全纯向量丛

本书暂定概念：尚未进入正式概念库。下面是本书给出的定义原文。

定义 · 定义 10.6

设 $E$ 与 $M$ 为复流形，$\pi :E\to M$ 为全纯满射。若下列公理成立，则称流形 $E$ 为一个全纯向量丛(holomorphic vector bundle)。

(i) 典型纤维为 ${\mathbb{C}}^k$，结构群为 $\mathrm{GL}(k,\mathbb{C})$。
(ii) 局部平凡化 ${\varphi }_i:U_i\times {\mathbb{C}}^k\to {\pi }^{-1}(U_i)$ 是双全纯映射。

(iii) 过渡函数 $t_{ij}:U_i\cap U_j\to G=\mathrm{GL}(k,\mathbb{C})$ 是全纯映射。

例如，设 $M$ 为复流形，且 ${\dim }_{\mathbb{C}}M=m$。则全纯切丛 $T{M}^{+}\equiv {\bigcup }_{p\in M}{T}_p{M}^{+}$ 是一个全纯向量丛，其典型纤维为 ${\mathbb{C}}^m$，局部基为 $\{\partial /\partial z^{\mu }\}$。

设 $h$ 为全纯向量丛上的一个内积，它在 $p\in M$ 处的作用为 $h_p:{\pi }^{-1}(p)\times {\pi }^{-1}(p)\to \mathbb{C}$。最自然的内积是满足下列条件的厄米结构：

(i) $h_p(u,av+bw)=a{h}_p(u,v)+b{h}_p(u,w)$，其中 $u,v,w\in {\pi }^{-1}(p),a,b\in \mathbb{C};$
(ii) $h_p(u,v)=\overline{h_p(v,u)},u,v\in {\pi }^{-1}(p);$
(iii) $h_p(u,u)\ge 0$，且 $h_p(u,u)=0$ 当且仅当 $u={\varphi }_i(p,0)$；以及
(iv) $h(s_1,s_2)\in \mathcal{F}(M{)}^{\mathbb{C}}$，其中 $s_1,s_2\in \Gamma (M,E)$

称一组截面 $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ 为酉标架(unitary frame)，若

$$h\left({\hat{e}}_i,{\hat{e}}_j\right)={\delta }_{ij}.\text{\hspace{1em}(10.76)}$$

酉标架丛 $LM$ 不是全纯向量丛，因为结构群 $\mathrm{U}(m)$ 不是复流形。

给定厄米结构 $h$，我们定义一个与 $h$ 相容的联络。所谓厄米联络 $\nabla$，是指满足下列条件的线性映射 $\Gamma (M,E)\to \Gamma (M,E\otimes T^{\ast }{M}^{\mathbb{C}})$：

(i) $\nabla (fs)=(\mathrm{d}f)s+f\nabla s,f\in \mathcal{F}(M{)}^{\mathbb{C}},s\in \Gamma (M,E);$
(ii) $\mathrm{d}[h(s_1,s_2)]=h(\nabla s_1,s_2)+h(s_1,\nabla s_2);$；以及
(iii) 按作用结果的不同，我们将 $\nabla$ 的作用分解为 $\nabla s=\mathrm{D}s+\bar{\mathrm{D}}s$，其中 $\mathrm{D}s$、$\bar{\mathrm{D}}s$（$\bar{\mathrm{D}}s$）分别是取值于截面的 $(1,0)$-形式（(0,1)-形式）。我们要求 $\bar{\mathrm{D}}=\bar{\partial }$。

可以证明，给定 $E$ 与厄米度量 $h$，存在唯一的厄米联络 $\nabla$。曲率由厄米联络定义。设 $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ 为酉标架，并如下定义局部联络形式 ${\mathcal{A}}_{\alpha }^{\beta }$：

$$\nabla {\hat{e}}_{\alpha }={\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.77)}$$

场强定义为

$$\mathcal{F}\equiv \mathrm{d}\mathcal{A}+\mathcal{A}\wedge \mathcal{A}.\text{\hspace{1em}(10.78)}$$

我们来验证

$$\nabla \nabla {\hat{e}}_{\alpha }=\nabla \left({\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }\right)={\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.79)}$$

