Subsection 8.3.2. (0PIG) · ScalingStacks

8.3.2. Setting

Let $\sigma:R_{\xi_{2}^{-}}(T,-)\otimes R_{\xi_{1}^{-}}(-,S)\to R_{\xi_{1}^{-}}(T,-)\otimes R_{\xi_{2}^{-}}(-,S)$ be defined as in (4.4.1).

0PDU

Lemma 8.3.2. The morphism $\sigma$ is invertible. Given $\alpha\in R_{\xi_{2}^{-}}^{\bullet}(T,U)$ and $\beta\in R_{\xi_{1}^{-}}^{\bullet}(U,S)$ , we have

\sigma(\alpha\otimes\beta)=\delta_{\alpha_{|U}\cdot\beta\neq 0}\bigl(\operatorname{id}\nolimits\boxtimes(\alpha_{\chi(\beta)(\xi_{1}^{-}(-1))}\cdot\beta_{\xi_{1}^{-}(-1)})\bigr)\otimes\bigl(\alpha_{\xi_{2}^{-}(-1)}\boxtimes(\alpha_{|U\setminus\{\chi(\beta)(\xi_{1}^{-}(-1))\}}\cdot\beta_{|S})\bigr).

Given $\alpha^{\prime}\in R_{\xi_{1}^{-}}^{\bullet}(T,U^{\prime})$ and $\beta^{\prime}\in R_{\xi_{2}^{-}}^{\bullet}(U^{\prime},S)$ , we have

\sigma^{-1}(\alpha^{\prime}\otimes\beta^{\prime})=\delta_{\alpha^{\prime}_{|U^{\prime}}\cdot\beta^{\prime}\neq 0}\bigl(\operatorname{id}\nolimits\boxtimes(\alpha^{\prime}_{\chi(\beta^{\prime})(\xi_{2}^{-}(-1))}\cdot\beta^{\prime}_{\xi_{2}^{-}(-1)})\bigr)\otimes\bigl(\alpha^{\prime}_{\xi_{1}^{-}(-1)}\boxtimes(\alpha^{\prime}_{|U^{\prime}\setminus\chi(\beta^{\prime})(\xi_{2}^{-}(-1))}\circ\beta^{\prime}_{|S})\bigr).

0PDV

Proof. We have

\sigma=(R_{\xi_{1}^{-}}\circ\textrm{mult})\circ(R_{\xi_{1}^{-}}\otimes R_{\xi_{2}^{-}}\otimes\varepsilon_{L_{\xi_{1}^{+}},R_{\xi_{1}^{-}}})\circ(R_{\xi_{1}^{-}}\otimes\lambda\otimes R_{\xi_{1}^{-}})\circ(\eta_{L_{\xi_{1}^{+}},R_{\xi_{1}^{-}}}\otimes\operatorname{id}\nolimits).

We have $\alpha=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\cdot(\alpha_{|U}\boxtimes\operatorname{id}\nolimits)$ , hence $\alpha\otimes\beta=(\operatorname{id}\nolimits\boxtimes\alpha_{\xi_{2}^{-}(-1)})\otimes(\alpha_{|U}\cdot\beta)$ . As a consequence, it is enough to prove the first statement of the lemma assuming that $\alpha_{|U}=\operatorname{id}\nolimits_{U}$ . In that case, the composition above is given by

	$\displaystyle\alpha\otimes\beta$	$\displaystyle\mapsto\sum_{x\in\tilde{\xi}_{1}^{-1}(T)}(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([-1\to x]))\otimes(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([x\to 1]))\otimes\alpha\otimes\beta$
		$\displaystyle\mapsto\sum_{x\in\tilde{\xi}_{1}^{-1}(T)}(\operatorname{id}\nolimits_{T\setminus\{\tilde{\xi}_{1}(x)\}}\boxtimes\tilde{\xi}_{1}([-1\to x]))\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([x\to 1]))\otimes\beta$
		$\displaystyle\mapsto(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\beta_{\|S}$
		$\displaystyle\mapsto(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\otimes(\alpha_{\xi_{2}^{-}(-1)}\boxtimes\beta_{\|S}).$

