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Note.tex
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\begin{document}
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\title{Notes for Articles}
\subtitle{}
\edition{First Edition}
\bookseries{Research Notes}
\author{Ethan Lu}
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\frontmatter
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\chapter*{Preface}
Here is a comprehensive list of the academic papers that I have perused.
\hfill
\begin{tabular}{lr}
&-- Ethan Lu\\
&2024-07-06
\end{tabular}
\clearpage}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\mainmatter
\pagenumbering{arabic}
\partabstract{\fontsize{15pt}{15pt}\selectfont This notes are mainly about some vanishing theorems and their proofs.}
\part{Research Notes For Articles}
\chapter{Vector Bundles with Semidefinite Curvature and Cohomology Vanishing Theorems}\index{Vector Bundles with Semidefinite Curvature and Cohomology Vanishing Theorems}
\section{Preliminaries}\index{Preliminaries}
\begin{theorem}[][Kodaira vanishing theorem\cite[P196]{Huybrechts2010Complex}][thm:kodaira-vanishing]
Let $\mL$ be a positive (ample) line bundle on a compact K\"ahler manifold $X$ with $\dim X=n$. Then
\begin{equation}
H^q (X,\Omega_X^p \otimes \mL)=0,\quad \text{for }p+q>n.
\end{equation}
\end{theorem}
\begin{definition}[][Chern classes\cite[P196]{Huybrechts2010Complex}][def:Chern classes]
Let $\{\widetilde{P}_k\}$ be the homogeneous polynomials with $\deg (\widetilde{P}_k)=k$ defined by
\[\det(\textrm{Id}+B)=1+\widetilde{P}_1(B)+\ldots+\widetilde{P}_r(B).\]
Clearly, these $\widetilde{P}_k$ are \textit{invariant}.
The Chern classes of a complex manifold $X$ are \[c_k(X):=c_k(\mT_X)\in H^{2k}(X,\bR),\]
where $\mT_X$ is the holomorphic tangent bundle.
\end{definition}
\begin{definition}[][Chern forms\cite[P196]{Huybrechts2010Complex}][def:Chern forms]
The \textit{Chern forms} of a vector bundle $E$ of rank $k$ endowed with a connection $\nabla$ are \[c_k(E,\nabla):=\widetilde{P}_k\br{\frac{i}{2\pi}F_\nabla}\in\mA_C^{2k}(M).\]
The \textit{$k$-th Chern class} of the vector bundle $E$ is induced cohomology class \[c_k(E):=[c_k(E,\nabla)]\in H^{2k}(M,\bC).\]
\end{definition}
\begin{theorem}[][Another description of Kodaira vanishing theorem\cite[Introduction]{gigante1981vector}]
Let $F$ be a holomorphic line bundle over a compact K\"ahler manifold $M$. If the Chern class $\mC_{\bR}(F)\in H^2(M,\bR)$ contains a negative definite form $\mX (\mX<0)$ then all cohomology groups $$H^q(M,\Omega(F))=0$$ when $q\leqslant n-1$.
\end{theorem}
\begin{theorem}[][Akizuki-Nakano Vanishing Theorem\cite[Introduction]{gigante1981vector}]
\textbf{(The generalization of Kodaira vanishing theorem by Akizuki-Nakano)}
If $\mX<0$, then
\[H^q(M,\Omega^p(F))=0\]
when $p+q\leqslant n-1$.
\end{theorem}
\begin{theorem}[][Vesentini vanishing theorem\cite[Introduction]{gigante1981vector}]
If $\mX$ is semi-definite of rank, $k$ (i.e. $\mX\leqslant 0$ and $\mX$ has $k$ negative eigenvalues at each point of $M$) then
\[H^q(M,\Omega(F))=0 \quad \And \quad H^0(M,\Omega^q(F))=0\]
when $q\leqslant k-1$.
\end{theorem}
\begin{problem}[][\cite[Introduction]{gigante1981vector}][prob:Main problem]
If $\mX\leqslant 0$ with rank $k$, then
\[H^q(M,\Omega^p(F))=0\]
when $p+q\leqslant k-1$?
\end{problem}
\subsection{Correlation Techniques}
\begin{fancybox}
Problems in holomorphic vector bundles can be often reduced to similar problems in line bundles, by means of constructing the projective bundle $PE$ over $M$ and the tautological line bundle $LE^{-1}$.
\end{fancybox}
By using the technique, we shall generalize to vector bundles some of the results of the following two section.
\section{The K\"ahler Case}
\newcommand{\tmcolor}[2]{{\color{#1}{#2}}}
\newcommand{\tmverbatim}[1]{\text{{\ttfamily{#1}}}}
\newcommand{\tmop}[1]{\ensuremath{\operatorname{#1}}}
\newcommand{\tmstrong}[1]{\textbf{#1}}
\providecommand{\xequal}[2][]{\mathop{=}\limits_{#1}^{#2}}
The first result says that let $M$ be a
compact K\"ahler manifold and let $F$ be a holomorphic line bundle over $M$. If
$C_{\mathbb{R}} (F)$ contains a form $\mathcal{X}$ whose associated hermitian
form is negative semidefinite of rank $k$ at each point of $M$, then
\[ H^t (M, \Omega^s (F)) = H^s (M, \Omega^t (F)) = 0, \quad \tmop{for} \quad
s + t \leqslant k - 1. \]
The proof mainly depends on the \textit{Akizuki-Nakano Inequality} that given any harmonic $(p,
q)$-form $\varphi$ with values in $E$, then
\begin{equation}
([\Lambda, \tmop{ie} (\Theta)] \varphi, \varphi) \geqslant 0, \label{1}
\end{equation}
where $e(\Theta)$ denotes the exterior multiplication of the matrix of local $(1,1)$-form with the column vector $\varphi$. And in \cite[]{calabi1960compact}, we have known that $e(\Theta)=(\bd_E \bdd+ \bdd\bd_E)\varphi$.
Here is a simplified version.
In the K\"ahler case, let $F \rightarrow M$ be a holomorphic line bundle over a
compact K\"ahler manifold of dimension $n$. Take $(U ; (z^1, \cdots, z^n))$ be a
local coordinate system on $M$ and
\[ \varphi = \sum \varphi_{A \bar{B} } \tmop{dz}^A \wedge
\dd \bar z^{B}, \]
where $A=(\alpha_1,\ldots,\alpha_p) \; (\alpha_1<\cdots<\alpha_p), B=(\beta_1,\ldots,\beta_q) \; (\beta_1<\cdots<\beta_q)$.
Denoting by $X_1 \leqslant X_2 \leqslant \cdots \leqslant X_n$ the eigenvalues
of the hermitian form associated to $\mathcal{X}$, for any $z \in U$,
compatible coordinates centered at $z$ can be chosen in $U$ in such a way that
the fundamental form $\omega$ of the K\"ahler metric is given at $z$ by
\[ \omega = i \sum \tmop{dz}^{\alpha} \wedge \overline{\tmop{dz}^{\alpha}} \]
and
\begin{eqnarray*}
\mathcal{X} & = \dfrac{i}{2 \pi} \Theta & = \frac{i}{2 \pi} \sum_{\alpha}
X_{\alpha} \tmop{dz}^{\alpha} \wedge \overline{\tmop{dz}^{\alpha}} .
\end{eqnarray*}
\textit{Note that the expression of $\omega$ above is the consequence of diagonalization. And one can easily obtain these two equations by using simultaneously diagonalization for $\omega$ and $\mX$.}
Then we compute the formula of Akizuki-Nakano Inequality by
\begin{eqnarray*}
&&([\Lambda, \tmop{ie} (\Theta)] \varphi)_{A \bar{B}} (z) \\
& = & [\Lambda,
\tmop{ie} (\Theta)]_{A \bar{B}} (z) \cdot \varphi_{A \bar{B}} (z)\\
& = & [\Lambda, 2 \pi i\mathcal{X}]_{A \bar{B}} (z) \cdot \varphi_{A
\bar{B}} (z)\\
& = & - \left[ \sum_{\alpha} X_{\alpha} \tmop{dz}^{\alpha} \wedge
\overline{\tmop{dz}^{\alpha}}, \Lambda \right]_{A \bar{B}} (z) \cdot
\varphi_{A \bar{B}} (z)\\
& = & - \left[ \left( \sum_{\alpha} X_{\alpha} \tmop{dz}^{\alpha} \wedge
\overline{\tmop{dz}^{\alpha}} \right) \wedge (\Lambda \varphi_{A \bar{B}})
\right] (z) + \Lambda \left[ \left( \sum_{\alpha} X_{\alpha}
\tmop{dz}^{\alpha} \wedge \overline{\tmop{dz}^{\alpha}} \right) \wedge
\varphi_{A \bar{B}} \right] (z)\\
& \xequal{\star} & - \sum_{\alpha \in A \cap B} X_{\alpha} (z)
\varphi_{A \bar{B}} (z) + \sum_{\alpha \not{\in} A \cup B} X_{\alpha} (z)
\varphi_{A \bar{B}} (z),
\end{eqnarray*}
where
\begin{align*}
\left[ \left( \sum_{\alpha} X_{\alpha} \tmop{dz}^{\alpha} \wedge
\overline{\tmop{dz}^{\alpha}} \right) \wedge (\Lambda \varphi_{A \bar{B}})
\right] =
\begin{cases}
\sum X_{\alpha} \varphi_{A \bar{B}}, &\alpha \in A \cap B,\\
0, & \alpha \not\in A \cap B,
\end{cases}
\end{align*}
and
\begin{align*}
\Lambda \left[ \left( \sum_{\alpha} X_{\alpha} \tmop{dz}^{\alpha} \wedge
\overline{\tmop{dz}^{\alpha}} \right) \wedge \varphi_{A \bar{B}} \right]
=\begin{cases}
\sum X_{\alpha} \varphi_{A \bar{B}}, &\alpha \not\in A \cup
B,\\
0, &\alpha \in A \cup B.
