# Monday March 30th

Reminder of what we did already: we started on chapter 4, going into more detail on the structure of Verma modules and morphisms between them.
We showed that the socle is an irreducible Verma modules, any nonzero morphism is injective, and the dimension of the hom space is at most 1.
We ended showing a proposition about how to commute elements.

Proposition (Key Result: Containments of Vermas When Applying Weyl Elements)
:   Let $\lambda, \mu \in \lieh\dual$ and $\alpha\in\Delta$ be simple.
    Assume that $n\definedas (\lambda + \rho, \alpha\dual) \in \ZZ$ and $M(s_\alpha \cdot \mu) \subset M(\mu) \subset M(\lambda)$.
    Then either

    a. $n\leq 0$ and $M(\lambda) \subset M(s_\alpha \cdot \lambda)$, or
    b. $n>0$ and $M(s_\alpha \cdot \mu) \subset M(s_\alpha \cdot \lambda) \subset M(\lambda)$.

    In either case, $M(s_\alpha \cdot \mu) \subset M(s_\alpha \cdot \lambda)$.

Proof (of (a))
: Use proposition 1.4 (exchanging $\lambda$ and $s_\alpha \cdot \lambda$).

## Proof (of (b))

Assume $n>0$.
Then $M(s_\alpha \cdot \lambda) \subset M(\lambda)$ by proposition 1.4.
Set $s = (\mu + \rho, \alpha\dual) \in \ZZ^+$.
Denote maximal vectors as follows:

![Image](figures/2020-03-30-09:29.png)\

Apply the lemma about nilpotent lie algebras to $\lien^-, y_\alpha, u$, and $n$, then there exists a $t>0$ such that $y_\alpha^t u \in U(\lien^-) y_\alpha^n$.
Then
\begin{align*}\label{star}
y_\alpha^t \cdot v_\lambda^+ = y_\alpha^t u \cdot v_\lambda^+ \in U(\lien^-) y_\alpha^n \cdot v_\lambda^+ \subseteq M(s_\alpha \cdot \lambda)
.\end{align*}

Now there are two cases.

**Case 1**:

If $t\leq s$, we can apply $y_\alpha^{s-t}$ to equation star to obtain $y_\alpha^s \cdot v_\lambda^+ \in M(s_\alpha \cdot \lambda)$.
Thus we have the containment we wanted to prove.

**Case 2**:

Suppose $t > s$.
We can't divide in the enveloping algebra, but recall the identity in lemma 1.4(c):
\begin{align*}
[x_\alpha y_\alpha^t] = t y_\alpha^{t-1} \qty{ h_\alpha - t + 1}
.\end{align*}

Thus
\begin{align*}
[x_\alpha y_\alpha^t] \cdot v_\mu^+ = t(s-t) y_\alpha^{t-1} \cdot v_\mu^+
.\end{align*}

Calculating the bracket another way, the LHS is equal to $x_\alpha y_\alpha^t \cdot v_\mu^+ - y_\alpha^t x_\alpha \cdot v_\mu^+$ and the second term is zero, so this is in $M(s_\alpha \cdot \lambda)$ by equation star.
We can then iterate if  $t-1 > s$, reducing the power of $y_\alpha$ until we get down to $y_\alpha^s \cdot v_\mu^+ \in M(s_\alpha \cdot \lambda)$, in which case we are done by case 1.

$\qed$

## 4.6: Existence of Embeddings

Theorem (Verma's Thesis: Existence of Embeddings)
: Let $\lambda \in \lieh\dual$ and $\alpha\in\Phi^+$ and assume $\mu \definedas s_\alpha \cdot \lambda \leq \lambda$.
  Then $M(\mu) \subset M(\lambda)$.

### Proof

Assume $\lambda \in \Lambda$ is integral and $\mu$ is linked to $\lambda$, all weights involved are integral.
Without loss of generality, $\mu < \lambda$, since we can apply a Weyl group element to place it in the dominant Weyl chamber.

1. Since $\mu$ is integral, choose $w\in W$ such that $\mu' \definedas w\inv \cdot \mu \in \Lambda^+ - \rho$.
    Following the notation in proposition 4.3, write $w = s_n \cdots s_1, \mu_k = s_k \cdots s_1 \cdot \mu'$.
    Then $\mu' = \mu_0 \geq \mu_1 \geq \cdots \geq \mu_n = \mu$, which yields a chain of inclusions of Verma modules $M(\mu_0) \supset M(\mu_1) \supset \cdots$.

2. Set $\lambda' = w\inv \lambda$ and $\lambda_k = s_k \cdots s_1 \cdot \lambda'$ so $\lambda_0 = \lambda'$ and $\lambda_n = \lambda$.
    Note that since $\mu \neq \lambda$, we have $\mu_k \neq \lambda_k$.

3. How are $\mu_k$ and $\lambda_k$ related? Set $w_k = s_n \cdots s_{k+1}$.
    It can be checked that $\mu_k = w_k\inv s_\alpha w_k \cdot \lambda_k = s_{\beta_k} \lambda_k$ where $\beta_k = \abs{w_k\inv} \in \Phi^+$ is the choice of whichever is positive by Humphreys 1, Lemma 9.2.
    In particular, $\mu_k - \lambda_k \in \ZZ \beta_k$.

4. We have $\mu' = \mu_0 \geq \cdots \geq \mu_k \geq \mu_{k+1} \geq \cdots \geq \mu_n = \mu$.
    Since $\lambda'<\mu'$ (because $\mu'$ is the unique dominant weight in $W\cdot \lambda$ but $\mu < \lambda$, so the inequalities must switch at some $k$.
    So say $\lambda_k < \mu_k$ but $\lambda_{k+1} > \mu_{k+1}$, where $k$ is chosen to be the smallest index for which this happens.
    Note that all of the weights are linked to $\lambda$.

5. We want to show that $M(\mu_{k+1}) \subset M(\lambda_{k+1}), \cdots, M(\mu) = M(\mu_n) \subset M(\lambda_n) = M(\lambda)$.

6. First, $\mu_{k+1} - \lambda_{k+1} = s_{k+1} \cdot \mu_k - s_{k+1} \cdot \lambda_k$, where the LHS is some negative times $\beta_{k+1}$, and the RHS is equal to $s_{k+1} \qty{ \mu_k - \lambda_k }$, which is a positive times $\beta_k$ by exercise 1.8.
    Since $s_{k+1}$ permutes the positive roots other than $\alpha_{k+1}$, this forces $\beta_k = \beta_{k+1} = \alpha_{k+1}$.
    So we have $\mu_{k+1} = s_{\alpha_{k+1}} \lambda_{k+1}$ which by proposition 1.4 implies that $M(\mu_{k+1}) \subset M(\lambda_{k+1})$.

7. Combining 1 and 6, we have $M(\mu_{k+2}) = M(s_{k+2} \cdot \mu_{k+1}) \subset M(\mu_{k+1}) \subset M(\lambda_{k+1})$.
    This is the setup of proposition 4.5 and wither alternative leads to $M(\mu_{k+2}) \subset M(\lambda_{k+2}) = M(s_{\alpha_{k+2}} \cdot \lambda_{k+1})$,

8. Since this increases the index, we can iterate step 7 to complete step 5 and get the desired containment.