# Künneth formulas/theorem for bordism groups and cobordisms?

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:

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

. 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.

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 (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 $$i\neq 0,1$$.

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

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 $$Y$$.

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