 Very interesting conference and I'm glad to be back again. So today I want to talk about joint work with Mark Coller at the University of Illinois, Chicago and the background for what I'm going to tell you about was described by Steve Boyer Earlier in this conference and those of you who are last week heard talks by Josh Green sort of related to the conjecture that motivates this but still just to Recap let me give the basic definition that I want to talk about and I will Remind you what the conjecture says so a group Should always think accountable group G Is left orderable if it has a total order, which is invariant under left multiplication So where if I have two elements a and b where a is less than b this implies That c times a is less than c times b for all a b and c in the group and by convention I want The trivial group not to be left orderable. So let me Let me exclude that so examples Sure, you've seen but I'll me to remind you something like the integers Under addition with the usual order on the integers is an example something like a free group It's less obvious, but true another example would be Kind of thing I'm going to be interested in today's three manifold. So for example the fundamental group of a prime Three manifold with positive first Betty number Will always have orderable fundamental group and some non examples I want to think in some sense of a grouping left orderables. It's sort of a big group any group with torsion torsion elements Cause problems here or another example of a again a three manifold example. Let's say the fundamental group of the weeks manifold Smallest volume close type of all so the it's the definition of An orderable group and what I'm going to be interested in today is when is the fundamental group of a three manifold left? Orderable, so let me just So I don't have to keep saying fundamental group of let me just define Three manifold why I'll call it orderable when it's fundamental group Is left order? So The conjecture that motivates what I'm going to tell you today Is suppose we have a rational homology three sphere some and Let me make let me assume it's irreducible Then the conjecture posits that the following three things are equivalent the first is that The fundamental group of why is left orderable? the second is that The manifold has some non-trivial Let's say hangar floor homology So it's what's called why is not an L space? and then three that the fundamental that sorry that The manifold why has a co-orientable taut foliation The conjecture is that these things are equivalent the only implication that we currently know amongst them is that Having a taut foliation implies that you're not an L space So these three things I'm not going to tell you about these Think you already heard something about that, but Our priority these things things don't look like they're connected this conjecture is Surprising to me and to be honest with you. I don't believe it But despite that Show how wrong I am a lot of evidence has accumulated over the past decade for this conjecture, so Maybe just to mention two things combining work of Boyer and clay with work of Hanselman Rasmussen Rasmussen and Watson this conjecture is true for all graph manifolds So anything that's built up solely of cypher fiber pieces Another piece of evidence I've done some experimental studies of this question, so it's true for at least a hundred thousand Hyperbolic close hyperbolic three manifolds in some some senses or another So in fact although I my work here was motivated by a desire to disprove this conjecture What I'm going to tell you about today is further evidence for this conjecture in particular I'm going to tell you about ways to produce A lot of orders. I'm going to give you to describe a tool for producing Producing orders on three manifold groups and then use that to prove some some theorems which provide evidence for this conjecture Are the questions so far? so the setting for today is we're going to have Let's say we have a rational homology three sphere and We have some not and I'm going to be interested in the exterior of this They'll take M to be Y minus an open regular neighborhood of K and Always going to assume that M here is irreducible And what I'm going to be interested in is Dane filling on this manifold M Right the manifold M has boundary which is a torus So you can take a solid torus and glue that solid torus to M along their boundary different ways you can do that and what I'd like to Explore is which Dane fillings on M are orderable if you believe this conjecture That's Steve. I'm sure talked about earlier. This tells you certain things about how you expect what kind of explain a minute how you parameterize the Dane fillings but sort of Orderable fillings should come in kind of families and they should behave in certain ways and What I'm going to do today is give two theorems which are compatible with that those predictions and I'm going to introduce so I'm going to state the two theorems and Then I'm going to spend the rest of the time describing the technique behind them, which is hardly new The technique I'll use for producing these left orders is something that's been used for going back to work of of Eisenbud Hirsch and and Neumann side from Fiber case. That's almost 40 years But the new Contribution will be sort of organizing this This technique into a picture Which we can argue about so after giving the theorems Explain how the theorems are derived from pictures from pictures such as these Okay, so The first theorem I want to talk about Said this all joint work with Mark Caller. This theorem was also discovered independently by Steve Boyer So there'll be some technical hypotheses that I'll have to define but first. Let me just write down the statement so Throughout my