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wofsy
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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 said: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?)
Hurkyl said: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?
Hurkyl said: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
Hurkyl said: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.
Hurkyl said: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!
OrderOfThings said:A [itex]Z_2[/itex]-valued 1-form [itex]\Omega[/itex] can be visualized as an n-1-dimensional submanifold N. The value of [itex]\Omega[/itex] on a curve [itex]\gamma[/itex] is given by
[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.
OrderOfThings said:A [itex]Z_2[/itex]-valued 1-form [itex]\Omega[/itex] can be visualized as an n-1-dimensional submanifold N. The value of [itex]\Omega[/itex] on a curve [itex]\gamma[/itex] is given by
[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.
OrderOfThings said: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?
zhentil said: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.
OrderOfThings said: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.
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 said:This argument seems right but I think you are just restating the problem in terms of bundles.
zhentil said: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).
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...wofsy said:This workd if the cut creates two boundaries. But what if it does not?
Hurkyl said: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.
Now I see what your are saying. So it is like the classical thing of cutting a Moebius band down the middle and getting a twisted cylinder of twice the length.OrderOfThings said:That's right. Just glue the two manifolds together. (But to be able to visualize this you might have turn one of the manifolds inside out.) When you cut along the nonbounding cycle, the resulting boundary is a two fold cover of the cycle. If the cycle is orientable then the cover is disconnected and if it is not orientable the cover will be connected.
Example. Start with P3 and cut along a P2-hypersurface. This will result in a ball with S2-boundary, say [itex]\{|(x,y,z)|\leq 1\}[/itex]. Take the other manifold to be the ball [itex]\{|(x,y,z)|\geq 1\}\cup \{\infty\}[/itex]. Gluing them together results in S3. (See http://sketchesoftopology.wordpress.com/2009/07/25/two-balls/" [Broken] for a nice animation.)
Yet another way to think of the gluing, is to again start with two copies of the manifold and consider the nonbounding cycle to be a "magic membrane". If you pass through it you do not arrive on the other side, but instead the other side of the membrane in the other manifold.
A 2-fold cover is a mathematical concept that refers to a covering space of a topological space, where each point in the original space is covered by two points in the covering space. In other words, it is a map that takes each point in a space and "covers" it with two points in a different space.
Constructing a 2-fold cover of a space can provide insight into the topology and geometry of the original space. It can also be used to study the fundamental group of the space, which is a fundamental concept in algebraic topology.
H^1(M;Z2) refers to the first cohomology group of the space M with coefficients in the group Z2. In simpler terms, it is a mathematical tool used to study the topology of M by assigning algebraic structures to it.
Constructing a 2-fold cover of a space using H^1(M;Z2) is a technique commonly used in algebraic topology to study the fundamental group and other algebraic structures of a space. It allows for a deeper understanding of the topology and geometry of a space through algebraic methods.
Yes, the concept of constructing a 2-fold cover using H^1(M;Z2) can be applied to other spaces as well. In fact, it is a common technique used in algebraic topology to study various spaces and their fundamental groups.