Complex Geometry and Kahler Geometry_20130227p1

Vanabel/ 3月 1, 2013/ Complex and Kahler Geometry/ 3 comments

0.1. Holomorphic Map

定义 1. A complex-valued function $f(z)$ defined on a connected open domain $W\subseteq\C^n$ is called holomorphic if for each $a=(a_1,\ldots, a_n)\in W$, $f(z)$ can be represented as a power series
\sum_{k_1\geq0,\cdots,k_n\geq0}^\infty C_{k_1\cdots k_n}(z_1-a_1)^{k_1}\cdots (z_n-a_n)^k_n
in some neighbourhood of $a$.

We have the following equivalent definition
定义 2. Let $f(z)$ be a (continuously) differentiable function on an open set $W\subseteq\C^n$. Then $f(z)$ is holomorphic iff $\frac{\pt{f}}{\pt\bar z_\nu}=0$, $1\leq\nu\leq n$.

证明 . The proof is based on the Cauchy integral theorem and called “Osgood Theorem”, cf. [1] for detail.

Recall that a complex-valued function of $n$ complex variables can be considered as a function of $2n$ real variables, since $\C^n\simeq \R^{2n}$, related by $z_\nu=x_\nu+i y_\nu$, $i=\sqrt{-1}$, $x_\nu,y_\nu\in\R$. We have the following
\rd z_\nu&=\rd x_\nu+i\rd y_\nu,
\rd \bar z_\nu&=\rd x_\nu-i\rd y_\nu\\
\ppt{\bar z_\nu}&=\frac{1}{2}\left(\ppt{x_\nu}+i\ppt{y_\nu}\right)\\
\rd x_\nu&=\frac{1}{2}(\rd z_\nu+\rd\bar z_\nu),
\rd y_\nu&=\frac{1}{2i}(\rd z_\nu-\rd\bar z_\nu)\\
\ppt{x_\nu}&=\ppt{z_\nu}+\ppt{\bar z_\nu},
\ppt{y_\nu}&=i\left(\ppt{z_\nu}-\ppt{\bar z_\nu}\right).
With the help of the above relation, the total differential of a complex-$n$ valued function $f$ is
\rd f &=\frac{\pt f}{\pt x_\nu}\rd x_\nu+\frac{\pt f}{\pt y_\nu}\rd y_\nu\\
\frac{\pt f}{\pt z_\nu}+\frac{\pt f}{\pt \bar z_\nu}\right)\frac{1}{2}(\rd z_\nu+\rd\bar z_\nu)+
\frac{\pt f}{\pt z_\nu}-\frac{\pt f}{\pt \bar z_\nu}\right)
\frac{1}{2i}(\rd z_\nu-\rd\bar z_\nu)\\
&=\frac{\pt f}{\pt z_\nu}\rd z_\nu+\frac{\pt f}{\pt \bar z_\nu}\rd \bar z_\nu\\
&\eqdef \pt f+\bar\pt f.
We shall call $\pt f$ and $\bar\pt f$ the holomorphic part and Anti-holomorphic part of $f$, respectively.
注记 1. $\bar\pt f=0$ is the Cauchy-Riemann Equation.

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    1. mathjax已经失效!

      1. 有时可能是mathjax服务器的问题. 现在已经恢复正常了.

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