talk. I'm going to have this setup so I have some Not exterior in a oh Sorry, I said rational homologies here. Let's Can work with that, but let's just for simplicity. Let me say an integral Homology three sphere so you can think about it as just the ordinary three sphere, but it is more general than that So if the Alexander polynomial of the manifold has a simple root on the unit circle And an additional technical hypothesis. I'll explain a minute called being lean then a Lot of Dane fillings on this manifold Are orderable So then there exists some epsilon greater than zero such that If we look at a Dane filled manifold So I have to describe my notation for parameterizing these Dane filling. Sorry get to that in just a second But the Dane fillings will be parameterized by a rational number So that this Dane filling is orderable for all choices of R In let's say the interval from minus epsilon to epsilon so R is going to be a rational number So saying that if you're in this interval Then your Dane filling is always always work and that so that's a statement And then if you combine this theorem with results of Rachel Roberts then You learn that in fact the conjecture here is Holds for these manifolds I'll just put it here more over though specifically Roberts constructs taught Foliations on all Dane fillings in this interval and so You're in the situation where you do have a top Foliations hence by this theorem You're not no space and what we're proving is that the manifold is over Well, I mean yes in theory But actually I mean In Roberts results you can actually take epsilon to be one so Epsilon's less than once other questions Okay, so so I need to tell you what lean is and also fix my notation for Dane filling right so here Done the usual thing we fixed the basis called mu and lambda for the Homology of the boundary of M Where here talk is not going to survive. I Hope no one after me is giving a chalk talk. There's only this whole box here So we're here you as a meridian for the not K and Land does the homological longitude so in other words lambda becomes zero in The homology of M itself the usual just as an S3 meridian longitude framing and then if we take Rational number a over B the Dane filling M of a of B This is M union a solid torus attached So that the curve a mu plus B lambda that Curve on the boundary torus bounds the meridian disc so that specifies the Dane filling completely and then I guess the Remaining thing I have to tell you about is this Definition of lean This is somewhat technical and you can replace it with an even more technical hypothesis, which is weaker. Anyway M is lean means that if you look at the Zero Dane filling on M. So this is the only day in filling which has positive Betty number means that this guy is prime and The only incompressible surfaces in this thing are fibers and vibrations over the circle The only incompressible if you prefer the term essential in this closed manifold is a fiber over this so This manifold again the first bay number is positive. So it does contain an incompressible surface There aren't very many. So I think that's now a complete statement of the theorem Questions. Oh, so maybe before Going on to the second result. I should talk about How restrictive these hypotheses are? It turns out that having a simple route On the unit circle is very common So for example, if you look at the 1.6 million knots with less than 16 crossings some 80% of them have Simple routes on the unit circle have Alexander polynomials with simple routes on the unit circle And we just did sort of a random sampling of a hundred thousand other knots with between ten and sorry between a hundred and a thousand crossings and Those more than 99% of them had Alexander polynomials with simple routes on the unit circle So I think it's reasonable to view this this hypothesis as Generic it's a very weak hypothesis This hypothesis is probably not generic the thing we actually replace it with in the guts of the proof might well be generic, but I It's sort of hard to investigate experimentally But certainly this applies to many many examples And the second theorem that I wanted to talk about before explaining to you the Ideas the machinery that describes the pretty pictures you can look at up there. It's a following so our second result is suppose That our manifold M here again as denoted there is hyperbolic So it's the interior as a complete hyperbolic metric of finite volume then so Describe more in a minute. There's a a number field that's associated with a hyperbolic manifold called its trace field and The hypotheses of this theorem are on that trace field. So if the trace field of M so some number field Some finite extension of Q has a real embedding again, we're going to get some kind of Going to show a bunch of Dane fillings are Orderable so in particular The first conclusion is that there is an interval Though it's not sort of symmetric about zero like the first one, but maybe it's of the form minus infinity up to a or possibly from a up to positive infinity where every Dane filling is Orderable and second result concerns Looking at branch covers right, so if we look at the branch cover of this integral homology sphere Branched over the knot K then it turns out you can also show that those guys are orderable so for a large N The n-fold Cover of this integral homology sphere why branched over K is order. So Yes, those are two Two of our results. Are there any questions? Ah So Dave asked how restrictive is that the trace field having a real embedding? I think that's almost certainly generic so Matias Garner has computed the trace field for about sixty thousand Hyperbolic three manifolds everything that could be triangulated with at most nine ideal tetrahedra and of those 95 percent of them had trace