A combinatorial Thurston theory
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Winarski, Rebecca
Thurston proved that a post-critically finite branched cover of the plane is either equivalent to a polynomial (that is: conjugate via a mapping class) or it has a topological obstruction. We use topological techniques — adapting tools used tostudy mapping class groups — to produce an algorithm that determines when a branched cover is equivalent to a polynomial, and if it is, determines which polynomial a branched cover is equivalent to. This is joint work with Jim Belk, Justin Lanier, and Dan Margalit.
 So, while I was preparing this talk, I realized that the title that I submitted, Characterizing Thirst and Maps by Lifting Trees, was actually not quite as accurate as the title, A Combinatorial Thirst and Theory, and I thought that that also fit better within the mini course on Thirst and Theory, but maybe to be modest, I should say that this is a special case of a combinatorial Thirst and Theory. In particular, what we're interested in is advancing bridges between complex dynamics, or within complex dynamics, that's the theme of the conference, but a bridge that I'd also like to emphasize is the bridge between complex dynamics and low-dimensional topology and geometric group theory. So, the bridge that I'm most interested in is that one. And I'll also maybe just give a plug this top image. Here is an image from ICERM, which is having a semester in braids at the same time as MSRI, but for one week, both Dan Margoliet, my collaborator, and I will be speaking about this related work at ICERM during MSRI, just to advertise that. All right. So, we're focused on Thirst and Theory. I said we'll be talking about a special case, and we have this dichotomy of William Thurston that says if we have a post-critically finite topological polynomial, then either it's equivalent to a polynomial or it has a levy cycle. And here, by the way, what I mean by topological polynomial is a branch cover from the complex plane to itself. In the spirit of this being a mini course, I should also define what I mean by equivalence because I haven't seen that elsewhere. So, we say that branch self-covers of the complex plane to itself that preserve a set of points P. Those are our topological polynomials, and we say that they're equivalent if they differ by a change of coordinates of the complex plane or, more precisely, two branched covers with the same post-critical set. F and G are equivalent if I can draw this commutative diagram, that is, if they differ by orientation-projerving homeomorphisms, H1 and H2, that are isotopic, rel, the proper set. All right. So, back to the idea of Thurston's theorem. Thurston's theorem gives us this dichotomy of post-critically finite topological polynomials as either being equivalent to a polynomial or being obstructed in the special way of having a levy cycle. And one interesting problem is to effectivize this theorem. That is, to give an algorithm to determine whether or not a given topological polynomial is, in fact, equivalent to a polynomial or whether it's obstructed. And in the case where it is equivalent to a polynomial to determine which polynomial it's equivalent to. So, in the spirit of this being a combinatorial algorithm, we're going to use the characterization of a polynomial as the existence of a Hubbard tree. So this equivalence, one direction is due to Doherty and Hubbard, the other can be attributed through a different set of conditions due to Poillet, and that's what we'll focus on later. But what we give is a combinatorial algorithm to decide whether a post-critically finite topological polynomial is a polynomial or not. And so what Belk linear Margoliet and myself find or do is we find this algorithm that determines the Hubbard tree if a topological polynomial is equivalent to a polynomial or it finds an obstruction and the obstruction that it finds specifically is the canonical obstruction. So the strategy that we use is one motivated from geometric group theory that we build a simplicial complex, we define a simplicial map on that complex, and then what we show is that by iterating that simplicial map, we converge to something, and that something is either a finite set in the case of polynomials or it's an infinite set with a specific structure in the case of obstructed topological polynomials. And once we've found that set to which we converge, we simply check a neighborhood within that set to find either the Hubbard tree or the canonical obstruction. This is an overview of what we'll be doing. And by the way, this procedure might sound familiar and that's rightly so because we are contributing a new bridge in already rich theory of algorithms effectivizing Thurston's theory. So I've listed names, I won't say all of them out loud. Okay. What we are really introducing here is an approach to mapping class groups that hasn't previously been applied in the context of complex dynamics. So there's many ways of approaching the mapping class group, but two of them of specific interests to understanding topological polynomials are that of geometric group theory and that of combinatorial topology. So Bartoljnick or Shevich use iterated monodromy groups. Specifically they're viewing the mapping class group as the outer automorphisms of the fundamental group of in this case a punctured surface. And this is really a geometric group theoretic perspective on the mapping class group. In contrast, what Belclanier, Margoliet and myself use is we use trees specifically through the Alexander method approach to the mapping class group. And so this is more of a topological approach to understanding the mapping class group. And so what I mean by Alexander method for mapping classes is if I have a surface and I want to look at a homeomorphism of that surface. What I can do is I can look at curves. And the reason it's enough to look at curves is this Alexander method says if I have two homeomorphisms of the surface and the image of a sufficient set of curves, namely a filling set of curves and maybe a few others, if the image of that set of curves is isotopic under these two maps, the two maps themselves are isotopic. And so we can reduce the isotope problem of maps of homeomorphisms to an isotope problem of curves. Similarly, we'll use an Alexander method for branch covers. So the version that I'll state is due to Belclanier, Margoliet and myself, but it's also a special case of or the quadratic case is also a special case of Shepletewska and Timmerin, which says if I take a tree and now instead of looking at its image, I look at its pre-image under two branched covers. If the two branched covers or if the two pre-images are isotopic as trees, the branched covers themselves are isotopic. I also need some additional data. I need to know how the map works, maps between the pre-image. I need the actual action of edges between the pre-image and the original tree. However, the information of a tree, its pre-image, and the action of the branched cover suffices to determine a topological polynomial. Suffices has enough information to tell you what map we're referring to. Okay, so as a summary, this Alexander method is a combinatorial topological approach to understanding mapping classes and then further branched covers when viewed as higher dimensional mapping classes. So specifically what we're doing is we're viewing a topological polynomial as a tree map. This holds some themes to what Dylan was talking about on Tuesday, and we can view the equivalence class of a topological polynomial in correspondence with the equivalence class of a pre-image of a tree. And here I'm specifically saying equivalence class because there's two possible equivalence classes that we might be interested in. We might be interested in the isotopy class in order to say that two branched covers are isotopic, but we might also be interested in the Thurston equivalence class or the homeomorphism class. And the correspondence holds for both types of equivalence classes. Okay, so this Alexander method gives us a combinatorial topological approach to mapping class groups, which we're then translating to a combinatorial topological approach to branched covers. Okay, so back to Thurston's theorem that a post critically finite topological polynomial is either equivalent to a polynomial or it has a levy cycle. In order to understand this combinatorial topological approach to Thurston theory, we'll first focus on this case where f is equivalent to a polynomial. And as a reminder, the idea that we're going to use there is that our topological polynomial has a Hubbard tree. Okay. So a fact is that every, a fact due to Doherty and Hubbard is that every polynomial has a distinct Hubbard tree. And what I mean by this is the Hubbard tree suffices to distinguish the topological polynomial. For instance, the Hubbard tree for the rabbit is shown here within the Julia set. And the features that will be important to us are that the Hubbard tree is invariant under a lifting operation. So what I mean by this lifting operation is I first take the preimage and I've shown what this means both abstractly as an abstract embedding in the complex plane and also as an embedding in the Julia set. We take the preimage, but then we forget any edges that are not part of paths between the post critical set. I call this operation taking the hull. So the two operations here together of taking the preimage and then taking the hull are what I'll call the lifting map of a tree. I'd also like to point out that we need extra data that is the action of the topological polynomial on the tree because the tripod that I've shown here is both the Hubbard tree of the rabbit and of the co-rabbit. But in the case of the rabbit, the preimage rotates the edges clockwise and in the case of the co-rabbit, the preimage rotates the edges counterclockwise. So we need additionally the data of how the topological polynomial maps the preimage to the original tree. And by the way, many of you may be familiar with the Hubbard tree of the airplane polynomial as being a path of length two thought of in the real axis. But what we're doing here is we're allowing for abstract embeddings in the complex plane. So I could choose three marked points in the complex plane to not fall along the real line and the Hubbard tree for the airplane polynomial would still be a path of length two. It would just look a little bit different. And so one of the advantages is that we're not constrained by the actual numerical data or I see that as an advantage. Okay. So the key feature here is that the Hubbard tree is invariant under the lifting map and we'll use as the definition of the Hubbard tree. This characterization by Poirier that says that the Hubbard tree for a polynomial is the unique tree that has an