Tilting modules

Theorem 1. For each $\lambda\in \Lambda$, there exists at most 1 indecomposible parity complex $\mathscr{E}_ {\lambda}$ supported on $\overline{X}_ {\lambda}$, such that $\mathscr{E}_ {\lambda}|_ {X_ {\lambda}} \cong \mathscr{L}_ {\lambda}[\text{dim}X_ {\lambda}]$. Moreover, any indecomposible parity complex is isomorphic to $\mathscr{E}_ {\lambda}[n]$ for some $\lambda\in \Lambda$ and $n\in \mathbb{Z}$.

Proof.

First, let’s proof uniqueness. Suppose that $\mathscr{E}_ {\lambda}$ and $\mathscr{E}_ {\lambda’}$ are indecomposible supported on $\overline{X}_ {\lambda}$m such that $\mathscr{E}|_ {\lambda}\cong \mathscr{L}_ {\lambda}[\text{dim}X_ {\lambda}]\cong \mathscr{E}_ {\lambda}‘|_ {X_ {\lambda}}$. Then $$\mathrm{Hom}(\mathscr{E}_ {\lambda},\mathscr{E}_ {\lambda’})\twoheadrightarrow \mathrm{Hom}(\mathscr{E}_ {\lambda}|_ {X_ {\lambda}},\mathscr{E}_ {\lambda’}|_ {X_ {\lambda}})\cong\mathbb{k}.$$ There is $f: \mathscr{E}_ {\lambda}\rightarrow \mathscr{E}_ {\lambda}‘$ such that $f|_ {X_ {\lambda}}$ is an isomorphism, and $g: \mathscr{E}_ {\lambda}’\rightarrow \mathscr{E}_ {\lambda}$ such that $g|_ {X_ {\lambda}}$ is an isomorphism. Note that $g\circ f\in \mathrm{End}(\mathscr{E}_ {\lambda})$, which is a local ring and $g\circ f$ is not nilpotent (since its restriction is not). Therefore, $g\circ f$ is invertible. Similarly, $f\circ g$ is invertible. Therefore, both $f$ and $g$ are isomorphisms and hence $\mathscr{E}_ {\lambda}\cong \mathscr{E}_ {\lambda}’$.

Now let $\mathscr{F}$ be some indecomposible parity complex and $Y$ be its support. There is a unique $\lambda$ such that $X_ {\lambda}$ is open in $Y$ (otherwise if $X_ {\lambda}\cup X_ {\mu}$ is open in $Y$, then $\mathscr{F}|_ {X_ {\lambda}\cup X_ {\mu}}\cong \mathscr{F}|_ {X_ {\lambda}}\opuls \mathscr{F}|_ {X_ {\mu}}$). So $Y=\overline{X}_ {\lambda}$ and hence by uniqueness we see that $\mathscr{F}$ is some shift of $\mathscr{E}_ {\lambda}$.

Remark 2. In cases we will consider, $\mathscr{E}_ {\lambda}$ exists. If $\mathbb{F}=\mathbb{C}$, and if each $X_ {\lambda}$ is contractible (which also implies that $\mathscr{L}_ {\lambda}\cong \underline{\mathbb{k}}_ {X_ {\lambda}}$), then the existence is guaranteed.

The case of affine flag variety

Let $\mathbb{F}$ be an algebraically closed field and $G$ be a connected reductive group over $\mathbb{F}$. Fix $G\supseteq B\supseteq T$. To $G$, we associate two functors

  • $LG$ and $L^{+}G$ from the category of $\mathbb{F}$-algebras to $\mathrm{Set}$ by setting $LG(R)=G(R(\!(t)\!))$ and $L^{+}G(R):=G(R[\![t]\!])$. Then $L^{+}G$ is representable by a group scheme over $\mathbb{F}$ and $L^{+}G$ is representable by a group ind-scheme over $\mathbb{F}$.
  • To each subset $A$ of simple affine reflections $S_ {\text{aff}}$, we can associate a parahoric subgroup $Q_ {A}\subseteq LG$, and consider the functor $R\mapsto LG(R)/Q_ {A}(R)$, and let $\mathrm{FL}_ {A}$ be its fppf sheafification.
  • If $A=\emptyset$, then $Q_ {A}$ = the standard Iwahori subgroup of $LG$, and we write $\mathrm{FL}$ for $\mathrm{FL}_ {A}$.
  • If $A=S_ {\text{aff}}$, then $Q_ {S_ {\text{aff}}}=L^{+}G$, and we weite $\mathrm{FL}_ {S_ {\text{aff}}}=\mathrm{Gr}$ the affine Grassmannian.

