# Kac-Moody Groups (Monday, August 30)

:::{.remark}
See exercises in first two sections, 1.1 and 1.2.
See also the proof of the Borel-Weil theorem.
:::

## 1.2: Root Space Decompositions

:::{.remark}
Starting with a generalized Cartan matrix $A$, we produced a Lie algebra $\tilde\lieg(A)$ by taking the free Lie algebra and modding out by certain relations.
This algebra only depended on the realization of $A$, namely $(\lieg, \pi, \pi\dual)$, which we thought of as $(\lieg,\lieh\dual, \lieh)$, yielding $\lieg(A)$ modulo Serre relations.
:::

:::{.definition title="Root Lattice"}
Define
\[
Q &\da \ZZ\pi \subseteq \lieh\dual && \text{the root lattice }\\
Q^+ &\da \ZZ_{\geq 0} \pi \subseteq \lieh\dual \\
\lieg_\alpha &\da \ts{x\in g \st \forall h\in\lieh, [hx] = \alpha(h) x} && \text{for }\alpha\in\lieh\dual
.\]
:::

:::{.theorem title="?"}
\envlist

1. $\lieg = \lien^- \oplus \lieh \oplus \lien$, which are all nonzero.
2. $\lien^{\pm \alpha} = \bigoplus_{\alpha\in Q^+\smts{0}} \lieg_{\pm \alpha}$.
3. $\dim_\CC \lieg_\alpha < \infty$.
4. $\lien \da \gens{e_1,\cdots, e_\ell}$ subject *only* to the Serre relations, i.e. no additional relations are needed for this subalgebra.

:::

:::{.proof title="?"}
First step: prove for $\tilde \lieg$ and put a tilde on everything appearing in the theorem statement.

Let $\ts{v_1, \cdots, v_\ell}$ be a basis for $V$ and fix $\lambda \in \lieh\dual$.
Define an action of generators of $\tilde \lieg$ on $T(V)$ in the following way:

1. $\alpha:$ Set $f_i(\alpha) \da v_i\tensor a$ for $a\in T(V)$
2. $\beta:$ set $h(1) \da \inner{\lambda}{h}1 \da \lambda(h)\cdot 1$, and inductively on $s$ set 
\[
h(v_j \tensor a) \da - \inner{\alpha_j}{h} v_j \tensor a + v_j \tensor h(a) && a\in T^{s-1}(V), h\in \lieh, 1\leq j \leq \ell
.\]
3. $\gamma:$ Set $e_i(1) \da 0$ to kill constants, and inductively on $s$,
\[
e_i(v_j \tensor a) = \delta_{ij}\alpha\dual_{i}(a) + v_j\tensor e_i(a) && a\in T^{s-1}(V), 1\leq j \leq \ell
.\]

One should show that these define a representation by checking the Serre relations.
Consider instead how this works in the $\lieg = \liesl_2$ case:


:::{.example title="?"}
For $\liesl_2$, take the realization $(\CC, \ts{\alpha}, \ts{ \alpha\dual } )$ corresponds to the matrix $A = (s)$.
Here $T(V) = \CC[x]$, and since there are no Serre relations, $\tilde \lieg = \lieg$.
We have $e = [0,1; 0,0], f = [0,0; 1,0]$ which generate the positive/negative unipotent parts respectively.
Then $h = \ts{ \diag(h, -h) }$.
Checking the action:

1. $\alpha: \matt 0 0 1 0 p = xp$ which raises degree by 1.
M
2. 
\[
\beta:
h(1) = \lambda(h)1
\implies
h(xp) = -\alpha(h)xp + x(hp)
,\]
where $p\in \CC[x]_{g-1}$.
  For example, 
  \[
  h(x) &= -\alpha(x) + x \lambda(h) = (\lambda - \alpha)(h)x \\
  h(x^2) &= (\lambda - 2\alpha)(h)x^2
  ,\] 
  so this acts diagonally and preserves degree.

3. Check 
\[
\matt 0 1 0 0 \cdot (1) &= 0 \\
\matt 0 1 0 0 \cdot (xp) &= p + x \matt 0 1 0 0 p
.\]

  Check that $ex = 1+0$ and $ex^2 = x+x = 2x$, so $e$ acts by differentiation. 


:::

Note that $\lieh$ forms a subalgebra since it's a nondegenerate map.
This follows from the fact that we get a representation $\rho_{\lambda}$ of $\tilde\lieg$ on $T(V)$, which for each $h$ acts nontrivially on some $T(V)$.
So use $\rho_{\lambda}$ to deduce the theorem for $\tilde\lieg$:
\[
\ts{ [ x, y] \st x,y\in \lieh = \gens{\ts{e_i, f_i}_{i=1}^\ell} }  \subseteq 
\tilde \lien^- + \lieh + \tilde\lien
= \lieg
,\]
we'll show this sum is direct.

