I am trying to calculate $H^*(\mathbb{R}P^3 \times \mathbb{C}P^5,\mathbb{Z})$ as a cohomology ring.

I know that

$$H^*(\mathbb{R}P^3,\mathbb{Z}) = \frac{\mathbb{Z}[\alpha,\beta]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta)}$$

with $\alpha$ the generator of $H^2(\mathbb{R}P^3)$ and $\beta$ generating $H^3(\mathbb{R}P^3)$


$$H^*(\mathbb{C}P^5) = \frac{\mathbb{Z}[\gamma]}{(\gamma^6)}$$

with $\gamma$ the generator of $H^2(\mathbb{C}P^5)$

Initially I thought the cross-product would just give an isomorphism, but the presence of the $\mathbb{Z}/2\mathbb{Z}$ term in the cohomology of $\mathbb{R}P^3$ kills that idea.

Now I could calculate the cohomology groups since both space have a nice CW structure, it would be possible to use

$$H^n(X \times Y) = \sum_{i+j=n}H^i(X) \otimes H^j(Y) \oplus \sum_{p+q=n+1} \operatorname{Tor}(H^p(X),H^q(Y))$$

but how can we deduce the ring/cup product structure?

(I am guessing the answer is to somehow use the cross product, but I just can't see it)


Hatcher has as Theorem $3.16$ that the cross product is an isomorphism if $X$ and $Y$ are CW complexes and $H^k(Y)$ is finitely generated free for all $k$. So, you should still be able to apply it.

  • 1
    $\begingroup$ @Qwirk but $H^*(\mathbb{C}P^5;\mathbb{Z})$ is. $\endgroup$ – wckronholm Jun 15 '11 at 3:48
  • $\begingroup$ @wckronholm - oh I completely misread the Lemma! (apologies to Joe) $\endgroup$ – Juan S Jun 15 '11 at 4:02
  • $\begingroup$ So can we simplify $$\frac{\mathbb{Z}[\gamma]}{(\gamma^6)} \otimes \frac{\mathbb{Z}[\alpha,\beta]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta)} = \frac{\mathbb{Z}[\alpha,\beta,\gamma]}{(2 \alpha, \alpha^2,\beta^2,\alpha \beta,\gamma^6)}$$ or do I need to check the cohomology groups and resulting product? $\endgroup$ – Juan S Jun 15 '11 at 4:05
  • 1
    $\begingroup$ The theorem tells you the resulting cohomology ring is the tensor product of the two cohomology rings you mentioned, which is the ring you describe in the comment above. I should add that the isomorphism is not just as rings, but as graded rings. It would be good to note the degree of each of $\alpha$, $\beta$, and $\gamma$. $\endgroup$ – wckronholm Jun 15 '11 at 4:11
  • $\begingroup$ @wrkronholf - thanks again. $\endgroup$ – Juan S Jun 15 '11 at 4:26

Have you seen this paper? He starts by giving an example showing that the cohomology ring isn't necessarily determined by the integral cohomology rings of the factors, but analyzes conditions under which you can compute the cohomology ring of the product from knowledge of the cohomology of the factors.

  • $\begingroup$ thanks, I tuned out when he started talking about Bockstein homomorphism and cohomology spectrums, but I see in Section 6 there might be some way to do it. (This is from an old problem sheet so it should be relatively do-able) $\endgroup$ – Juan S Jun 15 '11 at 1:12

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