# Curvature formulas and curvature strict positivity of direct image bundles

Xiangsen Qin  Xiangsen Qin: School of Mathematical Sciences, University of
Chinese Academy of Sciences
Beijing 100049, P. R. China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we consider the curvature strict positivity of direct image
bundles (vector bundles) associated to a strictly pseudoconvex family of
bounded domains. The main result is that the curvature of the direct image
bundle (vector bundles) associated to a strictly pseudoconvex family of
bounded domains is strictly positive in the sense of Nakano even if the
curvature of the original vector bundle is just Nakano positive. Based on our
(I and my coauthors) previous results, this result further demonstrates that
strictly pseudoconvex domains and pseudoconvex domains have very different
geometric properties. To consider the curvature strict positivity, we will
first construct the curvature formulas for vector bundles, then the curvature
strict positivity of a strictly pseudoconvex family of bounded domains will be
a simple consequence of it. As applications to convex analysis, we get a
corresponding version of Prékopa’s Theorem, i.e., we get the strict positivity
of a strictly convex family of bounded domains.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
    1. 2.1 Notations and conventions
    2. 2.2 Curvature positivity of Hermitian holomorphic vector bundles
  3. 3 Curvature formulas of direct image bundles
  4. 4 The proof of Theorem 1.2
  5. 5 The proof of Theorem 1.4

##  1\. Introduction

Let $U\subset\mathbb{C}^{n}$ and $D\subset\mathbb{C}^{m}$ be pseudoconvex
bounded domains, and let $\varphi$ be a smooth plurisubharmonic function
defined on some open neighborhood of the closure of $\Omega:=U\times D$ in
$\mathbb{C}^{n}\times\mathbb{C}^{m}$. For $t\in U$, we define the Hilbert
space

$E_{t}:=\\{f\in\mathcal{O}(D)|\
\|f\|^{2}_{t}:=\int_{D}|f(z)|^{2}e^{-\varphi(t,z)}d\lambda(z)<\infty\\},$

where $\mathcal{O}(D)$ is the space of holomorphic functions on $D$, and
$d\lambda(z)$ is the Lebesgue measure on $\mathbb{C}^{m}$. When $t$ varies in
$U$, $E_{t}$ is invariant as a vector space, but the underlying inner product
defined by the above norm varies if $\varphi$ is not constant with respect to
$t$. Set $E:=\cup_{t\in U}E_{t}$ and view it as a trivial vector bundle over
$U$ (of infinite rank) in a natural way, then $E$ is a holomorphic vector
bundle with a Hermitian metric $h$ given by

$h_{t}(f,g):=\int_{D}f(z)\overline{g(z)}e^{-\varphi(t,z)}d\lambda(z),\ \forall
f,g\in E_{t}.$

A fundamental result of Berndtsson is as follows.

######  Theorem 1.1 ([2, Theorem 1.1]).

With the above notations and assumptions, the curvature of $(E,h)$ is positive
in the sense of Nakano, and is strictly positive in the sense of Nakano if
$\varphi$ is strictly plurisubharmonic.

To generalize Theorem 1.1, we introduce some notions and notations which have
been used in [3] with a few slight modifications.
Let $p\colon\mathbb{C}^{n}\times\mathbb{C}^{m}\rightarrow\mathbb{C}^{n}$ be
the natural projection, and for a set
$A\subset\mathbb{C}^{n}\times\mathbb{C}^{m}$, we denote $p^{-1}(t)\cap A$ by
$A_{t}$, which is called the fiber of $A$ over $t$. Of course we can view
$A_{t}$ as a family of subsets in $\mathbb{C}^{m}$ depending on the parameter
$t$.

######  Definition 1.1.

Let $U\subset\mathbb{C}^{n}$ be a domain.

  * (1)

A _family of domains_ of dimension $m$ over $U$ is a domain $\Omega\subset
U\times\mathbb{C}^{m}$ such that $p(\Omega)=U$; such a family is called a
_family of bounded domains_ if all fibers $\Omega_{t}\subset\mathbb{C}^{m}\
(t\in U)$ are bounded.

  * (2)

A family of domains $\Omega$ over $U$ has _ $C^{k}$ ($k\geq 1$) boundary _if
there exists a $C^{k}$-function $\rho(t,z)$ defined on
$\mathbb{C}^{n}\times\mathbb{C}^{m}$ such that $\Omega=\\{(t,z)\in
U\times\mathbb{C}^{m}|\ \rho(t,z)<0\\}$ and $d(\rho|_{\Omega_{t}})\neq 0$ on
$\partial\Omega_{t}$ for all $t\in U$. Such a function $\rho$ is called a
_defining function_ of $\Omega$.

  * (3)

A family of domains $\Omega\subset U\times\mathbb{C}^{m}$ with
$C^{4}$-boundary is said to _have a plurisubharmonic defining function_ if it
admits a $C^{4}$-boundary defining function $\rho$ that is plurisubharmonic on
some neighborhood of $\overline{\Omega}$ (the closure of $\Omega$) and
$\rho(t,\cdot)$ is _strictly plurisubharmonic_ on $\overline{\Omega}_{t}$ for
any $t\in U$. It is called _strictly pseudoconvex_ if it admits a
$C^{4}$-boundary defining function that is strictly plurisubharmonic on some
neighborhood of $\overline{\Omega}$ .

Then we may generalize Theorem 1.1 to the following

######  Theorem 1.2.

Let $\Omega\subset U\times\mathbb{C}^{m}$ be a family of bounded domains which
has a plurisubharmonic defining function over a domain
$U\subset\mathbb{C}^{n},$ and let $(F,h^{F})$ be a Hermitian holomorphic
trivial vector bundle of finite rank $r$ defined on some neighborhood of
$\overline{\Omega}$. Let $\\{e_{1},\cdots,e_{r}\\}$ be the canonical
holomorphic frame of $F$. For any $t\in U$, set

$E_{t}:=\\{u:=\sum_{\lambda=1}^{r}u_{\lambda}dz_{1}\wedge\cdots\wedge
dz_{m}\otimes e_{\lambda}|\ u_{\lambda}\in\mathcal{O}(\Omega_{t})\cap
L^{2}(\Omega_{t})\text{ for all }\lambda\\}$

with an inner product $h_{t}$ given by

$h_{t}^{E}(u,v):=\int_{\Omega_{t}}\sum_{\lambda,\mu=1}^{r}u_{\lambda}(z)\overline{v_{\mu}(z)}h^{F}_{(t,z)}(e_{\lambda},e_{\mu})d\lambda(z),\
\forall u,v\in E_{t}.$

We set $E:=\cup_{t\in U}E_{t}$, then $(E,h^{E})$ is Nakano positive if
$(F,h^{F})$ is Nakano positive on $\Omega$, and it is strictly positive in the
sense of Nakano if $(F,h^{F})$ is Nakano strictly positive on $\Omega$.

For the definition of Nakano positivity (resp. strict positivity), see
Definition 2.1 and Definition 3.3. To prove Theorem 1.2, we construct a
curvature formula of $E$. More explicitly, we get the following

######  Theorem 1.3 (see Theorem 3.3).

Assumptions of $\Omega,F,E$ as in Theorem 1.2. For any $u_{1},\cdots,u_{n}\in
C^{2}(U,E)$, we have (locally)

$\displaystyle\quad h^{E}(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k})$ (1.1)
$\displaystyle=\sum_{\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}\left(u_{j},u_{k}\right)d\lambda+\sum_{p}\int_{\partial\Omega_{t}}H(\rho)_{j\bar{k}}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|},$

where

$L_{j}u_{j}:=D_{t_{j}}^{F}u_{j}+\sum_{p}D_{z_{p}}^{F}(\nu_{j}^{p}u_{j}),$
$H(h^{F})_{jk\lambda\mu}:=A_{jk\lambda\mu}+\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu},$
$H(\rho)_{j\bar{k}}:=\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)+\frac{(\rho_{j}-\sum_{s}\rho_{j\bar{s}}\rho^{s})(\rho_{\bar{k}}-\sum_{p}\overline{\rho_{k\bar{p}}}\rho^{\bar{p}})}{|\partial\rho|^{2}}.$

All notations and related concepts in Theorem 1.3 will be explained in Section
3. When $F$ is a line bundle, the curvature formula has been constructed in
[8], see [8, Theorem 2.11]. Different from the method using in [8] which
depends strongly on the horizontal lift of a vector field, the method we adopt
is more direct and we didn’t make any additional requirements on the vector
bundle $F$ (comparing formula (2.26) of [8]). After we construct the curvature
formula for vector bundles, Theorem 1.2 is then a direct consequence of it.
Note that to make things simple and to make the main idea clear, we only
consider sections of $E$ which are smooth up to the boundary of $\Omega$.
As mentioned in the abstract, our main purpose is to study strict positivity
of curvature of direct image bundles. Now we can state the main result as
follows.

######  Theorem 1.4.

Assumptions of $\Omega,F,E$ as in Theorem 1.2. We moreover assume $\Omega$ is
a strictly pseudoconvex family of bounded domains over $U$, then $(E,h^{E})$
is strictly positive in the sense of Nakano if $(F,h^{F})$ is just Nakano
positive on $\Omega$.

Note that we didn’t assume $(F,h^{F})$ is Nakano strictly positive, so we can
think that the strict positivity of the curvature of $(E,h^{E})$ comes from
the strict pseudoconvexity of $\Omega$.
Before we state the main idea of the proof of Theorem 1.4, let us first give
some applications of it to complex analysis and convex analysis.
By Theorem 1.4, Corollary 1.4 in [3] holds for an arbitrary strictly
pseudoconvex family of bounded domains, without symmetry. The proof is similar
to that proof, so we omit it here.

######  Corollary 1.5.

Let $\Omega\subset U\times\mathbb{C}^{m}$ be a strictly pseudoconvex family of
bounded domains over a domain $U\subset\mathbb{C}^{n}$ and let $\varphi$ be a
$C^{2}$ plurisubharmonic function which is continuous to the boundary of
$\Omega$. For any $t\in U$, let $K(t,z)$ be the weighted Bergman kernel of
$\Omega_{t}$ with weight $\varphi_{t},$ then $\ln K(t,z)$ is a strictly
plurisubharmonic function on $\Omega$.

Recall a domain $D\subset\mathbb{C}^{m}$ is called a circular domain if it is
invariant under the action of $\mathbb{S}^{1}$ given by

$e^{i\theta}\cdot(z_{1},\cdots,z_{m})=\left(e^{i\theta}z_{1},\cdots,e^{i\theta}z_{m}\right),\
\forall\theta\in\mathbb{R},$

and is called a Reinhardt domain if it is invariant under the action of
$\mathbb{T}^{m}$ given by

$\left(e^{i\theta_{1}},\cdots,e^{i\theta_{m}}\right)\cdot(z_{1},\cdots,z_{m})=\left(e^{i\theta_{1}}z_{1},\cdots,e^{i\theta_{m}}z_{m}\right),\
\forall\theta_{1},\cdots,\theta_{m}\in\mathbb{R}.$

It is clear that [4, Theorem 1.4] is a simple consequence of Theorem 1.4 (see
[3, Lemma 3.1] for the details). Note that we didn’t require $(F,h^{F})$ is
Nakano positive on some neighborhood of $\overline{\Omega}$ in Theorem 1.4.
However, the requirement $(F,h^{F})$ is Nakano positive on some larger domains
is indispensable in the proof of [4, Theorem 1.4]. For convenience, we state
the precise statement as follows. The proof is a simple modification of [3,
Lemma 3.1], so we omit here.

