# Monday October 28

## Algebra Generated by a Cartan Matrix

Last time:
The claim was that for a Cartan matrix $A$, there is a lie algebra $\lieg(A)$ that is semisimple with CSA $\lieh$ and a root system $\Phi$ that defines that Cartan matrix $A$.

The algebra $\lieg$ is generated by $\theset{e_i, f_i, h_i \mid i\in I = \theset{1, 2, \cdots \ell}}$, with relations

\begin{align*}
[h_i, h_j] &= 0 \\
[h_i, e_j] &= a_{ji} e_j \\
[e_i, f_j] &= \delta_{ij} h_i \\
[h_i, f_j] &= -a_{ji} f_j,
\end{align*}

along with the Serre relations (which only appear in higher degrees):

\begin{align*}
s^+_{ij} &\coloneqq \ad(e_i)^{1 - a_{ji}} (e_j) = 0 \quad \text{ if } i\neq j\\
s^-_{ij} &\coloneqq \ad(f_i)^{1 - a_{ji}} (f_j) = 0 \quad \text{ if } i\neq j \\
.\end{align*}

*Proof:*

1. Show that $\theset{e_i, f_i, h_i}$ generates $\lieg$.

The subalgebra $\lieh$ is spanned by $\theset{ t_{\alpha_i} \mid i\in I }$ and hence spanned by $\theset{h_i \mid i\in I}$.
So it suffices to show that $\lieg_{\alpha} \subseteq \generators{e_i}$ for all $\alpha -in \Phi^+$.

Write $\alpha = \alpha_i + \beta$ for each $i\in I, \beta \in \Phi^+$. 
Then $[\lieg_{\alpha_i}, \lieg_\beta] = \lieg_\alpha = \CC e_\alpha$, so $e_\alpha = [e_i, e_\beta]$ for some nonzero $e_\beta \in \lieg_\beta$.

By repeating this argument, we find that $e_\alpha = [ [ \cdots [e_{i_1}, e_{i_2}], e_{i_3}] \cdots ], \cdots e_{i_k} ]$.

2. Verify the relations

We need to check that $s^+_{ij} = 0$.
The $\alpha_i$ root string through $\alpha_j$ is given by 
$$
\alpha_j + p\alpha_i \to \cdots \to \alpha_j + q \alpha_i
$$

where $p\neq 0$ because $\alpha_j - \alpha_i \not\in\Phi$ for any $i$, so the smallest root must be $\alpha_j \in \Phi$.
By prop 8.4d, this means that $-q = \alpha_j(h_i) = \alpha_{ji}$.

Thus $\ad(e_i)^{1-\alpha_{ji}} (e_j) = \ad(e_i)^{1+q} \in \lieg_{\alpha_j + (q+1)\alpha_i} = \theset{0}$. 

## The Lie Algebra $\tilde \lieg(A)$

Fix a Cartan matrix $A = (a_ij)_{i, j \in I}$where $I = \theset{1, \cdots, \ell}$.
Let $\tilde J \normal \mathcal{FL}(\theset{e_i, f_i, h_i \mid i\in I})$ generated by

- $[h_i, h_j]$,
- $[h_i, e_j] - a_{ji}e_j$,
- $[e_i, f_j] - \delta_{ij} h_i$
- $[h_i, f_j] + a_{ji} f_j$.

Then let $J$ be the same ideal with the additional relations $s^+, s^-$, and set 

- $\tilde \lieg(A) = \mathcal{FL}(\theset{e_i, f_i, h_i}) / \tilde J$,
- $\lieg(A) = \mathcal{FL}(\theset{e_i, f_i, h_i}) / J$.

