 All right, thanks Dick. Yeah, sorry. I'm late So All right, so I talk about The reefers condition for Congress covers So last time I was here at ICTP, I guess I remember standing in the lunch line with Alan Reed and He was asking me whether whether the Bianchi groups might Might be virtually fired and whether the sort of techniques we had used to analyze the the LERF property for Bianchi groups might might be useful for that and I Guess out of that suggestion. I think a year or two later. I came up with this condition for three manifolds to be virtually fired and and that's Then led to this this this reefers condition that I'll explain And so now This semester we've been at the Institute for Advanced Study and he asked me well, can you Essentially the question. Can you can you do this? similar thing for Congress covers at least of Bianchi groups and so That's that's what I wanted to talk about today is sort of what progress I've made in that question It's a little more specific question than that. So I'll get to the more details. So so the plan for the talk is to First talk about the reefers condition and virtual firing and then and talk about Bianchi groups and then a little bit about well, then so So I'll describe the Bianchi groups and then about congruent subgroups and then about the I Guess that's our theory how you can apply it to to get certain Bianchi groups which which have Congress covers that that are that are fibered okay, so So first about the reefers condition virtual firing Okay, so here we'll have and a compact manifold with battery and Possibly closed any other characteristic equals zero. So all the boundary components if they exist are Torra, I guess I should say orientable to three-man fold The examples would be non-split link complements. Oh, sorry. I forgot to put a spherical here as well. So Here's a cover is contractable or it's Or it's Equivalent to a variety of other things like pi pi two is zero and pi three is zero And And hyperbolic three-man folds, okay, so then Another way of Constructing such things is via fibrations. So I'll talk more about this. I guess on Wednesday The M is fibered if there's a Fiber bundle we have a surface compact surface vibration over over the circle So Equivalently there's It's a mapping torus there's a Self-homomorphism a surface with M is homomorphic to the the mapping torus a fee so it's Sigma Times we multiply it by an interval and we identify Endpoints and And we get a manifold now where the boundary components correspond to the Boundary components of sigma Cross to zero when then endpoints glued together if it's all orientable then those will all be will all be Torah so So example would just be I guess the product manifold and Figure eight not compliment so if we take a regular neighborhood of the figure eight not in the three-stere and Take its compliment. It's well known to be So to be fibered over the over the circle So is the is the white head link I guess So this Fibers with a genus one once punctured surface fiber This fiber is actually in inflating many different ways Which it can do since it's its first bedding number is two All right, so so then M is virtually fibered if There's a fine sheet of cover and told of them such that I'm told of fibers Can everyone see okay? All right, so Well the goal them will be to discuss three manifolds that have a Fine sheet of cover that fibers over the circle. Let's first look at a non-example for comparison We take a closed surface of oil a characteristic lesson zero And then take the unit tangent bundle like a genus to surface Then that's a three manifold It's actually also a vibration over the over the surface but with with the fibers or or these circles and This doesn't fiber and it has no fine sheet of cover that that fibers either so Anyway, so And this is this is also Satisfies the other conditions orientable a spherical So that's that's an example of a of a cypher fibered space so The Thurston's Thurston asks this question then our hyperbolic three manifolds virtually fibered so So I answered this question. Yes, it's actually more general than that applies to many a spherical manifolds as long as you excludes Certain cypher fibered spaces and some other some other things analogous to that certain graph manifolds but otherwise by work of Schuzitzky and and wise and myself then we know that most Most most three manifolds a circle three manifolds are virtually fibered if they if they admit a metric of non positive curvature It's a cool condition. Okay, so but I want to talk about some very specific examples today. So let's talk about the reefers condition then That's used to prove this So a group G is Residually finite rational solvable This is reefers for short that if There exists a filtration of the group by subgroups to find index which Are some sense of solvable Filtration so that I guess I won't I wasn't gonna explain the terminology really but the this comes sort of from the rational derived series But I'll just describe it more directly. So yeah g not G1 So that the intersection