# Monday February 17th

## Character Formulas

*Last time:*
The second character formula (equation (2)),

\begin{align*}
\ch L(\lambda) =  \ch M(\lambda) + \sum_{\mu < \lambda, ~~ \mu \in W\cdot \lambda} b_{\lambda, \mu} ~\ch M(\mu)
.\end{align*}

Note that $b_{\lambda, \mu} \in \ZZ$, and this formula comes from inverting the previous one.

> Holy grail: characters of simple modules!

We can write $M(\lambda) \cong U(\lien^-) \tensor_\CC \CC_\lambda$ as a $\lieh\dash$module.
Define $p: \lieh\dual \to \ZZ$ where $p(\gamma)$ is the number of tuples $(t_\beta)_{\beta\in\Phi^+}$ where $t_\beta \in \ZZ^+$ and $\gamma = - \sum_{\beta \in \Phi^+} t_\beta \beta$.
We have $\supp(p) = - \ZZ^+ \Phi^+$, which gives us something like a negative quadrant of the lattice.

The function $p$ is essentially the *Kostant partition function*.
The advantage here is that $p \in \mcx$ (defined last time, support is less than some finite weights?).

*Observation:*
$p = \ch_{M(0)}$ since $U(\lien^-) \tensor \CC_{\lambda = 0}$ has PBW basis
$$
\theset{ \prod_{\beta\in\Phi^+} y_\beta^{t_\beta} \tensor 1_{\lambda = 0} \suchthat t_\beta \in \ZZ^+ }.
$$

*Example:*
Let $\lieg = \liesl(3)$, then $\Phi^+ = \theset{\alpha_1, \alpha_2, \alpha_1 + \alpha_2}$.
Then $\gamma = -\qty{\alpha_1 + 2\alpha_2}$ corresponds to $(1,2,0), (0,1,1)$ so $p(\gamma) = 2$.
If $\gamma = -\qty{2\alpha_1 + 2\alpha_2}$, this corresponds to $(2,2,0), (1,1,1), (0,0,2)$ so $p(\gamma) = 3$.

> Note: just the number of ways of obtaining $-\gamma$ as a linear combinations of roots.

In general, $\dim M(0)_\gamma = p(\gamma)$.

Proposition (Characters as Convolution Products)
:   For any $\lambda \in \lieh\dual$, we have $\ch_{M_\lambda} = p\ast e_\gamma$, taking the convolution product.

In particular, $\ch_{M(0)} = p$.

Proof (of Proposition)
:   We have the following computation:
    \begin{align*}
    (p\ast e_\lambda)(\lambda+\gamma)
    &= p(\gamma) e_\lambda(\lambda) \\
    &= p(\gamma) 1 \\
    &= p(\gamma) \\
    &= \dim M(\lambda)_{\lambda + \gamma} \quad\text{ as a weight space }
    .\end{align*}

Note that we can also write equation (2) as

\begin{align*}
\ch L(\lambda) = \sum_{w\cdot \lambda \leq \lambda} b_{\lambda, w} \ch M(w\cdot \lambda)
.\end{align*}

Here $b_{\lambda, w} \in \ZZ$ and in fact $b_{\lambda, 1} = 1$.

*Example:*
Let $\lieg = \liesl(2)$.
We know

\begin{align*}
\ch M(\lambda) &= \ch L(\lambda) + \ch L(s_\alpha \cdot \lambda) \\
\ch M(s_\alpha \cdot \lambda) &= \ch L(s_\alpha \cdot \lambda)
.\end{align*}

We can think of this pictorially as the 'head' on top of the socle:

\begin{align*}
M(\lambda) = \frac{L(\lambda)}{L(s_\alpha \cdot \lambda)}
.\end{align*}

The formula above corresponds to the matrix
$$
\left[\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right]
$$

We can invert the formula to get equation (2), which corresponds to inverting this matrix:

\begin{align*}
\ch L(\lambda) &= \ch M(\lambda) - \ch M(s_\alpha \cdot \lambda) \\
\ch L(s_\alpha \cdot \lambda) &= \ch M(s_\alpha \cdot \lambda)
.\end{align*}

Note that the coefficients $b_{w, \lambda} \in \theset{0, \pm 1}$ in this equation are independent of $\lambda \in \Lambda^+$.

If $\lambda \not\in\Lambda^+$, then $\ch L(\lambda) = \ch M(\lambda)$ and $b_{\lambda, 1} = 1, b_{\lambda, s_\alpha} = 0$ are again independent of $\lambda \in \lieh\dual \setminus \Lambda^+$.

**Question:**
To what extent to $b_{\lambda, w}$ depend on $\lambda$?
The answer is seemingly "not much".

## Category $\OO$ Methods

> Note: skipping chapter 2 since we're focusing on infinite dimensional representations.

