I'm wondering if anyone can point me to a reference on how the various Lefschetz-Poincare dual torsion pairings of a manifold with boundary fit together.

To explain in more detail, consider a connected compact oriented n-manifold $M$ with boundary. Then we have the various dualities with rational coefficients $H_i(M;Q)=H_{n-i}(M, \partial M; Q)$ and $H_i(\partial M;Q)=H_{n-i-1}(\partial M; Q)$, but we also obtain, from some easy tinkering with the long exact sequence of the pair and other basic observations that, e.g. $im(H_i(M)\to H_i(M, \partial M))$ is dual to $im(H_{n-i}(M)\to H_{n-i}(M, \partial M))$ (in fact this is how we define signatures on manifolds with boundary) and similar results hold for the other "shared" terms in the long exact sequence.

I'd like to know more about the generalization of this over Z and, in particular, about what torsion pairings to Q/Z exist on the various images, cokernels, etc. of the long exact sequence and which are nonsingular. Does anyone know of any place in the literature where this is worked out?


Updated: Thanks, Tom, for your response. I've been thinking about it, and while I agree that the situation is much murkier over Z, I think there are still some things that can be said.

For example, it's true that the map $H_n(M)\to H_n(M, bd M)$ (let's assume $n$ is the middle dimension to simplify the discussion), only yields a nondegenerate pairing on the image (which you call $A$) mod torsion and not a perfect (nonsingular) pairing. But now since $H_n(M)\to A/\text{torsion}$ has a free group as its image, we have a (non-unique) splitting that lets us consider $A/\text{torsion}$ as a direct summand of $H_n(M)$. Let's fix this summand for now (the choice turns out not to matter). Since $H_n(M)/\text{torsion}$ and $H_n(M, bd M)/\text{torsion}$ are (perfectly) Z-dual, $A/\text{torsion}$ must be dual to something in $H_n(M, bd M)/\text{torsion}$, and I claim that the thing it's dual to is isomorphic to the kernel of $H_n(M,bd M)\to H_{n-1}(bd M)/\text{torsion}$ (mod torsion). Notice that any non-torsion element of this kernel has a multiple represented by a cycle in $M$, and so it must have non-zero intersection number with any cycle in the boundary, and this shows that as far as intersection pairings are concerned, the choice of $A/\text{torsion}$ as a summand of $H_n(M)$ doesn't matter. I don't really want to get into the details of the rest of my claim here, but the basic idea should be that these are the things that should be dual once we tensor with Q but over Z we need to make sure we have all the appropriate Z-primitives, which I'm pretty sure this does (since if a multiple of $x\in H_n(M,bd M)$ is in $A$, then $x$ becomes torsion in $H_{n-1}(M)$).

So what's the point of all this? Well, even over Z we can say that there is a nonsingular pairing between a certain cokernel and a certain kernel, as long as we mod out the torsion in the right places. I suspect that there's a more subtle version of this for the torsion linking pairings where instead of quotienting out all torsion we just kill certain torsion subgroups. I think I've seen things somewhat of this nature in papers relating Witt groups of Z pairings to Witt groups of Q/Z pairings.

