双次数

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 8.2

设 $M$ 是满足 ${\dim }_{\mathbb{C}}M=m$ 的复流形。设 $\omega \in {\Omega }_p^q(M{)}^{\mathbb{C}}$（$q\le 2m$）以及 $r,s$ 为正整数，且 $r+s=q$。设 $V_i\in T_p{M}^{\mathbb{C}}$（$1\le i\le q$）是 $T_p{M}^{+}$ 或 $T_p{M}^{-}$ 中的向量。若只要 $r$ 个 $V_i$ 属于 $T_p{M}^{+}$ 且 $s$ 个 $V_i$ 属于 $T_p{M}^{-}$ 就有 $\omega (V_1,\dots ,V_q)=0$，则称 $\omega$ 具有双次数(bidegree) $(r,s)$，或简称为 $(r,s)$-形式。在点 $p$ 处的 $(r,s)$-形式的集合记作 ${\Omega }_p^{r,s}(M)$。若一个 $(r,s)$-形式在 $M$ 的每一点处都光滑地给出，就得到定义在 $M$ 上的一个 $(r,s)$-形式。$M$ 上 $(r,s)$-形式的集合记作 ${\Omega }^{r,s}(M)$。

取一个坐标卡 $(U,\phi )$，其复坐标为 $\phi (p)=z^{\mu }$。我们取如(8.13)所示的基作为切空间 $T_p{M}^{\pm }$ 的基，其对偶基由(8.14)给出。注意 $\mathrm{d}{z}^{\mu }$ 具有双次数 $(1,0)$，因为 $⟨\mathrm{d}{z}^{\mu },\partial /\partial {\overline{z}}^{\nu }⟩=0$，而 $\mathrm{d}{\overline{z}}^{\mu }$ 具有双次数 $(0,1)$。在这些基下，双次数为 $(r,s)$ 的形式 $\omega$ 可写成

$$\omega =\frac{1}{r!s!}{\omega }_{{\mu }_1\dots {\mu }_r{v}_1\dots v_s}d{z}^{{\mu }_1}\wedge \cdots \wedge d{z}^{{\mu }_r}\wedge d{\bar{z}}^{v_1}\wedge \cdots \wedge d{\bar{z}}^{v_s}.\text{\hspace{1em}(8.35)}$$

集合 $\{\mathrm{d}{z}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{z}^{{\mu }_r}\wedge \mathrm{d}{\overline{z}}^{{\nu }_1}\wedge \dots \wedge \mathrm{d}{\overline{z}}^{{\nu }_s}\}$ 是 ${\Omega }_p^{r,s}(M)$ 的基。其分量分别关于 $\mu$ 与 $\nu$ 全反对称。设 $z^{\mu }$ 与 $w^{\mu }$ 是两个相互重叠的坐标系。读者应验证：在 $z^{\mu }$ 坐标系中的 $(r,s)$-形式，在 $w^{\nu }$ 坐标系中同样是 $(r,s)$-形式。

英文原文

Let $M$ be a complex manifold with ${\dim }_{\mathbb{C}}M=m$ . Let $\omega \in {\Omega }_p^q(M{)}^{\mathbb{C}}$ ( $q\le 2m$ ) and $r,s$ be positive integers such that $r+s=q$ . Let $V_i\in T_p{M}^{\mathbb{C}}$ ( $1\le i\le q$ ) be vectors in either $T_p{M}^{+}$ or $T_p{M}^{-}$ . If $\omega (V_1,\dots ,V_q)=0$ unless $r$ of the $V_i$ are in $T_p{M}^{+}$ and $s$ of the $V_i$ are in $T_p{M}^{-}$ , $\omega$ is said to be of bidegree $(r,s)$ or simply an $(r,s)$ -form. The set of $(r,s)$ -forms at $p$ is denoted by ${\Omega }_p^{r,s}(M)$ . If an $(r,s)$ -form is assigned smoothly at each point of $M$ , we have an $(r,s)$ -form defined over $M$ . The set of $(r,s)$ -forms over $M$ is denoted by ${\Omega }^{r,s}(M)$ .

Take a chart $(U,\phi )$ with the complex coordinates $\phi (p)=z^{\mu }$ . We take the bases (8.13) for the tangent spaces $T_p{M}^{\pm }$ . The dual bases are given by (8.14). Note that $\mathrm{d}{z}^{\mu }$ is of bidegree $(1,0)$ since $⟨\mathrm{d}{z}^{\mu },\partial /\partial {\overline{z}}^{\nu }⟩=0$ and $\mathrm{d}{\overline{z}}^{\mu }$ is of bidegree $(0,1)$ . With these bases, a form $\omega$ of bidegree $(r,s)$ is written as

$$\omega =\frac{1}{r!s!}{\omega }_{{\mu }_1\dots {\mu }_r{v}_1\dots v_s}d{z}^{{\mu }_1}\wedge \cdots \wedge d{z}^{{\mu }_r}\wedge d{\bar{z}}^{v_1}\wedge \cdots \wedge d{\bar{z}}^{v_s}.\text{\hspace{1em}(8.35)}$$

The set $\{\mathrm{d}{z}^{{\mu }_1}\wedge \dots \wedge \mathrm{d}{z}^{{\mu }_r}\wedge \mathrm{d}{\overline{z}}^{{\nu }_1}\wedge \dots \wedge \mathrm{d}{\overline{z}}^{{\nu }_s}\}$ is the basis of ${\Omega }_p^{r,s}(M)$ . The components are totally anti-symmetric in the $\mu$ and $\nu$ separately. Let $z^{\mu }$ and $w^{\mu }$ be two overlapping coordinates. The reader should verify that an $(r,s)$ -form in the $z^{\mu }$ coordinate system is also an $(r,s)$ -form in the $w^{\nu }$ system.

出处：8.3 复微分形式 · node a-ce1bc18b7171bbc2fa2209fd · 源节点 SHA-256 2fef415da62b4ef6…