I am trying to show that $S^2 \times S^4$ is not homotopy equivalent to $\mathbb{C}P^3$ using cohomology rings.

I know that $H^*{\mathbb{C}P^3} \simeq \mathbb{Z}[\lambda]/(\lambda^4)$ as a graded ring with $|\lambda|=2$.

By the Künneth formula, we have

$H^*(S^2 \times S^4) \simeq H^2(S^2) \otimes H^4(S^4) \simeq \mathbb{Z}[\alpha]/(\alpha^2) \otimes \mathbb{Z}[\beta]/(\beta^2)$ where $|\alpha|=2$ and $|\beta|=4$.

What I am stuck at now is showing how they have a different cup product structure. I think I can somehow use the fact that multiplication on the basis of the graded tensor product is defined by $(a \otimes b)(c \otimes d) = (-1)^{|b||c|}(ac \otimes bd)$.

  • $\begingroup$ Can't you just notice that the generator of $H^4(\Bbb CP^3)$ is the square of the generator of $H^2(\Bbb CP^3)$? Is that true of $S^2\times S^4$? $\endgroup$ – Ted Shifrin Sep 16 '18 at 20:59
  • $\begingroup$ @TedShifrin Thanks Ted, I corrected the typos. Let me think about what you said about the generators. $\endgroup$ – TuoTuo Sep 16 '18 at 21:01

You should first decipher the isomorphism $$H^*(S^2 \times S^4) \cong \mathbb{Z}[\alpha, \beta]/(\alpha^2, \beta^2, (\alpha \cup \beta)^2)$$ where $|\alpha|=2$ and $|\beta|=4$. Thus, if you take a generator of $H^*(S^2 \times S^4)$, then its cube is clearly zero.

But in the cohomology ring $H^*(\mathbb C P^3)$ it is not the case.


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