# Can a simply connected manifold satisfy $𝑀\simeq 𝑀\times 𝑀$?

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.)

• 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. Jul 24, 2023 at 22:00

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.

• So is there a counterexample in the non-simply-connected case? Jul 25, 2023 at 8:49
• @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 $G\cong G\times G?$ Jul 25, 2023 at 22:20
• 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. Jul 26, 2023 at 16:07
• @DaveBenson I am using that the tensor product of nontrivial vector spaces is nontrivial. Jul 26, 2023 at 19:45
• 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. 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 $$M\times M\simeq M$$.

• The lemma must be in a book somewhere, but I don't know where. Jul 25, 2023 at 8:43
• Probably the same book where this problem is a homework problem. :) Jul 25, 2023 at 20:29
• @DaveBenson This is a nice answer, thank you. Jul 26, 2023 at 4:50
• @RyanBudney Thanks for your initial comment. If you do find this problem in a textbook I would be interested in knowing which one. Jul 26, 2023 at 5:49
• @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. Jul 26, 2023 at 17:50