ScalingStacks

Consider two injective morphisms of curves $\xi_{1}^{+}:{\mathbf{R}}_{>0}\to Z$ and $\xi_{2}^{-}:{\mathbf{R}}_{<0}\to Z$ where ${\mathbf{R}}_{<0}$ and ${\mathbf{R}}_{>0}$ are unoriented. We assume that $\xi_{1}^{+}$ is outgoing for $Z$, that $\xi_{2}^{-}$ is incoming for $Z$ and that $\xi_{1}^{+}({\mathbf{R}}_{>0})\cap\xi_{2}^{-}({\mathbf{R}}_{<0})=\emptyset$. We write $r$ instead of $\xi_{1}^{+}(r)$ and $-r$ instead of $\xi_{2}^{-}(-r)$, for $r\in{\mathbf{Z}}_{>0}$.

Let $M$ be a subset of $Z\setminus(\xi_{1}^{+}({\mathbf{R}}_{\geq 1})\sqcup\xi_{2}^{-}({\mathbf{R}}_{\leq-1}))$.

Fix an oriented diffeomorphism ${\mathbf{R}}_{>0}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathbf{R}}_{<-1}$ and let $i_{+}:{\mathbf{R}}_{>0}\to{\mathbf{R}}$ be its composition with the inclusion map. Similarly, fix an oriented diffeomorphism ${\mathbf{R}}_{<0}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathbf{R}}_{>1}$ and let $i_{-}:{\mathbf{R}}_{<0}\to{\mathbf{R}}$ be its composition with the inclusion map.

Consider $m,n\geq 0$. Let $E_{m,n}$ be the $({\mathcal{S}}_{M}^{\bullet}(Z),{\mathcal{S}}_{M}^{\bullet}(Z))$-bimodule given by

Note that $E_{0,1}=R^{\bullet}_{\xi_{2}^{-}}$ and $E_{1,0}=L^{\bullet}_{\xi_{1}^{+}}$, but $E_{m,n}$ is not isomorphic to $(R^{\bullet}_{\xi_{2}^{-}})^{n}(L^{\bullet}_{\xi_{1}^{+}})^{m}$ in general.

There is an action of $H_{m}^{\bullet}\wedge H_{n}^{\bullet}$ on $E_{m,n}$ given by

for $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S\sqcup(-n,-1),T\sqcup(1,m))$, $a\in{\mathfrak{S}}_{m}$ and $b\in{\mathfrak{S}}_{n}$.

There is a map $\ast:E_{m,n}E_{m^{\prime},n^{\prime}}\to E_{m+m^{\prime},n+n^{\prime}}$ given by

This map is compatible with the action of $(H_{m}^{\bullet}\wedge H_{n}^{\bullet})\wedge(H_{m^{\prime}}^{\bullet}\wedge H_{n^{\prime}}^{\bullet})$ via the canonical embeddings $H_{m}^{\bullet}H_{m^{\prime}}^{\bullet}\to H_{m+m^{\prime}}^{\bullet}$ and $H_{n}^{\bullet}H_{n^{\prime}}^{\bullet}\to H_{n+n^{\prime}}^{\bullet}$. We have $(\alpha\ast\beta)\ast\gamma=\alpha\ast(\beta\ast\gamma)$.

So, we have defined a bimodule lax bi-$2$-representation on ${\mathcal{S}}_{M}^{\bullet}(Z)$.

