# I Contour integration over Riemann surface

1. Jul 23, 2016

Cauchy integral theorem states that the contour integration of a complex harmonic function along a closed simply connected path=0. What if this simply connected path is drawn over a Riemann surface of function like $f(z)=\sqrt z$. Will that be possible in the first place? and will the integration over that path yield zero?

2. Jul 27, 2016

### FactChecker

As long as the winding number of the path around the point z=0 is 0, the integral will be 0.

3. Jul 27, 2016

Is the winding number really 0? If we move counterclockwise over Riemann surface representing the real part of $f(z)=\sqrt z$ we will end on an infinite looping around $z=0$ switching from the positive to the negative roots.

4. Jul 27, 2016

### FactChecker

It is not guaranteed that the winding number is 0. That is essential. For closed paths that do not wind around zero, the integral is 0.

5. Jul 27, 2016

I don`t get it. If a closed path is taken around $z=0$ in the complex flat plane, the integral of $f(z)=\sqrt z$ vanishes. But in my case, what is the value of the integration with a closed path winding around $z=0$ in Riemann surface that describe the same function?

6. Jul 27, 2016

### andrewkirk

I think the reference to a 'closed path' is to a closed path in the Riemann surface, not in the domain of the 'function' (by the way $z\mapsto \sqrt z$ is not a function because it doesn't give unique results). I am not sure that there is any closed path on the Riemann surface in $\mathbb C^2$ generated by the relation $z_2{}^2=z_1$ (which is a more careful way of expressing the square root notion), whose $z_1$ coordinate winds around 0.

If that's correct then the integral of any sufficiently nice function along a closed path on that Riemann surface will be zero.

7. Jul 28, 2016

### Ssnow

Hi, I agree with the ideas of @andrewkirk because a closed path that contains the zero on the riemann surface $f=z^2$ seems not viable. I want to be precise, the $0$ must be in the interior of the support of the path and I don't think this is possible.
In the end the case where $0$ is not inside the path was explained before ...

8. Jul 28, 2016

### andrewkirk

It occurs to me that the meaning of the integral is not clear in the question. The Riemann surface $R$ is a one-dimensional complex manifold that is embedded in $\mathbb C^2$. We can define a function $f:R\to\mathbb C$ and we can define a simple, closed curve in $R$ as a function $\gamma:[0,1]\to R$ that is continuous when we interpret $R$ as a two-dimensional subset of 4D Euclidean space, such that $\gamma(0)=\gamma(1)$. But I cannot see any obvious interpretation of the notion of a 'line integral' of $f$ along $\gamma$. Note that $f$ takes two complex arguments, rather than one.

For a loop integral of $g:\mathbb C\to\mathbb C$ around a loop $\eta$ in the complex plane we have the simple formulation
$$\oint_\eta g(z)\,dz=\int_0^1 g(\eta(t))\eta'(t)\,dt$$

There is no immediately obvious (to me) equivalent of this when the function to be integrated has a domain in $\mathbb C^2$ rather than $\mathbb C$. Some form of complex Jacobean may be needed. But then I have never studied complex manifolds. Perhaps there is a standard definition of a line integral of a function on a complex manifold. Presumably it would use the metric on the manifold. What would the standard metric on $\mathbb C^2$ be? Would it just be
$$ds^2=dz_1{}^2+dz_2{}^2$$

9. Jul 29, 2016

### Ssnow

Sorry I correct myself $f=\sqrt{z}$

10. Jul 29, 2016

### vanhees71

That's tricky ;-). If you integrate over a closed path in the Riemann surface, it gives 0. To demonstrate that on your example take the usual Riemann surface for the square-root function by cutting along the negative real axis. Now a closed curve around 0 must run over the two sheets ("double cover" of the complex plane). Let's take a circle. The first sheet is starting from $\text{arg} z=-\pi$ and runs to $\text{arg} z=\pi$. On this sheet you have along the unit circle
$$\sqrt{z}=\sqrt{\exp(\mathrm{i} \varphi)}=\exp(\mathrm{i} \varphi/2), \quad \varphi = \text{arg} z \in (-\pi,\pi)$$
Then you enter the 2nd sheet of the Riemann surface, where you have
$$\sqrt{z}=-\sqrt{\exp(\mathrm{i} \varphi)}=-\exp(\mathrm{i} \varphi/2) \quad \varphi =\text{arg} z \in (-\pi,\pi)$$
Only then you close the circle and thus you get trivially 0.

Equivalently you can also integrate $\exp(\mathrm{i} \varphi/2)$ along $\varphi \in (-\pi,3 \pi)$, which covers also the two branches of the square-root function, and it gives of course also 0.

11. Jul 30, 2016

### aheight

My understanding is that Riemann conceived of his idea in order to devise a coordinate system in which multi-valued functions become single-valued so that the foundation of Complex Analysis of single-valued functions could be applied to multivalued functions. The normal Riemann surface for the square root is a Riemann sphere with a double covering of the complex plane mapped onto it conformally with every point on the sphere equal to $(z,\sqrt{z})$. And therefore for the function $f(z)=\sqrt{z}$ over this new coordinate system, we have the simple linear function $f(z,w)=w$ which is an entire function zero over any closed contour over its Riemann surface. Contrast this with $f(z)=\sqrt{z}+\frac{k}{z}$. What do you suppose the integral of this function over a closed contour (winding number of one) about the point $(0,\infty)$ over it's Riemann surface is? $4k\pi i$? What about $f(z)=\sqrt[n]{z}+\frac{k}{z}$? $2nk\pi i$? What might the Riemann surface of $\sqrt[n]{z}$ even look like? I imagine it like one of those cantelopes with deep furrows with each section between the furrows corresponding to one copy of the z-plane. Once this perspective is taken, we can then re-cast the Residue Theorem for multivalued functions over their normal Riemann surfaces, then Laurent's Theorem and then all the other theorems for single-valued functions.
For example, the Laurent series for $\sqrt{z}$ is $\sqrt{z}$, the series for $\sqrt{z}+\frac{1}{z}$ is $\frac{1}{\left(\sqrt{z}\right)^2}+\sqrt{z}$. That was easy. What would the Laurent series for $\frac{1}{\sqrt{z(z-1)(z-2)}}$ be about the point $(0,0)$ over its Riemann surface? But if you know that all the theorems of single-valued functions apply to multivalued functions over their Riemann surface, how might you calculate the coefficients for this series expansion? Use Laurent's Theorem.

Last edited: Jul 30, 2016