ICM2014 — Ian Agol plenary lecture

On the second day of the congress I hauled myself out of bed in time, I hoped, to have a shower and find some breakfast before the first plenary lecture of the congress started at 9am. The previous day in the evening I had chanced upon a large underground shopping mall directly underneath the conference centre, so I thought I’d see if I could find some kind of café there. However, at 8:30 in the morning it was more or less deserted, and I found myself wandering down very long empty passages, constantly looking at my watch and worrying that I wouldn’t have time to retrace my steps, find somewhere I could have breakfast, have breakfast, and walk the surprisingly long distance it would be to the main hall, all by 9am.

Eventually I just made it, by going back to a place that was semi-above ground (meaning that it was below ground but you entered it a sunken area that was not covered by a roof) that I had earlier rejected on the grounds that it didn’t have a satisfactory food option, and just had an espresso. Thus fortified, I made my way to the talk and arrived just in time, which didn’t stop me getting a seat near the front. That was to be the case at all talks — if I marched to the front, I could get a seat. I think part of the reason was that there were “Reserved” stickers on several seats, which had been there for the opening ceremony and not been removed. But maybe it was also because some people like to sit some way back so that they can zone out of the talk if they want to, maybe even getting out their laptops. (However, although wireless was in theory available throughout the conference centre, in practice it was very hard to connect.)

The first talk was by Ian Agol. I was told before the talk that I would be unlikely to understand it — the comment was about Agol rather than about me — and the result of this lowering of my expectations was that I enjoyed the talk. In fact, I might even have enjoyed it without the lowering of expectations. Having said that, I did hear one criticism afterwards that I will try to explain, since it provides a good introduction to the content of the lecture.

When I first heard of Thurston’s famous geometrization conjecture, I thought of it as the ultimate aim of the study of 3-manifolds: what more could you want than a complete classification? However, this view was not correct. Although a proof of the geometrization conjecture would be (and later was) a massive step forward, it wouldn’t by itself answer all the questions that people really wanted to answer about 3-manifolds. But some very important work by Agol and others since Perelman’s breakthrough has, in some sense that I don’t understand, finished off some big programme in the subject. The criticism I heard was that Agol didn’t really explain what this programme was. I hadn’t really noticed that as a problem during the talk — I just took it on trust that the work Agol was describing was considered very important by the experts (and I was well aware of Agol’s reputation) — but perhaps he could have done a little more scene setting.

What he actually did by way of introduction was to mention two questions from a famous 1982 paper of Thurston (Three-dimensional manifolds, Kleinian groups and hyperbolic geometry) in which he asked 24 questions. The ones Agol mentioned were questions 16-18. I’ve just had a look at the Thurston paper, and it’s well worth a browse, as it’s a relatively gentle survey written for the Bulletin of the AMS. It also has lots of nice pictures. I didn’t get a sense from my skim through it that questions 16-18 were significantly more important than the others (apart from the geometrization conjecture), but perhaps the story is that when the dust had settled after Perelman’s work, it was those questions that were still hard. Maybe someone who knows what they’re talking about can give a better explanation in a comment.

One definition I learned from the lecture is this: a 3-manifold is said to have a property P virtually if it has a finite-sheeted cover with property P. I presume that a finite-sheeted cover is another 3-manifold and a suitable surjection to the first one such that each point in the first has k preimages for some finite k (that doesn’t depend on the point).

Thurston’s question 16 asks whether every aspherical 3-manifold (I presume that just means that it isn’t a 3-sphere) is virtually Haken.

A little later in the talk, Agol told us what “Haken” meant, other than being the name of a very well-known mathematician. Here’s the definition he gave, which left me with very little intuitive understanding of the concept. A compact 3-manifold with hyperbolic interior is Haken if it contains an embedded \pi_1-injective surface. An example, if my understanding of my rapidly scrawled notes is correct, is a knot complement, one of the standard ways of constructing interesting 3-manifolds. If you take the complement of a knot in \mathbb{R}^3 you get a 3-manifold, and if you take a tubular neighbourhood of that knot, then its boundary will be your \pi_1-injective surface. (I’m only pretending to know what \pi_1-injective means here.)