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  • $\begingroup$ It is easier to understand what happens to the torsion if you write Poincaré duality as an isomorphism between homology and cohomology; $H^i(M;\mathbb Z)=H_{n-i}(M,\partial M;\mathbb Z)$. To get a formula involving only cohomology you then combine that with the universal coefficient formula. If you do that for $\mathbb Z$-coefficients, the torsion moves around a little bit as it comes from an $\mathrm{Ext}^1$. (cont'd) $\endgroup$ – Torsten Ekedahl Jun 8 '10 at 7:01
  • $\begingroup$ (cont'd) If you do it for $\mathbb Q/\mathbb Z$-coefficients it doesn't but you lose the precise integral information (as you cannot canonically recover a finitely generated $\mathbb Z$-module from its $\mathbb Q/\mathbb Z$-dual. $\endgroup$ – Torsten Ekedahl Jun 8 '10 at 7:02
  • $\begingroup$ @Torsten: could you elaborate a bit more on the fact that torsion comes from an $Ext^1$ and how it moves around? $\endgroup$ – John Jiang Jun 8 '10 at 7:07
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    $\begingroup$ The universal coefficient formula for cohomology says that $H^i(X,M)$ is isomorphic to $\mathrm{Hom}(H_i(X,\mathbb Z),M)\bigoplus\mathrm{Ext}^1(H_{i-1}(X,\mathbb Z)$. For $M=\mathbb Z$ that means that the torsion free part of $H^i(X,\mathbb Z)$ is dual (in the sense of $\mathrm{Hom}(-,\mathbb Z)$ to the torsion free part of $H_i(X,\mathbb Z)$ whereas the torsion of $H^i(X,\mathbb Z)$ is dual (in the sense of $\mathrm{Hom}(-,\mathbb Q/\mathbb Z)$ to the torsion part of $H_{i-1}(X,\mathbb Z)$. $\endgroup$ – Torsten Ekedahl Jun 8 '10 at 8:06
  • $\begingroup$ @Torsten: Just an aesthetic point; the splitting you describe always exists but is not natural. Thus it might be preferable to say, "There is a split short exact sequence...." $\endgroup$ – Daniel Litt Jul 1 '10 at 21:52

I don't know a reference, but I have some thoughts. I'm going to tackle this using just homology. I'll ignore gradings in the name of legibility.

There are three kinds of homology in this story: $H(M)$, $H(M,b M)$, and $H(b M)$. The last of these is self-dual; the other two are dual to each other. What I just said is literally true over $Q$, with 'dual to' meaning 'canonically isomorphic to the vector-space dual of'. Over $Z$ of course it's only true in a derived sense. One consequence of the derived statement is that, of the free abelian groups $H(M)/{torsion}$, $H(M,b M)/{torsion}$, and $H(b M)/{torsion}$, the last one is self-dual in the $Hom(-,Z)$ sense while the other two are dual to each other. Another is that the torsion part of $H(b M)$ is self-dual in the $Hom(-,Q/Z)$ sense and the torsion parts of the others are dual to each other in the same sense.

Then there are the other three players: the images of the three maps $H(M)\rightarrow H(M,b M)\rightarrow H(b M)\rightarrow H(M)$. Call them $A$, $B$, and $C$. (Each of the three can also be described as a kernel, or as a cokernel.) Here the rational story is very nice: $A$ is self-dual (more precisely the perfect pairing between $H(M)$ and $H(M, b M)$ when restricted to be a pairing between $H(M)$ and $A$ yields a pairing between $H(M)/ker=A$ and $A$ which is perfect). And $B$ and $C$ are dual to each other (more precisely, in the perfect self-pairing of $H(b M)$ the subspace $B$ is its own orthogonal complement, so that $B$ and $C=H(b M)/B$ are dual).

Over $Z$ things become murkier for $A$, $B$, and $C$. It's true that we get a pairing of $A$ with itself, and that this yields a nondegenerate pairing between the free abelian group $A/{torsion}$ and itself. But it's not perfect; it just injects $A$ into $Hom(A,Z)$. The same goes for pairing $B/{tors}$ with $C/{tors}$. And as for pairings of the torsion parts into $Q/Z$, the images just don't seem to behave well in general.

Update in response to Greg's update to the question: I think the following is basically what you're saying, but it can be said just algebraically and has nothing to do with choosing a splitting (i. e. identifying a quotient of $H(M)/tors$ with a subgroup). The sequence of free abelian groups $H(M)/tors\to H(M,bM)/tors\to H(bM)/tors$ is not exact. The image of $H(M)/tors\to H(M,bM)/tors$, which is the same as my $A/tors$, is not a summand in general; the kernel of $H(M,bM)/tors\to H(bM)/tors$ is a summand and consists of all elements such that some multiple is in $A/tors$; call this $A'$. So $A'$ contains $A/tors$ with finite index. And we get a perfect Z-pairing between $A/tors$ and $A'$. Likewise we get $B'$ and $C'$ containing $B/tors$ and $C/tors$ with finite index, and Z-dual to $C/tors$ and $B/tors$. I don't know what to say about all of this in relation to pairings into $Q/Z$.

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