Subsection 7.4.2. (0PHJ) · ScalingStacks

7.4.2. Degree

Consider $\theta:I\to J$ a braid. We put

i(\theta)=\frac{1}{2}\sum_{\Omega\in\pi_{0}(Z)}\sum_{\begin{subarray}{c}s\neq s^{\prime}\in I\cap\Omega\end{subarray}}i(\theta_{s},\theta_{s^{\prime}})e_{\Omega}\in({\mathbf{Z}}_{\geq 0})^{\pi_{0}(Z)}

We define $\llbracket\theta\rrbracket=\sum_{s\in I}\llbracket\theta_{s}\rrbracket\in R(Z)$ and

m(\theta)=\sum_{s\in I}\sum_{c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))}m_{c}(\llbracket\theta\rrbracket)e_{c}\in L(Z).

Finally, we define $\deg^{\prime}(\theta)\in\Gamma(Z)$ by

\deg^{\prime}(\theta)=(-i(\theta),(-m(\theta),-\llbracket\theta\rrbracket)).

Given $D\subset T(Z)$ with $D\cap\iota(D)=\emptyset$ , we denote by $\deg_{D}(\theta)$ the image of $\deg^{\prime}(\theta)$ in $\Gamma(Z,D)$ . Note that if $D^{\prime}\subset D$ , then $\deg_{D}(\theta)$ is the image of $\deg_{D^{\prime}}(\theta)\in\Gamma(Z,D^{\prime})$ in $\Gamma(Z,D)$ .

We put $\deg(\theta)=\deg_{Z_{exc}^{+}}(\theta)$ and we denote by $\overline{\deg}(\theta)$ (resp. $\overline{\deg}_{D}(\theta)$ ) the image of $\deg(\theta)$ (resp. $\deg_{D}(\theta)$ ) in $\bar{\Gamma}(Z,Z_{exc}^{+})$ (resp. $\bar{\Gamma}(Z,D)$ ).

0PAS

Lemma 7.4.7. Let $\theta:I\to J$ be a braid in $Z$ . Let $E$ be a subset of $\{s\in I\cap Z_{o}\ |\ \theta_{s}=\operatorname{id}\nolimits_{s}\}$ and let $\bar{\theta}=(\theta_{s})_{s\in I-E}$ . We have $\deg_{E^{+}}(\theta)=\deg_{E^{+}}(\bar{\theta})$ .

0PAT

Proof. Note that $\llbracket\theta\rrbracket=\llbracket\bar{\theta}\rrbracket$ . Let $s\in E$ . We have

\sum_{c\in C(s)}m_{c}(\llbracket\theta\rrbracket)e_{c}=\sum_{c\in C(s)^{+}}\sum_{s^{\prime}\in I,\ s^{\prime}\neq s}(m_{c}-m_{\iota(c)})(\llbracket\theta_{s^{\prime}}\rrbracket)e_{c}\xrightarrow{e_{c}\to 1}2\sum_{s^{\prime}\in I,\ s^{\prime}\neq s}i(\operatorname{id}\nolimits_{s},\theta_{s^{\prime}})

by Lemma 7.3.21. The lemma follows. ∎

0PAU

Remark 7.4.8. Note that $i(\theta)=\sum_{I^{\prime}\subset I,\ |I^{\prime}|=2}i(\theta_{|I^{\prime}})$ .

The next lemma shows that the failure of multiplicativity of $\deg$ and $i$ coincide up to terms involving points in $Z_{exc}$ .

0PAV

Lemma 7.4.9. Let $\theta:I\to J$ and $\theta^{\prime}:I^{\prime}\to I$ be two braids such that $\theta\circ\theta^{\prime}$ is a braid. The element $\deg(\theta)\cdot\deg(\theta^{\prime})\cdot\deg(\theta\circ\theta^{\prime})^{-1}$ of $\Gamma(Z,Z_{exc}^{+})$ is in $\bigoplus_{\Omega}\frac{1}{2}{\mathbf{Z}}e_{\Omega}$ and it is equal to

i(\theta\circ\theta^{\prime})-i(\theta)-i(\theta^{\prime})\\ +\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,\ s^{\prime}\in I^{\prime}\cap Z_{exc}\cap\Omega\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})^{+}\setminus\theta_{s^{\prime}}(0+)\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket)e_{\Omega}+\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,\ s^{\prime}\in I^{\prime}\cap\Omega\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits,\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))^{+}\setminus\iota(\theta^{\prime}_{s^{\prime}}(1-))\\ \theta^{\prime}_{s^{\prime}}(1)\in Z_{exc}\end{subarray}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{\Omega}.