######  Corollary 1.6.

Let $\Omega\subset U\times\mathbb{C}^{m}$ be a strictly pseudoconvex family of
bounded domains over a domain $U\subset\mathbb{C}^{n},$ and let $(F,h^{F})$ be
a trivial vector bundle of finite rank $r$ defined on some neighborhood of
$\overline{\Omega}$. Let $\\{e_{1},\cdots,e_{r}\\}$ be the canonical
holomorphic frame of $F$. We assume that all fibers $\Omega_{t}\ (t\in U)$ are
(connected) Reinhardt domains and $h^{F}_{(t,z)}$ is
$\mathbb{T}^{n}$-invariant with respect to $z$ for any $t\in U$. For any $t\in
U$, set $E_{t}:=\mathbb{C}^{r}$, with an inner product given by

$h_{t}^{E}(u,v):=\int_{\Omega_{t}}\sum_{\lambda,\mu=1}^{r}u_{\lambda}\overline{v_{\mu}}h^{F}_{(t,z)}(e_{\lambda},e_{\mu})d\lambda(z),\
\forall u,v\in E_{t}.$

We set $E:=\cup_{t\in U}E_{t}$ and view it as a Hermitian holomorphic
(trivial) vector bundle over $U$, then the curvature of $(E,h^{E})$ is
strictly positive in the sense of Nakano if $(F,h^{F})$ is Nakano positive on
$\Omega$.

Of course, similar to Theorem 1.2 and Theorem 1.3 of [4], we may replace
$E_{t}$ by general homogeneous polynomials, or we can replace Reinhardt
domains by circular domains.
Using Corollary 1.6 and the same method in the proof of [4, Corollary 1.5], we
can get the strict positivity of a strict convex family of bounded domains
(the definition is similar to Definition 1.1, see [3, Page 6] or [4,
Definition 2.3]).

######  Corollary 1.7.

Let $D\subset U_{0}\times\mathbb{R}^{m}$ be a strictly convex family of
bounded domains (with connected fibers) over a domain
$U_{0}\subset\mathbb{R}^{n},$ and let $(F,g^{F})$ be a trivial vector bundle
of finite rank $r$ defined on some neighborhood of $\overline{D}$. Let
$\\{e_{1},\cdots,e_{r}\\}$ be the canonical frame of $F$. For any $t\in
U_{0}$, set $E_{t}:=\mathbb{R}^{r}$, with an inner product $g_{t}^{E}$ given
by

$g_{t}^{E}(u,v):=\int_{D_{t}}\sum_{\lambda,\mu=1}^{r}u_{\lambda}v_{\mu}g^{F}_{(t,x)}(e_{\lambda},e_{\mu})d\lambda(x)$

for all

$u:=(u_{1},\cdots,u_{r})\in E_{t},\ v:=(v_{1},\cdots,v_{n})\in E_{t}.$

We set $E:=\cup_{t\in U_{0}}E_{t}$ and view it as a Riemannian (trivial)
vector bundle over $U_{0}$, then the curvature of $(E,g^{E})$ is strictly
positive in the sense of Nakano if $(F,g^{F})$ is Nakano positive on $D$.

Corollary 1.7 is a corresponding result of Prékopa’s Theorem. For the
classical version of Prékopa’s Theorem, see [6]. For the matrix valued
version, see [7]. Note that different from that two papers, we here emphasize
the strict positivity of $(E,g^{E})$ when $D$ is a strictly convex family of
bounded domains.
Now we state the main idea of the proof of Theorem 1.4. Based on curvature
formulas of vector bundles, the first three terms in the right hand side of
Formula (1.3) is nonnegative after we summing all $j,k$. As for the last term
of Formula (1.3), which will produce a positive lower bound after summing all
$j,k$, by the following simple observation.

######  Lemma 1.8.

[5, Corolllary 1.7] Let $D\subset\mathbb{C}^{m}$ be a bounded domain with
$C^{1}$-boundary, then there is a constant $\delta:=\delta(D)>0$ such that

$\delta\int_{D}|f|^{2}d\lambda\leq\int_{\partial D}|f|^{2}dS,\ \forall
f\in\mathcal{O}(D)\cap C^{0}(\overline{D}),$

where $dS$ is the surface measure of $\partial D$, $C^{0}(\overline{D})$
denotes the space of complex valued functions that are continuous to the
boundary of $D$.

Lemma 1.8 plays an essential role in the proof of Theorem 1.4, see [5] for
more related inequalities. When $\Omega$ is strictly pseudoconvex with
$C^{2}$-boundary, Lemma 1.8 have appeared implicitly in the proof of [1,
Proposition 2.2.2].

######  Remark 1.1.

By Lemma 6.6 of [5], we may give a concrete value of the above $\delta$, and
then we can give a concrete lower bound for the curvature strict positivity of
$(E,h^{E})$, which only depends on $|\Omega_{t}|,|\partial\Omega_{t}|,m$ and
the boundary defining function of $\partial\Omega_{t}$.

### Acknowledgements

The author is grateful to Professor Fusheng Deng, his Ph.D. advisor, for
valuable discussions on related topics.

##  2\. Preliminaries

In this section, we fix some notations, conventions and collect some knowledge
that are needed in our discussions.

###  2.1. Notations and conventions

Our convection for $\mathbb{N}$ is that $\mathbb{N}:=\\{1,2,3,\cdots\\}$, and
set $\mathbb{C}^{*}:=\\{z\in\mathbb{C}|\ z\neq 0\\}.$ Let $\mathbb{R}_{+}$
denote the set of positive real numbers. For any complex number
$z\in\mathbb{C}$, let $\operatorname{Re}(z)$ (resp. $\operatorname{Im}(z)$)
denote the real part (rep. imaginary part) of $z$. For any
$z:=(z_{1},\cdots,z_{n})\in\mathbb{C}^{n}$, set

$\operatorname{Re}(z):=(\operatorname{Re}(z_{1}),\cdots,\operatorname{Re}(z_{n})),\
\operatorname{Im}(z):=(\operatorname{Im}(z_{1}),\cdots,\operatorname{Im}(z_{n})).$

For any subset $A\subset\mathbb{R}^{n}$, let $\overline{A}$ denote the closure
of $A$ in $\mathbb{R}^{n}$, and let $\partial A:=\overline{A}\setminus A$. Any
neighborhood of a subset $A\subset\mathbb{R}^{n}$ is open by default. We say
$U$ is a domain of $\mathbb{R}^{n}$ if it is a connected open subset of
$\mathbb{R}^{n}$.
For any $k\in\mathbb{N}\cup\\{0,\infty\\}$. If $U\subset\mathbb{R}^{n}$ is an
open subset, then we let $C^{k}(U)$ be the space of complex valued functions
which are of class $C^{k}$. For any $f\in C^{1}(U)$, let $\nabla f$ denote its
gradient. We write $A\subset\subset U$ if $\overline{A}$ is a compact subset
of $U$.
If $U\subset\mathbb{C}^{n}$ is a open subset, let $\mathcal{O}(U)$ denote the
set of holomorphic functions in $U$. We recall an upper semi-continuous
function $f\colon U\rightarrow[-\infty,\infty)$ is plurisubharmonic if its
restriction to every complex line in $U$ is subharmonic, i.e. satisfying the
submean value inequality. A real valued function $\varphi\in C^{2}(U)$ is
strictly plurisubharmonic if its complex Hessian
$\left(\varphi_{j\bar{k}}\right)_{1\leq j,k\leq n}$ is positive definite for
any $z\in U$, where

$\varphi_{j\bar{k}}:=\frac{\partial^{2}\varphi}{\partial z_{j}\partial
z_{k}},\ 1\leq j,k\leq n,$

and we let $\left(\varphi^{j\bar{k}}\right)_{1\leq j,k\leq n}$ be the inverse
matrix of its complex Hessian. We recall that $U$ is strictly pseudoconvex if
it has $C^{2}$-boundary, and it has a $C^{2}$-boundary defining function which
is strictly plurisubahrmonic in a neighborhood of $\overline{U}$.
For any $k\in\mathbb{N}\cup\\{0,\infty\\}$. If $A\subset\mathbb{R}^{n}$ is an
arbitrary subset, let $C^{k}(A)$ denote the space of complex valued functions
which is of class $C^{k}$ in an open neighborhood of $A$.
Let $d\lambda$ denote the Lebesgue measure on $\mathbb{R}^{n}$. Note that all
measurable sets, measurable maps, measurable differential forms will be
Lebesgue measurable. If $A\subset\mathbb{R}^{n}$ is a measurable subset, we
also write $|A|$ for the measure of $A.$ If an open subset
$U\subset\mathbb{R}^{n}$ has $C^{1}$-boundary, let $dS$ denote the surface
measure on $\partial U$.
For any open subset $U\subset\mathbb{R}^{n}$, let $L^{2}(U)$ denote the set of
$L^{2}$-integrable complex valued functions. For any $f\in L^{2}(U)$, let
$\|f\|_{L^{2}(U)}$ denote the $L^{2}$-norm of $f$.
A metric on a vector bundle will be of class $C^{2}$, all sections of a vector
bundle will be global unless stated otherwise. Let $(E,h^{E})$ be a Hermitian
holomorphic vector bundle of finite rank $r$ over an open subset
$U\subset\mathbb{C}^{n}$, and let $e_{1},\cdots,e_{r}$ be a local holomorphic
frame of $E$. For any $x\in U$, let $h^{E}_{x}$ denote the inner product on
the fiber $E_{x}$ of $x$. The metric $h^{E}$ will always be complex linear
with respect to its first entry. For any $1\leq\alpha,\beta\leq r,$ set

$h^{E}_{\alpha\bar{\beta}}:=h^{E}(e_{\alpha},e_{\beta}),$

and let $\left((h^{E})^{\alpha\bar{\beta}}\right)_{1\leq\alpha,\beta\leq n}$
be the inverse matrix of
$\left(h^{E}_{\alpha\bar{\beta}}\right)_{1\leq\alpha,\beta\leq r}.$ Let $u,v$
be any measurable sections of $E$, write

$u=\sum_{\lambda=1}^{r}u_{\lambda}e_{\lambda},\
v=\sum_{\mu=1}^{r}v_{\mu}e_{\mu},$

then we let

$|u|^{2}:=\sum_{\lambda=1}^{r}|u_{\lambda}|^{2},\ \langle
u,v\rangle_{h^{E}}:=h^{E}(u,v)=\sum_{\lambda,\mu=1}^{r}u_{\lambda}\overline{v_{\mu}}h^{E}(e_{\lambda},e_{\mu}),$
$|u|_{h}^{2}:=\langle u,u\rangle_{h},\ (u,v)_{h}:=\int_{U}\langle
u,v\rangle_{h^{E}}d\lambda,\ \|u\|_{h^{E}}^{2}:=(u,u)_{h^{E}}.$

Let $L^{2}(U,E)$ be the set of measurable $E$-valued sections which are
$L^{2}$-integrable, i.e.

$L^{2}(U,E):=\\{u\text{ is a measurable }E\text{-valued section of }E|\
\|u\|_{h^{E}}^{2}<\infty\\}.$

For any $k\in\mathbb{N}\cup\\{0,\infty\\}$, let $C^{k}(U,E)$ be the set of
$E$-valued sections which are of class $C^{k}$.