invariant angle assignment under lifting. Non-zero angle assignment under lifting. So what I mean by that is I can assign to every angle in my tree in angle. I can lift that or I can take the preimage which naturally assigns an angle to every angle of my preimage. And then when I forget I have to combine angles so that means that I add some of them together. And Poirier says that a tree is a Hubbard tree for the given topological polynomial if it has an invariant angle assignment. In this case we see that the angles just rotate so we have two sort of separate systems or separate equalities within this system of linear equations. One of which determines that the angles are pi over three around that central vertex and one of which determines that the angles along the outer vertices are all two pi. Okay. So that's an example of a tree that does have an invariant non-zero angle assignment. Let's look at a tree that does not. So here I have a tree that's invariant under some topological polynomial and it suffices to tell you the topological polynomial that I'm talking about by telling you its preimage. That's what the Alexander method says. So I'm telling you this topological polynomial by first finding the preimage. I then eliminate any of the edges that are not part of the hull among the post-critical set. And when I combine the angles I have to add their measures and I end up with this system of linear equations and you might notice that I have these two equations that are in conflict namely this equation here and this equation here cannot both hold with non-zero angle measure. So if theta 3 equals one half theta 5 and theta 5 equals theta 3 they must both be 0. Therefore this tree does not have a non-zero angle, invariant angle assignment. The other condition of Poirier is that there are no periodic edges between Julia vertices. And so again what we're using here is this Alexander method that says that a tree its preimage and the action between the two suffices to determine not just a polynomial but also a topological polynomial. This is what the Alexander method says. But when we can say more, when we can find a Hubbard tree, that Hubbard tree will then suffice to determine the equivalence class of our topological polynomial and we need to consider trees up to equivalence where here I just mean up to orientation preserving homeomorphism and isotope. So for instance while these two trees are not isotopic they are in fact homeomorphic, they're both paths of length too. And so if I discover two different topological polynomials each of which has one of these trees as its Hubbard tree those two topological polynomials will themselves be equivalent. So our goal here is to find the homeomorphism class of the Hubbard tree. And so the natural question is how do we do it? And this is where we appeal to geometric group theory. We build a simplicial complex. And the simplicial complex that we'll build we call the tree complex. We fix a set P of marked points in the complex plane and we'll define the tree complex, relp, where the vertices correspond to isotopic classes of trees and the simplices correspond to sub-forest collapses or alternatively sub-forest expansions. So what I mean by this is given a set of four marked points in the complex plane these two trees are both correspond to vertices of the simplicial complex but there's this edge E here where if I contract it in the tree on the left it will yield the tree on the right therefore these two trees are adjacent in the tree complex. Basically we could think of adding in an edge to go from the tree from the right to the tree on the left and that's what I mean by sub-forest expansion. Okay, so for example if we have three marked points in the complex plane the portion of the tree complex is this well the tree complex is the 3-2 bipartite regular tree and the portion of it looks like this here in the center you can see the hubbard tree for the rabbit polynomial and adjacent to it you see these three paths of length two each of which you obtain by contracting one of the three edges. And so a couple facts that will be useful about this tree complex the first is that the tree complex is locally finite so for every vertex it corresponds to an isotope class of trees there's only finitely many edges that you can contract there's also only finitely many edges that you could add without ending up outside of the hull of the marked set so the tree complex is locally finite. An equally important but harder theorem or proposition is that the tree complex is connected and in fact it's simply connected but this can be seen due to work of Hubbard and Mazer or alternatively Penner that this tree complex is dual to a triangulation of tecmalar space. So for example we have our tree complex here with three marked points and the dual triangulation of tecmalar space is as such. Okay. But what we'll be now interested in doing is defining a simplicial map on this simplicial complex and the simplicial map that we'll use is actually the lifting map defined for individual trees. So if we have our post critically finite topological polynomial F we can lift any given tree and remember what I mean by lifting is I take the preimage and then I remove any edges that are not part of the hull among the marked points that's our lifting map and then we can extend that to the entire tree complex. It's known that there is a fixed point namely the Hubbard tree is a fixed point of this lifting map and our strategy will be to lift any tree until you end up at the Hubbard tree. So for example for the airplane polynomial we can start with any tree we want such as this one and our algorithm says you take the preimage, some of the edges of the preimage are not part of the regulated hull among the post critical set and so you forget those edges. The tree that you obtain is not the same as the tree that you started with therefore you repeat this process. You take the preimage, some of the edges are not part of the hull of the marked set so we take that hull, we simplify and the tree that we get some of you may recognize as the Hubbard tree for the airplane polynomial but if we didn't know that we were using the airplane polynomial all we know is that it's not the same as the tree that we started with so we have to repeat this process. We take the preimage, some of the edges are not part of the hull, we forget those edges and we in fact obtain the same tree therefore we've landed on an invariant tree and our hope is that this process will always land on the Hubbard tree, that we found an invariant tree that's great but in fact that's not true. For instance for simply the rabbit polynomial we have cyclic permutation of the three trees that are adjacent to the Hubbard tree in the tree complex so what I mean by this is here in the center we have the Hubbard tree and the three vertices that are distance one from it just permute of cycle three and so it's too much to hope for that there's this globally attracting fixed point so the next best possibility would be that there's a finite nucleus that is that there's a finite set of vertices that underlifting we end up in that finite set and what we prove and what makes our algorithm work is that that's exactly what happens so if we have an unobstructed post critically finite topological polynomial this lifting map will converge to a finite set as a finite nucleus and this finite nucleus contains the Hubbard tree. So let's look at a proof of this theorem, I'll do it in two steps, the first step will be that there is a nucleus and the second step will be that it's finite. So the nucleus I propose are the periodic trees and the existence of this nucleus comes from three basic facts, one is that the lifting map is simplicial and how we check that the lifting map is simplicial is we need to know that vertices go to vertices but all that says is that when you lift a tree you get another tree. The harder condition to check is that we have to check that simplices go to simplices but remember what adjacency means in our simplicial complex is that if two trees differ by a contraction they're adjacent and so what we need here is that contraction of a tree lifts to contraction of another tree. But we can see that this happens if we just highlight the edges we plan on contracting their pre-images will be edges in the pre-image and we can contract those so contraction in fact lifts to contraction. So f star is a simplicial map and in particular that means that it's distance non-increasing, we need that it's distance increasing but we know that our simplicial map has a fixed point and we know that our simplicial complex is locally finite and these three facts together are enough to say that all trees eventually are in some periodic cycle. In other words under the lifting map all trees are eventually are pre-periodic. So that tells us what our nucleus is. Now we're going to study our nucleus to determine that it's finite. So what we're going to show is in fact that every periodic tree is distance at most two from the Hubbard tree. So to do this we'll use Poiller's conditions again. Recalling that Poiller's conditions say that a tree is a Hubbard tree if it has a non-zero invariant angle assignment and that it has no periodic Julia edges or no periodic edges between Julia vertices. And we'll use these conditions to fix any periodic or invariant tree to bring it closer to the Hubbard tree and we'll show that we can do this in two steps. So first of all we know that our hull consists of periodic trees. We're going to take some power of the lifting map just to make a periodic tree invariant so that we can work with invariant trees. And what we'll then do is we'll fix the tree so that it satisfies Poiller's conditions. We'll first fix the angle structure. And what I mean by that is if we start with this tree that does not have an invariant non-zero angle assignment from the previous example, we saw that there was this system of linear equations and there were two equations in particular that required that theta 3 and theta 5 had to be zero. What we'll do to fix those angles is we'll just fold so we'll take the zero angles, we'll actually make them zero. And what I mean by this is I highlight the angles, I collapse the edges until I identify them as or I identify the first half of the edges to be equivalent edges. So I go from this path of length 3 to this four pod on the right. And I'd like to highlight that this process of folding the edges gave me something that was adjacent in the tree complex because in particular I can contract the two blue edges in the tree on the right to obtain the tree on the left. Alternatively, if I started with the tree on the left, I could expand at the angles by adding in an edge. By the way, this is one way to go from the tree on the right to the tree on