Thurston, in the paper mentioned earlier, describes Haken manifolds in a different, and for me more helpful, way. Let me approach the concept in top-down fashion: that is, I’ll define it in terms of other mysterious concepts, then work backwards through Thurston’s paper until everything is defined (to my satisfaction at least).

Thurston writes, “A 3-manifold M^3 is called a Haken manifold if it is prime and it contains a 2-sided incompressible surface (whose boundary, if any, is on \partial M) which is not a 2-sphere.”

Incidentally, one thing I picked up during Agol’s talk is that it seems to be conventional to refer to a 3-manifold as M^3 the first time you mention it and as M thereafter.

Now we need to know what “prime” and “incompressible” mean. The following paragraph of Thurston defines “prime” very nicely.

The decomposition referred to really has two stages. The first stage is the prime decomposition, obtained by repeatedly cutting a 3-manifold M^3 along 2-spheres embedded in M^3 so that they separate the manifold into two parts neither of which is a 3-ball, and then gluing 3-balls to the resulting boundary components, thus obtaining closed 3-manifolds which are “simpler”. Kneser proved that this process terminates after a finite number of steps. The resulting pieces, called the prime summands of M^3 , are uniquely determined by M^3 up to homeomorphism.

Hmm, perhaps the rule is more general: you refer to it as M^3 to start with and after that it’s sort of up to you whether you want to call it M^3 or M.

The equivalent process in two dimensions could be used to simplify a two-holed torus. You first identify a circle that cuts it into two pieces and doesn’t bound a disc: basically what you get if you chop the surface into two with one hole on each side. Then you have two surfaces with circles as boundaries. You fill in those circles with discs and then you have two tori. At this point you can’t chop the surface in two in a non-trivial way, so a torus is prime. Unless my intuition is all wrong, that’s more or less telling us that the prime decomposition of an arbitrary orientable surface (without boundary) is into tori, one for each hole, except that the sphere would be prime.

What about “incompressible”? Thurston offers us this.

A surface N^2 embedded in a 3-manifold M^3 is two-sided if N^2 cuts a regular neighborhood of N^2 into two pieces, i.e., the normal bundle to N^2 is oriented. Since we are assuming that M^3 is oriented, this is equivalent to the condition that N^2 is oriented. A two-sided surface is incompressible if every simple curve on N^2 which bounds a disk in M^3 with interior disjoint from N^2 also bounds a disk on N^2.

I think we can forget the first part there: just assume that everything in sight is oriented. Let’s try to think what it would mean for an embedded surface not to be incompressible. Consider for example a copy of the torus embedded in the 3-sphere. Then a loop that goes round the torus bounds a disc in the 3-sphere with no problem, but it doesn’t bound a disc in the torus. So that torus fails to be incompressible. But suppose we embedded the torus into a 3-dimensional torus in a natural way, by taking the 3D torus to be the quotient of \mathbb{R}^3 by \mathbb{Z}^3 and the 2D torus to be the set of all points with x-coordinate an (equivalence class of an) integer. Then the loops that don’t bound discs in the 2-torus don’t bound discs in the 3-torus either, so that surface is — again if what seems likely to be true actually is true — incompressible. It seems that an incompressible surface sort of spans the 3-manifold in an essential way rather than sitting inside a boring part of the 3-manifold and pretending that it isn’t boring.

OK, that’s what Haken manifolds are, but for the non-expert that’s not enough. We want to know why we should care about them. Thurston gives us an answer to this too. Here is a very useful paragraph about them.

It is hard to say how general the class of Haken manifolds is. There are many closed manifolds which are Haken and many which are not. Haken manifolds can be analyzed by inductive processes, because as Haken proved, a Haken manifold can be cut successively along incompressible surfaces until one is left with a collection of 3-balls. The condition that a 3-manifold has an incompressible surface is useful in proving that it has a hyperbolic structure (when it does), but intuitively it really seems to have little to do with the question of existence of a hyperbolic structure.

