# Wednesday April 8th

Recall that we are setting up Jantzen's filtration.
Let $A$ be a PID, $\mfp \in A$ prime, $\bar A = A/pA$, $M$ a free $A\dash$module of rank $r$, with a nondegenerate symmetric bilinear form $(\wait ,\wait)$ having nonzero determinant wrt some basis of $M$.
Define $M(n), \bar{M(n)}$ as before

Lemma
:   \hfill

    1. $v_p(D) = \sum_{n > 0} \dim_{\bar A} \bar{M(n)}$
    2. For each $n$, the modified bilinear form induces a nondegenerate form on $\bar{M(n)} / p\bar{M(n)}$?

## Proof of Jantzen's Theorem

Let $A = \CC[T]$ and $K = \CC(T)$ its fraction field, and let $\lieg_A = A \tensor_\CC \lieg$ and $\lieg_K = K \tensor_\CC \lieg$, which is a Lie algebra that is split over $K$, i.e. every $h\in \lieh_K = K \tensor_\CC \lieh$ has all eigenvalues of $\ad h$ in $K$.

The theory we need carries over to the extended setting.
The plan is the following:

- Construct and look at basic properties of Verma modules (1.3-1.4)
- Look at properties of their contravariant forms (3.14 - 3.15)
- Find a simplicity criterion (4.8)

We'll use Lemma 5.6 to construct filtrations on the weight spaces of the extended Verma module, then reduce mod $T$ (using evaluation morphisms) to assemble the Jantzen filtration for the original $\CC\dash$module.

Given $\lambda \in \lieh\dual$, set $\lambda_T = \lambda + T_\rho \in \lieh_K\dual$.
For all $\alpha\in \Phi$, we have
\begin{align*}
(\lambda_T  + \rho, \alpha\dual) =
(\lambda + \rho , \alpha\dual) + T(\rho, \alpha\dual) \not\in \ZZ
,\end{align*}
since this is a linear polynomial.
So $\lambda_T$ is antidominant.

Therefore $M(\lambda_T)$ is simple, and equivalently (unique up to scalars) its nonzero contravariant form is nondegenerate.

The module $U(\lieg_A) \cong A \tensor_\CC U(\lieg)$ is a natural "$A\dash$form" of $U(\lieg_K) \cong K \tensor_\CC U(\lieg)$.
This yields $M(\lambda_T)_A$, an $A\dash$form of $M(\lambda_T)$, where each weight space is an $A\dash$module of finite rank.

Some remarks about contravariant forms on highest weight modules: given $M$ and such a form $(\wait, \wait): M\cross M \to \CC$, the transpose serves as an adjoint and we have $(uv, v') = (v, \tau(u) v')$.

Distinct weight spaces are orthogonal, i.e. $(M_\mu, M_\nu) = 0$ since
\begin{align*}
(hv, v') = \mu(h) (v, v') \\
= (v, hv') = \nu(h) (v, v') \\
\end{align*}
where $\mu(h) \neq \nu(h)$, implying $(v, v') = 0$.

We can compute
\begin{align*}
(u v^+ \in M_\mu, u' v^+ \in M_\mu) = (v^+, \tau(u) u' v') = a (v^+, v^+)
\end{align*}

for some $a\in A$, since $\lambda_T = \lambda + T_p$ maps $\lieh_A \to A$.
We can use the decomposition $U(\lieg) = U(\lieh) \oplus (\lien^- U(\lieg) + U(\lieg) \lien)$, where $\lien^+$ kills $v^+$ and $\lien^-$ lowers into an orthogonal weight space, and so this pairing only depends on $(v^+, v^+)$.

> Note that the radical of this bilinear form is a maximal submodule.

The weights are of the form $\lambda_T - \nu$ for $\nu \in \lien^+ \Phi^+ = \Lambda_r^+$.
Apply lemma 5.6 to the $A\dash$form $M_{\lambda - \nu}$ of $M(\lambda_T)_{\lambda_T - \nu}$ to get a decreasing finite filtration of $A\dash$submodules

\begin{align*}
M_{\lambda-\nu}(0)=M_{\lambda_{T}-v}(1) \supset \cdots
\end{align*}

where $M_{\lambda_T - \nu}(i) = \theset{e\in M_{\lambda_T - \nu} \suchthat (e, M_{\lambda_T - \nu}) \subset     T^i A}$.

