## Milnor wins 2011 Abel Prize

I have just completed one of the more difficult assignments of my mathematical life: to give a popular presentation of the work of John Milnor immediately following the formal announcement that he was the winner of this year’s Abel Prize. Of course, in one way the task is very straightforward, since Milnor is a mathematical giant and has a large number of fascinating theorems to his name. However, these theorems are not in my field, the talk was supposed to last fifteen minutes, and my immediate audience included not necessarily mathematical journalists who were supposed to understand what I was saying. If you go to the Abel Prize website, you will find a webcast of the whole announcement, including my talk (which includes a telephone interview with Milnor himself), and also a link to a written version of the talk, in which I go into more detail. But if you are a mathematician, then be warned that even the more detailed version is more about the background to Milnor’s results than to the results themselves. And since I was obliged to prepare the talk in secret, I cannot rule out that some of what I have said is wrong, or gives the wrong emphasis.

As I started to prepare the talk, I realized I had a problem. Although Milnor is famous for many results, the one for which he is best known is his construction of a seven-dimensional “exotic sphere”. This means a differentiable 7-manifold that is homeomorphic to a 7-sphere but not diffeomorphic to a 7-sphere. My problem with this result was that it felt false: given a homeomorphism from the 7-sphere to another differentiable manifold, surely one can just “iron out the kinks” or “smooth off the corners”, or however one wants to put it. This was not the only false-seeming result I had to contend with. Another was Milnor’s famous counterexample to the Hauptvermutung: he found two different triangulations of a triangulable space (that is, a topological space that is homeomorphic to a simplicial complex) that have no common refinement. Later it was shown by Casson and Sullivan that you could do this for manifolds as well. And Freedman discovered a 4-manifold that cannot be triangulated at all. My intuition tells me that you can just scatter a large number of points fairly randomly into a manifold and then join them up to form a triangulation. Certainly, this works in two dimensions, and similar arguments show that any two triangulations have a common refinement.

The best explanation I have at the moment for why you can’t just iron out the non-differentiability is that the nature of the singularities is much more complicated than it is in any situation that one can visualize. (I myself have trouble visualizing non-differentiable maps beyond maps from $\mathbb{R}^2$ to $\mathbb{R}.$) In a higher dimension, you can find that the set of singularities forms a submanifold that you can’t just iron away: you can push the crease from one place to another, but you still have a crease. If something like that is correct, then you would expect that the existence or otherwise of exotic differentiable structures would depend very much on subtle topological properties of the spheres, and this I know to be true: the sequence of numbers of differentiable structures on spheres of various different dimensions starts 1, 1, 1, ?, 1, 1, 28, 2, 8, 6, 992, 1, 3, 2, 16256, 2, 16, 16, 523264, 24 (the question mark being because nobody knows whether there are exotic 4-spheres, or whether, if so, the number of exotic 4-spheres is finite), and these incredible numbers, worked out by Kervaire and Milnor, are, I read, related to homotopy groups of spheres. And I presume something similar is true for triangulations: that there are local difficulties to triangulating a manifold that cause every triangulation to pick out some privileged direction, or orientation, or something topological at any rate, in such a way that one cannot make a continuous choice. If anybody can enlighten me further, I’d be very interested and grateful …

### 22 Responses to “Milnor wins 2011 Abel Prize”

1. Greg Kuperberg Says:

A while ago I heard a good explanation of this business from Shmuel Weinberger; the explanation ultimately comes from important papers by Munkres.

Imagine the task of smoothing a piecewise linear manifold. (Topological manifolds are a decidedly harder starting point.) What do you really do? A PL manifold, since it comes with a triangulation, is already smooth on an open dense set, but at other points it is locally cone-like, like the tip of a sharpened pencil or crinkled aluminum. So the task at hand is to smooth the cone-like points. And to do that, you should:

(1) Work your way down from codimension 1, the facets, to codimension 0, the vertices. After all, most of a k-dimensional face is far away from the (k-1)-skeleton of the manifold, so whatever you do on the lower-dimensional kinks won’t affect what you must do there; on the other hand, what you do to the edges CAN effect what you do to the vertices, etc.

(2) Smooth a conical structure, at a vertex say, by removing a neighborhood of the apex and gluing in a ball smoothly. Or if it is not at a vertex, it is locally the Cartesian product of such an apex and a vector space. This is like smoothing the tip of a pencil; what was a point before becomes a disk that has some small but non-zero area.

(3) Now suppose that everything works as you expected it to up to dimension 6. (The suppose is true, by the way.) And suppose that you have a 7-manifold and you do smooth every part of the triangulation except at the end for the vertices. To smooth a vertex, remove an open neighborhood to leave a 6-sphere boundary; you want to glue in a 7-ball along that 6-sphere. Two different gluings will give the same answer if their composition is a diffeomorphism that extends to the 7-ball. So okay, Diff(S^6) is a group, and it has a subgroup Diff_0(S^6) of diffeomorphisms that extend to the 7-ball. It turns out that they are not the same and the quotient has 28 elements. They could in principle give you 28 different smooth structures on the same PL 7-manifold, and in the case of the 7-sphere, they actually do.

