Proof of Proposition 5.2 . [03IB]

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Proof of Proposition 5.2.

We work on the compact manifold $M$ . Let $S$ be a holomorphic section of $K_{M}^{-1}$ with $S^{-1}(0)=D$ , and let $h$ be a smooth hermitian metric on $K_{M}^{-1}$ whose curvature form $\omega_{h}$ is a Kähler form on $M$ with positive Ricci curvature. By Theorem 3.3 near $D$ we have

(5.6)

C^{-1}\sqrt{-1}\partial\bar{\partial}(-{\log|S|^{2}_{h}})^{3/2}\leq\omega_{TY}\leq C\sqrt{-1}\partial\bar{\partial}(-{\log|S|^{2}_{h}})^{3/2}.

By a straightforward computation this implies that

(5.7)

\omega_{TY}\leq C|S|_{h}^{-2}\omega_{h}

and hence, trivially,

(5.8)

|\gamma|_{\omega_{h}}\leq C^{-1}|S|_{h}^{-1}|\gamma|_{\omega_{TY}}=O(|S|_{h}^{-1-\epsilon})

for any $\epsilon>0$ . Define $\alpha=\gamma\otimes S$ . This is a section of $\Lambda_{M}^{0,1}\otimes K_{M}^{-1}$ which lies in $L^{p}_{\omega_{h}}(M,\Lambda^{0,1}_{M}\otimes K_{M}^{-1})$ for all $p\geq 1$ . Since $\bar{\partial}\gamma=0$ , one can directly check that $\bar{\partial}\alpha=0$ in the distributional sense. Now notice that $H^{1}(M,K_{M}^{-1})=H^{1}(M,K_{M}\otimes L)=0$ by the Kodaira vanishing theorem applied to the ample line bundle $L=K_{M}^{-2}$ . Thus, we can define $\beta=\bar{\partial}^{*}\Delta_{\bar{\partial}}^{-1}\alpha$ with respect to $\omega_{h}$ . It follows from elliptic regularity that $\beta\in W^{1,p}_{\omega_{h}}(M,K_{M}^{-1})$ for all $p\geq 1$ , so that $\beta\in C^{\alpha}_{\omega_{h}}(M,K_{M}^{-1})$ for all $\alpha<1$ . Moreover by local regularity we know $\beta$ is smooth outside $D$ and $\bar{\partial}\beta=\alpha$ . Let $f=\beta\otimes S^{-1}$ , then on $X$ we have $\bar{\partial}f=\gamma$ . The immediate estimate we get is that for some constant $C>0$ ,

(5.9)

f=O(|S|_{h}^{-1})=O(e^{Cz^{2}}).

The lemma below allows us to improve (5.9) to the growth order $e^{\epsilon z^{2}}$ for any $\epsilon>0$ . The key point is that the estimate (5.7) can be improved to almost $O(1)$ in directions tangential to $D$ .

Lemma 5.4.

Denote $\beta_{0}:=\beta|_{D}$ , then $\bar{\partial}\beta_{0}=0$ , i.e. $\beta_{0}$ is a holomorphic section of $K_{M}^{-1}|_{D}$ .

Proof.

We choose a finite cover $D=\bigcup_{k=1}^{N_{0}}O_{k}$ such that for each $k$ there exists a local holomorphic coordinate system $(z,w)$ on some domain $U_{k}\subset M$ such that $U_{k}\cap D=O_{k}=\{w=0\}$ . We will show that $\bar{\partial}\beta_{0}=0$ in every $O_{k}\subset D$ in the distributional sense. Let $\psi$ be a smooth section of $\Lambda^{0,1}_{D}\otimes(K_{M}^{-1}|_{D})$ with compact support in $O_{k}$ . It suffices to show that $\langle\beta_{0},\bar{\partial}^{*}\psi\rangle_{O_{k}}=0$ . To this end, write $\psi(z)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1}$ for some smooth function $\sigma\in C^{\infty}_{0}(O_{k},\mathbb{C})$ and use this to define the trivial extension $\hat{\psi}(z,w)=\sigma(z)d\overline{z}\otimes(dz\wedge dw)^{-1}$ for all $(z,w)\in U_{k}$ . Denote by $O_{k}(\tau)$ the slice $\{w=\tau\}$ in $U_{k}$ , which is a complex submanifold of $M$ , and equip $O_{k}(\tau)$ with the restriction of the Kähler metric $\omega_{h}$ from $M$ . Notice that $\hat{\psi}$ restricts to a smooth section of $\Lambda_{O_{k}(\tau)}^{0,1}\otimes(K_{M}^{-1}|_{O_{k}(\tau)})$ with compact support in $O_{k}(\tau)$ . Since $\bar{\partial}\beta=\alpha$ and $\beta\in W^{1,p}\cap C^{\alpha}$ for any $p\geq 1$ , it follows that

(5.10)

\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)

|\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 $\alpha=\gamma\otimes S$ , it then follows that $|\alpha(\partial_{\bar{z}})|=O(|S|_{h}^{1-\epsilon})\rightarrow 0$ uniformly as $w\rightarrow 0$ . Using (5.10), it follows that

(5.12)

\langle\beta_{0},\bar{\partial}^{*}{\psi}\rangle_{O_{k}}=0,

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

Since $M$ is Fano we have $H^{1}(M,\mathcal{O}_{M})=0$ so by a standard exact sequence ([GH94, p.139]) the restriction map $H^{0}(M,K_{M}^{-1})\rightarrow H^{0}(D,K_{M}^{-1}|_{D})$ is surjective. This means we can find some $\beta_{1}$ $\in$ $H^{0}(M,K_{M}^{-1})$ such that $\beta_{1}|_{D}=\beta_{0}|_{D}$ . Let $f=(\beta-\beta_{1})\otimes S^{-1}$ . Then we still have $\bar{\partial}f=\gamma$ on $X$ but now since $\beta-\beta_{1}=0$ on $D$ and $\beta\in C^{\alpha}_{\omega_{h}}(M,\mathbb{C})$ for all $\alpha<1$ , we finally obtain Proposition 5.2.∎