Cauchy's integral theorem says that the integral of any holomorphic function over a closed path in a simply connected domain is always $0$.
Is there a version of Cauch's integral theorem for several complex variables?
I mean, if $f:\Omega\subset\mathbb{C}^n\rightarrow\mathbb{C}$ is holomorphic and $\Omega$ simply connected, then for any closed path $\gamma\subset\Omega$, it holds
$$\int_\gamma f=0.$$
Maybe it's a trivial question, but I've never found it anywhere, so I'm starting to think it's wrong.
Best Answer
In one variable, you can define the integral $\int_\gamma f(z)\,dz$ by $$\int_a^bf(\gamma(t))\cdot \gamma'(t)\,dt$$ You can approximate it by a Riemann sum $$\sum_{k=0}^{n-1} f(z_k)(z_{k+1}-z_k)$$ If you have a primitive function $F(z)$, you can compute the integral as $F(b)-F(a)$, where $a$ and $b$ are the end points of the curve $\gamma$.
In several variables, all this fail because of dimensional considerations. Then $\gamma'(t)$ and $z_{k+1}-z_k$ are vectors instead of complex numbers, and there is no notion of a primitive function without specifying which variable you are differentiating with respect to. Hence the expression $\int_\gamma f$, integrating a function along a curve, is not even well defined. This is why we integrate 1-forms instead of functions. In contrast to the situation in one variable, there is no universal 1-form to use here.
Example
In ${\bf C}^2$, use the coordinates $(z,w)$ where $z=x+iy$ and $w=u+iv$. Consider the closed curve $\gamma$ given by $x^2+u^2=1$. Note that $\gamma$ is contained in ${\bf R}\times {\bf R}\subset {\bf C}\times {\bf C}$, and that we have no complex structure on ${\bf R}\times {\bf R}$. We can compute $\int_\gamma f(z,w)\,dz$ and $\int_\gamma f(z,w)\,dw$, but they reduce to ordinary real line integrals, and there is no reason why Cauchy's integral theorem should hold here. For instance, if $f(z,w)=w$ and we use the parametrization $z=\cos(t)$, $w=\sin(t)$, we get $$\int_\gamma f(z,w)\,dz=\int_\gamma u\,dx = -\int_0^{2\pi}\sin^2(t)\,dt=-\pi$$