###  2.2. Curvature positivity of Hermitian holomorphic vector bundles

Let $U\subset\mathbb{C}_{t}^{n}$ be a domain, and let $(E,h^{E})$ be a
Hermitian holomorphic trivial vector bundle over $U$ of finite rank $r$. The
Chern connection is now given by a collection of differential operators
$\\{D_{t_{j}}^{E}\\}_{1\leq j\leq n}$, which satisfies

$\partial_{t_{j}}h^{E}(u,v)=h^{E}(D_{t_{j}}^{E}u,v)+h^{E}(u,\bar{\partial}_{t_{j}}v),\
\forall u,v\in C^{2}(U,E),$

where $\partial_{t_{j}}:=\frac{\partial}{\partial t_{j}}$ and
$\bar{\partial}_{t_{j}}:=\frac{\partial}{\partial\bar{t}_{j}}$ for any $1\leq
j\leq n.$ For any $1\leq j,k\leq n,$ let
$\Theta^{(E,h^{E})}_{jk}:=[D_{t_{j}}^{E},\bar{\partial}_{t_{k}}]$ (Lie
bracket), then the Chern curvature of $(E,h^{E})$ is given by

$\Theta^{(E,h^{E})}=\sum_{j,k=1}^{n}\Theta^{(E,h^{E})}_{jk}dt_{j}\wedge
d\bar{t}_{k}.$

######  Definition 2.1.

The curvature of $(E,h^{E})$ is said to be positive (resp. strictly positive)
in the sense of Nakano if for any nonzero $n$-tuple $(u_{1},\cdots,u_{n})\\\
\in C^{\infty}(U,E),$ we have

$\sum_{j,k=1}^{n}h^{E}\left(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k}\right)\geq 0\
(\text{resp.}>0).$

##  3\. Curvature formulas of direct image bundles

In this section, we will construct curvature formulas for vector bundles. To
do this, let us firstly fix some notations and introduce some basic concepts.
Let $\Omega\subset U\times\mathbb{C}^{m}$ be a family of bounded domains which
has a plurisubharmonic defining function over a domain
$U\subset\mathbb{C}^{n},$ let $\rho$ be a $C^{4}$-boundary defining function
of it, and let $(F,h^{F})$ be a Hermitian holomorphic trivial vector bundle of
finite rank $r$ defined on some neighborhood of $\overline{\Omega}$. Let
$j,k=1,\cdots,n$ represent the indices of the components of
$t=(t_{1},\cdots,t_{n})$, $p,q,s,a=1,\cdots,m$ represent the indices of the
components of $z=(z_{1},\cdots,z_{m})$,
$\lambda,\mu,\alpha,\beta,\gamma=1,\cdots,r$ represent the indices of the
components of $F$, and let $\\{e_{1},\cdots,e_{r}\\}$ be the canonical
holomorphic frame of $F$. For any $t\in U$, set

$E_{t}:=\\{u:=\sum_{\lambda}u_{\lambda}dz_{1}\wedge\cdots\wedge dz_{m}\otimes
e_{\lambda}|\ u_{\lambda}\in\mathcal{O}(\Omega_{t})\cap
L^{2}(\Omega_{t})\text{ for all }\lambda\\}$

with an inner product $h_{t}^{E}$ given by

$h_{t}^{E}(u,v):=\int_{\Omega_{t}}\sum_{\lambda,\mu}u_{\lambda}(z)\overline{v_{\mu}(z)}h^{F}_{(t,z)}(e_{\lambda},e_{\mu})d\lambda(z),\
\forall u,v\in E_{t}.$

We set $E:=\cup_{t\in U}E_{t}$, and view it as a Hermitian holomorphic
(trivial) vector bundle over $U$ in some sense (which we don’t pursue in this
direction). To simplify the notation, we always write $dz$ for
$dz_{1}\wedge\cdots\wedge dz_{m}$.

######  Definition 3.1.

We call $u\colon U\rightarrow E,\ t\mapsto
u(t):=\sum_{\lambda=1}^{r}u_{\lambda}(t,z)dz\otimes e_{\lambda}\in E_{t}$ a
$C^{k}$-section ($0\leq k\leq\infty$) of $E$ if $u_{\lambda}\in
C^{k}(\overline{\Omega},\mathbb{C})$ for any $1\leq\lambda\leq r$. Let
$C^{k}(U,E)$ denote the space of all $C^{k}$-sections of $E$.

Note that any $C^{k}$-section of $E$ can be naturally regarded as a
$C^{k}$-section of $F.$
To give the definition of Chern curvature, let us recall some knowledge that
have been used in [8]. Set

$\rho_{j}:=\frac{\partial\rho}{\partial t_{j}},\
\rho_{\bar{j}}:=\overline{\rho_{j}},\ \rho_{p}:=\frac{\partial\rho}{\partial
z_{p}},\ \rho_{\bar{p}}:=\overline{\rho_{p}},$
$\rho^{p}:=\sum_{q}\rho^{p\bar{q}}\rho_{q},\
\rho^{\bar{p}}:=\overline{\rho^{p}},\
|\partial\rho|^{2}:=\sum_{p}\rho_{\bar{p}}\rho^{p},$
$\nu_{j}^{p}:=\frac{\rho_{j}\rho^{\bar{p}}}{\rho-|\partial\rho|^{2}}-\sum_{q}\left(\rho_{j\bar{q}}\rho^{q\bar{p}}+\frac{\rho_{j\bar{q}}\rho^{q}\rho^{\bar{p}}}{\rho-|\partial\rho|^{2}}\right),$
$V_{j}:=\frac{\partial}{\partial
t_{j}}+\sum_{p}\nu_{j}^{p}\frac{\partial}{\partial z_{p}},\
V_{\bar{j}}:=\overline{V_{j}}.$

It is clear that $\nu_{j}^{p}\in C^{2}(\overline{\Omega})$. A direct
computation shows that $V_{j}(\rho)|_{\partial\Omega}=0.$ By Corollary A.4 and
Lemma 2.9 (and its proof) of [8], we have the following

######  Lemma 3.1.

Let $f\in C^{1}(\overline{\Omega})$, then for any $t_{0}\in U$, there is an
open neighborhood $V\subset U$ of $t_{0}$ such that, for any $t\in V$, we have

$\frac{\partial}{\partial
t_{j}}\int_{\Omega_{t}}f(t,z)d\lambda(z)=\int_{\Omega_{t}}\left(\frac{\partial
f}{\partial t_{j}}(t,z)+\sum_{p}\frac{\partial}{\partial
z_{p}}(\nu_{j}^{p}f)(t,z)\right)d\lambda(z).$

Let us also recall the definition of $\bar{\partial}$-operator on $E$. For any
$u\in C^{1}(U,E)$, write

$u:=\sum_{\lambda}u_{\lambda}dz\otimes e_{\lambda},\
\bar{\partial}_{t_{j}}u:=\frac{\partial
u_{\lambda}}{\partial\bar{t}_{j}}dz\otimes e_{\lambda},$

then

$\bar{\partial}u:=\sum_{j=1}^{n}d\bar{t}_{j}\wedge\bar{\partial}_{t_{j}}u$

is the $\bar{\partial}$-operator of $E$.
Now we give the definition of Chern curvature of $E$.

######  Definition 3.2.

The Chern connection on $E$ is a collection of differential operators
$\\{D_{t_{j}}^{E}\\}_{1\leq j\leq n}$ on $E$, which satisfies

$\partial_{t_{j}}h^{E}(u,v)=h^{E}(D_{t_{j}}^{E}u,v)+h^{E}(u,\bar{\partial}_{t_{j}}v),\
\forall u,v\in C^{1}(U,E).$

For any $t\in U$, set

$G_{t}:=\\{u:=\sum_{\lambda}u_{\lambda}dz\otimes e_{\lambda}|\ u_{\lambda}\in
L^{2}(\Omega_{t})\text{ for all }\lambda\\}$

with an inner product $h^{G}_{t}$ given by

$h_{t}^{G}(u,v):=\int_{\Omega_{t}}\sum_{\lambda,\mu}u_{\lambda}(z)\overline{v_{\mu}(z)}h^{F}_{(t,z)}(e_{\lambda},e_{\mu})d\lambda(z),\
\forall u,v\in G_{t},$

then $G:=\cup_{t\in U}G_{t}$ can be viewed as a Hermitian holomorphic
(trivial) vector bundle over $U.$ Let $\pi\colon G\rightarrow E$ be the
fiberwise orthogonal projection, and let $\pi_{\perp}$ be the orthogonal
projection on the orthogonal complement of $E$ in $G.$
Now we prove the existence and uniqueness of the Chern connection over $E$.

######  Theorem 3.2.

There exists a unique Chern connection on $E$, and it is (locally) given by

$D_{t_{j}}^{E}u=\pi\left(D_{t_{j}}^{F}u+\sum_{p}\nu_{j}^{p}D_{z_{p}}^{F}u+\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}u\right),\ \forall u\in C^{1}(U,E),$

where if $u:=\sum_{\lambda}u_{\lambda}dz\otimes e_{\lambda}$, then

$D_{z_{p}}^{F}u:=\sum_{\lambda}\frac{\partial u_{\lambda}}{\partial
z_{p}}dz\otimes e_{\lambda}+\sum_{\lambda}u_{\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}.$

###### Proof..

By Lemma 3.1, we know

$\displaystyle\quad h^{E}(D_{t_{j}}^{E}u,v)+h^{E}(u,\frac{\partial
v}{\partial\bar{t}_{j}})$ $\displaystyle=\frac{\partial}{\partial
t_{j}}h^{E}(u,v)$ $\displaystyle=\int_{\Omega_{t}}\frac{\partial}{\partial
t_{j}}h^{F}(u,v)d\lambda+\sum_{p}\int_{\Omega_{t}}\frac{\partial}{\partial
z_{p}}(\nu_{j}^{p}h^{F}(u,v))d\lambda$
$\displaystyle=\int_{\Omega_{t}}h^{F}(D_{t_{j}}^{F}u,v)d\lambda+\int_{\Omega_{t}}h^{F}(u,\frac{\partial
v}{\partial\bar{t}_{j}})d\lambda+\int_{\Omega_{t}}h^{F}(D_{z_{p}}^{F}(\nu_{j}^{p}u),v)d\lambda.$

Since $v$ was arbitrary, the proof is completed. ∎

Let $\Theta^{(E,h^{E})}_{jk}:=[D_{t_{j}}^{E},\bar{\partial}_{t_{k}}]$, then
the Chern curvature of $(E,h^{E})$ is given by

$\Theta^{(E,h^{E})}:=\sum_{j,k=1}^{n}\Theta^{(E,h^{E})}_{jk}dt_{j}\wedge
d\bar{t}_{k}.$

We give the definition of Nakano positivity of $(E,h^{E})$. This is similar to
Definition 2.1.

######  Definition 3.3.

The curvature of $(E,h^{E})$ is said to be positive (resp. strictly positive)
in the sense of Nakano if for any nonzero $n$-tuple $(u_{1},\cdots,u_{n})\\\
\in C^{\infty}(U,E),$ we have

$\sum_{j,k=1}^{n}h^{E}(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k})\geq 0\
(\text{resp.}>0).$

Now we compute Chern curvature of $(E,h^{E})$. Write

$D_{t_{j}}^{F}e_{\lambda}=\sum_{\mu}\Gamma_{j\lambda}^{\mu}dt_{j}\otimes
e_{\mu},\
D_{z_{p}}^{F}e_{\lambda}=\sum_{\mu}\Gamma_{p\lambda}^{\mu}dz_{p}\otimes
e_{\mu}.$

We adopt the following notations

$\left(\begin{array}[]{cc}(A_{pq\lambda}^{\alpha})&(A_{pj\lambda}^{\alpha})\\\
(A_{jp\lambda}^{\alpha})&(A_{jk\lambda}^{\alpha})\end{array}\right):=\left(\begin{array}[]{cc}\left(-\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{z}_{q}}\right)&\left(-\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{t}_{j}}\right)\\\
\left(-\frac{\partial\Gamma_{j\lambda}}{\partial\bar{z}_{p}}\right)&\left(-\frac{\partial\Gamma_{j\lambda}^{\alpha}}{\partial\bar{t}_{k}}\right)\end{array}\right),$
$\left(\begin{array}[]{cc}(A_{pq\lambda\mu})&(A_{pj\lambda\mu})\\\
(A_{jp\mu\lambda})&(A_{jk\lambda\mu})\end{array}\right):=\left(\begin{array}[]{cc}\left(-\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{z}_{q}}h^{F}_{\alpha\mu}\right)&\left(-\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{t}_{j}}h^{F}_{\alpha\mu}\right)\\\
\left(-\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{t}_{j}}h^{F}_{\alpha\mu}\right)^{*}&\left(-\frac{\partial\Gamma_{j\lambda}^{\alpha}}{\partial\bar{t}_{k}}h^{F}_{\alpha\mu}\right)\end{array}\right),$

where ∗ represents the conjugate transpose of a matrix. It is clear that

$\Theta_{jk}^{(F,h^{F})}=[D_{t_{j}}^{F},\bar{\partial}_{t_{k}}]=\sum_{\lambda,\alpha}A_{jk\lambda}^{\alpha}e^{\lambda}\otimes
e_{\alpha},$

where $e^{1},\cdots,e^{n}$ is the dual basis of $e_{1},\cdots,e_{n}$.
Now we can state the curvature formula for vector bundles. Since it involves
many computations, its proof is given in the appendix Appendix.

