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STEVE: I'm glad you all made it back from lunch. And I guess if you didn't, you're not here. For my second act, it's my pleasure to [INAUDIBLE] Alan Reid from UT at Austin to tell us about arithmetic hyperbolic manifolds.
ALAN REID: Thanks very much, Steve. And let me begin by thanking the organizers for the kind invitation to speak at this Thurston Legacy Conference. It's a real privilege to be here, to be able to talk at this conference.
It's perhaps a little bit intimidating, as [INAUDIBLE] were saying, to talk about what's next, because usually I've got no idea what's next, next day, never mind what's next. But maybe the good thing is we can talk a little bit, because [INAUDIBLE] hasn't arrived yet. So maybe we can put things out there before [INAUDIBLE] actually shows up.
So I'm going to talk about arithmetic hyperbolic manifolds. And perhaps this talk will be notable for not having the definition of an arithmetic hyperbolic manifold. What I intend to do is really develop a story about ongoing work with Mark Baker to understand certain arithmetic hyperbolic manifolds, and try to understand their topology, and try and interweave that story with some email correspondence that I had with Thurston about these particular manifolds and some other related topics.
So the first thing is that Thurston really have a long-term interest in understanding these arithmetic hyperbolic manifolds, whatever they may well be. So in his famous 1982 bulletin article, he posed as, well, not really a question, but perhaps just as an idea. He says, "find topological and geometric properties of quotient spaces of arithmetic subgroups of PSL(2,C). These manifolds often seem to have a special beauty."
So this talk's really about some of these manifolds that have special beauty. And I think, in part-- I mean, I'm not 100% sure about this, but part of the beauty, I think perhaps, for Thurston was that when he was developing the ideas about geometric structures on 3-manifolds, some of the key examples in the development of the theory turned out to be arithmetic hyperbolic 3-manifolds. So for example, the figure-eight complement, the Whitehead link complement, the Borromean ring complement, and the so-called Magic manifold. So here are these.
So this is Whitehead. This is the Borromeans. This is the figure-eight. And this is the--
[INTERPOSING VOICES]
[LAUGHTER]
So this is my first [? Beamer ?] talk, so bear-- now, there's some gadget here that I'm-- oh, there we go. So we all know what these are. So this is so-called Magic--
[LAUGHTER]
--the so-called Magic manifold that is, again, very interesting. So all of these manifolds, as I said, have been interesting in the development of the understanding of geometric structures and 3-manifolds. And they've got lots of interesting other properties, connections with exceptional [INAUDIBLE] problems, et cetera.
So I think to some extent, Thurston was kind of intrigued by these kind of examples, and more generally, about arithmetic hyperbolic manifolds. So before getting into the story that I'd like to get to, what I thought I'd do is kind of motivate part of this story by dropping down a dimension and talk about a parallel story in dimension two.
So whatever an arithmetic group is, the modular group is the key example. So this is PSL(2,Z) It's 2 by 2 matrices with integer entry. It's determinant one moderated by the center, plus or minus the identity.
So there it is, a great example of a discrete group acting on a hyperbolic plane. It's this finite co-area. We've all seen pictures of the fundamental domain for this. And so really, one can take this as a definition perhaps that if we take any non-cocompact finite co-area arithmetic Fuchsian group, this is something that is commensurable with it. So i.e., up to maybe conjugating, there's a finite index subgroup of the modular group that's shared with this other arithmetic Fuchsian group.
OK, so allow me to talk a little bit about the structure of the commensurability class of this group. It's very beautiful and been very well studied for all sorts of reasons. And perhaps some of the most intriguing examples are the congruent subgroups inside PSL(2,Z).
So what are the congruence subgroups? So there's kind of a class of groups that are given to you by God, if you're any kind of linear group. And these come as follows. In the case of the modular group, we have these natural epimorphisms from PSL(2,Z) to PSL(2,Z/nZ). And we'll call the kernel of that guy gamma n. So that's the so-called principal congruence subgroup.
And any subgroup of finite index in the modular group that contains such a guy is defined to be a consequence subgroup. And this integer n here is called the level. So these have being well-studied for all sorts of reasons, in particular from the number [INAUDIBLE] aspect.
And one knows a fair bit about these, of genus, torsion, number of cusps. All of these have been well-studied. So sometimes they have torsion, sometimes they don't. But in particular, the principal congruence subgroups, it's fairly easy to see that these are always torsion-free.
So these are nice examples of finite-sheeted covers of the modular orbifold that are going to be Riemann surfaces with punctures. So there's a conjecture due to Rademacher, which I think was maybe the late '50s, early '60s, that said that that there should only be finitely many congruence subgroups of genus 0, or of a particular fixed genus. So this was proved by a guy called Denin in the '70s after a sequence of papers, basically, using the structure theory of Fuchsian groups, I'm going to give a slightly different take on the proof a little bit later.
So the finiteness was proved for genus 0 back in the '70s. Now, let me just talk about an easy case, perhaps, that if we look at the principal congruence subgroups of genus 0, and ask ourselves, well, how often can one have genus 0 principal congruence subgroups? And the easy answer is that there's only finitely many. And in fact, it's not so hard to see that, in fact, the only levels that can have genus 0-- so again, we've got this punctured Riemann surface-- the only cases when one can have genus 0 is when n equals 2, 3, 4, and 5.
And of course, the experts will know why we've stopped at 6, but let me just say a little word about one way that one might see that, well, depending on how time permits, it may be relevant later, this particular train of thought. So again, these principal congruence subgroups, these are torsion-free. So this guy is going to have genus 0, i.e, the quotient space will have genus 0, if and only if this guy is generated by parabolic elements, OK?
So there was an element of gamma n, maybe 1, n, 0, 1, that's visibly in there. It's congruent to the identity mode n. And that's the stabilizer of the cusp at infinity inside the principal congruence subgroup. All right, so this is a normal subgroup of the modular group, so if I take the normal closure of that cyclic subgroup generated by 1, n, 0, 1, then this is a subgroup of gamma n.
This is the smallest normal subgroup that's generated here. So this is a normal subgroup, so it's contained inside gamma n. Now, notice that the quotient of PSL(2,Z) by this normal closure, well, what is it? Well, we've got this element in the modular group. It looks like 1, 1, 0, 1.
