ScalingStacks

Proof. [03AG]

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Proof.

We start by recalling the construction of Δ𝒳\Delta_{\mathscr{X}} and p𝒳p_{\mathscr{X}} from [BFJ16a]. Note that, in loc. cit. the residue field kk is of characteristic zero, but once we assume that the model 𝒳\mathscr{X} is an SNC model, using the results of [MN15, § 3.1] it is possible to extend the presentation of [BFJ16a] to the case of positive and mixed characteristic.

Let Div0⁡(𝒳)\Div_{0}(\mathscr{X}) be the group of vertical Cartier divisors on 𝒳\mathscr{X}. Denote Div0⁡(𝒳)ℝ=Div0⁡(𝒳)⊗ℝ\Div_{0}(\mathscr{X})_{\mathbb{R}}=\Div_{0}(\mathscr{X})\otimes\mathbb{R} and let Div0⁡(𝒳)ℝ∗\Div_{0}(\mathscr{X})_{\mathbb{R}}^{\ast} be the dual. As explained in 2.2, each D∈Div0⁡(𝒳)D\in\Div_{0}(\mathscr{X}) determines a model function φD\varphi_{D}. The map D↦φDD\mapsto\varphi_{D} is linear in DD and can be extended by linearity to a map Div0⁡(𝒳)ℝ→C0​(Xan)\Div_{0}(\mathscr{X})_{\mathbb{R}}\to C^{0}(X^{{\mathrm{an}}}).

There is a map {ev}𝒳:Xan→Div0⁡(𝒳)ℝ∗\ev_{\mathscr{X}}\colon{X^{{\mathrm{an}}}}\to\Div_{0}(\mathscr{X})_{\mathbb{R}}^{\ast} determined by

(A.7) ⟨D,{ev}𝒳⁡(x)⟩=φD​(x).\langle D,\ev_{\mathscr{X}}(x)\rangle=\varphi_{D}(x).

Let D1,…,DℓD_{1},\dots,D_{\ell} be the components of the special fiber 𝒳s\mathscr{X}_{s}. Each DiD_{i}, i=1,…,ℓi=1,\dots,\ell, determines a divisorial point xi∈Xanx_{i}\in X^{{\mathrm{an}}} and we denote by ei={ev}𝒳⁡(xi)e_{i}=\ev_{\mathscr{X}}(x_{i}). For each J⊂{1,…,ℓ}J\subset\{1,\dots,\ell\} we write DJ=⋂j∈JDjD_{J}=\bigcap_{j\in J}D_{j} and σJ={conv}⁡(ej,j∈J)\sigma_{J}=\conv(e_{j},j\in J). Then the abstract skeleton of 𝒳\mathscr{X} is

Δ𝒳{abs}=⋃J⊂{1,…,ℓ}DJ≠∅σJ⊂Div0⁡(𝒳)ℝ∗.\Delta^{\abs}_{\mathscr{X}}=\bigcup_{\begin{subarray}{c}J\subset\{1,\dots,\ell\}\\ D_{J}\not=\emptyset\end{subarray}}\sigma_{J}\subset\Div_{0}(\mathscr{X})_{\mathbb{R}}^{\ast}.

By [BFJ16a, Thm. 3.1], the image of {ev}𝒳\ev_{\mathscr{X}} is Δ𝒳{abs}\Delta^{\abs}_{\mathscr{X}} and there exists a unique function {emb}𝒳:Δ𝒳{abs}→Xan\emb_{\mathscr{X}}\colon\Delta^{\abs}_{\mathscr{X}}\to X^{{\mathrm{an}}} such that

  1. (i)

    {ev}𝒳∘{emb}𝒳=IdΔ𝒳{abs}\ev_{\mathscr{X}}\circ\emb_{\mathscr{X}}=\Id_{\Delta^{\abs}_{\mathscr{X}}};

  2. (ii)

    for each s∈Δ𝒳{abs}s\in\Delta_{\mathscr{X}}^{\abs}, if s∈{relint}⁡(σJ)s\in\rint(\sigma_{J}), then red⁡({emb}𝒳⁡(s))=ξDJ{\mathrm{red}}(\emb_{\mathscr{X}}(s))=\xi_{D_{J}}, where ξDJ\xi_{D_{J}} is the generic point of DJD_{J}.

Then the skeleton and the retraction are given by

Δ𝒳={emb}𝒳⁡(Δ𝒳{abs})p𝒳={emb}𝒳∘{ev}𝒳.\Delta_{\mathscr{X}}=\emb_{\mathscr{X}}(\Delta^{\abs}_{\mathscr{X}})\quad\quad p_{\mathscr{X}}=\emb_{\mathscr{X}}\circ\ev_{\mathscr{X}}.

We now go back to the regular toric case. In particular, XX is a toric smooth projective variety over KK and 𝒳\mathscr{X} is a toric projective SNC model. Then all the divisors of Div0⁡(𝒳)\Div_{0}(\mathscr{X}) are toric divisors. Therefore, for D∈Div0⁡(𝒳)ℝD\in\Div_{0}(\mathscr{X})_{\mathbb{R}}, the function φD\varphi_{D} is invariant under the action of the compact torus 𝕊={val}K−1⁡(0)\mathbb{S}=\Val_{K}^{-1}(0). The restriction of φD\varphi_{D} to 𝕋an\mathbb{T}^{{\mathrm{an}}} factorizes as

(A.8) φD∣𝕋an=−ϕD∘{val}K,\varphi_{D}\mid_{\mathbb{T}^{{\mathrm{an}}}}=-\phi_{D}\circ\Val_{K},

where ϕD\phi_{D} is the function from [BPS14, Def. 4.3.6] corresponding to the trivial line bundle 𝒪X\mathcal{O}_{X} with the metric determined by DD and the section 11.