Let $Z_{\xi}=Z\sqcup_{{\mathbf{R}}_{>0}\sqcup{\mathbf{R}}_{<0}}{\mathbf{R}}$, where the gluing is done along the maps $\xi_{1}^{+}\sqcup\xi_{2}^{-}:{\mathbf{R}}_{>0}\sqcup{\mathbf{R}}_{<0}\to Z$ and $i_{+}\sqcup i_{-}:{\mathbf{R}}_{>0}\sqcup{\mathbf{R}}_{<0}\to{\mathbf{R}}$. Note that $Z_{\xi}$ is a $1$-dimensional space and it comes with an injective open morphism of $1$-dimensional spaces $\xi:{\mathbf{R}}\to Z_{\xi}$. We endow ${\mathbf{R}}$ with a curve structure by setting ${\mathbf{R}}_{u}={\mathbf{R}}_{\leq-1}\sqcup{\mathbf{R}}_{\geq 1}$ and by endowing $(-1,1)$ with its usual orientation. We extend the curve structure on $Z$ by endowing $\xi({\mathbf{R}})$ with the curve structure of ${\mathbf{R}}$. Note that $(Z_{\xi})_{u}=\overline{Z_{u}}$.

Given $\varepsilon,\varepsilon^{\prime}\in\{+,-\}$ and $a\in{\mathbf{R}}_{\varepsilon}$, $b\in{\mathbf{R}}_{\varepsilon^{\prime}}$, we put $[a\to b]=\xi([i_{\varepsilon}(a),i_{\varepsilon^{\prime}}(b)])$.

We consider the differential pointed category $T_{{\mathcal{S}}^{\bullet}_{M}(Z)}(R_{\xi_{2}^{-}}^{\bullet}L_{\xi_{1}^{+}}^{\bullet})$ with objects those of ${\mathcal{S}}^{\bullet}_{M}(Z)$ and with

We define a differential pointed functor $\tilde{\Xi}:T_{{\mathcal{S}}^{\bullet}_{M}(Z)}(R_{\xi_{2}^{-}}^{\bullet}L_{\xi_{1}^{+}}^{\bullet})\to{\mathcal{S}}^{\bullet}_{M}(Z_{\xi})$. It is the identity on objects and defined on maps by

The sections §8.2.2-8.2.4 below are devoted to the proof of Theorem 8.2.1.

If $\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z_{\xi})^{\bullet}}(\{-1\},\{1\})\neq 0$, then there is $\kappa^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z_{\xi})^{\bullet}}(\{-1\},\{1\})$ such that $\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z_{\xi})^{\bullet}}(\{-1\},\{1\})=\{\kappa^{n}\cdot\kappa^{\prime}\}_{n\geq 0}$, where $\kappa=\kappa^{\prime}\cdot[1\to-1]$.

When $\operatorname{Hom}\nolimits_{{\mathcal{S}}(Z_{\xi})^{\bullet}}(\{-1\},\{1\})=0$, we put $\kappa=\operatorname{id}\nolimits_{1}$.

We define a partial order on the component $Z^{\prime}$ of $Z$ containing $1$. We define $s<s^{\prime}$ if there exists an admissible path $\zeta:s^{\prime}\to 1$ in $Z^{\prime}$ whose support does not contain $s$.

We consider the map $\mu$ of §7.4.6 for the curve $Z_{\xi}$ and its point $z_{0}=0$.

Given $n\geq 0$, we put $G_{n}=E_{n,n}$.

There is a map $\nu_{n}:R_{\xi_{2}^{-}}^{\bullet}(-,-,e^{n})L^{\bullet}_{\xi_{1}^{+}}(-,-,e^{n})\to G_{n}$ given by

We have

for $a,b\in{\mathfrak{S}}_{n}$.

The multiplication map on $E$ defines a map $\mu_{n}:(R_{\xi_{2}^{-}}^{\bullet}L^{\bullet}_{\xi_{1}^{+}})^{n}=(E_{0,1}E_{1,0})^{n}\to E_{n,n}=G_{n}$, hence gives a morphism $T^{*}(R_{\xi_{2}^{-}}^{\bullet}L^{\bullet}_{\xi_{1}^{+}})\to G=\bigvee_{n\geq 0}G_{n}$ compatible with multiplication.

