Two fold coverings

  • Thread starter wofsy
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  • #26
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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).
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 -
 
  • #27
Hurkyl
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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....
 
  • #28
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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.
 
  • #29
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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
 
  • #30
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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.
 
  • #31
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.
 
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  • #32
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Oh bah -- I confused myself by trying to picture things, didn't I? One only needs to specify the gluing locally, so the fact that the cut locally splits the manifold into two pieces is all you need. Alas, there's a few technical details I would to work out (e.g. that the cut really can be covered by open balls that are cut in half -- I'm worried about the possibility of some pathological behavior).

That's why I really liked the universal covering space approach -- I really want to take advantage of something whose details have already been worked out (and is in the domain of things I know a little about). But I suppose I ought to be able to understand this way too....
 
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  • #33
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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.


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.

This makes me think that the general situation is: start with a manifold with boundary and a fixed point free involution of the boundary. The quotient space by the involution contains a hypersurface that is covered twice by the boundary. the hypersurface's Poincare dual is the corresponding element of H^1(quotient;Z2).

It would be interesting to do some examples. If you know of any good ones I'd like to see them.

For starters I am wondering what you get from involutions of a torus as the boundary of a solid torus. Suppose I rotate in 1 direction by 180 degrees and reflect in the other. The seam manifold will be a Klein bottle. What if I just rotate 180 degrees in one direction?

What if I take a solid Klein bottle and take the involution that rotates 180 degrees along its fiber direction?
 
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