It is immediate to check that the formula for $\sigma^{-1}$ does produce an inverse. ∎

Consider the map $\rho:L_{\xi_{1}^{+}}(T,-)\otimes R_{\xi_{1}^{-}}(-,S)\to R_{\xi_{1}^{-}}(T,-)\otimes L_{\xi_{1}}^{-}(-,S)$ defined in §4.4.2.

0PDW

Lemma 8.3.3. Given $\alpha\in L_{\xi_{1}^{+}}^{\bullet}(T,U)$ and $\beta\in R_{\xi_{1}^{-}}^{\bullet}(U,S)$ , we have

\rho(\alpha\otimes\beta)=\delta_{1}\bigr(\alpha_{|U\setminus\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}\cdot(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\bigl)\otimes\bigl((\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}\boxtimes\operatorname{id}\nolimits)\cdot\beta_{|S}\bigr)

where $\delta_{1}=1$ if $\chi(\alpha\circ\beta)(\xi_{1}^{-}(-1))\neq\xi_{1}^{+}(1)$ and $(\operatorname{id}\nolimits_{\chi(\beta)(\xi_{1}^{-}(-1))}\boxtimes\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))})\cdot(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))})\neq 0$ and $\delta_{1}=0$ otherwise.

0PDX

Proof. Assume first $\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}=\operatorname{id}\nolimits$ and $\beta_{|S}=\operatorname{id}\nolimits$ . We have

	$\displaystyle\rho(\alpha\otimes\beta)$	$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\circ L_{\xi_{1}^{+}}\tau L_{\xi_{1}^{+}}(\alpha\otimes\beta\otimes\eta_{1}(\operatorname{id}\nolimits_{S}))$
		$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in\tilde{\xi}_{1}^{-1}(S)}\alpha\otimes\tau\Bigl((\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\Bigr)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr)$
		$\displaystyle=\varepsilon_{1}R_{\xi_{1}^{-}}L_{\xi_{1}^{+}}\biggl(\sum_{x\in I}\alpha\otimes\bigl(\operatorname{id}\nolimits\boxtimes\tilde{\xi}_{1}([-1\to x])\bigr)\otimes(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes\bigl(\tilde{\xi}_{1}([x\to 1])\boxtimes\operatorname{id}\nolimits\bigr)\biggr)$
		$\displaystyle=\delta_{1}(\beta_{\xi_{1}^{-}(-1)}\boxtimes\operatorname{id}\nolimits)\otimes(\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))}\boxtimes\operatorname{id}\nolimits)$

where $I=\{x\in\tilde{\xi}_{1}^{-1}(S)\ |\ (\tilde{\xi}_{1}([-1\to x])\boxtimes(\beta_{\xi_{1}^{-}(-1)}\circ[\xi_{1}^{-}(-2)\to\xi_{1}^{-}(-1)])\cdot\tau\neq 0\}$ .

Since $\rho$ is a morphism of $({\mathcal{S}}_{M}(Z),{\mathcal{S}}_{M}(Z))$ -bimodules, the general result follows using the decompositions $\alpha=(\alpha_{|U\setminus\{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))\}}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{+}(1)})\cdot(\operatorname{id}\nolimits\boxtimes\alpha_{\chi(\alpha)^{-1}(\xi_{1}^{+}(1))})$ and $\beta=(\operatorname{id}\nolimits\boxtimes\beta_{\xi_{1}^{-}(-1)})\cdot(\beta_{|S}\boxtimes\operatorname{id}\nolimits_{\xi_{1}^{-}(-1)})$ . ∎

Andrew Manion, Raphael Rouquier

Original source: arXiv:2009.09627v2