Subsection 6.2.4. (0PGQ) · ScalingStacks

6.2.4. Non-commutative degree

Let us consider the free abelian groups $R_{n}=\bigoplus_{a\in{\mathbf{Z}}/n}{\mathbf{Z}}\alpha_{a}$ and $L_{n}=\bigoplus_{a\in{\mathbf{Z}}/n}{\mathbf{Z}}\varepsilon_{a}$ . We define a linear map $\rho:R_{n}\to L_{n}$ by $\rho(\alpha_{a})=\varepsilon_{a+1}-\varepsilon_{a}$ and a representation of the group $R_{n}$ on $L_{n}$ given by

\alpha_{a}\cdot\varepsilon_{b}=(\delta_{a,b}+\delta_{a+1,b})\varepsilon_{b}.

Note that $\delta=\sum_{a\in{\mathbf{Z}}/n}\alpha_{a}\in\ker\rho$ and $\delta\cdot\varepsilon_{b}=2\varepsilon_{b}$ for all $b$ .

We define a bilinear map

\langle-,-\rangle:R_{n}\times R_{n}\to L_{n},\ \langle\alpha,\alpha^{\prime}\rangle=\alpha\cdot\rho(\alpha^{\prime}).

Let $\Gamma_{n}^{\prime}=L_{n}\times R_{n}$ . We define a group structure on $\Gamma_{n}^{\prime}$ by

(l,\alpha)\cdot(l^{\prime},\alpha^{\prime})=(l+l^{\prime}+\langle\alpha,\alpha^{\prime}\rangle,\alpha+\alpha^{\prime}).

Given $I\subset{\mathbf{Z}}/n$ , we put $\varepsilon_{I}=\sum_{a\in I}\varepsilon_{a}\in L_{n}$ . Given $i,j\in{\mathbf{Z}}$ , we put

\alpha_{i,j}=\sum_{i\leq r<j}\alpha_{r+n{\mathbf{Z}}}-\sum_{j\leq r<i}\alpha_{r+n{\mathbf{Z}}}.

Note that $\alpha_{i,i+1}=\alpha_{i+n{\mathbf{Z}}}$ , $\alpha_{i+n,j+n}=\alpha_{i,j}$ and $\alpha_{i,j}+\alpha_{j,k}=\alpha_{i,k}$ for all $i,j,k\in{\mathbf{Z}}$ . Note also that $\delta=\alpha_{i,i+n}$ for all $i\in{\mathbf{Z}}$ . Note finally that $\rho(\alpha_{i,j})=\varepsilon_{j+n{\mathbf{Z}}}-\varepsilon_{i+n{\mathbf{Z}}}$ .

Consider $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ . We put

\llbracket\sigma\rrbracket=\sum_{i\in\tilde{I}\cap[1,n]}\alpha_{i,\sigma(i)}\in R_{n}.

Note that $\rho(\llbracket\sigma\rrbracket)=\varepsilon_{J}-\varepsilon_{I}$ and $\llbracket\sigma^{\prime}\circ\sigma\rrbracket=\llbracket\sigma^{\prime}\rrbracket+\llbracket\sigma\rrbracket$ for any $\sigma^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,K)$ .

We define

m(\sigma)=\llbracket\sigma\rrbracket\cdot\varepsilon_{I}\in L_{n}\text{ and }\operatorname{dm}\nolimits(\sigma)=(-m(\sigma),-\llbracket\sigma\rrbracket)\in\Gamma^{\prime}_{n}.

0P7K

Lemma 6.2.4. Let $w\in W_{|I|}$ , $m\in{\mathbf{Z}}$ and let $\sigma=F_{I}(wc^{m})$ be the element of $\operatorname{End}\nolimits_{{\mathcal{S}}_{n}}(I)$ corresponding to $wc^{m}$ . We have $\ell(\sigma)=\ell(w)$ , $\llbracket\sigma\rrbracket=m\cdot\delta$ and $m(\sigma)=2m\varepsilon_{I}$ .

0P7L

Proof. The first statement follows from the fact that $F_{I}$ preserves lengths (cf the discussion before Lemma 6.2.2).

Note that $\llbracket s_{i,j}\rrbracket=0$ for $i,j\in\tilde{I}$ with $i-j{\not\in}n{\mathbf{Z}}$ , while $\llbracket F_{I}(c)\rrbracket=\delta$ . We deduce that $\llbracket\sigma\rrbracket=m\cdot\delta$ .

The last statement of the lemma is immediate. ∎

0P7M

Lemma 6.2.5. Consider $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ and $\sigma^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(J,K)$ . We have $\operatorname{dm}\nolimits(\sigma^{\prime}\circ\sigma)=\operatorname{dm}\nolimits(\sigma^{\prime})\cdot\operatorname{dm}\nolimits(\sigma)$ .

