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

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?

• Do you know the cohomology rings for both these spaces? Not just as graded groups, but also the cup product structure? Mar 20, 2020 at 14:30
• 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? Mar 20, 2020 at 14:33
• 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 $H^*(\mathbb{C}P^3)$. Mar 20, 2020 at 14:54

## 1 Answer

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 $$CP^3$$