Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Maybe it's an easy question: How can we find the universal covering of the connected sum of tori or projective planes? Is there a general method?

share|cite|improve this question
See this answer by Ryan Budney. – Juan S Aug 29 '12 at 8:46
For the surfaces of genus $g > 0$ the universal cover is the hyperbolic plane, this is a classical result. I know that this is not really an answer to your question so I stated it as a comment. – mland Aug 29 '12 at 10:00
Have you a reference? – Seirios Aug 29 '12 at 15:36
up vote 3 down vote accepted

Let $X_1$ (resp. $X_2$) be a sphere and $D_1$, $D_1'$ (resp. $D_2$, $D_2'$) two diametrically opposed disks on $X_1$ (resp. $X_2$). Then, set $\tilde{X}_1= X_1 \backslash (D_1 \cup D_1')$ and $\tilde{X}_2=X_2 \backslash (D_2 \cup D_2')$. Finally, define $\tilde{X}$ by $(\tilde{X}_1 \coprod \tilde{X}_2 )/ \sim$, where $\sim$ identifies $D_1$ with $D_2$, and $D_1'$ with $D_2'$.

Then, $X_i$ is the universal covering of a projective plane (and the projection is given by identifying two diametrically opposed points on $X_i$) so $\tilde{X}$ is a $2$-sheeted covering space of the connected sum of two projective planes. But it is easy to see that $\tilde{X}$ is homeomorphic to $\mathbb{T}$, so the universal covering of the sum of two projective planes is homeomorphic to $\mathbb{R}^2$. (In fact, the result doesn't depend on the number of projective planes we add.)

However, the projection from $\mathbb{R}^2$ onto the connected sum is difficult to explicit because the homeomorphism between $\tilde{X}$ and the torus is itself difficult to explicit. But this method shows that the fundamental group of the connected sum of two projective plans contains a subgroup of index $2$ isomorphic to $\mathbb{Z}^2$.

It seems to be more difficult for a connected sum of tori, because the universal covering has infinitely many sheets.

Added: From John Lee's beautiful book Geometry of surfaces, we are able to prove

Theorem: Let $S \neq \mathbb{S}^2, \ \mathbb{R}^2P, \ \mathbb{T}^2$ be a connected compact surface and let $\Pi \subset \mathbb{H}^2$ denote a regular $4n$-gon such that $S$ is homeomorphic to some identification space $S_{\Pi}$. Then there exists a regular tesselation of $\mathbb{H}^2$ by $\Pi$ inducing a covering $\mathbb{H}^2 \to S$.

A proof follows from the theorems at pages 37, 123 and 129.

share|cite|improve this answer

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.