\end{cases}
\end{align*}
Thus, if the following relations are satisfied at each point $z \in M$
\[ (X_{t_1} + \cdots + X_{t_s}) - (X_{j_1} + \cdots + X_{j_{n - t}}) > 0 \]
for each choice of $i_1 < \cdots < i_s$ and $j_1 < \cdots j_{n - t}$ and for
all $s \leqslant t$, then
\begin{equation}
( [\Lambda, \tmop{ie} (\Theta)] \varphi, \varphi) (z) \leqslant 0. \label{2}
\end{equation}
\begin{definition}[][Positive Hermitian line bundle][def:Positive Hermitian vector bundle]
A hermitian holomorphic line bundle $E$ on $X$ is said to be positive (negative/semi-positive/semi-negative) if the hermitian matrix (Component) of its Chern curvature form
\[i\Theta(E)=i\sum_{1\leqslant j,k\leqslant n}c_{jk}(z)\dd z_j\wedge\dd \bar{z}_k\]
is positive (negative/semi-positive/semi-negative) definite at every point $z\in X$.
\end{definition}
In\cite[P334]{CACA}, by Prop VI-8.3, we gain that
\begin{align*}
\left\langle [i\Theta(E),\Lambda]u,u\right\rangle &= \sum_{J,K} (\sum_{j\in J}\gamma_j+\sum_{j\in K}\gamma_j-\sum_{1\leqslant j\leqslant n}\gamma_j)|u_{J,K}|^2\\
&\geqslant (\gamma_1+\cdots+\gamma_q-\gamma_{p+1}-\cdots-\gamma_n)|u_{J,K}|^2
\end{align*}
for any form $u=\sum_{J,K} u_{J,K} \zeta_J\wedge\bar{\zeta}_K\in \Lambda^{p,q}T^*(X)$.
Then we have
\begin{align*}
( [\Lambda, \tmop{ie} (\Theta)] \varphi, \varphi) (z) &=-\left\langle [i\Theta(E),\Lambda]u,u\right\rangle \\
& \leqslant (\gamma_1+\cdots+\gamma_q-\gamma_{p+1}-\cdots-\gamma_n)|u_{J,K}|^2\\
& \leqslant \br{(X_{j_1}+\cdots+X_{j_{n-t}}) -(X_{i_1}+\cdots+X_{i_s})}|u_{J,K}|^2\\
& \leqslant 0.
\end{align*}
By (\ref{2}), consequently, in view of inequality (\ref{1}), any harmonic $(s,
t)$-form vanishes identically, i.e. the Lemma 1.1 of the paper.
\begin{remark}
By (\ref{1}) and (\ref{2}), we have \cite[Lemma2,P483]{calabi1960compact}
\begin{eqnarray*}
( [\Lambda, \tmop{ie} (\Theta)] \varphi, \varphi) (z) = 0 &
\Longleftrightarrow & {\varphi \in H'}^{p, q} (M, F) {\cap H''}^{p, q} (M,
F)
\end{eqnarray*}
As ${H'}^{p, q} (M, F) \cong H_{\bar{\partial}}^{p, q} (M, F)$ and ${H'}^{p,
q} (M, F) \cong H_{\partial}^{p, q} (M, F)$, one has
\begin{equation}
{H'}^{p, q} (M, F) {\cap H''}^{p, q} (M, F) \cong H_{\bar{\partial}}^{p, q}
(M, F) \cap H_{\partial}^{p, q} (M, F) = \{ 0 \} . \label{3}
\end{equation}
Thus $\varphi \in \{ 0 \} \Longrightarrow \varphi \equiv 0$, i.e.
{\tmstrong{any harmonic $(p, q)$-form $\varphi$ vanishes identically}}.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%
Suppose $\mathcal{X} \leqslant 0$ of rank $k$. Now,
\tmcolor{red}{for any $\mu \in \mathbb{R}^+$, the form $(\omega - \mu
\mathcal{X})$ is still a K\"ahler form on $M$}. Denoting
{\color[HTML]{800080}the eigenvalues of $\mathcal{X}$ with respect to the
metric induced by $\omega$ as before}, {\color[HTML]{B4005A}the eigenvalues of
$\mathcal{X}$ with respect to the metric induced by $(\omega - \mu
\mathcal{X})$ are }
\[ {\color[HTML]{B4005A}\frac{X_i}{(1 - \mu X_i)}, \quad i = 1, \ldots, n.} \]
Now, if $\mu$ is sufficient large, we have
\begin{equation}
\tmcolor{DarkBlue}{\sum_{i = 1}^s \frac{X_i}{(1 - \mu X_i)} > \sum_{j = t + 1}^n
\frac{X_j}{(1 - \mu X_j)} .} \label{4}
\end{equation}
The computation should be the following.
Firstly, we gain the first inequality
\[ \sum_{i = 1}^s \frac{(X_i - X_{t + i})}{(1 - \mu X_i) (1 - \mu X_{t + i})}
\geqslant \frac{s (X_1 - X_k)}{(1 - \mu X_1) (1 - \mu X_k)} = s \left(
\frac{X_1}{(1 - \mu X_1)} - \frac{X_k}{(1 - \mu X_k)} \right) . \]
As $X_1 \leqslant X_2 \leqslant \cdots \leqslant X_n$, then
\[ \frac{X_i}{(1 - \mu X_i)} \leqslant \frac{X_j}{(1 - \mu X_j)}, \quad
\tmop{where} \quad 1 \leqslant i \leqslant j \leqslant n. \]
\tmcolor{red}{But for the associated hermitian form $\Theta$ is negative
semidefinite of rank $k$, there are $k$ negative eigenvalues of $\mathcal{X}$,
which can be written as $X_1 \leqslant \cdots \leqslant X_k < 0 \leqslant X_{k
+ 1} \leqslant \cdots \leqslant X_n$ without loosing generality.} ($X_l
\geqslant 0, l = k + 1, \ldots, n$)
Thus, we have
\begin{align*}
\sum_{i = 1}^s \frac{(X_i - X_{t + i})}{(1 - \mu X_i) (1 - \mu X_{t + i})}
&= \sum_{i = 1}^s \left[ \frac{X_i}{(1 - \mu X_i)} \downarrow - \frac{X_{t +
i}}{(1 - \mu X_{t + i})} \uparrow \right] \\
&\geqslant s \left( \frac{X_1}{(1
- \mu X_1)} - \frac{X_k}{(1 - \mu X_k)} \right) .
\end{align*}
Secondly, we abtain the second inequality
\begin{align*}
\sum_{j = s + t + 1}^k \frac{X_j}{(1 - \mu X_j)} &= \sum_{j = s + t + 1}^n
\frac{X_j}{(1 - \mu X_j)} \\
&= \sum_{j = s + 1}^{n - t} \frac{X_{j + t}}{(1 -
\mu X_{j + t})} \leqslant [k - (s + t + 1)] \frac{X_k}{(1 - \mu X_k)}
\end{align*}
by
\[ \frac{X_j}{(1 - \mu X_j)} \leqslant \frac{X_k}{(1 - \mu X_k)}, \quad
\tmop{where} X_1 \leqslant \cdots \leqslant X_k \leqslant 0
\leqslant\underbrace{ X_{k + 1} \leqslant \cdots \leqslant X_n}_{n - k} .
\]
Then, for each $z \in M$, we can find $\bar{\mu} \in \mathbb{R}^+$ so
that for any $\mu \geqslant \bar{\mu}$ the hermitian form associated to
$\Theta$ satisfies condition of Lemma 1.1 for any pair $(s, t)$ such that $s +
t \leqslant k - 1$, with respect to the K\"ahler metric $(\omega - \mu
\mathcal{X})$. \itbf{This means that for any K\"ahler form (Not just specific one), by lemma 1.1 of paper, any harmonic $(s,t)$-form vanishes identically.} Eventually, we gain the Theorem 2.1, i.e., the result we have written in the beginning of this section.