The n-th power of that lies inside this congruence subgroup of level n. So when we take the normal closure of what we see here, it's just simply the 2, 3, n triangle group. So notice that this is finite when we get n from, say, 2 through 5. And from 6 onwards, it's an infinite triangle group.
So then the claim is that gamma n is generated by parabolics if and only if N is equal to that. So the normal closure is equal to the principal congruence subgroup. So given that, then we're in good shape, because then this has finite index only when we're in the cases where n is 2, 3, 4, and 5.
OK, so the key thing is to show that the normal closure of this parabolics is the whole of the principal congruence subgroup. And let me just sort of quickly say, well, if n is equal to the principal congruence subgroup-- and I pointed at the screen, all of these-- so one direction is clear. So if N is equal to gamma n, then it's generated by parabolic elements and we're in business.
And then for the other case, well, we know that H2/PSL(2,Z) has one cusp, so we've got this normal subgroup, this regular cover of this orbifold here. And so if we're generated by parabolics, then all these cusps have to be identified down to the single cusp of the module orbifold. So that just means that these parabolic generators have to be PSL(2,Z) conjugate into that normal closure.
And therein is the proof. So that's a very easy example to show that we get a finiteness for these principal congruence subgroups of genus 0. The finiteness statement that Denin proved is a lot more involved. And what I now want to talk about is a generalization of the statement that I made earlier, and talk a little bit about a proof of a more general statement.
So there's a generalization of this Rademacher conjecture that I'm now going to talk about.
AUDIENCE: Hey, Alan.
ALAN REID: Yeah.
AUDIENCE: [INAUDIBLE]
ALAN REID: I'm just coming to this. I'm just coming to that. Thank you, Kurt. He's already ahead of me.
So now let's define congruence to be a little bit more general. Instead of just these subgroups of the modular group, we'll define something to be congruence if it's commensurable with PSL(2,Z) and it contains gamma n. So that's what I'm going to define is congruence more broadly inside PSL(2,R).
So you may think, well, what do these congruence subgroups look like? Well, here's a construction of one that's very important that illustrates a lot of the beauty of arithmeticity in a way that I'll mention later. So if we take some n bigger than 1 and we now define a particular congruence subgroup, the guy gamma0n, so all your number theorist friends, that's their favorite congruence subgroup.
And that's the guy that's congruent to the Borel subgroup modulo n. So it's going to have 0 in the 2, 1 entry, and we don't care about anything else. So if you just scribble on a piece of paper, this little element here, tau sub n, this normalizes again as 0 n.
So that means that this little guy here, this group generated by gamma0n and this little element of order 2, that's living inside the normalizer of this discrete group, which is a discrete group. So we've got this element here that's normalizing gamma0n. And it's, well-- so this means that this guy here, this group, the normalizer-- or this guy here-- it's commensurable with PSL(2,Z).
This is visibly not a subgroup of PSL(2,Z) because we've got the square root-- one of our square root n here. OK, so this is a guy that's commensurable, that's outside the modular group, and it contains gamma n. So this is an example of a congruence subgroup of PSL(2,R) that's not inside the modular group.
Now, these normalizers of gamma0n are really very important examples, he said boldly. So it turns out that if you look at the commensurability class of the modular group, then it's fantastically complicated. There are infinitely many maximal Fuchsian groups in that commensurability. And these arise, essentially, as normalizers of some of these gamma0n's.
So joining these involutions is kind of getting you up to interesting maximal groups in the commensurability class. And these involutions that are constructed in these normalizers, this is a really important theme in arithmeticity, that there's lots of hidden symmetry, i.e., those involutions were invisible to the modular group, but they were visible on some finite index subgroup. And they lived in this normalizer.
So this kind of hidden symmetries is a very powerful thing about-- Dylan's sneaking by here. They're switching me off.
[LAUGHTER]
Thanks, Dylan. I have no idea what he did, but [INAUDIBLE].
OK, so let me talk about this more generalized theorem that Thompson proved, and independently Zograf proved. And I'm going to talk a little bit about this kind of Zograf approach to this finiteness. So there's finitely many congruence Fuchsian groups commensurable with the modular group of genus 0, and indeed, of any particular fixed genus.
So the thing about these congruence groups is that they have a so-called spectral gap. So what does that mean? Well, I mean, let's not think about the first known zero eigenvalue Laplacian. Just let's think about what that means.
It means that if we interpret things in terms, say, of the Cheeger constants of these manifolds, that these congruence manifolds should be big and round. There should be no little things like this appearing. There'll be no bottlenecks. There should be no way of easily disconnecting these manifolds into two pieces.
So this spectral gap idea is kind of saying that these examples are big and round manifolds. That's what these congruence-- that's what this spectral gap for congruence subgroups is saying. Now, on the other hand, Zograf proved this theorem that is part of a lineage of theorems that goes back to [? Hirsch ?] and Lee and Yao and Yang and Yao, about estimating lambda 1 in terms of area.
And he says, well, in fact, basically that lambda 1 can be bounded above by this little quantity on the right-hand side involving genus and area. So this gamma here can have elements of finite order in the group as well. OK, so let's prove this theorem of Zograf and Thompson by just playing off these two statements, OK?
So we take a sequence of congruence subgroups of genus 0. Now, it's a fact that there are only finitely many conjugacy classes of arithmetic Fuchsian groups of bounded area, so if we've got this infinite sequence, the area of these guys is going to infinity. So in particular, the area's going to infinity, so Zograf then tells us that lambda 1 is going to go to 0. If we fix the genus at 0-- so we fix the genus at 0 there, then lambda 1 will go to 0, because the area is going to go to infinity.
But now, on the other hand, there's spectral gap from Selberg, so that's a contradiction. OK, we've got these guys for which lambda 1's going to 0, but they're congruent, so there is a spectral gap. And that's where you get the finiteness statements. You cannot have these infinite sequences of guys of genus 0 that are congruent. [INAUDIBLE]
So now let me get to--
AUDIENCE: Alan.
ALAN REID: Yeah.
AUDIENCE: Can we go back to [INAUDIBLE] finiteness statement, [INAUDIBLE].
ALAN REID: No, this is just using number theory so that you can estimate areas using the number theory data.
AUDIENCE: The arithmetic [INAUDIBLE].
ALAN REID: That's right, but there are explicit volume formulas for these groups. And we can talk about it later. But it's really just number theory. There's no geometric limit argument in this.
