# Wednesday, November 09

:::{.remark}
Progress so far: 
we start with the data of an irreducible root system $\Phi \contains \Delta = \tsl \alpha 1 \ell$ and Cartan matrix $A = (\inp{ \alpha_i}{\alpha_j})$ and Weyl group $W$.
We set $L_0 \da {\gens{x_i, y_i, h_i \st 1\leq i\leq \ell} \over \gens{\text{S1, S2, S3}}} = Y \oplus H \oplus X$.
Letting $h\in H$ act by $\ad_h$, we get weight spaces $(L_0)_{ \lambda} \da \ts{v\in L_0 \st [hv] = \lambda(h) v \, \forall h\in H}$.
:::

## $\S$ 18.3: Serre's Theorem

:::{.remark}
For $i\neq j$, set
\[
y_{ij} \da (\ad_{y_i})^{- \inp{\alpha_j}{ \alpha_i} + 1}(y_j)
,\]
and similarly for $x_{ij}$.
Recall \( \alpha_i(h_j) \da \inner{ \alpha_i }{ \alpha_j }  \).
:::

:::{.lemma title="A"}
\[
\ad_{x_k}(y_{ij}) = 0 \qquad \forall i\neq j, \forall k
.\]
:::

:::{.proof title="?"}

**Case 1**:
$k\neq i$.

In this case, $[x_k y_i] = 0$ and thus
\[
(\ad_{x_k}) (\ad_{y_i})^{- \inner{ \alpha_j }{ ga_i } +1 }(y_j) = (\ad_{y_i})^{- \inner{ \alpha_j }{ \alpha_i } +1 } (\ad_{x_k})(y_j)
.\]

- Case i, $k\neq j$: then $(\ad_{x_k})(y_j) = 0$.
- Case ii, $k=j$: then $(\ad_{x_j})(y_j) = h_j$ and $(\ad_{y_i})(h_j) = \inner{ \alpha_i }{ \alpha_j } y_i$.
  - Case a, \( \inner{ \alpha_j }{ \alpha_i } \neq 0  \), then
  \[
  (\ad_{y_i})^{- \inner{ \alpha_j }{ \alpha_i } +1 } (\ad_{x_j}) (y_j) = \inner{ \alpha_i }{ \alpha_j } (\ad_{y_i})^{- \inner{ \alpha_j }{ \alpha_i }  }(y_i) = 0 
  .\]
  - Case b, \( \inner{ \alpha_j }{ \alpha_i } = 0  \), then \( \inner{ \alpha_i }{ \alpha_j } =0  \).
    In this case we have $(\ad_{y_i})^1 (h_j) = \inner{ \alpha_i }{ \alpha_j } y_i = 0$. 

**Case 2**:
$k=i$.

In this case, we saw that for any fixed $i$, $\ts{x_i, h_i, y_i}$ spans a standard $\liesl_2$ triple in $L_0$, so consider the $\liesl_2\dash$submodule of $Y_J \leq L_0$ generated by $y_j$.
Since $i\neq j$, we know $[x_i y_j] = 0$, so $y_j$ is a maximal vector for $Y_j$ with weight $m \da - \inner{ \alpha_j }{ \alpha_i }$. 

One can show by induction on $t$ that the following formula holds:
\[
(\ad_{x_i}) (\ad_{y_i})^t (y_j) = t (m-t+1) (\ad_{y_i})^{t-1} (y_j) \qquad t\geq 1
.\]
So in particular $(\ad_{x_i}) (\ad_{y_i})^{m+1}(y_j) = 0$, and the LHS is $y_{ij}$.
:::

:::{.definition title="Locally nilpotent and the exponential"}
An endomorphism $x\in \Endo(V)$ is **locally nilpotent** if for all $v\in V$ there exists some $n$ depending on $v$ such that $x^n \cdot v = 0$.
If $x$ is locally nilpotent, then define the **exponential** as
\[
\exp(x) = \sum_{k\geq 0} {1\over k!} x^k = 1 + x + {1\over 2}x^2 + \cdots \qquad \in \Endo(V)
,\]
which is in fact an automorphism of $V$ since its inverse is $\exp(-x)$.
:::

:::{.lemma title="B"}
Suppose $\ad_{x_i}, \ad_{y_i}$ are locally nilpotent on $L_0$ and define
\[
\tau_i \da \exp(\ad_{x_i})\circ \exp(\ad_{-y_i}) \circ \exp(\ad_{x_i})
.\]
Then $\tau_i((L_0)_\lambda) = (L_0)_{s_i (\lambda)}$ where $s_i \da s_{ \alpha_i} \in W$ for $\alpha_i\in \Phi$.
Here \( \lambda\in H\dual\cong \CC\gens{ \alpha_1, \cdots, \alpha_\ell} \) since \( H = \CC\gens{h_1,\cdots, h_{\ell}} \), using that $A$ is invertible.
We use the formula \( s_{\alpha_i}(\alpha_j) = \alpha_j - \inner{ \alpha_j }{ \alpha_i } \alpha_i  \) and extending linearly to $H\dual$ as done previously.
:::

