# Monday, November 07

:::{.remark}
Last time: constructing a semisimple Lie algebra that has a given root system.
Setup:

- $\Delta = \ts{\alpha_1,\cdots, \alpha_l}$.
- $\hat L$ the free Lie algebra on $\ts{\hat x_i, \hat h_i, \hat y_i}_{1\leq i\leq l}$.
- $\hat K$ the ideal generated by the Serre relations.
- $L_0 \da \hat L/\hat K$ with quotient map $\pi$.
- $\hat \phi: \hat L\to \liegl(V)$ a representation we constructed.
- $\hat H$ the free Lie algebra on $\ts{h_i}$.

:::

:::{.theorem title="?"}
$\hat K \subseteq \ker \hat\phi$, so $\hat\phi$ induces a representation $\phi$ of $L_0$ on $\liegl(V)$

\begin{tikzcd}
	{\hat L} && {\liegl(V)} \\
	\\
	& {L_0}
	\arrow["\hat\phi", from=1-1, to=1-3]
	\arrow["\pi"', from=1-1, to=3-2]
	\arrow["{\exists \phi}"', dashed, from=3-2, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGhhdCBMIl0sWzIsMCwiXFxsaWVnbChWKSJdLFsxLDIsIkxfMCJdLFswLDEsIlxcaGF0XFxwaGkiXSxbMCwyLCJcXHBpIiwyXSxbMiwxLCJcXGV4aXN0cyBcXHBoaSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

:::

:::{.proof title="?"}
Straightforward but tedious checking of all relations, e.g. 
\[
\hat\phi(\hat h_i) \circ \hat\phi(\hat x_j) - \hat \phi(\hat x_j )\hat\phi(\hat h_i) = \inp{ \alpha_j}{\alpha_i} \hat \phi(\hat x_j)
.\]
:::

:::{.theorem title="?"}
In $L_0$, the $h_i$ form a basis for an $\ell\dash$dimensional abelian subalgebra $H$ of $L_0$, and moreover $L_0 = Y \oplus H \oplus X$ where $Y, X$ are the subalgebras generated by the $x_i$ and $y_i$ respectively.
:::

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

**Steps 1 and 2**:
:::{.claim}
$\pi(\hat H) = H$ is $\ell\dash$dimensional.
:::

Clearly the $\hat h_i$ span an $\ell\dash$dimensional subspace of $\hat L$, so we need to show that $\pi$ restricts to an isomorphism $\pi: \hat H\iso H$.
Suppose $\hat h \da \sum_{j=1}^\ell c_j \hat h_j \in \ker \pi$, so $\hat\phi (\hat h) = 0$.
Thus
\[
0 = \hat h \cdot v_i = \sum_{j} c_j \hat{h}_j \cdot v_i = - \sum_j c_j \inp{ \alpha_i}{\alpha_j}
= - \sum_j a_{ij} c_j \qquad \forall i
,\]
so $A\vector c = 0$ where $A$ is the Cartan matrix, and so $\vector c = \vector 0$ since $A$ is invertible (since it was essentially a Gram matrix).

**Step 3**:
Now $\sum \FF x_i + \sum \FF h_i + \sum \FF y_i \mapsvia{\pi} L_0$ maps isomorphically to $L_0$, and S2, S3 show that for each $i$.
Then $\FF x_i + \FF h_i + \FF y_i$ is a homomorphic image of $\liesl_2$, which is simple if $\characteristic \FF \neq 2$.
Note $\pi( \hat h_i) = h_i \neq 0$ in $L_0$ by (1), so this subspace of $L_0$ is isomorphic to $\liesl_2(\FF)$.
In particular $\ts{x_i, h_i, y_i}$ is linearly independent in $L_0$ for each fixed $i$.
Supposing $0 = \sum_{j=1}^\ell (a_j x_j + b_j h_j + c_j y_j)$, applying $\ad_{L_0, h_i}$ for each $i$ to obtain
\[
0 = \sum_{j=1}^\ell \qty{ a_j\inp{\alpha_j}{\alpha_i} x_j + b_j 0 - c_j \inner{ \alpha_j }{ \alpha_i } y_j  }
= \sum_{j=1}^\ell \inner{ \alpha_j }{ \alpha_i } (a_j x_j - x_y y_j) 
,\]
and by invertibility of $A$ we have $a_j x_j - c_j y_j = 0$ for each $j$.
So $a_j = c_j = 0$ for all $j$, and $\sum b_j h_j = 0$ implies $b_j = 0$ for all $j$ from (1).

