 So, let me just start by telling you what's going to happen. So, first I'm going to talk about a property of groups called left order ability. Then I'll tell you what this might say about topology, specifically three manifold topology through what's called the L space conjecture. I'll tell you about a specific construction of three manifolds called branch covers. And then all the results I'll talk about will actually focus on a very special subfamily of these manifolds. And then finally, I think maybe my results raise more questions than they actually answer so I'll end with some of those. Okay, so what is left order ability. So, you have a group and some comparison of the elements. And you want to ask that that comparison is preserved under the group operation. So that means if I take any element and multiply on the left by that element this comparison still makes sense. So, a group that you know and love, the integers is left orderable, right so if I compare to distinct integers and I add some number to both sides of this some inequality between them then that inequality still holds. But there are many groups that are not left orderable. So, for example, if a group has torsion, it can't be left orderable so we'll go through this cute little argument that shows that. So, if I assume that I have an element, it's not the identity and some power of it is the identity. Well, one thing I can say is since it's not the identity it's either bigger than one or smaller than one. So we'll just assume it's bigger than one. It's smaller than one is exactly the same. So, if I take this inequality that I already start with multiply on the left by G, I get the statement that G squared is bigger than G, which I know is bigger than one. So I get the same in here, I can follow along doing a similar argument and I end up with the same and that do the end is bigger than one. And now we're really in trouble. So this gives us a contradiction so whenever group has torsion it can't be left orderable, but in general it's really hard to tell what a group is or is not left orderable. And for us will focus on the case when our group is actually the fundamental group of a three manifold. And sometimes I'll be sloppy and I'll say that the three manifold is left orderable and I really mean that it's fundamental group is left orderable. Okay, so what does this have to do with topology. Well, maybe nothing we don't really know that much about it but there is this tantalizing conjecture out there. It might be true that the following three properties of a three manifold are equivalent with some extra adjectives to that three manifold that I won't focus on. So the first line is saying, basically that M has a really nice decomposition into surfaces. The second one we talked about so the same the fundamental group is left orderable. The third line is the hardest thing to explain in a 15 minute talk but let me just say that Hagrid-Fleur homology is a three manifold invariant. It's very powerful, very mysterious. So we want to know what is it measuring about three manifolds. So there's a rigorous sense in which a manifold either has simple Hagrid-Fleur homology or not. The manifold has this whatever this means simple Hagrid-Fleur homology will call it an L space. So what this line is saying is that M does not have simple Hagrid-Fleur homology. So the idea is that somehow maybe all of these properties are capturing some complexity of M and if this conjecture is true then they're measuring kind of the same kind of complexity. So let's say here there's been a lot of work on three manifolds that are surgeries on knots. So I won't say what that means but it's a special construction of three manifolds. That's not to say that all questions have been answered there, but if if a three manifold is not surgery on a knot then you know knowing when any of these three properties hold is even harder. I say that just to point out why maybe one of the reasons why the three manifolds I'm about to talk about are somewhat interesting. So I'll tell you how they're defined in the second, but let me just say that most of them in some sense are not surgery on a knot. So that's kind of interesting. So we're kind of in new territory in some sense. So how are they defined? So I have this downstairs manifold M and an upstairs manifold M and an action upstairs by a cyclic group that acts super nicely. So it's acting by orientation preserving diffeomorphisms. So I can take the quotient of this manifold by this action. In the case that I get M then I'll say that N is the branch cyclic is a branch cyclic cover of M. So basically these are almost covering spaces, but they're they're slightly bad at some places. So in particular, if this action is actually free then this is actually a covering space. So for us we're going to assume that there's some point that is fixed by something. And in this case you can show that the fixed set upstairs is a one manifold. So that's really nice. Okay, so this allows us to kind of refine our definition of a branch cover. So if I take the fixed set upstairs, and I look at its image downstairs and M, I'm going to call that L. And I'll say that now I'll say that N is a branched gamma cover of M over L. So now we're capturing the group in this definition also the fixed set downstairs. Okay. And I really want to focus on a very, very specific case. So if the manifold downstairs is a one sub manifold that's connected. I'll call it K now because in particular if I'm in S3, this is what we would call a knot. Okay, so my manifold downstairs is S3 and I have a knot inside of it. And if I pick a cyclic group so there's some N associated to it and it really only makes sense if N is at least two. There's some unique manifold that is the branched gamma cover over S3 branching over this. It's not K. So we'll write that as Sigma sub N of K. So this is keeping track of which cyclic group we're talking about the fixed set downstairs. It doesn't keep track of the fact that we're in S3 so we just have to remember that we're talking about knots in S3. And we'll call it the branched cyclic cover of K of index N. So the takeaway from this slide is that if I have a knot in S3, then for any integer bigger than or equal to two, there's some associated manifold to it. Okay. So it's kind of hard to imagine what these are, but they have some nice properties that help us understand certain questions about them. Okay, so here's a specific knot. So we're not really going to be able to prove anything in the time that I have, but just to give a picture of some of the things that I'm looking at. So this is a diagram. These boxes inside of them are twists. So the twists are defined by these, these are two integers. So this is an example if the integers were two and three. We're in the