the left is through this single edge. If we're not careful, we could also add a tree, an intermediate tree in between. But that's sort of irrelevant to my point because my point is that these two trees are adjacent, not that there isn't a longer path. Okay, so this is how I fix the angle structure. I can think of fixing the angle structure as a force expansion of the tree that I started with. I add in some edges. Next we're going to focus on eliminating periodic Julia edges. That's a little bit easier. So if I have this invariant tree, the marked, the red vertices are my post-critical set, and I'm also considering them to be my fo-to set in this particular example. Or my fo-to vertices in this particular example. So the vertices that aren't marked are Julia vertices, and the edge between them is what we call a Julia edge. And this tree is invariant under the same map that I was looking at previously. The map, by the way, happens to be the three-yeared rabbit map, but that's irrelevant to understanding the example. So this tree is invariant under my previous lifting map. However, I notice that the edges permute. So in this case, all of the edges are periodic, but only one of them is a Julia edge, and that's this middle edge. And we want to eliminate all periodic Julia edges, and we simply do that by contracting them in the original tree and then contracting their pre-image in the lift. So when we do the contraction, we have now this lift of the forepod to the forepod. So eliminating periodic Julia edges is just a force contraction. And what I'm claiming here is that by starting with a periodic tree, we can move by the sequence of expanding at certain angles and then contracting at certain edges to obtain the Hubbard tree. What we do need in the process is we need that the second step, the contractions, don't introduce zero angles, but in Poirier's conditions, he actually only requires that you have an invariant angle, an invariant non-zero angle assignment on the fo-2 vertices. And since the contractions are along Julia edges, they don't affect the angle assignment that matters. Okay, so that's fine. So just to summarize this proof, if we have an unobstructed post-critically finite topological polynomial, it has a finite nucleus, that finite nucleus contains the Hubbard tree, the nucleus consists of periodic trees. We take a power so that each tree is in there or that a tree of interest is invariant, and that tree we show is distance to that it that is an expansion and a contraction from the Hubbard tree. This tells us that the nucleus is contained in a two neighborhood of the Hubbard tree. So let's look at a couple nuclei. The nucleus for the rabbit polynomial exactly consists of the Hubbard tree and these three trees that are distance one away from it that cycle periodically. And earlier we looked at an example of the airplane where we lifted and actually ended up at the Hubbard tree, and that's always what happens in the case of the airplane because the nucleus is exactly the Hubbard tree. So all trees will eventually lift to the Hubbard tree for the airplane. Okay. As an application of this, Belklinger, Margoliet and myself observed that we're making a statement that trees converge to a finite set. We can also make a statement that curves converge to a finite set. And so this answers in the case of post-critically finite topological polynomials, a question of pilgrim that curves have a finite global attractor. And what that means is if I start with any multi-curve and lift, it will eventually end up in some finite set of multi-curves. And the proof is as follows. I start with any multi-curve. I find a tree such that that multi-curve surrounds a neighborhood of a sub-forest of the tree. I know from our theorem that if I lift that tree enough, I'll end up in some finite set of trees. But the curve will have to surround sub-forests of trees in that finite set. And those are all finite trees. So the possible sub-forest is also finite, or is it also a finite set. And therefore, I end up in some finite set of multi-curves. Okay. So I've been focusing on the unobstructed case. Maybe now is a good time to see if there are questions before I move on to the obstructed case. So you used Thurston's theorem to prove the existence of your fixed point. You didn't reprove it. The claim that there was an invariant nucleus depended on the statement that there actually was a fixed point. Yes. And in particular, Poillet's conditions require Thurston's theorem. I have one more question. Do you know in unobstructed case, maybe after iteration, this map you're considering is homotopic to a strictly contracting map, metrically? I do not know that falls with it. So we don't understand this lifting map very well yet. And so that's maybe a next direction for us or for other people. Okay. So let's move on to the obstructed case. So in the spirit of this being a mini-course, I'll remind us what levy cycles are. So if we have a post-critically finite topological polynomial, a levy cycle is a multi-curve with the property that the components of the multi-curve cyclically permute with degree 1. And they map to each other with degree 1. And a theorem due to Thurston, Burstein, Levy, Shishikura, Tan, and the only place that I've seen it written in generalities in Hubbard's book is that my post-critically finite topological polynomial is equivalent to a polynomial if and only if it does not have a levy cycle. The proof of this goes through Thurston's theorem and uses the pullback map on Teichmeler space. I just wanted to advertise this, though it's not at all used in what we're doing. It does somewhat mimic what we are lifting map. And I think that this is, again, an interesting question to consider in the future. Okay. So what we will focus on are not just levy cycles, but on canonical obstructions. First defined by Pilgrim, where Pilgrim says that an obstructed topological polynomial has what's called a canonical obstruction. He defines it as the curves that converge to having length zero under this lifting process. But Sellinger reformulates it as the canonical obstruction is the minimal multi-curve with the property that the exterior of that multi-curve is actually a polynomial, or the first return map on the exterior is actually a polynomial. Okay. So with this in mind, we're going to generalize the notion of trees to that of bubble trees. And we'll use bubble trees to help us find the canonical obstruction. Okay. So bubble tree is a generalization of a tree in that now we allow vertices, line segments, or curve edges, and then also simple closed curves that surround two or more marked points. And like with trees, we can lift them by taking their pre-images, or we can take their pre-image, and then we can also take a hull. Or taking a hull means we remove any edges or curves that are not part of paths between marked points or essential bubbles, where an essential bubble is one that contains two or bounds of disc containing two or more marked points. And the bubble trees will be in the boundary of our tree complex. So here I've drawn it in tight, Mueller space because I think it's a little bit easier to see, but an example of a bubble tree lies along this boundary. Okay. And a proposition of Belk linear Margoliet and myself is that every unobstructed topological polynomial has a Hubbard bubble tree. Thank you. Every obstructed topological polynomial has a Hubbard bubble tree. It's also true that every unobstructed topological polynomial has a Hubbard bubble tree. It's just the Hubbard tree with trivial bubbles. So my other statement was also correct. Thank you. And again, these bubble trees lie in the boundary of tight Mueller space, but the theorem that is actually central to the obstructed side of our algorithm is that a similar process to what we used to sort of find the Hubbard tree within a nucleus works for obstructed polynomials, but because we're looking for something on the boundary, a neighborhood of the Hubbard bubble tree is an infinite set. So if we lift, we'll eventually end up in a two neighborhood of the Hubbard bubble tree, except in this case, that two neighborhood is an infinite set. So for instance, this is what a twisting of z squared plus i looks like when we lift it. We sort of swap between these two paths of, or there is a path in the tree complex, and we sort of skip adjacent vertices in this lifting process. Okay. But our two theorems together that within a two neighborhood of any invariant tree is a either Hubbard tree or Hubbard bubble tree gives us an algorithm to effectivize Thurston's theorem, which says you start with any tree, you apply the lifting map. At any stage, check a two neighborhood of that lifting map, either for a tree that satisfies Poirier's conditions, or for a tree that shows you a canonical obstruction. And by shows you, I mean has bubbles that are the canonical obstruction. If you don't find a Hubbard tree or a canonical obstruction, just repeat the process with your previous tree. Okay. Actually, let's go back before I move on to another application. Yes. But you said that the two neighborhoods is an infinite set, so you check everything in an infinite set for Poirier conditions and canonical obstructions. Is it right? No. So you're checking for the tree at the lifting stage that you're at, and because your tree complex is locally finite, anything you get from lifting a tree will be another tree in your tree complex. And so the two neighborhood of any tree along the process is a finite set. So at any stage, you only have to check a finite set. Okay. Thank you. Thanks. Okay. So I will now discuss an application of this theorem to the twisted rabbit problem. Many of you in the room are experts, but for those of you who aren't, I'll just give a brief overview that we start with the rabbit polynomial. It has a post-critical point that is three periodic. We mark that set within the complex plane and draw a curve surrounding the ears of the rabbit. We then post-compose the rabbit polynomial with a dain twist around that curve. And due to bursting and levy, we know that this post-composition of the rabbit polynomial will be equivalent to a polynomial. The three options are the rabbit, the co-rabbit, and the airplane. And we want to know which one. And Hubbard made this into a broader problem where instead of looking at just a twisting by one power of the dain twist, he asked, can we give a function where we raise the dain twist to a power k and the output is the polynomial that the resulting map is equivalent to? And Bartolj and Neckar-Shevich originally solved this problem in 2006. And this was huge because the problem had been open for 25 years. What we do is we give a different solution using our techniques that allow us to answer similar twisting questions for higher numbers of