OK, so now let me get to the discussion that Kurt was alluding to earlier. So if you look at these maximal groups-- so again, recall that these normalizers that are constructed, that they are basically these maximal groups in the commensurability class. So if you look at these maximal guys and you try to figure out which ones of these guys have got genus 0, well in fact, there's been a long industry of doing this.
So these prime levels are these guys are here, 2, 3, 5, blah, blah, blah, blah, blah, blah. And then there's the non-prime levels that are here. Now, as Kurt was alluding to, these numbers that are the prime levels, Ogg noticed back in the '70s that these primes were exactly the prime values that were the divisors of the order of the Monster simple group.
So this is part of a whole industry that kind of relates to so-called monstrous moonshine that connects the genus 0 guys to various properties of the Monster simple group. So there's a very beautiful story here that I'm not going to say too much more about. But anyway, that's where Kurt was alluding.
OK, and let me just say a few more words. OK, there's a finiteness statement that was done back in the '70s. It was only kind of proved more recently, about 2004. So I don't know these fellows here. There's 132 guys of genus 0 inside PSL(2,Z) up to conjugacy in PSL(2,Z).
And of these, 26 of these guys are torsion-free. So this is just for reference later, OK? So you don't care about that. I actually do care about that, but that's why you're hearing it.
And here's the list that [? Selberg ?] produced. So the ones in red here, these are the guys that are principal congruence of genus 0. So we can see the gamma5, gamma4, gamma3, and gamma2. The rest of these are these congruence subgroups of genus 0 that are torsion-free.
OK, so now, let's move on to dimension 3. So that's the story. OK, so what's the takeaway here? There's a story that says that genus 0 congruence guys are finite in number. And the principal congruence ones are even more finite in number.
And that's the theme I'd like to take up for the rest of the talk in dimension 3. So there's a class of groups in dimension 3 that are the analogs of PSL(2,Z) and PSL(2,C). And for that, what we're going to do is we're going to take d square-free positive integer. Od is going to be the ring of integers inside this number field.
And then the Bianchi groups are going to be these guys. Now, Od-- this is a quadratic imaginary number ring. That's a discrete subring of the complex numbers. And so this gives you a discrete subgroup of PSL(2,C). And it turns out that these, in fact, have got finite volume.
And these guys here, the quotient spaces that we get, these are so-called Bianchi orbifolds. Now, they are really orbifolds. They always contain torsion. For example, inside here, we obviously see a copy of a PSL(2,Z) sitting inside PSL(2) of Od. So you do get orbifolds.
Now, what do these orbifolds look like? Well, they are quite complicated in general. But here's a quick picture of some for some small values of d. So one of my other goals in this talk is to see how many members of the Cornell faculty I can squeeze into a talk, both past and present.
OK, so I've said Thurston. Now I've got Hatcher. So Hatcher has a paper where he computes the orbifolds for some small values of d. Now, what he does in that paper is, it's quite the Bianchi groups. It's H3/PGL(2,Od) that he really does. But in any event, one can do that to do some computations of these orbifolds for the PSL2 version.
So notice just a couple of things here. So here we've got this guy. d is 1 and d is 3. So this is meant to be telling us that the orbifold here's got underlying space [INAUDIBLE], and then there's this garbage here, some singular locus. But notice here, the cusp that we see here, it's a 2, 2, 2, Euclidean crystallographic group. And here we see a 3, 3, 3, Euclidean crytallographic group.
And then the rest of these you see, well, we've got solid tori, solid tori, solid tori. And there's a little torus drilled out of here. And the reason for having those little bits of torsion at infinity here is because these number rings, when the d is 1 and d is 3, these have non-trivial units, whereas all the other ones don't have any interesting units, just plus or minus 1. So in this case, you get this kind of solid torus as the underlying orbifold-- base space of the orbifold.
And so here, this other little torus that I've mentioned, so this is really saying-- and these guys here, this d equal 1, d equals 2, d equals 3, and d equals 7, there's just one cusp. But in this guy here, there's two cusps. There's kind of the black solid torus and there's a little red torus here.
And more generally, the number of cusps of these orbifolds is just the class number of these number rings. So that's completely well-understood. And the topology of these things is something that people have looked at quite extensively for many years, in part because one would like to develop a theory of these Bianchi groups and these Bianchi orbifolds that's kind of mimicking what's known for the modular group. And in particular, this is one of the things that the number theorists are very keen to do.
OK, now again, we can maybe take a definition here that if we wanted to talk about non-cocompact arithmetic Kleinian groups, well, these are guys that are just commensurable with PSL(2,Od). So this is a consequence of the theory of arithmetic Kleinian groups. But the non-cocompact ones are precisely the guys that are given up to commensurability by these Bianchi groups. So they're a beautiful class of examples.
And so we think back to the discussion in the modular group, we were thinking about, well, how do we get genus 0 punctured surfaces? Covering the modular orbifold as non-congruence subgroups is pretty straightforward. But we could ask ourselves, as a first step, a natural generalization of this genus 0 surface groups are link groups in the 3-sphere.
So we can then ask, well, which of these Bianchi orbifolds are covered by link complements in the 3-sphere? Well, so here's another Cornell person, maybe.
AUDIENCE: [INAUDIBLE]
ALAN REID: Well, it is, but she's not German, at least the last time she spoke to me she wasn't. OK, so I'm being a little bit flippant here. The lots of Germans include [? Harder, ?] [? Rolfes, ?] [INAUDIBLE] and perhaps other, but that's basically it. But in any event, what the theorem says is that if you have a link complement covering one of these Bianchi orbifolds, then d must be one of these guys.
And so basically, what's proven is that link complements, all their homology, first homology, comes from the boundary. So what's going to prove to you is that these Bianchi orbifolds typically are awash in what's called cuspidal cohomology, but probably to people in this room, it's non-peripheral homology. It's homology that doesn't come from the cusp. So that's really what this theorem proves, that these are the only guys that are legal from the point of view of having link complements as covers.
Now, having said that, you will maybe be asking, well, wait a moment. Are there link complements at every such example? And in fact, this is a theorem of Mark Baker. So for each d on this list, there is a link so that the link complement covers that orbifold.
And we've already seen some of these, because I alluded to that at the start of the lecture. Namely, things like the figure-eight knot here. Sorry, I'm pointing again. That's a figure-eight. No, that's not. I know that. That's the figure-eight knot. The whitehead again, there's Magic.