Is it so hard to believe that not every diffeomorphism of S^6 extends to B^7? That which is hard to prove in one case, might well be outright false in another case. Can you prove that every diffeomorphism of a circle extends to a disk? Well, there is a proof that relaxes a diffeomorphism of a circle gradually using the heat equation, but that’s hardly a trivial proof. Can you prove that every diffeomorphism of a 2-sphere extends to a 3-ball? That’s also true, but it’s not easy to prove either. (It’s a theorem of Smale and Munkres.) You have to first accept that this non-trivial work is the real work that you have to do smooth a PL manifold. And second, it is yet harder in higher dimensions — it is really a formidable theorem of Cerf and Hatcher that every diffeomorphism of a 3-sphere extends to a 4-ball — and eventually it’s false.

The non-existence of a smooth structure is a similar story. Imagine Munkres procedure again in 8 dimensions. Without any claims to uniqueness, you can smooth the triangulation somehow down to the vertices. But, wrench in the works, the link of a vertex could be one of Milnor’s exotic 7-spheres. So that is NOT the boundary of a smooth 8-ball (which only has one smooth structure, by the way) and there does not exist a way to blunt that vertex. You might wonder if there is some way to compensate by altering the earlier gluings, but it is sometimes an unavoidable obstruction.

• gowers Says:

Thank you very much for this extremely nice explanation — just the kind of thing I was hoping for.

• David Speyer Says:

Thanks Greg! This is really excellent.

• Greg Kuperberg Says:

You’re both quite welcome, and thanks for the praise! A small correction: I exaggerated the difficulty of proving that every diffeomorphism of a circle extends to a disk. If you consider radius-preserving diffeomorphisms, it suffices to show that every diffeomorphism of the circle is isotopic to an isometry. (In a high enough dimension, not every extension to the ball can be realized by an isotopy leading to a radius-preserving diffeomorphism.) Yes, you could use the heat equation, but a better idea is to convert to a diffeomorphism of an interval and then simply gradually average it with the identity. Still, even though it is a very easy case of the result, you can already see a shadow of some of the difficulties later on.

2. Alon Says:

Two tiny comments on the written version:

– on p.14, the number of non-cancellable words of length n is of course 4.3^(n-1) and not as written.

– A subtler problem (subtle to the point that it’s arguable if it’s a problem at all) is the comment about 6 square roots of -1 in the ring of quaternions. One introduces these roots to define the quaternions, but once this is done we find that -1 actually has infinitely many distinct square roots, not just those 6. It’s possible to misinterpret this paragraph as saying that those 6 are the only square roots of -1.

• gowers Says:

Thanks for those remarks. I have to confess that I had never really thought about the question of how many square roots -1 had in the quaternions (though a similar problem arose in the Princeton Companion, where what I wrote suggested that it had three and someone objected and said what about -i, -j, -k — this is what underlies what I wrote in this article), though as soon as one does think about it it is sort of obvious that a single polynomial equation like that should have infinitely many roots in a three-dimensional space. (At some point I’ll actually do the exercise of calculating what they are.)

As it happens, the version available on their website was already supposed to have been updated to one where I improved the wording in a few places (e.g. not writing “quite exceptional” twice in the first paragraph of Section 2). In the light of your remarks, I have updated it further and asked them to put the new version on the website instead. I hope that will happen soon. (If I can’t get them to do it, then I’ll simply put it up here.)

Edit: they’ve put a corrected version there. In this one I don’t imply so strongly that there are only finitely many square roots of -1 (though the innocent reader might assume this because I don’t explicitly point out that there are more than just i, j and k), and I’ve got my 3 and 4 the right way round …

• gowers Says:

Oops, I wrote something stupid there. The equation $x^2=-1$ is a codimension-three condition rather than a codimension-one condition of course (if we restrict attention to the 3-sphere, that is) so it’s not obvious on dimension grounds. So I presume it’s related to the fact that there are many half turns in SO(2), double covers, etc. I will do the exercise just to lodge this properly where it ought to have been in my brain.

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6. Harald Hanche-Olsen Says:

I am sorry I cannot enlighten you, but at least I can congratulate you with a very nice presentation. I saw it streamed online, as I have seen all of these presentations since the Abel prizes began, and I’d say yours was one of the better ones. Of course it helps a lot that you can talk of spheres and triangulations in a way that the lay person can at least believe they understand – some of the other prize winners must have been much harder to present. But even so, I think it was a job very well done.

7. Richard Séguin Says:

I haven’t yet seen the video, but I have to say that the written version was also a very nice presentation, and it went down well with dinner tonight. I always find well written expository articles like this food for thought even if I’m not working directly in whatever area(s) of mathematics being discussed.

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9. jonas Says:

Let’s continue with a puzzle. Milnor is the third person to have the grand slam: all three of the Abel Prize, the Wolf Prize, and the Fields Medal. Who are the other two?

• dude Says:

Hmm, the Wolf Prize… how likely is it that a mathematician of Arab descent would accept a prize that is presented by the President of the state of Israel?

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