post-critical points and even higher degree polynomials. And so I'll demonstrate how that works. We start with our twisted rabbit. And we guess what the Hubbard tree might be. We'll then form a lifting map. And in this case, because the post-critically finite topological polynomial is a composition of the rabbit polynomial and a dain twist, we'll undo those in the opposite order. So with our guess, we'll first untwist, then we'll lift or pull back to get a different tree. And because this tree is different, it is not the Hubbard tree. The first one is not the Hubbard tree for this topological polynomial. So we perform the lifting algorithm with this resulting tree. We untwist, then we pull back. And the resulting tree is the same or at least isotopic to the original. Or this path of length two is the Hubbard tree for the topological polynomial, which tells me that in fact this twisted map is equivalent to the airplane polynomial. And so the solution to the original twisted rabbit problem as given by Bartoljnick or Shevich is that it's a four attic solution. But we chose to look at the twisted more-eared rabbit problem, which is still four attic, but only barely. And it works as follows. If I twist the, in this case, three-eared rabbit by the dain twist gamma, powers of the dain twist gamma, the topological polynomial that it's equivalent to depends only on the congruence class of the power mod four, unless it's a power of four, then you divide by all powers of four until you obtain one, two, or three mod four. And so the resulting topological polynomial is either equivalent to an airbus, a cocopelly, or a basilica tuned with basilica. And it turns out this answer is true for a twisted rabbit with any number of ears. So here I've shown where you need to add the ears in and where additional post-critical points would be added in in the solution maps, but you can twist an n-eared rabbit and the solution to the twisted rabbit problem will depend only on the power of k mod four, and it will be a generalization of an airbus, a generalization of a cocopelly, or a generalization of the basilica tuned with basilica. Oh, I think so. And I think that's true. Exactly. I think that's true. And the reason is, is because there's more relations in the mapping class group of a punctured sphere that has more than four punctures. I'll just repeat the question. Oh, yes. Okay. So thank you. So Dierick's question was, is the original, so what we're showing is that the higher mark point cases are easier than the original that Bartolj and Nekorchevich only did the only hard case. And I said yes. I think they did the only hard case. Yes, I think it's, it's really very close, in fact, because if for k equals zero, mod four, then you have to do recursively. The same thing exactly as for the classical rabbit. Now it so happens that if you just have three points, then there aren't three cases, airbus, cocopelly, and basilica. There are just two other cases. So it gets folded a little bit. So the case three, mod four is also something you have to do a recursion on. But really, it's the same kind of argument anyways. Now the way we answered the question is that we considered t to the four k composed with p, and we showed that it's really the same thing as p composed with t to the k. Yes. And we use the similar method. So probably you also, yes, you can see by these graph lifting. Yes, exactly. That you can get rid of multiple of four by doing it in the other direction. Yes, exactly. The difference, or one thing that we were able to make easier in the higher postcritical case is we were able to choose our post set representative differently for different situations where there's only three, where there's only two cosets, there's not a lot of choice of post set representatives or not a lot of need for choice for post set representatives, but we use different post set representatives for the different cases or different stages of lifting even. Yes, thank you. It's very similar. I guess I needed to include my joke, begin an airbus, a cocopelly. For here I've shown only a basilica, we actually need a basilica tuned with a basilica. And maybe I'll just advertise that our method can also be used for pre-periodic polynomials. Bartolian Necker-Shavage also did this. Looking at a twisted z squared plus i, our methods similarly work for such a, for similar cases. So the solution to the twisted z squared plus i was given by Bartolian Necker-Shavage, but like with rabbits we can also generalize to higher numbers of postcritical points that are still pre-periodic. And it turns out that in that case we actually get a periodic solution to the twisted generalized z squared plus i polynomial question when there's four or more postcritical points. Also Justin Lanier and I have been working on some additional twisted, twisting problems. So a couple years ago, this was more impressive two years ago when I gave this talk, but a couple years ago, Howard visited Michigan and asked what happens when you twist a cocopelly. And at some point we thought it was definitely, it definitely had a 16 adic solution, but there's two cases where we're really having trouble to finishing and that's why this is not out yet. But we think it's, well we know it's at least a 16 adic solution and there's two cases that might be even more. Justin and I have also solved a twisted cubic rabbit problem where there the solution is nine adic. Okay, thank you. Thank you. Any questions from the online audience? I have a few questions. I suppose you have an algorithm, so I start with this. So you have an algorithm, so maybe there's no hope for this, but do you have any estimate on the time that it would take it to converge and how one might be able to bound this? And is this something, how easy is it to implement and have you implemented it and how does it work in practice? Okay, so there's two questions here that I'll address separately. We have not considered the speed of the algorithm in part because we think that this is a great undergraduate project and we're just waiting for the right set of undergraduates to approach it. And for effectivizing the algorithm, actually Will Warden and with some help from Mark Bell are working on programming it. Thank you. So suppose that you have a rational function, which is post critically finite, but some, are some genuinely pre-periodic points. You can act on this by the mapping class group of the complement of the post-critical set. Do you have any idea what the structure in that mapping class group of the set of obstructed or the set leading to any particular polynomial might be? I really am trying to get it, not just powers of Dane Twiss, but a subset of the mapping class group. No, we also have not thought about it. That's a great direction to think about. One thing I'd be interested in that I think our algorithm might give us some power to consider is what's at, I think, the core of your question, which is what sorts of mapping classes might you apply to obtain an obstructed map? And therefore coming up with some sort of condition on what it takes to obstruct a map or what sorts of twists can cause a map to become obstructed. I think that that's a really interesting question and we haven't thought about it much. Thank you. I have two questions, one question or one answer. So first for what Hamel just asked. So there is such a thing called the EDT0L languages. And that's really saying that you iterate some substitution. Well, you have a finite collection of substitutions that you iterate on finite initial data and then you do some projection. So that's the general structure. If you take an invariant cyclic subgroup, such as the twist of the ES, you're iterating some number of times, multiply by 4 or multiply by 4 plus 3, and then you're projecting to get a rabbit, co-rabbit, or airplane. So I don't think you'll be able to see more than that, but that's still a reasonably small class and you can do it for the whole mapping class group. You're also encoded by generators, for example, and substitutions. This substitution is exactly the inverse of the, well, it's the expanding map on modular space or the expanding correspondence on modular space. Now I had a question for Becker, which is why do you do this with trees and not general graphs so as to address rational maps? It seems you can also lift graphs and project and do iteration entirely symbolically in this way. Well, certainly one of the limitations is that the Hubbard tree is already known, it's already known how the dictionary between Hubbard trees and branch covers work, but with rational maps there's a number of difficulties that we found in different cases. I need a moment to remember what these difficulties are. Well, first of all, you don't have this reduction. You can lift the graph and then what? For trees you lift and then you can cut off all the extra stuff. In general, you won't be able to do it for graphs. So can you hear me? Yes. Thank you for the great talk. I have kind of a related question to what Laster was asking. So you pull back the trees and so I guess we learned from Le Bon, I guess it was mainly because last week that the bicep is sort of, well, let's say, I don't know, an algebraic way of encoding, well, pulling back curves. So can you encode pulling back the edges that you use to pull back your trees and sort of, I mean, can you use a bicep to encode that or to make an algorithm that you can, let's say, implement in the computer? I don't think we had considered using biceps to encode the edges. Or maybe something else. I mean, some algebraic or what? I'm not sure. Okay. Yeah. Thank you. I have another question. So when you have a hover tree, in a way, the information that you really need is the tree on the singular set on the critical values. So the critical points that are kind of not in that tree already, they're kind of really only included to tell you which Hobbits class you're in. And if you know, I think, which Hobbits class you're in, if you have your covering, then you can always reconstruct the pre-image of that one. So I was wondering, in your algorithm, if you only did iteration on trees that includes the critical values, but not necessarily all of the critical points, would that still work? Is that enough? So do you mean contain the critical values and all post-critical points? Yes, yes. Sorry, all post-critical points. So the whole Hobbits, forward Hobbits of the critical values themselves, but not necessarily the critical points, that sits out of the way somewhere. We do not require that our trees contain the critical points if they're pre-periodic. Okay. Okay. Thank you. Yeah. Okay. We are running out of time. So we'll ask questions during breaks. So let's thank the speaker again. Thank you.
