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Let $M$ be a compact orientable manifold which is homeomorphic to its connected sum with itself $M\# M$. Must $M$ be homeomorphic to a sphere?

I will explain why I am interested (at the risk of being outright foolish or overambitious). The 'equation' $M = M\#M$ is the simplest conceivable equation in the category of compact topological manifolds, just as $x+x=x$ characterises $0$ in an abelian group. Although I am suspicious if $M$ could be a homology sphere, I hope to get a characterisation of spheres.

Thanks for any comments or suggested references.

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Let us suppose that $dim(M)\geq 3$ then we have that:

  • $\pi_1(M \# M)\cong \pi_1(M)*\pi_1(M)$,
  • $H_*(M;\mathbb{Z})\cong H_*(M;\mathbb{Z})\oplus H_*(M;\mathbb{Z})$ when $*< dim(M)$.

As $M$ is compact this implies that $\pi_1(M)$ is finitely presented thus that $M$ is simply connected (*). Together with the second point it implies that $M$ has the homotopy type of a sphere. And you conclude using the Poincaré conjecture that $M$ is homeomorphic to a sphere.

Edit (*): we have to use a non-trivial theorem in group theory about the minimal number of generators $m(G)$ of a group $G$. Namely we have $m(G*H)=m(G)+ m(H)$ (conjectured by Levi and proved independently by Gruschko and Neumann).

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  • $\begingroup$ Thank you very much for the answer. I would like to know if there is an analogue of the first relation (Van Kampen theorem) you mentioned for higher homotopy groups. Such an analogue might help us solve the 'equation' M = X#Y for compact manifolds. $\endgroup$ – Chaitanya Oct 31 '17 at 12:44
  • $\begingroup$ @Chaitanya: your follow-up question was asked at mathoverflow.net/questions/93282/… $\endgroup$ – Mark Grant Oct 31 '17 at 13:08
  • $\begingroup$ Is there a more elementary solution which avoid the Poincare conjecture? $\endgroup$ – A. Chu Nov 5 '17 at 10:16
  • $\begingroup$ Jason, I am not aware of such a proof. I will be very happy to learn a more elementary proof. $\endgroup$ – David C Nov 5 '17 at 10:41

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