# Wednesday, November 16

:::{.remark}
Let $L\in \Lie\Alg^{\fd, \semisimple}$ containing $H$ with $\Phi, \Delta, W$ as usual.
Recall that $V\in \mods{L}\implies V = \bigoplus _{\lambda\in H\dual} V_{ \lambda}$ where $V_{ \lambda} \da \ts{v\in V\st h.v = \lambda(h)v\, \forall h\in H}$, which we call a **weight space** when \( \lambda\neq 0 \).
Note that if if $V$ is *any* representation of $V$, even finite-dimensional, $V' \da \bigoplus _{ \lambda\in H\dual} V_{ \lambda} \leq V$ is always an $L\dash$submodule.
The sum is still direct since the terms correspond to eigenspaces with distinct eigenvalues.
Note that if $h\in H, x\in L_{\alpha}, v\in V_{ \lambda}$, then
\[
h.(x.v) 
&= x.(h.v) + [hx].v \\
&= \lambda(h) x.v + \alpha(h) x.v \\
&= (\lambda+ \alpha)(h) x.v
,\]
so $L_{ \alpha} V_{ \lambda} \subseteq V_{ \lambda+ \alpha}$.
:::

:::{.lemma title="?"}
Let $V\in \mods{L}$, then

a. $L_{\alpha}$ maps $V_{ \lambda}$ into $V_{ \lambda + \alpha}$,
b. The sum $V' \da \sum _{ \lambda\in H\dual} V_{ \lambda}$ is direct and $V' \leq V$ is an $L\dash$submodule,
c. If $\dim V < \infty$ then $V = V'$.

:::

## $\S 20.2$: Highest weight modules

:::{.definition title="Maximal vectors"}
A **maximal vector** in an $L\dash$module $V$ is a nonzero weight vector $v\in V_{ \lambda}$ such that $L_ \alpha.v = 0$ for all positive roots \( \alpha \in \Phi^+ \).
Equivalently, $L_ \alpha .v = 0$ for all \( \alpha\in \Delta \).
:::

:::{.definition title="Highest weight vectors"}
A **highest weight vector** is a nonzero $v\in V_{ \lambda}$ where \( \lambda \) is maximal among all weights of $V$ with respect to the ordering $\leq$ corresponding to the choice of $\Delta$.
:::

:::{.observation}
If $v$ is a highest weight vector then $v$ is necessarily a maximal vector, since \( \lambda+ \alpha > \lambda \), but the converse is not necessarily true.
:::

:::{.warnings}
I.e., the weight of a highest weight vector need not be maximal.
:::

:::{.example title="?"}
In $\S 18$, $L$ is constructed using the Serre relations to get $L_0 \surjects L$ where $L$ involved $(S_{ij}^{\pm})$ and $L_0$ involved S1-S3.
Recalling $y_i^{m+1} y_j = y_{ij}$, since $x_k . y_{ij} = 0$, $y_{ij}$ is a maximal vector in $L_0$ as an $L\dash$module but is *not* a highest weight vector since $\mathrm{wt} y_{ij} = (-m+1) \alpha_i - \alpha_j < - \alpha_j = \mathrm{wt}(y_j)$ and the weight is not maximal.
:::

:::{.example title="?"}
View $L\in \mods{L}$ via $\ad_L$, then $\S 10.4$ shows that there is a unique highest root $\tilde \alpha$ satisfying \( \tilde \alpha \geq \alpha \) for all \( \alpha\in \Phi \).
Any nonzero $v\in L_{\tilde \alpha}$ is a highest weight vector for the adjoint representation.
:::

:::{.definition title="Borel subalgebras"}
A **Borel subalgebra** of $L$ is a maximal solvable subalgebra $B\leq L$.
:::