######  Theorem 3.3.

For any $u_{1},\cdots,u_{n}\in C^{2}(U,E)$, we have (locally)

$\displaystyle\quad h^{E}\left(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k}\right)$
$\displaystyle=\sum_{\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}\left(u_{j},u_{k}\right)d\lambda+\sum_{p}\int_{\partial\Omega_{t}}H(\rho)_{j\bar{k}}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|},$

where

$L_{j}u_{j}:=D_{t_{j}}^{F}u_{j}+\sum_{p}D_{z_{p}}^{F}(\nu_{j}^{p}u_{j}),$
$H(h^{F})_{jk\lambda\mu}:=A_{jk\lambda\mu}+\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu},$
$H(\rho)_{j\bar{k}}:=\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)+\frac{(\rho_{j}-\sum_{s}\rho_{j\bar{s}}\rho^{s})(\rho_{\bar{k}}-\sum_{p}\overline{\rho_{k\bar{p}}}\rho^{\bar{p}})}{|\partial\rho|^{2}}.$

In particular, the curvature operator $\Theta_{jk}^{(E,h^{E})}$ is pointwise
defined, i.e. it does not involve derivatives with respect to $t\in U$.

##  4\. The proof of Theorem 1.2

Let us first give a lemma that will be used in the proof of Theorem 1.2. Its
proof is also given in the appendix Appendix.

######  Lemma 4.1.

If $F$ is Nakano strictly positive on $\Omega$, then for any
$u_{1},\cdots,u_{n}\in C^{2}(U,E)$, we have (locally)

$\displaystyle\quad\sum_{j,k,\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\sum_{j,k}\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{j,k,p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}(u_{j},u_{k})d\lambda$
$\displaystyle\geq\int_{\Omega_{t}}\sum_{j,k,\beta,\gamma}(A_{jk\beta\gamma}-\sum_{p,q,\lambda,\mu}B_{pq\lambda\mu}A_{jq\beta\mu}A_{pk\lambda\gamma})u_{j\beta}\overline{u_{k\gamma}}d\lambda,$

where $B$ is uniquely determined by $A$ and satisfies

$\sum_{q,\beta}A_{pq\lambda\beta}B_{sq\alpha\beta}=\delta_{ps}\delta_{\lambda\alpha}.$

By the usual approximation technique, we easily have

######  Corollary 4.2.

If $F$ is just Nakano positive on $\Omega$, then for any
$u_{1},\cdots,u_{n}\in C^{2}(U,E)$, we have (locally)

$\displaystyle\quad\sum_{j,k,\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\sum_{j,k}\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{j,k,p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}(u_{j},u_{k})d\lambda$ $\displaystyle\geq 0.$

Now we can prove Theorem 1.2. For convenience, we restate it here.

######  Theorem 4.3 (= Theorem 1.2).

Let $\Omega\subset U\times\mathbb{C}^{m}$ be a family of bounded domains which
has a plurisubharmonic defining function over a domain
$U\subset\mathbb{C}^{n},$ and let $(F,h^{F})$ be a Hermitian holomorphic
trivial vector bundle of finite rank $r$ defined on some neighborhood of
$\overline{\Omega}$. Let $\\{e_{1},\cdots,e_{r}\\}$ be the canonical
holomorphic frame of $F$. For any $t\in U$, set

$E_{t}:=\\{u:=\sum_{\lambda}u_{\lambda}dz_{1}\wedge\cdots\wedge dz_{m}\otimes
e_{\lambda}|\ u_{\lambda}\in\mathcal{O}(\Omega_{t})\cap
L^{2}(\Omega_{t})\text{ for all }\lambda\\}$

with an inner product $h_{t}$ given by

$h_{t}^{E}(u,v):=\int_{\Omega_{t}}\sum_{\lambda,\mu}u_{\lambda}(z)\overline{v_{\mu}(z)}h^{F}_{(t,z)}(e_{\lambda},e_{\mu})d\lambda(z),\
\forall u,v\in E_{t}.$

We set $E:=\cup_{t\in U}E_{t}$, then $(E,h^{E})$ is Nakano positive if
$(F,h^{F})$ is Nakano positive on $\Omega$, and it is strictly positive in the
sense of Nakano if $(F,h^{F})$ is Nakano strictly positive on $\Omega$.

###### Proof..

Notation as above and as in Appendix Appendix. To prove the theorem, we only
need to consider the case $(F,h^{F})$ is Nakano strictly positive by the usual
approximation technique, then we can use Lemma 5.3. By Theorem 3.3, Lemma 4.1,
Lemma 5.3, and Corollary 5.4, it suffices to prove that

$\int_{\Omega_{t}}\sum_{j,k,\beta,\gamma}(A_{jk\beta\gamma}-\sum_{p,q,\lambda,\mu}B_{pq\lambda\mu}A_{jq\beta\mu}A_{pk\lambda\gamma})u_{j\beta}\overline{u_{k\gamma}}d\lambda>0$

for any nonzero smooth sections $u_{1},\cdots,u_{n}$ of $E$.
Let $(W,\langle-,-\rangle)$ be the (complex) inner product space whose
orthonormal basis is given by $\\{d\bar{z}_{p}\otimes e_{\lambda},\
d\bar{t}_{j}\otimes e_{\mu}|\ \forall p,j,\lambda,\mu\\}$, and let (for the
simplicity of notations, we omit $dz$ here)

$U:=\operatorname{span}_{\mathbb{C}}\\{d\bar{z}_{p}\otimes e_{\lambda}|\
\forall p,\lambda\\},\
V:=\operatorname{span}_{\mathbb{C}}\\{d\bar{t}_{j}\otimes e_{\lambda}|\
\forall j,\lambda\\}.$

We claim that the linear map

$S:=\left(\begin{array}[]{cc}S_{1}&S_{2}\\\ S_{3}&S_{4}\end{array}\right)$

is positive define. In fact, we have

$\displaystyle\quad S(\sum_{p,\lambda}u_{p\lambda}d\bar{z}_{p}\otimes
e_{\lambda}+\sum_{j,\lambda}u_{j\lambda}d\bar{t}_{j}\otimes e_{\lambda})$
$\displaystyle=\sum_{p,q,\lambda,\mu}A_{pq\lambda\mu}u_{p\lambda}d\bar{z}_{q}\otimes
e_{\mu}+\sum_{j,q,\lambda,\mu}A_{jq\lambda\mu}u_{j\lambda}d\bar{z}_{q}\otimes
e_{\mu}$
$\displaystyle\quad+\sum_{p,k,\lambda,\mu}A_{pk\lambda\mu}u_{p\lambda}d\bar{t}_{k}\otimes
e_{\mu}+\sum_{j,k,\lambda,\mu}A_{jk\lambda\mu}u_{j\lambda}d\bar{t}_{k}\otimes
e_{\mu}$

and

$\displaystyle\quad\langle S(\sum_{p,\lambda}u_{p\lambda}d\bar{z}_{p}\otimes
e_{\lambda}+\sum_{j,\lambda}u_{j\lambda}d\bar{t}_{j}\otimes
e_{\lambda},\sum_{p,\lambda}u_{p\lambda}d\bar{z}_{p}\otimes
e_{\lambda}+\sum_{j,\lambda}u_{j\lambda}d\bar{t}_{j}\otimes
e_{\lambda}\rangle$
$\displaystyle=\sum_{p,q,\lambda,\mu}A_{pq\lambda\mu}u_{p\lambda}\overline{u_{q\mu}}+\sum_{j,q,\lambda,\mu}A_{jq\lambda\mu}u_{j\lambda}\overline{u_{q\mu}}$
$\displaystyle\quad+\sum_{k,p,\lambda,\mu}A_{pk\lambda\mu}u_{p\lambda}\overline{u_{k\mu}}+\sum_{j,k,\lambda,\mu}A_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}$

Since $(F,h^{F})$ is Nakano strictly positive, then we know $S$ is positive
define, and then we know $S_{4}-S_{3}S_{1}^{-1}S_{2}$ is positive definite by
Lemma 5.3.
However, from

$\displaystyle\quad(S_{4}-S_{3}S_{1}^{-1}S_{2})(d\bar{t}_{j}\otimes
e_{\lambda})$ $\displaystyle=(S_{4}-S_{3}BS_{2})(d\bar{t}_{j}\otimes
e_{\lambda})$ $\displaystyle=\sum_{k,\mu}A_{jk\lambda\mu}d\bar{t}_{k}\otimes
e_{\mu}-\sum_{p,q,k,\mu,\alpha,\beta}A_{jq\lambda\mu}B_{pq\alpha\mu}A_{pk\alpha\beta}d\bar{t}_{k}\otimes
e_{\beta}$

we know

$\displaystyle\quad\langle(S_{4}-S_{3}S_{1}^{-1}S_{2})(\sum_{j,\lambda}u_{j\lambda}d\bar{t}_{j}\otimes
e_{\lambda}),\sum_{j,\lambda}u_{j\lambda}d\bar{t}_{j}\otimes
e_{\lambda}\rangle$
$\displaystyle=\sum_{j,k,\lambda,\mu}A_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}-\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}A_{jq\lambda\mu}B_{pq\alpha\mu}A_{pk\alpha\beta}u_{j\lambda}\overline{u_{k\beta}}$
$\displaystyle=\sum_{j,k,\beta,\gamma}\left(A_{jk\beta\gamma}-\sum_{p,q,\lambda,\mu}B_{pq\lambda\mu}A_{jq\beta\mu}A_{pk\lambda\gamma}\right)u_{j\beta}\overline{u_{k\gamma}}$

so the positive definiteness of $S_{4}-S_{3}S_{1}^{-1}S_{2}$ implies what we
wanted. ∎

##  5\. The proof of Theorem 1.4

Now we can prove Theorem 1.4, which is a direct consequence of Theorem 3.3.

######  Theorem 5.1 (= Theorem 1.4).

Assumptions of $\Omega,F,E$ as in Theorem 1.2. We moreover assume $\Omega$ is
a strictly pseudoconvex family of bounded domains over $U$, then $(E,h^{E})$
is strictly positive in the sense of Nakano if $(F,h^{F})$ is just Nakano
positive on $\Omega$.

