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 $G \times G$?](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.


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



Can a simply connected manifold satisfy $\simeq \times $?
Thanks to [Dave Benson](https://mathoverflow.net/questions/451401/can-a-simply-connected-manifold-satisfy-simeq-%c3%97#comment1167394_451420) 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](https://pi.math.cornell.edu/%7Ehatcher/AT/AT.pdf)).


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.


