# Wednesday March 4th

## Projective Generators and Finite Dimensional Algebras

:::{.remark}
Let $\OO_\chi = \OO_{\chi_\lambda}$ for $\lambda \in \lieh\dual$ antidominant.
Consider
$$
\bigoplus_{w\in W/W_{\lambda + \rho}} n_w P(w\cdot \lambda)
$$
where each $n_w > 0$; then $\theset{P(w\cdot \lambda)}_{w}$ 
represents all isomorphism classes of indecomposable projectives in $\OO_\chi$.
This implies that for all $M\in \OO_\chi$, there exists an $n>0$ such that $P^{\oplus n} \surjects M$.
The objective $P$ is called a **projective generator** of the category $\OO_{\chi}$.

Define $A \definedas \Endo_\OO(P)$, which is a finite-dimensional algebra by theorem 1.11.
:::

:::{.proposition title="?"}
The functor $\hom_\OO(P, \wait): \OO_\chi \mapsvia{\sim} \modsright{A}$, and this is an equivalence of categories.
:::

:::{.proof title="?"}
This is a standard argument in Morita theory.
The right $A\dash$module structure on $\hom(P, \wait)$ is given taking fixing some module $M$, taking $\phi \in \hom(P, M)$ and $\eta \in \Endo_\OO(P)$, then $\phi \circ \eta: P \to P \to M$ yields a new map $P\to M$.
:::

:::{.remark}
Note that we're skipping 3.14, 3.15, which are important for non-integral theory.
We'll focus on the integral case so we can get to Kazhdan-Lusztig theory in chapter 8.
:::
 
## Chapter 4: Highest Modules I

:::{.question}
\envlist

- What is $\hom_\OO(M(\mu), M(\lambda))$?
- When is $M(\lambda) = L(\lambda)$
- When is $[M(\lambda): L(\mu)] \neq 0$? 
  - Really, what are these multiplicities?
- To what extent do these properties only depend on $W$?
:::

:::{.remark}
If we know the multiplicities, we can reduce to Verma modules, which are known.
This allows us to get the characters to all simple modules.
:::

### 4.1: Simples Submodules of Verma Modules

:::{.remark}
We know that $M(\lambda) \mapstofrom U(\lien^-)$, which is Noetherian and has no zero divisors.

:::

:::{.lemma title="?"}
Let $R$ be a left Noetherian ring, $x\neq 0 \in R$ not a right zero divisor, and $I$ any nonzero left ideal in $R$.
Then $Rx \intersect I \neq 0$, if $R$ has no right zero divisors, then any two nonzero left ideals intersect nontrivially.
:::

:::{.proof title="?"}
Suppose $Rx \intersect I = \{0\}$, then there is a chain of left ideals
$$
0 \subset I \subset I + Ix \subset I+ Ix + Ix^2 \subset \cdots
$$
We claim all of these sums are direct: Suppose $a_0 + a_1 x + \cdots + a_n x^n = 0$ for $a_i \in I$.
We'll show that all $a_i = 0$ by induction on $n$.
Write $a_0 = -(a_1 + \cdots + a_n x^{n-1})x \in I\intersect Rx = 0$, then since $a$ is not a left zero divisor, the remaining term is zero and of smaller degree.
But then the chain above is *strictly* increasing, contradicting the fact that $R$ is left-Noetherian.
:::

Applying the proposition to $R = U(\lien^-)$, we obtain the following:

:::{.proposition title="?"}
$M(\lambda)$ has a unique simple submodule, so $\text{socle}(M(\lambda)) = L(\mu)$ for some $\lambda \geq \mu \in W\cdot \lambda$.
:::

:::{.proof title="?"}
If $M(\lambda)$ had distinct simple submodules $L, L'$, then $L \intersect L' = 0$.
On the other hand, $M(\lambda) \cong U(\lien^-)$ as $U(\lien^-)\dash$modules, and submodule correspond to left ideals.
But by the proposition, $L \intersect L' \neq 0$.
:::

:::{.question}
Which single $L(\mu)$ is the socle[^socle_reminder_1]
for $M(\lambda)$.

