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For what manifolds $M$ is the suspension $\Sigma M$ also a manifold?

By the suspension of a topological space $X$ (not necessarily a manifold), I mean the space $$\Sigma X = (X \times [0,1])/{\sim}$$ where $\sim$ is the equivalence relation which glues $X \times \{0\}$ to a point $p$ and $X \times \{1\}$ to a point $q$. The topology is of course given by the quotient topology.

If $M = S^n$ is a sphere, then it is easy to see that $\Sigma M$ is homeomorphic to $S^{n+1}$, and hence a manifold.

Claim: The only possibility is that $M$ is a sphere.

I can prove this statement, but it relies on some pretty hefty results. So my question is the following:

Is there an easier proof than the following? In particular, can we remove the use of the double suspension theorem or the $h$-cobordism theorem / Poincaré Conjecture?

Proof: Suppose $M$ and $\Sigma M$ are manifolds of dimensions $n$ and $n+1$ respectively. This is obviously true locally everywhere except at $p$ and $q$, so we need only consider that some neighbourhood of $p$ is homeomorphic to a disk (by symmetry, this implies the same result for $q$).

Pick a neighbourhood $U = (M\times [0,\epsilon))/{\sim}$ of $p$. By excision and since $\Sigma M$ is a manifold, we have $$\widetilde{H_*}(S^{n+1}) \cong H_*(\Sigma M, \Sigma M \setminus \{p\}) \cong H_*(U,U \setminus \{p\}).$$ Now, $U \setminus \{p\} = M \times (0,\epsilon)$ deformation retracts to $M$, and so we can compute this by taking the LES for the pair $(U,M)$. Since $U$ is contractible, every third term of this sequence (except along the $0^{\mathrm{th}}$ cohomology) is $0$, and we easily can conclude that $$H_{k+1}(\Sigma M, \Sigma M \setminus \{p\}) \cong H_k(M)$$ for $k > 0$. Therefore, $H_k(M) = H_{k+1}(S^{n+1})$ for $k > 0$. And we know $H_0$ is just $\mathbb{Z}$ (since $H_n \cong \mathbb{Z}$, Poincaré duality implies $H^0 \cong \mathbb{Z}$ and so there is only one component). In particular, this proves that $M$ is a homology sphere.

That's all relatively standard, and all well and good. Here's the part that gets a little bit crazy. Since $M$ is a homology sphere, the double suspension theorem yields that $\Sigma \Sigma M$ is a sphere. So we need that $\Sigma M$ suspends to a sphere. By this MO question, the answer of which appears to rely either on $h$-cobordism or the Poincaré Conjecture (choose your poison), it follows that $\Sigma M$ is therefore a sphere, and applying this same reasoning to $\Sigma M$, it follows that $M$ is a sphere.

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KSackel
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  • No, there's something a bit more subtle in that last paragraph. In fact, if a homology sphere is not a sphere, then $\Sigma M$ is not a manifold, and so we cannot apply the reasoning of the last paragraph. (See the first comments on the MO question.) The point is that we needed $\Sigma M$ to be a manifold to conclude that it was a sphere, and then for $M$ after that. – KSackel May 07 '14 at 13:35
  • Yes, you're right, I realized afterwards. Anyway since MO is supposed to be "higher level" than MSE, you might get better answers there, considering you quote results from an MO question... – Najib Idrissi May 07 '14 at 13:36
  • I was more curious if there was some more elementary solution. I figured MSE would be a better place to start for that, but I suppose it could be moved later. – KSackel May 07 '14 at 13:37
  • I changed (X\times[0,1])/\sim to (X\times[0,1])/{\sim}. Sincee "\sim" is used as a binary relation symbol like "=" or "<", it has preceding and following spaces that should not be there when it's not used as a binary relation symbol. Thus: $(X\times[0,1])/\sim$ versus $(X\times[0,1])/{\sim}$. ${}\qquad{}$ – Michael Hardy May 07 '14 at 15:50
  • Cheers! TeXing was never my strong point. – KSackel May 07 '14 at 16:44
  • Could you perhaps say that a neighborhood of $p$ or $q$ is a cone on $X$ and homeomorphic to a disk, and that a homeomorphism of manifolds with boundary restricts to a homeomorphism of the boundaries? – Kevin Carlson May 27 '14 at 10:14
  • I'm not quite sure I see it. Although the cones on $X$ do form a neighbourhood basis for the topology around $p$ and $q$, it is not obvious to me that this means they are necessarily homeomorphic to a disk. Maybe a priori $\Sigma X$ could be a manifold, and a disk around $p$ or $q$ looks kind of funky in the $X \times [0,1]$? – KSackel Jun 02 '14 at 23:31

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You don't need the double suspension theorem, because $M$ is a homotopy sphere (hence, by the Poincaré conjecture, it is homeomorphic to the sphere). To prove that, it is enough to show that $M$ is simply-connected (assume that $\dim M \geq 2$ as otherwise there is nothing to prove). Then it follows from your computation of homology and the Whitehead theorem that a degree one map $M \rightarrow S^n$ is a homotopy equivalence.

The suspension $\Sigma M$ is simply-connected by the van Kampen theorem, because it is a union of two contractible open cones over $M$, whose intersection $M \times (-\epsilon,\epsilon)$ is connected. Since $\Sigma M$ is a manifold of dimension at least $3$, removing a finite number of points from $\Sigma M$ doesn't change its fundamental group (again, this is a consequence of the van Kampen theorem and the fact that $S^k$ is simply-connected for $k \geq 2$). In particular, $\Sigma M \setminus \{p,q\} = M \times (0,1)$ is simply-connected, and so is $M$.

Aleksander Doan
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