Alternative approaches to the universal coefficient theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:48:22Z http://mathoverflow.net/feeds/question/74097 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74097/alternative-approaches-to-the-universal-coefficient-theorem Alternative approaches to the universal coefficient theorem Greg Friedman 2011-08-30T19:51:11Z 2011-08-30T19:51:11Z <p>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. </p> <p>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 $(\ast)$.</p> <p>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 $I$. </p> <p>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 $H^\ast(Hom^\ast(A,I)) \to Hom(T^{-\ast+1}(A),Q(R)/R)$). </p> <p>Does anyone have any ideas or know of some references where this approach to the universal coefficient theorem has been taken before?</p> <p>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. </p> <p>Thanks.</p>