\subsection{An application of semidefinite theorem in K\"ahler case}
\begin{theorem}[][Lefschetz theorem on hyperplane sections]
If the Chern class $C_{\mathbb{R}} ([S]) \in H^2 (M, \mathbb{R})$ of $[S]$
contains a form $\mathcal{X} \geqslant 0$ of rank $k$, then
\[ \rho^{\ast}_S \text{ is an isomorphism for $s \leqslant k - 2$} \]
and
\[ \rho^{\ast}_S \text{ is injective for $s = k - 1$.} \]
\end{theorem}
\begin{proof}
The main body remains the same as [1], and now we provide the kernel of the
proof. We gain the {\tmstrong{exact cohomology sequence}} from above
\[ \cdots \rightarrow H^q (M, \Omega^p ([S]^{- 1})) \rightarrow H^q \left(
{M, \Omega'}^p \right) \rightarrow H^q (S, \Omega^{p - 1} ([S]^{- 1} |_S
)) \rightarrow \cdots . \]
Applying the vanishing theorem in Kahler case, we have
\[ H^q \left( {M, \Omega'}^p \right) = 0, \quad \tmop{whenever} \quad p + q
\leqslant k - 1. \]
{\tmstrong{The exact Cohomology sequence}} yields the
{\tmstrong{isomorphisms}}
\begin{equation}
H^q (M, \Omega^q) \cong H^q (S, \Omega^p), \quad \tmop{for} \quad p + q
\leqslant k - 2 = (k - 1) - 1 \label{5}
\end{equation}
and the {\tmstrong{injectivity}} of the map
\[ H^q (M, \Omega^p) \rightarrow H^q (S, \Omega^p), \quad \tmop{whenever}
\quad p + q = k - 2. \]
Since
\[ H^s (M, \mathbb{C}) \cong \oplus_{p + q = s} H^q (M, \Omega^p) \]
and a similar decomposition holds for $S$, the conclusion follows.
\end{proof}
\section{General case}
In the general case, the proof is very different, which is mainly depends on the \textit{local expression of the Laplace-Beltrami operator and on a result about hermitian forms}, due to E. Calabi and \textit{Lectures on Convexity of Complex manifolds and Cohomology vanishing theorems} by E. Vesentini.
If the hermitian metric on $X$ is a K\"ahler metric \cite[P72]{vesentini1967lectures}, then
\begin{equation*}
\bdd=\tilde{\partial}, \quad \vartheta=\tilde{\vartheta}, \quad \square=\tilde{\square} .
\end{equation*}
For $\varphi \in C^{p q}(X, E)$
\begin{equation*}
(\tilde{\vartheta} \varphi)_{\overline{A B^{\prime}}}^a=(-1)^{p-1} \nabla_\alpha \varphi_A^a \alpha_{\overline{B^{\prime}}},
\end{equation*}
so that, exactly as in the case of the Laplacian $\triangle$ in Chapter 2, we have
\begin{equation}\label{eq:3.11of vesentininote}
(\tilde{\square} \varphi)_{\overline{A B}}^a=-\nabla_\alpha \nabla^\alpha \varphi \frac{a}{A B}+\sum_{r=1}^q(-1)^{r-1}\left(\nabla_\alpha \nabla_{\overline{\beta_r}}-\nabla_{\overline{\beta_r}} \nabla_\alpha\right) \varphi_A^a \alpha_{\overline{B_r^{\prime}}}
\end{equation}
where
\begin{equation*}
\nabla^\alpha=g^{\alpha \bar{\beta}} \nabla_{\bar{\beta}},
\end{equation*}
and $\quad A=\left(\alpha_1, \ldots \alpha_p\right), B=\left(\beta_1, \ldots \beta_q\right), B_r^{\prime}=\left(\beta_1, \ldots , \hat{\beta}_r, \ldots , \beta_q\right)$.
In view of the Ricci identity, the summand of \textbf{\eqref{eq:3.11of vesentininote}} can be expressed by
\begin{equation}\label{eq:3.12of vesentininote}
\sum_{r=1}^q(-1)^{r-1}\left(\nabla_\alpha \nabla_{\overline{\beta_r}}-\nabla_{\overline{\beta_r}} \nabla_\alpha\right) \varphi_A^a \alpha_{\overline{B_r^{\prime}}}=(\tilde{\mathcal{K}} \varphi)^a A B
\end{equation}
where $\tilde{\mathcal{K}}$ is a mapping
\begin{equation*}
\tilde{\mathcal{K}}: C^{p q}(X, E) \rightarrow C^{p q}(X, E),
\end{equation*}
which is linear over $C^{\infty}$ functions, whose local expression involves linearly (with integral coefficients) only the coefficients of the curvature forms, $s$ and $L$, of $E$ and $\Theta_{\circ}$.
By \cite[Remark (3) after Lemma 3.2]{vesentini1967lectures}, we have
\begin{equation}\label{eq:3.13of vesentininote}
\left(\tilde{\mathcal{K}}_{\varphi}\right)_{A \bar{B}}^a=\sum_{r=1}^q(-1)^r s_{b \bar{\beta}_r \alpha}^a \varphi_A^b \alpha_{\overline{B_r^{\prime}}}+\left(\tilde{\mathcal{K}}_{\circ} \varphi\right)_{A \bar{B}}^a,
\end{equation}
where $\tilde{\mathcal{K}}_{\circ}$ involves only the curvature tensor of $\Theta_{\circ}$, and is completely independent of $E$.
Formula \textbf{\eqref{eq:3.11of vesentininote}} can be also written as
\begin{equation}\label{eq:3.14of vesentininote}
(\tilde{\square} \varphi)_{A \bar{B}}^a=-\nabla_\alpha \nabla^\alpha \varphi_{A \bar{B}}^a+(\tilde{\mathcal{K}} \varphi)_{A \bar{B}}^a .
\end{equation}
Now
\begin{equation*}
\begin{aligned}
\square & =(\bar{\partial} \vartheta+\vartheta \bar{\partial})=(\bar{\partial}+S)(\tilde{\vartheta}+T)+(\tilde{\vartheta}+T)(\tilde{\partial}+S) \\
& =\tilde{\square}+\tilde{\partial} T+T \tilde{\partial}+\tilde{\vartheta} S+S \tilde{\vartheta}+S T+T S .
\end{aligned}
\end{equation*}
It follows that
\begin{lemma}[][\cite[Lemma 3.4.]{vesentini1967lectures}][lem:lemma3.4ofvesentininote]
For any $\varphi \in C^{p q}(X, E)$
\begin{equation}\label{eq:local-expression-of-Beltrami-operator}
(\square \varphi)_{A \bar{B}}^a=(\tilde{\square} \varphi)_{A \bar{B}}^a+\br{\left(F_1 \varphi\right)_{A \bar{B}}^a+\left(F_2 \nabla^{\prime} \varphi\right)_{A \bar{B}}^a+\left(F_3 \nabla^{\prime \prime} \varphi\right)_{A \bar{B}}^a}
\end{equation}
where
\begin{equation*}
\begin{aligned}
& F_1: C^{p q}(X, E) \rightarrow C^{p q}(X, E), \\
& F_2: C^{p q}\left(X, E \otimes \Theta_o^*\right) \rightarrow C^{p q}(X, E), \\
& F_3: C^{p q}\left(X, E \otimes \Theta_o^*\right) \rightarrow C^{p q}(X, E),
\end{aligned}
\end{equation*}
are linear over $C^{\infty}$ functions. Their local expression involves the tensor product and its first derivatives.
\end{lemma}
Then the Laplace-Beltrami operator $\square$ on $(0,q)$-forms with coefficients in $F\otimes D$ is given locally by
\begin{equation}
\label{Local expression of Laplace-Beltrami operator}
(\square\varphi)_{\overline{B}}=-\nabla_\alpha\nabla^\alpha\varphi_{\overline{B}}+(\mK\varphi)_{\overline{B}}+\sum g^{\overline{\alpha}\beta}\Gamma_{\overline{\alpha}\beta}^{\overline{\lambda}}\nabla_{\overline{\lambda}}\varphi_{\overline{B}},
\end{equation}
where $B$ is a set of $q$ indices $B=(\beta_1,\cdots,\beta_q)$ and $\Gamma_{\overline{\alpha}\beta}^{\overline{\lambda}}$ are components of the Riemann-Christoffel connection defined by the hermitian metric on $M$ and $\mK$ is a linear operator on $(0,q)$-forms which splits as the sum of $\mK=\mK_0+\mK_\mX$.
\begin{lemma}[][\cite[Lemma 2.2 by E. Calabi, P80]{vesentini1967lectures}][lem:Lemma2.2]
Let $H$ be a hermitian quadratic differential form on $X$ and $G$ a hermitian metric on $X$. Assume that $H$ has at least $p$ positive eigen values. Let $\varepsilon_1(x), \ldots, \varepsilon_n(x)$ be the eigen values of $H$ (w.r.t.
