Inductions

🔗Spectral Parabolic Induction

Let $P\subseteq G$ be a parabolic with levi $M$. Then on the dual group side, we have the corresponding ${}^cG\supseteq P\twoheadrightarrow {}^cM$.

Then we obtain the following correspondence

Facts:

• $\mathcal{X}_ { {}^cG,F}$ and $\mathcal{X}_ { {}^cM,F}$ are classical algebraic stacks, while $\mathcal{X}_ { {}^cP,F}$ is in general derived.
• $\mathcal{X}_ { {}^cP, F}$ is quasi-smooth, i.e. its cotangent complex is perfect and concentrated on degree $[-1,1]$ (This can be viewed as a derived version of l.c.i, or intuitively visualized as a space cut out by equations inside a smooth space.).
• The map $r$ is schematic, proper and hence we have $r_ *:\mathrm{Coh}(\mathcal{X}_ { {}^cP,F})\rightarrow \mathrm{Coh}(\mathcal{X}_ { {}^cG,F})$.
• The map $\pi$ is quasi-smooth (It means that the relative cotangent complex is perfect and concentrated on degree $[-1,1]$.) and hence we have a functor $\pi^!:\mathrm{Coh}(\mathcal{X}_ { {}^cM,F})\rightarrow\mathrm{Coh}(\mathcal{X}_ { {}^cP,F})$.

Now we assume that $G$ splits over a tamely ramified extension. The spectral parabolic induction naturally restricts to the tame part, i.e. we have correspondence

Now we take parabolic $P$ to be Borel, i.e. ${}^cP={}^cB$ and ${}^cM={}^cT=\hat{T}\rtimes(\mathbb{G}_ m\times\Gamma_ {\widetilde{F}/F})$.

Note that unipotent part for $T$ is the same as unramified part, bacause requiring $I_ T$ to map into unipotent part of $T$ is the same as requiring $I_ T$ maps to 1 (any unipotent subgroup of $T$ is trivial).

In particular, $\mathcal{X}_ {{}^cT,F}^{\textrm{unr}}=\hat{T}\overline{\sigma}/\hat{T}\hookrightarrow\mathcal{X}_ {{}^cT,F}^{\textrm{tame}}$.

Let $G$ be a reductive group over $F$ as before. Assume that $G$ is unramified, i.e. $G$ has an integral model over $\mathcal{O}_ F$. Then we define $H_ {\textrm{sph}}:=C_ c(G(\mathcal{O})\backslash G(F)/G(\mathcal{O}))$. We have $\mathcal{X}_ {{}^cG,F}^{\textrm{unr}}\cong\hat{G}\overline{\sigma}/\hat{G}$.

🔗A Magical Correspondence

Consider the following general framework.

• $\phi_ x:\mathcal{X}\rightarrow \mathcal{X}$ and $\phi_ y:\mathcal{Y}\rightarrow \mathcal{Y}$ two endomorphisms of two spaces.
• $f:\mathcal{X}\rightarrow \mathcal{Y}$ a morphism compatible with $\phi_ x,\phi_ y$, i.e. $\phi_ y\circ f=f\circ\phi_ x$.

Note that $\mathcal{L}_ {\phi}\mathcal{Y}:=\mathcal{Y}\times_ {{}^{\Gamma_ {\phi_ y}}\mathcal{Y}\times \mathcal{Y}^\Delta} \mathcal{Y}$ is the (derived) $\phi_ y$-fixed points

Then we have a fundamental correspondence:

Let me explain the construction in detail. First, note that $\mathcal{X}\times_ {\mathcal{Y}}\mathcal{X} = (\mathcal{X}\times \mathcal{X})\times_ {\mathcal{Y}\times \mathcal{Y}^\Delta}\mathcal{Y}$, where

Next, we define

Then $f\times \textrm{id}: \mathcal{X}\times_ {\mathcal{Y}}\mathcal{L}_ \phi\mathcal{Y}\rightarrow \mathcal{L}_ \phi\mathcal{Y}$ gives the definition of $\pi$ in the correspondence above.

Finally, Since $f\circ\phi_ x=\phi_ y\circ f$, then $({\textrm{id}_ \mathcal{X}},\phi_ x)\times\textrm{id}_ {\mathcal{Y}}:\mathcal{X}\times_ {\mathcal{Y}\times\mathcal{Y}^\Delta}\mathcal{Y}\rightarrow (\mathcal{X}\times\mathcal{X})\times_ {\mathcal{Y}\times\mathcal{Y}^\Delta}\mathcal{Y}$ gives the definition of $\delta$ in the correspondence.

