# Friday April 24th

Chapter 11: Tilting Modules.

Recall that these are defined by modules with both a standard and a costandard filtration.

Theorem (7.14, Translation Off the Walls 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.

The SES here represents starting at the RHS, translating to get to a wall to get the middle term, then translating off the wall and picking up an $s$ term.

To see that standard and costandard filtrations exist, we can consider:

![](figures/image_2020-04-24-09-25-20.png)\


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.

Theorem (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 $M(\lambda) \injects T(\lambda)$.

      > Note that this implies that $T(\lambda)$ must lie in the single block $\OO_{\chi_\lambda}$, since it has a Verma $M(\lambda)$ and is indecomposable.

    b. Every indecomposable tilting module in $\OO_{\text{int}}$ is isomorphic to $T(\lambda)$ for some $\lambda \in \Lambda$.
    c. $T(\lambda)$ is the only tilting module up to isomorphicin $\OO_{\text{int}}$ having the properties in (a).
    d. $\theset{[T(\lambda)]}_{\lambda \in \Lambda}$ is a basis for $\mck = K(\OO_{\text{int}})$.


Proof (of (a))
:   Existence is by induction on length in $W$ along with translation to and from walls.
    Fix $\lambda \in \Lambda$ to be $\rho\dash$regular and antidominant.
    Consider the linked weights $w\cdot \lambda$ and their translates to walls.
    To start, set $T(\lambda) \definedas M(\lambda)$ which has the properties in (a).
    Likewise for $\mu \in \bar C$, for $C$ the antidominant chamber, $\mu$ is still antidominant, $T_\lambda^\mu M(\lambda) = M(\mu)$ again irreducible and seta $T(\mu) \definedas M(\mu)$

    For the inductive step, assume $T(w\cdot \mu)$ has been constructed for all $\mu \in \bar{w\cdot C}$ to $T(w\cdot \lambda)$ is defined.
    If $s$ is a simple reflection with $ws > w$ choose an antidominant integral weight $\mu$ with $W_{\mu}^0 = \theset{1, s}$ so we have defined $T(w\cdot \mu) T(ws \cdot \mu)$.
    Apply the exact functor $T_\mu^\lambda$ and use theorem 7.14: $T_\mu^\lambda T(w\cdot \mu)$ has a two-step filtration with sections
    \begin{align*}
    N: {M(w\cdot \lambda) \over M(ws\cdot \lambda)  }
    .\end{align*}
    where the top term is a non-split extension and the bottom is indecomposable with $\hd = L(w\cdot \lambda)$.

    The other sections $M(x\cdot \mu)$ of $T(w\cdot \mu)$ have $x\cdot \mu < w\cdot \mu$.
    Applying $T_\mu^\lambda$ to these produces 2-step modules like $N$ but with highest weights either $xs \cdot \lambda < ws\cdot \lambda$, or $x\cdot \lambda$, whichever is greater.
    
    \
    
    Thus $ws \cdot \lambda$ is the unique largest weight (occurring with multiplicity one) in the titling module $T_\mu^\lambda T(w\cdot \mu)$ by Prop 11.1e.
    Set $T(ws\cdot \lambda)$ to be the indecomposable summand of this involving the weight $ws\cdot \lambda$, this has the required properties in (a).
    
    \
    
    We can now translate $T(ws\cdot \lambda)$ to the walls of $\bar{ws\cdot C}$, which were not already walls of $\bar{w\cdot C}$.
    For weights $\nu$ in such walls, the translated module with have a 1-dim $\nu\dash$weight space $M$ (Theorem 7.6), so we can take the indecomposable summand containing $M$ to be $M(\nu)$, completing the inductive step.

Proof (of (b))
:   Let $T$ be any indecomposable tilting module having $\lambda$ as one of its maximal weights.
    By the remark concerning Prop 3.7a, there is an embedding $M(\lambda) \injects T$ at the bottom of a standard filtration and $T/M(\lambda)$ has a standard filtration.
    Thus $\ext^1(T/M(\lambda, T(\lambda))) = 0$, using Prop 11.1f and Exercise 6.12.

    Applying $\hom(\wait, T(\lambda))$ to $$0 \to M(\lambda) \mapsvia{f} T \to T/M(\lambda) \to 0$$ yields $$\hom(T, T(\lambda)) \mapsvia{f^*} \hom(M(\lambda), T(\lambda)) \to \ext^1(T/M(\lambda), T(\lambda)) = 0 \to \cdots.$$
    Thus $f_*$ is surjective, and the embedding $M(\lambda) \injects T(\lambda)$ lifts to a map $\phi: T\to T(\lambda)$:
    \begin{center}
    \begin{tikzcd}
    M(\lambda)
    & T \ar[d, dotted, "\phi"]  \\
    M(\lambda) \ar[r] 
    & T(\lambda) \ar[u, dotted, "\psi"]
    \end{tikzcd}
    \end{center}

    Similarly, reversing $T, T(\lambda)$ we get the map $\psi$ with is the identity on the specified submodules of $M(\lambda)$ in each.
    This we get endomorphisms $\phi \circ \psi$ of $T(\lambda)$, which is an isomorphism on the $\lambda\dash$weight space $T(\lambda)_\lambda$ and of $\psi\circ \phi(T)$, which is an isomorphism *at least* on the $\CC\dash$ span of $v_\lambda^+ \in M(\lambda) \subset T$ (although maybe not on all of $T_\lambda$ as claim in Humphreys).

    By Fitting's lemma (?) an endomorphism of a finite length indecomposable module is either nilpotent or invertible.
    But the two compositions can not be nilpotent since they are isomorphisms on $\CC v_\lambda^+$ viewed in $T(\lambda)$ and in $T$.

Proof (of (c))
: Take $T$ to be a tilting module...