And then there's this guy here in d equals 2 that I just threw in for good measure. So there's a kind of nice-- you can believe these nice, beautiful examples are just living kind of naturally. They should be arithmetic.
Well, there's some more exciting ones when d starts to get larger. So that's taken from a paper of Mark Baker's, when he gives the constructions of the existence of link complements, and all of these covering these Bianchi orbifolds. So this is for d equals 31.
And if you've ever seen Mark Baker in action, he gets out rope in his office. And he's down on the floor desingularizing Bianchi orbifolds. It's kind of an impressive feat. And just to show another one, so this is 47.
Now, you put that up on the screen there and you've got no reason to believe that is commensurable with a Bianchi. But the good thing is SnapPy is actually quite helpful to convince you that, well, you're probably right. Yeah.
AUDIENCE: [INAUDIBLE]
ALAN REID: No. No. So all that we would have as input data is the Bianchi orbifolds have got some number of cusps. And so therefore, a link complement is going to have at least that number of cusps. And so some of these, like 47, I think, has got class number 4. I'm just guessing, but it's not 1. That much I know.
So that's the only thing. And there are infinitely many links that are R arithmetic. And so the number of components can go to infinity, or can be quite small.
AUDIENCE: [INAUDIBLE]
ALAN REID: So how you get to that is really what I just said, that you look at the Bianchi orbifold and you start to try to desingularize this. So if you look at these orbifolds-- maybe if I can scroll back-- the all do. And who knows what that means?
However, I will say that every link appears as a sublink of an arithmetic link complement. So there are arithmetic guys that have got knotted components. But the congruence ones-- well, maybe that's congruence, maybe it isn't. But they tend to have unknotted components.
So if I can just-- let's see if I can get back quickly here. Yeah. So here's this picture here. So if you wanted to find a link complement covering, say, d equals 7, you would look at that singular locus and you'll see Z2's and Z3's. And what Baker really does is goes in there sort of starts, first of all, taking maybe a threefold cyclic cover and unwraps that three's torsion. Then he'll get some orbifold with single locus of just two. Then he'll get down on the floor and have a bit of rope. And then he'll unbranch that two-fold cover.
So that's how you get these diagrams. Now, when d is 71, I don't think any person has seen a link complement. You need a soccer field. And Thurston actually, in an email with Mark, had an idea about how one might really picture a link and [INAUDIBLE] and covering the Bianchi orbifold with d as 71, which is kind of [INAUDIBLE] that's [INAUDIBLE].
AUDIENCE: Question. So one way you could think of this corresponds [INAUDIBLE].
ALAN REID: The Z plus torsion [INAUDIBLE].
AUDIENCE: The Z plus torsion of the other. So is it know that there's only [INAUDIBLE] in Bianchi orbifolds that are perfect that occur [INAUDIBLE]?
ALAN REID: Yeah. So it's only these guys. Everything else has homology. It all comes from these units at infinity. I mean, they're not quite perfect. I mean, there may be Z2's and Z3's homology. That's the only one with finite H1.
OK, so there is some interesting examples of arithmetic link complements. And it's a real challenge to build these, and in fact, a bit of an industry. OK, so let's call a link arithmetic if the complement's arithmetic. So it looks like this.
Now, it's a theorem that the figure-eight is the only arithmetic knot. And as we were just discussing, it turns out, in fact, there are infinitely many arithmetic links that even have two components. So if you go back to one of the examples that I drew before, when d was equal to 2, there was a kind of unknot-- well, I'm not going to go back. There's an unknot with a little kind of guy through it. If you do the right cyclic covers when you stated the linking numbers, you get complements with just two components.
OK, so there's infinitely many arithmetic links, but again, if we think back to the task at hand comparing with the modular group, they were much more special. The congruence of genus 0 and the principal congruence of genus 0 were very special examples. OK, so let's define, first of all, what we mean by congruence in this case. And again, it's hopefully going to be straightforward.
We'll define the principal congruence subgroup in the way that's analogous to what we did before. We take the kernel of this homomorphism here. And that's called the gamma I. And then its congruence, that's the principal congruence guy. And then it's congruence if it contains one of these guys. So that's what it's going to mean for congruence group or congruence manifold in the obvious definition.
OK, so the question I'd like to raise here is an analog of Rademacher's conjecture, are there only finitely many congruence link complements in the 3-sphere? And this is still open. And another question that I'm not going to say too much about, but perhaps is very interesting independently is, is there some analog of Ogg's observation that related these primes that were the maximal groups of prime level that had genus 0 and related to prime divisors of the order of the [INAUDIBLE] simple group?
And so this seems to be a very interesting question about what the cuspidal cohomology for these maximal groups is doing, even for prime level. And I think that's a lot of interesting-- there's a lot of things to think about here that the number of theorists would like to know, but sadly so far [INAUDIBLE].
OK, so let me just talk about the analog of the theorem that we mentioned earlier. There were finitely many principal congruence link complements. So there's was finitely many genus 0 congruence surfaces. So let's just talk about the proof here.
So first of all, by Vogtmann's result, there's only finitely many possible d's for link complements in the 3-sphere. So then if we think about what these look like, so we've got these principal congruence subgroups for some ideal I. But now because we are a link complement in a 3-sphere, at least one of these cusp tori has to be small.
The shortest curve is going to be length less than 6. So the 6 theorem tells us this. However, the peripheral subgroups here-- [INAUDIBLE]. The peripheral subgroups, these have entries in the ideal I. But as the norm of the ideal I grows, so that's just the cardinality of the quotient ring, then elements in I have big absolute values.
In particular, you get past 6. And so you can only have finitely many possible ideals allowing the 6 theorem to give you S3. So this is a very easy consequence of, really, the 6 theorem.
Now, let me just mention an alternative approach that also proves a finiteness theorem. So Adams and I proved, way back in the day, that if you have a link L in a closed orientable 3-manifold and it doesn't admit a Riemannian metric of negative curvature, then there's an upper bound in the systole. OK, this is just a length of the shortest closed geodesic. So this is not too hard to prove, but I'm not going to say anything about it.