我们证明 $\mathcal{A}$ 与 $\mathcal{F}$ 都是反厄米的：

$${\bar{\mathcal{A}}}_{\alpha }^{\beta }+{\mathcal{A}}_{\beta }^{\alpha }=h(\nabla {\hat{e}}_{\alpha },{\hat{e}}_{\beta })+h({\hat{e}}_{\alpha },\nabla {\hat{e}}_{\beta })=\mathrm{d}h({\hat{e}}_{\alpha },{\hat{e}}_{\beta })=\mathrm{d}{\delta }_{\alpha \beta }=0$$ $$\begin{array}{rl}{\mathcal{F}}_{\alpha }^{\beta }+{\bar{\mathcal{F}}}_{\beta }^{\alpha }& =\mathrm{d}{A}_{\alpha }^{\beta }+A_{\gamma }^{\beta }\wedge A_{\alpha }^{\gamma }+\mathrm{d}{\bar{A}}_{\beta }^{\alpha }+{\bar{A}}_{\gamma }^{\alpha }\wedge {\bar{A}}_{\alpha }^{\gamma }\\ & =\mathrm{d}\left({\mathcal{A}}_{\alpha }^{\beta }-{\mathcal{A}}_{\alpha }^{\beta }\right)+{\mathcal{A}}_{\gamma }^{\beta }\wedge {\mathcal{A}}_{\alpha }^{\gamma }+{\mathcal{A}}_{\alpha }^{\gamma }\wedge {\mathcal{A}}_{\gamma }^{\alpha }=0.\end{array}$$

于是我们证明了

$${\mathcal{A}}_{\beta }^{\alpha }=-{\bar{\mathcal{A}}}_{\alpha }^{\beta }{\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }.\text{\hspace{1em}(10.80)}$$

下面我们证明 $\mathcal{F}$ 是一个 $(1,1)$-形式。设 $\{{\hat{e}}_{\alpha }\}$ 为酉标架。$\mathcal{F}$ 不可能含有双次数为 $(0,2)$ 的分量，因为

$${\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }=\nabla \nabla {\hat{e}}_{\alpha }=(\mathrm{D}+\bar{\partial })(\mathrm{D}+\bar{\partial }){\hat{e}}_{\alpha }=\mathrm{DD}{\hat{e}}_{\alpha }+(\mathrm{D}\bar{\partial }+\bar{\partial }\mathrm{D}){\hat{e}}_{\alpha }.$$

由 ${\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }^{\alpha }$ 可知，$\bar{\mathcal{F}}$ 不含双次数为 (0, 2) 的分量，从而 $\mathcal{F}$ 也不含双次数为 (2, 0) 的分量。因此 ${\mathcal{F}}_{\alpha }^{\beta }$ 是双次数为 (1, 1) 的二次形式。

来源与版本

正文：中文 · 机器译稿 · 待校对

核对状态：机器译稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

译稿产物 SHA-256：f121cf742ae1e0e36a1ecfe3b41c825801ee5eace3a28b9df9065d4d74d9e293

英文原文

Let $E$ and $M$ be complex manifolds and $\pi :E\to M$ a holomorphic surjection. The manifold $E$ is a holomorphic vector bundle if the following axioms are fulfilled.

(i) The typical fibre is ${\mathbb{C}}^k$ and the structure group is $\mathrm{GL}(k,\mathbb{C})$ .
(ii) The local trivialization ${\varphi }_i:U_i\times {\mathbb{C}}^k\to {\pi }^{-1}(U_i)$ is a biholomorphism.

(iii) The transition function $t_{ij}:U_i\cap U_j\to G=\mathrm{GL}(k,\mathbb{C})$ is a holomorphic map.

For example, let $M$ be a complex manifold with ${\dim }_{\mathbb{C}}M=m$ . The holomorphic tangent bundle $T{M}^{+}\equiv {\bigcup }_{p\in M}{T}_p{M}^{+}$ is a holomorphic vector bundle. The typical fibre is ${\mathbb{C}}^m$ and the local basis is $\{\partial /\partial z^{\mu }\}$ .

Let $h$ be an inner product on a holomorphic vector bundle whose action at $p\in M$ is $h_p:{\pi }^{-1}(p)\times {\pi }^{-1}(p)\to \mathbb{C}$ . The most natural inner product is a Hermitian structure which satisfies:

(i) $h_p(u,av+bw)=a{h}_p(u,v)+b{h}_p(u,w)$ , for $u,v,w\in {\pi }^{-1}(p),a,b\in \mathbb{C};$
(ii) $h_p(u,v)=\overline{h_p(v,u)},u,v\in {\pi }^{-1}(p);$
(iii) $h_p(u,u)\ge 0$ , $h_p(u,u)=0$ if and only if $u={\varphi }_i(p,0)$ ; and
(iv) $h(s_1,s_2)\in \mathcal{F}(M{)}^{\mathbb{C}}$ for $s_1,s_2\in \Gamma (M,E)$

A set of sections $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ is a unitary frame if

$$h\left({\hat{e}}_i,{\hat{e}}_j\right)={\delta }_{ij}.\text{\hspace{1em}(10.76)}$$

The unitary frame bundle $LM$ is not a holomorphic vector bundle since the structure group $\mathrm{U}(m)$ is not a complex manifold.

