Construct 2-Fold Cover of M Using H^1(M;Z2)

[wofsy]

if H^1(M;Z2) has a non-zero element then one can use this to construct a 2 fold cover of M. How does this work?

 

[Hurkyl]

Try starting with M = circle, and see what you can manage.

(P.S. What choices of 'definition' of H¹(M; Z₂) do you have available?)

 

[wofsy]

Try starting with M = circle, and see what you can manage.

(P.S. What choices of 'definition' of H¹(M; Z₂) do you have available?)

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.

 

[Hurkyl]

So you meant H₁, not H¹?

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?

 

[wofsy]

So you meant H₁, not H¹?

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?

Well either - but I meant cohomology.

 

[Hurkyl]

Ah. So then, the elements of H¹(M, Z₂) are named by functions from curves on M to Z₂ that have certain properties.

(Ponders a bit to think of there's an easy way to directly turn such a function into a topological space, and then to wonder if I can make that into a cover)


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 Z₂ seems like an obvious choice) Do you have any mental pictures of covering spaces that work in that fashion?

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.

 

[wofsy]

If the fundamental group is abelian then a non-zero element of H^1(M:Z2) corresponds to a subgroup of index 2, the kernel of the homomorphism into Z2. The universal covering space mod this subgroup of index 2 gives you the 2 fold covering space.

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.

 

[Hurkyl]

Hah! I didn't even think of doing something like that -- I had a much more lowbrow construction in mind. I bet your approach is easier if you have the relevant facts fresh in your mind.


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

 

[Hurkyl]

Actually -- yes, I think that is the construction I had in mind. I was just trying to construct it in a much more complicated fashion.

Spoiler

You can use the cohomology element directly to define the equivalence relation on the universal cover, rather than trying to do it in a roundabout way by constructing subgroups of the fundamental group.

Spoiler

The cohomology element tels you when two paths from the same basepoint in the universal cover ought to lead to the same point in the double cover

 

[wofsy]

Actually -- yes, I think that is the construction I had in mind. I was just trying to construct it in a much more complicated fashion.

Spoiler

You can use the cohomology element directly to define the equivalence relation on the universal cover, rather than trying to do it in a roundabout way by constructing subgroups of the fundamental group.

Spoiler

The cohomology element tels you when two paths from the same basepoint in the universal cover ought to lead to the same point in the double cover

but I think your more complicated construction will be needed for the whole proof.

 

[Hurkyl]

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. Rather than trying to build my own, I can just start with that space!

 

[wofsy]

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.

Right. I think your point that a subgroup of index 2 must be normal may be the key.

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

 

[wofsy]

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. Rather than trying to build my own, I can just start with that space!

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

 

[OrderOfThings]

A Z_2-valued 1-form \Omega can be visualized as an n-1-dimensional submanifold N. The value of \Omega on a curve \gamma is given by

\Omega(\gamma) = 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 H^1(M,Z_2) 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.

 

[wofsy]

A Z_2-valued 1-form \Omega can be visualized as an n-1-dimensional submanifold N. The value of \Omega on a curve \gamma is given by

\Omega(\gamma) = 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 H^1(M,Z_2) 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.

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.

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?

 

[wofsy]

A Z_2-valued 1-form \Omega can be visualized as an n-1-dimensional submanifold N. The value of \Omega on a curve \gamma is given by

\Omega(\gamma) = 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 H^1(M,Z_2) 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.

How would you do this construction for a circle?

 

[OrderOfThings]

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?

 

[wofsy]

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?

can you elaborate this idea?

 

[OrderOfThings]

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

 

[zhentil]

There may be a characteristic classes approach to this. If I recall correctly, the presence of such an element guarantees the existence of a non-trivial line bundle over M. Since a non-trivial line bundle is necessarily non-orientable, you can take the oriented double cover and the zero section will be a double cover of M.

 

[wofsy]

There may be a characteristic classes approach to this. If I recall correctly, the presence of such an element guarantees the existence of a non-trivial line bundle over M. Since a non-trivial line bundle is necessarily non-orientable, you can take the oriented double cover and the zero section will be a double cover of M.

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

 

[wofsy]

Since we are speaking of bundles - here is a related problem. Suppose I have a circle bundle over some space. When is there a 2 fold covering by another circle bundle in which fibers are mapped to fibers?

 

[OrderOfThings]

That would be possible if (and only if?) the base space has a two fold covering.

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.

 

[wofsy]

That would be possible if (and only if?) the base space has a two fold covering.

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.

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

 

[zhentil]

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

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).

 

[wofsy]

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).

Yes - I thought you were trying to restate the gluing argument - my mistake - I also think that the bundle argument which is very neat does not have an independent proof - you need to derive it from the other arguments - if I am wrong about this I would love to see the direct bundle argument - it would be very cool -

 

[Hurkyl]

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

I was wondering that too. It's clear (?) that locally the cut has two different sides, and in special cases it's obvious this holds true globally -- but it's not immediately obvious to me how to prove it in general. I want to show that if I construct a path from one side to the other while remaining in neighborhoods of the boundary, then either I must pass through the cut or the cut actually has a boundary...

 

[Hurkyl]

For example, consider the projective plane. Any line is a cycle that is not a boundary -- however it only has one side, and so OrderOfThings's construction can't work.

So, it is not enough to simply have a hypersurface that is a cycle and not a boundary -- the fact this cycle has an associated element of H¹(M, Z₂) has to be used in an essential way.

I assume orientability is the key -- I think if the manifold and hypersurface are orientable, that immediately separates defines two distinct sides to the cut.

 

[wofsy]

For example, consider the projective plane. Any line is a cycle that is not a boundary -- however it only has one side, and so OrderOfThings's construction can't work.

So, it is not enough to simply have a hypersurface that is a cycle and not a boundary -- the fact this cycle has an associated element of H¹(M, Z₂) has to be used in an essential way.

I assume orientability is the key -- I think if the manifold and hypersurface are orientable, that immediately separates defines two distinct sides to the cut.

this seems right

 

[Hurkyl]

On the other hand... (silly me for not noticing this yesterday)

I think the fact the projective plane's cut has one side allows a slightly simpler gluing construction: just attach another copy on the other side.

My suspicion now is that in every example, the cut is a disjoint union of one-sided and two-sided cuts.