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I am trying to analyze whether $S^2 \times S^2$ are diffeomorphic to $S^1 \times S^3$. First of all, the dimension matches because they are all four-dimensional manifolds. Then I tried thinking about techniques to prove whether manifolds are diffeomorphic. The only thing that I could come up was to compute their de Rahm cohomology, but I do not have any background in that other than the definitions.

I found some random notes online hinting that one should think about the extension of smooth maps on the sphere to the closed ball. That does not make any sense to me though.

penny
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2 Answers2

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They're not homeomorphic, so they won't be diffeomorphic. To see that they aren't homeomorphic, we compute their fundamental groups. Recalling that $$ \pi_1(S^1) = \mathbb{Z} \quad \text{and} \quad \pi_1(S^2) = \pi_1(S^3) = 1$$ and $\pi_1(X\times Y) \cong \pi_1(X) \oplus \pi_1(Y)$, we find $$\pi_1(S^2 \times S^2) = 1 \oplus 1 = 1$$ while $$\pi_1(S^1 \times S^3) = \mathbb{Z} \oplus 1 \cong \mathbb{Z} $$ Since their fundamental groups aren't isomorphic, they aren't homeomorphic, hence not diffeomorphic.

Hayden
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  • My apologies, but I am actually looking for a differential topological method since I am self-studying differential topology and do not have any background in algebraic topology. I know most of the basic techniques from differential topology except de Rham cohomology. If there are no any other answers, I will accept yours. My sincere apologies again and thank you for your answer. – penny Nov 17 '18 at 01:50
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    @penny That's understandable. A differential topology proof that comes to mind is that $S^2 \times S^2$ has no smooth field of non-zero tangent vectors, while $S^1\times S^3$ does, but I'll admit my differential topology is rusty. – Hayden Nov 17 '18 at 02:04
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I would have used the fundamental group as @Hayden, but you want a differential argument, so here it goes.
Since the tangent bundles $T\mathbb{S}^1$, $T\mathbb{S}^3$ are trivial (you can check this in Spivak's Volume 1, e.g, but the fact that $T\mathbb{S}^1$ is trivial is clear from the looks, and $\mathbb{S}^3$ is a Lie group, so it's parallelizable), then $T(\mathbb{S}^1\times\mathbb{S}^3)\cong T\mathbb{S}^1\times T\mathbb{S}^3$ is again trivial.
This is not the case for $\mathbb{S}^2\times\mathbb{S}^2$.
To see this, observe that since $\mathbb{S}^2\times\mathbb{S}^2$ is a $CW$-complex, we can use any of its $CW$ structures to compute its Euler characteristic. For example, use the one for $\mathbb{S}^2$ with one $0$-cell and one $2$-cell, to get one in $\mathbb{S}^2\times\mathbb{S}^2$ with one $0$-cell, two $2$-cells and one $4$-cell. Then its Euler characteristic is 4, now use the Poincaré-Hopf Index Theorem to prove that any vector field on $\mathbb{S}^2\times\mathbb{S}^2$ must have a zero, contradicting a possible triviallity of its tangent bundle.
$\textbf{Credit}:$ to Mike Miller for the final argument using the Euler characteristic.

Laz
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  • It is not clear to me how you intend the argument in the final line to go. I do not know how to prove that if $E$ and $F$ are nontrivial vector bundles, their Cartesian product is nontrivial. I am not sure I believe that it's true. Perhaps what you wanted to point out is that $\chi(S^2 \times S^2) = 4$, and so it carries no nonvanishing field of tangent vectors. –  Nov 19 '18 at 19:30
  • Thanks @MikeMiller, you are totally right about my sloppyness. I just happened to know the non-triviallity of $T(\mathbb{S}^2\times\mathbb{S}^2)$ but I was going for an argument that was going to take me way to long to prove it. I corrected it using your argument which is a thousand times finer. I just want to point out that in this case $E$, $F$ are not any, are the same sphere, of even dimension. – Laz Nov 19 '18 at 19:56
  • I don't need credit, I was just curious if there was a different argument that worked. Thank you though! –  Nov 19 '18 at 19:59
  • You totally got it, thanks again! – Laz Nov 19 '18 at 19:59