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Hurkyl

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(P.S. What choices of 'definition' of H

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For the circle that's easy but it's fundamental group is abelian. I'll take any definition of H^1(M;Z2) that you like. How about singular homology.

(P.S. What choices of 'definition' of H^{1}(M; Z_{2}) do you have available?)

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Hurkyl

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Sure it's easy to construct a 2-fold cover of a circle. But can you do so in a way that relates to the generator of the homology group?

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Well either - but I meant cohomology._{1}, not H^{1}?

Sure it's easy to construct a 2-fold cover of a circle. But can you do so in a way that relates to the generator of the homology group?

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Hurkyl

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(Ponders a bit to think of there's an easy way to

Anyways, what sort of mental picture do you have about covering spaces? For this problem, in order to construct something, we would probably want to name points in the covering space by points in the base space + some additional data. (An element of Z

If the circle is too trivial for ya, other familiar examples you might consider are Riemann surfaces -- e.g. the double-valued complex sqrt function. (Or maybe the complex log function, although that has infinite degree, rather than being a double cover)

I suppose, maybe a different approach is to start thinking about how to build things out of simplices.

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If the fundamental group is not abelian I am not sure. I guess what you are implying is that it is essentially the same argument. Let me think about this.

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Hurkyl

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P.S. If I recall correctly, in any group (even nonabelian ones), any subgroup of index 2 is normal.

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but I think your more complicated construction will be needed for the whole proof.isthe construction I had in mind. I was just trying to construct it in a much more complicated fashion.

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Hurkyl

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Right. I think your point that a subgroup of index 2 must be normal may be the key.Well, my complicated construction was trying to build the space from scratch. After choosing a basepoint, I expected that I should somehow be able to name points in the double cover by paths in the base space. I hadn't worked out exactly what data was needed to name points, exactly how to tell what data corresponded to the same point, and exactly how to specify the topology. But once you brought the universal cover into it, the way forward becomes clearer, because someone already worked the details out for us in that case.

You know that there is a subgroup of index 2 in the fundamental group abelianized. Its inverse image in the entire fundamental group is a subgroup of index 2.

By inverse image I mean in the exact sequence 1 -> Commutator subgroup of fundamental group - > fundamental group -> H1(M;Z) -> 0

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BTW I want to thank you for guiding this discussion. Much appreciated.

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[tex]\Omega(\gamma) =[/tex] number of crossings with N mod 2.

If the form is closed then the corresponding submanifold is a cycle. If the form is exact, the submanifold is a boundary.

Since [itex]H^1(M,Z_2)[/itex] is nontrivial, there is a 1-form which is closed but not exact and the corresponding submanifold is a cycle that does not bound. Now take two copies of M, cut them along such a nonbounding cycle and glue them together. The resulting manifold is a two fold cover.

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you appeal to a lot of machinery. How do you know that every element of H^1(M;Z2) determines a hypersurface? I guess you mean that the Poincare dual is a hypersurface.

[tex]\Omega(\gamma) =[/tex] number of crossings with N mod 2.

If the form is closed then the corresponding submanifold is a cycle. If the form is exact, the submanifold is a boundary.

Since [itex]H^1(M,Z_2)[/itex] is nontrivial, there is a 1-form which is closed but not exact and the corresponding submanifold is a cycle that does not bound. Now take two copies of M, cut them along such a nonbounding cycle and glue them together. The resulting manifold is a two fold cover.

I also don't see how the covering is necessarily two fold along the gluing boundary. Here it seems that it can be 1 fold. For instance, if I cut a Klein bottle along a Moebius band circle I get a manifold with connected boundary. What happens if you cut RP3 along an equatorial (quotient of an equatorial sphere in S3) projective plane. Isn't this also a manifold with connected boundary?

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How would you do this construction for a circle?

[tex]\Omega(\gamma) =[/tex] number of crossings with N mod 2.

If the form is closed then the corresponding submanifold is a cycle. If the form is exact, the submanifold is a boundary.

Since [itex]H^1(M,Z_2)[/itex] is nontrivial, there is a 1-form which is closed but not exact and the corresponding submanifold is a cycle that does not bound. Now take two copies of M, cut them along such a nonbounding cycle and glue them together. The resulting manifold is a two fold cover.

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Yes, it seems the gluing most be done cross-wise. Maybe it is possible to reason like this: The nonbounding cycle will also have nontrivial homotopy and will admit an antipodal fibration. This fibration gives a cross-wise gluing?

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can you elaborate this idea?Yes, it seems the gluing most be done cross-wise. Maybe it is possible to reason like this: The nonbounding cycle will also have nontrivial homotopy and will admit an antipodal fibration. This fibration gives a cross-wise gluing?

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The two manifolds should be glued together by a two fold covering of the nonbounding cycle.

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This argument seems right but I think you are just restating the problem in terms of bundles.

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But was the gluing in constructing the two fold cover clear enough? It should be done cross-wise: Cut the two manifolds along the nonbounding cycle. Label the two ends A1 and B1 on the first manifold and A2, B2 on the second. Then glue A1-B2 and B1-A2.

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This workd if the cut creates two boundaries. But what if it does not?

But was the gluing in constructing the two fold cover clear enough? It should be done cross-wise: Cut the two manifolds along the nonbounding cycle. Label the two ends A1 and B1 on the first manifold and A2, B2 on the second. Then glue A1-B2 and B1-A2.

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I thought you had already figured it out - I was trying to rephrase it terms you might be more familiar with. The natural projection from the fundamental group gives you an index 2 subgroup in the fundamental group, which a fortiori is normal. In this case, modding out the universal cover by this subgroup gives you a two-sheeted covering space of M. (cf. Hatcher's algebraic topology book, pp. 70-80).This argument seems right but I think you are just restating the problem in terms of bundles.

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