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I have a question about an argument using the Künneth Theorem for cohomologies: Let $X, Y$ topological spaces and $X$ has finite homology groups. Then Künneth provides for every degree $n \ge 0$ we have exact sequence

(*) $$0 \to \bigoplus_{p,q: p+q=n} H^p(X;R) \otimes H^q(Y;R) \to H^n(X \times Y; R) \to \bigoplus_{p,q: p+q=n+1}Tor^1 (H^p(X;R), H^q(Y;R)) \to 0 $$

For our purposes we can assume $R = \mathbb{Z}$.

Futhermore, the cohomology ring $\bigoplus_n H^n(X; R)$ is endowed with graded multiplication, the cup-product $\cup$ induced in every grade by

(**) $$H^k(X;R) \otimes H^m(X;R) \to H^{k+m}(X \times X; R) \to H^{k+m}(X; R)$$

where the right map is induced on cohomology by diagonal map $X \to X \times X$.

Now my question: For arbitrary $n,m \ge0$ it is stated by Künneth that for complex projective spaces $\mathbb{CP}^n$ their graded rings holds

$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = H^*(\mathbb{CP}^n , \mathbb{Z}) \otimes H^*(\mathbb{CP}^m , \mathbb{Z})$$

We know (using the fact that $\mathbb{CP}^n$ is compact, orientable manifold) that the graded ring structure with respect the $\cup$-product is:

$$H^*(\mathbb{CP}^n , \mathbb{Z}) = \mathbb{Z}[x]/(x^{n+1})$$

with $x \in H^2(\mathbb{CP}^n, \mathbb{Z})$. The question is: Why Künneth states that

$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = \mathbb{Z}[x]/(x^{n+1}) \otimes \mathbb{Z}[y]/(y^{m+1}) = \mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$$

is graded ring with respect to the $\cup$-product ?

Indeed, the formula (*) just states that (since here $Tor$-terms vanish) in each grade $k$ the $k$-th kohomology can be identified as $\mathbb{Z}$-modules via

$$H^k(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z})= \bigoplus_{p,q: 2p+2q=k} \mathbb{Z} \cdot x^p y^q$$

But from here I think that Künneth tells nothing about their $\cup$-multiplication structure. But the "identification"

$$H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = \mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$$

would state that here $ax^k \cup by^l= ab x^k y^l$. Finally, that's true, but I don't see why it should follow from Künneth's (*).

I think here we need an extra argument. Could anybody explain it with Künneth provides also that for this example the $\cup$-product coinsides indeed with naivve multiplication $ax^k \cup by^l= ab x^k y^l$ in ring $\mathbb{Z}[x,y]/(x^{n+1},y^{m+1})$?

Equivalently: Why $H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = H^*(\mathbb{CP}^n , \mathbb{Z}) \otimes H^*(\mathbb{CP}^m , \mathbb{Z})$ holds as algebras and not only $\mathbb{Z}$-modules?

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There is a ring map from $H^*(X;\Bbb Z) \otimes H^*(Y;\Bbb Z)$ to $H^*(X \times Y;\Bbb Z)$, the cross product map. This is an algebra homomorphism. The Kunneth theorem (in your case) guarantees that it is an isomorphism.

This is Theorem 3.15 of Hatcher, stated for coeffients in a ring $R$ and spaces $X, Y$ with $H^*(X;R)$ and $H^*(Y;R)$ both free and finitely generated over $R$.

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  • $\begingroup$ Yes, this is indeed the first map from (*). Let call it "$\times$". and if we take $X=Y$ this gives arise for the $\cup$ product. And yes, "$\times$" arises from "naive" multiplication of cocomplex elements and then modulo cohomology. The problem is why $\cup$ "coinsides" with this "multiplication"? Yes, Hatcher, skips the proof :(. Could you maybe sketch the idea to show it? $\endgroup$ – KarlPeter Oct 16 '18 at 23:09
  • $\begingroup$ The proof is directly above the statement of the theorem, if I recall. $\endgroup$ – user98602 Oct 16 '18 at 23:10
  • $\begingroup$ So the clou is to make implicitely identifications $x = pr_1^*(x)$ and $y = pr_1^*(y)$? where $pr_i^*$ are unduced by canonical projections. $\endgroup$ – KarlPeter Oct 16 '18 at 23:13
  • $\begingroup$ @KarlPeter It's worth noting that the proof of this version of the Kunneth theorem is not to just apply the classical version of the Kunneth theorem you stated. Instead, Hatcher shows that one may define two cohomology theories in $X$, one being $H^*(X;R) \otimes_R H^*(Y;R)$, and the other $H^*(X \times Y;R)$, where $Y$ is a fixed space. He then recalls the fact that a map of cohomology theories which induces an isomorphism on the 1-point space $X = *$ induces an isomorphism for all $X$. The map described here is a ring map, and this proves that it is an isomorphism for all $X$. $\endgroup$ – user98602 Oct 16 '18 at 23:19
  • $\begingroup$ Ah ok, this seems to work also. By the way I guess that if I use the identifications explained in my last comment I can deduce the desired (stronger) statement that $H^*(\mathbb{CP}^n \times \mathbb{CP}^m, \mathbb{Z}) = H^*(\mathbb{CP}^n , \mathbb{Z}) \otimes H^*(\mathbb{CP}^m , \mathbb{Z})$ as rings from the "weaker" Künneth as I introduced above by observing that $pr_1^*(x) \cup pr_2^*(y) = x \times y$. As $\mathbb{Z}$-modules the are isomorphic in each grade. Does this argument here suffice? $\endgroup$ – KarlPeter Oct 16 '18 at 23:31

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