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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.

JLA
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  • Isn't the dimension of a manifold a homotopy invariant? So having $M\cong M\times M$ for an d-dimensional manifold would only be possible for $d=2d$ and hence $d=0$? –  Jul 20 '23 at 06:56
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    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 – Thomas Rot Jul 20 '23 at 08:52
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    @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 $\Bbb R^0$). – Sassatelli Giulio Jul 20 '23 at 09:28
  • Can you think of a finite dimensional manifold with homotopy group that isn't finitely generated? – Daniel Teixeira Jul 20 '23 at 14:46
  • @ThomasRot doesn't every manifold have the homotopy type of a finite CW complex? – Daniel Teixeira Jul 20 '23 at 14:57
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    @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). – Thomas Rot Jul 20 '23 at 14:59
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    @DanielTeixeira: Take a neighborhood of S^2\vee S^1 in R^3 – Thomas Rot Jul 20 '23 at 15:01
  • This has a non-finitely generated second homotopy group (its universal cover is homotopy equivalent to a countable wedge of S^2's – Thomas Rot Jul 20 '23 at 15:02
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    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. – Ryan Budney Jul 24 '23 at 21:58
  • Suppose we ask the same question about manifolds with boundary. What about, e.g., a regular n-dimensional neighborhood of the countable one-point union of 2-spheres? – Dan Asimov Jul 24 '23 at 22:46
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    That's a funny use of $\cong$. Typically $\simeq$ is for homotopy equivalence – FShrike Jul 25 '23 at 13:33
  • @FShrike You're right, I fixed it. – JLA Jul 25 '23 at 21:05
  • Cf. this – J. W. Tanner Jul 25 '23 at 21:23
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    Duplicate question on MO has been answered. https://mathoverflow.net/questions/451401 – Ryan Budney Jul 26 '23 at 17:51

2 Answers2

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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 $M\times M\simeq M$.

Dave Benson
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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 $H_{2n}(M \times M;\mathbb Z_2) = \mathbb Z_2$.

Paul Frost
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