2
$\begingroup$

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 $(a \otimes b)(c \otimes d) = (-1)^{|b||c|}(ac \otimes bd)$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.