### Hom and Ext

Recall that $\hom(\wait, \wait)$ is left exact but not exact, and is either covariant or contravariant depending on which variable is fixed.
So taking $\hom$ of a SES yields a LES involving the derived functors $\ext^n$.
Convention: $\ext^0 \definedas \hom$ and $\ext^1 \definedas \ext$.

Let $A, C$ be $U(\lieg)\dash$modules.
Consider two short exact sequences

\begin{align*}
0 &\to A \to B \to C \to 0 \\
0 &\to A \to B' \to C \to 0
.\end{align*}

where $B, B'$ are extensions of $C$ by $A$.

We say two such sequences are equivalent iff there is an isomorphism making this diagram commute:

\begin{center}
\begin{tikzcd}
                        & B \arrow[rd] \arrow[dd, "\cong"] &   \\
A \arrow[rd] \arrow[ru] &                                  & C \\
                        & B' \arrow[ru]                    &
\end{tikzcd}
\end{center}

The set $\ext_{U(\lieg)}(C, A)$ of equivalence classes of extensions is a group under an operation called "Baer sum" (see Wikipedia) in which the identity is the class of the split SES
$$
0 \to A \to A\oplus C \to C \to 0.
$$

It turns out that the first right-derived functor of $\hom$ defined using projective resolutions, namely $\ext_1$, is isomorphic to $\ext$.
In particular, each SES leads to a pair of LESs given by applying $\hom(\wait, D)$ and $\hom(E, \wait)$ for $D, E \in U(\lieg)\dash$mod.

**Warning:**
Even if $A, C\in \OO$, there's no guarantee $B\in \OO$ for $B$ an extension.
In this case, we define $\ext_\OO(C, A)$ to be only those extensions lying in $\OO$.

Proposition (Homs and Exts for Vermas and Quotients)
:   Let $\lambda, \mu \in \lieh\dual$.

    a. If $M$ is a highest weight module of highest weight $\mu$ and $\lambda \not< \mu$, then $\ext_\OO(M(\lambda), M) = 0$.
      Contrapositive: nontrivial extensions force the strict inequality $\mu < \lambda$.
      In particular, $\ext_\OO(M(\lambda), X) = 0$ for $X = L(\lambda), M(\lambda)$.

    b. If $\mu \leq \lambda$, then $\ext_\OO(M(\lambda), L(\mu)) = 0$.

    c. If $\mu < \lambda$, then $\ext_\OO(L(\lambda), L(\mu)) \cong \hom_\OO(N(\lambda), L(\mu))$.

        > (c) is useful, homs can be easier to compute. Can just look at radical structure of $N$, i.e. just the head.

    d. $\ext_\OO(L(\lambda), L(\lambda)) = 0$.

Proof (of (a))
:   Given an extension $0 \to M \mapsvia{f} E \mapsvia{g} M(\lambda) \to 0$ where $M$ is a highest weight module of highest weight $\mu \not< \lambda$.
    We want to show it splits.

    *Claim:*
    Let $v^+$ be a maximal vector of $M(\lambda)$, let $v$ be its preimage under $g$, then $v$ is a maximal vector of weight $\lambda$ in $E$.
    For $x\in \lien$, we can think of the RHS as a quotient and identify

    \begin{align*}
    x\cdot v + M
    &= x\cdot (v+M) \\
    &= x\cdot v^+ \\
    &= 0 \\
    &= 0 + M
    ,\end{align*}

    and for these to be equal this implies $x\cdot v \in M$.
    But $x\cdot v$ has weight $> \lambda$; since $\mu \not> \lambda$, $M$ has no such weights.
    So we must have $x\cdot v = 0\in E$, and $v$ is a maximal vector.

    It's also the case that $U(\lien^-)$ acts freely on $v$, since it acts freely on its image in the quotient $M(\lambda)$.
    So $v$ generates a submodule $\generators{v} \leq E$ isomorphic to $M(\lambda)$.
    This defines a splitting (because of the freeness of this action) given by $h(v^+) = v$.


Proof (of (b))
: Follows from (a).

Proof (of (c))
:   Look at the SES $0\to N(\lambda) \to M(\lambda) \to L(\lambda) \to 0$.
    Apply $\hom_\OO(\wait, L(\mu))$ to get the LES

    \begin{align*}
    \cdots &\to \hom_\OO(M(\lambda), L(\mu)) \to \hom_\OO( N(\lambda), L(\mu)  ) \\
    &\to \ext_\OO( L(\lambda), L(\mu)  ) \to \ext_\OO( M(\lambda), L(\mu)  ) \to \cdots
    \end{align*}

    and since $L(\lambda)$ is the only simple quotient of $M(\lambda)$, so the first $\hom$ is zero.
    Similarly, the last $\ext_\OO$ is zero by (b), and the middle is an isomorphism.

Proof (of (d))
:   Replace $\mu$ by $\lambda$ in the LES, now term 2 above is zero since $\Pi(L(\lambda)) < \lambda$.
    Term 4 is zero by (b), and thus term 3 is zero.

Next section: duality in category $\OO$.