intersection of gi this trivial that exhausts G and G I plus one is equal to The the the kernel of a map from Gi That so we map gi to z and then we then we map it to z mod nz So it's a finite cyclic cover. Well, so the so the corresponding three manifold site, I should say mi then is equal to By when am I is equal to the gi gives us covering spaces So mi plus one is a is a finite cyclic cover of Mi that's actually coming from sort of the this factorization through z means that it's It's it's coming from sort of the non-torsion part of homology and that's where the term rational Comes from the rational derived series Yeah, there's some embedded two-sided orientable surface in there, which is non-separating. So that's Responsible for some map of pi 1m to maps to z and then we take so then that maps to z mod nz and That gives us a cyclic cover of n sheets. So we'll see You know no more copies of this surface here And we've sort of unwrapped the manifold a little bit So that's M sub one and then we find some other surface inside of there. So I don't know. It's it's usually Gonna be some different surface and then we we kind of unwrap around that so we'll get a You're just schematically. I guess we'll get another cover of that which We'll have some number of less of this other sort of dual to to this some other surface with like an end and one fold cover to give M2 And then you you kind of repeat that process and you can this this condition in the intersections So these groups is trivial if we choose a base point and lift it at each stage then the injectivity rate is there is Getting arbitrarily large. I guess if we If we maybe I should say Injectivity rates if there's like a non-positive curve metric so the theorem then to prove This virtual firing So what what what's happening here? I should also say unless you started with something that's like the three-dimensional torus Product of three circles if you did if you did this in three torus case Each of these covers the the first bay number is always gonna be remaining one He always get a torus three torus cover But in general what will happen is that the each when you pass the sequence of covers the first bay number Get starts getting larger and larger And you get more homology to work with The theorem is that if As above all those conditions is Reefers then Ember to the fibers in fact Mi is fibered for some I all right, so The proof of this Makes use of sutured manifold there and I for a second time. I didn't really want to go go into it today But the roughly the idea is that there's a There's a kind of Obstruction to to fibering so if we if we have a surface like so this is not separating it's not a fiber then there's some There's some non-product piece in here Part that That's an obstruction to this complement from being a product and The idea is that if you choose the the surface right here choose the Co-homology class do it correctly then You can make it so this non-product region also maps trivially through when you map any when you map to Z so there's The the criterion is that the Co-homology class should be on a top dimensional phase so the thirst of norm unit ball if you know what that is but then Then that then that will satisfy this property and so that means that these non-product regions will kind of lift these are called the the guts These will lift to To these covers and What'll happen so they'll keep lifting to a cover until You get to a cover where they become homologiated non-trivial so maybe it happens in this M1 cover and then it turns out that your first bay number will increase and When you pass to a different co-homology class you'll have cut up this guts in a non-trivial way and there's a complexity that reduces and you get a smaller Collection of guts in some sense I think there's a variety of ways in which one could make that complexity precise, but So because of this Cofinality of these groups then if I take any loop in the guts here it'll in one of these covers it'll eventually be homologiated on trivial and then you'll you'll see that some sort of suture manifold decomposition has to happen and These complexity has a property that it can only be a finite length So at some point at some stage you'll get you'll get a surface with trivial guts Which means that it's fired over the circle. So that's sort of a brief outline of the proof Okay, so So now When I First came up with this condition I didn't know of any examples of Reefers groups except for like the free group and the free billion groups and a few other examples. So the initial examples They're like Free and free billion groups I guess I suppose you're finally generated is really the the case that we're interested in and free products and direct products and then there's a There's a class of groups that are closed under these are Include these and enclose in these operations. So right angled art and groups Which I'm not really going to talk about today, but except to point out that they're And then subgroups of Which is key here, so so if I if I take Some