###### Proof..

Since the curvature operator of $(E,h^{E})$ is pointwise defined, we may prove
the Theorem for any fixed $t_{0}\in U$. For any smooth sections
$u_{1},\cdots,u_{n}$ of $E$. By Theorem 3.3 and Corollary 4.2, we know

$h^{E}\left(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k}\right)\geq\sum_{j,k}\int_{\partial\Omega_{t_{0}}}H_{0}(\rho)_{j\bar{k}}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|},$

where

$H_{0}(\rho)_{jk}:=\rho_{j\bar{k}}-\sum_{p,q}\rho_{j\bar{p}}\rho^{p\bar{q}}\rho_{q\bar{k}}.$

By Corollary 5.4, there is a constant $\delta_{1}>0$ such that
$\frac{H_{0}(\rho)(z)}{|\nabla_{z}\rho(t_{0},z)|}-\delta_{1}I$ is positive
definite (as matrices) for any $z\in\partial\Omega_{t_{0}}$, where $I$ is the
identity matrix. Recall for any positive definite matrix $A$, there is a
positive matrix $B$, such that $A=B^{*}B,$ where $B^{*}$ is the conjugate
transpose of $B$. Thus, as $h$ is also positive definite, there is a constant
$\delta_{2}>0$ such that

$\displaystyle\sum_{j,k}\int_{\partial\Omega_{t_{0}}}H_{0}(\rho)_{j\bar{k}}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|}$
$\displaystyle\geq\delta_{1}\int_{\partial\Omega_{t_{0}}}\sum_{j}h^{F}(u_{j},u_{j})dS$
$\displaystyle\geq\delta_{2}\sum_{j}\int_{\partial\Omega_{t_{0}}}|u_{j}|^{2}dS.$

By Lemma 1.8, we know there is a constant $\delta_{3}>0$ such that

$\sum_{j}\int_{\partial\Omega_{t_{0}}}|u_{j}|^{2}dS\geq\delta_{3}\sum_{j}\int_{\Omega_{t_{0}}}|u_{j}|^{2}d\lambda,$

which completes the proof. ∎

## References

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  * [2] B. Berndtsson, _Curvature of vector bundles associated to holomorphic fibrations_ , Ann. of Math. (2) 169, 531-560(2009)
  * [3] F. Deng, F. Hu, X. Qin, _Curvature strict positivity of direct image bundles associated to pseudoconvex families of domains_ , Arxiv: 2301.00160, 2022
  * [4] F. Deng, F. Hu, W. Jiang, X. Qin, _A bridge connecting convex analysis and complex analysis and $L^{2}$-estimate of $d$ and $\bar{\partial}$_, Preprint
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## Appendix

In this section, we give several long computations. All notations are the same
with Section 3. Firstly, we give the proof of Theorem 3.3.

######  Theorem 5.2 (= Theorem 3.3).

For any $u_{1},\cdots,u_{n}\in C^{2}(U,E)$, we have (locally)

$\displaystyle\quad h^{E}\left(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k}\right)$
$\displaystyle=\sum_{\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}\left(u_{j},u_{k}\right)d\lambda+\sum_{p}\int_{\partial\Omega_{t}}H(\rho)_{j\bar{k}}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|},$

where

$L_{j}u_{j}:=D_{t_{j}}^{F}u_{j}+\sum_{p}D_{z_{p}}^{F}(\nu_{j}^{p}u_{j}),$
$H(h^{F})_{jk\lambda\mu}:=A_{jk\lambda\mu}+\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu},$
$H(\rho)_{j\bar{k}}:=\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)+\frac{(\rho_{j}-\sum_{s}\rho_{j\bar{s}}\rho^{s})(\rho_{\bar{k}}-\sum_{p}\overline{\rho_{k\bar{p}}}\rho^{\bar{p}})}{|\partial\rho|^{2}}.$

In particular, the curvature operator $\Theta_{jk}^{(E,h^{E})}$ is pointwise
defined, i.e. it does not involve derivatives with respect to $t\in U$.

###### Proof..

For any $u\in C^{1}(U,F)$, set

$N_{j}u:=D_{z_{p}}^{F}(\nu_{j}^{p}u)=\sum_{p}\nu_{j}^{p}D_{z_{p}}^{F}u+\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}u.$

By Lemma 3.1, we have

$\displaystyle\quad\frac{\partial}{\partial\bar{t}_{k}}\int_{\Omega_{t}}h^{F}(L_{j}u_{j},u_{k})d\lambda$
$\displaystyle=\int_{\Omega_{t}}\frac{\partial}{\partial\bar{t}_{k}}h^{F}(L_{j}u_{j},u_{k})d\lambda+\sum_{q}\int_{\Omega_{t}}\frac{\partial}{\partial\bar{z}_{q}}(\overline{\nu_{k}^{q}}h^{F}(L_{j}u_{j},u_{k}))d\lambda$
$\displaystyle=\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{t}_{k}}(L_{j}u_{j}),u_{k}\right)d\lambda+\int_{\Omega_{t}}h^{F}(L_{j}u_{j},D_{t_{k}}^{F}u_{k})d\lambda$
$\displaystyle\quad+\sum_{q}\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k}\right)d\lambda+\sum_{q}\int_{\Omega_{t}}h^{F}(L_{j}u_{j},D_{z_{q}}^{F}(\nu_{k}^{q}u_{k}))d\lambda$
$\displaystyle=\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{t}_{k}}(L_{j}u_{j}),u_{k}\right)d\lambda+\int_{\Omega_{t}}h^{F}(L_{j}u_{j},L_{k}u_{k})d\lambda$
$\displaystyle\quad+\sum_{q}\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k}\right)d\lambda,$

so we get

(5.1) $\displaystyle\quad
h^{E}\left(\Theta_{jk}^{(E,h^{E})}u_{j},u_{k}\right)$
$\displaystyle=h^{E}\left(D_{t_{j}}^{E}\left(\frac{\partial
u_{j}}{\partial\bar{t}_{k}}\right),u_{k}\right)-h^{E}\left(\frac{\partial}{\partial\bar{t}_{k}}(D_{t_{j}}^{E}u_{j}),u_{k}\right)$
$\displaystyle=h^{E}\left(D_{t_{j}}^{E}\left(\frac{\partial
u_{j}}{\partial\bar{t}_{k}}\right),u_{k}\right)-\frac{\partial}{\partial\bar{t}_{k}}h^{E}(D_{t_{j}}^{E}u_{j},u_{k})+h^{E}(D_{t_{j}}^{E}u_{j},D_{t_{k}}^{E}u_{k})$
$\displaystyle=h^{E}\left(\pi\left(L_{j}\left(\frac{\partial
u_{j}}{\partial\bar{t}_{k}}\right)\right),u_{k}\right)-\frac{\partial}{\partial\bar{t}_{k}}h^{E}(\pi(L_{j}u_{j}),u_{k})+h^{E}(D_{t_{j}}^{E}u_{j},D_{t_{k}}^{E}u_{k})$
$\displaystyle=\int_{\Omega_{t}}h^{F}\left(L_{j}\left(\frac{\partial
u_{j}}{\partial\bar{t}_{k}}\right),u_{k}\right)d\lambda-\frac{\partial}{\partial\bar{t}_{k}}\int_{\Omega_{t}}h^{F}(L_{j}u_{j},u_{k})d\lambda+h^{E}(D_{t_{j}}^{E}u_{j},D_{t_{k}}^{E}u_{k})$
$\displaystyle=\int_{\Omega_{t}}h^{F}\left([L_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j},u_{k}\right)d\lambda+h^{E}(D_{t_{j}}^{E}u_{j},D_{t_{k}}^{E}u_{k})$
$\displaystyle\quad-\int_{\Omega_{t}}h^{F}(L_{j}u_{j},L_{k}u_{k})d\lambda-\sum_{q}\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k}\right)d\lambda$
$\displaystyle=\int_{\Omega_{t}}h^{F}\left(\Theta_{jk}^{(F,h^{F})}u_{j},u_{k}\right)d\lambda-\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\int_{\Omega_{t}}h^{F}\left([N_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j},u_{k}\right)d\lambda-\sum_{q}\int_{\Omega_{t}}h^{F}\left(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k}\right)d\lambda$
$\displaystyle=\sum_{\lambda,\mu}\int_{\Omega_{t}}A_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\int_{\Omega_{t}}h^{F}\left([N_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j},u_{k}\right)d\lambda-\sum_{q}\int_{\Omega_{t}}h^{F}(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k})d\lambda.$

Note that

$\displaystyle\quad[N_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j}$
$\displaystyle=\sum_{p}\nu_{j}^{p}D_{z_{p}}^{F}\left(\frac{\partial
u_{j}}{\partial\bar{t}_{k}}\right)+\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}\frac{\partial
u_{j}}{\partial\bar{t}_{k}}-\sum_{p}\frac{\partial}{\partial\bar{t}_{k}}\left(\nu_{j}^{p}D_{z_{p}}^{F}u_{j}+\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}u_{j}\right)$
$\displaystyle=\sum_{p,\lambda}\nu_{j}^{p}\left(\frac{\partial^{2}u_{j\lambda}}{\partial
z_{p}\partial\bar{t}_{k}}dz\otimes e_{\lambda}+\frac{\partial
u_{j\lambda}}{\partial\bar{t}_{k}}dz\otimes D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad-\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}D_{z_{p}}^{F}u_{j}-\sum_{p}\nu_{j}^{p}\frac{\partial}{\partial\bar{t}_{k}}(D_{z_{p}}^{F}u_{j})-\sum_{p}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j}$
$\displaystyle=\sum_{p,\lambda}\nu_{j}^{p}\left(\frac{\partial^{2}u_{j\lambda}}{\partial
z_{p}\partial\bar{t}_{k}}dz\otimes e_{\lambda}+\frac{\partial
u_{j\lambda}}{\partial\bar{t}_{k}}dz\otimes D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad-\sum_{p,\lambda}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes
e_{\lambda}+\sum_{p,\lambda}u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad-\sum_{p,\lambda}\nu_{j}^{p}\frac{\partial}{\partial\bar{t}_{k}}\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes
e_{\lambda}+\sum_{p,\lambda}u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}\right)-\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j}$
$\displaystyle=-\sum_{p,\lambda}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes e_{\lambda}+u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad-\sum_{p,\lambda}\nu_{j}^{p}u_{j\lambda}dz\otimes\frac{\partial}{\partial\bar{t}_{k}}(D_{z_{p}}^{F}e_{\lambda})-\sum_{p}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j}$
$\displaystyle=-\sum_{p,\alpha}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\left(\frac{\partial
u_{j\alpha}}{\partial
z_{p}}+\sum_{\lambda}u_{j\lambda}\Gamma_{p\lambda}^{\alpha}\right)dz\otimes
e_{\alpha}$
$\displaystyle\quad-\sum_{p,\alpha}\left(\sum_{\lambda}\nu_{j}^{p}u_{j\lambda}\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{t}_{k}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j\alpha}\right)dz\otimes e_{\alpha},$

then we get

(5.2) $\displaystyle\quad
h^{F}([N_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j},u_{k})$
$\displaystyle=-\sum_{p,\alpha,\beta}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\left(\frac{\partial
u_{j\alpha}}{\partial
z_{p}}+\sum_{\lambda}u_{j\lambda}\Gamma_{p\lambda}^{\alpha}\right)\overline{u_{k\beta}}h_{\alpha\bar{\beta}}^{F}$
$\displaystyle\quad-\sum_{p,\alpha,\beta}\left(\sum_{\lambda}\nu_{j}^{p}u_{j\lambda}\frac{\partial\Gamma_{p\lambda}^{\alpha}}{\partial\bar{t}_{k}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j\alpha}\right)\overline{u_{k\beta}}h_{\alpha\bar{\beta}}^{F}$
$\displaystyle=-\sum_{p,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\frac{\partial
u_{j\lambda}}{\partial
z_{p}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}-\sum_{p,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}u_{j\lambda}\overline{u_{k\mu}}\frac{\partial
h_{\lambda\bar{\mu}}^{F}}{\partial z_{p}}$
$\displaystyle\quad+\sum_{p,\lambda,\mu}\nu_{j}^{p}A_{pk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}-\sum_{p,\lambda,\mu}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}.$

