# Friday January 17th

Let $M$ 

1. Be finitely generated, 
2. Semisimple $M = \oplus_{\lambda \in \lieh\dual} M_\lambda$, 
3. Locally finite 
4. $\dim M_\mu < \infty$ for all $\mu \in \lieh\dual$,
5. Satisfy the forest condition for weights.

:::{.theorem title="Properties of $\OO$"}
\envlist

a. $\mathcal O$ is Noetherian [^def_noetherian]


b. $\mathcal O$ is closed under quotients, submodules, finite direct sums

c. $\mathcal O$ is abelian (similar to a category of modules)

d. If $M\in \mathcal O$, $\dim L < \infty$, then $L \tensor M \in \mathcal O$ and the endofunctor $M \mapsto L\tensor M$ is exact

e. If $M\in \mathcal O$, then $M$ is locally $Z(\lieg)\dash$finite [^recall_center]

f. $M\in \mathcal O$ is a finitely generated $U(\lien^-)\dash$module.

[^def_noetherian]: Ascending chain condition on submodules, i.e. no infinite filtrations by submodules.

[^recall_center]: Recall: this is the center of $U(\lieg)$), i.e. $\dim\spanof ~Z(\lieg)v < \infty$ for all $v\in M$.

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:::{.proof title="of a and b"}
See BA II, page 103.
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:::{.proof title="of c"}
Implied by (b), BA II Page 330.
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:::{.proof title="of d"}
Can check that $L\tensor M$ satisfies 2 and 3 above.
Need to check first condition.
Take a basis $\theset{v_i}$ for $L$ and $\theset{w_j}$ a finite set of generators for $M$.
The claim is that $B = \theset{v_i \tensor w_j}$ generates $L\tensor M$.
Let $N$ be the submodule generated by $B$.

For any $v\in V$, $v\tensor w_j \in N$.
For arbitrary $x\in \lieg$, compute $$x\cdot(v\tensor w_j) = (x\cdot v) \tensor w_j + x\tensor (v\cdot w_j).$$
Since the LHS is in $N$ and the first term on the RHS is in $N$, the entire RHS is in $N$.
By iterating, we find that $v\tensor (u\cdot w_j) \in N$ for all PBW monomials $u$.
So $L\tensor M \in \mathcal O$.
:::

:::{.proof title="of e"}
Since $v\in M$is a sum of weight vectors, wlog we can assume $v \in M_\lambda$ is a weight vector (where $\lambda \in \lieh\dual$).
For any central element $z\in Z(\lieg)$, we can compute $$h\cdot(z\cdot v) = z \cdot (h\cdot v) = z \cdot \lambda(h) v = \lambda(h)z \cdot v.$$ 
Thus $z\cdot v\in M_\lambda$.
By (4), we know that $\dim M_\lambda < \infty$, so $\dim \spanof ~Z(\lieg) v < \infty$ as well.

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:::{.proof title="of f"}
By 5, $M$ is generated by a finite dimensional $U(\mathfrak b)$ submodule $N$.
Since we have a triangular decomposition $U(\lieg) = U(\lien^-) U(\mathfrak b)$, there is a basis of weight vectors for $N$ that generates $M$ as a $U(\lien^-)$ module.
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## Highest Weight Modules

:::{.definition title="Maximal Vector"}
A **maximal vector** $v^+ \in M \in \mathcal O$ is a nonzero vector such that $\lien \cdot v^+ = 0$.
:::

:::{.remark}
By properties 2 and 3, every nonzero $M\in \mathcal O$ has a maximal vector.
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:::{.definition title="Highest Weight Modules"}
A **highest weight module** $M$ of highest weight $\lambda$ is a module generated by a maximal vector of weight $\lambda$, i.e. 
$$M = U(\lieg) v^+ = U(\lien^-) U(\lieh) U(\lien) v^+ = U(\lien^-) v^+$$
:::

:::{.theorem title="Properties of Highest Weight Modules"}
Let $M = U(\lien^-)v^+$ be a highest weight module, where $v^+ \in M_\lambda$.
Fix $\Phi^+ = \theset{\beta_1, \cdots, \beta_n}$ with root vectors $y_i \in \lieg_{\beta_i}$.

a. $M$ is the $\CC\dash$span of PBW monomials $\generators{ y_1^{t_1} \cdots y_m^{t_m}}$ of weight $\lambda - \sum t_i \beta_i$.
  Thus $M$ is a module.

b. All weights $\mu$ of $M$ satisfy $\mu \leq \lambda$

c. $\dim M_\mu < \infty$ for all $\mu \in T(M)$, and $\dim M_\lambda = 1$.
  In particular, property (3) holds and $M \in \mathcal O$.

d. Every nonzero quotient of $M$ is a highest-weight module of highest weight $\lambda$.

e. Every submodule of $M$ is a weight module, and any submodule generated by a maximal vector with $\mu < \lambda$ is proper.
  If $M$ is semisimple, then the set of maximal weight vectors equals $\CC\units v^+$.

f. $M$ has a unique maximal submodule $N$ and a unique simple quotient $L$, thus $M$ is indecomposable.

g. All simple highest weight modules of highest weight $\lambda$ are isomorphic.

:::

   
:::{.remark}
For such $M$, $\dim \endo(M) = 1$. (Category $\mathcal O$ version of Schur's Lemma, generalizes to infinite dimensional case)
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:::{.proof title="a through e"}
Either obvious or follows from previous results.
First few imply $M$ is in $\mathcal O$, and we know the latter hold for such modules.
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:::{.proof title="of f"}
$N$ is a sum of submodules, so $N = \sum M_i$, proper submodules of $M$.
So take $L = M/N$.
To see indecomposability, there exists a better proof in section 1.3.
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:::{.proof title="of g"}
Let $M_1 = U(\lien^-)v_1^+$ and $M_2$ be define similarly, where the $v_i \in (M_i)_\lambda$ have the same weight.
Then $M_0 \definedas M_1 \oplus M_2$ implies that $v^+ \definedas (v_1^+, v_2^+)$ is a maximal vector for $M_0$.
So $N \definedas U(\lien^-) v^+$ is a highest weight module of highest weight $\lambda$.

We have the following diagram:


\begin{tikzcd}
               &  & N \arrow[dd, hook] \arrow[llddd, tail] \arrow[rrddd, tail] &  & \\
               &  &                                                            &  & \\
               &  & M_1 \oplus M_2 \arrow[lld] \arrow[rrd]                     &  & \\
M_1            &  &                                                            &  & M_2
\end{tikzcd}


and since e.g. $N \to M_1$ is not the zero map, it is a surjection.

By (f), $N$ is a unique simple quotient, so this forces $M_1 \cong M_2$.
Since $M$ is simple, any nonzero $\lieg\dash$endomorphism $\phi$ must be an isomorphism, and so we take $v^+ \mapsto cv^+$ for some $c\neq 0$.
Note that since $\phi$ is also a $\lieh\dash$morphism, we have $\dim M_\lambda = 1$.

Since $v^+$ generates $M$ and 
\[
\phi(u\cdot v^+) = u \phi(v^+) = cu\cdot v^+
,\]
$\phi$ is multiplication by a constant.
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