 So, while I was preparing this talk, I realized that the title that I submitted, Characterizing  Thirst and Maps by Lifting Trees, was actually not quite as accurate as the title, A Combinatorial  Thirst and Theory, and I thought that that also fit better within the mini course on  Thirst and Theory, but maybe to be modest, I should say that this is a special case of  a combinatorial Thirst and Theory.  In particular, what we're interested in is advancing bridges between complex dynamics,  or within complex dynamics, that's the theme of the conference, but a bridge that I'd also  like to emphasize is the bridge between complex dynamics and low-dimensional topology and  geometric group theory.  So, the bridge that I'm most interested in is that one.  And I'll also maybe just give a plug this top image.  Here is an image from ICERM, which is having a semester in braids at the same time as MSRI,  but for one week, both Dan Margoliet, my collaborator, and I will be speaking about this related  work at ICERM during MSRI, just to advertise that.  All right.  So, we're focused on Thirst and Theory.  I said we'll be talking about a special case, and we have this dichotomy of William Thurston  that says if we have a post-critically finite topological polynomial, then either it's  equivalent to a polynomial or it has a levy cycle.  And here, by the way, what I mean by topological polynomial is a branch cover from the complex  plane to itself.  In the spirit of this being a mini course, I should also define what I mean by equivalence  because I haven't seen that elsewhere.  So, we say that branch self-covers of the complex plane to itself that preserve a set  of points P. Those are our topological polynomials, and we say that they're equivalent if they  differ by a change of coordinates of the complex plane or, more precisely, two branched covers  with the same post-critical set.  F and G are equivalent if I can draw this commutative diagram, that is, if they differ  by orientation-projerving homeomorphisms, H1 and H2, that are isotopic, rel, the proper  set.  All right.  So, back to the idea of Thurston's theorem.  Thurston's theorem gives us this dichotomy of post-critically finite topological polynomials  as either being equivalent to a polynomial or being obstructed in the special way of having  a levy cycle.  And one interesting problem is to effectivize this theorem.  That is, to give an algorithm to determine whether or not a given topological polynomial  is, in fact, equivalent to a polynomial or whether it's obstructed.  And in the case where it is equivalent to a polynomial to determine which polynomial  it's equivalent to.  So, in the spirit of this being a combinatorial algorithm, we're going to use the characterization  of a polynomial as the existence of a Hubbard tree.  So this equivalence, one direction is due to Doherty and Hubbard, the other can be attributed  through a different set of conditions due to Poillet, and that's what we'll focus on later.  But what we give is a combinatorial algorithm to decide whether a post-critically finite  topological polynomial is a polynomial or not.  And so what Belk linear Margoliet and myself find or do is we find this algorithm that  determines the Hubbard tree if a topological polynomial is equivalent to a polynomial or  it finds an obstruction and the obstruction that it finds specifically is the canonical  obstruction.  So the strategy that we use is one motivated from geometric group theory that we build  a simplicial complex, we define a simplicial map on that complex, and then what we show  is that by iterating that simplicial map, we converge to something, and that something  is either a finite set in the case of polynomials or it's an infinite set with a specific structure  in the case of obstructed topological polynomials.  And once we've found that set to which we converge, we simply check a neighborhood within  that set to find either the Hubbard tree or the canonical obstruction.  This is an overview of what we'll be doing.  And by the way, this procedure might sound familiar and that's rightly so because we  are contributing a new bridge in already rich theory of algorithms effectivizing Thurston's  theory.  So I've listed names, I won't say all of them out loud.  Okay.  What we are really introducing here is an approach to mapping class groups that hasn't  previously been applied in the context of complex dynamics.  So there's many ways of approaching the mapping class group, but two of them of specific interests  to understanding topological polynomials are that of geometric group theory and that of  combinatorial topology.  So Bartoljnick or Shevich use iterated monodromy groups.  Specifically they're viewing the mapping class group as the outer automorphisms of the fundamental  group of in this case a punctured surface.  And this is really a geometric group theoretic perspective on the mapping class group.  In contrast, what Belclanier, Margoliet and myself use is we use trees specifically through  the Alexander method approach to the mapping class group.  And so this is more of a topological approach to understanding the mapping class group.  And so what I mean by Alexander method for mapping classes is if I have a surface and  I want to look at a homeomorphism of that surface.  What I can do is I can look at curves.  And the reason it's enough to look at curves is this Alexander method says if I have two  homeomorphisms of the surface and the image of a sufficient set of curves, namely a filling  set of curves and maybe a few others, if the image of that set of curves is isotopic under  these two maps, the two maps themselves are isotopic.  And so we can reduce the isotope problem of maps of homeomorphisms to an isotope problem  of curves.  Similarly, we'll use an Alexander method for branch covers.  So the version that I'll state is due to Belclanier, Margoliet and myself, but it's also a special  case of or the quadratic case is also a special case of Shepletewska and Timmerin, which  says if I take a tree and now instead of looking at its image, I look at its pre-image under  two branched covers.  If the two branched covers or if the two pre-images are isotopic as trees, the branched covers  themselves are isotopic.  I also need some additional data.  I need to know how the map works, maps between the pre-image.  I need the actual action of edges between the pre-image and the original tree.  However, the information of a tree, its pre-image, and the action of the branched cover suffices  to determine a topological polynomial.  Suffices has enough information to tell you what map we're referring to.  Okay, so as a summary, this Alexander method is a combinatorial topological approach to  understanding mapping classes and then further branched covers when viewed as higher dimensional  mapping classes.  So specifically what we're doing is we're viewing a topological polynomial as a tree  map.  This holds some themes to what Dylan was talking about on Tuesday, and we can view the equivalence  class of a topological polynomial in correspondence with the equivalence class of a pre-image  of a tree.  And here I'm specifically saying equivalence class because there's two possible equivalence  classes that we might be interested in.  We might be interested in the isotopy class in order to say that two branched covers are  isotopic, but we might also be interested in the Thurston equivalence class or the homeomorphism  class.  And the correspondence holds for both types of equivalence classes.  Okay, so this Alexander method gives us a combinatorial topological approach to mapping  class groups, which we're then translating to a combinatorial topological approach to  branched covers.  Okay, so back to Thurston's theorem that a post critically finite topological polynomial  is either equivalent to a polynomial or it has a levy cycle.  In order to understand this combinatorial topological approach to Thurston theory, we'll first focus  on this case where f is equivalent to a polynomial.  And as a reminder, the idea that we're going to use there is that our topological polynomial  has a Hubbard tree.  Okay.  So a fact is that every, a fact due to Doherty and Hubbard is that every polynomial has a  distinct Hubbard tree.  And what I mean by this is the Hubbard tree suffices to distinguish the topological polynomial.  For instance, the Hubbard tree for the rabbit is shown here within the Julia set.  And the features that will be important to us are that the Hubbard tree is invariant  under a lifting operation.  So what I mean by this lifting operation is I first take the preimage and I've shown  what this means both abstractly as an abstract embedding in the complex plane and also as  an embedding in the Julia set.  We take the preimage, but then we forget any edges that are not part of paths between the  post critical set.  I call this operation taking the hull.  So the two operations here together of taking the preimage and then taking the hull are  what I'll call the lifting map of a tree.  I'd also like to point out that we need extra data that is the action of the topological  polynomial on the tree because the tripod that I've shown here is both the Hubbard tree  of the rabbit and of the co-rabbit.  But in the case of the rabbit, the preimage rotates the edges clockwise and in the case  of the co-rabbit, the preimage rotates the edges counterclockwise.  So we need additionally the data of how the topological polynomial maps the preimage to  the original tree.  And by the way, many of you may be familiar with the Hubbard tree of the airplane polynomial  as being a path of length two thought of in the real axis.  But what we're doing here is we're allowing for abstract embeddings in the complex plane.  So I could choose three marked points in the complex plane to not fall along the real line  and the Hubbard tree for the airplane polynomial would still be a path of length two.  It would just look a little bit different.  And so one of the advantages is that we're not constrained by the actual numerical data  or I see that as an advantage.  Okay.  So the key feature here is that the Hubbard tree is invariant under the lifting map and  we'll use as the definition of the Hubbard tree.  This characterization by Poirier that says that the Hubbard tree for a polynomial is  the unique tree that has an invariant angle assignment under lifting.  Non-zero angle assignment under lifting.  So what I mean by that is I can assign to every angle in my tree in angle.  I can lift that or I can take the preimage which naturally assigns an angle to every  angle of my preimage.  And then when I forget I have to combine angles so that means that I add some of them together.  And Poirier says that a tree is a Hubbard tree for the given topological polynomial  if it has an invariant angle assignment.  