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

Here's the situation I'm in.

I have a map from the (closure of) Upper Half Plane ($\mathbb{U}$) into the punctured (closed) disk ($\mathbb{D}$) called $q$ that satisfies

$q(0) = 1$, and $q(z+1) = q(z)$.

EDIT: And $q$ is injective on the interval $(0,1)$ on the real line.

I also know that $q$ is a holomorphic covering map.

From this information, I can show that $q$ has to be $q(z) = exp(2\pi i z)$.The argument is as follows:

Let $p(z) = exp(2\pi i z)$. Let $h$ be the lift of $q$ through $p$, and I choose it so that $h(0) = 0$. By taking the lift of $p$ through $q$, one can see that $h \in PSL(2,\mathbb{R})$.

Because $h$ is a lift, it follows that $h(z+1) = h(z) + n$ for some integer $n$. This argument shows that $h(\infty) = \infty$. It follows that $h(z) = a*z$, and $a=1$.

I feel like I'm being very stupid here. Does the fact that $q(z) = exp(2\pi i z)$ follow from some more general principle?

For instance, if I have a (simply connected) space $M$, and two covering maps $p$ and $q$ into $S$, can one say anything about $p$ and $q$?

Or maybe what I'm missing is complex analysis. Is there some kind of a uniqueness statement about holomorphic maps that I could've used to conclude that $q$ has to be the exponential map?

share|cite|improve this question
You don't think $q(z)$ could be, say, $\exp(4\pi i z)$? – WimC Mar 24 '12 at 19:18
Oops. I should have said that $q$ is injective on $(0,1)$. – Braindead Mar 29 '12 at 21:27

Your Answer


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

Browse other questions tagged or ask your own question.