Affine Grassmannian and geometric Satake

The $L^{+}G$-orbits on $\mathrm{Gr}_ {G}$ is parametrized by $\mathbb{X}_ {\ast}(T)^{+}$ dominant cocharacters of $T$.

  • We call $\mathrm{Gr}^{\lambda}$ the $L^{+}G$-orbit associated to $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$, and $\mathrm{Gr}^{\lambda}$ for $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$ is a stratification.
  • We also have that $\mathrm{Gr}^{\lambda} = \langle 2\rho,\lambda \rangle$.
  • We have that $\overline{\mathrm{Gr}^{\lambda}}\subseteq \overline{\mathrm{Gr}^{\mu}}$ if and only if $\mu-\lambda$ is a $\mathbb{Z}_ {>0}$-sum of positive coroots.
  • Connected components of $\mathrm{Gr}$ is in canonical bijection with $\pi_ {1}(G):=\mathbb{X}_ {\ast}(T)/\mathbb{Z}R^{\vee}$.
  • Two strata $\mathrm{Gr}^{\lambda}$ and $\mathrm{Gr}^{\mu}$ lies in the same connected component if and only if their dimension have the same parity.
  • We will consider $D_ {L^{+}G}^{b}(\mathrm{Gr},\mathbb{k})$. There is a convolution product $(-)\ast_ {L^{+G}}(-)$ on it and the category is monoidal.
  • Let $\mathrm{Perv}_ {L^{+}G}(\mathrm{Gr},\mathbb{k})$ be the heart of the perverse $t$-structure, which is stable under $\ast$.
  • For each $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$, $j^{\lambda}:\mathrm{Gr}^{\lambda}\hookrightarrow \mathrm{Gr}$, we have 3 natural objects $J_ {!}(\lambda):={}^{p}H^{0}(j^{\lambda}_ {!} \underline{\mathbb{k}}_ {\mathrm{Gr}^{\lambda}}[\langle 2\rho, \lambda \rangle])$ and similar for $J_ {\ast}(\lambda)$, and there is a canonical morphism $J_ {!}(\lambda)\rightarrow J_ {\ast}(\lambda)$ and let $J_ {!\ast}(\lambda)$ be the image.

Let’s now denote by $\check{G}$ the connected reductive group over $\mathbb{k}$ Langlands dual to $G$.

Theorem 3. There is a (symmetric) monoidal equivalence $$ (\mathrm{Perv}_ {L^{+}G}(\mathrm{Gr},\mathbb{k}),\ast) \cong (\mathrm{Rep}(\check{G}),\otimes).$$ called geometric Satake, which sends for each $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$ \begin{align} J_ {!}(\lambda) &\mapsto M(\lambda)\\ J_ {\ast}(\lambda) &\mapsto N(\lambda)\\ J_ {!\ast}(\lambda) &\mapsto L(\lambda).
\end{align}

We have just defined $\mathscr{E}_ {\lambda}$ for $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$.

Question 4.

  1. Are they perverse?
  2. If they are, where do they go under geometric Satake?

Theorem 5 ([Juteau-Mautner-Williamson]). If $\text{char}(\mathbb{k})$ is good for $G$, then $\mathscr{E}_ {\lambda}$ is perverse for any $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$.

Proposition 6. Let $\lambda\in\mathbb{X}_ {\ast}(T)^{+}$. If $\mathscr{E}_ {\lambda}$ is perverse, then its image under geometric Satake is tilting and denoted by $T(\lambda)$.

Proof.