:::{.proof title="?"}
Omitted. See $\S 14.3$ and $\S 2.3$ for a very similar calculation.
:::

:::{.theorem title="Serre"}
The Lie algebra $L$ generated by the $3\ell$ elements $\ts{x_i, h_i, y_i}_{1\leq i\leq \ell}$ subject to relations S1-S3 and the remaining two relations $S_{ij}^{\pm}$ (which hold in any finite dimensional semisimple Lie algebra) is a finite dimensional semisimple Lie algebra with maximal torus spanned by $\ts{h_i}_{1\leq i\leq \ell}$ and with corresponding root system $\Phi$.
:::

## Proof of Serre's Theorem

:::{.proof title="?"}
By definition, $L \da L_0/K$ where $K\normal L_0$ is generated by the elements $x_{ij}, y_{ij}$ where $i\neq j$.
Recall that $X, Y\leq L_0$ are the subalgebras generated by the $x_i$ and $y_i$ respectively, so let $I$ (resp. $J$) be the ideal in $X$ (resp. $Y$) generated by the $x_{ij}$ (resp. $y_{ij}$) for $i\neq j$.
Clearly $I, J \subseteq K$.

:::{.claim title="Step 1"}
\[
I, J \normal L_0
.\]
:::

:::{.proof title="of Step 1"}
We'll prove this for $J$, and $I$ is similar.
Note $J\normal Y$ and write $J = \gens{y_{ij} \st i\neq j}$.
Fix $1\leq k\leq \ell$, then $(\ad_{y_k}) (y_{ij}) \in J$ by definition.
Recall $y_{ij} = (\ad_{y_i})^{- \inner{ \alpha_i }{ \alpha_j } +1 }(y_{ij})$.
Note $(\ad_{h_k})(y_{ij}) = c_{ijk} y_{ij}$ for some constant $c_{ijk} \in \ZZ$, and $(\ad_{x_k})(y_{ij}) = 0$ by lemma A above.
Since $x_k, h_k, y_k$ generate $L_0$, we have $[L_0, y_{ij}] \subseteq J$.
Using the Jacobi identity and that $\ad_z$ is a Lie algebra derivation for $z\in L_0$, it follows that $[L_0, J] \subseteq J$.

> This essentially follows from $[h_\ell, Y] \subseteq Y$ and $[x_\ell, Y] \subseteq H + Y$, and bracketing these against $y_{ij}$ lands in $J$.

:::

:::{.claim title="Step 2"}
\[
K = I + J
.\]
:::

:::{.proof title="of Step 2"}
We have $I+J \subseteq K$, but $I+J \normal L_0$ by claim 1 and it contains the generators of $K$ -- since $K$ is the smallest such ideal, $K \subseteq I+J$.

:::

:::{.observation title="Step 3"}
We have a decomposition $L_0 = Y \oplus H \oplus X$ as modules under $\ad_H$, and $K = J \oplus 0 \oplus I$.
Taking the quotient yields $L \da L_0/K = Y/J \oplus H \oplus X/I \da N^- \oplus H \oplus N^+$.
:::

:::{.observation title="Step 4"}
As in the proof last time, $\ts{x_i, h_i, y_i} \subseteq L$ spans a copy of $\liesl_2$.
We deduce that $\sum_{1\leq i \leq \ell} \FF x_i + \FF h_i + \FF y_i \subseteq L_0$ maps isomorphically into $L$, so we can identify $x_i, h_i, y_i$ with their images in $L$, which are still linearly independent and still generate $L$ as a Lie algebra.
:::

:::{.observation title="Step 5"}
For \( \lambda\in H\dual \), set \( L_ \lambda\da \ts{z\in L \st [hz] = \lambda(h) z\, \forall h\in H } \) and write \( \lambda > 0 \iff \lambda \in \ZZ_{\geq 0} \Delta \) and similarly define $\lambda < 0$.
View \( \alpha_i \in H\dual \), extended linearly as before.
Note $H = L_0, N^+ = \sum_{\lambda> 0} L_{\lambda}, N^- = \sum_{\lambda<0} L_ \lambda$, and thus
\[
L = N^- \oplus H \oplus N^+
,\]
which is a direct sum since the eigenvalues in different parts are distinct.
:::

:::