Undergrad Understanding Kunneth Formula and Tensor Product in r-Forms

[Silviu]

Hello! Kunneth fromula states that for 3 manifolds such that $M=M_1 \times M_2$ we have $H^r(M)=\oplus_{p+q=r}[H^p(M_1)\otimes H^q(M_2)]$. Can someone explain to me how does the tensor product acts here? I am a bit confused of the fact that we work with r-forms, which are by construction antisymmetric, but that tensor product seems to break this anti-symmetry. I would have expected something like this $H^r(M)=\sum_{p+q=r}[H^p(M_1)\wedge H^q(M_2)]$ (actually when he does some examples he uses the wedge product for individual terms in the computations). Can someone clarify this for me please? Thank you!

 

[mathwonk]

This surprising fact arises already just in the algebra of exterior multiplication. Here is the basic fact:

It is the last thing discussed in these algebra notes, 845-3, p.56:

http://alpha.math.uga.edu/%7Eroy/845-3.pdf

 

[lavinia]

Hello! Kunneth fromula states that for 3 manifolds such that $M=M_1 \times M_2$ we have $H^r(M)=\oplus_{p+q=r}[H^p(M_1)\otimes H^q(M_2)]$. Can someone explain to me how does the tensor product acts here? I am a bit confused of the fact that we work with r-forms, which are by construction antisymmetric, but that tensor product seems to break this anti-symmetry. I would have expected something like this $H^r(M)=\sum_{p+q=r}[H^p(M_1)\wedge H^q(M_2)]$ (actually when he does some examples he uses the wedge product for individual terms in the computations). Can someone clarify this for me please? Thank you!

If $π_{M}$ is the projection map of $M×N$ onto $M$ then $π_{M}^{*}$ maps $H^{*}(M)$ into $H^{*}(M×N)$. Similary $π_{N}^{*}$ maps $H^{*}(N)$ into $H^{*}(M×N)$.

These pullback maps determine a bilinear mapping of $H^{*}(M)×H^{*}(N)→H^{*}(M×N)$ by $([α],[β])→[(π_{M}^{*}α)∧π_{N}^{*}β]$. Writing this in terms of the grading of cohomology dimensions, one has maps $Σ_{i+j=k}H^{i}(M)⊗H^{j}(N)→H^{k}(M×N)$ for each dimension $k$.

BTW: The Kunneth Theorem applies to any Cartesian product of manifolds not just to 3 manifolds that are Cartesian products..

- The cohomology determined by differential forms is called De Rham cohomology. It is a theory that is defined only for differentiable manifolds. Singular cohomology is another cohomology theory. It is defined for all topological spaces. De Rham cohomology is isomorphic to singular cohomology with real numbers as coefficients. The idea of the proof is to view differential forms as homomorphisms of the groups of smooth n- simplexes into the real numbers - or into the complex numbers.