I was in a talk and the speaker mentioned that it is easy to write down (real or complex?) polynomials whose zero loci are contractible connected manifolds but not homeomorphic to $\mathbb{R}^n$. Can someone give me some examples with proof of contractibility?
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2 Answers
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Here is what I know about the question which leads to the conclusion that while examples mentioned by the speaker do exist, they are by no means easy; I do not know any explicit examples (it does not mean that there aren't any).
Definition. A topological manifold $M$ is called tame if $M$ is homeomorphic to the interior of a compact manifold $N$ with boundary.
The Whitehead manifold (mentioned in Thomas Rot's answer) is the first (and the most famous) example of a non-tame contractible manifold. The relevance of this definition to real algebraic sets (sets given by polynomial equations in $R^k$ for some $k$) is that every real algebraic manifold is tame. Already this is not easy, it follows from the Lojasiewicz's theorem that every compact real algebraic set admits a triangulation.
Definition. A tame manifold $M$ is simply connected at infinity if it admits a compactification $M\subset N$ as above such that $\partial N$ is simply connected.
This is not the standard definition (which is a bit tricky, I can give one if you like), but one can show it to be equivalent to the standard one; in particular, simple connectivity at infinity (which is defined not only for tame manifolds) is independent of the compactification. The following theorem is worth a triple of Fields medals (Smale, Freedman and Perelman):
Theorem. (J. Stallings, M. Freedman, G. Perelman) A contractible $n$-dimensional manifold is homeomorphic to $R^n$ if and only if it is simply connected at infinity.
Thus, in order to construct examples mentioned by the speakers, one is looking for smooth compact contractible manifolds with boundary $N$ such that $\partial N$ is not simply connected. Such manifolds $N$ do not exist in dimensions $\le 3$ (I can explain why if you like), but exist in all dimensions $n\ge 4$. (Every smooth homology $n-1$-sphere bounds a smooth compact contractible manifold, as long as $n\ge 5$; in dimension $n=4$ the examples are called Mazur manifolds, they were first constructed independently by Mazur and Poenaru around 1960.) But you need more than that: you want examples where $int(N)$ is algebraic. Akbulut and King in:
S. Akbulut and H. King, The topology of real algebraic sets with isolated singularities, Annals of Math. 113 (1981) 425-446.
proved that the interior of any smooth compact manifold (with boundary) is diffeomorphic to a nonsingular real algebraic subset of some $R^k$.
By putting it all together we obtain that for every $n\ge 4$ there exist contractible nonsingular real algebraic sets which are smooth $n$-dimensional manifolds not homeomorphic to $R^n$. On the other hand, such examples do not exist for $n\le 3$.
answered Oct 28, 2016 at 18:16
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$\begingroup$ Would Michael Davis's manifolds be an example of smooth, contractible manifolds that are not simply connected at infinity? people.math.osu.edu/davis.12/old_papers/aspherical87.dmj.pdf I'm told they're fairly explicitly constructed, but I've (still) never sat down to go through their construction $\endgroup$ Nov 29, 2016 at 4:27
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$\begingroup$ @JeffreyRolland I know these examples well, they are not tame and hence useless for this question. $\endgroup$ Nov 29, 2016 at 9:14
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$\begingroup$ @JeffreyRolland Another thing: Finding explicit tame contractible manifolds is not the real problem, the issue is finding explicit algebraic equations. $\endgroup$ Nov 29, 2016 at 9:23
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$\begingroup$ @MoisheCohen OK, thanks; sorry I wasn't more helpful. Thanks for helping the OP as well $\endgroup$ Nov 30, 2016 at 22:34
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1$\begingroup$ @Dino: This is nontrivial even in the case $n=2$: Every simply-connected noncompact manifold is diffeomorphic to $R^2$. In this case $n=3$ this follows from the Poincare Conjecture (Perelman's theorem): Every tame open contractible 3-manifold is diffeomorphic to $R^3$. The reason is that tameness implies that the 1-point compactification is a simply-connected manifold. $\endgroup$ Mar 7, 2019 at 16:13
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You are probably looking for the Whitehead manifold. I don't know of an explicit polynomial whose zero locus is the Whitehead manifold, but I'm (edit NOT)-sure that it exists.
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3$\begingroup$ The complement of the Whitehead manifold is an infinite union of disjoint connected sets. But by o-minimality of the theory of real-closed fields, if the manifold were algebraic (i.e., the zero locus of a polynomial), its complement would be a finite cell complex. $\endgroup$ Oct 26, 2016 at 19:26
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$\begingroup$ @RobArthan: Thanks, this is a lot of stuff that I know nothing about. $\endgroup$ Oct 26, 2016 at 19:45
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$\begingroup$ Did you appreciate that it means your polynomial does not exist? $\endgroup$ Oct 27, 2016 at 0:28
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2$\begingroup$ @MoisheCohen: Your very informative answer confirms my suspicion that the Whitehead manifold is not homeomorphic to an algebraic set, doesn't it? So the polynomial originally conjectured by Thomas Rot in the above answer does not exist. $\endgroup$ Oct 28, 2016 at 19:37
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1$\begingroup$ @RobArthan: Of course, this follows from Lojasiewicz's theorem (a very nontrivial result, by the way). $\endgroup$ Oct 28, 2016 at 19:53