To put it more vaguely, Haken manifolds are good because they can be chopped into pieces in a way that makes them easy to understand. So I’d guess that the importance of showing that every aspherical 3-manifold is virtually Haken is that finite-sheeted coverings are sufficiently nice that even knowing that a manifold is virtually Haken means that in some sense you understand it.

One very nice thing Agol did was give us some basic examples of 3-manifolds, by which I mean not things like the 3-sphere, but examples of the kind that one wouldn’t immediately think of and that improve one’s intuition about what a typical 3-manifold looks like.

The first one was a (solid) dodecahedron with opposite faces identified — with a twist. I meant the word “twist” literally, but I suppose you could say that the twist is that there is a twist, meaning that given two opposite faces, you don’t identify each vertex with the one opposite it, but rather you first rotate one of the faces through 2\pi/5 and then identify opposite vertices. (Obviously you’ll have to do that in a consistent way somehow.)

There are some questions here that I can’t answer in my head. For example, if you take a vertex of the dodecahedron, then it belongs to three faces. Each of these faces is identified in a twisty way with the opposite face, so if we want to understand what’s going on near the vertex, then we should glue three more dodecahedra to our original one at those faces, keeping track of the various identifications. Now do the identifications mean that those dodecahedra all join up nicely so that the point is at the intersection of four copies of the dodecahedron? Or do we have to do some more gluing before everything starts to join together? One thing we don’t have to worry about is that there isn’t room for all those dodecahedra, which in a certain sense would be the case if the solid angle at a vertex is greater than 1. (I’m defining, I hope standardly, the solid angle of a cone to be the size of the intersection of that cone with a unit sphere centred at the apex, or whatever one calls it. Since a unit sphere has surface area 4\pi, the largest possible solid angle is 4\pi.)

Anyhow, as I said, this doesn’t matter. Indeed, far from mattering, it is to be positively welcomed, since if the solid angles of the dodecahedra that meet at a point add up to more than 4\pi, then it indicates that the geometry of the resulting manifold will be hyperbolic, which is exactly what we want. I presume that another way of defining the example is to start with a tiling of hyperbolic 3-space by regular dodecahedra and then identify neighbouring dodecahedra using little twists. I’m guessing here, but opposite faces of a dodecahedron are parallel, while not being translates of one another. So maybe as you come out of a face, you give it the smallest (anticlockwise, say) twist you can to make it a translate of the opposite face, which will be a rotation by an angle of \pi/5, and then re-enter the opposite face by the corresponding translated point. But it’s not clear to me that that is a consistent definition. (I haven’t said which dodecahedral tiling I’m even taking. Perhaps the one where all the pentagons have right angles at their vertices.)

The other example was actually a pair of examples. One was a figure-of-eight-knot complement, and the other was the complement of the Whitehead link. Agol showed us drawings of the knot and link: I’ll leave you to Google for them if you are interested.

How does a knot complement give you a 3-manifold? I’m not entirely sure. One thing that’s clear is that it gives you a 3-manifold with boundary, since you can take a tubular neighbourhood of the knot/link and take the complement of that, which will be a 3D region whose boundary is homeomorphic to a torus but sits in \mathbb{R}^3 in a knotted way. I also know (from Thurston, but I’ve seen it before) that you can produce lots of 3-manifolds by defining some non-trivial homeomorphism from a torus to itself, removing a tubular neighbourhood of a knot from \mathbb{R}^n and gluing it back in again, but only after applying the homeomorphism to the boundary. That is, given your solid knot and your solid-knot-shaped hole, you identify the boundary of the knot with the boundary of the hole, but not in the obvious way. This process is called Dehn surgery, and in fact can be used to create all 3-manifolds.

But I still find myself unable to explain how a knot complement is itself a 3-manifold, unless it is a 3-manifold with boundary, or one compactifies it somehow, or something. So I had the illusion of understanding during the talk but am found out now.

The twisted-dodecahedron example was discovered by Seifert and Weber, and is interesting because it is a non-Haken manifold (a discovery of Burton, Rubinstein and Tillmann) that is virtually Haken.

