# Wednesday January 22nd


:::{.exercise title="?"}
Try problems 1.1 and 1.3* in Humphreys.
:::


**Recall:**
In category $\mathcal O$, we have finite dimensional, semisimple modules over $\CC$ with triangular decompositions.

If $M$ is any $U(\lieg)$ module than a $v^+ \in M_\lambda$ a weight vector (so $\lambda \in \lieh\dual$) is primitive iff $\lien \cdot v^+ = 0$.
Note: it doesn't have to be of maximal weight.
$M$ is a highest weight module of highest weight $\lambda$ iff it's generated over $U(\lieg)$ as an associative algebra by a maximal vector $v^+$ of weight $\lambda$.
Then $M = U(\lieg) \cdot v^+$.

See structure of highest weight modules, and irreducibility.

:::{.corollary title="?"}
If $0 \neq M\in\mathcal O$, then $M$ has a finite filtration with quotients highest weight modules, i.e. $M_0 \subset M_1 \subset \cdots \subset M_n = M$ with $M_i/M_{i-1}$ highest weight modules.
:::
Note that the quotients are not necessarily simple, so this isn't a composition series, although we'll show such a series exists later.

:::{.theorem title="Module Span of Weight Vectors is Finite Dimensional"}
Let $V$ be the $\lien$ submodule of $M$ generated by a finite set of weight vectors which generate $M$ as a $U(\lieg)$ module,
i.e. take the finite set of weight vectors and act on them by $U(\lien)$.
Then $\dim_\CC V < \infty$ since $M$ is locally $\lien\dash$finite.
:::

Note that $\lien$ increases weights.

:::{.proof title="?"}
Induction on $n = \dim V$.
If $n=1$, $M$ itself is a highest weight module.
For $n > 1$, choose a weight vector $v_1 \in V$ of weight $\lambda$ which is maximal among all weights of $V$.
Set $M_1 \definedas U(\lieg) v_1$; this is a highest weight submodule of $M$ of highest weight $\lambda$.
($\lien$ has to kill v_1, otherwise it increases weight and $v_1$ wouldn't be maximal.)

Let $\bar M = M/M_1 \in \mathcal O$, this is generated by the image of $\bar V$ of $V$ and thus $\dim \bar V < \dim V$.
By the IH, $\bar M$ has the desired filtration, say $$0 \subset \bar M_2 \subset \bar M_{n-1} \subset \bar M_n = \bar M.$$ 
Let $\pi: M \to M/M_1$, then just take the preimages $\pi\inv(\bar M_i)$ to be the filtration on $M$.
:::


:::{.remark}
By isomorphism theorems, the quotients in the series for $M$ are isomorphic to the quotients for $\bar M$.
:::


## Verma and Simple Modules

Constructing *universal* highest weight modules using "algebraic induction".
Start with a nice subalgebra of $\lieg$ then "induce" via $\tensor$ to a module for $\lieg$.

Recall $\lieg = \lien^- \oplus \lieh \oplus \lien$, here $\lieh \oplus \lien$ is the Borel subalgebra $\lieb$, and $\lien$ corresponds to a fixed choice of positive roots in $\Phi^+$ with basis $\Delta$.
Then $U(\lieg) = U(\lien^-) \tensor_\CC U(\lieb)$.
Given any $\lambda \in \lieh\dual$, let $\CC_\lambda$ be the 1-dimensional $\lieh\dash$module (i.e. 1-dimensional $\CC\dash$vector space)on which $\lieh$ acts by $\lambda$.

Let $\theset{1}$ be the basis for $\CC$, so $h \cdot 1 = \lambda(h)1$ for all $h\in \lieh$.
Then there is a map $\lieb \to \lieb / \lien \cong \lieh$, so make $C_\lambda$ a $\lieb\dash$module via this map.
This "inflate" $C_\lambda$ into a 1-dimensional $\lieb\dash$module.

Note that $\lieh$ is solvable, and by Lie's Theorem, every finite dimensional irreducible $\lieb\dash$module is of the form $\CC_\lambda$ for some $\lambda \in \lieh\dual$.

:::{.definition title="Verma Modules"}

\[
M(\lambda) \da U(\lieg) \tensor_{U(\lieb)} \CC_\lambda \definedas \ind_\lieb^\lieg \CC_\lambda
\]
is the *Verma module of highest weight $\lambda$*.
:::


This process is called **algebraic/tensor induction**.
This is a $U(\lieg)$ module via left multiplication, i.e. acting on the first tensor factor.

