3 added 25 characters in body

I have editted this question from the previous version which did not obtain much attention.

Suppose I have two diagrams of chain complexes:

$A^* \rightarrow C^* \leftarrow B^*$

$\tilde{A}^* \rightarrow \tilde{C}^* \leftarrow \tilde{B}^*$

We can form the tensor product diagram

$(A\otimes \tilde{A})^* \rightarrow(C\otimes \tilde{C})^* \leftarrow(B\otimes \tilde{B})^*$.

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?

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

$\Omega_{\mathbb{Z}}^k(M)\rightarrow H^k(M,\mathbb{R})\leftarrow H^k(M,\mathbb{Z})$,

where the subscript $\mathbb{Z}$ denotes forms with integral periods. 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 Somehow we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from

$(...\rightarrow\Omega ^k_\mathbb{Z}\stackrel{0}{\rightarrow}\Omega^{k+1}_\mathbb{Z}\rightarrow...)\rightarrow(\mathrm{de Rham complex})\leftarrow(\mathrm{singular complex})$,

which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential

$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.

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.

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.

Thanks!

2 added 500 characters in body; edited title

# KunnethformulaindifferentialcohomologyOR homotopy pullbacks of tensor product diagramschaincomplexes(towardsKunnethformulaindiffcohomology)

Hello

I have editted this question from the previous version which did not obtain much attention.

Suppose I have two diagrams of chain complexes:

$A^* \rightarrow C^* \leftarrow B^*$

$\tilde{A}^* \rightarrow \tilde{C}^* \leftarrow \tilde{B}^*$

We can form the tensor product diagram

$(A\otimes \tilde{A})^* \rightarrow(C\otimes \tilde{C})^* \leftarrow(B\otimes \tilde{B})^*$.

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?

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

$\Omega_{\mathbb{Z}}^k(M)\rightarrow H^k(M,\mathbb{R})\leftarrow H^k(M,\mathbb{Z})$,

where the subscript $\mathbb{Z}$ denotes forms with integral periods. 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 we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from

$(...\rightarrow\Omega ^k_\mathbb{Z}\stackrel{0}{\rightarrow}\Omega^{k+1}_\mathbb{Z}\rightarrow...)\rightarrow(\mathrm{de Rham complex})\leftarrow(\mathrm{singular complex})$,

which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential

$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.

My question:

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 that something nice about the homotopy pullbackcomplex has the same property and hence a Kunneth theorem, but I am not sure . Direct argument with the universal property doesn't seem to work because the maps go the wrong waywhat.

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.

Thanks!

1

# Kunneth formula in differential cohomology OR homotopy pullbacks of tensor product diagrams

Hello,

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

$\Omega_{\mathbb{Z}}^k(M)\rightarrow H^k(M,\mathbb{R})\leftarrow H^k(M,\mathbb{Z})$,

where the subscript $\mathbb{Z}$ denotes forms with integral periods. 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 we can from the above description produce a chain complex whose homology is ordinary differential cohomology. That is, the above diagram comes from

$(...\rightarrow\Omega ^k_\mathbb{Z}\stackrel{0}{\rightarrow}\Omega^{k+1}_\mathbb{Z}\rightarrow...)\rightarrow(\mathrm{de Rham complex})\leftarrow(\mathrm{singular complex})$,

which has as homotopy pullback the chain complex $\Omega^k_\mathbb{Z}\times C^{k-1}_\mathrm{dR}\times C^k$ with differential

$(F,\mu,c)\rightarrow (0,F-c+d\mu,dc)$.

My question: 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 that the homotopy pullback complex has the same property and hence a Kunneth theorem, but I am not sure. Direct argument with the universal property doesn't seem to work because the maps go the wrong way.

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.

Thanks!