And so in particular, if you couple that statement with having this global bound of systole, and you think about what hyperbolic elements are doing inside these principal congruence subgroups, again, it's a fairly easy deduction that the trace of gamma looks like it's congruent 2 modulo I squared, in fact. It's even better than modulo I. So this observation here tells you that the absolute values of these traces are getting large. So the systole, the length of these guys, has to be getting large with the norm of the ideal.
So the systole bounds give you very interesting approaches to enumerating these principal congruence link complements. And using this idea of relating the systoles to get bounds, there's a more recent result of Grant Lakeland and Chris Leininger that proves an analogous statement. There are only finitely many principal congruence link complements in any fixed closed orientable 3-manifold.
And so here they got to relate systole length to volume of the ambient manifold M that you remove the link from. So one could really ask for an even more general version of Rademacher's conjecture, for a fixed closed orientable 3-manifold, are there only finitely many congruence? So finiteness is true for principal congruence, but what about finitely many congruence linked complements. And again, this is pretty much open.
Now, as I mentioned at the start, Thurston was always interested in arithmetic link complements. And he would occasionally send me an email. And I would sort of lie on the sofa for a while panicking, because I'm thinking, why would he ask me? Then I would sort of get up the courage to respond to him.
And so in 2009, he wrote to me to say, look, there are infinitely many arithmetic link complements. There are only finitely many that come from principal congruence subgroups. And again, he can appoint so-- some of these examples are really very interesting for all sorts of reasons. So Thurston understood these principal congruence manifolds, and he asked, well, what's the list of principal congruence link complements in the 3-sphere?
And it just so happened about this time that Baker and I had actually started thinking about enumerating these principal congruence link complements in the 3-sphere. And so we were a tad more motivated to start to do this when Thurston emailed us to say that, well, maybe his dog needed to know what principal congruence link complements were. But anyway, you know we took that as motivation to work harder on understanding these principal congruence link complements.
OK, so let's think about some examples. So from Baker's thesis, he constructed lots of examples of principal congruence. And so the ones I'm going to talk about here, the levels are all 2's. The ideal is just the ideal generated by 2 in the integer ring. So here d equals 1.It's this nice-- however many components-- 1, 2, 3, 4, 5, 6 chain link.
Then there's this more complicated guys for d equals 2. And then there's other ones that I'll put up here. So there's d equals 3. So this is kind of a famous one, it's a five chain link. And d equals 7, well, there's a handy looking dude here. And again, Baker constructed all of these by hand.
And then the Magic manifold that we talked about earlier, this three chain link, that's also principal congruence. It turns out to be principal congruence of level this ideal, 1 plus the square root of minus 7 over 2. So in this number ring, 2 splits as a product of that guy times its complex conjugate. And so this guy here turns out to give us the Magic manifold.
So again, Thurston knew this, that it was principal congruence. And again, he was intrigued by why these examples were principal congruence. So what's going on with these examples?
OK, so let me just throw up some other examples. This is due to Baker again, and some other people, where you look at d equals 15 and d equals 23. So again, we're always awash in unknotted components for these examples here.
Now, Thurston, in another email, talked about this particular example. It's a congruence link when d is equal to 3. And this guy here is the ideal, so 5 plus root minus 3 over 2. If you look at the norm of this, so it's 25 plus 3, 28, over 4, that's 7.
So it's 7 splits in this ring. That times its complex conjugate. And this guy here was found by Thurston. And how he found out was he was looking at this seven chain guy right here.
So if you forget about this component that's weaseling its way through here, there's this seven chain link here. And he looked at the complement and he looked at it in SnapPy and saw that the horrible diagram and the cusp diagram, it was trying hard to be regular. It was trying hard to look like something that came from d equals 3.
And something that Thurston liked to do was drill, because he had this picture of crystallizing manifolds back to arithmetic things. And so what he did do was he just thought, OK, which curve should I drill to see a principal congruence link? Well, he just found this.
He just did this one. Now, I don't know how long it took him to find that guy, but anyway, he drilled out this guy here. And magically, he found this principal congruence structure on this 8-component link. Yeah.
AUDIENCE: [INAUDIBLE]
ALAN REID: It could be. I haven't checked on that. And so what he also noticed here, so this principal congruence link, because the level is-- sorry, I'm pointing at that again. The level is this, is norm 7. So in fact, again, what Thurston was intrigued by was that, in fact, this induces a cover of the modular orbifold.
And the cover that you get turns out to be a punctured Klein quartic that's immersed inside here in a very explicit way that comes from that tessellation. And Bill could have spotted this as part of his analysis. So he says in the email of 2009, one of the most intriguing congruence cover right now is the ideal blah blah blah, which is an 8-component in the 3-sphere.
And part of this intriguement was-- Dylan, OK-- I think the connection to the Klein quartic that I've alluded to. OK, now in fact, even in his notes, he saw examples of principal congruence links. Here again, we have this 3-component guy with, again, this kind of geodesic meandering through all these components. Now, he didn't say it was explicitly principal congruence, but in fact, he probably knew.
And one can check directly that it is. And there's a similar kind of example that, again, this has another-- that's a double mark for Hatcher in the Cornell faculty. So Hatcher in a paper in the Journal of LMS from the '80s, constructed many arithmetic link complements. And in fact, this one here, one can check. This guy is also principal congruence of some level. And the level being this guy here, 1 plus root minus 11 over 2.
OK, so what I'm just going to mention-- I'm going to put it up here. It's probably too much to take in these slides. It's the following theorem that contains all the known principal congruence link complements in the 3-sphere. So it contains all examples a la Baker. It contains new ones of myself and Baker.
And it also contains examples of Matthias Goerner who completed this in his 2011 Berkeley PhD thesis. And he also has a very beautiful pre-print called regular tessellation links that just appeared a couple of weeks ago on the archive. So your eyes will glaze over. I apologize for that, but you can take it for what it is.
So there's a list here. So d equals 1, there are few. d equals 2, there are a few. d equals 3, there are a few. Then there's this guy here, d equals 5, d equals 7, et cetera.
So there's a list that we know of of these principal congruence links. Now, there's still several d's that are missing from this. And I should say it as well that Goerner has proven that this is a complete list in the cases of d equals 1 and 3.
And I believe, in fact, that he has explicit examples of links in the 3-sphere for all the examples for d equals 1 and 3. So if you go back to d equals 3 here, so this guy is 80-- so the norm of this ideal here is 81 plus 3, that's 84, divided by 4, that's 21.
