Simple connectivity and finite coverings

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In summary, my advisor said that simply connected spaces don't have any finite covers because if there were, then the fibres connecting the points would not be sets of finite points.
  • #1
cduston
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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 fundamental 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?
 
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  • #2
cduston said:
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 fundamental 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?

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
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
 
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  • #4
cduston said:
but I don't really understand her "finiteness" remark.
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.
 
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  • #5
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/" :wink:)
 
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What is simple connectivity?

Simple connectivity refers to the property of a topological space where any loop in the space can be continuously shrunk to a single point without leaving the space. In other words, there are no holes or gaps in the space that cannot be filled by a continuous path.

What are finite coverings?

Finite coverings are collections of sets that cover a given space, where each set in the collection is a subset of the space. These coverings are finite, meaning there are a limited number of sets in the collection, and they are used to study the topology of a space.

Why is simple connectivity important in topology?

Simple connectivity is important in topology because it allows for the classification of spaces into different categories based on their properties. It also helps to determine the relationship between different spaces and how they can be transformed into one another.

What are some examples of simple connectivity?

A simple example of simple connectivity is a circle, where any loop on the circle can be continuously shrunk to a single point. Other examples include spheres, tori, and any convex shape without holes or gaps.

How is simple connectivity related to finite coverings?

Simple connectivity is related to finite coverings because it can be used to prove that a space is simply connected by constructing a finite covering with certain properties. This allows for a more systematic approach to studying the topology of a space.

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