ScalingStacks

8.3.3. Diagonal bimodule

Recall that we have a (Δλ​𝒮M​(Z),Δλ​𝒮M​(Z))(\Delta_{\lambda}{\mathcal{S}}_{M}(Z),\Delta_{\lambda}{\mathcal{S}}_{M}(Z))-bimodule EE. Its restriction to a (𝒮M​(Z),Δλ​𝒮M​(Z))({\mathcal{S}}_{M}(Z),\Delta_{\lambda}{\mathcal{S}}_{M}(Z))-bimodule is the cone of π:Rξ2−⊗𝒮M​(Z)IdΔλ​𝒮M​(Z)→Rξ1−⊗𝒮M​(Z)IdΔλ​𝒮M​(Z)\pi:R_{\xi_{2}^{-}}\otimes_{{\mathcal{S}}_{M}(Z)}\operatorname{Id}\nolimits_{\Delta_{\lambda}{\mathcal{S}}_{M}(Z)}\to R_{\xi_{1}^{-}}\otimes_{{\mathcal{S}}_{M}(Z)}\operatorname{Id}\nolimits_{\Delta_{\lambda}{\mathcal{S}}_{M}(Z)}.

The (𝒮M​(Z),𝒮M​(Zξ))({\mathcal{S}}_{M}(Z),{\mathcal{S}}_{M}(Z_{\xi}))-bimodule E′=E∘(1⊗Ξ′−1)E^{\prime}=E\circ(1\otimes\Xi^{\prime-1}) is the cone of the map uu defined as follows.

Given α∈Rξ2−∙​(T,U)\alpha\in R_{\xi_{2}^{-}}^{\bullet}(T,U) with α|U=id\alpha_{|U}=\operatorname{id}\nolimits and given β∈Hom𝒮∙​(Zξ)⁡(S,U)\beta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,U), we have

u(α⊗β)=∑x∈ξ~1−1​(T)(id⊠ξ~1([−1→x]))⊗((αξ2−​(−1)⋅[ξ1+(1)→ξ2−(−1)]⋅ξ~1([x→1]))⊠id)⋅β.u(\alpha\otimes\beta)=\sum_{x\in\tilde{\xi}_{1}^{-1}(T)}\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\otimes\Bigl(\bigl(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\boxtimes\operatorname{id}\nolimits\Bigr)\cdot\beta.

We construct now an isomorphism between E′E^{\prime} and the restriction of Rξ−R_{\xi^{-}} to a (𝒮M​(Z),𝒮M​(Zξ))({\mathcal{S}}_{M}(Z),{\mathcal{S}}_{M}(Z_{\xi}))-bimodule.

We define two morphisms of pointed sets

f1:Rξ1−∙​(T,−)∧Hom𝒮∙​(Zξ)⁡(S,−)\displaystyle f_{1}:R_{\xi_{1}^{-}}^{\bullet}(T,-)\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,-) →Rξ−∙​(T,S)\displaystyle\to R_{\xi^{-}}^{\bullet}(T,S)
(α:U⊔{ξ1−(−1)}→T)∧(β:S→U)\displaystyle(\alpha:U\sqcup\{\xi_{1}^{-}(-1)\}\to T)\wedge(\beta:S\to U) ↦α⋅(β⊠idξ1−​(−1))=(α|U⋅β)⊠αξ1−​(−1)\displaystyle\mapsto\alpha\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})=(\alpha_{|U}\cdot\beta)\boxtimes\alpha_{\xi_{1}^{-}(-1)}

and

f2:Rξ2−∙​(T,−)∧Hom𝒮∙​(Zξ)⁡(S,−)\displaystyle f_{2}:R_{\xi_{2}^{-}}^{\bullet}(T,-)\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,-) →Rξ−∙​(T,S)\displaystyle\to R_{\xi^{-}}^{\bullet}(T,S)
(α:U⊔{ξ2−(−1)}→T)∧(β:S→U)\displaystyle(\alpha:U\sqcup\{\xi_{2}^{-}(-1)\}\to T)\wedge(\beta:S\to U) ↦α⋅(β⊠[ξ1−(−1)→ξ2−(−1)])\displaystyle\mapsto\alpha\cdot(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])
=(α|U⋅β)⊠(αξ2−​(−1)⋅[ξ1−(−1)→ξ2−(−1)]).\displaystyle\ \ \ \ =(\alpha_{|U}\cdot\beta)\boxtimes(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]).

