Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.