# What's wrong with my "proof" that branch cuts are not arbitrary?

Recently, I've been thinking about contour integrals around branch cuts in the complex plane. Now clearly the choice of contour is arbitrary, so long as you don't deform past any poles or cuts, but I've been going back and forth on whether or not the choice of cut is arbitrary or not. I have a hand-wavy argument that it's actually not arbitrary, but I'm sure there's a hole somewhere.

Say we have two branch points, we connect them with a straight line, and integrate a multi-valued function around the cut. This function is such that it "jumps" as you cross the cut. Now, if the branch cut is straight, you can bring the branch points closer and closer together, and consequently shrink your contour as well. As the points near collision, the integral should be computed by some residue at the point, or so.

However, if we imagine choosing the branch cut such that the points are connected by some giant, arcing cut, with the contour tightly encircling it, we can then do the same thing and imagine the branch points getting closer and closer together. As they near collision, in this case, the contour is essentially an inner loop oriented in one direction and an outer loop oriented in the other. Now, the inner loop can just be shrunk to zero, and thus doesn't contribute. All we have left is the outer loop which cannot be shrunk.

But there's no way this second choice of branch will yield the same value for the integral as the first choice of branch is there? I mean, you're integrating over a vastly different region in the complex plane. It seems like the result of a contour integral absolutely will depend on your choice of a branch cut.

EDIT: I've included a picture below. Sorry, I should have done this originally! The branch points are bolden, the choice of branch cuts are in red, and some contour with orientation is shown as dotted line. I hope this makes it more clear what I meant by colliding branch points, especially in the second case. • can you make a drawing ? and your example is $f(z) = \log(z) +\log(1-z)$ ? I don't get it. Jul 11, 2016 at 1:36
• @user1952009 Thanks, I've added a picture in an edit. The function I'm interested in is one defining a Riemann surface. So if $y^{2} = f(x)$, with $f$ a polynomial, then I want to integrate $y(x) dx$ on the contours indicated. Sorry, I wasn't sure the exact function would be crucial here. Jul 11, 2016 at 2:28
• can you take an example and explain how the branch points get closer and closer ? (and poles are not branch points, please correct it in § 2) Jul 11, 2016 at 2:32
• @user1952009 Thanks, typo fixed. So for example, a hyperelliptic curve $y^{2} = f(x)$ is given by choosing $2n$ branch points on $\mathbb{P}^{1}$, where the degree of $f$ is $2n+1$. The so-called periods of this Riemann surface are given by contour integrals of $y(x)dx$ around branch cuts arising by connecting the $2n$ branch points pairwise. In other words, the "way" two branch point might approach each other is simply that you're looking at different hyperelliptic curves, which are becoming closer to being singular. Jul 11, 2016 at 3:05

• Thanks! Very helpful. Although I'm not sure I understand completely what you mean by choosing the branches of the function to be the same on the contour. Is this something that's always possible to achieve for something like a double-cover on $\mathbb{P}^{1}$? Jul 11, 2016 at 3:36
• @spietro : your second picture is not correct, there is always a piece of the contour between the two branch points. To me the idea is that at first you have a multi-valued function $f(z)$, you choose a closed contour $\gamma$ enclosing the two branch points, such that $f(z)$ is holomorphic outside $\gamma$ (that's what you have drawn). Then not matter of where the branch cut pass, what matters is that $f(z)$ is holomorphic outside, since clearly $\displaystyle\int_\gamma f(z)dz$ doesn't care of what is inside (the shape of the branch cut) Jul 11, 2016 at 4:34
• @user1952009 Right, I was stupidly thinking that as the branch points approached one another, the contour would asymptote to what I drew below. I see now that doesn't make sense. The statement (that choosing a contour $\gamma$ outside of which $f(x)$ is holomorphic doesn't care about the shape of the cut, only the location of the branchpoints) is what I was after. I was hoping I had not found a counterexample, which clearly I did not. Thanks for all your help! Makes sense now. Jul 11, 2016 at 5:13