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Let $M$ be a simply connected, (finite dimensional) smooth manifold. Is it possible that $M$ is homotopy equivalent to $M\times M,$ without $M$ being contractible? This would imply $\pi_n(M)\times\pi_n(M)\cong \pi_n(M)$ fo all $n\in\mathbb{N}.$ I know there are groups which satisfy $G\cong G\times G,$ however I don't know if it can happen for homotopy groups of manifolds.

According to https://mathoverflow.net/questions/43805/when-is-g-isomorphic-to-g-times-g, if even one nontrivial homotopy group is finitely generated, this is impossible.

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    $\begingroup$ This cannot happen if $M$ is closed. If $M$ is simply connected and has the homotopy type of a finite CW complex then the higher homotopy groups are finitely generated. As the higher homotopy groups are abelian, this is also not possible in this case. So you are looking at quite complicated things $\endgroup$
    – Thomas Rot
    Jul 20, 2023 at 8:52
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    $\begingroup$ @B.Hueber No, the dimension of a manifold isn't a homotopy invariant. For instance, all the $\Bbb R^n$-s are homotopically equivalent to one another (i.e. to $\Bbb R^0$). $\endgroup$ Jul 20, 2023 at 9:28
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    $\begingroup$ @DanielTeixeira: Non compact ones don't have to. Think of an infinitely often punctured plane for example. Or an infinite set of points (a zero dim manifold). $\endgroup$
    – Thomas Rot
    Jul 20, 2023 at 14:59
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    $\begingroup$ I don't think such a manifold exists. The idea would be to ask what is the cup-length in your manifold, i.e. could it be finite? I think your homotopy equivalence together with the Kunneth theorem gets you into trouble. $\endgroup$ Jul 24, 2023 at 21:58
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    $\begingroup$ That's a funny use of $\cong$. Typically $\simeq$ is for homotopy equivalence $\endgroup$
    – FShrike
    Jul 25, 2023 at 13:33

2 Answers 2

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I've been encouraged to share my MathOverflow answer here, so here goes:

Lemma. If $A$ is an abelian group satisfying $A\otimes A=0$ and $\mathop{\rm Tor}(A,A)=0$ then $A=0$.

Proof. Since $\mathop{\rm Tor}$ is left exact on abelian groups, an inclusion of a finite cyclic group $C$ in $A$ gives an injection $\mathop{\rm Tor}(C,C)\to \mathop{\rm Tor}(A,A)$. So if $\mathop{\rm Tor}(A,A)=0$ then $A$ is torsion free. Then $A$ embeds in $\mathbb{Q}\otimes A$, so if $A\otimes A=0$ then $(\mathbb{Q}\otimes A)\otimes(\mathbb{Q}\otimes A)=\mathbb{Q}\otimes(A\otimes A)=0$. So $\mathbb{Q}\otimes A=0$ and then $A=0$.

Now given your manifold $M$, since it is smooth, simply connected, and finite dimensional, it has the homotopy type of a finite dimensional CW complex. So by the Whitehead theorem, if it's not contractible then it has some non-vanishing homology group in degree $\geqslant 2$. Let $H_k(M)\ne 0$ with $k\geqslant 2$ as large as possible (so $k$ is at most the dimension of $M$). Then by the Künneth theorem and the lemma, either $H_{2k}(M\times M)\ne 0$ or $H_{2k+1}(M\times M)\ne 0$. So we have a contradiction if $M\times M\simeq M$.

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This is just an extended comment to Dave Benson's perfect answer.

For a closed topological $n$-manifold it can never happen that $M \simeq M \times M$.

  1. If $M$ is not connected, then it is fairly obvious that $M \simeq M \times M$ is false. In fact, since $M$ is compact, it has a finite number $k$ of path components so that $M \times M$ has $k \times k$ path components. But the number of path components is an invariant of homotopy type.

  2. If $M$ is connected, we know that $H_n(M;\mathbb Z_2) = \mathbb Z_2$ and $H_m(M;\mathbb Z_2) = 0$ for $m > n$. But $M \times M$ is a closed $2n$-manifold so that $H_{2n}(M \times M;\mathbb Z_2) = \mathbb Z_2$.

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