homotopy pullbacks of tensor product chain complexes (towards Kunneth formula in diff cohomology) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:21:18Z http://mathoverflow.net/feeds/question/107820 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107820/homotopy-pullbacks-of-tensor-product-chain-complexes-towards-kunneth-formula-in homotopy pullbacks of tensor product chain complexes (towards Kunneth formula in diff cohomology) Ryan Thorngren 2012-09-22T07:09:31Z 2012-09-24T20:04:38Z <p>I have editted this question from the previous version which did not obtain much attention.</p> <p>Suppose I have two diagrams of chain complexes:</p> <p>$A^* \rightarrow C^* \leftarrow B^*$</p> <p>$\tilde{A}^* \rightarrow \tilde{C}^* \leftarrow \tilde{B}^*$</p> <p>We can form the tensor product diagram</p> <p>$(A\otimes \tilde{A})^* \rightarrow(C\otimes \tilde{C})^* \leftarrow(B\otimes \tilde{B})^*$.</p> <blockquote> <p>How can we express the homotopy pullback of the tensor product diagram (more importantly, its homology) in terms of the homotopy pullbacks of the first two diagrams?</p> </blockquote> <p>Now I'll give some motivation, because maybe someone knows the answer to the question the above seeks to find. I am interested in whether there is a Kunneth formula for (mainly ordinary) differential cohomology. The model of ordinary differential cohomology I am currently working with is as the homotopy pullback of</p> <p>$\Omega_{\mathbb{Z}}^k(M)\rightarrow H^k(M,\mathbb{R})\leftarrow H^k(M,\mathbb{Z})$,</p> <p>where the subscript $\mathbb{Z}$ denotes forms with integral periods. <strike>Taking homology commutes with taking homotopy pullbacks since we can replace anything by something quasi-isomorphic without affecting the quasi-isomorphism type (so I assume...), so</strike> Somehow we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from</p> <p>$(...\rightarrow\Omega ^k_\mathbb{Z}\stackrel{0}{\rightarrow}\Omega^{k+1}_\mathbb{Z}\rightarrow...)\rightarrow(\mathrm{de Rham complex})\leftarrow(\mathrm{singular complex})$,</p> <p>which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential</p> <p>$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.</p> <p>We have Eilenberg-Zilber q-isoms $\Omega_\mathbb{Z}(M\times N)\rightarrow\Omega_\mathbb{Z}(M)\otimes \Omega_\mathbb{Z}(N)$ etc for each of the three chain complexes above such that the diagram for $M\times N$ becomes the tensor product of the diagrams for $M$ and $N$. I expect that this implies something nice about the pullback, but I am not sure what.</p> <p>I would also appreciate any answers using another model of differential cohomology (eg. Deligne cohomology), being ultimately interested in geometrical application, but I would like to understand this piece of homological algebra.</p> <p>Thanks!</p>