Simple connectivity and finite coveringsby cduston Tags: algebraic topology, covering spaces, projective plane 

#1
Jul708, 10:06 AM

P: 9

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
Jul808, 04:54 AM

P: 234

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
Jul808, 06:11 AM

P: 255

Actually, no simplyconnected, Hausdorff, arcconnected, and locally arcwise 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 



#4
Jul808, 06:20 AM

Sci Advisor
HW Helper
P: 9,398

Simple connectivity and finite coveringsHere'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 preimage 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 nontrivial finitetoone 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 nontrivial in the sense that S^3 is not S^2 x S^1. 



#5
Jul808, 09:14 AM

P: 9

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/mathhelp/ ) 


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