Going back to the question of why the geometrization conjecture didn’t just finish off the subject, my guess is that it is probably possible to construct lots of complicated 3-manifolds that obviously satisfy the geometrization conjecture because they are already hyperbolic, but that are not by virtue of that fact alone easy to understand. What Agol appeared to say is that the role of the geometrization conjecture is essentially to reduce the whole problem of understanding 3-manifolds to that of understanding hyperbolic 3-manifolds. He also said something that is more or less a compulsory remark in a general lecture on 3-manifolds, namely that although they are topological objects, they are studied by geometrical means. (The corresponding compulsory remark for 4-manifolds is that 4D is the odd dimension out, where lots of weird things happen.)

As I’ve said, Agol discussed two other problems. I think the virtual Haken conjecture was the big one (after all, that was the title of his lecture), but the other two were, as he put it, stronger statements that were easier to think about. Question 17 asks whether every aspherical 3-manifold virtually has positive first Betti number, and question 18 asks whether it virtually fibres over the circle. I’ll pass straight to the second of these questions.

A 3-manifold M^3 fibres over the circle if there is a (suitably nice) map \eta:M^3\to S^1 such that the preimage of every point in S^1 is a surface S (the fibre at that point).

Let me state Agol’s main results without saying what they mean. In 2008 he proved that if M^3 is virtually special cubulated, then it is virtually fibred. In 2012 he proved that cubulations with hyperbolic fundamental group are virtually special, answering a 2011 conjecture of Wise. A corollary is that every closed hyperbolic 3-manifold virtually fibres over the circle, which answers questions 16-18.

There appears to be a missing step there, namely to show that every closed hyperbolic 3-manifold has a cubulation with hyperbolic fundamental group. That I think must have been the main message of what he said in a fairly long discussion about cubulations that preceded the statements of these big results, and about which I did not take detailed notes.

What I remember about the discussion was a number of pictures of cube complexes made up of cubes of different dimensions. An important aspect of these complexes was a kind of avoidance of positive curvature, which worked something like this. (I’ll discuss a low-dimensional situation, but it generalizes.) Suppose you have three squares that meet at a vertex just as they do if they are faces of a cube. Then at that vertex you’ve got some positive curvature, which is what you want to avoid. So to avoid it, you’re obliged to fill in the entire cube, and now the positive curvature is rendered harmless because it’s just the surface of some bit of 3D stuff. (This feels a bit like the way we don’t pay attention to embedded surfaces unless they are incompressible.)

I haven’t given the definition because I don’t remember it. The term CAT(0) came up a lot. At the time I felt I was following what was going on reasonably well, helped by the fact that I had seen an excellent talk by my former colleague Vlad Markovic on similar topics. (Markovic was mentioned in Agol’s talk, and himself was an invited speaker at the ICM.) The main message I remember now is that there is some kind of dictionary between cube complexes and 3-manifolds, so you try to find “cubulations” with particular properties that will enable you to prove that your 3-manifolds have corresponding properties. Note that although the manifolds are three-dimensional, the cubes in the corresponding cube complexes are not limited to three dimensions.

That’s about all I can remember, even with the help of notes. In case I have given the wrong impression, let me make clear that I very much enjoyed this lecture and thought it got the “working” part of the congress off to a great start. And it’s clear that the results of Agol and others are a big achievement. If you want to watch the lecture for yourself, it can be found here.

Update. I have found a series of three nice-looking blog posts by Danny Calegari about the virtual Haken conjecture and Agol’s proof. Here are the links: part 1, part 2 and part 3.

6 Responses to “ICM2014 — Ian Agol plenary lecture”

  1. Exciting News on Three Dimensional Manifolds | Combinatorics and more Says:

    […]   An earlier post on Wise’s work; A post VHC post; Update (August ’14): Here is a post by Tim Gowers on Agol’s lecture at ICM2014. The videotaped lecture can be found here. Ian’s ICM […]

  2. patrick_g Says:

    Excellent article relevant to this subject : http://www.simonsfoundation.org/quanta/20121002-getting-into-shapes-from-hyperbolic-geometry-to-cube-complexes-and-back/

  3. Qiaochu Yuan Says:

    “Aspherical” means that the higher homotopy groups vanish (so, roughly speaking, there aren’t any interesting ways to put higher-dimensional spheres into the space). This is equivalent to the universal cover being (weakly) contractible, and so in particular all closed hyperbolic manifolds are aspherical.

