Positivity of holomorphic vector bundles in terms of $L^p$-estimates for $\bar{\partial }$

Abstract

In this paper, we introduce a new notion of the Hermitian holomorphic vector bundles satisfying the optimal $L^2$-estimate, and give a characterization of Nakano positivity for Hermitian holomorphic vector bundles via the notion. As an application, we provide a new method to obtain Nakano positivity of direct image sheaves of twisted relative canonical bundles associated to holomorphic families of complex manifolds. We also present a comprehensive picture about converses of $L^p$-estimates and $L^p$-extension for $\bar{\partial }$.

Appendix

We prove a result used in the proof of Theorem 1.6, which seems to be already known.

Lemma 4.7

Let $U\subset {\mathbb {C}}^n$ be a domain, $\omega _1$, $\omega _2$ be any two Hermitian forms on U, and $E=U\times {\mathbb {C}}^r$ be trivial vector bundle on U with a Hermitian metric. Let $\Theta \in C^0(X,\Lambda ^{1,1}T^*_X\otimes End(E))$ such that $\Theta ^*=-\Theta$. Then

$$\begin{aligned} Im[i\Theta ,\Lambda _{\omega _1}]=Im [i\Theta ,\Lambda _{\omega _2}], \end{aligned}$$

and for any E-valued (n, 1) form $u\in Im[i\Theta ,\Lambda _{\omega _1}]$,

$$\begin{aligned} \langle [i\Theta ,\Lambda _{\omega _1}]^{-1}u,u\rangle _{\omega _1}dV_{\omega _1} =\langle [i\Theta ,\Lambda _{\omega _2}]^{-1}u,u\rangle _{\omega _2}d V_{\omega _2}. \end{aligned}$$

Proof

For any $z_0\in U$, after a linearly transformation, we may assume $\omega _1=i\sum _{j=1}^{n}dz_j\wedge d\bar{z}_j$ and $\omega _2=i\sum _{j=1}^{n}\lambda _j^2 dz_j\wedge d\bar{z}_j$ at $z_0$ with $\lambda _j>0$. Let $w_j=\lambda _j z_j$ for $j=1,2,\ldots ,n$, then $\omega _2=i\sum _{j=1}^{n}dw_j\wedge d\bar{w}_j$. We may write

$$\begin{aligned} i\Theta =i\sum _{jk\alpha \beta }c_{jk\alpha \beta }dz_j\wedge d\bar{z}_k \otimes e^*_\alpha \otimes e_\beta =i\sum _{jk\alpha \beta } c'_{jk\alpha \beta }dw_j\wedge d\bar{w}_k\otimes e^*_\alpha \otimes e_\beta \end{aligned}$$

(15)

with $c'_{jk\alpha \beta }=\frac{c_{jk\alpha \beta }}{\lambda _j\lambda _k}$.

Denote $\lambda =\prod _{j=1}^{n}\lambda _j$. Let $u=\sum _{j,\alpha }u_{j\alpha }dz\wedge d\bar{z}_j\otimes e_\alpha$, then $u=\sum _{j,\alpha }u'_{j\alpha }dw\wedge d\bar{w}_j\otimes e_\alpha$ with $u'_{j\alpha }=\frac{u_{j\alpha }}{\lambda \lambda _j}$. Note that

$$\begin{aligned} {[}i\Theta ,\Lambda _{\omega _1}]u=\sum _{jk\alpha \beta } u_{j\alpha }c_{jk\alpha \beta }dz\wedge d\bar{z}_k\otimes e_{\beta }, \end{aligned}$$

(16)

and

$$\begin{aligned} {[}i\Theta ,\Lambda _{\omega _2}]u=\sum _{jk\alpha \beta } u'_{j\alpha }c'_{jk\alpha \beta }dw\wedge d\bar{w}_k\otimes e_{\beta }. \end{aligned}$$

(17)

So it is easy to see $Im[i\Theta ,\Lambda _{\omega _1}] =Im[i\Theta ,\Lambda _{\omega _2}]$. We write

$$\begin{aligned}&{[}i\Theta ,\Lambda _{\omega _1}]^{-1}u=\sum _{jk\alpha \beta } u_{j\alpha }d_{jk\alpha \beta }dz\wedge d\bar{z}_k\otimes e_{\beta },\\&{[}i\Theta ,\Lambda _{\omega _2}]^{-1}u=\sum _{jk\alpha \beta } u'_{j\alpha }d'_{jk\alpha \beta }dw\wedge d\bar{w}_k\otimes e_{\beta }, \end{aligned}$$

Then from Eqs. (15),(16), (17), we can get

$$\begin{aligned} d'_{jk\alpha \beta }=\lambda _j\lambda _k d_{jk\alpha \beta }. \end{aligned}$$

We now assume that $\{e_\alpha \}$ are orthonormal at $z_0$. Then

$$\begin{aligned}&\langle [i\Theta ,\Lambda _{\omega _1}]^{-1}u,u\rangle _{\omega _1}d V_{\omega _1}=\sum _{jk\alpha \beta }d_{jk\alpha \beta }u_{j\alpha } \bar{u}_{k\beta }c_ndz\wedge d\bar{z},\\&\langle [i\Theta ,\Lambda _{\omega _2}]^{-1}u,u\rangle _{\omega _2}d V_{\omega _2}= \sum _{jk\alpha \beta }d'_{jk\alpha \beta }u'_{j\alpha } \bar{u'}_{k\beta }c_ndw\wedge d\bar{w}. \end{aligned}$$

Note also that

$$\begin{aligned} c_ndw\wedge d\bar{w}=\lambda ^2c_ndz\wedge d\bar{z}, \end{aligned}$$

We get

$$\begin{aligned} \langle [i\Theta ,\Lambda _{\omega _1}]^{-1}u,u\rangle _{\omega _1}d V_{\omega _1} =\langle [i\Theta ,\Lambda _{\omega _2}]^{-1}u, u\rangle _{\omega _2}dV_{\omega _2}. \end{aligned}$$

$\square$