and is also equal to

\frac{1}{2}\sum_{\begin{subarray}{c}\Omega,(s^{\prime}_{1},s^{\prime}_{2})\in(I^{\prime}\cap\Omega)^{2}\\ (s^{\prime}_{1},s^{\prime}_{2}){\not\in}E\cup E^{\prime}\end{subarray}}\bigl(i(\theta_{s_{1}}\circ\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})\bigr)e_{\Omega}+\\ +\sum_{\Omega,\ (s^{\prime}_{1},s^{\prime}_{2})\in E\cap\Omega}\bigl(i(\theta_{s_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}})-m_{\theta_{s_{1}}(0+)}^{+}(\theta^{\prime}_{s^{\prime}_{2}})\bigr)e_{\Omega}+\\ +\sum_{\Omega,\ (s^{\prime}_{1},s^{\prime}_{2})\in E^{\prime}\cap\Omega}\bigl(i(\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})-m_{\theta^{\prime}_{s^{\prime}_{1}}(1-)}^{-}(\theta_{s_{2}})\bigr)e_{\Omega}

where

•

given $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime 2}$ , we put $s_{i}=\theta^{\prime}_{s^{\prime}_{i}}(1)$
•

$E$ is the set of pairs $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime}\times I^{\prime}$ with $s^{\prime}_{1}\in\ Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}=\operatorname{id}\nolimits$ , $\theta_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ , $\theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits$
•

$E^{\prime}$ is the set of pairs $(s^{\prime}_{1},s^{\prime}_{2})\in I^{\prime}\times I^{\prime}$ with $s_{1}\in\ Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ , $\theta_{s_{1}}=\operatorname{id}\nolimits$ and $\theta_{s_{2}}\neq\operatorname{id}\nolimits$ .

0PAW

Proof. Given $s^{\prime}\in I^{\prime}$ and $s=\theta^{\prime}_{s^{\prime}}(1)$ , the class $\theta_{s}\circ\theta^{\prime}_{s^{\prime}}$ is admissible, hence $\theta_{s}(0+)\cup\iota(\theta_{s}(0+))=\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))$ unless $s\in Z_{exc}$ and one of $\theta_{s}$ and $\theta_{s^{\prime}}$ is the identity, but not the other.

Given $c\in T(Z)$ , we put

v_{c}=(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}=m_{c}(\llbracket\theta\rrbracket)e_{c}+m_{\iota(c)}(\llbracket\theta\rrbracket)e_{\iota(c)}=v_{\iota(c)}.

Let

a=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})\setminus\bigl((\theta_{s^{\prime}}(0+)\cup\iota(\theta_{s^{\prime}}(0+))\bigr)\end{subarray}}m_{c^{\prime}}(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\cap Z_{exc}\\ \theta^{\prime}_{s^{\prime}}=\operatorname{id}\nolimits,\ \theta_{s^{\prime}}\neq\operatorname{id}\nolimits\\ c^{\prime}\in C(s^{\prime})^{+}\setminus\theta_{s^{\prime}}(0+)\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket)e_{c^{\prime}}.

We have

m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=

=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in(\theta\circ\theta^{\prime})_{s^{\prime}}(0+)\cup\iota((\theta\circ\theta^{\prime})_{s^{\prime}}(0+))\end{subarray}}m_{c^{\prime}}(\llbracket\theta\rrbracket)e_{c^{\prime}}-\sum_{\begin{subarray}{c}s\in I\\ c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))\end{subarray}}m_{c}(\llbracket\theta\rrbracket)e_{c}-a

=\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))\end{subarray}}v_{c}-a

Using (7.3.1), we find

	$\displaystyle\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle$	$\displaystyle=-\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c^{\prime}\in\theta^{\prime}_{s^{\prime}}(0+)\cup\iota(\theta^{\prime}_{s^{\prime}}(0+))\end{subarray}}v_{c^{\prime}}+\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ c\in\theta^{\prime}_{s^{\prime}}(1-)\cup\iota(\theta^{\prime}_{s^{\prime}}(1-))\end{subarray}}v_{c}$
		$\displaystyle=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(0+)}+\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\end{subarray}}v_{\theta^{\prime}_{s^{\prime}}(1-)}.$

We deduce that

\langle\llbracket\theta\rrbracket,\llbracket\theta^{\prime}\rrbracket\rangle+m(\theta\circ\theta^{\prime})-m(\theta)-m(\theta^{\prime})=-\sum_{\begin{subarray}{c}s^{\prime}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}}\neq\operatorname{id}\nolimits\\ \theta_{\theta^{\prime}_{s^{\prime}}(1)}=\operatorname{id}\nolimits\\ c\in C(\theta^{\prime}_{s^{\prime}}(1))^{+}\setminus\iota(\theta^{\prime}_{s^{\prime}}(1-))\\ \theta^{\prime}_{s^{\prime}}(1)\in Z_{exc}\end{subarray}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket)e_{c}-a

and the first equality of the lemma follows.

