# Universal covering of a connected sum

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?

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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

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.

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