We also know

$\displaystyle\quad\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j})$
$\displaystyle=\sum_{\lambda}\frac{\partial}{\partial\bar{z}_{q}}\left(\frac{\partial
u_{j\lambda}}{\partial t_{j}}+\sum_{p}\nu_{j}^{p}\frac{\partial
u_{j\lambda}}{\partial z_{p}}+\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}u_{j\lambda}\right)dz\otimes e_{\lambda}$
$\displaystyle\quad+\sum_{\lambda,\mu}\frac{\partial}{\partial\bar{z}_{q}}\left(u_{j\lambda}\Gamma_{j\lambda}^{\mu}+\sum_{p}\nu_{j}^{p}u_{j\lambda}\Gamma_{p\lambda}^{\mu}\right)dz\otimes
e_{\mu}$
$\displaystyle=\sum_{p,\lambda}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\lambda}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}\right)dz\otimes e_{\lambda}$
$\displaystyle\quad-\sum_{\lambda,\mu}u_{j\lambda}\left(A_{jq\lambda}^{\mu}-\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\Gamma_{p\lambda}^{\mu}+\sum_{p}\nu_{j}^{p}A_{pq\lambda}^{\mu}\right)dz\otimes
e_{\mu},$

then we have

$\displaystyle\quad\sum_{q}h^{F}\left(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k}\right)$
$\displaystyle=\sum_{p,q,\lambda,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\lambda}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}\right)\overline{\nu_{k}^{q}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle\quad-\sum_{q,\lambda,\mu,\alpha}u_{j\lambda}\left(A_{jq\lambda}^{\alpha}-\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\Gamma_{p\lambda}^{\alpha}+\sum_{p}\nu_{j}^{p}A_{pq\lambda}^{\alpha}\right)\overline{\nu_{k}^{q}}\overline{u_{k\mu}}h_{\alpha\bar{\mu}}^{F}$
$\displaystyle=\sum_{p,q,\lambda,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\lambda}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}\right)\overline{\nu_{k}^{q}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}-\sum_{q,\lambda,\mu}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}$
$\displaystyle\quad+\sum_{p,q,\lambda,\mu}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}u_{j\lambda}\overline{u_{k\mu}}\frac{\partial
h_{\lambda\bar{\mu}}^{F}}{\partial
z_{p}}-\sum_{p}\nu_{j}^{p}\overline{\nu_{k}^{q}}u_{j\lambda}\overline{u_{k\mu}}A_{pq\lambda\mu}.$

Thus, we have

$\displaystyle\quad
h^{F}([N_{j},\frac{\partial}{\partial\bar{t}_{k}}]u_{j},u_{k})-\sum_{q}h^{F}(\frac{\partial}{\partial\bar{z}_{q}}(L_{j}u_{j}),\nu_{k}^{q}u_{k})$
$\displaystyle=-\sum_{p,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}\frac{\partial
u_{j\lambda}}{\partial
z_{p}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}-\sum_{p,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}u_{j\lambda}\overline{u_{k\mu}}\frac{\partial
h_{\lambda\bar{\mu}}^{F}}{\partial z_{p}}$
$\displaystyle\quad+\sum_{p,\lambda,\mu}\nu_{j}^{p}A_{pk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}-\sum_{p,\lambda,\mu}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle\quad-\sum_{p,q,\lambda,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\lambda}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}\right)\overline{\nu_{k}^{q}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}+\sum_{q,\lambda,\mu}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}$
$\displaystyle\quad-\sum_{p,q,\lambda,\mu}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}u_{j\lambda}\overline{u_{k\mu}}\frac{\partial
h_{\lambda\bar{\mu}}^{F}}{\partial
z_{p}}+\sum_{p}\nu_{j}^{p}\overline{\nu_{k}^{q}}u_{j\lambda}\overline{u_{k\mu}}A_{pq\lambda\mu}$
(5.3)
$\displaystyle=\sum_{\lambda,\mu}\left(\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu}\right)u_{j\lambda}\overline{u_{k\mu}}$
$\displaystyle\quad-\sum_{p,\lambda,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\overline{\nu_{k}^{q}}\right)\frac{\partial
u_{j\lambda}}{\partial z_{p}}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle\quad-\sum_{p,\lambda,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)u_{j\lambda}\overline{u_{k\mu}}\frac{\partial
h_{\lambda\bar{\mu}}^{F}}{\partial z_{p}}$
$\displaystyle\quad-\sum_{p,\lambda,\mu}\left(\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{t}_{k}}+\sum_{q}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}\overline{\nu_{k}^{q}}+\sum_{q}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}\right)u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle\quad+\sum_{p,q,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle=\sum_{\lambda,\mu}\left(\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu}\right)u_{j\lambda}\overline{u_{k\mu}}$
$\displaystyle\quad-\sum_{\lambda,\mu}\frac{\partial}{\partial
z_{p}}\left(\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}\right)$
$\displaystyle\quad+\sum_{p,q,\lambda,\mu}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}u_{j\lambda}\overline{u_{k\mu}}h_{\lambda\bar{\mu}}^{F}$
$\displaystyle=\sum_{\lambda,\mu}\left(\sum_{p}\nu_{j}^{p}A_{pk\lambda\mu}+\sum_{q}\overline{\nu_{k}^{q}}A_{jq\lambda\mu}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}A_{pq\lambda\mu}\right)u_{j\lambda}\overline{u_{k\mu}}$
$\displaystyle\quad-\frac{\partial}{\partial
z_{p}}\left(\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)h^{F}(u_{j},u_{k})\right)+\sum_{p,q}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}(u_{j},u_{k}).$

We claim $V_{\bar{k}}(V_{j}(\rho))|_{\partial\Omega}=0$. In fact,

$\displaystyle\quad V_{\bar{k}}(V_{j}(\rho))|_{\partial\Omega_{t}}$
$\displaystyle=V_{\bar{k}}\left.\left(\rho_{j}+\sum_{p}\left(\frac{\rho_{j}\rho^{\bar{p}}}{\rho-|\partial\rho|^{2}}-\sum_{q}\left(\rho_{j\bar{q}}\rho^{q\bar{p}}+\frac{\rho_{j\bar{q}}\rho^{q}\rho^{\bar{p}}}{\rho-|\partial\rho|^{2}}\right)\right)\rho_{p}\right)\right|_{\partial\Omega_{t}}$
$\displaystyle=V_{\bar{k}}\left.\left(\frac{\rho\rho_{j}}{\rho-|\partial\rho|^{2}}-\sum_{q}\frac{\rho\rho_{j\bar{q}}\rho^{q}}{\rho-|\partial\rho|^{2}}\right)\right|_{\partial\Omega_{t}}$
$\displaystyle=\left(-\frac{\rho_{j}}{|\partial\rho|^{2}}+\sum_{q}\frac{\rho_{j\bar{q}}\rho^{q}}{|\partial\rho|^{2}}\right)V_{\bar{k}}(\rho)|_{\partial\Omega_{t}}$
$\displaystyle=0$

By the Stokes theorem, we know

(5.4) $\displaystyle\quad-\int_{\Omega_{t}}\frac{\partial}{\partial
z_{p}}\left(\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)h^{F}(u_{j},u_{k})\right)d\lambda$
$\displaystyle=\sum_{p}\int_{\partial\Omega_{t}}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)\rho_{p}h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|}$
$\displaystyle=\int_{\partial\Omega_{t}}\left(V_{\bar{k}}(V_{j}(\rho))-\sum_{p}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{t}_{k}}+\sum_{q}\overline{\nu_{k}^{q}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\right)\rho_{p}\right)h^{F}(u_{j},u_{k})\frac{dS}{|\nabla\rho|}$
$\displaystyle=\int_{\partial\Omega_{t}}\left(\rho_{j\bar{k}}+\sum_{p}\nu_{j}^{p}\overline{\rho_{k\bar{p}}}+\sum_{q}\overline{\nu_{k}^{q}}\rho_{j\bar{q}}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}\rho_{p\bar{q}}\right)\frac{dS}{|\nabla\rho|}.$

On $\partial\Omega_{t}$, we have

$\nu_{j}^{p}=-\frac{\rho_{j}\rho^{\bar{p}}}{|\partial\rho|^{2}}-\sum_{s}\left(\rho_{j\bar{s}}\rho^{s\bar{p}}-\frac{\rho_{j\bar{s}}\rho^{s}\rho^{\bar{p}}}{|\partial\rho|^{2}}\right),$

then we know

$\sum_{p}\nu_{j}^{p}\overline{\rho_{k\bar{p}}}=-\sum_{p}\frac{\rho_{j}\rho^{\bar{p}}\overline{\rho_{k\bar{p}}}}{|\partial\rho|^{2}}-\sum_{p,s}\left(\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}-\frac{\rho_{j\bar{s}}\rho^{s}\rho^{\bar{p}}\overline{\rho_{k\bar{p}}}}{|\partial\rho|^{2}}\right),$
$\sum_{q}\overline{\nu_{k}^{q}}\rho_{j\bar{q}}=-\sum_{q}\frac{\rho_{\bar{k}}\rho^{q}\rho_{j\bar{q}}}{|\partial\rho|^{2}}-\sum_{a}\left(\overline{\rho_{k\bar{a}}}\overline{\rho^{a\bar{q}}}\rho_{j\bar{q}}-\frac{\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}\rho^{q}\rho_{j\bar{q}}}{|\partial\rho|^{2}}\right),$

and

$\displaystyle\quad\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}\rho_{p\bar{q}}$
$\displaystyle=-\sum_{q}\overline{\nu_{k}^{q}}\left(\frac{\rho_{j}\rho_{\bar{q}}}{|\partial\rho|^{2}}+\rho_{j\bar{q}}-\sum_{s}\frac{\rho_{j\bar{s}}\rho^{s}\rho_{\bar{q}}}{|\partial\rho|^{2}}\right)$
$\displaystyle=\sum_{q}\left(\frac{\rho_{\bar{k}}\rho^{q}}{|\partial\rho|^{2}}+\sum_{a}\left(\overline{\rho_{k\bar{a}}}\overline{\rho^{a\bar{q}}}-\frac{\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}\rho^{q}}{|\partial\rho|^{2}}\right)\right)\left(\frac{\rho_{j}\rho_{\bar{q}}}{|\partial\rho|^{2}}+\rho_{j\bar{q}}-\sum_{s}\frac{\rho_{j\bar{s}}\rho^{s}\rho_{\bar{q}}}{|\partial\rho|^{2}}\right)$
$\displaystyle=\frac{\rho_{j}\rho_{\bar{k}}}{|\partial\rho|^{2}}+\sum_{q}\frac{\rho_{\bar{k}}\rho^{q}\rho_{j\bar{q}}}{|\partial\rho|^{2}}-\sum_{s}\frac{\rho_{j\bar{s}}\rho^{s}\rho_{\bar{k}}}{|\partial\rho|^{2}}+\sum_{a}\frac{\rho_{j}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}+\sum_{a}\overline{\rho_{k\bar{a}}}\overline{\rho^{a\bar{q}}}\rho_{j\bar{q}}$
$\displaystyle\quad-\sum_{s,a}\frac{\rho_{j\bar{s}}\rho^{s}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}-\sum_{a}\frac{\rho_{j}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}-\sum_{q,a}\frac{\rho_{j\bar{q}}\rho^{q}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}+\sum_{s,a}\frac{\rho_{j\bar{s}}\rho^{s}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}$
$\displaystyle=\frac{\rho_{j}\rho_{\bar{k}}}{|\partial\rho|^{2}}+\sum_{a}\overline{\rho_{k\bar{a}}}\overline{\rho^{a\bar{q}}}\rho_{j\bar{q}}-\sum_{s,a}\frac{\rho_{j\bar{s}}\rho^{s}\overline{\rho_{k\bar{a}}}\rho^{\bar{a}}}{|\partial\rho|^{2}}.$