🔗Spectral Deligne-Lusztig Induction

Back to our situation:

• take $\mathcal{X}=\mathcal{X}_ {{}^cB.\breve{F}}^{\textrm{tame}}={\rho: I_ F^t\rightarrow {}^cB,d\circ\rho=(\textrm{cycl}^{-1},\textrm{pr})}/\hat{B}$. Fix a lifting of $\mathrm{Frob}$ to $W_ F^t$, denoted by $\phi$.
• Take $\mathcal{Y}=\mathcal{X}_ {{}^cG,\breve{F}}^{\textrm{tame}}={\rho:I_ F^t\rightarrow {}^cG,d\circ\rho=(\textrm{cycl}^{-1},\textrm{pr})}/\hat{G}$, also acted by $\phi$.
• There is a natural $\phi$-equivariant map (induced by ${}^cB\rightarrow {}^cG$) $$f:\mathcal{X}\rightarrow \mathcal{Y}.$$

Using Prop. 2.9, we know that $\mathcal{L}_ \phi\mathcal{Y}=\mathcal{X}_ {{}^cG,F}^{\textrm{tame}}$. Now apply the general construction above, and we obtain

where

• $\widetilde{\mathcal{X}_ {{}^cG,F}^{\textrm{tame}}}:=\mathcal{X}_ {{}^cB,\breve{F}}^{\textrm{tame}}\times_ {\mathcal{X}_ {{}^cG,\breve{F}}^{\textrm{tame}}}\mathcal{X}_ {{}^cG,F}^{\textrm{tame}}$.
• $S_ {{}^cG,F}:=\mathcal{X}\times_ \mathcal{Y}\mathcal{X}$ is called Steinberg stack.

Fact:

• Steinberg stack $S_ {{}^cG,F}$ is classical.
• $\delta$ is quasi-smooth.
• $\pi$ is (ind-)proper.

Then we can make the following definition.

Fix $\tau\in I_ F^t$ a topological generator. Let $\hat{\mathbb{Z}}^p:=\prod_ {(\ell,p)=1}\mathbb{Z}_ \ell$ denote the prime-to-$p$ part. Then $I_ F^t\cong \hat{\mathbb{Z}}^p$.

Then $\mathcal{X}_ {{}^cB,\breve{F}}^{\textrm{tame}}=(\hat{B}\overline{\tau}/\hat{B})\times_ {\hat{T}\overline{\tau}/\hat{T}}(\hat{T}\overline{\tau}/\hat{T})^{\wedge,p}\hookrightarrow\hat{B}\overline{\tau}/\hat{B}$ induces isomorphisms on tangent spaces.

Similarly, we have maps $$\mathcal{X}_ {{}^cG,\breve{F}}^{\textrm{tame}}\hookrightarrow \hat{G}\overline{\tau}/\hat{G}.$$

We have Grothendieck-Springer alteration $$\hat{B}\overline{\tau}/\hat{B}\rightarrow \hat{G}\overline{\tau}/\hat{G},$$ and

In general, we always have

We define $$S_ {\hat{G}\tau,w}^\square:=\textrm{closure of }f^{-1}(\mathcal{O}_ w).$$ Then we have $$\overline{S_ {\hat{G}\tau,w}}\subseteq \bigcup_ {w’\leq w} S_ {\hat{G}\overline{\tau},w’},$$ which is not equality in general.

But for $w=e$ identity, $S_ {\hat{G}\overline{\tau},e}=\hat{B}\overline{\tau}/\hat{B}$ is closed in $S_ {\hat{G},\overline{\tau}}$, so the equality holds for $w=e$.

For simplicity, assume that $\overline{\tau}=1$ (the general case is in the same spirit but more complicated in computation). Then unwinding definitions, we have

where $\widetilde{\mathcal{X}^{\textrm{tame}}_ {{}^cG, F,w}}$ is defined by the pullback square and is called spectral Deligne-Lusztig stack. We have the following diagram

When $w=1$, the condition $\rho(\overline{\sigma})\hat{B}\rho(\sigma^{-1})=\hat{B}$ is same as $\rho(\overline{\sigma})\in\hat{B}$. So, ${\widetilde{\mathcal{X}_ {{}^cG, F,w}^{\textrm{tame}}}}=\mathcal{X}_ {{}^cB,F}^{\textrm{tame}}$ and spectral Deligne-Lusztig induction recovers spectral parabolic induction, similar to the classical Deligne-Lusztig induction.