  4. kuessner Says:

    It is a general fact that fundamental groups of closed hyperbolic manifolds are hyperbolic.

    About cubularion of hyperbolic 3-manifolds: this is a theorem of Bergeron-Wise which relies heavily on the work of Kahn-Markovic about existence of immersed pi_1-injective surfaces. (The construction of cube complexes from immersed surfaces in a 3-manifold is due to Sageev, I think.)

  5. Sam Nead Says:

    The geometrisation theorem is not a classification theorem; it is a decomposition theorem. As a loose analogy: the fundamental theorem of arithmetic states that all natural numbers can be decomposed into primes. But that doesn’t classify numbers, and it certainly doesn’t classify primes! Instead, it (sometimes) lets you ignore those boring composite numbers, and instead study the primes, which have more structure.

  6. ianagol Says:

    I just came across this post, and I’m impressed by your comprehension of the material, my talk notwithstanding…
    I’ll make some comments addressed solely to you (since I
    suspect not many other people will read this post again).

    I agree that it would have been helpful to set the problems in a more historical context, but I didn’t do that for the sake of time (the goal I set to myself was to state the theorem precisely, which required most of the talk to consist of definitions and examples, and didn’t allow for a historical overview). I’m still not sure how important the result is compared to other breakthroughs in Thurston’s “program”, and I should be clear, it pales in comparison to Perelman’s resolution of the geometrization conjecture. I think it got some attention since it was “last” of a list of problems of Thurston, which included the geometrization conjecture and the ending lamination conjecture (resolved by Brock-Canary-Minsky). Actually, there are still a few unresolved open-ended questions from this list – there is a nice survey of Thurston’s paper by Otal. http://link.springer.com/article/10.1365%2Fs13291-014-0079-5

    I chose to describe Haken manifolds as containing π_1-injective surfaces rather than incompressible surfaces since this is easier to describe (π_1 is taught in any introductory algebraic topology course). It is a non-trivial result that the two notions (for co-orientable surfaces) are equivalent, called the “loop theorem”. Again, I didn’t want to get into these things, because they take some time to describe, even though the loop theorem was one of the most important early results on 3-manifolds of the 20th century (by Pappakyriokopoulos).
    I’ll say that the virtual Haken conjecture had some historical importance (in the view of the 3-manifold community) because it would have implied the geometrization conjecture. However, as we know, the implication went the other direction (geometrization was used to prove virtual Haken). By the time geometrization was proved, however, the virtual Haken question was a well-studied open question – Freedman, Lackenby, and Thurston among others had worked on it extensively, and hence it gained some notoriety.
    I think the solution I gave (based on the visionary ideas of Wise and his collaborators) will be more important in the future for the applications to cubulated hyperbolic groups, rather than the implications for 3-manifolds (although it does have many applications in 3-manifold topology that were not foreseen when it was conjectured).

    The example of the Seifert-Weber dodecahedral space is as you say: taking a hyperbolic dodecahedron with dihedral angles 2π/5 along each edge, the gluing by twists will yield a single vertex with solid angle 4π (one may apply the Poincaré polyhedron theorem to determine the hyperbolic structure). In fact, the cell-decomposition of the Seifert-Weber space is self-dual: put a point in the dodecahedron, an edge through each face, and a pentagon dual to each edge, and the complementary cell structure is isometric.

    As for the complements of knots and links (usually in the 3-sphere, so R^3 compactified by a point), these are manifolds for trivial reasons: they are open subsets of a manifold. In fact, the hyperbolic metric is on the open subset obtained by removing the link, and the diameter of the metric is infinite, even though the volume is finite. I like to think of the link as being “pulled” off to infinity like taffy, which gives one a rough idea of what the hyperbolic metric looks like (technically, the end is a warped product of a torus and a ray, with exponential warping function).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 2,023 other followers

%d bloggers like this: