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I was wondering what the fundamental group of an infinitely sheeted covering space of say some surface might be?

I'm thinking it should be an infinite cyclic group, but this is more intuitively, i cannot seem to construct an argument for this. Thanks!

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Do you have anything more specific in mind? The fundamental group of a covering space of $X$ is a subgroup of $\pi_1(X)$, whose index is the number of sheets, so if I take your question literally, the groups you'll get are the infinite-index subgroups of surface groups. –  arkeet Dec 11 '12 at 8:12
    
Say of a connected sum of two tori? –  seria Dec 11 '12 at 22:29
    
Please add that information to the question itself, seria. It is certainly not true that every infinitely sheeted covering of a torus has cyclic group of covering transformations... –  Mariano Suárez-Alvarez Dec 12 '12 at 17:27
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I'm confused because there is more than one infinitely sheeted covering space. Different coverings have different fundamental groups, so do you mean a particular covering space? –  Jason DeVito Dec 12 '12 at 17:30
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3 Answers

Just for fun, here is an example of an infinite sheeted cover of a genus two surface. The covering map maps each copy of the cylinder with a handle to the rolled-up cylinder where the two ends are identified. You could further unwrap it to a plane with handles attached in a grid pattern to get a covering space with deck transformation group $\mathbb Z\times\mathbb Z$.

enter image description here

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For the case of a connected sum of two tori, I believe this is equivalent to showing that every non trivial subgroup of the free product $\mathbb{Z}^2\star\mathbb{Z}^2$ is infinite cyclic. Here I am using the fact that the fundamental group of a connected sum of two tori( manifolds) is the free product of their corresponding fundamental groups.

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The fundamental group of the connected sum of two manifolds is not the free product, however the wedge sum does give the free product. –  Brandon Carter Dec 12 '12 at 4:56
    
(What Brandon says is correct, but it may be worth pointing out the for $n$-manifolds with $n > 2$, it is the free product) –  Jason DeVito Dec 12 '12 at 5:53
    
Yes you are correct. My topology seems to be rusty...in this case wouldn't the proper subgroup of the fundamental group of the connected sum of tori be infinite cyclic then? Since this corresponds to some cover of the connected sum of tori then it would answer the op's question I think. –  Haruki Dec 12 '12 at 6:06
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It is quite clear that not every subgroup of $\mathbb Z^2\star\mathbb Z^2$ is infinite cylic, by the way: that group contains a copy of $\mathbb Z^2$! –  Mariano Suárez-Alvarez Dec 12 '12 at 17:28
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@Haruki: An infinite sheeted covering will not have a compact total space. –  Alexander Thumm Mar 25 '13 at 10:05
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An infinitely-sheeted (connected) cover of a surface is necessarily a noncompact surface. Every noncompact surface is homotopy-equivalent to a graph and, hence, has countable free fundamental group. Conversely, every countable free group can be realized as the fundamental group of an infinitely sheeted cover of some compact surface.

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