I am working on a problem to show that the cohomology graded rings of $\mathbb{C}P^3$ and $S^2$ x $S^4$ are not isomorphic (unreduced with integer coefficients)

I have already calculated the graded groups of both and shown that they are isomorphic and both have a $\mathbb{Z}$ in dimensions 0,2,4,6 and zeros elsewhere.

Now from what I have seen in class I know that the cohomology of $\mathbb{C}P^3$ is isomorphic to $\mathbb{Z} [x_2] / (x^4)$ where $x_2$ is the generator of $H^2(\mathbb{C}P^3)$ This tells me that the fourth dimension is generated by $x_2^2$ and the sixth dimension is generated by $x_2^3$.

However this is not the case for $S^2$ x $S^4$. Say that the second dimension is generated by $u$, the fourth by $v$ and the sixth by $w$. I gather from a hint we were given in class that $uv = w$ but I am unsure how to prove this. It also appears to me that I need to verify that $u^2 \neq v$, but I am unsure how to do this as well. The notion of multiplication in these rings is still a bit fuzzy to me.

Any help you could offer on this would be appreciated. I could benefit from some general advice on how to determine the generators in an arbitrary cohomology ring.

Thanks in advance!

up vote 3 down vote accepted

Try Künneth. When $\textrm{Tor}_1$ vanishes, as it will here since each graded piece of $H^*(S^n; \mathbb{Z})$ is free abelian for any $n$, Künneth gives you an isomorphism of algebras.

(There's no need to upvote this answer, should anyone happening across this feel so inspired. I'm simply echoing an observation I've seen Mariano make a few times on this site.)

  • Got it thanks. I had used the Kunneth theorem to calculate the group but I guess I didn't look carefully enough the first time, because it also tells us how the multiplication works. So for example when we end up with a $\mathbb{Z} \otimes \mathbb{Z}$ in some dimension, then the generator in that dimension is the cup product of the generators of the two copies of $\mathbb{Z}$ that are being tensored. – Elliot Feb 14 '14 at 22:16

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