# Wednesday April 1st

Exercise
: Work through the steps for $\liesl(3)$, due next Thursday.

Preview of next sections:

- 4.8: Simplicity Criterion, General Case
  - Now that we know $M(s_\alpha \cdot \lambda) \subset M(\lambda)$ whenever $s_\alpha \lambda \leq \lambda$ for $\alpha \in \Phi^+$ (and not just $\alpha \in \Delta$) we can easily complete the proof of theorem 4.4 by copying the argument from the integral case.
- 4.9: Blocks of $\OO$, revisited
  - Skip, mainly relevant for nonintegral weights (c.f. Proposition 1.13 for the description of integral blocks)
- 4.10: Example: Antidominant Projectives
  - Skip, at least for now

## 4.11: Application to $\liesl(3, \CC)$

The simplest nontrivial case, what can we say about the Verma modules?

![Image](figures/2020-04-01-09:29.png)\

![Image](figures/2020-04-01-09:29.png)\

We have $\Delta = \theset{\alpha, \beta}$ and $\Phi^+ = \theset{\alpha, \beta, \gamma\definedas \alpha+\beta}$.
The Weyl group is
$$
W = \theset{1, s_\alpha,s_\beta, s_\alpha s_\beta, s_\beta s_\alpha, w_0\definedas s_\alpha s_\beta s_\alpha = s_\beta s_\alpha s_\beta  }
.$$

We first consider an integral regular linkage class $W\cdot \lambda$, and we way choose an antidominant $\lambda$ and assume
$$
(\lambda + \rho, \alpha\dual) \in \ZZ^{< 0} \quad \forall \alpha \in \Phi^+ \quad \text{e.g. } \lambda = - 2\rho
$$

Then $W_\lambda = \theset{1}$, given by the stabilizer of the isotropy subgroup, and $\abs{W\cdot \lambda} = 6$.
So there are 6 Verma modules to understand, and we have the following diamond:


![Image](figures/2020-04-01-09:33.png)\

The middle two edges are $s_\gamma$, and each edge corresponds to an inclusion of Verma modules (with the inclusions going upward).
By Verma's theorem, the Bruhat order corresponds to these inclusions.

1. $M(\lambda) = L(\lambda)$ since $\lambda$ is antidominant by Theorem 4.4

2. By the same theorem, no other $M(w\cdot \lambda)$ is simple.
    Then by Proposition 4.18, Theorem 4.2c, we have $\soc M(w\cdot \lambda) = L(\lambda)$ for all $w\in W$

3. Consider $M(s_\alpha \cdot \lambda)$ and set $\mu \definedas s_\alpha \cdot \lambda$ and the only possible composition factors are $L(\mu)$ and $L(\lambda)$ and $[M(\mu): L(\mu) ] = 1$.
    If we use Theorem 4.10, this multiplicity is 1 in the socle and we're done.
    If we don't, could it be larger than 1? Since $\ext_\OO (L(\lambda), L(\lambda)) = 0$, we can not have the following situation:

    ![Image](figures/2020-04-01-09:36.png)\

    The first extension doesn't exist, since the higher $L(\lambda)$ would drop down to give the bottom diagram, which contradicts $\soc M(\mu) =  L(\lambda)$.

    So the only possibilities are multiplicity 1, and $M(s_\alpha \cdot \lambda) = L(s_\alpha \cdot \lambda)$ which lives over $L(\lambda)$, so $\ch L(s_\alpha \cdot \lambda) = \ch M(s_\alpha \cdot \lambda) - \ch M(\lambda)$.

    Similary for $M(s_\beta \cdot \lambda)$.

4. For the higher weights in the orbit, we need more theory.
    We know there are inclusions $x\leq w \implies M(x\cdot \lambda) \subset M(w\cdot \lambda)$ according to the Bruhat order - so every edge in the weight poset is a reflection, so use Verma's theorem.
    \begin{align*}
    [M(w\cdot \lambda): L(x \cdot \lambda)] = \begin{cases}
    \geq 1 & ? \\
    0? & ?
    \end{cases}
    .\end{align*}

We'll skip 4.12,4.13, 4.14.

