# How to “see” that double suspension of homology 3-sphere is homeomorphic to a sphere?

Is there a good way to think about/understand the result that the double suspension of a homology 3-sphere is homeomorphic to a sphere, to get intuition for why this is true? For instance, what sort of neighborhood must one take for one of the suspension vertices in the $S^0$ in order to see that a neighborhood of this point is homeomorphic to ${\mathbb R}^5$? Thanks!!

I found the following related MathOverflow question, which raises relevant points and is interesting, but doesn't seem to answer my question:

"If a manifold suspends to a sphere..."

My posting of this question today was partly inspired by thinking about the comments to the very recent MO question:

"Is a finite CW complex minus a point still homotopy equivalent to a finite CW complex?"

At some point, I asked the question I'm now posting to a topologist who is known for his incredible intuition, and he remarked that he believed it was better to think in terms of taking a join with $S^1$ rather than repeatedly taking a join with $S^0$, but he wasn't sure what else to say.

Thanks again for any help with this!

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The motivation is to have a local homotopy theoretic characterization of manifolds. All you need to check is the link of points. This doesn't answer the question of why the link of curves is not topologically definable. See my answer, mainly the third paragraph: mathoverflow.net/questions/67458/… –  Ben Wieland Jun 13 '12 at 17:12
Ok, thanks for the comment and the pointer! I'm just learning this stuff. –  Patricia Hersh Jun 13 '12 at 17:41

## 1 Answer

To get some intuition behind the double suspension theorem, you can try three pages with a lot of pictures in Ferry's notes (starting with p.166, in Chapter 26). He gives a rough sketch of proof in the case of one particular homology sphere, the one for which the theorem was first proved by Edwards.

The double suspension theorem boils down to showing that a certain non-manifold (namely, the single suspension over a homology sphere) becomes a manifold when multiplied by $\Bbb R$. When Milnor conjectured the double suspension theorem in early 60s, he must have been aware of the existence of other non-manifolds with this property (found earlier by Bing). It is fortunate that they also exist in a lower dimension so it's easier to visualize what's going on. One such example is $(S^3/W)\times\Bbb R\cong S^3\times\Bbb R$, where $W$ is the Whitehead continuum, and there is a rather explicit construction of this homeomorphism in Ferry's Chapter 4 (p.15).

Added later: Another example is $(M/D)\times\Bbb R\cong M\times\Bbb R$, where $D$ is a wild copy of the $n$-disk contained in the interior of the manifold $M$. The case $n=2$ is actually used the above-mentioned proof of the double suspension theorem in Ferry's notes, but is not proved there; a proof of the case $n=1$ with some pictures can be found in the Daverman-Venema book, Section 2.6.

To address the specific question about neighborhoods, the open star (in the original double-suspension triangulation) of any vertex in the suspension circle is homeomorphic to $\Bbb R^5$ $-$ at least in the case of one particular homology sphere, the boundary of the Mazur manifold $W$. Indeed, the closed star of this vertex is the suspension over $cone(\partial W)$. As explained in Ferry's notes, $cone(\partial W)\times\Bbb R$ is homeomorphic to $W\times\Bbb R$. Hence the open star of the vertex is homeomorphic to $(W\setminus\partial W)\times\Bbb R$. The latter can be identified with the interior of $W\times I$. But it is easy to see that $W\times I$ is homeomorphic to the $5$-ball. So its interior is homeomorphic to $\Bbb R^5$.

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Sergey, thank you for yet another helpful answer to one of my questions!! I imagine that (hopefully) once I study these examples enough, I should be able to answer for myself the above question about what sort of neighborhood will be homeomorphic to ${\mathbb R}^5$. Unlike in my earlier question, I plan to wait like a week this time before accepting any answer as "the answer", just in case people here do have different ways of thinking about this. –  Patricia Hersh Jun 13 '12 at 21:02
Patricia, thank you for your thoughtful questions! I now added some info on a specific euclidean neighborhood. I'm afraid I'm not proficient enough in these matters to formulate a clear motto on why the theorem should hold. –  Sergey Melikhov Jun 13 '12 at 23:52
Thanks again, Sergey! Now I've had some time to read part of the Ferry notes, etc. Your answer is extremely helpful! –  Patricia Hersh Jun 21 '12 at 10:41