We define $({\mathcal{S}}_{M}^{\bullet}(Z),{\mathcal{S}}_{M}^{\bullet}(Z))$-subbimodules $A_{n}$, $B_{n}$, $C_{n}$, $D_{n}$, $E_{n}$ and $F_{n}$ of $G_{n}$. Let $\sigma\in G_{n}(T,S)$.

We have

• •

  $\sigma\in A_{n}$ if $\sigma(-i)\in T\sqcup(1,n-i)$ for $1\leq i\leq n$

• •

  $\sigma\in B_{n}$ if there exists $1\leq j\leq i\leq n$ with $\sigma(-i)=n-j+1$

• •

  $\sigma\in C_{n}$ if it is in the image of $\mu_{n}$

• •

  $\sigma\in D_{n}$ if $\sigma(-i)\in T$ for $1\leq i\leq n$

• •

  $\sigma\in E_{n}$ if $\sigma\in A_{n}$ and $L(\sigma_{|(-n,-1)})=\emptyset$.

• •

  $\sigma\in F_{n}$ if $\sigma\in A_{n}$ and $L(\sigma_{|\sigma^{-1}(1,n)})=\emptyset$.

We put $A=\bigvee_{n\geq 0}A_{n}$, $B=\bigvee_{n\geq 0}B_{n}$, etc.

Note that $G_{n}=A_{n}\vee B_{n}$.

We have $C_{n},D_{n},E_{n},F_{n}\subset A_{n}$.

We define a morphism of $({\mathcal{S}}_{M}^{\bullet}(Z),{\mathcal{S}}_{M}^{\bullet}(Z))$-bimodules $q:G\to\operatorname{Id}\nolimits_{{\mathcal{S}}_{M}^{\bullet}(Z_{\xi})}$:

Let $\alpha\in A_{n}$. We put $T_{1}=\alpha(S)\cap T$ and $I_{1}=\alpha(S)\cap(1,n)$. We define inductively $T_{m}\subset T$ and $I_{m}\subset(1,n-m+1)$ for $1<m\leq n+1$ by $T_{m}=T_{m-1}\sqcup(\alpha(-n+I_{m-1}-1)\cap T)$ and $I_{m}=\alpha(-n+I_{m-1}-1)\cap(1,n)$.

Note that $-n+I_{m-1}-1\subset(-n,-m+1)$, hence $I_{m}\subset(1,n-m+1)$ since $\alpha\in A_{n}$.

Note that $T_{n+1}=T$ and $I_{n+1}=\emptyset$.

Define

for $1\leq m\leq n$. We define $q(\alpha)=\beta^{n}\cdot\beta^{n-1}\cdots\beta^{1}\cdot\alpha_{|S}$

We put $q(\alpha)=0$ if $\alpha\in B_{n}$.

Assume now $q(\alpha)\neq 0$, hence $\alpha\in A_{n}$. Let $S^{\prime}=S\cap\alpha^{-1}(T)$ and $T^{\prime}=T\cap\alpha(S)$. Let $S^{\prime\prime}=S-S^{\prime}$ and $T^{\prime\prime}=T-T^{\prime}$.

Given $s\in S^{\prime}$, we have $q(\alpha)_{s}=\alpha_{s}$.

Note in particular that $S^{\prime}=\{s\in S\ |\ \mu(q(\alpha)_{s})=0\}$.

Let $s\in S^{\prime\prime}$, $t=q(\alpha)(s)$ and $i=\alpha^{-1}(t)$. Put $d_{s}=\mu(q(\alpha)_{s})-1\geq 0$. We have

Given a decomposition $q(\alpha)_{s}=\xi\cdot[1\to-1]\cdot\kappa^{d_{s}}\cdot\xi^{\prime}$ with $\xi^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(\{s\},\{1\})$ and $\xi\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(\{-1\},\{t\})$, we have $\alpha_{i}=\xi\cdot[i\to-1]$ and $\alpha_{s}=[1\to\alpha(s)]\cdot\xi^{\prime}$.

The next lemma is immediate.