Subsection 8.1.8. (0PI4) · ScalingStacks

8.1.8. Action as functors

We explain here how the $2$ -representation constructed in §8.1.1 can be described using functors between strand categories of different curves.

Let $Z$ be a singular curve and $\xi:{\mathbf{R}}_{>0}\to Z$ an injective morphism of curves with $\xi({\mathbf{R}}_{\geq 1})$ closed and contained in $Z\setminus Z_{exc}$ .

Let $A=\bigoplus_{I,J\subset Z_{exc}}\operatorname{Hom}\nolimits_{{\mathcal{A}}(Z)^{{\operatorname{opp}\nolimits}}}(J,I)$ . We denote by $e_{I}\in A$ the idempotent corresponding to the projection on $I$ , so that $e_{I}Ae_{J}=\operatorname{Hom}\nolimits_{{\mathcal{A}}(Z)}(I,J)$ .

The equivalence $A\operatorname{\!-diff}\nolimits\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}{\mathcal{A}}(Z)^{{\operatorname{opp}\nolimits}}\operatorname{\!-diff}\nolimits$ restricts to an equivalence $(\bar{A})^{i}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\bar{{\mathcal{A}}}^{i}(Z)$ (cf §2.1.4).

We consider a new singular curve $\hat{Z}=Z\sqcup_{\xi(1)}(-1,1)$ obtained as the quotient of the disjoint union of $Z$ and the oriented interval $(-1,1)$ identifying $\xi(1)$ with $0$ . Note that $\hat{Z}_{exc}=Z_{exc}\cup\{\xi(1)\}$ .

We put $\hat{A}=\bigoplus_{I,J\subset\hat{Z}_{exc}}\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z})^{{\operatorname{opp}\nolimits}}}(J,I)$ . As before, we have idempotents $e_{I}\in\hat{A}$ for $I\in\hat{Z}_{exc}$ . We put $e=\sum_{I\subset Z_{exc}}e_{I\sqcup\{\xi(1)\}}$ .

The inclusion $i:Z\hookrightarrow\hat{Z}$ provides a fully faithful functor $\Xi:\bar{{\mathcal{A}}}^{i}(Z)\to\bar{{\mathcal{A}}}^{i}(\hat{Z})$ . This gives rise to an isomorphism of algebras $h:A\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}(1-e)\hat{A}(1-e)$ and we have a commutative diagram

where $A$ acts on the right on $\hat{A}(1-e)$ by right multiplication preceded by $h$ .

The inclusion $(1-e)\hat{A}(1-e)\hookrightarrow\hat{A}$ induces a surjective morphism of algebras $g:(1-e)\hat{A}(1-e)\twoheadrightarrow\hat{A}/\hat{A}e\hat{A}$ . We have $e\hat{A}(1-e)=0$ , hence $\hat{A}e\hat{A}\cap(1-e)\hat{A}(1-e)=0$ . It follows that $g$ is an isomorphism.

The right adjoint to $\hat{A}(1-e)\otimes_{A}-$ is $\operatorname{Hom}\nolimits_{\hat{A}}(\hat{A}(1-e),-)$ , which is canonically isomorphic to $(1-e)\hat{A}\otimes_{\hat{A}}-$ and we have a commutative diagram

where $\Gamma:\bar{{\mathcal{A}}}^{i}(\hat{Z})\to\bar{{\mathcal{A}}}^{i}(Z)$ is the right adjoint of $\Xi$ .

0PCU

Remark 8.1.21. There is a sequence of four adjoint functors between $A$ -modules and $\hat{A}$ -modules:

\bigl(A\otimes_{\hat{A}}-,\hat{A}(1-e)\otimes_{A}-,(1-e)\hat{A}\otimes_{\hat{A}}-,\operatorname{Hom}\nolimits_{A}((1-e)\hat{A},-)\bigr).