[^socle_reminder_1]: 
Recall that the socle is the direct sum of all simple submodules.

:::

### 4.2: Homomorphisms Between Verma Modules

:::{.remark}
Suppose $\phi: M(\mu) \to M(\lambda)$ with highest weight vectors $v_\mu, v_\lambda$ where $v_\mu \mapsto u \cdot v_{\lambda}$ with $u\in U(\lien^-)$.
Similarly, an arbitrary element in the domain is of the form $u' \cdot v_\mu$, which maps to $u'u\cdot v_\lambda$.
Thus $\phi$ is entirely determined by the image of the maximal vector $v_\mu$.
:::

:::{.theorem title="?"}
Let $\lambda, \mu \in \lieh\dual$, then

a. Any nonzero morphism $\phi: M(\mu) \to M(\lambda)$ is injective.

b. $\dim_\CC \hom_\OO(M(\mu), M(\lambda)) \leq 1$

c. $\soc M(\lambda)$ is a Verma module equal to $L(\mu) = M(\mu)$.

:::

:::{.proof title="of a"}
$\phi \neq 0 \iff u \neq 0$, but $u' u \neq 0$ for all $0\neq u'\in U(\lien^-)$ since $U(\lien^-)$ has no zero divisors.
Since $M(\lambda)$ is a free $U(\lien^-)\dash$module, $u' u \cdot v_\lambda \neq 0$ for every $u' \neq 0$, and thus $\ker \phi = 0$.
:::

:::{.proof title="of b"}
We want to show that any two nonzero morphisms are scalar multiples.

See diagram.

Note that $L$ is simple, as are $L_1, L_2$, but $M(\lambda)$ has a unique simple submodule, so $L' \da L_1 = L_2$ is unique.
Then $\phi_2\inv \phi_1: L\selfmap$, and by the analog of Schur's Lemma for $\OO$ (Theorem 1.3), $\Endo_\OO(L) \cong \CC$
So $\phi_1, \phi_2$ restrict to the same map on $L$.
But then $\restrictionof{\phi_1 - c\phi_2}{L} = 0_L$, whereas by (a) this must be injective or 0, so $\phi_1 = c\phi_2$.
:::

:::{.proof title="of c"}
Say $\soc M(\lambda) = L(\mu)$.
Then there is a nonzero 
\[
\phi: M(\mu) \surjects L(\mu) \injects M(\lambda)
,\]
which by (b) is injective, which forces $N(\mu) = 0$ and thus $M(\mu) = L(\mu)$.
:::

:::{.remark}
Whenever $\hom_\OO(M(\mu), M(\lambda)) \neq 0$, we can unambiguously write $M(\mu) \subset M(\lambda)$ since the highest weight vector must map to a scalar multiple.
:::

:::{.question}
When does $M(\mu) \subset M(\lambda)$ occur?
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:::{.remark}
We already know that if $(\lambda+\rho, \alpha\dual) \in \ZZ^{> 0}$ and $\alpha \in \Delta$ then $M(s_\alpha, \lambda) \subset M(\lambda)$.

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### Special Case: Dominant Integral Weights

:::{.remark}
Assume $\lambda + \rho \in \Lambda^+$, which occurs iff $(\lambda + \rho, \alpha\dual) \in \ZZ^+$ for all $\alpha\in \Delta$.
Note that $\lambda = -\rho$ is a simple example.
:::

:::{.proposition title="?"}
Suppose $\lambda + \rho \in \Lambda^+$. 
Then
a. $M(w\cdot \lambda) \subset M(\lambda)$ for all $w\in W$.
b. Hence $[M(\lambda): L(w\cdot \lambda)] > 0$.

More precisely, if $w = \prod s_{\alpha_i}$ is reduced and $\alpha_i \in \Delta$, defining $\lambda_k \definedas \prod_{i=1}^k s_{\alpha_i} \lambda$, then there is a composition series
$$
M(w\cdot \lambda) = M(\lambda_n) \subset M(\lambda_{n-1}) \subset \cdots \subset M(\lambda_0) = M(\lambda)
$$
and in fact $(\lambda_k + \rho, \alpha_{k+1}\dual) \in \ZZ^{>0}$.
:::