# Friday October 25

## PBW Theorem

**Theorem (PBW):**
The universal enveloping algebra $U(\lieg)$ has a basis consisting of the PBW monomials.
If we fix a basis $\theset{x_i \suchthat i\in I}$ of $\lieg$ with a total order, then
$\theset{y_{i_1}^{r_1} \cdots y_{i_n}^{r_n} \suchthat n\in \NN > 0,~ i_j \in I,~ r_i \geq 1}$.

We will construct a map

\begin{align*}
\iota: \lieg &\to U(\lieg) \\
x_i &\mapsto x_i + J \coloneqq y_i
,\end{align*}

where we can recall that $U(\lieg) \coloneqq T(\lieg)/ J$ where $J$ was an ideal of specific relations.

**Corollary:**

a. The map $\iota$ is injective.
b. The map $\iota$ has no *zero divisors*. 

> We will use property (b) to study properties of Verma modules

*Proof of (a):*
If $\sum c_i x_i \in \ker(\iota)$, then 

\begin{align*}
0 &= \iota(\sum c_i x_i) = \sum c_i y_i \\
&\implies c_i = 0\quad \forall i \text{ since } \theset{y_i} \subsetneq \theset{\text{ PBW monomials }}\\
&\implies \ker(\iota) = 0.
\end{align*}

*Proof of (b):*
An arbitrary element in $U(\lieg)$ is of the form

\begin{align*}
a &= \sum c^a_{\vector i, \vector r} y_{i_1}^{r_1} \cdots y_{i_n}^{r_n} \text{ for some } c\in F \\
\coloneqq f_a(\vector y) + \text{ terms with smaller total degree }
.\end{align*}

where $f$ is defined by picking out only those terms of highest total degree, e.g. $f(2y_1 + y_1y_2y_3 + y_2^2) = y_1y_2y_3$, which is of total degree 3.

We want to show that $a\neq 0$ and $b\neq 0$ then $ab \neq 0$, i.e. $(f_a(\vector y) + \cdots)(f_b(\vector y) + \cdots) \neq 0$.

Recall that $y_ay_b = y_by_a + \sum_{a,b \in I} \text{ degree 1 monomials }$. Thus $f_a(\vector y)(f_b(\vector y)) \coloneqq  f_a f_b(\vector y) + \sum\text{ terms of smaller total degree }$.

Here we define $f_a(\vector y) f_b(\vector y)$ by e.g. if $b = y_2$, then $f_b(\vector y) = y_2$, and $f_a(\vector y) f_b(\vector y) = y_1 y_2 y_3 y_2 = y_1 y_2^2 y_3 + y_1y_2[y_3, y_2]$. Note that the leading term is of total degree 4, and the remaining term is a sum of lower degree terms.

## Free Lie Algebra

Let $X \coloneqq \theset{x_i \suchthat i\in I}$ be a set. 
Define the *free associative algebra* $\mathcal{F}(X)$ as $\theset{\sum_k c_{\vector i} X_{\vector i} \mid \vector i = (i_1, \cdots, i_k) \in I^k,~ c_{\vector i} \in F}$.
Then the associated *free lie algebra* $\mathcal{FL}(x) = \intersect_{\lieg} \lieg$ where $X \subseteq \lieg \subseteq \mathcal{F}(X)$ is a containment of lie algebras.

Let $\iota: X \injects \mathcal{FL}(X)$.

Proposition:

a. $\mathcal{FL}(X)$ satisfies a universal property -- for any map $\theta: X \to \lieg$ a lie algebra, there exists a unique $\psi$ making the following diagram commute:

\begin{center}
\begin{tikzcd}
X \arrow[rr, "\iota"] \arrow[rrdd, "\theta"] &  & \mathcal{FL}(X) \arrow[dd, "\exists! \psi", dotted] \\
                                             &  &                                                     \\
                                             &  & \lieg                                              
\end{tikzcd}
\end{center}

b. $U(\mathcal{FL}(X)) = \mathcal{F}(X)$.

> Upshot: we can define a Lie algebra $\lieg$ using generators and relations, and define $\lieg \coloneqq \mathcal{FL}(X) / (R)$ for some set of relations $R$.

## Generators and Relations

Recall that we have a correspondence

\begin{align*}
\theset{\lieg \suchthat \lieg \text{ is a semisimple Lie Algebra }} & &\\ 
&\iff \theset{\Phi, \text{ root systems }} & & \\
&\iff \theset{\text{Dynkin diagrams (Cartan Matrices)}} \\
(\lieg, \lieh) &\to \Phi,\quad \theset{a_i} \subseteq \theset{a} \coloneqq \Pi \subseteq \Phi &\mapsto A_{i, j} = \inner{\alpha_i}{\alpha_j\dual} \\
\lieg(A) &<-_? \Phi &<- A
.\end{align*}

We had an explicit construction to go from Dynkin diagrams to root systems, and an existence theorem of Serre's will take root systems $\Phi$ and produce semisimple Lie algebras from them.
The question will be whether or not there is a one-to-one correspondence here, and that's what we'll spend the rest of the semester showing.

## Cartan/Serre Relations

Recall from (8.3):
For all $\alpha \in \Phi$, we have $e_\alpha \in \lieg_\alpha \setminus\theset{0}$, then there exists a unique $f_\alpha \in \lieg_{-\alpha}$ such that $[e_\alpha, f_\alpha] = h_\alpha \coloneqq \frac{2t_\alpha} {\kappa(t_\alpha, t_\alpha)}$,
where $t_\alpha \coloneqq \alpha = \kappa(t_\alpha, \wait)$.

Fix $\Pi = \theset{\alpha_i \mid i\in I}$, and write $h_i \coloneqq h_{\alpha_i},~ e_i = e_{\alpha_i}$ for each $i$.
Then $\alpha_i(h_j) = a_{ij}$.
Now fix $e_i \in \lieg_{\alpha_i}, f_i \in \lieg_{-\alpha}$ such that $[e_i, f_i] = h_i$ for every $i\in I$.

**Proposition:**
$\lieg$ is generated by $\theset{e_i, f_i, h_i \mid i\in I}$.

We have the Cartan relations for each $i, j\in I$:

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


as well as Serre relations for each $i\neq j$:

\begin{align*}
(\ad e_i)^{1-a_{ji}}(e_j) = 0 \quad (ad f_i)^{1-a_{ji}}(f_j) = 0
.\end{align*}

*Example:* 
$\lieg = \liesl(2, \CC) = \generators{e_1 \coloneqq e, f_1 \coloneqq f, h_1 \coloneqq h}$ satisfies $[h, e] = 2e$ and $[h, f] = -2f$, and since there are no higher order relation, there are no Serre relations.
So we get $A = (2)$ as a matrix.

*Example:* 
$\lieg = \liesp(4, \CC)$ is of type $C_2$, and is generated by $\generators{e_1, e_2, f_1, f_2, h_1, h_2}$ satisfying

- $[h_1, h_2] = 0$
- $[h_1, e_1] = 2e_1$
- $[h_1, e_2] = -2 e_2$
- $\cdots$

Then e.g. we have $(\ad e_1)^{1-a_{ij}}(e_2) = (\ad e_1)^3(e_2) = 0$.