colors for a second. But now if I look at the colors, so you should think of this as a surface that's bounded by the knot, and I want to point out one property of it. So let's say the red part is the front of the surface. Then when I go along here, I twist, I see the back. I see the front again. I'm on the back. And if I go over either of these twists. I'm going to go back to the front. So this picture, by coloring it, I've shown that it's a two-sided surface, so it's orientable. So we need this for this definition here. So when I say the genus of the knot, I mean the smallest genus of the surface like the one that I drew here. So a surface whose boundary is the knot, and that's orientable. And if I take the smallest of those genuses, I get a number and I'll denote that by G of K. So I just need this to state some theorems. So, let me see. So before I get to those results, let me say something special about three manifolds from the left-order ability perspective. So, if M, with some extra adjectives, has a PSL2R tilde rep, so PSL2R tilde is a lead group. It's the universal cover of PSL2R. So if I have such representation, then actually my three manifold is left-orderable. Great. So this is really exciting. We just have to find one of these representations. You know, in general, it's actually really hard to construct such a thing. But work of Yinghu shows that actually in some cases, instead of trying to find these PSL2R tilde representations in these manifolds upstairs, I can actually look downstairs, so inside of S3, but I remove the knot that I'm thinking of, the fixed set. So this is a non-compact three manifold. I can take its fundamental group and instead look for PSL2R reps of that group, which is in some sense easier because like for this particular knot, it has a, it's easy to find a nice presentation for this group. That's not always true, but in this case it is. So her results turn out to be really powerful. So immediately it shows for these knots that we're considering that for the branch covers, as long as the genus of the knot is not one, then we get left-order ability eventually. So when n gets large enough, but she doesn't give a quantitative bound on when that might happen. So a few years later, Chan looked at this and showed in the same situation that you do get left-order ability and you get a quantitative bound. So if n is bigger than some function of these twist parameters, and this number grew with R and S. So as R and S got very large, the bound that you get would be worse. And what my theorem shows is actually these techniques that they're using are really powerful, at least in this particular situation. So this is a really, remember a very special family of knots, but we get left-order ability. For example, if the genus is two. So remember, these are only defined for integers bigger than or equal to two. So maybe we don't know what's going on for two or three or four. But from five on, we have left-order ability. And then as the genus increases, we get better and better bounds. And here, when the genus is at least four, this is the generic case, I guess, we actually get best possible bounds. So I didn't write this, but it's known that when n is two, you don't have left-order ability. So this means in this situation, we actually know what's going on with every single branch cover of each of these specific knots. And one thing I want to point out about this result is that it doesn't depend on the parameters so much as the genus, which is kind of interesting. So what questions do these results raise? So, of course, we know left-order ability for a lot of these manifolds now. So what about the other aspects of the L-space conjecture? So can we show that these manifolds are not L-spaces? So for some of them, Boalow, Boyer, and Gordon did show that, but some of them we still don't know. And even harder, these should all have top-foliations. And I think with our current knowledge, we really don't have any idea how to methodically construct them in these manifolds. Another question you could ask is, let's say a fix a knot. I look at its branch covers so I can look at all of them from n equal to as high as I want. I check, is it left-orderable or not? Yes or no? I collect all the yeses. So that's a subset of the integers bigger than or equal to two. What does it look like? So is it a consecutive set of integers? And does it depend on the genus? So my results kind of indicate maybe yes, but it's a very specific case. So we don't really know. And finally, Xinghua Gao showed that probably, so this means assuming the L-space conjecture, so that's a big if, but in general maybe having a PSL2R tilde representation is stronger than being left-orderable. So it might be a different question. One, does a three-manifold admit such a representation? And I will stop there. Thanks, Anna. Does anybody have any questions? Can I ask a question about something you said earlier? You said that most maybe in view of these results that you presented, most three-manifolds are not surgery on a knot. But it's true that all three-manifolds are surgery on a link, right? Yes. So is there some way to understand why you can't just join the links up somehow? Are you asking why those manifolds aren't surgery on knots or why it's harder in that situation? Yeah, I would have thought naively that I could make the licorice wallace there a little bit stronger and just say that all three-manifolds are surgery on a knot. Well, one thing I think you can do is like look at homology. So like if you're surgery on a link, you can have more complicated homology than surgery on a knot could have. Because if you do myoreviatoris, surgery on a knot always is cyclic, for example. So already you get some things that just from homology can't be surgery on a knot. But I think it's hard in general to see why something can't be surgery on a knot. And that's what a lot, I mean, I think there's recent results of people constructing manifolds that are not surgery on a knot but can't be obstructed from being so by homology and showing by other reasons that they can't be surgery on a knot. So I think it's a good question. Thanks. Are there any other questions? Yeah, I got just a quick question. Is it easy to know what the genus of these JRS knots is? Yes. It's, I don't know if I can come up with the top of my head but basically because, let's see if I can go back to the picture. So basically, from the diagram from it, so these knots are all alternating. And so, if you look at an alternating diagram, there's an algorithm that gives you a surface and it's going to be the minimal genus one. So if you just like carefully draw this picture, it's just a function of the parameters what the genus will be. Okay, thanks. If that wasn't specific enough. No, that was good.