field with a real embedding and And I would only expect that number to go up So I didn't say really what the trace field was So You have your whole known me representation of this hyperbolic structure Map from the fundamental group of M here into PSL to see you look at all the traces of elements In the image of that representation because of local rigidity it turns out that those Traces those are all actually algebraic numbers and they generate a finite extension of Q called the trace field and Anyways just some number field associated with the three manifold and just to say has a real embedding just means you can embed that Abstract field in the real numbers in other words. There's some polynomial associated With this manifold that defines this trace field, and we're just saying does that polynomial have a real root so more complicated a manifold typically the more complicated this field is and You would expect or what happens in Garner's data is like something like a third of all the roots are real so Ian Sorry does it depend on? Yes, it does Yeah, so this manifold are just erased at M here a view as a knot in this integral homology Three-sphere Y, but it's actually a knot in several different well infinitely many different integral homology spheres And so you can actually apply this theorem for a bunch of different branch covers and N would be very Depending on that. No, it's arbitrarily large No, so there's no the conclusions. I mean our in both cases many Dane fillings Are orderable, but the hypotheses are completely different although in the end. They'll be proved by the same technique. Okay, so I Want to now go ahead and try to explain spend the rest of my time explaining where these pictures come from And I should say that I mean I already said I guess that the technique is not new and in particular especially for this Result here. There are a number of antecedents of our work due to in particular Trann Hakamata and Terrigaito Cameron Gordon proving results In some sense, I think I mean at least in some cases Not only do are we generalizing their result, but we're generalizing the proof And also I should mention the work of Cameron and Ty Lidman studying what happens for these branch covers Okay, so What's the technique? so One way to show a group is left-orderable is if it's a subgroup of the group of Orientation preserving homeomorphisms of the line and for countable groups such as though pre manifold fundamental groups are talking about here In fact, this is if and only if Usually, I actually just view this as the definition of Left-orderable means you're a you have a faithful action by orientation preserving homeomorphisms of the line And here's the technique that I'm going to use to produce to embed groups in In homeoplas are so I'm going to start with Very simply group. Let's look at PSL to our which of course I view as the Orientation preserving isometry group the height of a plane And I'm actually going to view this as a subgroup of homeomorphisms of the circle via the action of the group G on the circle at infinity so you can of course determine The element of G sure purely by its by its action on the boundary and now I'm going to look at the universal cover of The circle right so we have our usual picture So we have our circle. Here's this universal cover Let's put these guys at the integer point zero one two three minus one Called the covering map pie or something like this and I want to consider The subgroup of homeomorphisms of the universal cover which are lifts of these homeomorphisms G That are homeomorphisms of the circle and so if I look at I'm going to find G tilde to be equal to Whiffs to the universal cover of homeomorphisms in SL2R And so this is a nice subgroup of orientation preserving homeomorphisms of the line and Of course a number of different ways you can describe or think about G tilde Of course, it has an actual map to G which is surjective The kernel is just the covering translations right the integer shifts It's another way of thinking about G tilde is it's the universal central extension Universal cyclic infinite cyclic extension of G Another way to say it is that well this g group G is not simply connected It's homotopy equivalent to a circle so you can take its universal cover Which is of necessity a league group and that's G tilde so in particular the the thing I'm going to start considering like representations to G tilde and the thing that's different from The representations that have come up in most of the other talks here is that this is not it's a league group to be sure But it's not linear does it you can't embed it in SLNR for any end such as life so Let me just give an example of using this Yes, oh, that's right. I erased the thing that I put to remind me not to write. Thank you Cameron So just as an example Application of the philosophy I'll be using let me prove that the free group is left-orderable Right. Well, we certainly know We can embed Our free group into SL to our we can think about it as a lattice in there if we like if we embed it properly and then Well, our group is free right so there's no obstruction Lifting that to a map the homomorphism embedding here to a homomorphism here It will still be faithful because this thing is And so now we've embedded this free group into G tilde Which is a bunch of homeomorphisms and so that tells us that this is left-order And this is the only technique I'm going to use Today basically to prove To prove groups are left-orderable. I'm going to find homomorphisms to G tilde and to be able to do that in a systematic way to produce To prove the theorems that I put up I'm going to organize The set of all representations of the fundamental group of M are manifold with torus boundary into G tilde into a picture That looks like of