Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A$ be a chain complex of free $R$-modules over a PID $R$, and let's assume $A$ has finite cohomological type, by which I mean $H^\ast(A)$ is finitely generated in each dimension and $0$ for large enough $|\ast|$ (though this may be stronger than necessary for the question.

This is enough for the universal coefficient theorem to tell us that $$H^i(Hom^\ast(A,R))\cong Hom(H^{-i}(A),R)\oplus Ext(H^{-i+1}(A),R).$$ Some more basic homological algebra tells us that $$ Ext(H^{-i+1}(A),R)\cong Hom(T^{-i+1}(A),Q(R)/R),$$ where $T^\ast$ is the torsion subgroup of $H^\ast$ and $Q(R)$ is the field of fractions of $R$. So, $$H^i(Hom^\ast(A,R))\cong Hom(H^{-i}(A),R)\oplus Hom(T^{-i+1}(A),Q(R)/R).$$ Let's call this formula $(\ast)$.

Now let $I$ be the complex with $Q(R)$ in degree $0$, $Q(R)/R$ in degree 1, and the projection as the only non-trivial map. Then the obvious coaugmentation $R\to I$ (thinking of $R$ as a complex concentrated in degree $0$) is a quasi-isomorphism, and since $A$ is free, $Hom^\ast(A,R)$ should be quasi-isomorphic to $Hom^\ast(A,I)$. Here, of course, the $i$th cohomology groups of this latter complex are chain homotopy equivalence classes of degree $i$ chain maps from $A$ to $I$.

So my question is whether there might be a more direct homological algebra argument to get formula $(\ast)$ from this starting point. It's not hard to get a map $H^\ast(Hom^\ast(A,I))\to Hom(H^{-\ast}(A),R)$, so the hard parts are seeing that this is onto and working in the torsion pairing somehow (I haven't yet stumbled upon the correct map $H^\ast(Hom^\ast(A,I)) \to Hom(T^{-\ast+1}(A),Q(R)/R)$).

Does anyone have any ideas or know of some references where this approach to the universal coefficient theorem has been taken before?

Ultimately my interest is in topology and how linking pairings on manifolds arise algebraically from intersection pairings. Of course the linking pairing is well-established in the literature, but I'm interested in getting at it from this point of view with the goal of some ultimate applications to sheaf-theoretic versions of duality theorems.

Thanks.

share|improve this question
    
If there was a "correct" map $H^*(Hom(A,I)) \to Hom(T^{1-*}(A), Q(R)/R)$, then the splitting in the universal coefficient sequence would be natural, no? –  Tyler Lawson Aug 30 '11 at 22:21
    
Hmmmm. But even an "unnatural splitting" must somehow arise out of the data at hand, right? And the only data present is a (homotopy class of a) chain map from A to I. So the only information present in computing, say, $H^0$ is a homomorphism $A^0\to Q(R)$, a homomorphism $A^1\to Q(R)/R$ and the boundary maps into and out of $A^0$ and $A^1$. Somehow that data must assemble to these groups, naturally or not, so I'd expect there must be a way to write down "the answer" in terms of that data. –  Greg Friedman Aug 30 '11 at 22:51
    
The derived category of a PID is the semidirect product of the category of graded modules and the shifted $Ext^1$ –  Fernando Muro Aug 30 '11 at 23:08
    
@Fernando - I'm not sure what that tells me about my question, though. Could you expand on that? - Thanks –  Greg Friedman Aug 31 '11 at 16:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.