpha$.
Set $A' \da \tau(A)$, then $A'\normal L$ is necessarily an abelian ideal and $A' \intersect L_{\alpha_i} = L_{\alpha_i}$.

So we can replace $\alpha$ by a simple root and replace $A$ by $A'$.
Then $x_i\in L_{\alpha_i} \subseteq A' \normal L$, but $A'\ni -[y_i, x_i] = h_i$, but $[h_i, x_i] \neq 0$, contradicting that $A'$ is abelian. $\contradiction$.

Note $[A', L_{\alpha_j}] \subseteq A' \subseteq H$ and $[A', L_{\alpha_j}] \subseteq L_{\alpha_j}$, but $H\actson L_{\alpha_j}$ with eigenvalues \( \alpha_j \) and thus $A' \subseteq \intersect _{j=1}^\ell \ker \alpha_j = 0$ since the \( \alpha_j \) span $H\dual$.
So $A' = 0$ and $L$ is semisimple.
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:::{.remark title="Step 14"}
Since $L = H \oplus \bigoplus _{\alpha\in \Phi} L_{ \alpha}$, it's easy to check that $C_L(H) = H$ by considering what happens when bracketing against any nonzero element in \( \bigoplus L_ \alpha \).
Thus $H$ is a maximal toral subalgebra with corresponding root system $\Phi$.
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:::{.remark}
Next: part VI on representation theory, although we'll first cover $\S 13$ on weights, especially $\S 13.1, \S 13.2$.
Goal: Weyl's character formula.
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