We now define {ev}Π:Nℝ→Div0⁡(𝒳)∗\ev_{\Pi}\colon N_{\mathbb{R}}\to\Div_{0}(\mathscr{X})^{\ast} by

⟨D,{ev}Π⁡(u)⟩=−ϕD​(u).\langle D,\ev_{\Pi}(u)\rangle=-\phi_{D}(u).

By construction, the restriction of {ev}Π\ev_{\Pi} to each polyhedron Λ∈Π\Lambda\in\Pi is affine. Moreover, using (A.7) and (A.8) we deduce that

(A.9) {ev}𝒳∣𝕋an={ev}Π∘{val}K.\ev_{\mathscr{X}}\mid_{\mathbb{T}^{{\mathrm{an}}}}=\ev_{\Pi}\circ\Val_{K}.

As before let D1,…,DℓD_{1},\dots,D_{\ell} be the components of the special fiber 𝒳s\mathscr{X}_{s} and xix_{i} the divisorial point determined by DiD_{i}. Then the set of vertices of Π\Pi is Π0={u1,…,uℓ}\Pi^{0}=\{u_{1},\dots,u_{\ell}\}, where ui={val}K⁡(xi)u_{i}=\Val_{K}(x_{i}). Therefore {ev}Π⁡(ui)=ei\ev_{\Pi}(u_{i})=e_{i}. Since {ev}Π\ev_{\Pi} is affine in each polyhedron of Π\Pi we deduce that the image of {ev}Π\ev_{\Pi} is Δ𝒳{abs}\Delta^{\abs}_{\mathscr{X}} and that {ev}Π\ev_{\Pi} determines a homeomorphism ΔΠ→Δ𝒳{abs}\Delta_{\Pi}\to\Delta^{\abs}_{\mathscr{X}}. We define {emb}Π:ΔΠ{abs}→Nℝ\emb_{\Pi}\colon\Delta^{\abs}_{\Pi}\to N_{\mathbb{R}} as the composition of the inverse of this homeomorphism with the inclusion ΔΠ↪Nℝ\Delta_{\Pi}\hookrightarrow N_{\mathbb{R}}. Using equation (A.9) and Lemma A.1 one can check that ζK∘{emb}Π\zeta_{K}\circ\emb_{\Pi} satisfies the conditions (1) and (2) that characterize {emb}𝒳\emb_{\mathscr{X}}. Therefore

(A.10) {emb}𝒳=ζK∘{emb}Π.\emb_{\mathscr{X}}=\zeta_{K}\circ\emb_{\Pi}.

We next claim that pΠ={emb}Π∘{ev}Π.p_{\Pi}=\emb_{\Pi}\circ\ev_{\Pi}. Indeed, for every D∈Div0⁡(𝒳)D\in\Div_{0}(\mathscr{X}), since DD is a model of the trivial vector bundle, we know that {rec}⁡(ϕD)\rec(\phi_{D}) is the zero function. Therefore, writing any u∈Λ∈Πu\in\Lambda\in\Pi is as in (A.6), one can show that

ϕD=ϕD∘pΠ.\phi_{D}=\phi_{D}\circ p_{\Pi}.

This implies that {ev}Π={ev}Π∘pΠ\ev_{\Pi}=\ev_{\Pi}\circ\,p_{\Pi}. By construction {emb}Π∘{ev}Π\emb_{\Pi}\circ\ev_{\Pi} is the identity in the image of pΠp_{\Pi}. Therefore

(A.11) {emb}Π∘{ev}Π={emb}Π∘{ev}Π∘pΠ=pΠ.\emb_{\Pi}\circ\ev_{\Pi}=\emb_{\Pi}\circ\ev_{\Pi}\circ p_{\Pi}=p_{\Pi}.

Using equations (A.11) (A.10) and (A.9) we deduce that

Δ𝒳={emb}𝒳⁡(Δ𝒳{abs})=ζK​({emb}Π⁡(Δ𝒳{abs}))=ζK​(ΔΠ)\Delta_{\mathscr{X}}=\emb_{\mathscr{X}}(\Delta^{\abs}_{\mathscr{X}})=\zeta_{K}(\emb_{\Pi}(\Delta^{\abs}_{\mathscr{X}}))=\zeta_{K}(\Delta_{\Pi})

and

p𝒳|𝕋an={emb}𝒳∘{ev}𝒳|𝕋an=ζK∘{emb}Π∘{ev}Π∘{val}K=ζK∘pΠ∘{val}Kp_{\mathscr{X}}|_{\mathbb{T}^{{\mathrm{an}}}}=\emb_{\mathscr{X}}\circ\ev_{\mathscr{X}}|_{\mathbb{T}^{{\mathrm{an}}}}=\zeta_{K}\circ\emb_{\Pi}\circ\ev_{\Pi}\circ\Val_{K}=\zeta_{K}\circ p_{\Pi}\circ\Val_{K}

concluding the proof. ∎

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