subgroups of the G's then that'll correspond to some covering space here And when I pass these cyclic covers, it'll induce cyclic covers of the corresponding covering space and Of the same of the same kind of a flavor. So So that's So if I can So as far as I know all all the examples know I'm let's say finally generated a reefers groups Come from subgroups of right angled art groups. So that's that's essentially the only way that we that we really know how to Define to find reefers these reefers groups so So what I want to talk about today, then is sort of a new method of approving this This condition and by happy coincidence when When I discovered this condition around the same time Haglin and wise had found Methods for producing lots of subgroups of right angled art groups. No so-called special cube complexes. And so eventually we had lots of examples of these Of these kind of three manifold groups that had that had reefers on an all group. Okay, so So now I'll talk about the Anki groups. Is there any questions so far? No any subgroup. Yeah, because the property If I take a subgroup and I intersect it with each of these GIs then I get a sequence of a filtration of subgroups which Satisfies the same kind of property. Oh just if there's a sequence and I should have said yeah, sorry In the case we'll consider today actually the ni's will all be the same number Okay, so let's talk about Bianchi groups now so We have a quadratic imaginary field you would join the square root of D minus D Here D is a Square free integer, but the square root is not an integer And OD is the ring of integers in Q of square root of minus D. So it's either equal to See adjoin square root of minus D or See adjoin one plus Square root of minus D over two If certain congruence conditions hold and I can never remember what the condition is so what I do is I multiply I check to see when this is a algebraic integer. So I multiply it out one plus square root of minus D Times one minus square merit minus D over four. Let's see. Can anyone do the math there? I can never think of my feet So what do we get? All right, I guess I have to think of my feet. Okay one minus minus D over four. So that's when these congruent to minus one mod four Okay, so Okay, so we got this condition here and then this is if D is Not congruent to minus one month four All right, so Gamma D then all denotes the group of two by two matrices with entries in In OD and then we projectivize. So this is The space of plus or minus so you can think of this cosets of the center of PSL2 that's all to ABCD permanent one and The entries are in the the ring of integers of OD so this always contains PSL2z which some of you know and love and But it's usually so this is of infinite index in in these groups in general Okay, so an example of subgroups Well, let me so I forgot to say so so Gamma D is a Greek group of isometries of hyperbolic three space And so any well in general though it has torsion because PSL2z does a matrix of trace zero for example Will be ordered will be an involution But if I take a fine index torsion free subgroup then I get a non-compact hyperbolic manifold by quotient H3 by that subgroup and So for example of subgroups pi one of the the figure eight not complement is Is a subgroup of PSL2 of 03 Okay, so a congruent subgroup is obtained by It's basically a subgroup of PSL2c Let's say up to congealcy wide commensurability that contains a principal congruent subgroup of one of Of the Bianca group so We take an ideal in a ring of integers then the principal congruent subgroup will be The elements ABCD such that these are congruent to 1001 Modulo the ideal So then gamma Let's say commensurable with gamma d so I commensurable I mean it's a subgroup of PSL2c that has intersection With gamma d is a fine index subgroup of gamma d and gamma is also discreet It is this is congruence if it if it contains one of these principal congruent subgroups Okay, so Standard example of a congruent subgroup. So that's not just a subgroup So any subgroup of gamma d that contains one of these principal congruent subgroups will be will be congruence but If you take gamma one, so that's PSL2 of Z join I Then we take this element of PSL2c and again, I'm abusing notation with not Putting in plus or minus This is in pgl2c Then gamma If we take alpha and we adjoin it to gamma where we get something in discreet, but we if we take Alpha gamma one alpha inverse Intersected with gamma one This you can check as a fine index subgroup. It's a congruent subgroup of PS of this Bianchi group So this guy is is congruence and if we adjoin an element to it then we get something that's Congruence as well But this guy here Since alpha is an evolution to trace zero and pgl2c then Alpha normalizes this group So this is actually contains this guy with index to it to the discreet group and yet it doesn't lie inside of PSL2 Z adjoin I so It that's because here this doesn't have determinant One so you have to take divide by the square root and you move outside of the the field of definition So in