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

Question

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.

Answer 1

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 $M\times M\simeq M$.

Answer 2

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