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Topic: Top Homology of Manifold with Boundary is Zero. True? Why?
Replies: 2   Last Post: May 28, 2013 11:54 PM

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Bacle H

Posts: 283
Registered: 4/8/12
Re: Top Homology of Manifold with Boundary is Zero. True? Why?
Posted: May 28, 2013 11:54 PM
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On Tuesday, May 14, 2013 7:41:58 PM UTC-7, W. Dale Hall wrote:
> bacmail.com wrote:
>

> >
>
> > Hi, All:
>
> >
>
> > I'm trying to see if it is true that the top homology of
>
> >
>
> > an n-manifold with boundary is zero. I have tried some examples;
>
> >
>
> > a closed annulus ( homotopic to S^1 , so H^2(S^1)=0 , R^n with
>
> >
>
> > a boundary copy of some R^m m<n retracts to R^m , etc.)
>
> >
>
> > It seems strange that a manifold would "lose its orientability"
>
> >
>
> > if we capped-in a boundary. It seems like this boundary would bound
>
> >
>
> > all the n-cycles that had no boundary before the boundary was capped-in;
>
> >
>
> > otherwise, how is the quotient Cycles/Boundaries suddenly zero?
>
> >
>
> > I tried some MAyer-Vietoris, but got nowhere. Any Ideas?
>
> >
>
> > Thanks.
>
> >
>
>
>
> The n-dimensional homology of an orientable, *connected* n-dimensional
>
> manifold with boundary vanishes.
>
>
>
> Note that the n-dimensional homology of a connected, closed,
>
> orientable n-manifold is Z. If you remove a small disc to yield the
>
> manifold-with-boundary Mo, then the Mayer-Vietoris sequence for
>
> M = Mo U D^n looks like this in dimensions n & n-1:
>
>
>
> (this needs to be viewed in a fixed-spacing font to make sense)
>
>
>
> H_n(Mo)
>
> H_n(S^(n-1)) --> (+) ---> H_n(M) --,
>
> H_n(D^n) /
>
> /
>
> /
>
> __________________________________/
>
> \ (boundary)
>
> \ H_(n-1)(Mo)
>
> `--> H_(n-1)(S^(n-1)) --> (+) --> H_(n-1)(M) -->...
>
> H_(n-1)(D^n)
>
>
>
> Plugging in what values we know, we get this:
>
>
>
>
>
> H_n(Mo)
>
> 0 --> (+) ---> Z --------,
>
> 0 /
>
> /
>
> /
>
> __________________________________/
>
> \ (boundary)
>
> \ H_(n-1)(Mo)
>
> `----------> Z ----------> (+) --> H_(n-1)(M) -->...
>
> 0
>
>
>
> or
>
> (bdy)
>
> 0 --> H_n(Mo) --> Z --> Z --> H_(n-1)(Mo) --> H_(n-1)(M) --> ..
>
>
>
> Next, note that H_n(Mo) can't map into a proper subgroup of H_n(M),
>
> since the next homomorphism is into Z (the quotient group would have
>
> nonzero torsion, which doesn't exist in Z). That is, the only options
>
> are H_n(Mo) = 0, or H_n(Mo) maps *onto* H_n(M).
>
>
>
> Further, the homomorphism
>
>
>
> H_(n-1)(S^(n-1)) ---> H_(n-1)(Mo)
>
>
>
> must be zero, since S^(n-1) bounds in Mo (after all, it's the boundary
>
> of Mo). What we find, therefore, is that H_n(M) maps isomorphically
>
> onto H_(n-1)(S^(n-1)).
>
>
>
> Your query
>
>
>
> It seems strange that a manifold would "lose its
>
> orientability" if we capped-in a boundary. It seems
>
> like this boundary would bound all the n-cycles that
>
> had no boundary before the boundary was capped-in;
>
> otherwise, how is the quotient Cycles/Boundaries
>
> suddenly zero?
>
>
>
> seems a bit confused.
>
>
>
> In the first place, orientability isn't the issue. The issue is
>
> in finding n-dimensional cycles (i.e., chains with 0 boundary). For a
>
> connected orientable manifold, the fundamental class (that is,
>
> the chain that carries n-dimensional homology) *is*, in some sense,
>
> the full manifold. If it's simplicial homology, and your manifold
>
> is representable as a simplicial complex, then the class [M] can
>
> be represented as the sum of all top-dimensional simplices (with
>
> appropriate orientations). No proper subset of simplices will add up
>
> to a chain whose boundary vanishes.
>
>
>
> Further, I don't understand the phrase "if we capped-in a boundary".
>
> A manifold with boundary (for lack of better phraseology) actually
>
> *has a boundary*. Any sum of n-simplices will have nonzero boundary,
>
> so won't be cycles at all. The boundary operator from C_n(Mo) to
>
> C_(n-1)(Mo) has zero kernel. There are *no* n-dimensional cycles
>
> to begin with. Nothing suddenly becomes zero, since there was never
>
> anything there to begin with.
>
>
>
> For instance, take one of your examples, say, the annulus. Have a
>
> look at what the 2-chains represent geometrically. Note that you
>
> simply can't arrange 2-dimensional simplices in such a way as to
>
> cancel out the boundary. You can force the boundary to reside in
>
> the bounding circles of the annulus, but that's the best you can do.
>
>
>
> You might be heartened to note that the *relative* homology,
>
>
>
> H_n(M,dM) = Z
>
>
>
> for a connected orientable n-manifold M with boundary dM. In this
>
> case, all those n-chains you pushed to force their boundaries into
>
> dM now (in the relative case) become cycles, and you get n-cycles.
>
> Once you have cycles, you can see homology in dimension n.
>
>
>
> I hope I've cleared some confusion & not strewn any more of the same.
>
>
>
> Dale


Thanks, W.Dale; I will try to think things through before asking; thanks for your patience.



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