general though these These guys will be we'll lie inside of the the commensurator of gamma D So these will lie inside pgl2 of Q adjoin the square root of minus D. So here we allow matrices of arbitrary determinant and And then we get the commensurator. So the commensurator is the the set of elements that when you When you conjugate gamma D by them the intersection with gamma D is fine index Okay, so now Oh, I should have said this contains Contains gamma one of the ideal generated the principal ideal generated by 1 plus I Okay, so let's look at some other examples of Principal congruence groups. So what? P in OD be a prime ideal like this ideal here generated by 1 plus I and N of P is the norm of the ideal is you take the Index as a lattice. It'll be a lattice inside of OD so it'll be Some fine index lattice you take the order of the quotient or the index if you like and That there's a normal P and that will be some power of P where the norm so Where P divides The principal ideal generated by the by the the rational prime P Then we get an infinite sequence of Principal congruent subgroups by taking powers of this ideal. So We have gamma D Contains gamma D of P Contains gamma D P squared cetera, this is a co-final sequence and The claim is that Gamma D of P to the I Modular gamma D of PI plus one is Isomorphic to Z mod Pz the J for some for some J so we have a A sequence of well, you can see here that After the initial gamma D of P these guys will all be Fine index subgroups abelian subgroups. So you could actually insert a few Subgroups in between so that each of the quotients is Is coming from a cyclic subgroup. So we can refine it to a GI such that GI to GI plus one or let's say mod GI plus one Is isomorphic to Z mod Pz so The proof of this is Essentially the binomial theorem or at least the fact that every element this quotient has ordered P Yeah, so I'm using the same notation. So P divides divides P Okay, so The gamma D of P to the I again, it's really the sort of double cover sitting inside of SL2 But so be his notation. So this is content. This is contained in I plus M2 of P to the I is another way of writing it. So we take the identity matrix and then we add some Matrix whose entries are all in the ideal P to the I So let's take one plus a I plus a contained in here and you take I plus a to the P will be we expand that out so we get I plus P times a plus P choose two times a squared up to a to the P and we see that each of these is contained in P to the I plus one so there's some as as well and and that means that I plus a has ordered P inside of the quotient group when we when we quotient by Gamma D of P to the I plus one And then you also can just check that it's the quotient group is commutative as well Essentially because when you take the commutator of two such matrices the linear term will cancel out like in I guess in like in Lee theory okay, so That that quotient then is a is a free a billion P group and so but now we can get to the actual question of Baker and Reed When is Is this sequence or some I guess refinement of it and the nice thing here is that these are congruent subgroups so you're getting Concurrent subgroups of of these Bianchi groups if this holds that that eventually fiber over the circle, okay, so The theorem then is that that holds under a very simple hypothesis so so this holds if H1 of this principal congruent subgroup But take H1 with integral coefficients. Let's say I guess I'll put the coefficients in Has no p-torsion there's no no p-torsion homology of this Of this principal principal congruent subgroup Now You can you might you might get the idea of How this works here that if gamma D of P has no p-torsion and that means that when we pass to this index p to the j subgroup that It sort of has to be factoring through There's some map here From the torsion free part, but there'll be some map from Z to the J So that this is in the the kernel of that map And then then this goes to Z mod P to the J And this will be the kernel of some such map. Does that make sense connect? Sorry, can everyone see that? So so at least this this first stage here will be a Sort of first stage of a of a reefer sequence. Is there any questions so far? I Don't know how much number 30 people know of it anyways So this now gives many examples of Bianchi groups the virtually fiber which have a congruence cover the fibers For gamma D where D equals One two three seven five 11 15 19 23 31 47 71 so Anyone recognize the sequence so that they prove that Baker and Reed prove that for these These these are the only Bianchi groups which have these are the the only Bianchi groups with the principal congruence Link cover so they enumerated these in some paper and when you have a link complement there's no torsion in first homology so in particular no p-torsion and For some of these there's a variety of different prime ideals you can choose there they give you link covers So I didn't I'm not going to list out all the Ideals, but you can look it up in their paper okay, so I Likely it's it seems