Consider $s^{\prime}_{1}\neq s^{\prime}_{2}$ in $I^{\prime}$ .

If $s^{\prime}_{1}\in Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}=\operatorname{id}\nolimits_{s^{\prime}_{1}}$ and $\theta_{s_{1}}\neq\operatorname{id}\nolimits_{s_{1}}$ , it follows from Lemma 7.3.21 that

\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta^{\prime}_{s^{\prime}_{2}}\neq\operatorname{id}\nolimits_{s^{\prime}_{2}}\\ c^{\prime}\in C(s^{\prime}_{1})^{+}\end{subarray}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\theta^{\prime}_{s^{\prime}_{2}})=\frac{1}{2}\sum_{c^{\prime}\in C(s^{\prime}_{1})^{+}}(m_{c^{\prime}}-m_{\iota(c^{\prime})})(\llbracket\theta^{\prime}\rrbracket).

Similarly, if $s_{1}\in Z_{exc}$ , $\theta^{\prime}_{s^{\prime}_{1}}\neq\operatorname{id}\nolimits$ and $\theta_{s_{1}}=\operatorname{id}\nolimits$ , we have

\sum_{\begin{subarray}{c}s^{\prime}_{2}\in I^{\prime}\\ \theta_{s_{2}}\neq\operatorname{id}\nolimits\end{subarray}}i(\operatorname{id}\nolimits_{s_{1}},\theta_{s_{2}})=\frac{1}{2}\sum_{c\in C(s_{1})^{+}}(m_{c}-m_{\iota(c)})(\llbracket\theta\rrbracket).

The second equality of the lemma follows. ∎

0PAX

Example 7.4.10. The left (respectively second) side of the diagram below shows a typical instance where the left (respectively right) sum of Lemma 7.4.9 is nonzero.

0PAY

Remark 7.4.11. Let $\theta:I\to J$ and $\theta^{\prime}:I^{\prime}\to I$ be two braids such that $\theta\circ\theta^{\prime}$ is a braid. By Lemma 7.3.23, the terms $i(\theta_{s_{1}},\theta_{s_{2}})+i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}}\circ\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ , $i(\theta_{s_{1}},\theta_{s_{2}})+m_{\theta_{s_{1}}(0+)}^{+}(\theta^{\prime}_{s^{\prime}_{2}})-i(\theta_{s_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ and $i(\theta^{\prime}_{s^{\prime}_{1}},\theta^{\prime}_{s^{\prime}_{2}})+m_{\theta^{\prime}_{s^{\prime}_{1}}(1-)}^{-}(\theta_{s_{2}})-i(\theta^{\prime}_{s^{\prime}_{1}},\theta_{s_{2}}\circ\theta^{\prime}_{s^{\prime}_{2}})$ in Lemma 7.4.9 are all non-negative.

We deduce that the following assertions are equivalent:

•

$\deg(\theta)\cdot\deg(\theta^{\prime})=\deg(\theta\circ\theta^{\prime})$
•

$\deg(\theta_{|E})\cdot\deg(\theta^{\prime}_{|E^{\prime}})=\deg(\theta_{|E}\circ\theta^{\prime}_{|E^{\prime}})$ for any two-element subset $E^{\prime}\subset I^{\prime}$ , where $E=\chi(\theta^{\prime})(E^{\prime})$ .

If given $s\in I^{\prime}$ with $\theta^{\prime}_{s}=\operatorname{id}\nolimits$ or $\theta_{\chi(\theta^{\prime})(s)}=\operatorname{id}\nolimits$ , we have $s{\not\in}Z_{exc}$ , then $\deg(\theta)\cdot\deg(\theta^{\prime})=\deg(\theta\circ\theta^{\prime})$ if and only if $i(\theta_{\theta^{\prime}_{s}(1)},\theta_{\theta^{\prime}_{s^{\prime}}(1)})+i(\theta^{\prime}_{s},\theta^{\prime}_{s^{\prime}})=i((\theta\circ\theta^{\prime})_{s},(\theta\circ\theta^{\prime})_{s^{\prime}})$ for all $s\neq s^{\prime}$ in $I^{\prime}$ .