Hence, on $\partial\Omega_{t}$, we have

(5.5)
$\displaystyle\quad\rho_{j\bar{k}}+\sum_{p}\nu_{j}^{p}\overline{\rho_{k\bar{p}}}+\sum_{q}\overline{\nu_{k}^{q}}\rho_{j\bar{q}}+\sum_{p,q}\nu_{j}^{p}\overline{\nu_{k}^{q}}\rho_{p\bar{q}}$
$\displaystyle=\rho_{j\bar{k}}+\frac{\rho_{j}\rho_{\bar{k}}}{|\partial\rho|^{2}}-\sum_{p}\frac{\rho_{j}\rho^{\bar{p}}\overline{\rho_{k\bar{p}}}}{|\partial\rho|^{2}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}+\sum_{p,s}\frac{\rho_{j\bar{s}}\rho^{s}\rho^{\bar{p}}\overline{\rho_{k\bar{p}}}}{|\partial\rho|^{2}}-\sum_{q}\frac{\rho_{\bar{k}}\rho^{q}\rho_{j\bar{q}}}{|\partial\rho|^{2}}$
$\displaystyle=\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)+\frac{(\rho_{j}-\sum_{s}\rho_{j\bar{s}}\rho^{s})(\rho_{\bar{k}}-\overline{\rho_{k\bar{p}}}\rho^{\bar{p}})}{|\partial\rho|^{2}}.$

Combing Formulas (5.1), (5.2), (5.3), (5.4) and (5.5), we get the desired
result. ∎

We also give an elementary lemma of linear algebra. Although this lemma is
simple and well known, we contains its proof as it is important in the proof
of Theorem 1.2. For this, we recall some definitions.
Let $(W,\langle-,-\rangle)$ (the inner product is complex linear with respect
to the first entry) be a finite dimensional (complex) inner product space, and
let $T\colon W\rightarrow W$ be a linear map, then $T$ is positive definite
(resp. positive semi-definite) if and only if $\langle Tx,x\rangle>0$ (resp.
$\langle Tx,x\rangle\geq 0$) for any $x\in W\setminus\\{0\\}.$
Let $T\colon W_{1}\rightarrow W_{2}$ be a linear map between two finite
dimensional inner product spaces, then the adjoint map $T^{*}$ of $T$ is
defined by the following formula

$\langle Tx,y\rangle=\langle x,T^{*}y\rangle,\ \forall x\in W_{1},\ y\in
W_{2}.$

######  Lemma 5.3.

Let $(W,\langle-,-\rangle)$ be a finite dimensional (complex) inner product
space, $U$ be a linear subspace of $W$, and let $V$ be the orthogonal
complement of $U$ in $W$. Suppose we are given four linear maps

$T_{1}\colon U\rightarrow U,\ T_{2}\colon V\rightarrow U,\ T_{3}\colon
U\rightarrow V,\ T_{4}\colon V\rightarrow V$

such that

  * (1)

$T_{1}$ is invertible;

  * (2)

The linear map $T:=\left(\begin{array}[]{cc}T_{1}&T_{2}\\\
T_{3}&T_{4}\end{array}\right)$ (i.e. $T(u+v)=T_{1}u+T_{2}v+T_{3}u+T_{4}v$ for
any $u\in U,\ v\in V$) is positive definite (resp. positive semi-definite).

Then we know $T_{4}-T_{3}T_{1}^{-1}T_{2}\colon V\rightarrow V$ is also
positive definite (resp. positive semi-definite).

###### Proof..

We only need to consider the case that $T$ is positive definite. Let $S$ be
the linear map defined by $\left(\begin{array}[]{cc}I&-T_{1}^{-1}T_{2}\\\
0&I\end{array}\right),$ where $I$ is the identity map, then $S$ is invertible.
We claim that

$S^{*}=\left(\begin{array}[]{cc}I&0\\\
-(T_{1}^{-1}T_{2})^{*}&I\end{array}\right).$

In fact, for any $u_{1},u_{2}\in U,v_{1},v_{2}\in V$, we have (as $U\bot V$)

$\displaystyle\quad\langle u_{1}+v_{1},S^{*}(u_{2}+v_{2})\rangle$
$\displaystyle=\langle S(u_{1}+v_{1}),u_{2}+v_{2}\rangle$
$\displaystyle=\langle u_{1}-(T_{1}^{-1}T_{2})v_{1}+v_{1},u_{2}+v_{2}\rangle$
$\displaystyle=\langle u_{1}+v_{1},u_{2}+v_{2}\rangle-\langle
T_{1}^{-1}T_{2}v_{1},u_{2}+v_{2}\rangle$ $\displaystyle=\langle
u_{1}+v_{1},u_{2}+v_{2}\rangle-\langle T_{1}^{-1}T_{2}v_{1},u_{2}\rangle$
$\displaystyle=\langle u_{1}+v_{1},u_{2}+v_{2}\rangle-\langle
v_{1},(T_{1}^{-1}T_{2})^{*}u_{2}\rangle$ $\displaystyle=\langle
u_{1}+v_{1},u_{2}+v_{2}\rangle-\langle
u_{1}+v_{1},(T_{1}^{-1}T_{2})^{*}u_{2}\rangle$ $\displaystyle=\langle
u_{1}+v_{1},u_{2}-(T_{1}^{-1}T_{2})^{*}u_{2}+v_{2}\rangle$

Now for any $v\in V\setminus\\{0\\}$, we have

$\displaystyle 0$ $\displaystyle<\langle TSv,Sv\rangle$ $\displaystyle=\langle
S^{*}TSv,v\rangle$ $\displaystyle=\langle
S^{*}T(-T_{1}^{-1}T_{2}v+v),v\rangle$ $\displaystyle=\langle
S^{*}(-T_{1}(T_{1}^{-1}T_{2}v)+T_{2}v-T_{3}(T_{1}^{-1}T_{2}v)+T_{4}v),v\rangle$
$\displaystyle=\langle S^{*}(T_{4}v-T_{3}T_{1}^{-1}T_{2}v),v\rangle$
$\displaystyle=\langle(T_{4}-T_{3}T_{1}^{-1}T_{2})v,Sv\rangle$
$\displaystyle=\langle(T_{4}-T_{3}T_{1}^{-1}T_{2})v,v\rangle$

then we know $T_{4}-T_{3}T_{1}^{-1}T_{2}$ is positive definite. ∎

######  Corollary 5.4.

$\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)_{1\leq
j,k\leq n}$ is a positive semi-definite matrix for any
$z\in\overline{\Omega}$. If $\Omega$ is a strictly pseudoconvex family of
bounded domains over $U$, then
$\left(\rho_{j\bar{k}}-\sum_{p,s}\rho_{j\bar{s}}\rho^{s\bar{p}}\overline{\rho_{k\bar{p}}}\right)_{1\leq
j,k\leq n}$ is a positive definite matrix for any $z\in\overline{\Omega}$.

The following Lemma is a direct generalization of Theorem A.5 of [8].

######  Lemma 5.5.

If $F$ is Nakano strictly positive on $\Omega$. Fix $t_{0}\in U$, and suppose
there are an open neighborhood $V\subset U$ of $t_{0}$, $v_{1}\in C^{1}(V,F)$
and $v_{2}\in C^{0}(V,F)$ such that $v:=\sum_{s}dz_{s}\wedge
D_{z_{s}}^{F}v_{1}+v_{2}$ is $\bar{\partial}_{z}$-closed, then we may solve
$\bar{\partial}_{z}u=v$ on $\Omega_{t}$ ($t\in V$) with an estimate

$\int_{\Omega_{t}}h^{F}(u,u)d\lambda\leq\int_{\Omega_{t}}h^{F}(v_{1},v_{1})d\lambda+\int_{\Omega_{t}}h^{F}(A^{-1}v_{2},v_{2})d\lambda,$

provided the right hand side is finite.

Now we give the proof of Lemma 4.1.

######  Lemma 5.6 (= Lemma 4.1).

If $F$ is Nakano strictly positive on $\Omega$, then (locally)

(5.6)
$\displaystyle\quad\sum_{j,k,\lambda,\mu}\int_{\Omega_{t}}H(h^{F})_{jk\lambda\mu}u_{j\lambda}\overline{u_{k\mu}}d\lambda-\sum_{j,k}\int_{\Omega_{t}}h^{F}(\pi_{\perp}(L_{j}u_{j}),\pi_{\perp}(L_{k}u_{k}))d\lambda$
$\displaystyle\quad+\sum_{j,k,p,q}\int_{\Omega_{t}}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial\overline{\nu_{k}^{q}}}{\partial
z_{p}}h^{F}(u_{j},u_{k})d\lambda$
$\displaystyle\geq\int_{\Omega_{t}}\sum_{j,k,\beta,\gamma}(A_{jk\beta\gamma}-\sum_{p,q,\lambda,\mu}B_{pq\lambda\mu}A_{jq\beta\mu}A_{pk\lambda\gamma})u_{j\beta}\overline{u_{k\gamma}}d\lambda,$

where $B$ is uniquely determined by $A$ and satisfies

$\sum_{q,\beta}A_{pq\lambda\beta}B_{sq\alpha\beta}=\delta_{ps}\delta_{\lambda\alpha}.$

###### Proof..

Put

$v:=\sum_{j}\pi_{\perp}\left(L_{j}u_{j}\right).$

We will use Lemma 5.5 to give an estimate of
$\int_{\Omega_{t}}h^{F}(v,v)d\lambda$. Note that

