Divergence theorem in complex analysis

Question

I am revisiting my understanding of integration by parts in several complex variables, but I have run into an apparent contradiction. This shows my understanding is flawed, which is somewhat embarrassing, but I guess I must fess up and humbly ask for clarification here. I reproduce the problem specified to one complex variable to make it more generally accessible.

The differential $1$-form $\mathrm{d}z = \mathrm{d}x+i\mathrm{d}y$ on $\mathbb{C}$ eats a complexified tangent vector $a\frac{\partial}{\partial x}+b\frac{\partial}{\partial y}$, where $a,b \in \mathbb{C}$, and returns the complex number $a+ib$.

Let $r : \mathbb{C} \to \mathbb{R}$ be a continuously differentiable function. Assume that $\Omega = \{z \in \mathbb{C} : r(z)<0\}$ is smoothly bounded, and that $\nabla r = \frac{\partial r}{\partial x} \frac{\partial }{\partial x} + \frac{\partial r}{\partial y} \frac{\partial }{\partial y}$ has length $1$ on the boundary of $\Omega$. Note that here I am thinking of $\nabla r$ as a "real tangent vector", but these are canonically included in the complexified tangent bundle as well.

Since $\nabla r$ is perpendicular to the boundary of $\Omega$, then by geometry $J(\nabla r) = \frac{\partial r}{\partial x} \frac{\partial }{\partial y} - \frac{\partial r}{\partial y} \frac{\partial }{\partial x}$ must be tangent to the boundary, where $J$ is the complex structure tensor. This is also of length $1$, so any real tangent vector to the boundary can be written $v = \alpha J(\nabla r)$.

Noting that $\mathrm{d}S(v) = \alpha$, it seems that we have point wise equalities

$ \begin{align*} \mathrm{d}z (v) &= \mathrm{d}z (\alpha J(\nabla r)) \\ &=\alpha \mathrm{d}z \left(\frac{\partial r}{\partial x} \frac{\partial }{\partial y} - \frac{\partial r}{\partial y} \frac{\partial }{\partial x}\right)\\ &=\left( -\frac{\partial r}{\partial y}+i\frac{\partial r}{\partial x}\right) \mathrm{d}S(v)\\ &=2i\cdot \frac{1}{2} \left( \frac{\partial r}{\partial x}+i\frac{\partial r}{\partial y}\right)\mathrm{d}S(v)\\ &=2i\frac{\partial r}{\partial \overline{z}} \mathrm{d}S(v) \end{align*}$

So we can apparently conclude that $\mathrm{d}z = 2i\frac{\partial r}{\partial \overline{z}} \mathrm{d}S$ as $1$-forms on the boundary of $\Omega$ (at least for real tangent vectors which is all we ever plug into either form when integrating something).

Unfortunately, this does not pass the following sanity check. Let $r(z) = |z|^2-1 = z\bar{z}-1$, so that $\Omega$ is the unit disk. Then, attempting to use the above identity, we obtain

$ \begin{align*} \int_{b\Omega} \frac{1}{z} \mathrm{d}z &= \int_{b\Omega} 2i \frac{1}{z} \frac{\partial r}{\partial \overline{z}} \mathrm{d}S\\ &=2i \int_{b\Omega} \mathrm{d}S\\ &=4\pi i \end{align*} $

which is a factor of $2$ more than it should be.

I have scoured the above calculations for the $2$ at fault, but have not pinpointed it. It seems likely to stem from the $\frac{1}{2}$ in the definition of the Wirtinger derivative $\frac{\partial}{\partial \overline{z}}$, but that $\frac{1}{2}$ really does need to be there.

Can anyone help me out?

Answer 1

The solution is to set $r = \frac{1}{2}(z\bar{z}-1)$. This gives $\nabla r = x\frac{\partial}{\partial x} +y\frac{\partial}{\partial y}$ for which $|\nabla r|=1$ for $|z|^2=x^2+y^2=1$. Then the sanity check works out just fine since $\frac{\partial r}{\partial \bar{z}}=\frac{z}{2}$ and: $$ \begin{align*} \int_{b\Omega} \frac{1}{z} \mathrm{d}z &= \int_{b\Omega} 2i \frac{1}{z} \frac{\partial r}{\partial \overline{z}} \mathrm{d}S\\ &=i \int_{b\Omega} \mathrm{d}S\\ &=2\pi i. \end{align*} $$

Answer 2

This is not a very satisfying answer, but the defining function for the disk has not been normalized. In fact, its gradient has length two.

Also, I see that James Cook already answered this question.

Answer 3

I prefer to avoid evaluating on basis vectors if I can, so here we go: Your hypothesis that $|dr|=1$ tells us that $dr\wedge\star dr = 1 dx\wedge dy = \dfrac i2 dz\wedge d\bar z$. This in turn, by the way, tells us that $\left|\dfrac{\partial r}{\partial z}\right| = \dfrac12$ (since $r$ is real-valued). (Simply write out $dr = \dfrac{\partial r}{\partial z}dz + \dfrac{\partial r}{\partial\bar z}d\bar z$ and use $\dfrac{\partial r}{\partial\bar z} = \overline{\dfrac{\partial r}{\partial z}}$.)

Now, we've agreed that your desired $ds$ along the level curve $r=0$ should therefore be $\star dr$. So determining the restriction of a $1$-form $\omega$ to a level curve of $r$ is given by determining $\omega \mod{dr}$, i.e., $\omega\wedge dr$. So \begin{align*} dz\wedge dr &= dz\wedge \dfrac{\partial r}{\partial\bar z}d\bar z, \text{ whereas} \\ \star dr\wedge dr &= -\frac i2 dz \wedge d\bar z. \end{align*} So we conclude that $dz\equiv 2i\dfrac{\partial r}{\partial\bar z}{\star}dr, \mod{dr}$, as desired.