Proposition 6.2.11 (0P7Y) · ScalingStacks

0P7Y

Proposition 6.2.11. The maps $d$ equip the ${\mathbf{F}}_{2}$ -linear $\Gamma_{n}$ -graded category ${\mathbf{F}}_{2}[{\mathcal{H}}_{n}]$ with a differential $\Gamma_{n}$ -graded structure, hence equip ${\mathcal{H}}_{n}$ with a differential $\Gamma_{n}$ -graded pointed structure.

Given $I\subset{\mathbf{Z}}/n$ , the morphism $F_{I}$ induces an isomorphism of differential ${\mathbf{Z}}$ -graded pointed monoids

\hat{{\mathfrak{S}}}_{|I|}^{\operatorname{nil}\nolimits}\mathrel{\mathop{\kern 0.0pt\to}\limits^{\sim}}\operatorname{End}\nolimits_{{\mathcal{H}}_{n}}(I).

0P7Z

Proof. Note that Lemma 6.2.9 shows that $d$ is homogeneous of degree $1$ . The compatibility of $d$ with $F_{I}$ follows from Lemma 3.2.4.

Consider now $\sigma\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(I,J)$ non-zero. There exists $\tau\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(J,I)$ with $\ell(\tau)=0$ . We have $d(\tau\circ\sigma)=\tau\circ d(\sigma)$ , hence $d^{2}(\tau\circ\sigma)=\tau\circ d^{2}(\sigma)$ . The compatibility of $F_{I}$ with $d$ shows that $d^{2}(\tau\circ\sigma)=0$ . Since $\tau$ is invertible, we deduce that $d^{2}(\sigma)=0$ .

Consider finally $\sigma^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(J,K)$ and fix $\tau^{\prime}\in\operatorname{Hom}\nolimits_{{\mathcal{H}}_{n}}(K,J)$ with $\ell(\tau^{\prime})=0$ . We have $d(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau)=\tau^{\prime}\circ d(\sigma^{\prime}\circ\sigma)\circ\tau$ and it follows from the compatibility of $F_{J}$ with $d$ that

	$\displaystyle d(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau)$	$\displaystyle=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}\circ\sigma\circ\tau))\bigr)=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime})\circ F_{J}^{-1}(\sigma\circ\tau))\bigr)$
		$\displaystyle=F_{J}\bigl(d(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}))\circ F_{J}^{-1}(\sigma\circ\tau)\bigr)+F_{J}\bigl(F_{J}^{-1}(\tau^{\prime}\circ\sigma^{\prime}))\circ d(F_{J}^{-1}(\sigma\circ\tau)\bigr)$
		$\displaystyle=d(\tau^{\prime}\circ\sigma^{\prime})\circ\sigma\circ\tau+\tau^{\prime}\circ\sigma^{\prime}\circ d(\sigma\circ\tau).$

∎