Timeline for How can there be topological 4-manifolds with no differentiable structure?
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| Jul 22, 2018 at 4:02 | comment | added | Fan Zheng | @NeilStrickland Just for the record, if $M$ is a Jordan curve (i.e., a compact, closed, connected 1-manifold embedded) in $\mathbb R^2$, then the Riemann mapping from the interior of $M$ to the open unit disk extends continuously to $M$. This is en.wikipedia.org/wiki/…, whose proof uses complex analysis and only works in dimension 2 (rightly so, because of the Alexander horned sphere). | |
| Mar 11, 2011 at 9:40 | comment | added | Neil Strickland | I think that's right. I'm no expert on this, but if I recall correctly, it's common to have a non-smoothable manifold where $M\setminus\{x\}$ is smoothable for all $x\in M$; that makes the global nature of the problem quite visible. (However, I have heard that there are very special features in dimension $4$ that I know nothing about.) | |
| Mar 11, 2011 at 9:28 | comment | added | gowers | So if I understand you correctly, the reason there can be such manifolds is not local and analytic (that some manifolds look wild and fractal) but global and topological (that the set of singularities of any reasonably nice smooth structure has to be topologically non-trivial in some suitable sense). Is that correct? | |
| Mar 10, 2011 at 15:41 | comment | added | Neil Strickland | fit together, and it starts to seem plausible that there could be a homotopical obstruction theory. | |
| Mar 10, 2011 at 15:40 | comment | added | Neil Strickland | I'd agree that my answer does not get to the heart of why unsmoothable 4-manifolds exist. However, in thinking about trying to smooth an arbitrary 4-manifold $M$, you should bear in mind that $M$ might originally be given as some horrible fractal subset of $\mathbf{R}^n$. To make it smoother you probably want to just throw away the embedding and start trying to simplify the charts. Of course the chart transition functions could also be fractal, but the Alexander Trick gives you some leverage to deal with that. Now we've got some reasonably tame choices for each chart domain that we need to | |
| Mar 10, 2011 at 15:01 | comment | added | gowers | ... to the manifold to which one is trying to give a smooth structure (a bit like the successive stages in building the Koch snowflake, but smoothed off at the corners). If one could ensure that the sequences you get by starting with a point in the smooth manifold and following its images in the subsequent closer and closer approximations converged uniformly, then might one have got somewhere? This is a new wrong-headed attempt at a proof: what I'd really like is to see why any such attempt is hopeless. (As usual, the mere fact that I'm trying to contradict a known theorem is not enough.) | |
| Mar 10, 2011 at 14:53 | comment | added | gowers | Is it the case that the examples of topological 4-manifolds that do not have smooth structures are very fractal like this? In other words, is what you write at least the kind of reason that such manifolds exist? I ask simply because although your example certainly makes it clear that my "sketch proof" doesn't work, it isn't hard to give a smooth structure to the Koch snowflake, so I don't quite lose the feeling that there might be an elementary proof. For instance, perhaps one could pick a sequence of smooth manifolds, all homeomorphic and each one getting closer and closer ... | |
| Mar 10, 2011 at 12:54 | history | answered | Neil Strickland | CC BY-SA 2.5 |