Note that we have an isomorphism of pointed sets

f2∨f1:(Rξ2−∙​(T,−)∧Hom𝒮∙​(Zξ)⁡(S,−))∨((Rξ1−∙​(T,−)∧Hom𝒮∙​(Zξ)⁡(S,−))→∼Rξ−∙​(T,S)CLOSE.f_{2}\vee f_{1}:\bigl(R_{\xi_{2}^{-}}^{\bullet}(T,-)\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,-)\bigr)\vee\bigl((R_{\xi_{1}^{-}}^{\bullet}(T,-)\wedge\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,-)\bigr)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}R_{\xi^{-}}^{\bullet}(T,S).
0PDY

Lemma 8.3.4. We have d⁡(f1)=0d(f_{1})=0 and d⁡(f2)=f1∘ud(f_{2})=f_{1}\circ u. There is an isomorphism of differential modules

(f2,f1):E′​(T,S)→∼Rξ−​(T,S)(f_{2},f_{1}):E^{\prime}(T,S)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}R_{\xi^{-}}(T,S)

functorial in T∈𝒮M​(Z)T\in{\mathcal{S}}_{M}(Z) and S∈𝒮M​(Zξ)S\in{\mathcal{S}}_{M}(Z_{\xi}).

0PDZ

Proof. It is immediate that d⁡(f1)=0d(f_{1})=0. For the second equality, consider α∈Rξ2−∙​(T,U)\alpha\in R^{\bullet}_{\xi_{2}^{-}}(T,U) and β∈Hom𝒮∙​(Zξ)⁡(S,U)\beta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z_{\xi})}(S,U). Since α⊗β=(id⊠αξ2−​(−1))⊗(α|U⋅β)\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|U}\cdot\beta), we can assume that α|U=id\alpha_{|U}=\operatorname{id}\nolimits. We have

d(f2)(α⊗β)=α⋅(d(β⊠[ξ1−(−1)→ξ2−(−1)])+d(β)⊠[ξ1−(−1)→ξ2−(−1)]).d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\bigl(d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]\bigr).

We have

d(β⊠[ξ1−(−1)→ξ2−(−1)])\displaystyle d(\beta\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]) =d((id⊠[ξ1−(−1)→ξ2−(−1)])⋅(β⊠idξ1−​(−1)))\displaystyle=d\bigl((\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})\bigr)
=(id⊠[ξ1+(1)→ξ2−(−1)])⋅d(idU⊠[ξ1−(−1)→ξ1+(1)])⋅(β⊠idξ1−​(−1))\displaystyle=(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\cdot d(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{-}(-1)\to\xi_{1}^{+}(1)])\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})
+(id⊠[ξ1−(−1)→ξ2−(−1)])⋅(d(β)⊠idξ1−​(−1))\displaystyle\ \ \ \ +(\operatorname{id}\nolimits\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)])\cdot(d(\beta)\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})
=(idU⊠[ξ1+(1)→ξ2−(−1)])∑x∈ξ~1−1​(U)(id⊠ξ~1([x→1])⊠ξ~1([−1→x]))\displaystyle=(\operatorname{id}\nolimits_{U}\boxtimes[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)])\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([x\to 1])\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)
⋅(β⊠idξ1−​(−1))+d(β)⊠[ξ1−(−1)→ξ2−(−1)]\displaystyle\ \ \ \ \cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})+d(\beta)\boxtimes[\xi_{1}^{-}(-1)\to\xi_{2}^{-}(-1)]

hence

d(f2)(α⊗β)=α⋅∑x∈ξ~1−1​(U)(id⊠([ξ1+(1)→ξ2−(−1)]⋅ξ~1([x→1]))⊠ξ~1([−1→x]))⋅(β⊠idξ1−​(−1))d(f_{2})(\alpha\otimes\beta)=\alpha\cdot\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\bigl([\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\boxtimes\tilde{\xi}_{1}([-1\to x])\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})
=∑x∈ξ~1−1​(U)(id⊠ξ~1([−1→x])⊠(αξ2−​(−1)⋅[ξ1+(1)→ξ2−(−1)]⋅ξ~1([x→1])))⋅(β⊠idξ1−​(−1))\displaystyle=\sum_{x\in\tilde{\xi}_{1}^{-1}(U)}\Bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\boxtimes\bigl(\alpha_{\xi_{2}^{-}(-1)}\cdot[\xi_{1}^{+}(1)\to\xi_{2}^{-}(-1)]\cdot\tilde{\xi}_{1}([x\to 1])\bigr)\Bigr)\cdot(\beta\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})
=f1∘u⁡(α⊗β).\displaystyle=f_{1}\circ u(\alpha\otimes\beta).

The lemma follows. ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2