So that's pretty big, in fact. And somehow the number of cusps of this is kind of gigantic. But I haven't seen the example yet, but he claims to have this link complement in the 3-sphere that's got some gigantic number of cusps that he's computed.
OK, so the only d's for which we don't have examples of--
[CHIME]
Wow.
AUDIENCE: [INAUDIBLE]
ALAN REID: OK. OK, good.
[LAUGHTER]
OK, so d equals 6, 39, 47, 71. They're the only ones that are remaining. So more recently, it does appear that work with Baker, there will be no principal congruence examples when d is equal to 6. So we haven't proved that yet, but certainly it's suggestive that this is the case.
And there's another theorem that says more explicitly when the level is a natural number n, these are all the principal congruence link complements in the 3-sphere. And for these guys, there are pictures.
[PHONE TONE RINGING]
The child's still missing I take it.
[LAUGHTER]
OK, so I don't know how much I really want to get into the comments on the strategy that Baker and I use, but it's a kind of combination of experimentation and just thinking. And experimentation uses magma computation, as well as some trial and error experimentation and a little bit of thinking.
So how would you try to show that these principal congruence subgroups corresponded to link groups in the 3-sphere? So let's let L be the link with n components that I've given here. X L with the exterior, and then gamma is going to be the fundamental group.
So here are three basic facts about link groups at S3. So the first thing is that if you take the [INAUDIBLE] of this guy, it's torsion-free and it's got the rank n, where n is the number of components. So that's one fact.
Another fact is that it's generated by parabolic elements. Again, like the case where we had the genus 0 guys, again here, we can think of being generated by meridians of these links. And then the third thing is that for each component LI, there's a curve there so that when you Dehn fill along these curves, you get the 3-sphere back. It's a link complement in the 3-sphere.
So by Perelman, basically, this is just saying that, well, you can find curves so that you adjoin those guys to the group, you trivialize the group. And now we know that that is really the 3-sphere, not some [INAUDIBLE] of a 3-sphere. So that's the strategy that we're going to try and implement.
So the first thing is you would like to show that these groups are generated by parabolic elements. That's the first step. So let me just say a few words about that.
So if we look at this peripheral subgroup, as we did in the example of the modular group when we looked at the principal congruences of our genus 0 in the modular group, you look at this guy here that's going to be the peripheral subgroup fixing infinity inside gamma I. And if we take the normal closure, as we did previously, of this inside PSL(2,Od), then again, because this guy is the normal subgroup, this normal closure will live inside this group gamma I.
So if we had the good fortune to have that this guy was equal to there, then we'll know, again, that we're generated by parabolic elements. Now, it's also true that, in fact, in this case, the converse also holds when you have one cusp, for the same kinds of arguments that we had before. You have this regular cover, and all the cusps have to wrap up to give you one cusp downstairs. So basically, all the cusp subgroups are conjugate under PSL(2,Od) into P.
AUDIENCE: So [INAUDIBLE]?
ALAN REID: No. So we've already kind of abandoned the ideal to some extent, the norm of the ideal, by the LDL lemma that kind of said that the ideal couldn't be too big in norm because of the 6 theorem. So the idea of this one here is just simply to say that as a first obstruction to being a link complement in the 3-sphere, this guy best be generated by parabolic elements.
So we're trying to manufacture a way to tell that this guy, gamma I, is generated by parabolics. And it won't be for I very big, because we've already proved that. So how does one do this?
Well, one thing that one can do is you can go to magma. And the orders of these finite groups that you get, PSL(2,Od)/I, are known. So if you give magma to test whether gamma I is equal to that, if it gives you the right answer, i.e., this order, well, you're in good shape. Gamma I will be generated by parabolics.
Now unfortunately, there is no God, so that quite often doesn't work. And so then you're toast. Then one has to work harder. OK, well, but in many cases, this does work quite efficiently to generate-- to give you the fact that this congruence subgroup is generated by parabolics.
OK, so then the next thing is, well-- next thing is I--
AUDIENCE: [INAUDIBLE]
[LAUGHTER]
ALAN REID: Still hear me? Yeah. You don't want to hear me, but you can hear me. So we want to find parabolic elements in here so that, as above, trivializing these elements trivializes the group. Let me just go back one second here. So this being a link group in the 3-sphere is generated by parabolics.
And it would have really helped us enormously if that was a statement that could only hold for link complements into 3-sphere. And in fact, Matthias Goerner came up with an example that he found by searching through the census from SnapPy, and found a knot complement not in the 3-sphere for which it was normally generated by parabolic elements. So in fact, the our wish list was that that wouldn't happen so we could simplify the proof a little bit, and we wouldn't have to do what comes next. Because this is a-- next part's a major pain in the bottom here.
So one has to kind of really find by trial and error the right collection of parabolics that can be trivialized to give you the 3-sphere. So maybe I'll just run through this example quickly and point out salient facts.
So here's the case of d equals 1. And this guy here, gamma 2 plus I, is the congruence group. So the norm of this guy is 5. And so what's the index of this? Well, the quotient group that we see is PSL(2,F5).
So in fact, this guy here is a normal subgroup of index 60. Now, what is the image of-- so how many cusps do we have? I see six components here. And the reason for six is that if you look at the peripheral subgroup in PSL(2,O1) fixing infinity, apart from a kind of Z cross Z subgroup, there's also this little element I00 minus I that fixes infinity.
So that gives you a kind of dihedral group in the image when you reduce modulous ideal. So you get this guy of order 10 giving you six cusps. And splendidly enough, magma does tell us in this case that this guy is generated by these parabolics.
OK, I'm not going to talk about that. So you can just give this to magma, and this has got a little routine that's going to run here. It's just basic magma stuff. And you're going to get-- OK, here's what I said up there. You get index 60.
You get Abelian quotients is this. That's what we expect for a 6-component link in a 3-sphere. Then there is some computation that's needed to identify inequivalent cusps that are representing these six cusps and working these guys out. And then what we find is that, in fact, this guy here does generate the [INAUDIBLE] of this normal closure.
So these are a collection of elements that are parabolic. They're all conjugates of things that are parabolic living inside this congruence subgroup. And these generate this link group. So we've got no picture of this link, but we do know it's a link group in the 3-sphere.
AUDIENCE: So is this the computation that's [INAUDIBLE]?
