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This is a very naive question, and I'm hoping that it will be matched by a correspondingly elementary answer. It is well known that not every topological 4-manifold admits a smooth structure. So what's wrong with the following very sketchy proof that, actually, a topological 4-manifold does admit a smooth structure (apart from the sketchiness)?

Step 1: Embed the manifold into $\mathbb{R}^9$, which, as I understand it, can be done.

Step 2: "Iron out the kinks" in the embedded manifold.

Step 3: Once the embedded manifold looks nice enough, give it an obvious smooth structure coming from $\mathbb{R}^9$.

The second step looks the dodgiest to me, because my intuition comes from cases that are presumably much too special, such as a 2-dimensional manifold sitting in $\mathbb{R}^3$. Take, for instance, the surface of a cube. We can easily smooth off the corners and edges and obtain a smooth manifold. There are many smoothing methods around (such as convolving with nice objects). So why can't we find one that works in general?

When I try to think how I would actually go about it, then I do of course run into difficulties. For instance, in the cube case I could take all points outside the cube of some fixed small distance from the cube. That would give me a smoother version. But if I try a trick like that when the codimension is not 1, then I get a set of the wrong dimension. That suggests that I have to make a clever choice of direction, and I don't see an obvious way of doing that.

I have similar questions about other wacky (as they seem to me) facts about manifolds, such as the existence of topological manifolds that cannot be given piecewise linear triangulations. I'm not looking for an insight into why such results are true. All I want to understand is why they are not obviously false. Can anyone say anything that might be helpful?

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You could apply the same argument to topologically slice knots that aren't smoothly slice. There's no way to iron out the kinks, so you can't get to step 3. A topologically slice knot in $S^3$ bounds a tame topologically embedded $D^2$ in $D^4$, where "tame" means it has a tubular neighbourhood -- the disc has a neighbourhood in $D^4$ that is homeomorphic to $D^2 \times D^2$. Things like Whitehead doubles are topologically slice but generally not smoothly slice. –  Ryan Budney Mar 10 '11 at 15:47
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From the chart formulation of manifolds, what this boils down to is that humans have a very difficult time imagining homeomorphisms. Most homeomorphisms that people produce are either diffeomorphisms or piecewise diffeomorphisms, or they have such elementary singularities they're readily smoothed. This is why the smooth category is so attractive -- objects are far more constructible and readily imagined. –  Ryan Budney Mar 10 '11 at 15:50
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It's not clear to me that the kind of answer you're looking for would single out dimension 4. –  Qiaochu Yuan Mar 10 '11 at 17:00
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Even harder to understand is that there may exist manifolds $M$ that cannot be triangulated as a simplicial complex (not necessarily PL). For every point $x\in M$ choose a small neighborhood $N(x)$. By compactness, finitely many of these neighborhoods cover $M$. One would think that by choosing the neighborhoods small enough you could get intersections of these neighborhoods to be nice enough to build up a triangulation of $M$, but this is open. See mathoverflow.net/questions/44021/…. –  Richard Stanley Mar 11 '11 at 2:05
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So this question was a subtle hint about the Abel Prize? –  Maxime Bourrigan Mar 24 '11 at 13:58

4 Answers 4

  1. The usual convolution method for approximating continuous maps by smooth maps does not succeed in approximating invertible [resp. injective] continuous maps by invertible [resp. injective] smooth maps.

  2. (referring now to the "new wrong-headed attempt" in a comment at Neil's thread) A sequence of diffeomorphisms or smooth embeddings may converge uniformly to a non-smooth homeomorphism or non-smooth topological embedding.

  3. A topological $n$-manifold has its tangent microbundle. This is essentially a fiber bundle whose fibers are open $n$-disks. There is no chance of smoothing the manifold unless one can give a vector bundle structure to this topological disk bundle. The structure group for microbundles, or open disk bundles, called $Top_n$, is the topological group of homeomorphisms from $\mathbb R^n$ to itself, or germs of such at the origin. And (this has to do with 1 above) it really has a different homotopy type from $O_n$ for most values of $n$.