:::{.proposition title="?"}
$B \da H \oplus \bigoplus _{\alpha\in \Phi^+} L_ \alpha$ is a Borel subalgebra of $L$.
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:::{.proof title="?"}
If \( \alpha, \beta\in \Phi^+ \) then $[L_{ \alpha}, L_{ \beta}] = L_{\alpha + \beta}$ where \( \alpha + \beta\in \Phi^+ \) (if this is still a root), so $H\leq L$.
One has $B^{(i)} \da [B^{(i-1)}, B^{(i-1)}] \subseteq \sum_{\height( \beta) \geq 2^{i-1} } L_{\beta}$, since bracketing elements of $H$ together will vanish (since $H$ is abelian) and bracketing height 1 roots yields height 2, bracketing height 2 yields height 4, and so on.
Thus $B$ is a solvable subalgebra, since the height is uniformly bounded above by a finite number.
To see that its maximal, note that any subalgebra $B' \leq L$ containing $B$, it must also contain some $L_{ - \alpha}$ for some $\alpha\in \Phi^+$.
But then $B' \contains \liesl_2(\CC) = L_{ - \alpha} \oplus [ L_{ - \alpha }, L_{ \alpha}] \oplus L_{ \alpha}$ which is not solvable, so $B'$ can not be solvable.
:::

:::{.remark}
Let $V\in \lmod^\fd$, then $V\in \mods{B}$ by restriction and by Lie's theorem $V$ must have a common eigenvector $v$ for the action of $B$.
Since $B\contains H$, $v$ is a weight vector, $[B, B] = \bigoplus _{ \alpha\in \Phi^+} L_ \alpha$ acts by commutators of operators acting by scalars, which commute, and thus this acts by zero on $v$ and makes $v$ a maximal vector in $V$.
So any finite dimensional $L\dash$module as a maximal vector.
:::

:::{.definition title="Highest weight modules"}
A module $V\in \lmod$, possibly infinite dimensional, is a **highest weight module** if there exists a \( \lambda \in H\dual \) and a nonzero vector $v^+ \in V_{ \lambda}$ such that $V$ is generated as an $L\dash$module by $v^+$, i.e. $U(L).v^+ = V$.
:::

:::{.remark}
Let $x_ \alpha \in L_ \alpha, y_ \alpha\in L_{- \alpha}, h_{\alpha} = [x_ \alpha, y_ \alpha]$ be a fixed standard $\liesl_2$ triple in $L$.
:::

:::{.theorem title="?"}
Let $V\in \lmod$ be a highest weight module with maximal vector $v^+ \in V_{ \lambda}$.
Write \( \Phi^+ = \tsl \beta 1 m  \), \( \Delta = \tsl \alpha 1 \ell \), then

a. $V$ is spanned by the vectors $y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m} .v^+$ for $i_j\in \ZZ_{\geq 0}$.
  In particular, $V = \bigoplus _{ \mu\in H\dual} V_ \mu$.

b. The weights of $V$ are of the form $\mu = \lambda- \sum_{i=1}^\ell k_i \alpha_i$ with $k_i\in \ZZ_{ \geq 0}$, and all weights $\mu$ satisfy $\mu \leq \lambda$.

c. For each $\mu\in H\dual$, $\dim V_{ \mu} < \infty$, and for the highest weight \( \lambda \), one has $\dim V_{ \lambda} = 1$ spanned by $v^+$.

d. Each $L\dash$submodule $W$ of $V$ is a direct sum of its weight spaces.

e. $V$ is indecomposable in $\lmod$ with a unique maximal proper submodule and a corresponding unique irreducible quotient.

f. Every nonzero homomorphic image of $V$ is also a highest weight module of the same highest weight.

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

\envlist

a. Use the PBW theorem and extend a basis for any $B\leq L$ to a basis where the $B$ basis elements come second.
  Writing $L = N^ \oplus B$, one can decompose $U(L) = U(N^-) \tensor_\CC U(B)$ and get $U(L) v^+ = U(N^-)U(B)v^+ = U(N^-)U(H \oplus N^+)v^+$

b. Writing the $\beta$ in terms of $\alpha$ yields this expression.

c. Clear.

We'll finish the rest next time.
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