**Proposition:**

a. Let $V = \mathcal{F}(\theset{f_1, \cdots, f_\ell})$. Then $\pi: \tilde \lieg \to \liegl(V)$ is a *representation* with
  - $f_j: f_{i_1} \cdots f_{i_r} \mapsto f_j f_{i_1}$
  - $h_j: f_{i_1} \cdots f_{i_r} \mapsto (\alpha_{ji_1} + \cdots )f_{i_1} \cdots f_{i_r}$
  - $e_j: f_{i_1} \cdots f_{i_r} \mapsto (\sum \delta_{} \sum a)(\alpha_{ji_1} + \cdots )f_{i_1} \cdots f_{i_r}$

b. $\theset{h_1, \cdots h_\ell}$ is linearly independent set in $\tilde \lieg$.

For (a), it suffices to check $[\pi(h_i), \pi(h_j)] = 0$, $[\pi(h_i), \pi(e_j)] = a_{ji} \pi(e_j)$, etc.
For (b), it suffices to show that $\theset{\pi(h_i) \mid i\in I}$ is linearly independent.

Suppose $\sum c_i \pi(h_i) = 0$ in $\liegl(V)$. Then,

\begin{align*}
0 = \left( \sum_c c_i \pi(h_i) \right)(f_j) 
&= -\left( \sum_i c_i \alpha_{ji} \right) f_j \\
\implies \sum c_i \alpha_{ji} &= 0 \quad \forall ~j \\
\implies c_i &= 0 \quad \forall~i,
.\end{align*}

since $A$ is invertible.

Thus $\tilde \lieh \coloneqq \mathrm{span}_\CC\theset{h_i}$ is a lie subalgebra of $\tilde\lieg$..

**Theorem:**

a. $\tilde \lieg = \bigoplus_{u\in \tilde\lieh^*} \tilde \lieg_\mu$ as vector spaces, where
$$
\tilde\lieg_\mu \coloneqq \theset{x\in\tilde\lieg\suchthat [h, x] = \mu(h) x \quad \forall h\in\tilde\lieh}.
$$

b. $\tilde\lieg = \tilde n^- \oplus \tilde\lieh \oplus \tilde n$ as vector spaces, where $\tilde n^- \coloneqq \generators{f_i}$ and $\tilde n \coloneqq \generators{e_i}$.

*Proof of (a):*

It's easy to check that $[\tilde\lieg_\lambda, \tilde\lieg_\mu] \subseteq \lieg_{\lambda + \mu}$ for all $\lambda,\mu \in \tilde\lieh^*$.
Define $\alpha_i \in \tilde\lieh^*$ by $h_j \mapsto a_{ij}$. Then

- $e_i \in \tilde\lieg_{\alpha_i}, f_i \in \tilde\lieg_{-\alpha_i}, h_i \in\tilde\lieg_0$ for all $i$.
- Any $x\in \tilde\lieg$ lies in $\tilde\lieg_\mu$ for *some* $\mu$.
- $\tilde\lieg = \sum_\mu \tilde\lieg_\mu$.

We just need to show that the last sum is in fact a direct sum.

Suppose that $\exists x\neq 0$ such that $x\in \tilde\lieg_\mu, x = \sum_\nu x_\nu$ where $x_\nu \in \tilde\lieg_{\nu} - \theset{0}$ and $\nu$ runs over a finite set of weights that are not equal to $\mu$.

Then $[h, x] = \mu(h) x$, and so $(\ad h - \mu(h)) (x) = 0$. 
On the other hand, $\prod_\nu (\ad h - \nu(h)) (x_\nu) = 0$.
So pick some $h\in\tilde\lieh$ such that $\mu(h) \neq \nu(h)$ for all $\nu$.
Then the polynomials 
$t - \mu(h), \prod_\nu (t - \nu(h))$ 
are coprime, and so there exist $a,b$ such that
$$
a(t - \mu(h) + b \prod_\nu (t - \nu(h)) = 1, 
$$

Then evaluating at $t= \ad h$, we get 

\begin{align*}
x = 1(x) = a(\ad h) (\ad h - \mu(h))(x) + b(\ad h)(\prod_\nu \ad h - \nu(h))(x) = 0
,\end{align*}

and so $\tilde\lieg = \oplus_\nu \tilde\lieg_\mu$.