# $\S 18$: Generators and Relations for Simple Lie Algebras (Friday, November 04)

:::{.remark}
Recall that the free Lie algebra of a set $X$, $L(X)$ satisfies a universal property: 

\begin{tikzcd}
	X && {L(X)} \\
	\\
	&& M
	\arrow["\iota", hook, from=1-1, to=1-3]
	\arrow["{\forall \phi}"', from=1-1, to=3-3]
	\arrow["{\exists ! \tilde \phi}", dashed, from=1-3, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiTChYKSJdLFsyLDIsIk0iXSxbMCwxLCJcXGlvdGEiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDIsIlxcZm9yYWxsIFxccGhpIiwyXSxbMSwyLCJcXGV4aXN0cyAhIFxcdGlsZGUgXFxwaGkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

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## $\S 18.1$: Relations satisfied by $L$

:::{.definition title="Relations"}
Given an arbitrary $L\in \liealg$ and fix a set $X$ of generators for $L$ and form $LX()$, then there is a Lie algebra morphism $\pi: L(X) \surjects L$ which is surjective since $X$ generates $L$.
Defining $R\da \ker \pi$, one has $L \cong L(X)/R$, so $R$ is called the **ideal of relations**.
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:::{.remark}
Let $L \in \liealg^{\fd, \ss}_{\CC}$, let $H \subseteq L$ be a maximal toral subalgebra, and $\Phi$ its root system.
Fix a base $\Delta = \ts{ \alpha_1, \cdots, \alpha_\ell} \subseteq \Phi$.
Recall 
\[
\inp{ \alpha_j}{ \alpha_i} \da {2 (\alpha_j, \alpha_i) \over (\alpha_i, \alpha_i)} = \alpha_j(h_i),\qquad g_i \da h_{\alpha_i} = {2 \alpha_i \over (\alpha_i, \alpha_i)}
.\]
The root strings are of the form $\beta - r \alpha, \cdots, \beta, \cdots, \beta+ q \alpha$ where $r-q = \beta(h_ \alpha)$.
For any $i$ we can fix a standard $\liesl_2$ triple $\ts{x_i, h_i, y_i}$ such that $x_i\in L_{ \alpha_i}, y_i \in L_{- \alpha_i}, h_i\in [x_i y_i]$.
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:::{.proposition title="Serre relations"}
$L$ is generated as a Lie algebra by the $3\ell$ generators $X\da \ts{x_i, h_i, y_i \st 1\leq i\leq \ell}$ subject to at least the following relations:

- S1: $[h_i h_j] = 0$,
- S2: $[x_i y_j] = \delta_{ij} h_i$,
- S3: $[h_i x_j] = \inp{ \alpha_j}{\alpha_i} x_j$ and $[h_i y_j] = - \inp{ \alpha_j}{ \alpha_i} y_j$.
- $\text{S}_{ij}^+$: $\ad_{x_i}^{-\inp{ \alpha_j}{\alpha_i} + 1}(x_j) = 0$ for $i\neq j$
- $\text{S}_{ij}^-$: $\ad_{y_i}^{-\inp{ \alpha_j}{\alpha_i} + 1}(x_j) = 0$ for $i\neq j$

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:::{.proof title="?"}
Recall that differences of simple roots are never roots, since the coefficients have mixed signs.
Since \( \alpha_i - \alpha_j \not\in \Phi \), we have $[x_i y_j] = 0$ for $i\neq j$ since it would have to be in $L_{\alpha_i - \alpha_j}$.
Consider the $\alpha_i$ root string through $\alpha_j$: we have $r=0$ from above, and the string is
\[
\alpha, \alpha+ \alpha_i, \cdots, \alpha_j - \inp{\alpha_j}{\alpha_i} \alpha_i
\]
since $L_\beta = 0$ for $\beta \da \alpha_j - \qty{ \inp{\alpha_j}{ \alpha_i} + 1} \alpha_i$.
The relations for $S_{ij}^\pm$ follow similarly.
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:::{.remark}
Note that these relations are all described in a way that only involves the Cartan matrix of $\Phi$, noting that changing bases only permutes its rows and columns.
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:::{.theorem title="Serre"}
These five relations form a complete set of defining relations for $L$, i.e. $L \cong L(X)/R$ where $R$ is the ideal generated by the Serre relations above.
Moreover, given a root system $\Phi$ and a Cartan matrix, one can define a Lie algebra using these generators and relations that is finite-dimensional, simple, and has root system $\Phi$.
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## $\S 18.2$: Consequences of Serre relations S1, S2, S3