0PAZ

Lemma 7.4.12. Let $f:Z\to Z^{\prime}$ be a morphism of curves. Let $I$ and $J$ be two finite subsets of $Z$ such that $|f(I)|=|f(J)|=|I|=|J|$ . Let $\theta:I\to J$ be a braid in $Z$ . Let $E=\{s\in I\cap Z_{f}\ |\ \theta_{s}=\operatorname{id}\nolimits_{s}\}$ .

We have $f(\deg_{f^{-1}(f(E))^{+}}(\theta))=\deg_{f(E)^{+}}(f(\theta))$ .

0PB0

Proof. Assume first $E=\emptyset$ . Given $s\in I$ with $\theta_{s}=\operatorname{id}\nolimits_{s}$ , we have a bijection $C(s)\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}C(f(s))$ . It follows that

	$\displaystyle f(m(\theta))$	$\displaystyle=\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}\neq\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in\theta_{s}(0+)\cup\iota(\theta_{s}(0+))}m_{c}(\llbracket\theta\rrbracket)f(e_{c})+\sum_{\begin{subarray}{c}s\in I\\ \theta_{s}=\operatorname{id}\nolimits_{s}\end{subarray}}\sum_{c\in C(s)}m_{c}(\llbracket\theta\rrbracket)f(e_{c})$
		$\displaystyle=\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}\neq\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in f(\theta)_{s^{\prime}}(0+)\cup\iota(f(\theta)_{s^{\prime}}(0+))}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}+\sum_{\begin{subarray}{c}s^{\prime}\in f(I)\\ f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}\end{subarray}}\sum_{c^{\prime}\in C(s^{\prime})}m_{c^{\prime}}(\llbracket f(\theta)\rrbracket)e_{c^{\prime}}$
		$\displaystyle=m(f(\theta))$

by Lemma 7.1.24.

Given $s^{\prime}\in f(I)$ such that $f(\theta)_{s^{\prime}}=\operatorname{id}\nolimits_{s^{\prime}}$ , we have $s^{\prime}{\not\in}Z^{\prime}_{f}$ . We deduce that $i(\theta_{s},\theta_{t})=i(f(\theta)_{f(s)},f(\theta)_{f(t)})$ for all $s\neq t\in I$ by Lemma 7.3.22. So $f(i(\theta))=i(f(\theta))$ . We deduce that the lemma holds for $\theta$ .

Consider now the case where $E\neq\emptyset$ . Let $\bar{\theta}=(\theta_{s})_{s\in I-E}$ . We have $\deg_{E^{+}}(\theta)=\deg_{E^{+}}(\bar{\theta})$ by Lemma 7.4.7; taking quotients, we obtain $\deg_{f^{-1}(f(E))^{+}}(\theta)=\deg_{f^{-1}(f(E))^{+}}(\bar{\theta})$ . Since $f(\bar{\theta})=(f(\theta)_{t})_{t\in f(I)-f(E)}$ , it follows again from Lemma 7.4.7 that $\deg_{f(E)^{+}}(f(\theta))=\deg_{f(E)^{+}}(f(\bar{\theta}))$ . Since the lemma holds for $\bar{\theta}$ , we deduce that the lemma holds for $\theta$ . ∎

As a consequence of Lemma 7.4.12, we have the following result.

0PB1

Proposition 7.4.13. Let $f:Z\to Z^{\prime}$ be a morphism of curves and let $\theta^{\prime}$ be a non-zero map in ${\mathcal{P}}^{\bullet}(Z^{\prime})$ . Then $f^{\#}(\theta^{\prime})$ is a sum of maps $\theta$ such that $f(\deg_{Z_{f}^{+}}(\theta))=\deg_{f(Z_{f})^{+}}(\theta^{\prime})$ .

Let $Z_{1},\ldots,Z_{r}$ be the connected components of $Z$ . The isomorphism (7.4.1) is compatible with the degree function in the following sense. Given $\theta:I\to J$ a braid in $Z$ , let $\theta_{i}$ be the restriction of $\theta$ to $I\cap Z_{i}$ . The image of $(\deg(\theta_{1}),\ldots,\deg(\theta_{r}))$ in $\Gamma(Z)$ by the map of (7.3.3) is $\deg(\theta)$ .