$G)$ at $x$ in decreasing order: $\varepsilon_r(x) \geq \varepsilon_{r+1}(x)$
Then given $c_1, c_2>0, G$ can be so chosen that
\begin{equation*}
l_H(x)=c_1 \varepsilon_p(x)+c_2 \operatorname{Inf}\left(0, \varepsilon_n(x)\right)>0,\quad \text { for all } x \in X .
\end{equation*}
\end{lemma}
\begin{proof}
Let $G$ be any complete hermitian metric whatever. Let $\sigma_1(x), \ldots$, $\sigma_n(x)$ be the eigen-values of $H$ with reference to $G$ arranged in decreasing order. We construct now a metric $G$ on $X$ whose eigen values are functions of $\left\{\sigma_i(x)\right\}_{1 \leq i \leq n}$ as follows: let $\lambda: X \rightarrow \mathbb{R}$ be a $C^{\infty}-$ function (we will impose conditions on $\lambda$ letter); let $U$ be a coordinate open set in $X$ with holomorphic coordinates $\left(z^1, \ldots, z^n\right)$; in $U$, we have $G=G_{U \alpha \bar{\beta}} d z^\alpha d \bar{z}^\beta$ so that $\left(G_{U \alpha \bar{\beta}}\right)_{\alpha \beta}$ is a function whose values are positive definite hermitian matrices; then the matrix valued function $\widehat{G}_U=\left(\widehat{G}_{U \alpha \bar{\beta}}\right)_{\alpha \beta}$ where $\widehat{G}=\sum \widehat{G}_{U \alpha \bar{\beta}} d z^\alpha d z^\beta$ in $U$ is defined by
\begin{equation*}
\widehat{G}_U^{-1}=G_U^{-1} \sum_{r=0}^{\infty} \frac{\lambda(x)^r}{(r+1) !}\left(H_U G_U^{-1}\right)^r
\end{equation*}
where $H_U$ is the matrix valued function $\left(H_{U \alpha \beta}\right)$ defined by
\begin{equation*}
H=\sum H_{U \alpha \bar{\beta}} d z^\alpha d z^\beta
\end{equation*}
in $U$.
We now assert that $\widehat{G}_U$ define a global hermitian defferential form on $X$ and under a suitable choice of $\lambda$, it is positive definite. To see that $\widehat{G}_U$ defines a global hermitian differential form on $X$, we need only prove the following. Let $V$ be another coordinate open set with coordinates complex $\left(w^1, \ldots, w^n\right)$. Let $J=\frac{\partial\left(z^1, \ldots, z^n\right)}{\partial\left(w^1, \ldots, w^n\right)}$ be the Jacobian matrix. As before let $G_V=\left(G_{V \alpha \beta}\right)$ be defined by $G=\sum G_{V \alpha \bar{\beta}} d w^\alpha d \bar{w}^\beta$ in $V$. Then if $\widehat{G}_V$ is defined starting from $G_V$ as $\widehat{G}_U$ from $G_U$, we have
\begin{equation*}
J \widehat{G}_U^t \bar{J}=\widehat{G_V}
\end{equation*}
We have in fact, writing $J^*$ for $^t \bar{J}^{-1}$,
\begin{equation*}
J^* \widehat{G}_U^{-1} J^{-1}=G_V^{-1} J\left(\sum_{r=0}^{\infty} \frac{\lambda(x)^1}{(r+1) !}\left(H_U G_U^{-1}\right)^r\right) J^{-1} \text {, since } J G_U^t \bar{J}=G_V .
\end{equation*}
It follows from the above that
\begin{equation*}
\begin{aligned}
J^* \widehat{G}_U^{-1} J^{-1} & =G_V^{-1} \sum_{r=0}^{\infty} \frac{\lambda(x)^r}{(r+1) !}\left(J H_U G_U^{-1} J^{-1}\right)^r \\
& =G_V^{-1} \sum_{r=0}^{\infty} \frac{\lambda(x)^r}{(r=1) !}\left(H_V J^* G_U^{-1} J^{-1}\right)^r
\end{aligned}
\end{equation*}
since $J H_U^t \bar{J} H_V$. Hence we obtain
\begin{equation*}
\begin{aligned}
J^* \widehat{G}_U^{-1} J^{-1} & =G_V^{-1} \sum_{r=0} \frac{(\lambda(x))^r}{(r+1) !}\left(H_V G_V^{-1}\right)^r \\
& =\widehat{G}_V^{-1}
\end{aligned}
\end{equation*}
This proves that $\widehat{G}_U$ defines on $X$ a global hermitian differential. We next show that $\widehat{G}$ is positive definite. For this we look for the eigenvalues of $\widehat{G}$ with reference to $G$. TO compute these, we may assume, in the above formula for $\widehat{G}_U$, that $G_U$ is the identity matrix
Then we have
\begin{equation*}
\widehat{G}_U=\sum_{r=0}^{+\infty} \frac{\lambda(x)^r}{(r+1) !} H_U^r
\end{equation*}
It follows that the eigen values of $\widehat{G}_U$ are
\begin{equation*}
\left\{\sum_{r=0}^{+\infty} \frac{\lambda(x)^r}{(r+1) !} \sigma_q(x)^r\right\}_{1 \leq q \leq n} .
\end{equation*}
It is easily seen that these are all strictly greater than zero: this assertion simply means this: $f(t)=\frac{e^t-1}{t}=\sum_{r=0}^{\infty} \frac{t^r}{(r+1) !}$ for $t \neq 0, f(0)=1$ (which is continuous in $t$ ) is everywhere greater than $0$ .
We will now look for conditions on $\lambda$ such that $\widehat{G}$ satisfies our requirements. From the formula for $\widehat{G}$, we have
\begin{equation*}
H_U \widehat{G}_U^{-1}=\sum_{r=0}^{+\infty} \frac{\lambda(x)^r}{(r+1)}\left(H_U G_U^{-1}\right)^{r+1} .
\end{equation*}
Now the eigenvalues of $H$ with respect to $\widehat{G}$ (resp. $G$ ) are simply those of the matrix $H_U \widehat{G}_U^{-1}$ (resp. $H_U G_U^{-1}$ ), Hence these eigenvalues $\varepsilon_q(X)$ of $H$ with reference to $\widehat{G}$ are
\begin{equation*}
f\left(\lambda(x), \sigma_q(x)\right)
\end{equation*}
where $f(s, t)$ is the function on $\mathbb{R}^2$ defined by
\begin{equation*}
f(s, t)=\sum_{r=0}^{t \infty} \frac{S^r}{(r+1)} t^{r+1}
\end{equation*}
Since $\frac{\partial f(s, r)}{\partial t}=e^{s t}>0$ for any, $f(s, t)$ is monotone increasing in $t$. Hence we have
\begin{equation*}
\varepsilon_r(x) \geqslant \varepsilon_{r+1}(x) \text { for } 1 \leqslant r \leqslant n-1 .
\end{equation*}
Moreover $f(s, t) \geqslant t$ for $s \geqslant 0$. Thus, if we choose $\lambda(x) \geqslant 0$ for every $x \in X$, then $\varepsilon_q(x) \geqslant \sigma_q(x)>0$.
The choice of $\lambda(x)$ is now made as follows . Let, for every integer $v>0 . B_v=\left\{x \mid d\left(x, x_0\right) \leqslant \gamma\right\}$ for some $x_0 \in X$, the distance being $i$ the metric $G$. The $B_v$ are then compact. Let $b_v=\operatorname{Inf}_{x \in B_\gamma}\left(\sigma_p(x)\right)$.
Then $b_1 \geqslant b_2 \geqslant \ldots \geqslant b_{v+1} \geqslant \ldots$.
Let $b(x)$ be a $C^{\infty}$ function on $X$ such that $b(x)>0$ for $x \in X$ and $b(x)<b_v$ in $B_v-B_{v-1}$. Then clearly $b(x) \leq \sigma_p(x)$.
Finally let $\rho(x)$ be a $C^{\infty}$ function on $X$ such that $\rho(x) \geqslant d\left(x, x_0\right)$, and $k>\sqrt{\frac{C_2}{C_1}} b_1$ be a real constant. Set $\lambda(x)=\frac{2 k e^{\rho(x)}}{b^2(x)}$. We have then
\begin{equation*}
\varepsilon_q(x)=f\left(\lambda(x), \sigma_p(x)\right)=\sigma_p(x)+\frac{\lambda(x)}{2 !} \sigma_p(x)^2+\cdots
\end{equation*}
so that
\begin{equation*}
\varepsilon_p(x) \geqslant \frac{k e^{\rho(x)}}{b^2(x)} \sigma_p(x)^2 \geqslant k e^{\rho(x)} \geqslant k .
