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We are familiar with Künneth theorem:

The Kunneth formula is given by $R$ as a ring, $M,M'$ as the R-modules, $X,X'$ are some chain complex. The Kunneth formula shows the cohomology of a chain complex $X \times X'$ in terms of the cohomology of a chain complex $X$ and another chain complex $X'$. For topological cohomology $H^d$, we have $$ H^d(X\times X', M\otimes_R M') \simeq \Big[\oplus_{k=0}^d H^k(X,M)\otimes_R H^{d-k}(X',M')\Big]\oplus $$ $$\Big[\oplus_{k=0}^{d+1} \text{Tor}_1^R(H^k(X,M),H^{d-k+1}(X',M'))\Big] . $$

$$ H^d(X\times X',M) \simeq \Big[\oplus_{k=0}^d H^k(X,M)\otimes_{\mathbb Z} H^{d-k}(X',\mathbb{Z})\Big]\oplus $$ $$ \Big[\oplus_{k=0}^{d+1} \text{Tor}_1^{\mathbb Z}(H^k(X,{M}),H^{d-k+1}(X',\mathbb Z))\Big]. $$

The above is valid for both topological cohomology $H^d$ and group cohomology $\mathcal{H}^d$ (for $G'$ is a finite group): $$ \mathcal{H}^d(G\times G',M) \simeq \Big[\oplus_{k=0}^d \mathcal{H}^k(G,M)\otimes_{\mathbb Z} \mathcal{H}^{d-k}(G',\mathbb Z)\Big]\oplus $$ $$ \Big[\oplus_{k=0}^{d+1} \text{Tor}_1^{\mathbb Z}(\mathcal H^k(G,M),\mathcal H^{d-k+1}(G',\mathbb Z))\Big]. $$

  • Questions:

[1]. Do we have similar results of Künneth theorem for bordism groups $\Omega_d^{...}(...)$? Schematically, maybe something like $$ \Omega_d^{...}(...) \simeq \oplus_n \Omega_n^{...(1)}(...) \otimes \Omega_{d-n}^{...(2)}(...)? $$

[2]. Can we, and, how can we interpret the decompositions of bordism group generators as manifolds $\Sigma_d$: $$ \Sigma_d \sim \Sigma_{d-n}^{(1)} \times \Sigma_{n}^{(2)}? $$ where $d$-manifold generators are related to the $d-n$-manifold generator and $n$-manifold generator. Are these general, or are they only special cases?

Refs are welcome. Thanks.

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The Künneth formula for ordinary homology as you present it works only when $R$ is a PID (or more generally of cohomological dimension 1).

For a general well-behaved homology theory[1] (this includes both ordinary cohomology, K-theory and cobordism) there is a Künneth spectral sequence

$$E^2_{p,q}=\mathrm{Tor}_{p,q}^{E_*}(E_*X,E_*Y)\Rightarrow E_{p+q}(X\times Y)$$

(see for example theorem 4.1 in

Elmendorf, A. D.; Kříž, Igor; Mandell, Michael A.; May, J. P., Rings, modules, and algebras in stable homotopy theory. With an appendix by M. Cole, Mathematical Surveys and Monographs. 47. Providence, RI: American Mathematical Society (AMS). xi, 249 p. (1997). ZBL0894.55001.

with $M=E\wedge \Sigma^\infty_+X$ and $N=E\wedge \Sigma^\infty_+Y$).

When $X$ and $Y$ have the homotopy type of finite CW-complexes, you can use Spanier-Whitehead duality to obtain the same result for cohomology $E^*X=E_{-*}(\mathbb{D}X)$, with a bit of care due to the signs that appear from the duality. So under these hypotheses we obtain a spectral sequence

$$E_2^{p,q}=\mathrm{Tor}^{E_*}_{-p,-q}(E^{-*}X,E^{-*}Y)\Rightarrow E^{p+q}(X\times Y)\,.$$

The case of the Künneth formula corresponds to the degenerate situation in which $\mathrm{Tor}^{E_*}_i=0$ for $i\neq 0,1$.

[1] Precisely, I need the homology theory to be represented by an $E_1$-ring spectrum.

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If you work with the unoriented bordism groups $\Omega^O_*(X)=MO_*(X)$ then there is a Künneth isomorphism. This is just because there is a natural isomorphism $MO_*(X)=H_*(X;\mathbb{Z}/2)\otimes MO_*$, where $MO_*$ is a graded polynomial ring over $\mathbb{Z}/2$ with one generator $x_i$ in each degree $i>0$ not of the form $2^j-1$. I think that all of this was already known to Thom. The key point is that $H_*(MO;\mathbb{Z}/2)$ is a cofree comodule over the dual Steenrod algebra, by a fairly concrete and straightforward calculation.

There are also some interesting things to say if you are willing to work with the complex bordism groups $MU_*(X)=\Omega^U_*(X)$ instead of the real ones. Here it is known that $MU_*=\mathbb{Z}[a_1,a_2,a_3,\dotsc]$,so in particular this is a free abelian group. Consider the following conditions:

  1. $H_*(X)$ is free abelian
  2. $MU_*(X)$ is a free $MU_*$-module
  3. $MU_*(X)$ is a flat $MU_*$-module
  4. $MU_*(X)$ is Landweber exact
  5. $MU_*(X)$ is torsion-free
  6. The Künneth map $MU_*(X)\otimes_{MU_*}MU_*(Y)\to MU_*(X\wedge Y)$ is an isomorphism for all $Y$.

First, 1 implies 2. Indeed, there is an Atiyah-Hirzebruch spectral sequence $H_*(X)\otimes MU_*\Rightarrow MU_*(X)$, and the differentials are controlled by homotopy groups of spheres in nonzero dimensions so they are torsion-valued, so they must be zero. The spectral sequence therefore collapses and the claim follows easily. Once you have recalled the definition of Landweber exactness it is easy to see that 2 implies 3 implies 4 implies 5. The Landweber exact functor theorem says that 4 implies 6. By taking $Y$ to be a generalised Moore spectrum $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ one can also check that 6 implies 4.

Because $MU_*(X)$ is actually an $MU_*MU$-comodule rather than just an $MU_*$-module, I think it works out that 5 implies 4. However, I am surprised to find that this very natural question has never occurred to me before.

Now suppose we want to consider the oriented real bordism groups $\Omega^{SO}_*(X)=MSO_*(X)$. The coefficient ring $MSO_*$ is somewhat complicated: it is the direct sum of a polynomial algebra over $\mathbb{Z}$ with a module over $\mathbb{Z}/2$, and does not have any very simple presentation. However, there is a map $MU\to MSO$ of ring spectra, and this gives a map $MU_*\to MSO_*$ of coefficient rings, and thus a natural map $$ MSO_* \otimes_{MU_*} MU_*(X) \to MSO_*(X). $$ If $H_*(X)$ is free abelian then we have seen that $MU_*(X)$ is a free $MU_*$-module. It is then not hard to see that the above map is an isomorphism and so $MSO_*(X)$ is free over $MSO_*$. Assuming that, it is not hard to deduce that the Künneth map $$ MSO_*(X)\otimes_{MSO_*}MSO_*(Y)\to MSO_*(X\wedge Y) $$ is again an isomorphism for all $Y$.

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  • $\begingroup$ Thanks! +1, I was not aware of this before $\endgroup$ – wonderich Oct 24 '18 at 19:51

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