$\displaystyle v$
$\displaystyle=\sum_{j}\bar{\partial}_{z}(\pi_{\perp}(L_{j}u_{j}))$
$\displaystyle=\sum_{j}\bar{\partial}_{z}(L_{j}u_{j})$
$\displaystyle=\sum_{j,\lambda}\bar{\partial}_{z}(\frac{\partial
u_{j\lambda}}{\partial t_{j}}dz\otimes e_{\lambda}+u_{j\lambda}dz\otimes
D_{t_{j}}^{F}e_{\lambda})$
$\displaystyle\quad+\sum_{j,p,\lambda}\bar{\partial}_{z}\nu_{j}^{p}\wedge(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes e_{\lambda}+u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda})$
$\displaystyle\quad+\sum_{j}\nu_{j}^{p}\bar{\partial}_{z}\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes e_{\lambda}+u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}\right)+\sum_{j,p,\lambda}\bar{\partial}_{z}\left(\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}\right)\wedge u_{j}$
$\displaystyle=\sum_{j,\lambda}\bar{\partial}_{z}(u_{j\lambda}dz\otimes
D_{t_{j}}^{F}e_{\lambda})+\sum_{j,p,\lambda}\bar{\partial}_{z}\nu_{j}^{p}\wedge\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\otimes e_{\lambda}+u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad+\sum_{j}\nu_{j}^{p}\bar{\partial}_{z}(u_{j\lambda}dz\otimes
D_{z_{p}}^{F}e_{\lambda})+\sum_{j,p,\lambda}\bar{\partial}_{z}\left(\frac{\partial\nu_{j}^{p}}{\partial
z_{p}}\right)\wedge u_{j}$
$\displaystyle=(-1)^{m}\sum_{j,\lambda}u_{j\lambda}dz\wedge\bar{\partial}_{z}(D_{t_{j}}^{F}e_{\lambda})$
$\displaystyle\quad+(-1)^{m}\sum_{j,p,q,\lambda}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\left(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\wedge d\bar{z}_{q}\otimes
e_{\lambda}+u_{j\lambda}dz\wedge d\bar{z}_{q}\otimes
D_{z_{p}}^{F}e_{\lambda}\right)$
$\displaystyle\quad+(-1)^{m}\sum_{j}\nu_{j}^{p}u_{j\lambda}dz\wedge\bar{\partial}_{z}(D_{z_{p}}^{F}e_{\lambda})+(-1)^{m}\sum_{j,p,q,\lambda}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}dz\otimes d\bar{z}_{q}\otimes
e_{\lambda}$
$\displaystyle=(-1)^{m}\sum_{j,\lambda}u_{j\lambda}\frac{\partial\Gamma_{j\lambda}^{\mu}}{\partial\bar{z}_{q}}dz\wedge
d\bar{z}_{q}\otimes e_{\mu}$
$\displaystyle\quad+(-1)^{m}\sum_{j,p,q,\lambda}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}(\frac{\partial
u_{j\lambda}}{\partial z_{p}}dz\wedge d\bar{z}_{q}\otimes
e_{\lambda}+u_{j\lambda}\Gamma_{p\lambda}^{\mu}dz\wedge d\bar{z}_{q}\otimes
e_{\mu})$
$\displaystyle\quad+(-1)^{m}\sum_{j}\nu_{j}^{p}u_{j\lambda}\frac{\partial\Gamma_{p\lambda}^{\mu}}{\partial\bar{z}_{q}}dz\wedge
d\bar{z}_{q}\otimes
e_{\mu}+(-1)^{m}\sum_{j,p,q,\lambda}\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\lambda}dz\otimes d\bar{z}_{q}\otimes
e_{\lambda}$
$\displaystyle=(-1)^{m}\sum_{j,q,\lambda,\mu}\left(u_{j\lambda}\frac{\partial\Gamma_{j\lambda}^{\mu}}{\partial\bar{z}_{q}}+\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}u_{j\lambda}\Gamma_{p\lambda}^{\mu}+\sum_{p}\nu_{j}^{p}u_{j\lambda}\frac{\partial\Gamma_{p\lambda}^{\mu}}{\partial\bar{z}_{q}}\right)dz\wedge
d\bar{z}_{q}\otimes e_{\mu}$
$\displaystyle\quad+(-1)^{m}\sum_{j,p,q,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\mu}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\mu}\right)dz\wedge d\bar{z}_{q}\otimes e_{\mu}$
$\displaystyle=(-1)^{m+1}\sum_{j,q,\lambda,\mu}\left(u_{j\lambda}A_{jq\lambda}^{\mu}-\sum_{p}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}u_{j\lambda}\Gamma_{p\lambda}^{\mu}+-\sum_{p}\nu_{j}^{p}u_{j\lambda}A_{pq\mu}^{\mu}\right)dz\wedge
d\bar{z}_{q}\otimes e_{\mu}$
$\displaystyle\quad+(-1)^{m}\sum_{j,p,q,\mu}\left(\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}\frac{\partial
u_{j\mu}}{\partial z_{p}}+\frac{\partial^{2}\nu_{j}^{p}}{\partial
z_{p}\partial\bar{z}_{q}}u_{j\mu}\right)dz\wedge d\bar{z}_{q}\otimes e_{\mu}$
$\displaystyle=(-1)^{m+1}\sum_{j,p,\lambda,\alpha}u_{j\alpha}\left(A_{jp\alpha}^{\lambda}-\sum_{s}\frac{\partial\nu_{j}^{s}}{\partial\bar{z}_{p}}\Gamma_{s\alpha}^{\lambda}+\sum_{s}\nu_{j}^{s}A_{sp\alpha}^{\lambda}\right)dz\wedge
d\bar{z}_{p}\otimes e_{\lambda}$
$\displaystyle\quad+(-1)^{m}\sum_{j,p,s,\lambda}\left(\frac{\partial\nu_{j}^{s}}{\partial\bar{z}_{p}}\frac{\partial
u_{j\lambda}}{\partial z_{s}}+\frac{\partial^{2}\nu_{j}^{s}}{\partial
z_{s}\partial\bar{z}_{p}}u_{j\lambda}\right)dz\wedge d\bar{z}_{p}\otimes
e_{\lambda}$ $\displaystyle=\sum_{s}dz_{s}\wedge D_{z_{s}}^{F}v_{1}+v_{2},$

where

$v_{1}:=\sum_{j,p,q,\lambda}(-1)^{m+p-1}\frac{\partial\nu_{j}^{p}}{\partial\bar{z}_{q}}u_{j\lambda}dz_{1}\wedge\cdots\wedge\widehat{dz_{p}}\wedge\cdots\wedge
dz_{m}\wedge d\bar{z}_{q}\otimes e_{\lambda},$
$v_{2}:=(-1)^{m+1}\sum_{j,p,\lambda,\alpha}u_{j\alpha}(A_{jp\alpha}^{\lambda}+\sum_{s}\nu_{j}^{s}A_{sp\alpha}^{\lambda})dz\wedge
d\bar{z}_{p}\otimes e_{\lambda}.$

By Lemma 5.5, we know

$\int_{\Omega_{t}}h^{F}(v,v)d\lambda\leq\int_{\Omega_{t}}h^{F}(v_{1},v_{1})d\lambda+\int_{\Omega_{t}}h^{F}(A^{-1}v_{2},v_{2})d\lambda.$

To prove Inequality (5.6), we only need to verify

$\displaystyle\quad h^{F}(A^{-1}v_{2},v_{2})$
$\displaystyle=\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}B_{pq\lambda\mu}A_{jp\alpha\mu}A_{pk\lambda\beta}u_{j\alpha}\overline{u_{k\beta}}+\sum_{j,k,p,\alpha,\beta}\overline{\nu_{k}^{p}}A_{jp\alpha\beta}u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle\quad+\sum_{j,k,p,\alpha,\beta}\nu_{j}^{p}(A_{pk\alpha\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\alpha\beta})u_{j\alpha}\overline{u_{k\beta}}$

for some $B$ to be determined later.
For any $g:=\sum_{p,\lambda}g_{p\lambda}dz\wedge d\bar{z}_{p}\otimes
e_{\lambda},$ we have

$\displaystyle Ag$ $\displaystyle=i\Theta^{(F,h^{F})}(\Lambda_{\omega}g)$
$\displaystyle=(i\sum_{p,q,\lambda,\mu}A_{pq\lambda}^{\mu}dz_{p}\wedge
d\bar{z}_{q}\otimes e_{\lambda}^{*}\otimes e_{\mu})(\Lambda_{\omega}g)$
$\displaystyle=\sum_{j,p,q,\lambda,\mu}A_{pq\lambda}^{\mu}g_{p\lambda}dz\wedge
d\bar{z}_{q}\otimes e_{\mu}.$

Note that we may view $A$ as a linear map $T$ which satisfies

$T(d\bar{z}_{p}\otimes
e_{\lambda})=\sum_{q,\mu}A_{pq\lambda}^{\mu}d\bar{z}_{q}\otimes e_{\mu}.$

Since $(F,h^{F})$ is Nakano strictly positive, then for any nonzero matrices
$u_{p\lambda}$, we have

$\sum_{p,q,\lambda,\mu}A_{pq\lambda\mu}u_{p\lambda}\overline{u_{q\mu}}>0,$

and then $T$ is invertible. Let $T^{-1}$ be the inverse map of $T$, and write

$T^{-1}(d\bar{z}_{q}\otimes
e_{\mu})=\sum_{p,\lambda}B_{pq\lambda}^{\mu}d\bar{z}_{p}\otimes e_{\lambda},$

then we have

$d\bar{z}_{p}\otimes
e_{\lambda}=T^{-1}(\sum_{q,\mu}A_{pq\lambda}^{\mu}d\bar{z}_{q}\otimes
e_{\mu})=\sum_{q,s,\alpha,\mu}A_{pq\lambda}^{\mu}B_{sq\alpha}^{\mu}d\bar{z}_{s}\otimes
e_{\alpha},$

i.e.

$\sum_{q,\mu}A_{pq\lambda}^{\mu}B_{sq\alpha}^{\mu}=\delta_{ps}\delta_{\lambda\alpha}.$

Let

$B_{sq\alpha\beta}:=\sum_{\mu}B_{sq\alpha}^{\mu}(h^{F})^{\beta\mu},$

then we have

$\sum_{q,\beta}A_{pq\lambda\beta}B_{sq\alpha\beta}=\sum_{q,\mu}A_{pq\lambda}^{\mu}B_{sq\alpha}^{\mu}=\delta_{ps}\delta_{\lambda\alpha}.$

Define linear maps (for the simplicity of notations, we omit $dz$ here)

$S_{1}(d\bar{z}_{p}\otimes
e_{\lambda})=\sum_{q,\mu}A_{pq\lambda\mu}d\bar{z}_{q}\otimes e_{\mu},\
S_{2}(d\bar{t}_{j}\otimes
e_{\lambda})=\sum_{q,\mu}A_{jq\lambda\mu}d\bar{z}_{q}\otimes e_{\mu},$
$S_{3}(d\bar{z}_{p}\otimes
e_{\lambda})=\sum_{k,\mu}A_{pk\lambda\mu}d\bar{t}_{k}\otimes e_{\mu},\
S_{4}(d\bar{t}_{j}\otimes
e_{\lambda})=\sum_{k,\mu}A_{jk\lambda\mu}d\bar{t}_{k}\otimes e_{\mu},$
$B(d\bar{z}_{q}\otimes
e_{\mu})=\sum_{p,\lambda}B_{pq\lambda\mu}d\bar{z}_{p}\otimes e_{\lambda},$

then we have

$BS_{1}(d\bar{z}_{p}\otimes
e_{\lambda})=B(\sum_{q,\mu}A_{pq\lambda\mu}d\bar{z}_{q}\otimes
e_{\mu})=\sum_{q,s,\mu,\alpha}A_{pq\lambda\mu}B_{sq\alpha\mu}d\bar{z}_{s}\otimes
e_{\alpha}=d\bar{z}_{p}\otimes e_{\lambda},$

so $B=S_{1}^{-1}$.
Let

$(v_{2})_{p\lambda}:=(-1)^{m+1}\sum_{j,\alpha}u_{j\alpha}(A_{jp\alpha}^{\lambda}+\sum_{s}\nu_{j}^{s}A_{sp\alpha}^{\lambda}),$

then we know

$A^{-1}v_{2}=\sum_{p,q,\lambda,\mu}B_{pq\lambda}^{\mu}v_{q\mu}dz\wedge
d\bar{z}_{p}\otimes e_{\lambda},$

and

$\displaystyle\quad h^{F}(A^{-1}v_{2},v_{2})$
$\displaystyle=\sum_{j,k,p,q\lambda,\mu,\alpha,\beta,\gamma}B_{pq\lambda}^{\mu}(A_{jp\alpha}^{\mu}+\sum_{s}\nu_{j}^{s}A_{sq\alpha}^{\mu})\overline{(A_{kp\beta}^{\gamma}+\sum_{a}\nu_{k}^{a}A_{ap\beta}^{\gamma})}u_{j\alpha}\overline{u_{k\beta}}h_{\lambda\gamma}^{F}$
$\displaystyle=\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}B_{pq\lambda}^{\mu}(A_{jp\alpha}^{\mu}+\sum_{s}\nu_{j}^{s}A_{sq\alpha}^{\mu})(A_{pk\lambda\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\lambda\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle=\sum_{j,k,p,\lambda,\alpha,\beta}(\sum_{q,\mu}B_{pq\lambda}^{\mu}A_{jp\alpha}^{\mu}+\nu_{j}^{p}\delta_{\alpha\lambda})(A_{pk\lambda\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\lambda\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle=\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}B_{pq\lambda}^{\mu}A_{jp\alpha}^{\mu}(A_{pk\lambda\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\lambda\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle\quad+\sum_{j,k,p,\alpha,\beta}\nu_{j}^{p}(A_{pk\alpha\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\alpha\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle=\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}B_{pq\lambda\mu}A_{jp\alpha\mu}(A_{pk\lambda\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\lambda\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle\quad+\sum_{j,k,p,\alpha,\beta}\nu_{j}^{p}(A_{pk\alpha\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\alpha\beta})u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle=\sum_{j,k,p,q,\lambda,\mu,\alpha,\beta}B_{pq\lambda\mu}A_{jp\alpha\mu}A_{pk\lambda\beta}u_{j\alpha}\overline{u_{k\beta}}+\sum_{j,k,p,\alpha,\beta}\overline{\nu_{k}^{p}}A_{jp\alpha\beta}u_{j\alpha}\overline{u_{k\beta}}$
$\displaystyle\quad+\sum_{j,k,p,\alpha,\beta}\nu_{j}^{p}(A_{pk\alpha\beta}+\sum_{a}\overline{\nu_{k}^{a}}A_{pa\alpha\beta})u_{j\alpha}\overline{u_{k\beta}}.$

Thus the proof is completed. ∎