\end{equation*}
On the other hand,
\begin{equation*}
\begin{gathered}
\varepsilon_n(x)=f\left(\lambda(x), \sigma_n(x)\right)=\frac{1}{\lambda(x)}\left\{e^{\lambda(x) \sigma_n(x)_{-1}}\right\} \geqslant \frac{-1}{\lambda(x)}=-\frac{b^2(x)}{2 k e^{\rho(x)}} \geqslant-\frac{b_1^2}{k} \\
C_1 \varepsilon_p(x)+C_2 \operatorname{Inf}\left(0, \varepsilon_n(x)\right) \geqslant C_1 k-C_2 \frac{b_1^2}{k} \geqslant \frac{1}{k}\left(C_1 k^2-C_2 b_1^2\right)>0 .
\end{gathered}
\end{equation*}
\end{proof}
The general version of the theorem will be described as follows :
\begin{theorem}[][The general semidefinite vanishing theorem]
Let $M$ be a compact hermitian manifold. Let $F$ be a holomorphic line bundle and $D$ be a holomorphic vector bundle over $M$.
If $C_\bR (F)$ contains a form $\mX$ whose associated hermitian form has at least $k$ positive eigenvalues at each ponit of $M$, then there exists a positive integer $\mu_0$ such that
\[H^q(M,\Omega(F^\mu\otimes D))=0\]
for all $\mu\geqslant \mu_0$, and all $q\leqslant n-k+1$.
\end{theorem}
\begin{proof}
By \autoref{lem:Lemma2.2}, $M$ can be equipped with a new hermitian metric
in such a way that, denoting by $X_1 (z) \geqslant \cdots \geqslant X_n (z)$
the eigenvalues of $\mathcal{X}$ w.r.t. this metric, then $X_k (z) > 0$ and
\[ X_k (z) + n \cdot \inf (0, X_n (z)) > 0, \quad \tmop{at} \tmop{each}
\tmop{point} z \in M. \]
As $K = K_0 + K_{\mathcal{X}}$ is a linear operator, then $A (K \varphi,
\varphi) = A (K_0 \varphi, \varphi) + A (K_{\mathcal{X}} \varphi, \varphi)$.
Aaccording to a straightforward computation, we abtain
\[ A (K_{\mathcal{X}} \varphi, \varphi) (z) \geqslant (X_k (z) + n \cdot
\tmop{inf} (0, X_n (z))) A (\varphi, \varphi) (z) \]
by using the fact that for $q \geqslant n - k + 1$, there is at least one
of the indices $\beta \leqslant k$.
And since $M$ is compact, there exist a positive $C$ s.t. $A (K_0 \varphi,
\varphi) (z) \geqslant - \tmop{CA} (\varphi, \varphi) (z)$ at every point $z
\in M$.($K_0$ is a bounded linear operator on $M$ when $M$ is comapct.) Then
we choose a $\mu_0$ s.t.
\[ \mu_0 (X_k (z) + n \cdot \inf (0, X_n (z))) - (C + 1) > 0, \quad
\tmop{at} \tmop{each} \tmop{point} z \in M, \]
which implies $A (K \varphi, \varphi) \geqslant A (\varphi, \varphi)$. (
\begin{align*}
A (K \varphi, \varphi) (z) &\geqslant [ (X_k (z) + n \cdot
\tmop{ing} (0, X_n (z)) - C] A (\varphi, \varphi) (z) \\
&\geqslant \frac{[(1
- \mu_0) C + 1]}{\mu_0} \cdot A (\varphi, \varphi) (z) \geqslant A
(\varphi, \varphi) (z)
\end{align*}
)
Thus, if $\varphi$ is any harmonic $(0, q)$-form with coefficients in
$F^{\mu} \otimes D$, with $\mu \geqslant \mu_0$ and $q \geqslant n - k + 1$,
we have
\[ 0 \leqslant \| \varphi \|^2 = (\varphi, \varphi) \leqslant (K \varphi,
\varphi) \leqslant 0, \]
which means that $\varphi \equiv 0$.
\end{proof}
\section{On vector bundles}
\subsection{Terminologies}
\begin{definition}[][Hermitian metric]
On a holomorphic vector bundle with a hermitian metric $h$, there is a
unique connection compatible with $h$ and the complex structure. Namely, it
must be $\nabla = \partial + \bar{\partial}$, where \tmcolor{red}{$\partial
s = h^{- 1} \partial \tmop{hs}$}.
\end{definition}
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\begin{center}
\begin{tblr}[long,theme = fancy,
caption = {Terminologies Interpretation},
entry = {Interpretation},
label = {tblr:Terminologies Interpretation 1},
% note{a} = {第一个表注。},
% note{$\dag$} = {第二个长长长长长长长的表注。},
% remark{Attention!} = {For any \textit{fine sheaf} $\sS$, one has $H^q(X,\sS)=0$ for $q\geqslant 1$.},
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width = \linewidth,
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\textbf{Terminologies} & \textbf{Interpretations}\\
$\tmop{PE}$ & $= (E - 0) /\mathbb{C}^{\ast}$,\\
$(E - 0)$ & The bundle space $E$ minus its zero section,\\
Curvature form $\hat{\Theta}$ & $\hat{\Theta} = \sum_{i, j} \;
\Theta_{\tmop{ri} \bar{j}}^r \tmop{dz}^i \wedge \overline{\tmop{dz}}^j -
\sum_1^{r - 1} d \zeta^{\alpha} \wedge \overline{d \zeta^{\alpha}}$,\\
$D$ & A holomorphic vector bundle over $M$,\\
$S^k E$ & $k$-th symmetric tensor power of $E$,\\
$\tmop{LE}$ & the associated complex line bundle over $\tmop{PE}$.\\
\itbf{Terminologies} & \itbf{Interpretations}\\
\end{tblr}
\end{center}
We have the following homeomorphisms \cite[Theorem 2.1, P504 and P502]{kobayashi1970complex}:
\[ H^q (\tmop{PE}^{\ast}, \Omega ((\tmop{LE}^{\ast})^{- k} \otimes \pi^{\ast}
D)) \cong H^q (M, \Omega (S^k E \otimes D)) \]
and
\[ H^q (\tmop{PE}^{\ast}, \Omega^p (\tmop{LE}^{\ast})^{- 1}) \cong H^q (M,
\Omega^p (E)) . \]
\section{Vanishing theorems for positive semidefinite vector bundles of rank
k}
In order to define the following concept, it is necessary to calculate the
Hermitian quadratic form $\Theta (\zeta, \eta)$, which is strongly associated
with the definition. (cf [8],proof of Proposition 6.3)
\begin{definition}[][Vector bundle being positive-semidefinite of rank $k$]
$E$ is said to be {\emph{positive semidefinite of rank $k$}} if
there exists a hermitian metric on $E$ whose curvature tensor $\Theta$
satisfies the following condition: for all $\zeta \in \mathbb{C}^r - 0$, the
quadratic form on the variable $\eta$ : $\Theta (\zeta, \eta)$
{\emph{is positive semidefinite of rank $k$ at each point of $M$}}, where
\[ \Theta (\zeta, \eta) = \sum_{i, j} \Theta_{\sigma i \bar{j}}^{\rho}
\zeta^{\sigma} \bar{\zeta}^{\rho} \eta^i \bar{\eta}^j . \]
\end{definition}
\subsection{K\"ahler case over semidefinite vector bundles}
\begin{theorem}[][Semidefinite vanishing theorem for vector bundle of rank $k$ (K\"ahler Case)]
Let $E$ be positive semidefinite (or negative semidefinite) of rank $k$ at
each point $z$ of a compact K{\"a}hler manifold $M$. Then
\[ H^q (M, \Omega^p (E)) = 0 \quad \text{if } \quad p + q \geqslant 2 n - (k
- r) . \]
(respectively, $H^q (M, \Omega^p (E)) = 0$ if $p + q \leqslant k - r$. (By
Serre Duality Theorem.))
\end{theorem}
\begin{theorem}[][The generalization of Theorem 3.1][thm:The generalization of Theorem 3.1]
\begin{align*}
\begin{cases}
\rho_S^* \text{ is an isomorphism}, & \text{if } s\leqslant k-r-2,\\
\rho_S^* \text{ is injective}, & \text{if } s= k-r-1.\\
\end{cases}
\end{align*}
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%%%
This paper gives an application of its vanishing theorem of K\"ahler case over semidefinite vector bundle, which is the \textit{Lefschetz theorem on hyerplane sections}. Here, $S$ will be a non-singular complex submanifold of codimension $r$ regularly imbedded in the compact K\"ahler manifold $M$, and $S$ will be assumed to be the zero set of a holomorphic section $\zeta$ of a holomorphic $r$-vector bundle $E\to M$. Let $\rho^*_S : H^s(M,\bC)\to H^s(S,\bC)$. Let $\widehat{\zeta}\in H^0(PE^*,(LE^*)^{-1})$ be the holomorphic section of the line bundle $(LE^*)^{-1}$ corresponding to $\zeta$. Thus, $\widehat{\zeta}=\{p\colon \widehat{\zeta}(p)=0\}$ is a non-singular submanifold of codimension $1$ in $PE^*$.