Let $I$ and $J$ be two finite subsets of $Z$ and let $\theta:I\to J$ be a braid in $Z$ . We define

L(\theta)=\coprod_{i_{1}\neq i_{2}\in I}I(\theta_{i_{1}},\theta_{i_{2}}).

Note that $\zeta\mapsto\zeta^{-1}$ induces a fixed-point free involution $\mathrm{inv}$ on $L(\theta)$ .

Let $\zeta\in L(\theta)$ . Put $i_{1}=\zeta(0)$ and $i_{2}=\zeta(1)$ . We define $\theta^{\zeta}$ by $(\theta^{\zeta})_{i}=\theta_{i}$ if $i\in I-\{i_{1},i_{2}\}$ , $(\theta^{\zeta})_{i_{1}}=\theta_{i_{2}}\circ\zeta=\bar{\zeta}\circ\theta_{i_{1}}$ and $(\theta^{\zeta})_{i_{2}}=\theta_{i_{1}}\circ\zeta^{-1}=\bar{\zeta}^{-1}\circ\theta_{i_{2}}$ . Note that $\theta^{\zeta^{-1}}=\theta^{\zeta}$ .

Let $D(\theta)$ be the set of classes $\zeta$ in $L(\theta)$ such that

(a)

given a class of smooth paths $\zeta^{\prime}:\zeta(0)\to\zeta(1)$ such that $\zeta\circ\zeta^{\prime-1}$ and $\zeta^{\prime-1}\circ\zeta$ are smooth and have the same orientation as $\zeta$ and $\zeta^{\prime}$ , and given a class of smooth paths $\zeta^{\prime\prime}:\bar{\zeta}(0)\to\bar{\zeta}(1)$ such that $\bar{\zeta}\circ\zeta^{\prime\prime-1}$ and $\zeta^{\prime\prime-1}\circ\bar{\zeta}$ are smooth and have the same orientation as $\bar{\zeta}$ and $\zeta^{\prime\prime}$ , then $\zeta^{\prime}=\zeta$ or $\zeta^{\prime\prime}=\bar{\zeta}$ .
(b)

given $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ in $L(\theta)$ with $\zeta=\zeta^{\prime}\circ\zeta^{\prime\prime}$ , then $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations.

0PB2

Remark 7.4.14. Condition (a) above is automatically satisfied if the component of the support of $\zeta$ is not isomorphic to $S^{1}$ .

The subset $D(\theta)$ of $L(\theta)$ is stable under the involution $\mathrm{inv}$ .

The next lemma restricts the cases where condition (b) above needs to be checked.

0PB3

Lemma 7.4.15. Let $\zeta,\zeta^{\prime},\zeta^{\prime\prime}\in L(\theta)$ such that $\zeta=\zeta^{\prime}\circ\zeta^{\prime\prime}$ . If $\zeta^{\prime}(0)\in Z_{o}$ and $\theta_{\zeta^{\prime}(0)}=\operatorname{id}\nolimits$ , then $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations.

0PB4

Proof. Let $z=\zeta^{\prime}(0)=\zeta^{\prime\prime}(1)$ . We have $\zeta^{\prime}\in I(\operatorname{id}\nolimits_{z},\theta_{\zeta(1)})$ . Since $\bar{\zeta}^{\prime}=\theta_{\zeta(1)}\circ\zeta^{\prime}$ is smooth and has opposite orientation to $\zeta^{\prime}$ , it follows that $\zeta^{\prime}(0+)\in\iota(C(z)^{+})$ . Similarly, $\zeta^{\prime\prime}(1-)\in\iota(C(z)^{+})$ . We deduce that $\zeta^{\prime}$ and $\zeta^{\prime\prime}$ have opposite orientations. ∎

0PB5

Lemma 7.4.16. Let $I^{\prime}$ be a subset of $I$ such that $I-I^{\prime}\subset Z_{o}$ and $\theta_{i}=\operatorname{id}\nolimits$ for $i\in I-I^{\prime}$ .

We have $D(\theta_{|I^{\prime}})\subset D(\theta)$ .

0PB6

Proof. We have $L(\theta_{|I^{\prime}})\subset L(\theta)$ and Lemma 7.4.15 shows that $D(\theta_{|I^{\prime}})\subset D(\theta)$ . ∎

0PB7

Example 7.4.17. In the picture below, the left side shows a valid braid $\theta$ , for which the conclusion of Lemma 7.4.15 holds. For contrast, the right side shows a braid $\theta$ that is disallowed since $\theta_{i_{2}}$ is not oriented, and the conclusion of Lemma 7.4.15 fails.

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2