Are products of spheres $\prod_i\Bbb S^{n_i}$ different?

Question

If $(n_i)_{i\leq k}$ and $(n_i^\prime)_{i\leq k^\prime}$ are increasing (not necessarily strictly) sequences of non-zero integers, do we have the following?

$$\prod_{i\leq k}\Bbb S^{n_i}\simeq \prod_{i\leq k^\prime}\Bbb S^{n_i^\prime}\iff (n_i)_{i\leq k}=(n_i^\prime)_{i\leq k^\prime}$$

Here I am being a bit vague about "$\simeq$". This is because I am interested in two aspects of this problem.

• First aspect I am interested in (which has been answered):

Is this known to be true if "$\simeq$" denotes homotopy equivalence? If not, is this known to be true if "$\simeq$" denotes homeomorphic spaces?

I don't really know about higher homotopy groups and fiber bundles which I guess could be powerful tools to solve this, but feel free to use any of such tools, I don't mind having to do some research about these subjects. I am not really looking for a full proof, this is more for culture (a reference will do the job).

• Second aspect I am interested in (which hasn't been answered):

Is this known to be true if "$\simeq$" denotes diffeomorphic smooth manifolds, with canonical structures on spheres?

For this last meaning of $\simeq$, I am more interested in finding invariants from differential geometry to solve this in special cases (fixed dimension for example).

Some necessary conditions: If $\prod_{i\leq k}\Bbb S^{n_i}$ and $\prod_{i\leq k^\prime}\Bbb S^{n_i^\prime}$ are diffeomorphic manifolds, then their dimension are equal: $$\sum_{i\leq k}n_i=\sum_{i\leq k^\prime}n_i^\prime$$ Also using fundamental group we have $$\left\lbrace i\leq k~\vert~n_i=1\right\rbrace=\left\lbrace i\leq k^\prime~\vert~n_i^\prime=1\right\rbrace.$$

An example of what I am interested in the smooth case: The two conditions above can help use to solve the problem when the dimension is $4$. In this case we have $5$ spaces: $$\Bbb S^4,\quad\Bbb S^2\times \Bbb S^2,\quad\Bbb S^1\times\Bbb S^3,\quad\Bbb S^1\times \Bbb S^1\times \Bbb S^2,\quad\Bbb S^1\times \Bbb S^1\times\Bbb S^1\times \Bbb S^1.$$

Here fundamental group tells us that the last $3$ spaces are different from the other $4$, and $\pi_4(\Bbb S^4)\approx\Bbb Z$ but $\pi_4(\Bbb S^2\times \Bbb S^2)\approx \pi_4(\Bbb S^2)\times \pi_4(\Bbb S^2)\approx \Bbb Z_2\times \Bbb Z_2$ so all these space are different. But this answer doesn't satisfy me that much. What I find very nice is for example to prove that $\Bbb S^2\times \Bbb S^2$ and $\Bbb S^1\times\Bbb S^3$ are not diffeomorphic by using parallelism as done here. Is there any other similar examples to distinguish products of spheres? One thing I thought about is the minimal dimension in which we can embed the product of spheres in euclidean space, but I am pretty sure this will fail (see here). Maybe we can change "euclidean space" with something else?..

Let me know if the question is too vague and I'll try to modify it. Thanks in advance!

Answer 1

Yes and yes. You can use cohomology and Künneth formula to see it. The polynomial $p = \prod_{i=1}^k (1 + t^{n_i})$ is an invariant of the space $\prod_i^k S^{n_i}$ (see explanation after) and it's clear that one can reads the $n_i$ up to permutation from $p$.

More precisely, this polynomial is the Poincaré polynomial defined as $p_X(t) = \sum b_it^i$ where $b_i = \dim H^i(X, \Bbb Q)$ is the $i$-th Betti number. Künneth formula tells you that $p_X(t)p_Y(t) = p_{X \times Y}(t)$.