which those are examples But before I do that I have to Just define Tell you a couple things about the group G tilde so the first is that There's a translation number So this is a map from G tilde to R It's not a homomorphism Quasi-morphism. Anyway, it doesn't it's not so important The find as follows So if I take an element of G tilde you look at How far it translates stuff, let's say to the right On average specifically Going to be defined as some kind of limiting thing. So Take G tilde to the nth power apply this to some initial base point X zero look how far you've moved in n steps Divide that by n. So that's the average amount you move and it turns out that this is independent of the choice of an initial base point So that signs a number to every element in G tilde and To give you a feel of what these numbers look like let me Talk just for a second about the kind of elements you have in G tilde So as you all know the group SL to R. There are four kinds of elements You can partition it up into the identity element Then there are the elliptic elements These things fix a point in H2 and rotate about it. There's the parabolic guys Which don't have any fixed points on the interior, but a unique fixed point on the boundary and then the hyperbolic ones Translate along along some axis. So if we look at our map here From G tilde to G call that P or something we can take the pre-images of this Partition here and We'll think of G tilde as partition into the central elements So that corresponds to my Z over there The pre-image of the elliptic elements, which I'll call elliptic the pre-image of the parabolic ones and And the hyperbolic ones. So the translation numbers What is the translation number look on these different kind of elements the parabolic and hyperbolic ones These things have fixed points when acting on the circle downstairs And that turns out the force that if you look at the translation number of an element which is not elliptic It turns out that this is always an integer the translation numbers of the elliptic elements can be any real numbers And it relates to the geometric formulation of the of an elliptic element By the following formula. So if you look at the translation number of some elliptic element in G tilde Then if I view this modular one take the fractional part This is just the rotation angle Take this elliptic you push it down to G. It's just the rotation angle of the corresponding elliptic Acting on H2 about its fixed point divided by 2 pi one over 2 pi rotation angle of Let's call this elliptic element G tilde P of G tilde About fixed point. So in particular There's a subgroup if you think about our standard picture here of hyperbolic space if you look at Rotations about the center here in the Poincare disc model Then the pre-image of this up in G tilde is a copy of of the real numbers which just corresponds to Translating the universal cover here by by any particular element in R. So there's this sort of The subgroup here which contains the center, but then if I if I were to pick a different point I would get a different copy of R. Okay So we have this translation number So how am I going to try to use this to get a picture? We're all here for I presume so If we have the we have some representation From the fundamental group of of M. Remember, that's my manifold with torus boundary the thing I'm doing filling I want to consider the map So I want to focus on What this representation looks like on the fundamental group of the boundary? And in particular, I'm just going to record that via the translation number. So I'm going to consider the map Where I take we're here. Let's say Yoda is the map from fundamental group of the boundary Into the fundamental group of M So we'll go from here to the fundamental group of M to g tilde and the translation number So this is a map from pi one of boundary M R and You know your basic facts about homogeneous quasi morphisms like this thing You can see that because this group is a billion this map is actually a homomorphism. It's not completely obvious, but it's true So in particular I it's a homomorphism from Z squared to R, which I going to think of as an element of The first homology of the boundary So this is giving me something called called the translation along the boundary of row This is an element of H upper one of the boundary of M With real coefficients so I mean just Don't believe my mumbling about why this is a homomorphism. I can just make this concrete for you Right. We had our basis U and lambda for The homology of the boundary It's how we defined our day in filling you can look at the algebraic dual basis U star lambda star basis for The co-homology of the boundary and then in these coordinates the map What is the boundary translations of row? It's just you take row you apply it to mu you take the translation number of that That's some real number and then you do the same thing for the longitude so associated to each representation like this we get a Pair of real numbers is going to giving us some kind of points points in the plane as you see in the picture So in particular then now I think to find the basic object we use So let's first I want to look at The story kind of breaks up into pieces. I'm going to focus on one piece Right now. I want to look at all homomorphisms from the fundamental group of M into g tilde Where I mean, I'm looking at translation numbers here, and I said that The parabolic and hyperbolic guys always sort of have integer translation numbers So I've got to kind of ignore those representations So where let's say row of the fundamental group of the boundary Does not contain any hyperbolic elements that we're just interested in Representations once you're stricken the boundary there either these elliptic elements basically