# Two fold coverings

1. Oct 6, 2009

### 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?

2. Oct 6, 2009

### Hurkyl

Staff Emeritus
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?)

3. Oct 6, 2009

### wofsy

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.

4. Oct 6, 2009

### Hurkyl

Staff Emeritus
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?

5. Oct 6, 2009

### wofsy

Well either - but I meant cohomology.

6. Oct 6, 2009

### Hurkyl

Staff Emeritus
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.

7. Oct 6, 2009

### 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.

8. Oct 6, 2009

### Hurkyl

Staff Emeritus
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.

9. Oct 6, 2009

### Hurkyl

Staff Emeritus
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.

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.

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

10. Oct 6, 2009

### wofsy

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

11. Oct 6, 2009

### Hurkyl

Staff Emeritus
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!

12. Oct 6, 2009

### wofsy

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

13. Oct 7, 2009

### wofsy

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

14. Oct 8, 2009

### 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.

15. Oct 8, 2009

### wofsy

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?

16. Oct 8, 2009

### wofsy

How would you do this construction for a circle?

17. Oct 8, 2009

### 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?

Last edited: Oct 8, 2009
18. Oct 8, 2009

### wofsy

can you elaborate this idea?

19. Oct 8, 2009

### OrderOfThings

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

20. Oct 8, 2009

### 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.