0P7N

Proof. We have

m(\sigma^{\prime}\circ\sigma)=\llbracket\sigma^{\prime}\rrbracket\cdot\varepsilon_{I}+\llbracket\sigma\rrbracket\varepsilon_{I}=m(\sigma^{\prime})+m(\sigma)+\llbracket\sigma^{\prime}\rrbracket\cdot(\varepsilon_{I}-\varepsilon_{J}),

hence

m(\sigma^{\prime})+m(\sigma)-m(\sigma^{\prime}\circ\sigma)=\llbracket\sigma^{\prime}\rrbracket\cdot\rho(\llbracket\sigma\rrbracket).

The lemma follows. ∎

We put $\Gamma_{n}=\frac{1}{2}{\mathbf{Z}}\times\Gamma^{\prime}_{n}$ . We endow $\Gamma_{n}$ with a structure of ${\mathbf{Z}}$ -monoid by using the canonical embedding ${\mathbf{Z}}\hookrightarrow\frac{1}{2}{\mathbf{Z}}\hookrightarrow\Gamma_{n}$ .

Given $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ , we put $\deg(\sigma)=(-\ell(\sigma),\operatorname{dm}\nolimits(\sigma))\in\Gamma_{n}$ .

Let $D$ be a subset of $\{1,\ldots,n\}\times\{\pm 1\}$ that embeds in its projection on $\{1,\ldots,n\}$ . We denote by $\Gamma_{D}$ the quotient of $\Gamma_{n}$ by the subgroup generated by $(0,\varepsilon_{i+n{\mathbf{Z}}})+(\frac{1}{2}\nu_{i},0)$ , where $(i,\nu_{i})\in D$ . We identify $\frac{1}{2}{\mathbf{Z}}$ with the image of $\frac{1}{2}{\mathbf{Z}}\times 0$ in $\Gamma_{D}$ . We define a partial order on $\Gamma_{D}$ by $h\geq g$ if $hg^{-1}$ is in $\frac{1}{2}{\mathbf{Z}}_{\geq 0}$ . We denote by $\deg_{D}(\sigma)$ the image of $\deg(\sigma)$ in $\Gamma_{D}$ .

Given $E$ a subset of $\{1,\ldots,n\}$ , we put $E^{+}=\{(i,1)\ |\ i\in E\}$ .

By Lemmas 6.2.1 and 6.2.5, we obtain a $\Gamma_{D}$ -filtration on ${\mathcal{S}}_{n}$ by defining

\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}^{\geq g}}(I,J)=\{\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)\ |\ \deg_{D}(\sigma)\geq g\}.

It follows from Lemma 6.2.5 that the pointed category ${\mathcal{H}}_{n}$ is isomorphic to the graded pointed category associated to the $\Gamma_{D}$ -filtration of ${\mathcal{S}}_{n}$ (after forgetting the $\Gamma_{D}$ -grading to ${\mathbf{Z}}$ ).

Note that if $D=\emptyset$ , then $\Gamma_{D}=\Gamma_{n}$ , $\deg_{D}=\deg$ and the ${\mathbf{Z}}_{\leq 0}$ grading on ${\mathcal{H}}_{n}$ given by the length can be recovered from the $\Gamma_{n}$ -grading by using the quotient map $\Gamma_{n}\to\Gamma_{n}/\Gamma^{\prime}_{n}=\frac{1}{2}{\mathbf{Z}}$ .

This quotient map provides a ${\mathbf{Z}}$ -grading on the $\Gamma_{n}$ -graded pointed category associated to the $\Gamma_{n}$ -filtration of ${\mathcal{S}}_{n}$ . This ${\mathbf{Z}}$ -graded pointed category is isomorphic to ${\mathcal{H}}_{n}$ .

0P7P

Remark 6.2.6. The bilinear form $\langle\langle-,-\rangle\rangle:R_{n}\times R_{n}\to\frac{1}{2}{\mathbf{Z}}$ obtained from $\langle-,-\rangle$ by composing with the morphism $L_{n}\to\frac{1}{2}{\mathbf{Z}},\ \varepsilon_{i}\mapsto-\frac{1}{2}$ is given by $\langle\langle\alpha_{a},\alpha_{b}\rangle\rangle=\frac{1}{2}(\delta_{b,a+1}-\delta_{b+1,a})$ . It is antisymmetric.

