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 $\pi_n(M)\times\pi_n(M)\cong \pi_n(M)$ for all $n\in\mathbb{N}.$ I know there are groups which satisfy $G\times G\cong G,$ but this is a very strong condition, and this condition still seems much weaker than the condition in question.

According to When is $G$ isomorphic to $G \times G$?, if even one nontrivial homotopy group is finitely generated, this is impossible.

(I asked this on stackexchange and didn't get any responses.)

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    $\begingroup$ You should include a link to your identical thread: math.stackexchange.com/questions/4739282/… $\endgroup$ Jul 24, 2023 at 21:45
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    $\begingroup$ I put a sketch answer in the comments to your original thread. The basic idea is to use the Kunneth theorem and apply it to your homotopy-equivalence to ask what the cup-length is for your manifold. $\endgroup$ Jul 24, 2023 at 22:00

2 Answers 2


Thanks to Dave Benson for pointing out an algebra error in the first draft of this answer and a missing detail.

Suppose $X$ is homotopy equivalent to a finite dimensional, simply-connected, and noncontractible CW complex. Then for $k=\mathbb Q$ or $k = \mathbb Z/p$ for some $p$, there exists an $i>0$ such that $H_i(M; k)$ is nontrivial, otherwise $X$ would be contractible by the homology Whitehead theorem (see Corollary 3A.7 of Hatcher).

The Künneth theorem implies $H_{2i}(M \times M;k)$ is nontrivial which implies $M$ can't be homotopy equivalent to $M \times M$.

The simply-connectedness assumption is necessary since the homology could potentially vanish in all degrees.

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    $\begingroup$ So is there a counterexample in the non-simply-connected case? $\endgroup$
    – Zerox
    Jul 25, 2023 at 8:49
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    $\begingroup$ @Zerox If there is, it must be a space with contractible universal cover. The question is then equivalent to: is there a group $G$ which acts properly discontinuously on $\mathbb{R}^n$ such that $G\cong G\times G?$ $\endgroup$ Jul 25, 2023 at 22:20
  • $\begingroup$ I suppose you're using the fact that if $A$ is a non-zero abelian group then either $\mathbb{Q}\otimes A$ or $\mathop{\rm Tor}(\mathbb{Z}/p,A)$ for some $p$ must be non-zero. $\endgroup$ Jul 26, 2023 at 16:07
  • $\begingroup$ @DaveBenson I am using that the tensor product of nontrivial vector spaces is nontrivial. $\endgroup$ Jul 26, 2023 at 19:45
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    $\begingroup$ Exactly what I was saying. You're using the fact that if $A\ne 0$ then either $\mathbb{Q}\otimes A \ne 0$ or $\mathop{\rm Tor}(\mathbb{Z}/p,A)\ne 0$. This is the proof given in Hatcher. $\endgroup$ Jul 26, 2023 at 21:15

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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    $\begingroup$ The lemma must be in a book somewhere, but I don't know where. $\endgroup$ Jul 25, 2023 at 8:43
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    $\begingroup$ Probably the same book where this problem is a homework problem. :) $\endgroup$ Jul 25, 2023 at 20:29
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    $\begingroup$ @DaveBenson This is a nice answer, thank you. $\endgroup$ Jul 26, 2023 at 4:50
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    $\begingroup$ @RyanBudney Thanks for your initial comment. If you do find this problem in a textbook I would be interested in knowing which one. $\endgroup$ Jul 26, 2023 at 5:49
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    $\begingroup$ @JoshLackman: I have not looked in an algebraic topology textbook in some time. That said, I think you've found yourself a pretty textbook problem, if you were to write one. $\endgroup$ Jul 26, 2023 at 17:50

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