ALAN REID: Yeah, exactly. Exactly. So you've got to get the cusps. And then you've got to choose the right parabolics in each of those cusps.
AUDIENCE: And when you say the right, aren't there, in fact, usually many choices?
ALAN REID: Not for the 3-sphere [INAUDIBLE].
AUDIENCE: No, but the other link.
ALAN REID: So as long as we get some subset that we know gives a 3-sphere, then we're happy. As long as we've got--
AUDIENCE: I was wondering because [INAUDIBLE] there are any other choices available?
ALAN REID: Well, there can't be too many, I guess. I mean, because it is a 3-sphere. So there's a bound-- so if you think of the 6 theorem again, there's only going to be so many curves of length at most 6. And then these principal congruence subgroups, all the cusp shapes are the same for this particular example. So there aren't too many possibilities in this case.
OK, so how one can prove the finiteness statement. So how do we prove the analog of the Rademacher conjecture for link complements in the 3-sphere. Now, I said at the start that it was still open. So obviously, I haven't proved it yet.
And so how might we approach a proof of finiteness for these guys? Well, there is some partial progress that Baker and I did at certain levels, but it's nothing worth talking about, certainly in such esteemed company. But the idea would be, well, what do we do with Zograf?
There was spectral gap that we traded off, that lambda 1 was bounded away from zero by some constant, because there were congruence. But on the other hand, lambda 1 had to be forced to go to zero through this sequence of guys whose areas went to infinity. So is that a reasonable approach to prove finiteness here? Can you trade-off a kind of spectral gap for congruence subgroups, which we do have here-- this is a general phenomena-- and use this to build examples that prove finiteness for link complements that are congruence in the 3-sphere?
Well, the bad news is no. So it's a theorem of Lackenby and Souto that there exists an infinite family of links in S3 whose volumes go to infinity. So like these congruence examples, they're big. OK, they're like these kind of guys that are around, because lambda 1 is bigger than some fixed constant bigger than zero. And so there is no Zograf theorem in dimension 3.
So there are these link complements that have the spectral gap. And so you cannot use this kind of idea of using the spectral geometry directly, or Cheeger constant, or however one interprets it. So this is bad news. But again, that's OK. We're used to bad news in the math department.
And so we struggle on. So let's call these examples an expander family, for reasons that for those that know about expanders will be clear. And so, in fact, Lackenby showed that alternating links-- so I have no idea what these links really look like. I mean, there may be some chance Souto tells me that it can even be made into knots.
It should all be said that these examples of Lackenby and Souto, they've got nothing to do with each other as far as one can tell. They're not commensurable in any obvious way. They're just a sequence of examples that have this property that the lambda 1 is bounded away from zero by some constant. But who knows what they are.
So Lackenby proved that alternating links don't form expander families. So you don't have volume going to infinity and lambda 1 being bounded away from zero. So at least in particular, there are only finitely many congruence alternating link complements. And there's other work by Dave Futer and Kalfagianni and Purcell that give other examples of known expander families of link complements.
OK, then again, so how might we address this problem that we can't use in any reasonable way the kind of idea that we would have liked to have used? OK, before I get to that, here's an amusing-- well, I think it's kind of a funny and amusing side note.
So in another correspondence with Thurston, we had about expander link complements. And so I mentioned Lackenby's result to him in an email. He said, oh, I wasn't familiar with Lackenby's work. But alternating knots related in spirit to Reimannian metrics in 2-sphere, which don't admit an expander sequence of metrics. So he knew that alternating links weren't going to give you expander families, because for whatever reason, and I'm sure he knows, they're related to metrics on the 2-sphere. And again, this is kind of this [? Hirsch ?] result that says there are no ways that get lambda 1 being big on the 2-sphere.
And then he proceeded to outline a construction to produce expander families of link complements in this email, by first building a sequence of expander metrics in the 3-sphere, so which that doesn't happen in dimension 2. And then he says, by carefully choosing the links in the 3-sphere, the expander property will surely be preserved.
And he can kind of outlined how that might be done. And this is the essence of what Lackenby and Souto do in their paper. It's a careful construction of expander metrics in the 3-sphere using triangulations, and then carefully choosing links to build the expander family.
OK, so what else can one possibly use? And so here I'm going to go out on a limb here. And this is perhaps the what comes next part. So the expectation that's built out of a lot of work now that's going around, that perhaps goes back to Bergeron, Venkatesh, Frank Caligari and Venkatesh, and others, is that congruence manifolds, congruence subgroups, should quickly develop torsion in each one. So you shouldn't have these sequences of congruence subgroups that are devoid of torsion in H1.
That's kind of the conjectural picture that is out there. It's true for certain families of congruence subgroups. So this is maybe the thing that might be a better approach in these cases, to analyze these congruence subgroups via torsion and rule those guys out. So maybe Schulz has already done it. But anyway, who knows.
So let me just finish by again talking a little bit about what's next. So the idea is we'd like to build these manifolds that are kind of big and round. OK, where are these manifolds with big Cheeger or big lambda 1? Can one find these examples in commensurability classes? I think that's kind of one of the interesting things.
So there's work of Long and Lubotzky and myself a few years ago that basically give an analog of what's true for congruence subgroups, that we produce spectral gap for towers of descending normal subgroups of finite index in any old finite co-volume Kleinian group. The intersecting identity, he said, this looks like a congruence tower and you get spectral gap.
OK, so this comes out of the work of [INAUDIBLE] and others about understanding when Cayley graphs are expanders. So here is another possibility. We're going to forget about arithmeticity here. Can these kinds of constructions lead to link groups in the 3-sphere? So these towers that Long and Lubotzky and I constructed, I mean, can these guys be link groups in the 3-sphere?
And again, there's no number theory here. It's just topology at that point. Yep, still missing. So let me go back to dimension 2 for a minute to finish off.
so there are notions of congruence subgroups for co-compact fixing group. I haven't really talked about the co-compact world. It's a little bit more involved to define what an arithmetic co-compact fixing group is.
But it's an interesting question. And again, there are these guys that have spectral gap. So are there congruence surfaces of every genus? So here I mean it's genus g, closed, orientable surface. There's no torsion. So are the congruence surfaces of every genus?
Now again, as I mentioned, there's only finitely many conjugacy classes of arithmetic Fuchsian groups of genus g, so these are kind of special points in the moduli space. Now, if there are congruent surfaces of every genus, what I'm pretty sure is true is, in fact, that these guys will not lie in a fixed commensurability class.
