# Friday April 3rd

Recall from last time that we defined a new partial order for all positive roots generated by "reflecting down", namely *strong linkage*.
We had a theorem: $\mu \uparrow \lambda \implies M(\mu)$ occurs as a composition factor of $M(\lambda)$.
We also have a side-arrow notation $w' \mapsvia{s_\alpha} w$ indicates that $w' = s_\alpha w$ and $w'$ is shorter than $w$.
We conclude that $x\cdot \lambda \uparrow w\cdot \lambda \iff x\leq w$ for $x, w\in W$, where the RHS is the usual Bruhat order and is notably independent of $\lambda$.

Corollary
:   Let $\lambda \in \Lambda$ be antidominant and $\rho\dash$regular and $x, w\in W$.
    Then
    $$
    [M(w\cdot \lambda) : L(x\cdot \lambda)] \neq 0 \iff M(x\cdot \lambda) \injects M(w\cdot \lambda) \iff x \leq w
    $$
Note that this statement is why we use antidominant instead of dominant, since this equation now goes in the right direction.

## Jantzen Filtration

Theorem (The Jantzen Filtration and Sum Formula)
:   Given $\lambda \in \lieh\dual$, $M(\lambda)$ has a terminating descending filtration satisfying

    a. Each nonzero quotient has a certain nondegenerate contravariant form (3.14)

    b. $M(\lambda)^i = N(\lambda)$

    c. $$\sum_{i > 0} \ch M(\lambda)^i = \sum_{\alpha > 0, s_\alpha \cdot \lambda < \lambda} \ch M(s_\alpha \cdot \lambda)$$ (the Integer sum formula, very important)

Note that the sum on the RHS is over $\theset{ \alpha\in\Phi^+_{[\lambda]} \suchthat s_\alpha \cdot \lambda < \lambda } \definedas \Phi^+_\lambda$.

Fact
:   $\soc M(\lambda) = L(\mu)$ for the unique antidominant $\mu$ in $W_{[\lambda]}\cdot \lambda$.
    Moreover, $[M(\lambda) : L(\mu)] = 1$.

Now suppose $M(\lambda)^n \neq 0$ but $M(\lambda)^{n+1} = 0$.
Each $M(\lambda)^i \supset \soc M(\lambda) = L(\mu)$, since they're submodules, and each $M(s_\alpha \cdot \lambda) \supset L(\mu)$, using the uniqueness of $\mu$.
By looking at coefficients of $\ch L(\mu)$ on each side of the sum formula, we obtain $n = \abs{\Phi^+}$.

Exercise (5.3)
:   When $\lambda$ is antidominant, integral, and $\rho\dash$regular, then $n = \ell(w)$.
    More generally, for nonintegral, $n = \ell_\lambda(w)$ where $\ell_\lambda$ is the length function of the system $(W_{[\lambda]}, \Delta_{[\lambda]})$.

Some natural questions:

1. Is the Jantzen filtration unique for properties (a)-(c)?
2. What are the "layer multiplicities" $[M(\lambda): L(\mu)]$?
3. Are the layers $M(\lambda)$ semisimple?
    If so, is the Jantzen filtration the same as the canonical filtrations with semisimple quotients (the radical or socle filtrations)?
4. When $M(\mu) \subset M(\lambda)$, how do the respective Jantzen filtrations compare?

A guess for (4):
Assume $\mu \up \lambda$, set $r = \abs{\Phi^+_\lambda} - \abs{\Phi_\mu^+}$, which is the difference in lengths of the two Jantzen filtrations.

Is it true that:

![Image](figures/2020-04-03-09:28.png)\

with $M(\mu) \intersect M(\lambda)^i = M(\mu)^{i-r}$ for $i\geq r$?

This is called the *Jantzen conjecture* and turns out to be true.

> Thought equivalent to KL-conjecture, but turned out to be deeper.
> See decomposition theorem, sheaves on flag varieties, no simple algebraic proof until recently.
> See chapter 8.

Recall that we obtained a hexagon:

![Image](figures/2020-04-03-09:32.png)\

We have
\begin{align*}
\Phi_{w\cdot \lambda}^{+}=\left\{\gamma \in \Phi^{+} | s_{\gamma} \cdot(w\cdot \lambda)<w \cdot \lambda\right\}=\{\alpha, \alpha+\beta\}
\end{align*}

with corresponding weights $s_\gamma(w\cdot \lambda) = s_\beta \cdot \lambda, s_\alpha \cdot \lambda$.
Thus we have a two-step filtration, and we've worked out the characters of the pieces previously.

