ScalingStacks

Proof. [017R]

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Proof.

Pick a representative ℒ#=(ℒ,ψ0){\mathcal{L}}^{\#}=({\mathcal{L}},\psi_{0}) of ψ#\psi^{\#} such that ℒ{\mathcal{L}} is defined on a proper snc model 𝒳{\mathcal{X}}. Let 𝒳′{\mathcal{X}}^{\prime} be the normalized base change by t=t′m{t}={t}^{\prime m}.

Let σ\sigma be a dd-dimensional face of Δ⁡(ℒ)\Delta({\mathcal{L}}). By Lemma 5.13, p−1​(σ)p^{-1}(\sigma) is the union of gσg_{\sigma} distinct isomorphic faces σα′\sigma^{\prime}_{\alpha} of Δ⁡(𝒳)\Delta({\mathcal{X}}) such that

bσα′=bσ/gcd⁡(m,bσ)b_{\sigma^{\prime}_{\alpha}}=b_{\sigma}/\gcd(m,b_{\sigma}) (6.2)
Vol⁡(σα′)=md​Vol⁡(σ).\operatorname{Vol}(\sigma^{\prime}_{\alpha})=m^{d}\operatorname{Vol}(\sigma). (6.3)

Further, the induced map Yσα′′→YY^{\prime}_{\sigma^{\prime}_{\alpha}}\to Y is generically finite, of degree fσf_{\sigma} independent of α\alpha, and we have fσ​gσ=gcd⁡(m,bσ)f_{\sigma}g_{\sigma}=\gcd(m,b_{\sigma}). Pick a toroidal modification 𝒳′′→𝒳′{\mathcal{X}}^{\prime\prime}\to{\mathcal{X}}^{\prime} with 𝒳′′{\mathcal{X}}^{\prime\prime} snc, denote by ρ:𝒳′′→𝒳\rho\colon{\mathcal{X}}^{\prime\prime}\to{\mathcal{X}} the composition, and set ℒ′′:=ρ∗​ℒ{\mathcal{L}}^{\prime\prime}:=\rho^{*}{\mathcal{L}}.

Each face σα′\sigma^{\prime}_{\alpha} above is subdivided into simplices σα​β′′\sigma^{\prime\prime}_{\alpha\beta} of Δ⁡(ℒ′′)\Delta({\mathcal{L}}^{\prime\prime}) of dimension dd, each corresponding to a stratum Yα​β′′Y^{\prime\prime}_{\alpha\beta} of 𝒳0′′{\mathcal{X}}^{\prime\prime}_{0}, and ρ|Yα​β′′:Yα​β′′→Y\rho|_{Y^{\prime\prime}_{\alpha\beta}}\colon Y^{\prime\prime}_{\alpha\beta}\to Y is generically finite, of degree fσf_{\sigma}. Further, (6.2) implies that

bσα​β′′=bσα′=bσ/gcd⁡(m,bσ)for all α,β.b_{\sigma^{\prime\prime}_{\alpha\beta}}=b_{\sigma^{\prime}_{\alpha}}=b_{\sigma}/\gcd(m,b_{\sigma})\quad\text{for all $\alpha,\beta$}. (6.4)

We shall need the following result:

Lemma 6.8.

With notation as above, we have, for all α\alpha, β\beta:

ResYα​β′′(ℒ′′#)=gcd(m,bσ)−2ρ∗ResY(ℒ#).\operatorname{Res}_{Y^{\prime\prime}_{\alpha\beta}}({\mathcal{L}}^{\prime\prime\#})=\gcd(m,b_{\sigma})^{-2}\rho^{*}\operatorname{Res}_{Y}({\mathcal{L}}^{\#}).

Grant this result for the moment. Lemma 6.8 implies

∫Yα​β′′ResYα​β′′(ℒ′′#)=fσgcd(m,bσ)−2∫YResY(ℒ#),\int_{Y^{\prime\prime}_{\alpha\beta}}\operatorname{Res}_{Y^{\prime\prime}_{\alpha\beta}}({\mathcal{L}}^{\prime\prime\#})=f_{\sigma}\gcd(m,b_{\sigma})^{-2}\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#}),

and hence

(p∗μ′)(σ)=∑α,βμ′(σ′′α​β)=∑α,β(∫Yα​β′′ResYα​β′′(ℒ′′#))bσα​β′′−1Vol(σ′′α​β)=fσ​gcd⁡(m,bσ)−2​(∫YResY⁡(ℒ#))​bσ−1​gcd⁡(m,bσ)​∑αVol⁡(σα′)=md​(∫YResY⁡(ℒ#))​bσ−1​Vol⁡(σ)=md​μ​(σ),(p_{*}\mu^{\prime})(\sigma)=\sum_{\alpha,\beta}\mu^{\prime}(\sigma^{\prime\prime}_{\alpha\beta})=\sum_{\alpha,\beta}\left(\int_{Y^{\prime\prime}_{\alpha\beta}}\operatorname{Res}_{Y^{\prime\prime}_{\alpha\beta}}({\mathcal{L}}^{\prime\prime\#})\right)b_{\sigma^{\prime\prime}_{\alpha\beta}}^{-1}\operatorname{Vol}(\sigma^{\prime\prime}_{\alpha\beta})\\ =f_{\sigma}\gcd(m,b_{\sigma})^{-2}\left(\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})\right)b_{\sigma}^{-1}\gcd(m,b_{\sigma})\sum_{\alpha}\operatorname{Vol}(\sigma^{\prime}_{\alpha})\\ =m^{d}\left(\int_{Y}\operatorname{Res}_{Y}({\mathcal{L}}^{\#})\right)b_{\sigma}^{-1}\operatorname{Vol}(\sigma)=m^{d}\mu(\sigma),

thanks to (6.3) and (6.4). ∎

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