And indeed, I will be amazed if they come from trace-fields of bounded degree. For example, over q, perhaps if a gun were held to my head-- and I'm in Texas after all, so I'm in the right place for that. So if a gun were held to my head, it seems that genus 4 doesn't arise as a congruence genus over q. But it does arise from a field of degree 4 over q.
So who knows what that numerology means, but it is a challenge to do things, I think, when the things have got bounded degree. So you may well ask, well, what about just constructing examples of Riemann surfaces of big lambda 1? Well, there are constructions around.
The work of Bob Brooks and Makover that give a construction to cusp surfaces of large lambda 1 and tweaked those guys to produce closed examples with large lambda 1, whatever large might mean here. And then one might wonder, well, I was talking about these arithmetic surfaces or these congruence surfaces. These are kind of special guys in this moduli space of genus g surfaces.
So what, typically, is the behavior of these surfaces? Are they big, or do they have small lambda 1, do they have small Cheeger? Well, in fact-- and I don't know too much what I'm talking about next, so I'm just going to say to you-- that no, basically, [INAUDIBLE] showed that as g goes to infinity, basically the random surface has got spectral gap. So these are the typical things that one finds in high genus, so why can't we find congruence surfaces? OK, they're not random perhaps, but nevertheless, the indications are that they should be out there, because congruence stuff has this kind of property of being big lambda 1.
OK, so again, let me get back to the issue of can one build guys in a fixed commensurability class? I mean, I told you I believe that doesn't happen for congruence surfaces. And I kind of want to just do this. OK, build a sequence of surfaces of every genus in a fixed commensurability class that have spectral gap.
Well, I don't know how to do that. And well, with my final Cornell person, Kassabov. So Kassabov proved that Sn and An can be-- their Cayley graphs can be made expanders on some banded collection of generators. So if you can make these guys expanders on two generators, then you can be able to build surfaces that are expanders, and therefore, have this spectral gap. The end. Thank you very much.
[APPLAUSE]
STEVE: Is there a question for Alan? Yeah.
AUDIENCE: This idea that [INAUDIBLE] but this idea that the [INAUDIBLE] should [INAUDIBLE].
ALAN REID: No, no. So there's precise statements about what the expectation, that log of the order of the torsion part of the homology divided by index, that should go with precise asymptotics.
AUDIENCE: So [INAUDIBLE]?
ALAN REID: Well, because there's no torsion for link groups in the 3-sphere. So we have this--
AUDIENCE: So is that just in the 3-sphere?
ALAN REID: Well, I mean, so the prediction might be more generally if you fix the closed 3-manifold that you're looking at, then there's a fixed torsion there that you see from that fixed 3-manifold. In the case of the 3-sphere, it's zero. And so for a fixed closed 3-manifold, and you're trying to understand more generally, can you have infinitely many congruence links, then again, I would believe that torsion might well help there. Because there's only so much torsion that you see to start with. And you should quickly start to see torsion that's got nothing to do with the base manifold.
AUDIENCE: Just for my education, if you [INAUDIBLE].
ALAN REID: That's what we're doing.
AUDIENCE: Well, let's go back to the beginning where you--
[LAUGHTER]
AUDIENCE: --is it true [INAUDIBLE] that there are only [INAUDIBLE]?
ALAN REID: Yes, that's true. That's true. And it uses the same argument. It uses Zograf together with spectral gap. Because one still has spectral gap for these congruence co-compact examples.
AUDIENCE: So I'm just wondering if you can formulate an analogous question in dimension 3 which isn't so tied to links--
ALAN REID: Yes.
AUDIENCE: --that include both compact and non-compact.
ALAN REID: So one thing that one might say, for example, let's just fix attention on these Bianchi groups. And we could think of-- let's think of co-compact arithmetic lattice coming from quadratic imaginary number fields. Question-- are there infinitely many rational homology spheres in those commensurability classes? Because [INAUDIBLE] should allow you to relate that statement, suitably interpreted, to the cohomology of certain congruence subgroups of the Bianchi groups. So that's the kind of analogy that people have been thinking about, in fact.
AUDIENCE: [INAUDIBLE]
ALAN REID: Yeah, or not. But again, that gets back to this question about you look at these gamma 0 P's, where let's say P is a split prime in the case of these quadratic imaginary integer rings. It's unknown to the number theorists whether those guys have trivial cuspidal cohomology [INAUDIBLE] or non-trivial cuspidal coholmology [INAUDIBLE], i.e., other elliptic curves with the right blah blah blah.
AUDIENCE: Do you have an idea of what the size of this very explicit [INAUDIBLE]?
ALAN REID: Yes, it's something-- well, I may give you the Cheeger version. I think it's log 2 [INAUDIBLE].
AUDIENCE: [INAUDIBLE]
ALAN REID: So Mirzakhani has this kind of theorem that there's an explicit constant that's independent of the genus that says that the Cheeger constant, or lambda 1, is bigger than C. And I think for Cheeger, it might be log 2 over pi plus log 2. I mean, [INAUDIBLE].
AUDIENCE: So it's a number smaller than [INAUDIBLE].
ALAN REID: I can't do the math at this point of the day, John. You can figure it out.
AUDIENCE: How important is it?
ALAN REID: It is indeed very important. And it probably should be [INAUDIBLE].
AUDIENCE: [INAUDIBLE] is arbitrarily large [INAUDIBLE] and therefore, lambda 1 can't be bounded [INAUDIBLE].
STEVE: OK, so let's thank Alan again.
[APPLAUSE]
Thurston had a longstanding interest in understanding geometric and topological properties of arithmetic hyperbolic manifolds. For example in his 1982 Bulletin Article, as Qn 19 of the problem list, Thurston posed: "Find topological and geometric properties of quotient spaces of arithmetic subgroups of PSL(2,C). These manifolds often seem to have special beauty." Alan Reid of the University of Texas at Austin discussed this theme, including recent work and future directions, at the Bill Thurston Legacy Conference, June 23, 2014.
The conference, "What's Next? The mathematical legacy of Bill Thurston," held at Cornell June 23-27, 2014, brought together mathematicians from a broad spectrum of areas to describe recent advances and explore future directions motivated by Thurston's transformative ideas.