The first and fourth functors are not exact in general. Here,

•

$\hat{A}$ acts on the right on $A$ by right multiplication preceded by the composition

$\hat{A}\xrightarrow{{\mathrm{can}}}\hat{A}/\hat{A}e\hat{A}\xrightarrow[\sim]{g^{-1}}(1-e)\hat{A}(1-e)\xrightarrow[\sim]{h^{-1}}A$
•

$\bigl(\hat{A}(1-e)\otimes_{A}-\bigr)=\bigl((1-e)\hat{A}(1-e)\otimes_{A}-\bigr)\xrightarrow[\sim]{h^{-1}}\bigl(A\otimes_{A}-\bigr)=\operatorname{Hom}\nolimits_{A}(A,-)$
•

$\bigl((1-e)\hat{A}\otimes_{\hat{A}}-\bigr)\xrightarrow[\sim]{{\mathrm{can}}}\bigl(\operatorname{Hom}\nolimits_{\hat{A}}(\hat{A}(1-e),\hat{A})\otimes_{\hat{A}}-\bigr)\xrightarrow[\sim]{{\mathrm{can}}}\operatorname{Hom}\nolimits_{\hat{A}}(\hat{A}(1-e),-)$ .

There is also a fully faithful functor

\Upsilon:\bar{{\mathcal{A}}}^{i}(Z)\to\bar{{\mathcal{A}}}^{i}(\hat{Z}),\ T\mapsto T\sqcup\{\xi(1)\}

sending a braid $(\theta_{t})_{t\in T}$ to $(\theta_{t})_{t\in T}\sqcup(\mathrm{id}_{\xi(1)\}})$ . It gives rise to an isomorphism of algebras $u:A\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}e\hat{A}e$ and there is a commutative diagram

where the right action of $A$ on $\hat{A}e$ is by right multiplication preceded by $u$ .

We have

L(T,S)=\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z})}(\Xi(S),\Upsilon(T))\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\mathrm{Hom}_{{\mathcal{A}}(Z)}(S,\Gamma\Upsilon(T)).

Denote by $E=L\otimes_{\bar{{\mathcal{A}}}^{i}(Z)}-$ the endofunctor of $\bar{{\mathcal{A}}}^{i}(Z)$ induced by the bimodule $L$ . The isomorphism above gives rise to an isomorphism of functors $E\xrightarrow{\sim}\Gamma\Upsilon$ .

We put $\hat{Z}_{0}=Z$ and we define inductively $\hat{Z}_{r}=\hat{Z}_{r-1}\sqcup_{\xi(r)}(-1,1)$ for $r\geq 1$ , where $\xi(r)$ is identified with $0$ .

We denote by $\Xi_{r}:\bar{{\mathcal{A}}}^{i}(\hat{Z}_{r-1})\to\bar{{\mathcal{A}}}^{i}(\hat{Z}_{r}),\ I\mapsto I$ the functor associated with the inclusion $\hat{Z}_{r-1}\hookrightarrow\hat{Z}_{r}$ , defined as $\Xi$ above.

We denote by $\Upsilon_{r}:\bar{{\mathcal{A}}}^{i}(\hat{Z}_{r})\to\bar{{\mathcal{A}}}^{i}(\hat{Z}_{r}),\ T\mapsto T\sqcup\{\xi(r)\}$ the functor $\Upsilon$ for $Z$ replaced by $\hat{Z}_{r-1}$ .

Composition with $\operatorname{id}\nolimits_{T}\sqcup\{[\xi(1)\to\xi(r)]\}$ gives an isomorphism

L(T,S)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{r})}(\Xi_{r}\cdots\Xi_{1}(S),\Upsilon_{r}\Xi_{r-1}\cdots\Xi_{1}(T))=\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{r})}(S,T\sqcup\{\xi(r)\})

for $r\geq 1$ . Similarly, we have an isomorphism

f=(\operatorname{id}\nolimits_{T}\sqcup\{[\xi(1)\to\xi(2)],[\xi(2)\to\xi(3)]\})\circ-

L(T,S,2)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{3})}(\Xi_{3}\Xi_{2}\Xi_{1}(S),\Upsilon_{3}\Upsilon_{2}\Xi_{1}(T))=\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{3})}(S,T\sqcup\{\xi(2),\xi(3)\}).

We consider the morphism

g=(\operatorname{id}\nolimits_{T\sqcup\{\xi(2)\}}\sqcup\{[\xi(1)\to\xi(3)]\})\circ-

L(T,S,2)\to\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{3})}(\Xi_{3}\Xi_{2}\Xi_{1}(S),\Upsilon_{3}\Upsilon_{2}\Xi_{1}(T))=\operatorname{Hom}\nolimits_{{\mathcal{A}}(\hat{Z}_{3})}(S,T\sqcup\{\xi(2),\xi(3)\}).

The composition $f^{-1}\circ g$ is the endomorphism $\tau$ of $L(T,S,2)$ .

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2