:::{.remark}
Since $U(\lieg) \cong U(\lien^-) \tensor_\CC U(\lieh)$, we have $M(\lambda) \cong U(\lien^-) \tensor_\CC \CC_\lambda$, but at what level of structure?

- As a vector space (clear)
- As a $\lien^-\dash$module via left multiplication
- As a $\lieh^-\dash$module via the $\tensor$ action.

In particular, $M(\lambda)$ is a *free* $U(\lien^-)\dash$module of rank 1.
Note that this always happens when tensoring with a vector space.
:::


Consider $v^+ \definedas 1\tensor 1 \in M(\lambda)$.
Note that $U(\lien^-)$ is not homogeneous, so not graded, but does have a filtration.
Then $v^+$ is nonzero, and freely generates $M(\lambda)$ as a $U(\lien^-)\dash$module.
Moreover $\lien \cdot v^+ = 0$ since for $x\in \lieg_\beta$ for $\beta \in \Phi^+$, we have

\[
x(1\tensor 1) &= x\tensor 1  \\
&= 1\tensor x\cdot 1 \quad\text{since } x\in \lieb \\
&= 1 \tensor 0 \implies x\in \lien \\
&= 0
,\]

and for $h\in \lieh$,

\[
h(1\tensor 1) 
&= h1\tensor 1 \\
&= 1 \tensor h1\\
&=1 \tensor \lambda(h) 1 \\
&=\lambda(h) v^+
.\]

So $M(\lambda)$ is a highest weight module of highest weight $\lambda$, and thus $M(\lambda) \in \mathcal O$.

:::{.remark}
Any weight $\lambda \in \lieh\dual$ is the highest weight of some $M\in \mathcal O$.
Let $\Pi(M)$ denote the set of weights of a module, then $\Pi(M(\lambda)) = \lambda - \ZZ^+ \Phi^+$.

By PBW, we can obtain a basis for $M(\lambda)$ as $\theset{ y_1^{t_1} \cdots y_m^{t_m}v^+ \suchthat t_i \in \ZZ^+}$.
Taking a fixed ordering $\theset{\beta_1, \cdots, \beta_m} = \Phi^+$, then $0\neq y_i \in \lieg_{-\beta_i}$.
Then every weight of this form is a weight of some $M(\lambda)$, and every weight of $M(\lambda)$ is of this form: $\lambda - \sum t_i \beta_i$.
:::

:::{.remark}
The functor $\Ind_\lieh^\lieg(\wait) = U(\lieg) \tensor_{\lieb} \wait$ from the category of finite-dimensional $\lieg\dash$semisimple $\lieb\dash$modules to $\mathcal O$ is an exact functor, since it is naturally isomorphic to $U(\lien^-) \tensor_\CC \wait$ (which is clearly exact since we are tensoring a vector space over its ground field).
:::


:::{.proposition title="Alternate construction of $M(\lambda)$"}
Let $I$ by a left ideal of $U(\lieg)$ which annihilates $v^+$, so 
$$
I = \generators{\lien, h-\lambda(h)\cdot 1 \suchthat h\in\lieh}
.$$
Since $v^+$ generates $M(\lambda)$ as a $U(\lieg)\dash$module, then (by a standard ring theory result) $M(\lambda) = U(\lieg)/I$, since $I$ is the annihilator of $M(\lambda)$.
:::

:::{.theorem title="Universal Property of Verma Modules"}
Let $M$ be any highest weight module of highest weight $\lambda$ generated by $v$.
Then $I\cdot v = 0$, so $I$ is the annihilator of $v$ and thus $M$ is a quotient of $M(\lambda)$.
Thus $M(\lambda)$ is universal in the sense that every other highest weight module arises as a quotient of $M(\lambda)$.
:::

:::{.remark}
By theorem 1.2, $M(\lambda)$ has a unique maximal submodule $N(\lambda)$ (nonstandard notation) and a unique simple quotient $L(\lambda)$ (standard notation).
:::



:::{.theorem title="Characterization of Simple Modules and Schur's Lemma"}
Every simple module in $\mathcal O$ is isomorphic to $L(\lambda)$ for some $\lambda \in \lieh\dual$
and is determined uniquely up to isomorphism by its highest weight.
Moreover, there is an analog of Schur's lemma: $$\dim \hom_{\mathcal O}(L(\mu), L(\lambda)) = \delta_{\mu\lambda}$$, i.e. it's 1 iff $\lambda=\mu$ and 0 otherwise.
:::

:::{.remark}
Up to isomorphism, we've found all of the simple modules in $\mathcal O$, and most are finite-dimensional.
:::