In this case we see that the angles just rotate so we have two sort of separate systems or  separate equalities within this system of linear equations.  One of which determines that the angles are pi over three around that central vertex and  one of which determines that the angles along the outer vertices are all two pi.  Okay.  So that's an example of a tree that does have an invariant non-zero angle assignment.  Let's look at a tree that does not.  So here I have a tree that's invariant under some topological polynomial and it suffices  to tell you the topological polynomial that I'm talking about by telling you its preimage.  That's what the Alexander method says.  So I'm telling you this topological polynomial by first finding the preimage.  I then eliminate any of the edges that are not part of the hull among the post-critical  set.  And when I combine the angles I have to add their measures and I end up with this system  of linear equations and you might notice that I have these two equations that are in conflict  namely this equation here and this equation here cannot both hold with non-zero angle  measure.  So if theta 3 equals one half theta 5 and theta 5 equals theta 3 they must both be 0.  Therefore this tree does not have a non-zero angle, invariant angle assignment.  The other condition of Poirier is that there are no periodic edges between Julia vertices.  And so again what we're using here is this Alexander method that says that a tree its  preimage and the action between the two suffices to determine not just a polynomial but also  a topological polynomial.  This is what the Alexander method says.  But when we can say more, when we can find a Hubbard tree, that Hubbard tree will then  suffice to determine the equivalence class of our topological polynomial and we need  to consider trees up to equivalence where here I just mean up to orientation preserving  homeomorphism and isotope.  So for instance while these two trees are not isotopic they are in fact homeomorphic,  they're both paths of length too.  And so if I discover two different topological polynomials each of which has one of these  trees as its Hubbard tree those two topological polynomials will themselves be equivalent.  So our goal here is to find the homeomorphism class of the Hubbard tree.  And so the natural question is how do we do it?  And this is where we appeal to geometric group theory.  We build a simplicial complex.  And the simplicial complex that we'll build we call the tree complex.  We fix a set P of marked points in the complex plane and we'll define the tree complex,  relp, where the vertices correspond to isotopic classes of trees and the simplices correspond  to sub-forest collapses or alternatively sub-forest expansions.  So what I mean by this is given a set of four marked points in the complex plane these two  trees are both correspond to vertices of the simplicial complex but there's this edge  E here where if I contract it in the tree on the left it will yield the tree on the  right therefore these two trees are adjacent in the tree complex.  Basically we could think of adding in an edge to go from the tree from the right to  the tree on the left and that's what I mean by sub-forest expansion.  Okay, so for example if we have three marked points in the complex plane the portion of  the tree complex is this well the tree complex is the 3-2 bipartite regular tree and the  portion of it looks like this here in the center you can see the hubbard tree for the  rabbit polynomial and adjacent to it you see these three paths of length two each of which  you obtain by contracting one of the three edges.  And so a couple facts that will be useful about this tree complex the first is that  the tree complex is locally finite so for every vertex it corresponds to an isotope  class of trees there's only finitely many edges that you can contract there's also only  finitely many edges that you could add without ending up outside of the hull of the marked  set so the tree complex is locally finite.  An equally important but harder theorem or proposition is that the tree complex is connected  and in fact it's simply connected but this can be seen due to work of Hubbard and Mazer  or alternatively Penner that this tree complex is dual to a triangulation of tecmalar space.  So for example we have our tree complex here with three marked points and the dual triangulation  of tecmalar space is as such.  Okay.  But what we'll be now interested in doing is defining a simplicial map on this simplicial  complex and the simplicial map that we'll use is actually the lifting map defined for  individual trees.  So if we have our post critically finite topological polynomial F we can lift any given tree and  remember what I mean by lifting is I take the preimage and then I remove any edges that  are not part of the hull among the marked points that's our lifting map and then we  can extend that to the entire tree complex.  It's known that there is a fixed point namely the Hubbard tree is a fixed point of this  lifting map and our strategy will be to lift any tree until you end up at the Hubbard tree.  So for example for the airplane polynomial we can start with any tree we want such as  this one and our algorithm says you take the preimage, some of the edges of the preimage  are not part of the regulated hull among the post critical set and so you forget those  edges.  The tree that you obtain is not the same as the tree that you started with therefore  you repeat this process.  You take the preimage, some of the edges are not part of the hull of the marked set so  we take that hull, we simplify and the tree that we get some of you may recognize as the  Hubbard tree for the airplane polynomial but if we didn't know that we were using the airplane  polynomial all we know is that it's not the same as the tree that we started with so we  have to repeat this process.  We take the preimage, some of the edges are not part of the hull, we forget those edges  and we in fact obtain the same tree therefore we've landed on an invariant tree and our  hope is that this process will always land on the Hubbard tree, that we found an invariant  tree that's great but in fact that's not true.  For instance for simply the rabbit polynomial we have cyclic permutation of the three trees  that are adjacent to the Hubbard tree in the tree complex so what I mean by this is here  in the center we have the Hubbard tree and the three vertices that are distance one from  it just permute of cycle three and so it's too much to hope for that there's this globally  attracting fixed point so the next best possibility would be that there's a finite nucleus that  is that there's a finite set of vertices that underlifting we end up in that finite set  and what we prove and what makes our algorithm work is that that's exactly what happens so  if we have an unobstructed post critically finite topological polynomial this lifting  map will converge to a finite set as a finite nucleus and this finite nucleus contains the  Hubbard tree.  So let's look at a proof of this theorem, I'll do it in two steps, the first step will  be that there is a nucleus and the second step will be that it's finite.  So the nucleus I propose are the periodic trees and the existence of this nucleus comes  from three basic facts, one is that the lifting map is simplicial and how we check that the  lifting map is simplicial is we need to know that vertices go to vertices but all that  says is that when you lift a tree you get another tree.  The harder condition to check is that we have to check that simplices go to simplices but  remember what adjacency means in our simplicial complex is that if two trees differ by a contraction  they're adjacent and so what we need here is that contraction of a tree lifts to contraction  of another tree.  But we can see that this happens if we just highlight the edges we plan on contracting  their pre-images will be edges in the pre-image and we can contract those so contraction in  fact lifts to contraction.  So f star is a simplicial map and in particular that means that it's distance non-increasing,  we need that it's distance increasing but we know that our simplicial map has a fixed  point and we know that our simplicial complex is locally finite and these three facts together  are enough to say that all trees eventually are in some periodic cycle.  In other words under the lifting map all trees are eventually are pre-periodic.  So that tells us what our nucleus is.  Now we're going to study our nucleus to determine that it's finite.  So what we're going to show is in fact that every periodic tree is distance at most two  from the Hubbard tree.  So to do this we'll use Poiller's conditions again.  Recalling that Poiller's conditions say that a tree is a Hubbard tree if it has a non-zero  invariant angle assignment and that it has no periodic Julia edges or no periodic edges  between Julia vertices.  And we'll use these conditions to fix any periodic or invariant tree to bring it closer  to the Hubbard tree and we'll show that we can do this in two steps.  So first of all we know that our hull consists of periodic trees.  We're going to take some power of the lifting map just to make a periodic tree invariant  so that we can work with invariant trees.  And what we'll then do is we'll fix the tree so that it satisfies Poiller's conditions.  We'll first fix the angle structure.  And what I mean by that is if we start with this tree that does not have an invariant  non-zero angle assignment from the previous example, we saw that there was this system  of linear equations and there were two equations in particular that required that theta 3 and  theta 5 had to be zero.  What we'll do to fix those angles is we'll just fold so we'll take the zero angles,  we'll actually make them zero.  And what I mean by this is I highlight the angles, I collapse the edges until I identify  them as or I identify the first half of the edges to be equivalent edges.  So I go from this path of length 3 to this four pod on the right.  And I'd like to highlight that this process of folding the edges gave me something that  was adjacent in the tree complex because in particular I can contract the two blue edges  in the tree on the right to obtain the tree on the left.  Alternatively, if I started with the tree on the left, I could expand at the angles  by adding in an edge.  By the way, this is one way to go from the tree on the right to the tree on the left  is through this single edge.  If we're not careful, we could also add a tree, an intermediate tree in between.  But that's sort of irrelevant to my point because my point is that these two trees are  adjacent, not that there isn't a longer path.  Okay, so this is how I fix the angle structure.  I can think of fixing the angle structure as a force expansion of the tree that I started  with.  