translation like things with parabolic elements And then the main definition It's what we call the translation extension locus and this is It's just the image of all of these representations under This map here translation boundary M. So Each representation you look at how its boundary translates And then I'm going to take the closure of this The sitting inside H1 of the boundary of M a plane And let me call this I guess I should give it some notation. Let me call it yell of Out so it's images of representations or the name extension its points for which It's the homomorphisms from the bound pi one of boundary M to R Which can be extended to? Representations to g-tilt are the questions. All right, so now I think to Now I need to show the pictures clearly So and there's a basic basic tool that we provide in our paper is a certain structure theorem About this extension locus and sort of put it up incrementally as I sort of talk through Some some pictures So this is a picture of the translation extension locus for the exterior of the Minus 237 pretzel knot. Okay, so this is a picture in the first cohomology of the boundary of M R2 and so this direction here This is the mu star direction And this direction here. This is the lambda star direction so the first thing you notice is that my new star axis only goes from zero up to one and the reason I only drew that part is that one of the basic Properties of this translation extension locus is it's periodic with respect to horizontal shifts so the extension locus of M So okay, so first I should have said what is it in this picture? It's all the purple stuff That's the translation extension locus the purple stuff. So the first thing you notice is a bunch of arcs So this is a locally finite union of analytic arcs and isolated points It's always invariant Under some symmetries. So under a horizontal symmetry. So like a B Goes to a plus one B horizontal translations by integer amounts and also It's a second symmetry which in this picture Manifests itself by being if you rotate about this point here. It's the point one half on the X axis if you rotate about this by 180 degrees it sends it to itself and that's Equivalent to saying that if I drew the whole picture, which I didn't that you have a rotational symmetry about about this point So there's this infinite dihedral group acting on the picture Which allows me then to only draw this finite Finite vertical chunk. So you see the vertical direction you're not just going from zero to one or something this case We're going from minus six up to plus six But it turns out that in fact this thing can't extend arbitrarily far up or down So another property is that the height in either the positive Or negative direction is bounded by twice the genus Of this manifold genus of a cypher surface minus one The manifestation of the Milner-Wood inequality So once I mod out by the symmetry That effectively glues this side to this side My translation extension locus is actually just a compact thing. It's a finite union of analytic arcs Now there's some some points here On it which are marked In particular along this axis So the two special kinds of points that I have to mention One of which is I took the closure of this thing And so the points that I add on by taking that closure are the ideal points There are no ideal points in this picture or in any of the pictures. I'll show you and there's also This the representations we're looking at We excluded the ones where the boundary acts by hyperbolic elements So we either have ones where the boundaries acting by elliptic elements or by parabolic elements and the parabolic elements well Those things parabolic always have integer Integer translation numbers. So these dots here are half dots They're half dots you see because this side you should view as sort of being glued to this side So these these half dots here these Correspond to things that come from representations to g tilde where when you restrict to the boundary You're getting parabolic parabolic Elements whereas all these others when the translation numbers between 0 and 1 these are always corresponding to guys where the boundaries acting by elliptic elements so there's also these special parabolic points and One thing we show is only finitely many of these once you mod out by the translational symmetry and Okay, so then Let's see what else do I need to explain in this picture. Oh, there's the difference between the black and the green dots so in the second theorem there was an hypothesis about You know if the trace field had a real embedding, right? So if the trace field has a real embedding Then what that gives you is you can take the whole known me representation of the hyperbolic structure So that's a representation, which is where the boundary acts by parabolics And then you can Galois conjugate the whole thing into SL2 are and so you get a representation into the group G from this real place and what I should have said and neglected to is that In the situation we're in here The exterior of a not an integral homology sphere H upper 2 is 0 So the I didn't say it the if there's an obstruction To lifting a homomorphism into G to 1 and G tilde Which is just an oiler class that lives in H upper 2. That's the oiler class of the flat bundle Associated to the circle action of G so in our situation if you ever you have a real embedding of your trace field you get some point in this diagram And the convention here is those points which are geometric in this weird Galois sense. Those are the green points So in this case we have Well depends on how you count them essentially up to orbit up to action of of the symmetries one One of these what is a Galois conjugate of the whole only representation and the others are just random random guys okay, so That's a