:::{.remark}
Fix an irreducible root system $\Phi$ of rank $\ell$ with Cartan matrix $A$.
Let $\hat L \da L(\hat X)$ where $\hat X \da \ts{\hat x_i, \hat h_i, \hat y_i \st 1\leq i\leq \ell}$. 
Let $\hat K \normal \hat L$ be the 2-sided ideal generated by the relations S1, S2, S3.
Let $L_0 \da \hat L/\hat K$ and write $\pi$ for the quotient map $\hat L\to L_0$ -- note that $L_0$ is infinite-dimensional, although it's not yet clear that $L_0\neq 0$.
We'll study $L_0$ by defining a representation of it, which is essentially the adjoint representation of $L_0$ acting on $\ts{y_i}$.
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:::{.remark}
Recall that a representation of $M\in \liealg$ is a morphism $\phi\in \liealg(M, \liegl(V))$ for $V\in\fmod$.
This yields a diagram

\begin{tikzcd}
	{} && \Assoc\Alg \\
	M && {U(M)} \\
	\\
	&& {\Endo(V)}
	\arrow["\iota", hook, from=2-1, to=2-3]
	\arrow["\phi"', from=2-1, to=4-3]
	\arrow["{\exists !\phi}", dashed, from=2-3, to=4-3]
\end{tikzcd}
> [Link to diagram](https://q.uiver.app/?q=WzAsNSxbMCwxLCJNIl0sWzIsMSwiVShNKSJdLFsyLDMsIlxcZW5kbyhWKSJdLFsyLDAsIlxcQXNzb2NcXEFsZyJdLFswLDBdLFswLDEsIlxcaW90YSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMiwiXFxwaGkiLDJdLFsxLDIsIlxcZXhpc3RzICFcXHBoaSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==)

Conversely given an algebra morphism $\tilde \phi: U(M) \to \Endo(V)$, and restricting $\tilde \phi$ to $M \subseteq U(M)$ gives a Lie algebra morphism $\phi: M\to \Endo(V) = \liegl(V)$.
This representations of $M$ (using $\liegl(V)$) correspond to associative algebra representations of $U(M)$ (using $\Endo(V)$).
Since $U(M) = T(V(\hat X))$, using the various universal properties, having a representation $V$ of $\hat L$ is equivalent to having a set map $\hat X\to \Endo(V)$, i.e. elements of $\hat X$ should act linearly on $V$.
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:::{.remark}
Let $V$ be the tensor algebra on a vector space with basis $\ts{v_1,\cdots, v_\ell}$, thinking of each $v_i$ being associated to $\hat y_i$.
Write $v_1 v_2\cdots v_t \da v_1 \tensor v_2\tensor\cdots\tensor v_t$, and define elements of $\Endo(V)$ by

- $\hat h_j \cdot 1 \da 0$,
- $\hat h_j \cdot v_{i_1}\cdots v_{i_t} \da -\qty{ \inp{ \alpha_{i_1}}{\alpha_j} + \cdots \inp{\alpha_{i_t}}{\alpha_j} }v_{i_1}\cdots v_{i_t}$,
- $\hat{y}_j\cdot v_{i_1}\cdots v_{i_t} \da v_j v_{i_1} \cdots v_{i_t}$ for $t\geq 0$,
- $\hat x_j \cdot 1 \da 0$,
- $\hat x_j \cdot v_i \da 0$ for all $i$,
- $\hat x_j \cdot v_{i_1}\cdots v_{i_t} \da v_{i_1} \qty{ \hat x_j v_{i_2} \cdots v_{i_t}} - \delta_{i_1, j}\qty{\inp{\alpha_{i_2} }{\alpha_j} + \cdots + \inp{\alpha_{i_2} }{\alpha_j} }v_{i_2}\cdots v_{i_t}$ for $t\geq 2$.

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