ScalingStacks

Proof. [03ID]

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Proof.

We choose a finite cover D=⋃k=1N0OkD=\bigcup_{k=1}^{N_{0}}O_{k} such that for each kk there exists a local holomorphic coordinate system (z,w)(z,w) on some domain Uk⊂MU_{k}\subset M such that Uk∩D=Ok={w=0}U_{k}\cap D=O_{k}=\{w=0\}. We will show that ∂¯​β0=0\bar{\partial}\beta_{0}=0 in every Ok⊂DO_{k}\subset D in the distributional sense. Let ψ\psi be a smooth section of ΛD0,1⊗(KM−1|D)\Lambda^{0,1}_{D}\otimes(K_{M}^{-1}|_{D}) with compact support in OkO_{k}. It suffices to show that ⟨β0,∂¯∗​ψ⟩Ok=0\langle\beta_{0},\bar{\partial}^{*}\psi\rangle_{O_{k}}=0. To this end, write ψ⁡(z)=σ⁡(z)​d​z¯⊗(d​z∧d​w)−1\psi(z)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1} for some smooth function σ∈C0∞​(Ok,ℂ)\sigma\in C^{\infty}_{0}(O_{k},\mathbb{C}) and use this to define the trivial extension ψ^​(z,w)=σ⁡(z)​d​z¯⊗(d​z∧d​w)−1\hat{\psi}(z,w)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1} for all (z,w)∈Uk(z,w)\in U_{k}. Denote by Ok​(τ)O_{k}(\tau) the slice {w=τ}\{w=\tau\} in UkU_{k}, which is a complex submanifold of MM, and equip Ok​(τ)O_{k}(\tau) with the restriction of the Kähler metric ωh\omega_{h} from MM. Notice that ψ^\hat{\psi} restricts to a smooth section of ΛOk​(τ)0,1⊗(KM−1|Ok​(τ))\Lambda_{O_{k}(\tau)}^{0,1}\otimes(K_{M}^{-1}|_{O_{k}(\tau)}) with compact support in Ok​(τ)O_{k}(\tau). Since ∂¯​β=α\bar{\partial}\beta=\alpha and β∈W1,p∩Cα\beta\in W^{1,p}\cap C^{\alpha} for any p≥1p\geq 1, it follows that

(5.10) ⟨β0,∂¯∗​ψ⟩Ok=limτ→0⟨β,∂¯∗​ψ^⟩Ok​(τ)=limτ→0⟨∂¯​β,ψ^⟩Ok​(τ)=limτ→0⟨α,ψ^⟩Ok​(τ).\displaystyle\langle\beta_{0},\bar{\partial}^{*}{\psi}\rangle_{O_{k}}=\lim\limits_{\tau\to 0}\langle\beta,\bar{\partial}^{*}\hat{\psi}\rangle_{O_{k}(\tau)}=\lim\limits_{\tau\to 0}\langle\bar{\partial}\beta,\hat{\psi}\rangle_{O_{k}(\tau)}=\lim\limits_{\tau\to 0}\langle\alpha,\hat{\psi}\rangle_{O_{k}(\tau)}.

Notice that

(5.11) |γ(∂z¯)|≤|γ|ωT​Y|∂z¯|ωT​Y≤|γ|ωT​Y(−log|S|h2)14=O(|S|h−ϵ).|\gamma(\partial_{\bar{z}})|\leq|\gamma|_{\omega_{TY}}|\partial_{\bar{z}}|_{\omega_{TY}}\leq|\gamma|_{\omega_{TY}}(-\log|S|^{2}_{h})^{\frac{1}{4}}=O(|S|_{h}^{-\epsilon}).

Since α=γ⊗S\alpha=\gamma\otimes S, it then follows that |α(∂z¯)|=O(|S|h1−ϵ)→0|\alpha(\partial_{\bar{z}})|=O(|S|_{h}^{1-\epsilon})\rightarrow 0 uniformly as w→0w\rightarrow 0. Using (5.10), it follows that

(5.12) ⟨β0,∂¯∗​ψ⟩Ok=0,\langle\beta_{0},\bar{\partial}^{*}{\psi}\rangle_{O_{k}}=0,

as desired. By standard elliptic regularity, β0\beta_{0} is a holomorphic section. ∎

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