Subsection 8.2.2. (0PI8) · ScalingStacks

8.2.2. Bimodules

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 $G_{n}=E_{n,n}$ .

0PCZ

Lemma 8.2.4. Let $\alpha\in G_{n}-\{0\}$ .

Given $i\in(1,n-1)$ , the following assertions are equivalent

(1)

$\alpha(i-n-1)>\alpha(i-n)$
(2)

$L(\alpha_{|\{i-n-1,i-n\}})\neq\emptyset$
(3)

$[i-n-1\to i-n]\in D(\alpha)$
(4)

$\alpha\in G_{n}T_{i}$
(5)

$\alpha T_{i}=0$ .

There exists $i\in(1,n-1)$ such that $\alpha\in G_{n}T_{i}$ if and only if $L(\alpha_{|(-n,-1)})\neq\emptyset$ .

0PD0

Proof. The equivalence between (1) and (2) follows from Lemma 7.4.20.

Assume (2). We deduce that $[i-n-1\to i-n]\in L(\alpha)$ , hence $[i-n-1\to i-n]\in D(\alpha)$ . So (3) holds.

Assume (3). Writing $\alpha=\alpha\cdot 1$ , we deduce from Lemma 7.4.35 that (4) holds.

The implication (4) $\Rightarrow$ (5) is immediate.

Asssume (5). We have $\alpha_{|\{i-n-1,i-n\}}\cdot([i-n-1\to i-n]\boxtimes[i-n\to i-n-1])=0$ by Remark 7.4.11. Lemma 7.4.9 shows that $i(\alpha_{|\{i-n-1,i-n\}})\neq 0$ , hence (2) holds.

Assume now $L(\alpha_{|(-n,-1)})\neq\emptyset$ . It follows from Lemma 7.4.20 that there is $i\in(1,n-1)$ with $\alpha(i-n-1)>\alpha(i-n)$ , hence $\alpha\in G_{n}T_{i}$ . This shows the last statement of the lemma. ∎

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

\operatorname{Hom}\nolimits(-\sqcup(-n,-1),T)\wedge\operatorname{Hom}\nolimits(S,-\sqcup(1,n))\to\operatorname{Hom}\nolimits(S\sqcup(-n,-1),T\sqcup(1,n))

\beta\wedge\alpha\mapsto(\beta\boxtimes\operatorname{id}\nolimits_{(1,n)})\cdot(\alpha\boxtimes\operatorname{id}\nolimits_{(-n,-1)})

We have

\nu_{n}\bigl((\beta\cdot T_{b})\wedge(T_{a}\cdot\alpha)\bigr)=T_{a}\cdot\nu_{n}(\beta\wedge\alpha)\cdot\iota_{n}(T_{b})

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

0PD1

Lemma 8.2.5. We have an isomorphism $\nu_{n}:R_{\xi_{2}^{-}}^{\bullet}(-,-,e^{n})L^{\bullet}_{\xi_{1}^{+}}(-,-,e^{n})\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D_{n}$ .

In particular, we have an isomorphism $\nu_{1}:R_{\xi_{2}^{-}}^{\bullet}L^{\bullet}_{\xi_{1}^{+}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}D_{1}=A_{1}=C_{1}$ and $C_{n}=C_{1}^{\ast n}=A_{1}^{\ast n}$ .

0PD2

Proof. Let $\beta\wedge\alpha\in\operatorname{Hom}\nolimits(-\sqcup(-n,-1),T)\wedge\operatorname{Hom}\nolimits(S,-\sqcup(1,n))$ . We have $\beta\wedge\alpha=\beta^{\prime}\wedge\alpha^{\prime}$ where $\beta^{\prime}=\beta_{|(-n,-1)}\boxtimes\operatorname{id}\nolimits$ and $\alpha^{\prime}=(\beta_{|-}\boxtimes\operatorname{id}\nolimits_{(1,n)})\cdot\alpha$ . If $\nu_{n}(\beta^{\prime}\wedge\alpha^{\prime})=0$ , then $\beta^{\prime}=\alpha^{\prime}=0$ (cf the beginning of §7.4.10). Now $\nu_{n}$ has an inverse given by $\sigma\mapsto(\operatorname{id}\nolimits\boxtimes\sigma_{|(-n,-1)})\wedge\sigma_{|S}$ . ∎

0PD3

Remark 8.2.6. Consider $\bar{\xi}_{1}^{+}:{\mathbf{R}}_{>0}\to Z^{{\operatorname{opp}\nolimits}},\ x\mapsto\xi_{2}^{-}(-x)$ and $\bar{\xi}_{2}^{-}:{\mathbf{R}}_{<0}\to Z^{{\operatorname{opp}\nolimits}},\ x\mapsto\xi_{1}^{+}(-x)$ . There is an isomorphism $(Z^{\operatorname{opp}\nolimits})_{\bar{\xi}}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}(Z_{\xi})^{\operatorname{opp}\nolimits}$ that is the identity on $Z$ and $x\mapsto-x$ on ${\mathbf{R}}$ . This provides an isomorphism $({\mathcal{S}}^{\bullet}(Z_{\xi}))^{\operatorname{opp}\nolimits}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{S}}^{\bullet}(Z^{{\operatorname{opp}\nolimits}}_{\bar{\xi}})$ . It induces isomorphisms

\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z)}(S\sqcup(-n,-1),T\sqcup(1,n))\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}(Z^{\operatorname{opp}\nolimits})}(T\sqcup(-n,-1),S\sqcup(1,n)).

This restricts to isomorphisms between $A_{n}$ (resp. $B_{n}$ , $D_{n}$ , $E_{n}$ , $F_{n}$ ) for $Z$ and $A_{n}$ (resp. $B_{n}$ , $D_{n}$ , $F_{n}$ , $E_{n}$ ) for $Z^{\operatorname{opp}\nolimits}$ .

0PD4

•

Lemma 8.2.7. $B_{n}$ and $D_{n}$ are stable under the action of $H_{n}^{\bullet}\wedge(H_{n}^{\bullet})^{\operatorname{opp}\nolimits}$ .
•

$E_{n}$ is stable under the action of $H_{n}^{\bullet}$ and $F_{n}$ is stable under the action of $(H_{n}^{\bullet})^{\operatorname{opp}\nolimits}$ .
•

$A$ and $C$ are stable under multiplication
•

Given $\alpha\in B$ and $\beta\in G$ , we have $\alpha\ast\beta\in B$ and $\beta\ast\alpha\in B$ .

0PD5

Proof. Let $\sigma\in B_{n}$ and $r\in\{1,\ldots,n-1\}$ . Assume $\sigma T_{r}\neq 0$ .

If there is $1\leq j\leq i\leq n$ with $\sigma(-i)=n-j+1$ and $i\neq n+1-r$ , then $\sigma T_{r}\in B_{n}$ .

Assume now $\sigma(-i)\in T\sqcup(1,n-i)$ for all $i\neq n+1-r$ . We deduce that $L(\sigma_{|\{-(n+1-r),-(n-r)\}})\neq\emptyset$ , hence $\sigma T_{r}=0$ (cf Lemma 8.2.4), a contradiction.

Using Remark 8.2.6, we deduce that $T_{r}\sigma\in B_{n}$ .

The other assertions of the lemma are immediate. ∎

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2