# Monday January 27th

## Section 1.4

Fix $\Delta = \theset{\alpha_1, \cdots, \alpha_\ell}$, $x_i \in g_{\alpha_i}$ and $y_i \in g_{-\alpha_i}$ with $h_i = [x_i y_i]$.

Lemma
:   For $k\geq 0$ and $1 \leq i, j \leq \ell$,

    a. $[x_j y_i^{k+1}] = 0$ if $j\neq i$
    b. $[h_j y_i^{k+1}] = -(k+1) \alpha_i(h_j) y_i^{k+1}$
    c. $[x_i y_i^{k+1}] = (k+1) y_i^{k} (k\cdot 1 - h_i)$.

Proof (Sketch for (c))
:   By induction, where $k=0$ is clear.

    \begin{align*}
    [x+i y_i^{k+1}]
    &= [x_i y_i] y_i^k + y_i [x_i y_i^k] \\
    &=h_i y_i^k + y_i(-k y_i^{k-1} ((k-1)1 - h_i)) \quad\text{by I.H.} \\
    &= (k+1)y_i^k h_i - (k^2 -k + 2k)y_i^k \\
    &= -(k+1) y_i^k ( k\cdot 1 - h_i )
    .\end{align*}

Proposition (Existence of Morphisms of Verma Modules)
:   Suppose $\lambda \in \lieh\dual, \alpha \in \Delta$, and $n\definedas (\lambda, \alpha\dual) \in \ZZ^+$.
    Then in $M(\lambda)$, $y_\alpha^{n+1} v^+$ is a maximal weight vector of weight $\mu \definedas \lambda - (n+1)\alpha < \lambda$.

    > Note this is free as an $U(\lien^-)\dash$module, so $v^+ \neq 0$.
    > Note that $n = \lambda(h_\alpha)$.

    By the universal property, there is a nonzero homomorphism $M(\mu) \to M(\lambda)$ with image contained in $N(\lambda)$, the unique maximal proper submodules of $M(\lambda)$.

Proof
: Say $\alpha = \alpha_i$.
	Fix $j\neq i$.

	\begin{align*}
	x_i y_\alpha^{n+1} \tensor 1 
	&= [x_j y_i^{n+1}] \tensor 1 + y_i^{n+1} \tensor x_j \cdot 1 \\
	&= [x_j y_i^{n+1}] \tensor 1 + y_i^{n+1} \tensor 0 \quad\text{by a} \\
	&= 0
	.\end{align*}

	\begin{align*}
	x_i y_i^{n+1} \tensor 1 
	&= [x_i y_i^{n+1} \tensor 1] \\
	&= -(n+1) y_i^n (n\cdot 1 - h_i) \tensor 1 \\
	&= -(n+1) (n - \lambda(h_i)) 1 \tensor 1 \\
	&\definedas -(n+1) (\lambda(h_i) - \lambda(h_i)) 1 \tensor 1 \\
	&= 0
	.\end{align*}

	Since $g_{\alpha_j}$ generate $\lien$ as a Lie algebra, since $[\lieg_\alpha, \lieg_\beta] = \lieg_{\alpha + \beta}$.
	This shows that $\lien \cdot y_i^{n+1} v^+ = 0$, and the weight of $y_i^{n+1} v^+$ is $\lambda - (n+1)\alpha_i$.
	So $y_i^{n+1}$ is a maximal vector of weight $\mu$.
	The universal property implies there is a nonzero map $M(\mu) \to M(\lambda)$ sending highest weight vectors to highest weight vectors and preserving weights.
	The image is proper since all weights of $M_\mu$ are less than or equal to $\mu < \lambda$.

Consider $\liesl(2)$, then $M(1) \supset M(-3)$.
Note that reflecting through 0 doesn't send 1 to -3, but shifting the origin to $-1$ and reflecting about that with $s_\alpha \cdot$ fixes this problem.
Note that $L(1)$ is the quotient.

