Determine with justification, whether $S^2 \times S^4$ is homeomorphic to $\mathbb{C}P^{3}$.

I know that they are not, but I do not know how to justify it , I got a hint that squaring a generator in $S^2 \times S^4$ is zero but in $\mathbb{C}P^{3}$ is not . could anyone help me in understanding this hint please?

  • $\begingroup$ Do you know the cohomology rings for both these spaces? Not just as graded groups, but also the cup product structure? $\endgroup$
    – William
    Mar 20, 2020 at 14:30
  • $\begingroup$ I am using AT book and "Modern Classical Homotopy Theory" and they are difficult for me. so can you point out on which pages in any one of those books should I look please? $\endgroup$
    – Secretly
    Mar 20, 2020 at 14:33
  • 1
    $\begingroup$ For $H^*(S^2 \times S^4)$ look at Hatcher's section on the Künneth formula starting on 214, especially look at the end of 215/start of 216 for the ring structure on a tensor product of graded rings. Look at Theorem 3.19 on page 220 for $H^*(\mathbb{C}P^3)$. $\endgroup$
    – William
    Mar 20, 2020 at 14:54

1 Answer 1


If you know something about homotopy theory, you should know that $\pi _4(CP^3)=0$. To prove this, consider the fibration $S^7\to CP^3$, and look at the long exact sequence on homotopy. On the other hand $\pi _4(S^4)=\mathbb Z\neq 0$. So $S^2\times S^4$ contains a 4 sphere which is not homotopic to $0$, unlike $CP^3$


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.