Let $u = n^- + h + n^+ = 0$, then in $T(V)$ we have $u(1) = n^-(1) + \inner{\lambda}{h}1$, which forces $\inner{\lambda}{h} = 0$ for all \( \lambda\in\lieh\dual \) and thus $h = 0$.
Use the restriction $\tilde\lieg\to \tilde\lien$ to get a map $U(\tilde\lien^-)\to T(V)$ out of the enveloping algebra, using that $T(V)$ is an associative algebra.
Using $f_i \mapsto v_i$, this is surjective and in fact an isomorphism.
Sending $\lien^- \mapsto \lien^-(1)$ yields $\tilde\lien^- \subseteq U(\tilde\lien^-) = T(V)$.
This yields $n^- = 0$, making the sum direct.

We can write 
$\tilde\lien^- = F\gens{f_1,\cdots, f_\ell}$  
and 
$\tilde\lien = F\gens{e_1,\cdots, e_\ell}$ 
and by the PBW theorem, $\dim \tilde\lieg_\alpha < \infty$.
This uses that the weight spaces for $\tilde\lien^-$ are contained in $U(\tilde\lien^-)$.
Note that there is a *Cartan involution*
\[
\tilde\omega: \tilde\lieg &\selfmap \\
e_i &\mapsto -f_i \\
f_i &\mapsto -e_i \\
h &\mapsto -h
.\]

Now to prove the theorem for $\lieg$ itself, write $\tilde \lier \da \ker( \tilde\lieg \mapsvia{\gamma}  \lieg_\alpha) \normal \tilde\lieg$.
This is an ideal, and thus $\lieh\dash$stable.
We can thus write
\[
\tilde\lier = 
\qty{
\bigoplus_{\alpha\in Q^+ \smts{0}} \lier_{-\alpha}
}
\oplus 
\tilde\lier_0
\oplus 
\qty{
\bigoplus_{\alpha\in Q^+ \smts{0}} \lier_{\alpha}
}
\]
where $\tilde\lier_{\beta} \da \tilde\lier \intersect \tilde\lieg_{\beta}$ and $\tilde \lier_0 = \tilde\lier \intersect \lieh$.
We have ideals $\tilde\lier^{\pm} \normal \tilde\lien^{\pm}$, which are respectively generated by 
\[
\ts{e_{i, j} = (\ad e_i)^{1 - a_{i, j}} (e_{j}) \st i\neq j }
&&
\ts{f_{i, j} = (\ad f_i)^{1 - a_{i, j}} (f_{j}) \st i\neq j }
,\]
where $\ad f_k (e_{i, j}) = 0$ for all $k$ and $i\neq j$.
Skipping a few things that are spelled out in the book, e.g. that $\tilde \lier_0 = 0$, we conclude that $\tilde\lier = \tilde\lier^+ \oplus \tilde\lier^-$, both of which are ideals in $\tilde\lieg$.
Since $\tilde\lier_0 = 0$, we get $\lieh \subseteq \lieg$, and using that $\gamma$ is surjective we have an isomorphism of $\CC\dash$modules
\[
\lieg = \tilde\lieg/\tilde\lier = \tilde\lien^-/ \tilde\lier^- \oplus \lieh \oplus \tilde\lien / \tilde\lier^+
.\]

Write $\Delta \da \ts{\alpha\in Q\smts{0} \st \lieg_\alpha \neq 0}$ the *set of roots* and $\lieg_{\alpha}$ the *root space*, then set 
\[
\Delta^+ &\da \Delta \intersect Q^+ \\
\Delta^- &\da \Delta \intersect (-Q^+) \\
\Delta &\da \Delta^+ \union \Delta^-
.\]
Also for $Y \subseteq \ts{1,\cdots,\ell}$ write 
\[
\Delta_Y &\da \Delta \intersect \qty{\bigoplus_{i\in Y} \ZZ \alpha_i } \\
\lieg_Y &\da \lieh \oplus \qty{ \bigoplus_{\alpha\in \Delta_Y} \lieg_\alpha  }
.\]
We say $Y$ is *finite type* if $\lieg_Y$ is finite dimensional, and given $A$ we can associate some matrix $(a_{i, j})_{i, j\in Y}$.
:::

:::{.remark}
See Ch. 13 for how this generalizes the semisimple case.
:::