# Wednesday April 22nd

For $p\in \QQ[x]$, we'll denote $p[i]$ the $i$th coefficient.

For $M\in \OO$ or any module of finite length, we define its radical series: $$\rad^0 M = M,\quad \rad^{i+1} M = \rad(\rad^i M)$$
Note that the layers/subquotients are semisimple.

Dually, $\soc M$ is the largest semisimple submodule of $M$, and iterating this yields the *socle series*.
Denote $$\soc_i M \definedas \soc^{i+1} M / \soc^i M$$ the $i$th socle layer.
The corresponding layers are the same as in the radical filtration, just with reversed indexing.


For convenience, set $$Q_{x, w} = P_{w_0 w, w_0 x}(q)$$ the "inverse" KL polynomial.
Recall that a consequence of the KL conjecture is that $[M_w] = \sum_{x\leq w} Q_{x, w}(1) [L_x]$

The following theorem follows from the Jantzen conjecture

Theorem (Coefficients of Inverse KL Polynomials)
:   \hfill

    a. $$Q_{x, w}[i] = [ \rad_{\ell(xw) - 2i} M_w : L_x] = [\soc_{\ell(x) + 2i} M_w: L_x]$$ 
    
        > Note that Humphreys adds $+1$ here due to indexing.

    b. If $x < w$, then $\dim \ext^1 (L_x, L_w) = \mu(x, w)$.


That concludes the KL theory.

## Tilting Modules (Ch. 11)

The Lusztig conjecture is an analog of the KL conjecture for representations of algebraic groups in characteristic $p> 0$.
It gives the characters of simple modules in terms of characters of known "standard" modules. for $p  > h$ and the formulas are independent of $p$.

> Holy grail: characters of simple modules!

? showed that that Lusztig character formula is correct for $p \gg 0$ but the bounds are very large.
In 2016, Geordie Williamson found counterexamples to this conjecture for fairly large $p$.

Most recent work, suggests that more uniform formulas can be obtained using another family of indecomposable representations, the tilting modules.

Definition (Tilting Modules)
:   $M\in \OO$ is a *tilting module* if both $M, M\dual$ have standard filtrations (quotients are Vermas).
    Equivalently, adapts to settings in which there are *standard* and *co-standard* modules: $M$ is a tiling module iff $M$ has a standard filtration (for $\OO$, sections are Verma modules) and a costandard filtration (in $\OO$, sections are duals of Verma modules).

Note that a self-dual modules with a standard filtration is automatically tilting.

Example
: If $\lambda$ is antidominant, $M(\lambda) = L(\lambda) = L(\lambda)\dual$ is self-dual and thus tilting.

Example
:   If $\lambda + \rho \in \Lambda^+$ is dominant integral, so $w_0 \cdot \lambda$ is antidominant and integral, then $P(w_0 \cdot \lambda)$ is self-dual and hence tilting.
    Its standard filtration has each $M(w\cdot \lambda)$ as a section exactly once, see Theorem 4.10.

Proposition (Properties of Tilting Modules)
:   Let $M$ be a tilting module.

    a. $M\dual$ is tilting.
    b. For $N$ tilting, $M \oplus N$ is tilting.
    c. Any direct summand of $M$ is tilting.
    d. If $\dim L < \infty$, then $L \tensor M$ is tilting.
    e. $T_\lambda^\mu M$ is tilting.
    f. If $N$ is tilting then $\ext_\OO^n(M, N)$ for all $n>0$ (take projective resolution and apply hom)

Proof
:   \hfill

    a. Obvious since $(M\dual)\dual \cong M$.
    b. $M\oplus N$ has a standard filtration by section 3.7, so does $(M\oplus N)\dual \cong M\dual \oplus N\dual$ (Theorem 3.2d)
    c. From Proposition 3.7b direct summands also have standard filtrations, and the formula used in the proof applies to the dual module here.
    d. This follows from theorem 3.6 since $L \tensor M(\lambda)$ has a standard filtration and exercise 3.2 distributing duals through tensors.
    e. This follows from (e) and (d).
    f. In theorem 3.3d we proved $\ext_\OO^1 (M(\mu), M(\lambda)\dual) = 0$ for all $\mu, \lambda$.
      In section 6.12 this is extended to $\ext^n$ and thus $\ext^n(M, N\dual) = 0$ for any $M, N$ with standard filtrations.
    
From now on, for simplicity we work in the full subcategory $\OO_{\text{int}}$ of modules whose weights lie in $\Lambda$, but the results generalize to arbitrary weights.
Set $\mck = K(\OO_{\text{int}})$ which is a subgroup of $K(\OO)$.

In order to classify all tilting modules, it suffices to classify indecomposable by Proposition 11.1c.
These turn out to be classified by highest weight.

To prove existence, we'll use translation functors to move to and from walls.

Theorem (Translation Off the Wall for Antidominant Regular Weights)
:   Let $\lambda, \mu \in \Lambda$ be antidominant with $\lambda$ regular (so in the antidominant chamber $C$) and $\mu$ lies on a single simple root wall $H_\alpha \intersect \bar C$ (i.e. the stabilizers $W_\mu^0)$ of $\mu$ under the dot action is $\theset{1, s}$ with $s = s_\alpha$ for some $\alpha \in \Delta$.)\
    Assume that $w\in W$ with $ws > w$, then

    a. There is a SES (singular to regular, translation off the wall): $$ 0 \to M(ws \cdot \lambda ) \to T_{\mu}^\lambda M(w\cdot \mu) \to M(w\cdot \lambda) \to 0.$$
    b. The head is given by $\text{Head} T_{\mu}^\lambda M(w\cdot \mu) = L(w\cdot \lambda)$, and in particular the LHS is indecomposable and the sequence in (a) is non-split.

Also recall Proposition 3.7a: If $M\in \OO$ has a standard filtration and $\lambda \in \Pi(M)$ is maximal, then $M(\lambda) \injects M$ and $M/M(\lambda)$ has a standard filtration.

Proposition (Existence of "Highest Weight" Tilting Modules)
:   Let $\lambda \in \Lambda$:

    a. There exists an indecomposable tilting module $T(\lambda) \in \OO_{\text{int}}$ such that $\dim T(\lambda)_\lambda = 1$ and $\mu \in \Pi(T(\lambda)) \implies \mu \leq \lambda$.
      In particular, $( T(\lambda): M(\lambda) ) = 1$ and $M(\lambda) \injects T(\lambda)$.
    b. Every indecomposable tilting module in $\OO_{\text{int}}$ is isomorphic to $T(\lambda)$ for some $\lambda \in \Lambda$.