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Re: Top Homology of Manifold with Boundary is Zero. True? Why?
Posted:
May 14, 2013 10:41 PM


baclesback@gmail.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