likely that This could hold true in much more generality, so we've used a very strong condition here to verify the no torsion but On the other hand it seems like it might be hard to to check this for For infinitely many D's unless you have some kind of general of general theory I Also asked and I asked I asked Richard Taylor whether he thought this theorem might be interesting and he didn't seem to think so Or didn't I mean he didn't see any application maybe to number theory or whatever maybe So I'm not really sure what the what the point is I guess Except that it's a natural question. Well, so I guess part of the point is that it gives a new way of constructing Reefers groups so that means that That these these link groups in particular are satisfied the the reefers property, so we don't know in general I should also add a caveat that it's possible that all these principal congress covers they construct. They actually might be fibered So Which I haven't attempted to check. I think some of these they don't even have link projections for but in principle You can run some program and check it for all these so this theorem might be extremely trivial, but In any case this technique at least should hold should hold a much greater generality All right, so I'll sketch the proof I guess so I have how much time now like Few minutes. What's it? Oh? Okay, thanks, sorry Okay, so let's let's look at it so This uses bas-serre theory a special case of the of an example of a Baratouche building So what K be the right a Q of square root of minus D The P is the the P at a completion of With valuation The nonic medium valuation V sub P associated to the the primary DLP So I'll I'll give a specific example of this in the sector for you not familiar with it in certain cases like for for the ones the beyond groups of pasta or one you can actually see these Bas-serre tree is much more directly Okay, so O V is the K is the valuation ring and This actually will contain O sub D the ring of integers and then we can fix Pi and K with Evaluation of pi equal to one and then P will be contained in the Ideal generated by by pi this is an ideal and O V And all ideals when you so the nice thing about passing to this completion is that It becomes Well all the ideals are out of the form the end for some for some end Okay, so let's look at our example Oh one we have Z join I the Gaussian integers and then we can take pie in here it'll take our ideal to be the one we have before one plus I and Hi, we can actually just take to be one plus I since it's a principal ideal and Oh one O V modulo pi then will be isomorphic to Z mod 2 Z actually isomorphic to Oh one mod One plus I In inside of the ring of enders, so this sits inside the completion So this is the the pediatric integers inside of this number field with respect to this this valuation So what's K? Well, you can think of it as it's a generalization of paedic numbers if you're familiar with that we can think of it as sort of formal expansions in Our K is equal to the sum equals K to infinity epsilon J. Oh, I should have said here that this is represented By the coset zero and one we get epsilon J one plus I to the J Or epsilon J is contained in so these are sort of like binary expansions, but we're going in the positive direction And then the valuation of some Z and K is equal to the minimum of J such that epsilon J is not equal to zero and Here K is K is an arbitrary integer So these can be sort of Laurent expansions if you like so this so this contains Maybe not so obvious, but it contains the the number field You join the square root of minus D So it's sort of clear I think that it contains Oh one, okay, it certainly contains oh one Then we can divide by one over one plus One over one plus I so we can take so these will be precisely the sort of finite expansions here But in fact you can inside of this completion you can invert any element that's co-prime with one plus I and so you can Get all the other Elements of the number field sitting inside of there. So this is contained in In K as well Okay, so now the bas-ser-tree. Is there any questions on this the vertices are OV lattice season K squared up to homo 30. Maybe I shouldn't say lattice. I should say rank two modules and Or maybe I'll say yeah, they're sort of commensurable with OV squared. That's kind of what I mean by a lattice So up to up to homophedy Up to multiplication by elements of K Excluding the zero element Edges are Are such modules So we have say lambda one will be adjacent to lambda two if Lambda one contains lambda two and Lambda one mod lambda two is isomorphic to OV Modulo Modulo pi and then this is our Our finite residue field. So like in this in our example. This would be Z mod two Okay, so then note that this is This looks might look asymmetric but So But if we take lambda two then it contains By this condition here, it'll contain pi times lambda one and we'll have because it's ranked to the corresponding quotient lambda two mod pi lambda one Well, we'll satisfy the same condition So that's why we