Simple connectivity and finite coverings

1. Jul 7, 2008

cduston

Hey everyone,
First of all this is my first post and it's in regards to something I am (supposed to be) learning for my research. The topic is Algebraic Topology, so this was the closest general topic I could find.

The question is in regards to the connection between covering spaces and the fundamental group. I had a conversation with my advisor about CP_2 (complex projective plane in 2 D), and she said "Since CP_2 is simply connected, there cannot be any actual finite covering because these would correspond to normal subgroups of the fundemental group". Then she goes on to talk about branched coverings. Now, I understand that simple connectivity implies the fundamental group is trivial, and that a covering space p:X'->X means p(pi(X'))->pi(X) must be injective, but I don't really understand her "finiteness" remark. Does anyone have any thoughts on this?

2. Jul 8, 2008

eastside00_99

This is something I don't quite understand how to prove, but apparently, there is a one to one correspondence between the MINIMUM number of "copies" of your space X in your covering space with the number of elements in pi(X). For example, RP_1 can be thought of by identifying antipodal points on the circle S^2. If we let U and V be open semi-circles on S^2 such that U and V cover S^2. Then the disjoint union of U and V will form a covering space of RP_1. Thus, pi(RP_1) has two elements. There is only one such group: a cyclic group of order 2.

So, in your case, if you did have a finite and nontrivial open covering of CP_2, then you should be able to form a covering space of CP_2. These would tell you that pi(CP_2) has more than one element which as you already know it cannot. So, CP_2 can't have a finite covering.

3. Jul 8, 2008

Doodle Bob

Actually, no simply-connected, Hausdorff, arc-connected, and locally arc-wise connected space has a nontrivial covering. For a proof, see Section 3 (Covering Spaces) of Chapter III (Fundamental Group) of Bredon's "Topology and Geometry." Actually, any standard algebraic topology text should prove this result.

Also, you can look up at Prop 1.32 on page p. 61 of this: http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf

Last edited by a moderator: Apr 23, 2017
4. Jul 8, 2008

matt grime

Here's what she shaid: finite covers correspond to normal subgroups, so if there are no normal subgroups, then there are no finite covers. This doesn't tell you anything about covers where the fibres (a fibre is a pre-image of a point under the covering map) are not sets of finite points.

For example take the sphere S^2, this is simply connected, so has no non-trivial finite-to-one covers. But there is the Hopf fibration which is a covering map

S^3 --> S^2

and the fibre above each point is S^1. This is non-trivial in the sense that S^3 is not S^2 x S^1.

Last edited: Jul 8, 2008
5. Jul 8, 2008

cduston

Great work everyone, I think I'm beginning to get this. I found the chapter in Hatcher on the issue but I'm glad you (Doodle Bob) confirmed what I thought was the relevant result. But I think I understand what's going on, thanks very much everyone!

(and for the record, you guys were more helpful then http://www.mathhelpforum.com/math-help/" )

Last edited by a moderator: Apr 23, 2017