\subsection{General case over semidefinite vector bundles}
\begin{theorem}[][Semidefinite vanishing theorem for vector bundle of rank $k$ (General Case)][thm:Semidefinite vanishing theorem for vector bundle of rank $k$ (General Case)]
Let $M$ be a compact complex manifold, and assume that there exists a hermitian metric on $E$, whose curvature tensor $\Theta$ satisfies the following condition at each point of $M$.
For any $\zeta\in \bC^r-0$, the quadratic form in $\eta,\Theta(\zeta,\eta)$, has at least $n-k_1$ positive eigenvalues and at least $k_2$ negative eigenvalues. Then
\[H^q(M,\Omega(S^\mu E\otimes D))=0\]
for any $q\not\in (k_1,\ldots,k_2)$ if $\mu\geqslant 0$.
\end{theorem}
\section{The semi-curvature of a line bundle \texorpdfstring{$L(E)$}- over \texorpdfstring{$P(E)$}-}
\begin{proposition}[][\sc Proposition 6.1. of \cite{kobayashi1970complex}]
Given a point $o \in M$, there exist local holomorphic sections $s_1, \cdots, s_r$ around o such that
\begin{equation*}
h_{\alpha \bar{\beta}}=\delta_{\alpha \beta} \quad \text { and } \quad d h_{\alpha \bar{\beta}}=0 \quad \text { at } o .
\end{equation*}
\end{proposition}
\begin{proof}
Choose local holomorphic sections $t_1, \cdots, t_r$ around $o$ which are orthonormal at $o$. We set
\begin{equation*}
s_\alpha=\sum a_\alpha^\beta t_\beta \quad (a_\alpha^\beta: \text { holomorphic })
\end{equation*}
and try to find $\left(a_\alpha^\beta\right)$ such that $a_\alpha^\beta=\delta_\alpha^\beta$ at $o$ and $s_1, \cdots, s_r$ satisfy the required second condition. If we set
\begin{equation*}
g_{\alpha \bar{\beta}}=h\left(t_\alpha, \bar{t}_\beta\right),
\end{equation*}
then
\begin{equation*}
h_{\alpha \bar{\beta}}=h\left(s_\alpha, \bar{s}_\beta\right)=\sum a_\alpha^\gamma g_{r \bar{\delta}} \bar{a}_\beta^{\grave{\delta}},
\end{equation*}
or in matrix form
\begin{equation*}
H={ }^t A \cdot G \cdot \bar{A} .
\end{equation*}
We want to find $A$ such that $A=I$ at $o$ and $d H=0$ at $o$. Since
\begin{equation*}
\partial H=\partial^t A \cdot G \cdot \bar{A}+{ }^t A \cdot \partial G \cdot \bar{A},
\end{equation*}
it suffices to set
\begin{equation*}
a_\alpha^\beta=\delta_\alpha^\beta-\Sigma\left(\frac{\partial g_{\alpha \bar{\beta}}}{\partial z^j}\right)_0 \cdot z^j,
\end{equation*}
where $z^1, \cdots, z^n$ is a local coordinate system with origin $o$.
\end{proof}
\begin{proposition}[][Proposition 6.3. of \cite{kobayashi1970complex}][prop:6.3]
If $E$ is a hermitian vector bundle with negative curvature (resp. semi-negative curvature), then the line bundle $L(E)$ over $P(E)$ with the induced hermitian metric has negative (resp. semi-negative) curvature.
\end{proposition}
\begin{proof}
The naturally induced hermitian metric $\tilde{h}$ in $L(E)$ may be described as follows. Since $L(E)$ minus its zero section is naturally isomorphic to $E$ minus its zero section
\[(L(E)-0)\cong (E-0),\]
every nonzero element $X$ of $L(E)$ may be identified with an element of $E$, and
\begin{equation*}
\tilde{h}(X, X)=h(X, X) .
\end{equation*}
Fixing a point $o$ in the base manifold $M$, we choose holomorphic sections $s_1, \cdots, s_r$ in a neighborhood of $o$ with the properties stated in Proposition 6.1. Then we may write
\begin{equation*}
h(X, X)=\Sigma h_{\alpha \bar{\beta}} \xi^\alpha \bar{\xi}^\beta \quad \text { for } X=\Sigma \xi^\alpha s_\alpha .
\end{equation*}
We shall \textit{compute the Ricci tensor of the line bundle $L(E)$ at an arbitrarily fixed point of $P(E)$ which lies over $o \in M$.} This point is represented by a unit vector $X_0 \in E$. Applying a unitary transformation to $s_1, \cdots, s_r$, we may assume that $X_0=s_r(0)$. We take $z^1, \cdots, z^n, \xi^1, \cdots, \xi^{r-1}$ as a local coordinate system around $\left[X_0\right]$ in $P(E),\left[X_0\right]$ denotes the point of $P(E)$ represented by $X_0$. Then the components of the Ricci tensor of $L(E)$ at $\left[X_0\right]$ are given by
\begin{equation*}
\left(\begin{array}{l}
-\dfrac{\partial^2 \log h(X, X)}{\partial z^i \partial \bar{z}^j}-\dfrac{\partial^2 \log h(X, X)}{\partial z^i \partial \bar{\xi}^\beta} \\[1mm]
-\dfrac{\partial^2 \log h(X, X)}{\partial \xi^\alpha \partial \bar{z}^j}-\dfrac{\partial^2 \log h(X, X)}{\partial \xi^\alpha \partial \bar{\xi}^\beta}
\end{array}\right)=\left(\begin{array}{cr}
-\dfrac{\partial^2 \log h_{\alpha \bar{\beta}}}{\partial z^i \partial \bar{z}^j} & 0 \\[1mm]
0 & -\delta_\beta^\alpha
\end{array}\right)
\end{equation*}
where $i, j=1, \cdots, n$ and $\alpha, \beta=1, \cdots, r-1$. It is clear that this matrix is negative (semi-) definite if the curvature of $E$ is (semi-) negative.
\end{proof}
\begin{remark}
If $E$ has (semi-) positive curvature, its dual $E^*$ has (semi-) negative curvature by Proposition 6.2 and hence the line bundle $L(E *)$ over $P\left(E^*\right)$ has (semi-) negative curvature by Proposition 6.3 and its dual $L\left(E^*\right)^{-1}$ $=L\left(E^*\right)^*$ has (semi-) positive curvature. But $L(E)$ itself does not have (semi-) positive curvature.
\end{remark}
From \textit{Proposition 6.3}, we obtain immediately the following
\begin{theorem}[][\sc Teorem 6.4. of \cite{kobayashi1970complex}]
A hermitian vector bundle $E$ with negative (resp. semi-negative, positive, or semi-positive) curvature is negative (resp. semi-negative, positive, or semi-positive).
\end{theorem}
We do not know if the converse is true, e.g., if a negative vector bundle $E$ admits a hermitian metric with negative curvature. For a line bundle $E$, by definition $E$ is negative (resp. positive) if and only if it admits a hermitian metric with negative (resp. positive) curvature. It is, however, not clear if a semi-negative (resp. semi-positive) line bundle admits a hermitian metric with semi-negative (resp. semi-positive) curvature.
So form the proof of \autoref{prop:6.3} we get the curvature of $L(E)$ with respect to the induced hermitian metric $\tilde{h}$ at the point $[X_0]\in P(E)$
\begin{align*}
\Theta_{L(E)}([X_0]) &=\sum_{i,j} \br{-\dfrac{\partial^2 \log h_{\alpha \bar{\beta}}}{\partial z^i \partial \bar{z}^j} }\dd z^i\wedge \dd \bar{z}^j-\sum_{\alpha=1}^{r-1}\dd \xi^\alpha\wedge\dd \bar{\xi}^\alpha.\\
&=\sum_{i,j} \Theta_{ri\bar{j}}^r \dd z^i\wedge \dd \bar{z}^j-\sum_{\alpha=1}^{r-1}\dd \xi^\alpha\wedge\dd \bar{\xi}^\alpha.
\end{align*}
There is still an issue.
\begin{align*}
\Theta_{i\bar{j}} =\sum \Theta^\alpha_{\alpha i\bar{j}} =-\dfrac{\partial^2\log \det h_{\alpha\bar{\beta}}}{\partial z^i \partial\bar{z}^j}
?=?-\dfrac{\partial^2 \sum_{\alpha,\beta}\log h_{\alpha\bar{\beta}}}{\partial z^i \partial\bar{z}^j}
=\sum_{\alpha,\beta} \br{-\dfrac{\partial^2 \log h_{\alpha\bar{\beta}}}{\partial z^i \partial\bar{z}^j}}.
\end{align*}
\subsection{Computation of \texorpdfstring{$\Theta\wedge\varphi=0$}{}}
For P.52 of \cite{gigante1981vector}
\begin{fancybox}
It can be verified that using (1), if $E$ is negative semidefinite, then any harmonic $(p,0)$- or $(0,p)$- form with coefficients in $E$ has to satisfy the condition : $\Theta\wedge\varphi=0$ at each point of $M$.