I add in some edges.  Next we're going to focus on eliminating periodic Julia edges.  That's a little bit easier.  So if I have this invariant tree, the marked, the red vertices are my post-critical set,  and I'm also considering them to be my fo-to set in this particular example.  Or my fo-to vertices in this particular example.  So the vertices that aren't marked are Julia vertices, and the edge between them is what  we call a Julia edge.  And this tree is invariant under the same map that I was looking at previously.  The map, by the way, happens to be the three-yeared rabbit map, but that's irrelevant to understanding  the example.  So this tree is invariant under my previous lifting map.  However, I notice that the edges permute.  So in this case, all of the edges are periodic, but only one of them is a Julia edge, and  that's this middle edge.  And we want to eliminate all periodic Julia edges, and we simply do that by contracting  them in the original tree and then contracting their pre-image in the lift.  So when we do the contraction, we have now this lift of the forepod to the forepod.  So eliminating periodic Julia edges is just a force contraction.  And what I'm claiming here is that by starting with a periodic tree, we can move by the sequence  of expanding at certain angles and then contracting at certain edges to obtain the Hubbard tree.  What we do need in the process is we need that the second step, the contractions, don't  introduce zero angles, but in Poirier's conditions, he actually only requires that you have an  invariant angle, an invariant non-zero angle assignment on the fo-2 vertices.  And since the contractions are along Julia edges, they don't affect the angle assignment  that matters.  Okay, so that's fine.  So just to summarize this proof, if we have an unobstructed post-critically finite topological  polynomial, it has a finite nucleus, that finite nucleus contains the Hubbard tree, the  nucleus consists of periodic trees.  We take a power so that each tree is in there or that a tree of interest is invariant, and  that tree we show is distance to that it that is an expansion and a contraction from the  Hubbard tree.  This tells us that the nucleus is contained in a two neighborhood of the Hubbard tree.  So let's look at a couple nuclei.  The nucleus for the rabbit polynomial exactly consists of the Hubbard tree and these three  trees that are distance one away from it that cycle periodically.  And earlier we looked at an example of the airplane where we lifted and actually ended  up at the Hubbard tree, and that's always what happens in the case of the airplane because  the nucleus is exactly the Hubbard tree.  So all trees will eventually lift to the Hubbard tree for the airplane.  Okay.  As an application of this, Belklinger, Margoliet and myself observed that we're making a statement  that trees converge to a finite set.  We can also make a statement that curves converge to a finite set.  And so this answers in the case of post-critically finite topological polynomials, a question  of pilgrim that curves have a finite global attractor.  And what that means is if I start with any multi-curve and lift, it will eventually end  up in some finite set of multi-curves.  And the proof is as follows.  I start with any multi-curve.  I find a tree such that that multi-curve surrounds a neighborhood of a sub-forest of the tree.  I know from our theorem that if I lift that tree enough, I'll end up in some finite set  of trees.  But the curve will have to surround sub-forests of trees in that finite set.  And those are all finite trees.  So the possible sub-forest is also finite, or is it also a finite set.  And therefore, I end up in some finite set of multi-curves.  Okay.  So I've been focusing on the unobstructed case.  Maybe now is a good time to see if there are questions before I move on to the obstructed  case.  So you used Thurston's theorem to prove the existence of your fixed point.  You didn't reprove it.  The claim that there was an invariant nucleus depended on the statement that there actually  was a fixed point.  Yes.  And in particular, Poillet's conditions require Thurston's theorem.  I have one more question.  Do you know in unobstructed case, maybe after iteration, this map you're considering is  homotopic to a strictly contracting map, metrically?  I do not know that falls with it.  So we don't understand this lifting map very well yet.  And so that's maybe a next direction for us or for other people.  Okay.  So let's move on to the obstructed case.  So in the spirit of this being a mini-course, I'll remind us what levy cycles are.  So if we have a post-critically finite topological polynomial, a levy cycle is a multi-curve  with the property that the components of the multi-curve cyclically permute with degree  1.  And they map to each other with degree 1.  And a theorem due to Thurston, Burstein, Levy, Shishikura, Tan, and the only place that I've  seen it written in generalities in Hubbard's book is that my post-critically finite topological  polynomial is equivalent to a polynomial if and only if it does not have a levy cycle.  The proof of this goes through Thurston's theorem and uses the pullback map on Teichmeler  space.  I just wanted to advertise this, though it's not at all used in what we're doing.  It does somewhat mimic what we are lifting map.  And I think that this is, again, an interesting question to consider in the future.  Okay.  So what we will focus on are not just levy cycles, but on canonical obstructions.  First defined by Pilgrim, where Pilgrim says that an obstructed topological polynomial  has what's called a canonical obstruction.  He defines it as the curves that converge to having length zero under this lifting process.  But Sellinger reformulates it as the canonical obstruction is the minimal multi-curve with  the property that the exterior of that multi-curve is actually a polynomial, or the first return  map on the exterior is actually a polynomial.  Okay.  So with this in mind, we're going to generalize the notion of trees to that of bubble trees.  And we'll use bubble trees to help us find the canonical obstruction.  Okay.  So bubble tree is a generalization of a tree in that now we allow vertices, line segments,  or curve edges, and then also simple closed curves that surround two or more marked points.  And like with trees, we can lift them by taking their pre-images, or we can take their pre-image,  and then we can also take a hull.  Or taking a hull means we remove any edges or curves that are not part of paths between  marked points or essential bubbles, where an essential bubble is one that contains two  or bounds of disc containing two or more marked points.  And the bubble trees will be in the boundary of our tree complex.  So here I've drawn it in tight, Mueller space because I think it's a little bit easier to  see, but an example of a bubble tree lies along this boundary.  Okay.  And a proposition of Belk linear Margoliet and myself is that every unobstructed topological  polynomial has a Hubbard bubble tree.  Thank you.  Every obstructed topological polynomial has a Hubbard bubble tree.  It's also true that every unobstructed topological polynomial has a Hubbard bubble tree.  It's just the Hubbard tree with trivial bubbles.  So my other statement was also correct.  Thank you.  And again, these bubble trees lie in the boundary of tight Mueller space, but the theorem that  is actually central to the obstructed side of our algorithm is that a similar process  to what we used to sort of find the Hubbard tree within a nucleus works for obstructed  polynomials, but because we're looking for something on the boundary, a neighborhood  of the Hubbard bubble tree is an infinite set.  So if we lift, we'll eventually end up in a two neighborhood of the Hubbard bubble tree,  except in this case, that two neighborhood is an infinite set.  So for instance, this is what a twisting of z squared plus i looks like when we lift  it.  We sort of swap between these two paths of, or there is a path in the tree complex, and  we sort of skip adjacent vertices in this lifting process.  Okay.  But our two theorems together that within a two neighborhood of any invariant tree is  a either Hubbard tree or Hubbard bubble tree gives us an algorithm to effectivize Thurston's  theorem, which says you start with any tree, you apply the lifting map.  At any stage, check a two neighborhood of that lifting map, either for a tree that satisfies  Poirier's conditions, or for a tree that shows you a canonical obstruction.  And by shows you, I mean has bubbles that are the canonical obstruction.  If you don't find a Hubbard tree or a canonical obstruction, just repeat the process with  your previous tree.  Okay.  Actually, let's go back before I move on to another application.  Yes.  But you said that the two neighborhoods is an infinite set, so you check everything in  an infinite set for Poirier conditions and canonical obstructions.  Is it right?  No.  So you're checking for the tree at the lifting stage that you're at, and because your tree  complex is locally finite, anything you get from lifting a tree will be another tree in  your tree complex.  And so the two neighborhood of any tree along the process is a finite set.  So at any stage, you only have to check a finite set.  Okay.  Thank you.  Thanks.  Okay.  So I will now discuss an application of this theorem to the twisted rabbit problem.  Many of you in the room are experts, but for those of you who aren't, I'll just give  a brief overview that we start with the rabbit polynomial.  It has a post-critical point that is three periodic.  We mark that set within the complex plane and draw a curve surrounding the ears of the  rabbit.  We then post-compose the rabbit polynomial with a dain twist around that curve.  And due to bursting and levy, we know that this post-composition of the rabbit polynomial  will be equivalent to a polynomial.  The three options are the rabbit, the co-rabbit, and the airplane.  And we want to know which one.  And Hubbard made this into a broader problem where instead of looking at just a twisting  by one power of the dain twist, he asked, can we give a function where we raise the  dain twist to a power k and the output is the polynomial that the resulting map is equivalent  to?  And Bartolj and Neckar-Shevich originally solved this problem in 2006.  And this was huge because the problem had been open for 25 years.  What we do is we give a different solution using our techniques that allow us to answer  similar twisting questions for higher numbers of post-critical points and even higher degree  polynomials.  And so I'll demonstrate how that works.  We start with our twisted rabbit.  And we guess what the Hubbard tree might be.  