Assume $D=[1,n]^{+}$ . Composing with the quotient map $\Gamma_{n}\twoheadrightarrow\Gamma_{D}$ , the embedding $\frac{1}{2}{\mathbf{Z}}\hookrightarrow\Gamma_{n},\ r\mapsto(r,0)$ and the quotient map $\Gamma_{n}\twoheadrightarrow R_{n},\ (r,(l,\alpha))\mapsto\alpha$ induce an embedding of $\frac{1}{2}{\mathbf{Z}}$ as a central subgroup of $\Gamma_{D}$ with quotient map $\Gamma_{D}\twoheadrightarrow R_{n}$ . So, $\Gamma_{[1,n]^{+}}$ identifies with the set $\frac{1}{2}{\mathbf{Z}}\times R_{n}$ , with multiplication given by $(r,\alpha)\cdot(r^{\prime},\alpha^{\prime})=(r+r^{\prime}+\langle\langle\alpha,\alpha^{\prime}\rangle\rangle,\alpha+\alpha^{\prime})$ . When $n\geq 3$ , the group $\Gamma_{[1,n]^{+}}$ has a presentation with generators $z=(\frac{1}{2},0)$ , $g_{a}=(0,\alpha_{a})$ , $a\in{\mathbf{Z}}/n$ and relations

zg_{a}=g_{a}z,\ g_{a}g_{b}g_{a}^{-1}g_{b}^{-1}=\begin{cases}z&\text{ if }b=a+1\\ z^{-1}&\text{ if }b=a-1\\ 1&\text{ otherwise.}\end{cases}

We define a morphism of groups

\epsilon:\Gamma_{[1,n]^{+}}\to{\mathbf{Z}}/2,\ z\mapsto 1,\ g_{a}\mapsto 1.

0P7Q

Lemma 6.2.7. Given $(r,\sum_{a}v_{a}\alpha_{a})\in\Gamma_{[1,n]^{+}}$ , we have $\epsilon(r,\sum_{a}v_{a}\alpha_{a})=2r+\frac{1}{2}\bigl|\{a\in{\mathbf{Z}}/n\ |\ v_{a}+v_{a+1}\text{ odd}\}\bigr|$ .

Given $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{S}}_{n}}(I,J)$ , we have $\epsilon(\deg_{[1,n]^{+}}(\sigma))=0$ .

0P7R

Proof. Denote by $\tilde{\epsilon}$ the map defined by the right hand side of the equality of the lemma.

Let $N$ (resp. $N^{\prime}$ ) be the cardinality of the set of $a\in{\mathbf{Z}}/n$ such that $v_{a}+v_{a+1}$ (resp. $v^{\prime}_{a}+v^{\prime}_{a+1}$ ) is odd, where $v^{\prime}_{a}=v_{a}+\delta_{ab}$ . The integers $N$ and $N^{\prime}$ are even. We have

	$\displaystyle\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}((r,\sum_{a}v_{a}\alpha_{a})(s,\alpha_{b}))$	$\displaystyle=\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}(r+s+\frac{1}{2}(v_{b-1}-v_{b+1}),\alpha_{b}+\sum_{a}v_{a}\alpha_{a})$
		$\displaystyle=2s+v_{b+1}+v_{b-1}+\frac{1}{2}(N+N^{\prime}).$

We have

N^{\prime}=\begin{cases}N+2&\text{ if }v_{b-1},\ v_{b}\text{ and }v_{b+1}\text{ have the same parity}\\ N-2&\text{ if }v_{b-1},\ v_{b}+1\text{ and }v_{b+1}\text{ have the same parity}\\ N&\text{otherwise}.\end{cases}

It follows that

\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})+\tilde{\epsilon}((r,\sum_{a}v_{a}\alpha_{a})(s,\alpha_{b}))=2s+1.

We deduce by induction on $\sum_{a\in{\mathbf{Z}}/n}|v_{a}|$ that $\tilde{\epsilon}(r,\sum_{a}v_{a}\alpha_{a})=\epsilon(r,\sum_{a}v_{a}\alpha_{a})$ .

Given $a,b\in{\mathbf{Z}}/n$ and $i\in I$ , we have

\alpha_{a,b}\cdot\varepsilon_{i}=\delta_{\begin{subarray}{c}i\in\{a,b\}\\ a\neq b\end{subarray}}\ \varepsilon_{i}\mod 2L_{n}.

It follows that $m(\sigma)\equiv\sum_{i\in I\setminus(I\cap J)}\varepsilon_{i}\mod{2L_{n}}$ . Write $\llbracket\sigma\rrbracket=\sum_{a}v_{a}\alpha_{a}$ . Given $a\in{\mathbf{Z}}/n$ , the integer $v_{a}+v_{a+1}$ is odd if and only if $a\in I\Delta J$ , hence $\epsilon(0,\llbracket\sigma\rrbracket)=\frac{1}{2}|I\Delta J|=|I\setminus(I\cap J)|$ . It follows that $\epsilon(\deg_{[1,n]^{+}}(\sigma))=0$ . ∎

It follows from Lemma 6.2.7 that the $\Gamma_{n}$ -grading on ${\mathcal{H}}_{n}$ comes from a grading by the kernel of the composition $\Gamma_{n}\xrightarrow{{\mathrm{can}}}\Gamma_{[1,n]^{+}}\xrightarrow{\epsilon}{\mathbf{Z}}/2$ .

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2