take things up to homophedy so that it's this condition becomes symmetric and so if we so we can start with OV squared and then this will be adjacent so that will give us some Vertex of the bastard tree and then we can Sit inside of here, there's OV direct some high OV and they'll be High OV direct some OV sitting inside of there and these both clearly have this property that this they're both contained inside of here and This module of this will be Isomorphic to OV mod pi There's a There's another one we can get We can get the span of OV one say Minus one direct some OV of pi comma pi if I think if I did choose that right then that's um Anyways, that's that's an example so in in the case of our example at hand these are the the bastard tree has degree three in general It's the degree of the residue field plus one the the projective line over that residue field All right, so Now GL The claim I guess is that this is a tree But in general what we'll see here are the degree will be will correspond to FQ P1 Or FQ is the residue field a finite field of some prime power The claim is that this forms a tree and Sarah has a Beautiful book written about this called trees so you can read all about them there Okay, so the The nice thing about this tree is that GL to K acts on this tree So GL to K If I apply some linear map to one of these lattices then I get a lattice or one of these modules that has Will will be commensurable with oV squared so That's that's not that that hard to check and then it also preserves this sort of jason c property of these of these lattices In fact adjacent lattices are represented by By Frick and pollution so we let epsilon be 0 pi 1 0 Times oV squared What do we get there if I'm thinking of oV squared is like Represented by a column vectors. So this is an involution you get one of these three guys. Which one is it what's it? Second one. Yeah, thanks. Okay, so pi acts on the second coordinate. So we get pi oV Direct sum and then one sends this first coordinate to the second coordinate. So we get oV and then we take epsilon squared Well, when you square this it's an involution in the projective group so you get pi 0 0 pi you can just check that I want So we get pi oV Direct sum pi oV But this is equivalent up to homo 30 to oV squared So that represents so we have this involution that it changes these adjacent Lattices decent lattices here Okay, so and that's called a fricca involution I think or some generalization of fricca involution All right, so so I gotta finish up here. So now let me just Explain how you get this condition. So let's let theta be SL2 oV and we take SL2 oV and again we should projectivize everything here, but intersected SL2 oV to the epsilon This will be abcd such that Pi divides B So in other words the valuation of B is equal to 1. So this is If we then intersect it with SL2 oD, then we get a similar sort of thing here. So If I take SL2 z of i intersected SL2 z of i to the epsilon and this will be the 2 by 2 matrices such that B is congruent to 0 mod 1 plus i It turns out then if you intersect With all these adjacent guys conjugated by their corresponding fricca involutions Then you'll the intersection of all those will give you the principal congruent subgroup when you induce it on gamma D So we restrict to But actually we're going to restrict to Gamma of P this hypothetical principal congruent subgroup that is Horses and has no p-torsion in first homology like these link complements of Baker and read and This gamma of P Mod gamma of P Intersected with gamma P epsilon is A billion p-group and since by our hypothesis this is torsion-free then this will be this will be sort of a reefers Subgroup so now what you do is you sort of work your way outwards in the in the in the bas-ser-tree GL2 k squared x transily on the vertices you can check We're actually yeah, it's a GL2 of q a joint square to minus d will as well So we start with our vertex and we put it we can take gamma P and then we can conjugate it To the various vertices of the tree so here epsilon and then there'll be a gamma of P the G1 gamma P P2 and then you work your way outwards and Gamma of P contains gamma P intersected with Gamma P to the epsilon and contains gamma P Intersected gamma P to the epsilon intersected a gamma P to the G1 Etc and the claim is that this is a This is a reefers sequence and so the point is that Each of these edges are sort of conjugate to the to the the initial edge that we looked at where This subgroup is it is a P group so each time we intersect with a new G1 So we do this in such a way that all that we're always Taking subtrees and we adjoin on one leaf at a time So each time you adjoin a leaf you're intersecting that group This group with the next one which corresponds to an intersection of the two edge groups of that leaf Which is an abelian P subgroup factoring through Z and so it induces a P subgroup index subgroup, which is coming from Non-torsion non-P torsion of malady the infinite order homology So that's sort of a summary of the proof they all stop there