Subsection 7.4.5. (0PHM) · ScalingStacks

7.4.5. Generation

We equip $Z$ with a metric. Given $\xi$ a path in $Z$ , we denote by $|\xi|$ its length. Given $\zeta$ a homotopy class of paths in $Z$ , we put $|\zeta|=|\xi|$ , where $\xi$ is a minimal path in $\zeta$ . Given $\theta:I\to J$ a braid in $Z$ , we put $|\theta|=\sum_{s\in I}|\theta_{s}|$ .

Let $M$ be a finite subset of $Z$ .

0PBJ

Lemma 7.4.25. Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I,J)$ and $\theta^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I^{\prime},I)$ such that $\theta\circ\theta^{\prime}$ is a braid. Let $I_{0}$ be a finite subset of $M\setminus(I\cup I^{\prime}\cup J)$ .

If $|\theta|+|\theta^{\prime}|=|\theta\circ\theta^{\prime}|$ , then $(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})\cdot(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=(\theta\cdot\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}}$ .

0PBK

Proof. Let $s^{\prime}\in I^{\prime}$ . Since $|\theta_{\theta^{\prime}(s^{\prime})}|+|\theta^{\prime}_{s^{\prime}}|=|\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}}|$ , it follows that $i(\theta_{\theta^{\prime}(s^{\prime})},\operatorname{id}\nolimits_{i})+i(\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})=i(\theta_{\theta^{\prime}(s^{\prime})}\circ\theta^{\prime}_{s^{\prime}},\operatorname{id}\nolimits_{i})$ for all $i\in I_{0}$ . As a consequence,

i((\theta\circ\theta^{\prime})\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta\boxtimes\operatorname{id}\nolimits_{I_{0}})-i(\theta^{\prime}\boxtimes\operatorname{id}\nolimits_{I_{0}})=i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime}).

The lemma follows now from Lemma 7.4.9. ∎

Let $\theta\in\operatorname{Hom}\nolimits_{{\mathcal{S}}^{\bullet}_{M}(Z)}(I,J)$ be a non-zero braid.

Let $I_{0}=\{i\in I\ |\ \theta_{i}=\operatorname{id}\nolimits_{i}\}$ . Let $i\in I\setminus I_{0}$ . There is a (unique) decomposition $\theta_{i}=\beta^{i}\cdot\alpha^{i}$ in ${\mathcal{S}}^{\bullet}(Z,1)$ with

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$\alpha^{i}(1)\in M\setminus I_{0}$
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$|\theta_{i}|=|\alpha^{i}|+|\beta^{i}|$
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given a minimal path $\xi$ in $\alpha^{i}$ , we have $\xi((0,1))\cap M\subset I_{0}$ .

We define a quiver $\Gamma(\theta)$ with vertex set $I\setminus I_{0}$ . There is an arrow $i\to i^{\prime}$ if $\alpha^{i}(1)=i^{\prime}$ .

Note that there is at most one arrow with a given source (that arrow can be a loop).

0PBL

Lemma 7.4.26. Let $I^{\prime}$ be a non-empty finite subset of $I\setminus I_{0}$ such that

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if there is an arrow $i\to i^{\prime}$ in $\Gamma(\theta)$ with $i\in I^{\prime}$ , then $i^{\prime}\in I^{\prime}$
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given $i\neq i^{\prime}\in I^{\prime}$ , we have $\alpha^{i}(1)\neq\alpha^{i^{\prime}}(1)$ .

There is a (unique) decomposition $\theta=\theta^{u}\cdot u$ in ${\mathcal{S}}_{M}^{\bullet}(Z)$ , where $|\theta|=|\theta^{u}|+|u|$ and

u_{i}=\begin{cases}\alpha^{i}&\text{ if }i\in I^{\prime}\\ \operatorname{id}\nolimits_{i}&\text{ otherwise.}\end{cases}

0PBM

Proof. Note that the second assumption on $I^{\prime}$ shows that the full subquiver of $\Gamma(\theta)$ with vertex set $I^{\prime}$ is a disjoint union of oriented lines and oriented circles.