Given a Hermitian structure $h$ , we define a connection which is compatible with $h$ . The Hermitian connection $\nabla$ is a linear map $\Gamma (M,E)\to \Gamma (M,E\otimes T^{\ast }{M}^{\mathbb{C}})$ which satisfies:

(i) $\nabla (fs)=(\mathrm{d}f)s+f\nabla s,f\in \mathcal{F}(M{)}^{\mathbb{C}},s\in \Gamma (M,E);$
(ii) $\mathrm{d}[h(s_1,s_2)]=h(\nabla s_1,s_2)+h(s_1,\nabla s_2);$ and
(iii) according to the destination, we separate the action of $\nabla$ as $\nabla s=\mathrm{D}s+\bar{\mathrm{D}}s$ , $\mathrm{D}s$ ( $\bar{\mathrm{D}}s$ ) being a $(1,0)$ -form ((0,1)-form) valued section. We demand that $\bar{\mathrm{D}}=\bar{\partial }$ .

It can be shown that given $E$ and a Hermitian metric $h$ , there exists a unique Hermitian connection $\nabla$ . The curvature is defined from the Hermitian connection. Let $\{{\hat{e}}_1,\dots ,{\hat{e}}_k\}$ be a unitary frame and define the local connection form ${\mathcal{A}}_{\alpha }^{\beta }$ by

$$\nabla {\hat{e}}_{\alpha }={\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.77)}$$

The field strength is defined by

$$\mathcal{F}\equiv \mathrm{d}\mathcal{A}+\mathcal{A}\wedge \mathcal{A}.\text{\hspace{1em}(10.78)}$$

We verify that

$$\nabla \nabla {\hat{e}}_{\alpha }=\nabla \left({\hat{e}}_{\beta }{\mathcal{A}}_{\alpha }^{\beta }\right)={\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }.\text{\hspace{1em}(10.79)}$$

We prove that both $\mathcal{A}$ and $\mathcal{F}$ are skew Hermitian:

$${\bar{\mathcal{A}}}_{\alpha }^{\beta }+{\mathcal{A}}_{\beta }^{\alpha }=h(\nabla {\hat{e}}_{\alpha },{\hat{e}}_{\beta })+h({\hat{e}}_{\alpha },\nabla {\hat{e}}_{\beta })=\mathrm{d}h({\hat{e}}_{\alpha },{\hat{e}}_{\beta })=\mathrm{d}{\delta }_{\alpha \beta }=0$$ $$\begin{array}{rl}{\mathcal{F}}_{\alpha }^{\beta }+{\bar{\mathcal{F}}}_{\beta }^{\alpha }& =\mathrm{d}{A}_{\alpha }^{\beta }+A_{\gamma }^{\beta }\wedge A_{\alpha }^{\gamma }+\mathrm{d}{\bar{A}}_{\beta }^{\alpha }+{\bar{A}}_{\gamma }^{\alpha }\wedge {\bar{A}}_{\alpha }^{\gamma }\\ & =\mathrm{d}\left({\mathcal{A}}_{\alpha }^{\beta }-{\mathcal{A}}_{\alpha }^{\beta }\right)+{\mathcal{A}}_{\gamma }^{\beta }\wedge {\mathcal{A}}_{\alpha }^{\gamma }+{\mathcal{A}}_{\alpha }^{\gamma }\wedge {\mathcal{A}}_{\gamma }^{\alpha }=0.\end{array}$$

Thus, we have shown that

$${\mathcal{A}}_{\beta }^{\alpha }=-{\bar{\mathcal{A}}}_{\alpha }^{\beta }{\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }.\text{\hspace{1em}(10.80)}$$

Next we show that $\mathcal{F}$ is a $(1,1)$ -form. Let $\{{\hat{e}}_{\alpha }\}$ be a unitary frame. $\mathcal{F}$ cannot have a component of bidegree- $(0,2)$ since

$${\hat{e}}_{\beta }{\mathcal{F}}_{\alpha }^{\beta }=\nabla \nabla {\hat{e}}_{\alpha }=(\mathrm{D}+\bar{\partial })(\mathrm{D}+\bar{\partial }){\hat{e}}_{\alpha }=\mathrm{DD}{\hat{e}}_{\alpha }+(\mathrm{D}\bar{\partial }+\bar{\partial }\mathrm{D}){\hat{e}}_{\alpha }.$$

It follows from ${\mathcal{F}}_{\alpha }^{\beta }=-{\bar{\mathcal{F}}}_{\beta }^{\alpha }$ that $\bar{\mathcal{F}}$ has no component of bidegree-(0, 2), and, hence, $\mathcal{F}$ has no component of bidegree-(2, 0) either. Thus ${\mathcal{F}}_{\alpha }^{\beta }$ is a two-form of bidegree-(1, 1).

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：9a2c71f0a38326e2a18f5d3faa21fa36a935259f7b31ef0745addc53a7c16480

出处：10.4 伴随向量丛上的协变导数 · node a-08b2129aefb40660419f2012 · 源节点 SHA-256 3dcbb06e53a300c1…