For each $i \geq 0$, set $M(\lambda_T)_A^i = \sum_{\nu \in \Lambda_r^+} M_{\lambda_T - \nu}(i)$.
This yields a decreasing filtration of $A\dash$submodules.

Next we want to "set $T=0$": formally, pass to the quotient $\bar M = M/TM$ over the field $\bar A = A/TA \cong \CC$.
Since $\lambda_T = \lambda + T_\rho \mapsvia{T = 0} \lambda$, we have $M(\lambda_T)_A \cong M(\lambda)$ and the filtration becomes a decreasing filtration of $M(\lambda)$.

By the lemma, the sections of this filtration inherit nondegenerate contravariant forms, proving (a).
By the proof of that lemma, the filtration on each individual weight space terminates at 0.

Claim:
Some $M(\lambda)^{n+1} = 0$.

Proof
:   If not, since $M(\lambda)$ has finite length, we must have $0 \neq M(\lambda)^n = M(\lambda)^{n+1} = \cdots$ for some $n$.
    Choose some $u\in \lieh\dual$ such that $M(\lambda)_\mu^n = 0$, but then $0 \neq M(\lambda)_\mu^n = M(\lambda)_\mu^n = \cdots$, a contradiction.

For a proof of (c), we want to show $\sum_{i > 0} \ch M(\lambda)^i = \sum_{\alpha \in \Phi^+} \ch M(s_\alpha \cdot \lambda)$.
We can show that the LHS is given by
\begin{align*}
\cdots
&= \sum_{i > 0} \sum_{\nu \in \Lambda_r^+} \dim M(\lambda)_{\lambda-\nu}^i \\
&= \sum_{i > 0} \sum_\nu \dim\qty{\bar{M(\Lambda_T)_A^i}}_{\lambda_T - \nu} e(\lambda - v) \\
&= \sum_i \sum_\nu \dim \bar{M_{\lambda_T -\nu}(i)} e(\lambda - \nu) \\
&= \sum_\nu \sum_i \dim \bar{M_{\lambda_T -\nu}(i)} e(\lambda - \nu) \\
&= \sum_\nu v_T(D_\nu(\lambda_T)) e(\lambda - v) \quad\text{Lemma 5.6a}
.\end{align*}

where $D_\nu(\lambda_T)$ is the determinant of the matrix of the contravariant form on the $\lambda_T - \nu$ weight space of $M(\lambda_T)$.

Fact (Jantzen, Shapovalov):
Up to a nonzero scalar multiple depending on a choice of basis of $U(\lien^-)$,
\begin{align*}
D_\nu(\lambda_T) = \prod_{\alpha > 0} \prod_{r \in \ZZ^{>0}} \qty{ (\lambda_T, \rho, \alpha\dual) - r  }^{P(\nu - r_\alpha)}
\end{align*}
where $P$ is the Kostant partition function.

We can calculate $v_T$ of this, which doesn't depend on the scalar:
\begin{align*}
(\lambda_T + \rho, \alpha\dual) - r = (\lambda+\rho, \alpha\dual) -r + T(\rho, \alpha\dual)\\
\implies v_T( \cdots ) =
\begin{cases}
1 & (\lambda + \rho, \alpha\dual) = r > 0 \iff \alpha \in \Phi_\lambda^+ \\
0 & \text{else}
\end{cases}
.\end{align*}

We then have
\begin{align*}
v_T(D_\nu(\lambda_T)) = \sum_{\alpha \in \Phi^+_\lambda} P(\nu - (\lambda + \rho, \alpha\dual)\alpha)
.\end{align*}

Thus the LHS is given by
\begin{align*}
\cdots
&= \sum _{\nu \in \Lambda_r^+} \sum_{\alpha\in \Phi_\lambda^+} P(\nu - (\lambda + \rho, \alpha\dual) \alpha) e(\lambda - \nu) \\
&= \sum _{\alpha \in \Phi_\lambda^+} \sum_{\sigma \in \Lambda_r^+} P(\sigma) e(\lambda - (\lambda + \rho, \alpha\dual)\alpha - \sigma) \\
&= \sum_{\alpha \in \Phi_\lambda^+} \ch M(s_\alpha \cdot \lambda)
,\end{align*}

where we've used what we know about characters of Verma modules.

> Note that the proof of the determinant formula is technical.

We'll skip chapter 6 on KL theory.