We need to show that \begin{align} \mathrm{Hom}(\mathscr{E}_ {\lambda},J_ {\ast}(\mu)[1]) &=0, \\ \mathrm{Hom}(J_ {!}(\mu),\mathscr{E}_ {\lambda}[1]) &=0 . \end{align} We can assmue that $\langle 2\rho,\lambda \rangle$ and $\langle 2\rho,\mu \rangle$ have the same parity, say both even. Then $j_ {!}^{\lambda}\underline{\mathbb{k}}_ {\mathrm{Gr}^{\lambda}}[\langle 2\rho,\lambda \rangle]$ is concentrated in non-positive degrees. Let $A$ be the fiber $$A\rightarrow j_ {!}^{\mu}\underline{\mathbb{k}}_ {\mathrm{Gr}^{\mu}}[\langle 2\rho,\mu \rangle]\rightarrow J_ {!}(\mu)$$ and $A$ is concentrated in negative degrees. So we get $$0\mathrm{Hom}(A,\mathscr{E}_ {\lambda})\rightarrow \mathrm{Hom}(J_ {!}(\mu),\mathscr{E}_ {\lambda}[1])\rightarrow \mathrm{Hom}(j_ {!}^{\mu}\underline{\mathbb{k}}_ {\mathrm{Gr}^{\mu}}[\langle 2\rho,\mu \rangle],\mathscr{E}^{\lambda}[1]),$$ then $j_ {!}^{\mu}\underline{\mathbb{k}}_ {\mathrm{Gr}^{\mu}}[\langle 2\rho,\mu \rangle]$ is $\ast$-even, and $\mathscr{E}_ {\lambda}[1]$ is odd and therefore $!$-odd, so $\mathrm{Hom}(j_ {!}^{\mu}\underline{\mathbb{k}}_ {\mathrm{Gr}^{\mu}}[\langle 2\rho,\mu \rangle],\mathscr{E}^{\lambda}[1])=0$. Therefore, $\mathscr{E}_ {\lambda}$ is tilting.

Remark 7. Even if $\text{char}(\mathbb{k})$ is bad for $G$, each cohomology ${}^{p}H^{n}(\mathscr{E}_ {\lambda})$ is still tilting and they give all the tilting objects.

Tilting modules in $\mathrm{Rep}(G)$

Proposition 8. Let $M\in\mathrm{Rep}(G)$, the followings are equivalent:

  1. $M$ admits a costandard filtration,
  2. for any $\lambda\in \mathbb{X}^{+}$, and any $n>0$, we have vanishing $\mathrm{Ext}^{n}(M(\lambda), M)=0$,
  3. for any $\lambda\in \mathbb{X}^{+}$, we have vanishing $\mathrm{Ext}^{1}(M(\lambda), M)=0.$

Similarly, the followings are equivalent:

  1. $M$ admits a standard filtration,
  2. for any $\lambda\in \mathbb{X}^{+}$, and any $n>0$, we have vanishing $\mathrm{Ext}^{n}(M,N(\lambda))=0$,
  3. for any $\lambda\in \mathbb{X}^{+}$, we have vanishing $\mathrm{Ext}^{1}(M,N(\lambda))=0.$

Finally we have that the followings are equivalent:

  1. $M$ is tilting,
  2. for any $\lambda\in \mathbb{X}^{+}$, and any $n>0$, we have vanishing $\mathrm{Ext}^{n}(M,N(\lambda))=0$ and $\mathrm{Ext}^{n}(M(\lambda), M)=0$,
  3. for any $\lambda\in \mathbb{X}^{+}$, we have vanishing $\mathrm{Ext}^{1}(M(\lambda), M)=0$ and $\mathrm{Ext}^{1}(M,N(\lambda))=0.$

Remark 9. In literature, standard filtration is also called Weyl filtration, and costandard filtration is called good filtration.

Definition 10. Let $\mathrm{Tilt}(G)\subseteq \mathrm{Rep}(G)$ be the subcategory of tilting objects. If $M\in\mathrm{Tilt}(G)$, then write $[M:N(\lambda)]$ for the occurances of $N(\lambda)$ as a subquotient in a costandard filtration, and similarly for $[M:M(\lambda)]$. Then $[M:N(\lambda)]=\mathrm{dim}(\mathrm{Hom}(M(\lambda),M))$.

In particular, in $K$-groups we have $[M]=\sum_ {\lambda\in\mathbb{X}^{+}}[M:N(\lambda)] [N(\lambda)]$ with non-negative coefficients. Also recall that in $K_ {0}(\mathrm{Rep}(G))$, we have that $[M(\lambda)]=[N(\lambda)]$. So if $M,N$ are tilting, we have that $$\text{dim}\mathrm{Hom}(M,N) = \sum_ {\lambda\in\mathbb{X}^{+}}[M:N(\lambda)]\cdot[N:N(\lambda)].$$