# Topology of Branch Cuts and Elliptic Integrals

In reading these notes (elliptic curves starting from elliptic integrals) I came across a couple claims about the topology of some complex surfaces.

On page 4, they discuss the integral $$\phi(x) = \int_0 ^x \frac{dt}{\sqrt{1 - t^2}}$$ In order to define it on all of $\mathbb C$, you have to use a branch cut; they glue two copies of $\mathbb C$ together along $[-1,1]$, in the same crossing-over manner as you do when dealing with $\sqrt z$ (at least, I think). They then claim that this surface $C$ is homeomorphic to a cylinder. However, I'm having trouble seeing a way to explicitly bend $C$ into a cylinder. I think I might be missing some intuition on what a complex cylinder looks like.

I think I understand why $C$ would be homotopically equivalent (not sure if that's the best term) to a cylinder, because there's one set of loops from going around the branch cut, and if you avoid integrating across the branch cut the other ones are all null-homotopic. But why glue two copies of $\mathbb C$ together at all if you're going to avoid integration across the branch cut?

I also don't quite get they say that $C$ can also be defined as $\{ (x,y) \in \mathbb C ^2 : x^2 + y^2 = 1 \}$; is it just that we can integrate $dx/y$ on $C$, because that differential on $C$ looks like the differential in $\phi$?

They make similar claims a page later about $$\psi (x) = \int_0^x \frac{dt}{\sqrt{t(t-1)(t-\lambda)}}$$ but I think my issues are basically the same.

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Here is an explanation of what is going on; I haven't comparted it directly with the notes that you link to, but they will surely say something essentially equivalent:

The differential $dt/\sqrt{1-t^2}$ is not unambiguously defined on the $t$-plane, because the denominator $\sqrt{1-t^2}$ is not. If we make a cut along $[-1,1]$ then we obtain a region homeomorphic to a cylinder (as you can see by looking at it --- but if this is your point of confusion, feel free to say so in the comments), on which we can define two branches of $dt/\sqrt{1-t^2}$.

If we want to "glue" these two branches together into a single differential on a single Riemann surface, then we have to consider the Riemann surface $C:= \{(x,y) \, | \, y^2 = 1 - x^2\}$; on this surface, we identify $x$ with $t$, so that $y$ is then one the of the two choices of $\sqrt{1-t^2}$. The differential can then be written as $dx/y$.

If we define the map $\pi: C \to \mathbb C$ via $(x,y) \mapsto x$ then the restriction of $\pi$ to the preimage of $\mathbb C \setminus [-1,1]$ is $2$-to-$1$, and indeed this preimage is the disjoint union of two copies of $\mathbb C\setminus [-1,1]$; this is just a consequence of the fact that over the cut plane, we can choose $\sqrt{1-t^2}$ in two distinct well-defined ways.

The entire curve $C$ is however connected: it consists of the two copies of $\mathbb C \setminus [-1,1]$ that lie over $\mathbb C\setminus [-1,1]$, glued together along a circle (the preimage under $\pi$ of $[-1,1]$). Gluing two cylinders along a common boundary circle just gives another cylinder, and so $C$ is indeed homeomorphic to a cylinder.

Another way to think of it is as being homeomorphic to a sphere with two points removed; this is topologically the same as a cylinder.

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I think I was envisioning a more complex gluing, but if you look at the simple way to do it (and notice that the gap in $\mathbb C / [-1,1]$ looks like a circle) then it's obvious. Thanks. –  Calvin McPhail-Snyder Aug 16 '12 at 23:01
@CalvinMcPhail-Snyder: Dear Calvin, Exactly! Regards, –  Matt E Aug 17 '12 at 2:33
@CalvinMcPhail-Snyder: P.S. To see a more complex gluing, look at the elliptic integral example, which ultimately yields a torus (i.e. an elliptic curve). –  Matt E Aug 17 '12 at 2:34