Picture are the questions Yes So in the examples that I'll show you today you never Never get up to this genus bound and One can actually well Yeah, and in particular at least for the parabolic ones Like the fact that this so that I should say this thing has genus 5 Okay, so conceivably you could get up to 9 and we only get up here up to 6 And you can sort of show that you can't have a parabolic point up here at height 9 By using some ideas of Calgary Okay, so I think what I last thing I need to do before showing you the rest of the pictures highlight of the talk It's just to say how does this relate to The theorems that I started with So here's the key lemma of how we can use a picture like this to prove That Filling is audible so if we have What can a dain filling here? Corresponding to the rational number a over b if this is irreducible and the following line All this L of r so this consists of all elements of H upper 1 which I think of as homomorphisms from The homology of the boundary to R So such that they kill the slope corresponding to this a mu plus B lambda And so this is a line. It's the line of slope minus R. I apologize for that, but it's inevitable and this if if this is an irreducible manifold and this line meets the extension locus at a point which is not ideal Parabolic or the origin then This dain filling is order and the point is simply that The intersection between this line and the translation extension locus Corresponds to some homomorphism to G till to some non-trivial homomorphism now You should complain to me that maybe I don't know that it's faithful and to be honest. It's probably not but work of Boyer-Rolfson and Vist allows you to promote that non-trivial action to a faithful action All right, so then in my picture like this So I maybe I should have also said this horizontal axis will be in all the pictures this comes from the reducible reducible representations, so if I take a line Which impures here to have sort of negative slope something like this Then all of those guys there's a whole family of lines here So if I can I take a line whose slope is sort of between Six and minus infinity it will meet the translation extension locus And so what this thing about our course by to minus our from this picture You can see that you can use this lemma to order all the dain fillings for our in the interval between minus six and infinity And this is a big chunk of what you expect from the conjecture which is that any Dain filling in the interval from minus nine to infinity is a non-else space and The one other feature that I I want to mention here is these blue points these correspond to the So said this line here is the line corresponding from reducible representations things that factor through the Abelianization of the fundamental group M factor through a map to Z these correspond to roots of the Alexander polynomial that are on the unit circle so if you if you can ever deform from a reducible representation off into Irreducible representations that can only happen at a root of the Alexander polynomial And if the root of the Alexander polynomial is simple you in fact know you can actually always do this And so the theorem for the theorems, which of course I'm not going to give They are simply that the hypotheses ensure that there's diagram is sort of non-trivial Did you have something coming out of a blue dot or if you have some kind of green dot? And then you just argue from the picture that that means a large number of lines must meet the translation extension locus And then you then you profit those two advanced things Well, it's significance of the lens space surgeries. I Mentioned them because otherwise this convention is opposite of what most people would think So just the end just to show you a few few more pictures This is a picture. It's very similar to the first one except much more complicated. I think there's something like 74 arcs These purple arcs in my translation extension locus This is the exterior of a twisted torus knot which also has a lens space filling This pattern is very common the thing you saw for minus three three seven There's lots of lots of the lens space surgery not have this picture And you can anyway you can use this then to order all the Dane fillings in the interval minus 75 up to infinity And you know that the set of non L space slopes is minus 93 up to infinity So again, you get a lot of them, but you certainly don't get all of them Here's an example, which is just sort of a a little different in that you have a An arc leaving a an out a point correspond to the root of the Alexander polynomial, and then it's going back to a parabolic point On the boundary except now the parabolic points on the axis Here's something without any reducible interesting reducible representations. You just have an arc that goes from a Parabolic correspond to a Gala conjugate of the whole no me representation to some other random parabolic guy Here's an example, which exhibits both of these phenomena Here's a really complicated example This one's neat because it has some arcs In the translation extension look at the cross the diagram from a parabolic on this side to a parabolic on that side When you have this it means that you can use the lemma to show that Every Dane filling every non trivial Dane filling on this manifold has orderable fundamental group And if you apply results of Roberts You can see that in fact all those fillings also have taught foliations. So in particular the conjecture holds for all Dane fillings on this particular Some genus 6 fiber knot in the three sphere Here's another example. This is not a knot exterior in S3. It has some new phenomena. I'm out of time This has to do with when the anyway, this is a pretty picture. I'll stop there. Thank you for your attention