Thus
\begin{align*}
\sum_{i=1}^n \ch M(w\cdot \lambda)^i 
&= \ch M(s_\alpha \cdot \lambda) + \ch M(s_\beta \cdot \lambda) \\
&= \ch L(s_\alpha \cdot \lambda) + \ch L(s_\beta \cdot \lambda) + 2\ch L(\lambda)
\end{align*}

where we know the last expression explicitly.
Since $n$ has the be the number of $L(\lambda)$s occurring on the RHS, we must have $n=2$.

We can reason that $M(w\cdot \lambda)^2 = L(\lambda)$, since any composition factor in $M(w\cdot \lambda)^2$ recurs in $M(w\cdot \lambda)^1$, and so
\begin{align*}
\ch M(w\cdot \lambda)^1 
&= \ch N(w\cdot \lambda) \\
&= \ch L(s_\alpha \cdot \lambda) + \ch L(s_\beta \cdot \lambda) + \ch L(\lambda)
.\end{align*}

We then obtain the following structure on the sections/subquotients of the Jantzen filtration

![Image](figures/2020-04-03-09:39.png)\

where the subquotients move upward through the diagram, e.g. the middle is $M(w\cdot \lambda)^1 / M(w\cdot \lambda)^2$.

Exercise (Last Assignment)
:   \hfill
    
    1. Try to work on the Jantzen filtration sections for $M(w_0 \cdot \lambda)$.
      List completely any additional assumptions or facts needed to deduce $M(w_0 \cdot \lambda)^i$ uniquely.
    2. Continue 4.11 in the case where $\lambda$ is singular
      Does this allow you to deduce that structure of all $M(w\cdot \lambda)$ using the Jantzen sum formula?
    3. Work out the non-integral case for $\liesl(3, \CC)$.
      (There are four different cases to consider here.)

## Showing Jantzen Implies BGG

We'll prove that $[M(\lambda) : L(\mu)] \neq 0 \implies \mu \up \lambda$.

Proof
:   By induction on the number of linked weights $\mu \leq \lambda$

    If $\lambda$ is minimal in its linkage class, then $M(\lambda) = L(\lambda)$ so $\mu = \lambda$ and $\lambda \up \lambda$ trivially.

    Otherwise, inductively suppose $[M(\lambda): L(\mu)] > 0$ with $\mu < \lambda$.
    Then $[M(\lambda)^1: L(\mu)] > 0$ since $M(\lambda)^1 = N(\lambda)$.
    By the sum formula, $[M(s_\alpha \cdot \lambda) : L(\mu)] > 0$ for some $\alpha \in \Phi_\lambda^+$.
    Then $s_\alpha \cdot \lambda < \lambda$ so the number of linked weights $\nu \leq s_\alpha \cdot \lambda$ is *smaller* than for $\lambda$.

    So by induction,

    ![Image](figures/2020-04-03-09:47.png)\

    and $\mu \up \lambda$ as required.

Example:
$\liesl(4, \CC)$ with Dynkin diagram $\cdot \to \cdot \to \cdot$.

If $\lambda = (0, -1, 0) \in \Lambda^+ - \rho$ with coordinates with respect to the fundamental weight basis for $\Lambda$ or $\lieh\dual$.
Then take $w = s_2 s_3, x= s_3 s_2 s_3 s_1 s_2$, then $\mu = w\cdot \lambda = (1, -2, -1)$ and $\mu - x\cdot \mu = \alpha_1 + \alpha_3$ so $x\cdot \mu < \mu$.

However, Verma's direct calculations in $U(\liesl(4, \CC))$ showed that $M(x\cdot \mu) \not\subset M(\mu)$, so $x\cdot \mu \not\up \mu$.

The explanation (due to Verma) is that $x\cdot \mu = xw\cdot \lambda$, using the fact that $s_1, s_3$ commute,
\begin{align*}
xw &= (s_3 s_2 s_3 s_1 s_2) s_2 s_3 \\
&= s_3 s_2 s_3 s_3 s_1 \\
&= s_3 s_2 s_1
.\end{align*}

and $s_2 s_3, s_3 s_2 s_1$ are not related in the Bruhat order.

> This is because there is no root reflection relating the two.
> Note that this can be seen by considering subexpressions: $a < b$ iff $a$ occurs as some deleted subexpression of $b$.

So it's possible to have one weight less than another with no inclusion of the Verma modules.