\end{fancybox}
In fact, let $\varphi$ be a harmonic $(p,0)$- or $(0,p)$- form with coefficients in $E$, then clearly, we have $\Lambda\varphi\equiv 0$, which shows that (1) can be written as $(\sqrt{-1}\Lambda e(\Theta)\varphi,\varphi)\geqslant 0$. And the pointwise scalar product $A(\sqrt{-1}\Lambda e(\Theta)\varphi,\varphi)\leqslant 0$ at each point of $M$. Thus by (1) it must be $(\Lambda e(\Theta)\varphi,\varphi)=0$ and therefore, for (2), we have
\begin{align*}
&(\Lambda(\partial_E \bar{\partial}+\bar{\partial}\partial_E)\varphi,\varphi)=0\\
&\Longrightarrow \left.\begin{cases}
\Lambda\bar{\partial}\partial_E\varphi=0\rightarrow \bar{\partial}\varphi=0\rightarrow \partial_E\varphi=0, &\text{ when $\varphi$ is a $(0,p)$-form}\\
\Lambda\partial_E \bar{\partial}\varphi=0\rightarrow \partial_E\varphi=0, &\text{ when $\varphi$ is a $(p,0)$-form}\\
\end{cases}\right\}\implies \partial_E\varphi=0.
\end{align*}
Then by (2), $e(\Theta)\varphi=\Theta\wedge\varphi=0$.
For p.53. Since in the strong sense $\Theta=\Theta_{\sigma i\bar{j}}^\rho \tau^\sigma_i \bar{\tau}_j^\rho\geqslant 0$ if $E$ is positive semidefinite and let $\varphi=\sum \varphi_A \dd z^A$ be the local representation of the $(p,0)$-form, then we have
\begin{align*}
\Theta\wedge\varphi &=\sum_{i\in A}\br{\sum \Theta_{\sigma i\bar{j}}^\rho \;\tau^\sigma_i \bar{\tau}_j^\rho\wedge\varphi_{A(i)}\dd z^{A(i)}}
=\sum_{i\in A}\br{\sum \Theta_{\sigma i\bar{j}}^\rho \varphi_{A(i)} \;\tau^\sigma_i \wedge\bar{\tau}_j^\rho\wedge\dd z^{A(i)}}\\
&=\sum_\rho \br{\sum_{i\in A}\sum_\sigma \Theta_{\sigma i\bar{j}}^\rho \varphi_{A(i)}^\sigma (-1)^{p(i)}\;\tau^\sigma_i \wedge\dd z^{A(i)} \wedge\bar{\tau}_j^\rho}\\
&=\sum_\rho \br{\br{\sum_{i\in A}\sum_\sigma \Theta_{\sigma i\bar{j}}^\rho \varphi_{A(i)}^\sigma (-1)^{p(i)}}\;\tau^\sigma_i \wedge\dd z^{A(i)} \wedge\bar{\tau}_j^\rho}\\
&=0,
\end{align*}
which implices that
\[\sum_{i\in A}\sum_\sigma \Theta_{\sigma i\bar{j}}^\rho \varphi_{A(i)}^\sigma (-1)^{p(i)}=0.\]
\newcommand{\tmmathbf}[1]{\ensuremath{\boldsymbol{#1}}}
\newcommand{\tmscript}[1]{\text{\scriptsize{$#1$}}}
% \begin{equation}
% \Theta = - (2 \pi i\mathcal{X}) = - (\bar{\partial} \partial \log h) = -
% \sum_{\alpha, \beta} \frac{\partial^2 \log h}{\partial z^{\alpha} \partial
% \bar{z}^{\beta}} \tmop{dz}^{\alpha} \overline{\tmop{dz}^{\beta}} \label{6}
% \end{equation}
\subsection{Computation of the quadratic form \texorpdfstring{$\Theta (\zeta^0, \eta)$}{}}
For $L (T^{\ast} P)$ its Ricci curvature is
\[ \Theta (\xi, \eta) = \left( \begin{array}{c}
- \dfrac{\partial^2 \log h (X, X)}{\partial z^i \partial \bar{z}^j} -
\dfrac{\partial^2 \log h (X, X)}{\partial z^i \partial
\bar{\xi}^{\beta}}\\[1mm]
- \dfrac{\partial^2 \log h (X, X)}{\partial \xi^{\alpha} \partial
\bar{z}^j} - \dfrac{\partial^2 \log h (X, X)}{\partial \xi^{\alpha}
\partial \bar{\xi}^{\beta}}
\end{array} \right) = \left( \begin{array}{rr}
- \dfrac{\partial^2 \log h_{\alpha \beta}}{\partial z^i \partial
\bar{z}^j} & 0\\[1mm]
0 & - \delta_{\beta}^{\alpha}
\end{array} \right) \]
For the Fubini-Study metric on $P_n (\tmmathbf{C})$ with holomorphic sectional
curvature $c$, the curvature tensor is given by
\[ K_{i \bar{j} k \bar{l}} = - \frac{c}{2} (h_{i \bar{j}} h_{k \bar{l}} +
h_{i \bar{l}} h_{k \bar{j}}) . \]
Given a point $o$ in $P_n (\tmmathbf{C})$, we may always choose a local
coordinate system around $o$ so that the metric tensor $h_{i \bar{j}}$
coincides with $\delta_{ij}$ at $o$. Then the curvature of the cotangent
bundle is given by
\[ \frac{c}{2} (\delta_{ij} \delta_{kl} + \delta_{il} \delta_{kj}) . \]
Note that the sign changes when we pass from $TP$ to $T \ast P$. The matrix
representing the Ricci curvature of $L (T^{\ast} P)$ in the proof of
Proposition 6.3 reduces in this case to the following:
\[ \left( \begin{array}{cc}
- \frac{c}{2} (\delta_{ij} + \delta_{in} \delta_{jn}) & 0\\
0 & - \delta_{\alpha \beta}
\end{array} \right)_{\tmscript{\begin{array}{l}
i, j = 1, \cdots, n\\
\alpha, \beta = 1, \cdots, n - 1
\end{array}}} . \]
The Ricci curvature of $L (T^{\ast} P)^{- 1}$ is obtained from that of $L
(T^{\ast} P)$ by changing its sign. On the other hand, the Ricci curvature of
$P$ at $o$ (which is nothing but the Ricci curvature of $K_P^{- 1}
= H^{n + 1})$ is given by $K_{i \bar{j}} = - \frac{c}{2} (n + 1)
\delta_{ij}$. Hence, the Ricci curvature of $H$ is given by
\[ - \frac{1}{n + 1} K_{i \bar{j}} = \frac{c}{2} \delta_{ij} . \]
Consequently, the Ricci curvature of $L (T^{\ast} P)^{- k} \otimes \pi^{\ast}
H^m$ can be expressed by the following matrix:
\[ \left( \begin{array}{cc}
k \frac{c}{2} (\delta_{ij} + \delta_{in} \delta_{jn}) & 0\\
0 & k \delta_{\alpha \beta}
\end{array} \right) + \left( \begin{array}{cc}
m \frac{c}{2} \delta_{ij} & 0\\
0 & 0
\end{array} \right), \]
which is clearly positive if $m + k \geqq 1$ and $k \geqq 1$.
Now we compute the quadratic form $\Theta (\zeta^0, \eta)$ of $S^k
(\tmop{TP}) \otimes H^m$ as follow: \cite[P52]{gigante1981vector}
\begin{align*}
\Theta ( \eta,\zeta^0) & = \sum_{\alpha,\beta,i, j} \br{\left( \begin{array}{cc}
k \frac{c}{2} (\delta_{ij} + \delta_{in} \delta_{jn}) & 0\\
0 & k \delta_{\alpha \beta}
\end{array} \right) + \left( \begin{array}{cc}
m \frac{c}{2} \delta_{ij} & 0\\
0 & 0
\end{array} \right)} \begin{pmatrix}
\eta^i
\overline{\eta^j}\\ \zeta^{\alpha} \overline{\zeta^{\beta}}
\end{pmatrix} \\
(\delta_{\alpha \beta} \equiv 0) & =\sum_{\alpha,\beta,i, j}
\begin{pmatrix}
k \frac{c}{2} (\delta_{ij} + \delta_{in} \delta_{jn})+m \frac{c}{2} \delta_{ij} &0\\ 0&0
\end{pmatrix}
\begin{pmatrix}
\eta^i
\overline{\eta^j}\\ \zeta^{\alpha} \overline{\zeta^{\beta}}
\end{pmatrix} \\
& = \sum_{i, j}\left[ k \frac{c}{2} (\delta_{ij} + \delta_{in} \delta_{jn}) + m
\frac{c}{2} \delta_{ij} \right] (\eta^i \overline{\eta^j})\\
& = k \frac{c}{2} \left( \sum_{i, j \leqslant n - 1} \delta_{\tmop{ij}}
\eta^i \overline{\eta^j} + \sum_i \delta_{\tmop{in}} \eta^i
\overline{\eta^n} + \sum_j \delta_{\tmop{jn}} \eta^n \overline{\eta^j}
\right) + m \frac{c}{2} \sum_{i, j} \delta_{\tmop{ij}} \eta^i
\overline{\eta^j}\\
& = (k + m) \frac{c}{2} \sum_{i, j \leqslant n - 1} \delta_{\tmop{ij}}
\eta^i \overline{\eta^j} + (2 k + m) \frac{c}{2} \eta^n \overline{\eta^n}\\
& = (k + m) \frac{c}{2} \sum_{i \leqslant n - 1} \eta^i \overline{\eta^i}
+ (2 k + m) \frac{c}{2} \eta^n \overline{\eta^n}
\end{align*}
which is semi-positive if $m + k \geqq 1$ and $k \geqq 1$.
