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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


On Tuesday, May 14, 2013 7:41:58 PM UTC7, W. Dale Hall wrote: > bacmail.com wrote: > > > > > > Hi, All: > > > > > > I'm trying to see if it is true that the top homology of > > > > > > an nmanifold 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 cappedin a boundary. It seems like this boundary would bound > > > > > > all the ncycles that had no boundary before the boundary was cappedin; > > > > > > otherwise, how is the quotient Cycles/Boundaries suddenly zero? > > > > > > I tried some MAyerVietoris, but got nowhere. Any Ideas? > > > > > > Thanks. > > > > > > > The ndimensional homology of an orientable, *connected* ndimensional > > manifold with boundary vanishes. > > > > Note that the ndimensional homology of a connected, closed, > > orientable nmanifold is Z. If you remove a small disc to yield the > > manifoldwithboundary Mo, then the MayerVietoris sequence for > > M = Mo U D^n looks like this in dimensions n & n1: > > > > (this needs to be viewed in a fixedspacing font to make sense) > > > > H_n(Mo) > > H_n(S^(n1)) > (+) > H_n(M) , > > H_n(D^n) / > > / > > / > > __________________________________/ > > \ (boundary) > > \ H_(n1)(Mo) > > `> H_(n1)(S^(n1)) > (+) > H_(n1)(M) >... > > H_(n1)(D^n) > > > > Plugging in what values we know, we get this: > > > > > > H_n(Mo) > > 0 > (+) > Z , > > 0 / > > / > > / > > __________________________________/ > > \ (boundary) > > \ H_(n1)(Mo) > > `> Z > (+) > H_(n1)(M) >... > > 0 > > > > or > > (bdy) > > 0 > H_n(Mo) > Z > Z > H_(n1)(Mo) > H_(n1)(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_(n1)(S^(n1)) > H_(n1)(Mo) > > > > must be zero, since S^(n1) bounds in Mo (after all, it's the boundary > > of Mo). What we find, therefore, is that H_n(M) maps isomorphically > > onto H_(n1)(S^(n1)). > > > > Your query > > > > It seems strange that a manifold would "lose its > > orientability" if we cappedin a boundary. It seems > > like this boundary would bound all the ncycles that > > had no boundary before the boundary was cappedin; > > 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 ndimensional cycles (i.e., chains with 0 boundary). For a > > connected orientable manifold, the fundamental class (that is, > > the chain that carries ndimensional 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 topdimensional 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 cappedin a boundary". > > A manifold with boundary (for lack of better phraseology) actually > > *has a boundary*. Any sum of nsimplices will have nonzero boundary, > > so won't be cycles at all. The boundary operator from C_n(Mo) to > > C_(n1)(Mo) has zero kernel. There are *no* ndimensional 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 2chains represent geometrically. Note that you > > simply can't arrange 2dimensional 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 nmanifold M with boundary dM. In this > > case, all those nchains you pushed to force their boundaries into > > dM now (in the relative case) become cycles, and you get ncycles. > > 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.