We'll then form a lifting map.  And in this case, because the post-critically finite topological polynomial is a composition  of the rabbit polynomial and a dain twist, we'll undo those in the opposite order.  So with our guess, we'll first untwist, then we'll lift or pull back to get a different  tree.  And because this tree is different, it is not the Hubbard tree.  The first one is not the Hubbard tree for this topological polynomial.  So we perform the lifting algorithm with this resulting tree.  We untwist, then we pull back.  And the resulting tree is the same or at least isotopic to the original.  Or this path of length two is the Hubbard tree for the topological polynomial, which  tells me that in fact this twisted map is equivalent to the airplane polynomial.  And so the solution to the original twisted rabbit problem as given by Bartoljnick or  Shevich is that it's a four attic solution.  But we chose to look at the twisted more-eared rabbit problem, which is still four attic,  but only barely.  And it works as follows.  If I twist the, in this case, three-eared rabbit by the dain twist gamma, powers of the dain  twist gamma, the topological polynomial that it's equivalent to depends only on the congruence  class of the power mod four, unless it's a power of four, then you divide by all powers  of four until you obtain one, two, or three mod four.  And so the resulting topological polynomial is either equivalent to an airbus, a cocopelly,  or a basilica tuned with basilica.  And it turns out this answer is true for a twisted rabbit with any number of ears.  So here I've shown where you need to add the ears in and where additional post-critical  points would be added in in the solution maps, but you can twist an n-eared rabbit and the  solution to the twisted rabbit problem will depend only on the power of k mod four, and  it will be a generalization of an airbus, a generalization of a cocopelly, or a generalization  of the basilica tuned with basilica.  Oh, I think so.  And I think that's true.  Exactly.  I think that's true.  And the reason is, is because there's more relations in the mapping class group of a  punctured sphere that has more than four punctures.  I'll just repeat the question.  Oh, yes.  Okay.  So thank you.  So Dierick's question was, is the original, so what we're showing is that the higher mark  point cases are easier than the original that Bartolj and Nekorchevich only did the only  hard case.  And I said yes.  I think they did the only hard case.  Yes, I think it's, it's really very close, in fact, because if for k equals zero, mod  four, then you have to do recursively.  The same thing exactly as for the classical rabbit.  Now it so happens that if you just have three points, then there aren't three cases, airbus,  cocopelly, and basilica.  There are just two other cases.  So it gets folded a little bit.  So the case three, mod four is also something you have to do a recursion on.  But really, it's the same kind of argument anyways.  Now the way we answered the question is that we considered t to the four k composed with  p, and we showed that it's really the same thing as p composed with t to the k.  Yes.  And we use the similar method.  So probably you also, yes, you can see by these graph lifting.  Yes, exactly.  That you can get rid of multiple of four by doing it in the other direction.  Yes, exactly.  The difference, or one thing that we were able to make easier in the higher postcritical  case is we were able to choose our post set representative differently for different situations  where there's only three, where there's only two cosets, there's not a lot of choice of  post set representatives or not a lot of need for choice for post set representatives,  but we use different post set representatives for the different cases or different stages  of lifting even.  Yes, thank you.  It's very similar.  I guess I needed to include my joke, begin an airbus, a cocopelly.  For here I've shown only a basilica, we actually need a basilica tuned with a basilica.  And maybe I'll just advertise that our method can also be used for pre-periodic polynomials.  Bartolian Necker-Shavage also did this.  Looking at a twisted z squared plus i, our methods similarly work for such a, for similar  cases.  So the solution to the twisted z squared plus i was given by Bartolian Necker-Shavage, but  like with rabbits we can also generalize to higher numbers of postcritical points that  are still pre-periodic.  And it turns out that in that case we actually get a periodic solution to the twisted generalized  z squared plus i polynomial question when there's four or more postcritical points.  Also Justin Lanier and I have been working on some additional twisted, twisting problems.  So a couple years ago, this was more impressive two years ago when I gave this talk, but a  couple years ago, Howard visited Michigan and asked what happens when you twist a cocopelly.  And at some point we thought it was definitely, it definitely had a 16 adic solution, but  there's two cases where we're really having trouble to finishing and that's why this is  not out yet.  But we think it's, well we know it's at least a 16 adic solution and there's two cases  that might be even more.  Justin and I have also solved a twisted cubic rabbit problem where there the solution is  nine adic.  Okay, thank you.  Thank you.  Any questions from the online audience?  I have a few questions.  I suppose you have an algorithm, so I start with this.  So you have an algorithm, so maybe there's no hope for this, but do you have any estimate  on the time that it would take it to converge and how one might be able to bound this?  And is this something, how easy is it to implement and have you implemented it and how does it  work in practice?  Okay, so there's two questions here that I'll address separately.  We have not considered the speed of the algorithm in part because we think that this is a great  undergraduate project and we're just waiting for the right set of undergraduates to approach  it.  And for effectivizing the algorithm, actually Will Warden and with some help from Mark Bell  are working on programming it.  Thank you.  So suppose that you have a rational function, which is post critically finite, but some,  are some genuinely pre-periodic points.  You can act on this by the mapping class group of the complement of the post-critical set.  Do you have any idea what the structure in that mapping class group of the set of obstructed  or the set leading to any particular polynomial might be?  I really am trying to get it, not just powers of Dane Twiss, but a subset of the mapping  class group.  No, we also have not thought about it.  That's a great direction to think about.  One thing I'd be interested in that I think our algorithm might give us some power to  consider is what's at, I think, the core of your question, which is what sorts of mapping  classes might you apply to obtain an obstructed map?  And therefore coming up with some sort of condition on what it takes to obstruct a map  or what sorts of twists can cause a map to become obstructed.  I think that that's a really interesting question and we haven't thought about it much.  Thank you.  I have two questions, one question or one answer.  So first for what Hamel just asked.  So there is such a thing called the EDT0L languages.  And that's really saying that you iterate some substitution.  Well, you have a finite collection of substitutions that you iterate on finite initial data and  then you do some projection.  So that's the general structure.  If you take an invariant cyclic subgroup, such as the twist of the ES, you're iterating  some number of times, multiply by 4 or multiply by 4 plus 3, and then you're projecting to  get a rabbit, co-rabbit, or airplane.  So I don't think you'll be able to see more than that, but that's still a reasonably small  class and you can do it for the whole mapping class group.  You're also encoded by generators, for example, and substitutions.  This substitution is exactly the inverse of the, well, it's the expanding map on modular  space or the expanding correspondence on modular space.  Now I had a question for Becker, which is why do you do this with trees and not general  graphs so as to address rational maps?  It seems you can also lift graphs and project and do iteration entirely symbolically in  this way.  Well, certainly one of the limitations is that the Hubbard tree is already known, it's  already known how the dictionary between Hubbard trees and branch covers work, but with rational  maps there's a number of difficulties that we found in different cases.  I need a moment to remember what these difficulties are.  Well, first of all, you don't have this reduction.  You can lift the graph and then what?  For trees you lift and then you can cut off all the extra stuff.  In general, you won't be able to do it for graphs.  So can you hear me?  Yes.  Thank you for the great talk.  I have kind of a related question to what Laster was asking.  So you pull back the trees and so I guess we learned from Le Bon, I guess it was mainly  because last week that the bicep is sort of, well, let's say, I don't know, an algebraic  way of encoding, well, pulling back curves.  So can you encode pulling back the edges that you use to pull back your trees and sort of,  I mean, can you use a bicep to encode that or to make an algorithm that you can, let's  say, implement in the computer?  I don't think we had considered using biceps to encode the edges.  Or maybe something else.  I mean, some algebraic or what?  I'm not sure.  Okay.  Yeah.  Thank you.  I have another question.  So when you have a hover tree, in a way, the information that you really need is the tree  on the singular set on the critical values.  So the critical points that are kind of not in that tree already, they're kind of really  only included to tell you which Hobbits class you're in.  And if you know, I think, which Hobbits class you're in, if you have your covering, then  you can always reconstruct the pre-image of that one.  So I was wondering, in your algorithm, if you only did iteration on trees that includes  the critical values, but not necessarily all of the critical points, would that still work?  Is that enough?  So do you mean contain the critical values and all post-critical points?  Yes, yes.  Sorry, all post-critical points.  So the whole Hobbits, forward Hobbits of the critical values themselves, but not necessarily  the critical points, that sits out of the way somewhere.  We do not require that our trees contain the critical points if they're pre-periodic.  Okay.  Okay.  Thank you.  Yeah.  Okay.  We are running out of time.  So we'll ask questions during breaks.  So let's thank the speaker again.  Thank you.
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