**Step 4**:
$H = \sum_{j=1}^\ell \FF h_j$ is an $\ell\dash$dimensional abelian subalgebra of $L_0$ by (1) and S1.

**Step 5**:
Write $[x_{i_1}\cdots x_{i_t}] \da [x_{i_1} [ x_{i_2} [ \cdots [x_{i_{t-1}} x_{i_t}] \cdots ]] \in X$ for an iterated bracket, taken by convention to be bracketing from the right.
We have
\[
\ad_{L_0, h_j}([x_{i_1} \cdots x_{i_t}] ) = \qty{ \inner{ \alpha_{i_1} }{ \alpha_j } + \cdots + \inner{ \alpha_{i_t} }{ \alpha_j }    } [x_{i_1} \cdots x_{i_t}] \qquad t\geq 1
,\]
and similarly for $[y_{i_1}\cdots y_{i_t}]$.

**Step 6**:
For $t\geq 2$, $[y_j [ x_{i_1} \cdots x_{i_t} ] ] \in X$, and similarly with the roles of $x_i, y_i$ reversed.
This follows from the fact that $\ad_{L_0, y_j}$ acts by derivations, and using S2 and S3.

**Step 7**: It follows from steps 4 through 6 that $Y+H+X$ is a subalgebra of $L_0$.
One shows that $[[ x_{i_1} \cdots x_{i_t}], [y_{i_1} \cdots y_{i_t} ]]$, which comes down to the Jacobi identity and induction on $s+t$.
E.g. 
\[
[ [x_1 x_2], [y_3 y_4] ] = [x_1[ x_2 [y_3 y_4 ] ] ] - [x_2 [x_1 [y_3 y_4]] \in [x_1, \FF y_3 + \FF y_4] + \cdots \in H + \cdots
,\]
which lands in $H$ since there are as many $x_i$ as $y_i$, whereas if there are more $x_i$ than $y_i$ this lands in $X$, and so on.
Since $Y+H+X$ is a subalgebra that contains the generators $x_i, h_i, y_i$ of $L_0$, it must be equal to $L_0$.

**Step 8**:
The decomposition $L_0 = X + H + Y$ is a direct sum decomposition of $L_0$ into submodule for the adjoint action of $H$.
Use the computation in the previous step to see that every element of $X$ is a linear combination of elements $[x_{i_1}\cdots x_{i_t}]$ and similarly for $Y$.
These are eigenvectors for the action of $\ad_H \actson L_0$ by (5), and eigenfunctions for $X$ have the form $\lambda = \sum_{i=1}^\ell c_i \alpha_i$ with $c_i \in \ZZ_{\geq 0}$.
The \( \lambda_i \) is referred to as a **weight**, and $c_i$ is the number of times $i$ appears is an index in $i_1,\cdots, i_t$.
So every weight space $X_\lambda$ is finite-dimensional, and the weights of $Y$ are $-\lambda$.
Since the weights in $X, H, Y$ are all different, their intersections must be trivial and the sum is direct.
:::

:::{.remark}
$L_0 = Y \bigoplus H \bigoplus X$ is known as the **triangular decomposition**, where the $x_i$ are on the super diagonal and bracket to upper-triangular elements, and the $y_i$ are their transposes.
:::