\section*{hyperplane Theorem}
We note that the fibre of $S^k E^*$ over $x \in M$ is the space of homogeneous polynomials of degree $k$ on $E_x$.
\begin{theorem}[][Theorem 1.3. of \cite{gigante1981vector}]
Let $E$ be positive semidefinite (or negative semidefinite) of rank $k$ at each point $z$ of a compact K\"ahler manifold $M$. Then
\begin{equation*}
H^q\left(M, \Omega^p(E)\right)=0 \quad \text { if } \quad p+q \geqslant 2 n-(k-r).
\end{equation*}
(respectively, $H^q\left(M, \Omega^p(E)\right)=0$ if $p+q \leqslant k-r$. (By Serre Duality Theorem.))
\end{theorem}
\begin{theorem}[][Lefschetz theorem on hyperplane sections][thm:lefschetz]
If the Chern class $\mC_\bR ([S])\in H^2 (M,\bR)$ of $[S]$ contains a form $\mX\geqslant 0$ of rank $k$, then
\[\rho^*_S\text{ is an isomorphism for $s\leqslant k-2$}\]
and
\[\rho^*_S\text{ is injective for $s=k-1$}.\]
\end{theorem}
\begin{proof}
Let $\boldsymbol{S}^{n-1}$ be a non-singular analytic subvariety of $\boldsymbol{V}^n, \mathfrak{B}_s$ the restriction of the bundle $\mathfrak{B}$ on $S$, and the sequence
\begin{equation*}
0 \rightarrow \Omega^{\prime p}(\mathfrak{B}) \stackrel{i}{\rightarrow} \Omega^p(\mathfrak{B}) \stackrel{r}{\rightarrow} \Omega^p\left(\mathfrak{B}_s\right) \rightarrow 0
\end{equation*}
be exact. And let $\eta \in \Omega^{\prime p}(\mathfrak{B})$
\begin{equation*}
\eta=\sum_{\alpha_1<\cdots<\alpha_p} \eta_{\alpha_1 \ldots \alpha_p} d z^{\alpha_1} \ldots d z^{\alpha_p},
\end{equation*}
where we assume that $z^1=0$ the local equation of $S$. Then it is clear, that for $p \geqq 1$
\begin{equation*}
\eta^{\prime}=\sum_{1<\alpha_2<\cdots<\alpha_p}\left(\eta_{1 \alpha_2 \ldots \alpha_p}\right)_s d z^{\alpha_2} \ldots d z^{\alpha_p}
\end{equation*}
belongs to $\Omega^{p-1}\left\{(\mathfrak{B}-\{\boldsymbol{s}\})_s\right\}$. If we denote by $\bar{r}$ the mapping $\Omega^{\prime p}(\mathfrak{B})\rightarrow \Omega^{p-1}\left\{(\mathfrak{B}-\{\boldsymbol{s}\})_s\right\}$ such that
\begin{equation*}
\overline{\boldsymbol{r}}: \quad \eta \in \Omega^{\prime p}(\mathfrak{B}) \rightarrow \eta^{\prime} \in \Omega^{p-1}\left\{(\mathfrak{B}-\{\boldsymbol{s}\})_s\right\},
\end{equation*}
then we can easily prove that the sequence
\begin{equation*}
0 \rightarrow \Omega^p(\mathfrak{B}-\{\boldsymbol{s}\}) \rightarrow \Omega^{\prime p}(\mathfrak{B}) \rightarrow \Omega^{p-1}\left\{(\mathfrak{B}-\{\boldsymbol{s}\})_s\right\} \rightarrow 0
\end{equation*}
is exact. By taking the sequence of cohomology groups corresponding to (12), we obtain the exact sequence
\begin{equation*}
\rightarrow H^q\left(\Omega^p(\mathfrak{B}-\{\boldsymbol{s}\})\right) \rightarrow H^q\left(\Omega^{\prime p}(\mathfrak{B})\right) \rightarrow H^q\left(\Omega^{p-1}\left\{(\mathfrak{B}-\{\boldsymbol{s}\})_s\right\}\right) \rightarrow .
\end{equation*}
Now let us assume that $\boldsymbol{S}$ is so ample that $c(\mathfrak{B}-\{\boldsymbol{s}\})$ contains an everywhere negative definite form, then we see by theorem $1^{\prime \prime}$ that
\begin{equation*}
H^q\left(\Omega^p(\mathfrak{B}-\{\boldsymbol{s}\})\right) \simeq 0 \quad \text { for } \quad p+q \leqq n-1,
\end{equation*}
\begin{equation*}
H^q\left(\Omega^{p-1}(\mathfrak{B}-\{\boldsymbol{s}\})_s\right) \simeq 0 \quad \text { for } \quad p+q \leqq n-1
\end{equation*}
Putting these in (13), we have, if $p \geqq 1$, for $p+q \leqq n-1$
\begin{equation*}
H^\alpha\left(\Omega^{\prime p}(\mathfrak{B})\right) \simeq 0
\end{equation*}
Moreover it is also the case even when $p=0$, because
$$\Omega^{\prime 0}(\mathfrak{B})=\Omega^0(\mathfrak{B}-\{\boldsymbol{s}\}), \quad H^q\left(\Omega^0(\mathfrak{B}-\{\boldsymbol{s}\})\right) \simeq 0 \quad\text{ for }q \leqq n-1.$$
On the other hand, taking the sequence of cohomology groups of the sequence (11), we get the exact sequence
\begin{equation*}
\rightarrow H^q\left(\Omega^{\prime p}(\mathfrak{B})\right) \rightarrow H^q\left(\Omega^p(\mathfrak{B})\right) \rightarrow H^q\left(\Omega^p\left(\mathfrak{B}_s\right)\right) \rightarrow H^{q+1}\left(\Omega^{\prime p}(\mathfrak{B})\right) \rightarrow .
\end{equation*}
\begin{theorem}
If the divisor $\boldsymbol{S}^{n-1}$ is so ample such that $c(\mathfrak{B}-\{\boldsymbol{s}\})$ contains an everywhere negative definite form, then for $p+q \leqq n-1$ there exists the isomorphism
\begin{equation*}
H^q\left(\Omega^p(\mathfrak{B})\right) \rightarrow H^q\left(\Omega^p\left(\mathfrak{B}_s\right)\right)
\end{equation*}
and this is an isomorphism onto or into according as $p+q \leqq n-2$ or $p+q=n-1$.
\end{theorem}
Consider the special case, where $\boldsymbol{V}$ is a projective variety, $\boldsymbol{S}$ a generic hyperplane section of $\boldsymbol{V}$. Taking $\mathfrak{B}$ as $\{0\} \quad(\{0\}$ the trivial bundle), then clearly $\{0\}-\{\boldsymbol{S}\}$ contains an everywhere negative definite form, so the mapping
\begin{equation*}
H^q\left(\Omega^p(0)\right) \rightarrow H^q\left(\Omega^p\left(0_s\right)\right) \text { for } \quad p+q \leqq n-1
\end{equation*}
is isomorphic. But we see by the Dolbeault's theorem
\begin{equation*}
H^q\left(\Omega^p(0)\right) \simeq H^{p, q}(\boldsymbol{V}, C), \quad H^q\left(\Omega^p\left(0_s\right)\right) \simeq H^{p, q}(\boldsymbol{S}, C),
\end{equation*}
where $C$ is complex number field. Hence we have the Lefschetz theorem in the classical form:
\begin{theorem}
Let $\boldsymbol{V}$ be an algebraic variety of dim. $n$ without singularities immersed in a projective space, $\boldsymbol{S}$ be a generic hyperplane section of it (consequently $\boldsymbol{S}$ is irreducible and has no singularities), $H(\boldsymbol{V}, C)$ the cohomology group of degree $r$.
Then $H^r(\boldsymbol{V}, C)$ is isomorphic to $H^r(\boldsymbol{S}, C)$ if $r \leqq n-2$, and $H^{n-1}(\boldsymbol{V}, C)$ is isomorphic to a submodule of $H^{n-1}(\boldsymbol{S}, C)$.
\end{theorem}