ScalingStacks

Proof. [017Y]

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Proof.

Let 𝒳{\mathcal{X}} be a semistable model, i.e. 𝒳{\mathcal{X}} is snc with 𝒳0{\mathcal{X}}_{0} reduced. By (7.1), we have κmin∈ℤ\kappa_{\min}\in{\mathbb{Z}}. Since some non-empty EJE_{J} might have several components, the dual complex Δ⁡(𝒳)\Delta({\mathcal{X}}) is possibly not a triangulation of Sk⁡(X)\operatorname{Sk}(X). However, the barycentric subdivision Δ′\Delta^{\prime} of Δ⁡(𝒳)\Delta({\mathcal{X}}) is a triangulation; the corresponding toroidal modification 𝒳′{\mathcal{X}}^{\prime} is snc, with 𝒳0′{\mathcal{X}}^{\prime}_{0} is possibly non-reduced, but bσ=1b_{\sigma}=1 for each nn-simplex σ\sigma of Δ′\Delta^{\prime}. Applying the above discussion to 𝒳′{\mathcal{X}}^{\prime}, we infer

μlog⁡|η|#=∑σ|Resyσ⁡(ω)|2​λσ,\mu_{\log|\eta|^{\#}}=\sum_{\sigma}|\operatorname{Res}_{y_{\sigma}}(\omega)|^{2}\lambda_{\sigma},

with σ\sigma ranging over the nn-dimensional faces of Δ′\Delta^{\prime}, with corresponding strata yσ∈𝒳0′y_{\sigma}\in{\mathcal{X}}^{\prime}_{0} reduced to single points. It will thus be enough to show that |Resyσ⁡(ω)||\operatorname{Res}_{y_{\sigma}}(\omega)| is independent of σ\sigma.

By the strong connectedness property, any two nn-simplices σ\sigma, σ′\sigma^{\prime} of Δ′\Delta^{\prime} can be joined by a chain of nn-simplices σ=σ1,…,σN=σ′\sigma=\sigma_{1},\dots,\sigma_{N}=\sigma^{\prime} with σi\sigma_{i} and σi+1\sigma_{i+1} sharing a common (n−1)(n-1)-face τi\tau_{i}. Denoting by yi=yσiy_{i}=y_{\sigma_{i}} and Yi=YτiY_{i}=Y_{\tau_{i}} the corresponding strata in 𝒳′{\mathcal{X}}^{\prime}, we thus have yi,yi+1∈Yiy_{i},y_{i+1}\in Y_{i}. Further, the Poincaré residue ResYi⁡(ω)\operatorname{Res}_{Y_{i}}(\omega) has poles precisely at yi,yi+1y_{i},y_{i+1}, since any other pole would correspond to an nn-simplex of Δ′\Delta^{\prime} containing τi\tau_{i}, contradicting the non-branching property. Since Resyi⁡ResY⁡(ω)=Resyi⁡(ω)\operatorname{Res}_{y_{i}}\operatorname{Res}_{Y}(\omega)=\operatorname{Res}_{y_{i}}(\omega), the residue theorem applied to the Riemann surface YiY_{i} yields Resyi⁡(ω)+Resyi+1⁡(ω)=0\operatorname{Res}_{y_{i}}(\omega)+\operatorname{Res}_{y_{i+1}}(\omega)=0, and hence |Resy1⁡(ω)|=⋯=|ResyN⁡(ω)||\operatorname{Res}_{y_{1}}(\omega)|=\dots=|\operatorname{Res}_{y_{N}}(\omega)|. ∎

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