Is $S^2 \times S^2$ diffeomorphic to $S^1 \times S^3$?

Question

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.

Answer 1

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.

Answer 2

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.