## Chapter 5: Highest Weight Modules II

Development by BGG after 1970s, based on partly incorrect proof in Verma's thesis.

### 5.1: The BGG Theorem

Which simple modules occur as composition factors of $M(\lambda)$?

Definition (Strongly Linked Weights)
: For $\mu, \lambda \in \lieh\dual$, write $\mu \uparrow \lambda$ if $\mu = \lambda$ or there exists an $\alpha \in \Phi^+$ such that $\mu = s_\alpha \cdot \lambda < \lambda$, i.e. $(\lambda + \rho, \alpha\dual) \in \ZZ^{> 0}$.
  Extend this relation transitively: if there exists $\alpha_1, \cdots, \alpha_r \in \Phi^+$ such that $\mu = (s_{\alpha_1} \cdots s_{\alpha_r}) \cdot \lambda \uparrow (s_{\alpha_2} \cdots s_{\alpha_r} \uparrow \cdots \uparrow s_{\alpha_r} \lambda \uparrow \lambda$, we again write $\mu \uparrow\lambda$ and say $\mu$ is *strongly linked* to $\lambda$, yielding a partial order on $\lieh\dual$.

Theorem (Strong Linkage implies Verma Embedding)
:   Let $\mu, \lambda \in \lieh\dual$.
    
    - (Verma) If $\mu\uparrow \lambda$ then $M(\mu) \injects M(\lambda)$. In particular, $[M(\lambda): L(\mu)] > 0$.
    - ??? ???

Corollary
: $[M(\lambda): L(\mu)] \neq 0 \iff M(\mu) \injects M(\lambda)$.

The situation is not as straightforward as it might appear (and as Verma believed).
Namely, if $0 = M_0 \subset \cdots \subset M_n = M(\lambda)$ is a composition series and if $M_i / M_{i-1} \cong L(\mu) \ni \bar v_{\mu}^+$, there need not be any preimage of $v_\mu^+$ which is a maximal vector in $M(\lambda)$, leading to a map $M(\mu) \injects M(\lambda)$.

However, when this happens, there will always be some other $M_j/M_{j-1} \cong L(\mu)$ where a preimage of a maximal vector **is** maximal in $M(\lambda)$, leading to the required embedding.

### 5.2 Bruhat Ordering

In the case of "$\rho\dash$regular" integral weights, the BGG theorem has a nice reformulation in terms of $W$ and the Bruhat ordering.
Fix $\lambda \in \Lambda$ antidominant and $\rho\dash$regular, so $(\lambda + \rho, \alpha\dual) \in \ZZ^{< 0}$ for all $\alpha\in \Phi^+$.

As in the discussion of $\liesl(3)$, this means that $\abs{W\cdot \lambda} = \abs{W}$ and $[M(w\cdot \lambda) : L(\mu)] \neq 0$ implying that $\mu = x\cdot \lambda$ for some $x\in W$.
What can we say about the relative positions of $w$ and $x$?

Suppose that $w\in W, \alpha\in\Phi^+$ and $s_\alpha \cdot (w\cdot \lambda) < w\cdot \lambda$ so that $M(s_\alpha w \cdot \lambda) \injects M(w\cdot \lambda)$.
Our assumption is equivalent to
\begin{align*}
\ZZ^{>0} \ni (w\cdot \lambda + \rho, \alpha\dual) 
&= (w(\lambda+\rho), \alpha\dual)\\
&= (\lambda + \rho, w\inv \alpha\dual) \\
&= (\lambda + \rho, (w\inv\alpha)\dual) \\
&\iff w\inv \alpha \in \Phi^- \iff (w\inv s_\alpha) \alpha \in \Phi^+ \\
&\iff \ell(w) > \ell(w') \quad \text{where } w' = s_\alpha w \\
&\iff w' \mapsvia{s_\alpha} w
.\end{align*}