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My understanding of part (3) is that the difference between Top_4 and O_4 lives in pi_3. So that means (I think) that there should be a continuous automorphism of S^3 \times S^3, commuting with projection to the second factor, which does not deform to a smooth family of orthogonal symmetries. Is it possible to actually visualize this? –  David Speyer Mar 10 '11 at 16:46
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$Homeo(S^3)$ is the structure group for topological bundles of closed $4$-disks, which is in fact equivalent to $O_4$ (I believe); whereas $Top_4$ is the structure group for topological bundles of open $4$-disks. –  Tom Goodwillie Mar 10 '11 at 16:57
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I'm awfully ignorant in this: what would be a good reference to read more about part (3)? –  Willie Wong Mar 10 '11 at 17:11
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@willie : there's a paper by Milnor Milnor, J. Microbundles. I. Topology 3 1964 suppl. 1, 53–80. 57.30 (57.20) Usually when Milnor explains, things are easy... –  Pierre Mar 10 '11 at 19:36

Let's try doing this with a compact, closed, connected $1$-manifold $M$. Certainly I can choose a topological embedding $f:M\to\mathbb{R}^2$. However, the image might be very fractal, like a Koch snowflake for example. If I try to take a short piece of the image and straighten it out by projecting in some direction, the fractalness will ensure that I lose injectivity. If I consider all points at a fixed distance $\epsilon$ from $f(M)$ and try to retract back to $f(M)$, the same phenomenon will mean that the retraction is not well-defined.

If I instead try to smooth it out by convolving, then it seems that I need a convolution kernel $u:M\times M\to\mathbb{R}$ that is supported near the diagonal, together with a measure on $M$, so I can define $g(x)=\int u(x,y)f(y)dx$. If $M$ already has a smooth structure and $u$ and the measure are smooth, then $g$ will be smooth and close to $f$. It isn't obviously injective but perhaps that could be arranged. However, I can't see anything like this that you could do if $M$ did not already have a smooth structure.

In this two-dimensional case the Riemann mapping theorem gives a nearly unique conformal isomorphism $h$ from the open unit disc to the bounded component of $\mathbb{R}^2\setminus f(M)$. You could try to define $h_1:S^1\to f(M)$ by $h_1(z)=\lim_{t\to 1}h(tz)$. If I recall correctly, there are people in Finland who have thought about this a lot and decided that $h_1$ can be arbitrarily bad.

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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 ... –  gowers Mar 10 '11 at 14:53
    
... 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.) –  gowers Mar 10 '11 at 15:01
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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 –  Neil Strickland Mar 10 '11 at 15:40
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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? –  gowers Mar 11 '11 at 9:28
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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.) –  Neil Strickland Mar 11 '11 at 9:40

Alexander's horned sphere is a topological sphere in 3-space that cannot be "ironed out", otherwise we would get a smooth (or PL) 2-sphere having a complementary region which is not simply-connected, a fact which is excluded because every smooth (or PL) 2-sphere in 3-space is standard.

Another simple topological object that cannot be smoothly ironed is a 2-dimensional disc inside $D^4$, obtained by coning over a knot in $S^3$.

Maybe such pathological embeddings can be excluded a priori from the topological Whitney-embedding theorem, so it might be that these are not really an issue here.

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I am not sure how relevant it is, but here is something that made me better understand the possible problems when one tries to smooth out an embedded topological manifold.

Take your favorite knot $K$ in $S^3$ (or, if you are minimalist, take your second favorite knot so that it is not trivial), view $S^3$ as an equator in $S^4$, and consider the cone over $K$, based on the south pole of $S^4$ (using geodesics of the round metric, say). You get what is called a non-tame embedding of a disc in $S^4$. Now it is intuitive that if you deform continuously the embedding, then you will deform continuously its trace on a small sphere centered at the south pole (called, by Milnor I think, the link of the embedding at the point), which starts from a copy of $K$. So any close embedding must also be non-tame, while if it where differentiable it would have a trivial knot as its link.

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I agree with you that after thinking about that example it becomes clearer that being able to smooth things out is not obvious. –  gowers Mar 24 '11 at 9:53

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