Top Homology of Connected , Orientable Manifolds with Boundary

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WWGD
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Hi, I'm trying to show that if ## M^n ## is orientable and connected, with boundary (say with just one boundary component), then its top homology is zero. Sorry, I have not done much differential topology/geometry in a while.

I'm trying to avoid using Mayer-Vietoris, by using this argument; but I don't think this is too rigorous; please critique: let B be the boundary. Now, B is an (n-1)-submanifold of M, and so it must bound some n-dimensional manifold. Since M is orientable and n-dimensional, then B must bound M itself. But then M, as the top cycle is a cycle that bounds, so the top cycle is a trivial cycle, and so the homology is zero.
 

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So are you assuming the top homology (i.e. [itex]H_{n-1}[/itex]) of the boundary is non-zero?
 
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jgens
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Hi, I'm trying to show that if ## M^n ## is orientable and connected, with boundary (say with just one boundary component), then its top homology is zero.

There is an easy way to see this that does not require orientability. Note that attaching a collar to M does not affect its homotopy type. Since M+Collar is certainly noncompact it follows that the top homology group is zero.

I'm trying to avoid using Mayer-Vietoris, by using this argument; but I don't think this is too rigorous; please critique: let B be the boundary. Now, B is an (n-1)-submanifold of M, and so it must bound some n-dimensional manifold. Since M is orientable and n-dimensional, then B must bound M itself. But then M, as the top cycle is a cycle that bounds, so the top cycle is a trivial cycle, and so the homology is zero.

I am not sure what kind of cycles you mean here. Which homology theory are you using?
 
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WWGD
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So are you assuming the top homology (i.e. [itex]H_{n-1}[/itex]) of the boundary is non-zero?

No; not for now, I may if I think it is necessary, thanks.

jgens: I'm using Singular homology, to apply Mayer-Vietoris.
 
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jgens
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No; not for now, I may if I think it is necessary, thanks.

It's not. The argument I provided in my past post shows this.

jgens: I'm using Singular homology, to apply Mayer-Vietoris.

My opinion is that one should try to avoid arguments at the chain level. If you insist on doing so, then that is no problem, but I cannot make sense of your "proof" as currently written. I can offer the following simplification: Since B is the boundary of M, it trivially follows that M bounds B, so there is no need for orientability in that piece of the argument.
 
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WWGD
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It's not. The argument I provided in my past post shows this.



My opinion is that one should try to avoid arguments at the chain level. If you insist on doing so, then that is no problem, but I cannot make sense of your "proof" as currently written. I can offer the following simplification: Since B is the boundary of M, it trivially follows that M bounds B, so there is no need for orientability in that piece of the argument.

Someone I thought knew more than me insisted the manifold must be orientable. But, yes, my argument was basically that if B is a boundary, then, by basic dimension reasons, it must be the boundary of M; then M seen as a top cycle bounds, so it is , by definition, a trivial cycle, so it must be the zero cycle.
 
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jgens
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Someone I thought knew more than me insisted the manifold must be orientable.

Everyone makes mistakes. Read my previous argument through and see if you can find a hidden use of orientability. But everything there should check out.

But, yes, my argument was basically that if B is a boundary, then it must be the boundary of M; then M seen as a top cycle bounds, so it is , by definition, a trivial cycle, so it must be the zero cycle.

Well you defined B to be the boundary of M so that M bounds B is a triviality. I am not used to computing with singular chains, so there might be some obvious fixes here, but here are my hangups about the argument:
  1. You need to clarify what M is as a cycle and justify that M can be written this way.
  2. The whole argument is showing that a specific cycle bounds another. This is usually not enough to guarantee the nth homology group is zero. You need to show that all possible cycles are boundaries.
  3. If you want to show that the top homology of M is zero, then you need to establish that every singular cycle in M is a boundary. The argument you have currently looks like it is trying to show that B is a singular boundary. So this point needs clarification.
Anyway as mentioned before you can avoid descending to the chain level if you want. Something to really think about there.
 
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WWGD
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Thanks, jgens, you have been very helpful and patient, I appreciate it.
 
  • #9
lavinia
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This is not a proof but can be made into one with effort.

The idea for a triangulable manifold without boundary is that the top dimensional n-simplices share each of their n-1 faces with exactly one other n simplex and if the manifold is orientable the simplices can be oriented so that they cancel in pairs under the boundary operator. The Z homology class of this cycle can be shown to be unique up to sign, meaning that any attempt to create an n cycle from oriented n- simplices always leads to the same chain or its negative. It the manifold is not orientable then some of the n-1 faces get counted twice so there is no cycle. I think there is a proof in Spanier's Algebraic Topology.

For a manifold with boundary the n-1 simplices at the boundary are not paired off and so will remain uncancelled under the boundary operator. One can see this better by imagining gluing the manifold to itself along matching n-1 simplices along the boundary.

Also, and may this what what you were thinking about, if the manifold is orientable then the boundary of a properly oriented sum of the n simplices will be the fundamental cycle of the boundary manifold. So the fundamental cycle of the boundary manifold is homologous to zero in the larger manifold. If the manifold is not orientable then the boundary may not be homologous to zero as in the Mobius band.
 
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