Let $i_{1}\neq i_{2}\in I$ . If $i_{1},i_{2}\in I\setminus I^{\prime}$ , then $u(i_{1})\neq u(i_{2})$ . Assume now $i_{1}\in I^{\prime}$ and $i_{2}\in I\setminus I^{\prime}$ . Since $i_{1}\to i_{2}$ is not an arrow of the quiver, we have $u(i_{1})\neq i_{2}$ , hence $u(i_{1})\neq u(i_{2})$ . Finally if $i_{1},i_{2}\in I^{\prime}$ , then $u(i_{1})\neq u(i_{2})$ . We have shown that $u$ is a braid.

Note that there is a (unique) decomposition $\theta=\theta^{u}\circ u$ with $|\theta|=|\theta^{u}|+|u|$ . In order to show that $\theta^{u}\cdot u\neq 0$ , we can replace $\theta$ by $\theta_{|I\setminus I_{0}}$ and $M$ by $M\setminus I_{0}$ , thanks to Lemma 7.4.25. So, we assume now that $I_{0}=\emptyset$ .

Let $q:\tilde{Z}\to Z$ be a non-singular cover of $Z$ . Let $\tilde{M}=q^{-1}(M)$ . Let $\tilde{\theta}:\tilde{I}\to\tilde{J}$ be the unique lift of $\theta$ to $\tilde{Z}$ . We have a decomposition $\tilde{\theta}_{i}=\tilde{\beta}^{i}\cdot\tilde{\alpha}^{i}$ for $i\in\tilde{I}$ and $q(\tilde{\alpha}^{i})=\alpha^{q(i)}$ .

Let $\tilde{I}^{\prime}=q^{-1}(I^{\prime})\cap\tilde{I}$ . Note that $q$ induces a morphism of quivers $\Gamma(\tilde{\theta})\to\Gamma(\theta)$ , hence $\tilde{I}^{\prime}$ satisfies the assumptions of the lemma and we have a decomposition $\tilde{\theta}=\tilde{\theta}^{\tilde{u}}\circ\tilde{u}$ . Since $q(\tilde{u})=u$ , it follows that if the lemma holds for $\tilde{\theta}$ , then it holds for $\theta$ .

We assume now that $Z$ is non-singular. If the lemma holds for connected components of $Z$ , it will hold for $Z$ , hence it is enough to prove the lemma for $Z$ connected. Assume now $Z$ is connected. There is an injective morphism of curves $f:Z\to S^{1}$ , where $S^{1}$ is unoriented. It the lemma holds for $S^{1}$ , it holds for $Z$ .

We assume finally that $Z=S^{1}$ unoriented. Let $i_{1}\neq i_{2}\in I^{\prime}$ such that $i(u_{i_{1}},u_{i_{2}})\neq 0$ . Note that $u_{i_{1}}$ and $u_{i_{2}}$ have opposite directions and $i(u_{i_{1}},u_{i_{2}})=1$ . Furthermore, $\theta_{i_{r}}$ has the same direction as $u_{i_{r}}$ , hence $i(\theta_{i_{1}},\theta_{i_{2}})=i((\theta^{u})_{i_{1}},(\theta^{u})_{i_{2}})+1$ . Given $i_{1}\neq i_{2}\in I$ with $i_{1}{\not\in}I^{\prime}$ , we have $i(u_{i_{1}},u_{i_{2}})=0$ . It follows from Remark 7.4.11 that $\theta^{u}\cdot u\neq 0$ . This completes the proof of the lemma. ∎

Note that the length of a map in ${\mathcal{S}}^{\bullet}_{M}(Z)$ takes value in a finitely generated submonoid of ${\mathbf{R}}_{\geq 0}$ . So, a repeated application of the previous lemma provides a decomposition of any map $\theta$ of ${\mathcal{S}}_{M}^{\bullet}(Z)$ as a product $\theta=u_{n}\cdots u_{1}$ , where $u_{i}$ is a map $u$ as in the lemma.

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2