For $\lambda \in \lieh\dual$ and $\alpha \in \Delta$, we can compute $s_\alpha \cdot \lambda \definedas s_\alpha(\lambda + \rho) - \rho$ where $\rho = \sum_{j=1}^\ell e_i$.
Then $(w_j, \alpha_i\dual) = \delta_{ij}$ and $(\rho, \alpha_i\dual) = 1$.

\begin{align*}
s\alpha \cdot \lambda 
&= s_\alpha(\lambda + \rho) - \rho \\
&= (\lambda + \rho) - (\lambda + \rho, \alpha\dual)\alpha -\rho \\
&= \lambda + \rho - ((\lambda< \alpha\dual) +1)\alpha - \rho \\
&= \lambda - (n+1)\alpha \\
&= \mu
.\end{align*}

So this gives a well-defined, nonzero map $M(s_\alpha \cdot \lambda) \to M(\lambda)$ for $s_\alpha \cdot \lambda < \lambda$.

![Image](figures/2020-01-27-09:35.png)\

Corollary
: Let $\lambda, \alpha, n$ be as in the above proposition.
	Let $\bar v^+$ now be a maximal vector of weight $\lambda$ in $L(\lambda)$.
	Then $y_\alpha^{n+1} \bar v^+ = 0$.

Proof
: If not, then this would be a maximal vector, since it's the image of the vector $y_i^{n+1}v^+ \in M(\lambda)$ under the map $M(\lambda) \to L(\lambda)$ of weight $\mu < \lambda$.
	Then it would generate a proper submodules of $L(\lambda)$, but this is a contradiction since $L(\lambda)$ is irreducible.


## Section 1.5

Example: $\liesl(2)$.
What do Verma modules $M(\lambda)$ and their simple quotients $L(\lambda)$ look like?

Fix a Chevalley basis $\theset{y,h,x}$ and let $\lambda \in \lieh\dual \cong \CC$.

Fact 1
: For $v^+ = 1\tensor 1_\lambda$, we have $$M(\lambda) = U(\lien^-) v^+ = \CC \generators{y^i v^+ \suchthat i\in \ZZ^+}$$ is a basis for $M(\lambda)$ with weights $\lambda - 2i$ where $\alpha$ corresponds to 2.
	So the weights of $M(\lambda)$ are $\lambda, \lambda-2, \lambda-4, \cdots$ each with multiplicity 1.

	Letting $v_i = \frac 1 {i!} y^i v^+$ for $i\in \ZZ^+$; this is a basis for $M(\lambda)$.
	Using the lemma, we have

	\begin{align*}
	h\cdot v_i &= (\lambda - 2i) v_i \\
	y \cdot v_i &= (i+1) v_{i+1} \\
	x\cdot v_i &= (\lambda - i + 1)v_{i-1} 
	.\end{align*}

	Note that these are the same for *finite-dimensional* $\liesl(2)\dash$modules, see section 0.9.

Fact (2)
: We know from the proposition that if $\lambda \in \ZZ^+$, i.e. $(\lambda, \alpha\dual) \in \ZZ^+$, then $M(\lambda)$ has a maximal vector of weight $$\lambda - (n+1)\alpha = \lambda - (\lambda+1)2 = -\lambda-2 = s_\alpha \cdot \lambda.$$

Exercise
: Check that this maximal vector generates the maximal proper submodule $$N(\lambda) = M(-\lambda - 2).$$

	So the quotient $L(\lambda) = M(\lambda) / N(\lambda) = M(\lambda) / M(-\lambda - 2)$ has weights $\lambda, \lambda-2, \cdots, -\lambda+2, -\lambda$.
	So when $\lambda \in \ZZ^+$, $L(\lambda)$ is the familiar simple $\liesl(2)\dash$module of highest weight $\lambda$.

Fact (3)
: When $\lambda \not\in\ZZ^+$,

	- $N(\lambda) = \theset{0}$,
	- $M(\lambda) = L(\lambda)$,
	- $M(\lambda)$ is irreducible
	- $L(\lambda)$ is infinite dimensional.


Proof
: Argue by contradiction.
	If not, $M(\lambda) \supset M \neq 0$ is a proper submodule.
	So $M\in \OO$, and thus $M$ has a maximal weight vector $w^+$, and by the restriction of weights for modules in $\OO$, we know $w^+$ has height $\lambda - 2m$ for some $m\in \ZZ^+$.
	Then $w^+ = c v_i$ where $0\neq c \in \CC$, and taking $v_{-1} \definedas 0$ and $x\cdot v_i = (\lambda - i + 1)v_{i-1}$ for $i\geq 1$, so $\lambda = i-1 \implies \lambda \in \ZZ^+$.