 what conditions the driving measure must satisfy so that what comes out of the Lovner revolution has a geometric characterization. So in the end, the game always over here is kind of the interplay between the properties of the driving measure and the properties of the hull. And so the question we're really asking is what property must the driving measure have when it starts at a singular initial condition so that the hull will actually have a geometric property? Okay, it's a conformal mapping piece. And it turns out that what truly matters is that when points come into the singularity, so if you've got a branch point and you've got two points coming out of the singularity, they really must spread apart exactly at a square root of t. So somehow you can go from the geometry of what you want, so the geometry of how points emerge from a singularity, how the hull emerges from a singularity, to a time scaling on the driving measure. Okay, so this is not quite a standard fact in Lovner theory. Despite its simplicity. And this is sort of one of the main calculations that we've been did, and it's got some sort of interesting features which I can't get into. So the main point that comes out of this is that somehow if you drive a Lovner evolution with the forcing, which is like the Dyson forcing, or let's say just Coulombic repulsion, that's a better way of putting it. So if you choose points to sort of repel according to the Coulombic repulsion, then this is perfectly suited to Lovner revolution. Okay, Lovner revolution with branching. So it's a priority. I mean, it's really like somehow there's this miracle. So you need Coulombic repulsion for, so somehow the Lovner revolution has got a square root t scaling. Coulombic repulsion emerging from singular and initial conditions has got a square root t scaling. And you can put these two pieces together. So, you know, a priority, you wouldn't think that these two things have something to do with one other, but it just works. So it was really a lot of work before we finally converged on this. But, you know, once you get it, I mean, you can really go to town very easily. So women's theorem, one of them is basically the following. So, you know, once you know how to go past one singularity, you can go past a tree of singularities. So it really doesn't matter, you know, that you have many, many branch points over here. You're just doing it piecewise, and you need to be able to piece the solution across the branch points. And so, just by sort of using deterministic repulsion, she could prove a theorem that says that the solution to Lovner revolution driven by a repulsion of this form across the branch points will give you a graph embedding of the tree. So there are two constants over here. So one is sort of the size of the measure. And there's a second constant over there which is the time scaling of here. And the ratio of these two constants determines the angle at which these guys pop out. And so in our work, actually, we choose both of these constants to be the same. Okay, these will be going to zero. But it turns out that our approximations always have this property that the geometry is controlled. The ratio of the angles at which branching occurs always stays equal to one. Okay, so the proof basically relies on certain OD results. We haven't done truly the Dyson case, but it doesn't seem that hard. And the point is that you control the angles at which the branching occurs. So in some sense, the true theorem is really to do with evolutions of this nature. Okay, so these are trees which are generated by this mechanism. And the theorem is really that you can take any Galton Watson tree and that you can embed this Galton Watson tree in a natural way in the upper half plane. So this one looks different from the one I gave you before because there's no jitter. Okay, these things are not moving around. Okay, I feel like I'm talking too much, so I don't know. Maybe I should pause, ask if anyone has a question. Sigma is the variance of the branching mechanism that you choose. Okay, so the sigma is not related to the spatial motion anyway. So the picture is always the same. So somehow branching process theory has sort of worked out. And what I'm really doing is I'm taking branching process theory and I'm sort of asking how do I choose a driving measure for love and revolution that drives it. So let me sort of turn to the question of the scaling limit. And so let me actually just sort of explain what goes into the proof tightness and the form in which we actually obtain tightness and sort of the form in which we obtain the sort of super process limit. So the question that we really want to ask now is that now that we have a way of embedding Galton Watson trees using love and revolution, the question that we want to ask is how do I take scaling limits of this process? So what I want to do is I want to take a scaling limit over here where the tree is actually converging to a continuum tree and the spatial, the sort of driving measure over here is going to the super process. So what we really want, of course, is to say that I'm obtaining these embedded shapes in the complex plane. What I really want to say is that these embedded shapes actually have a limiting, that they have limiting geometric properties. But this is not, you know, this is kind of delicate and the step what we get is really that the scaling, what we really want to do is to identify the scaling limit of the corresponding sequence of random driving measures. So, you know, this involves choices of norming constants. So, you know, I'm going to take a sequence of measures. So the right hand side over here is an integer valued measure. I mean, it's got an integer number of particles, but then I'm scaling it in some way. And then I've got a time scaling in the repulsion, just pure Coulombic repulsion. Okay, so the Coulombic repulsion is really, you know, just because we want a complete picture. So we haven't really sort of put in the dice and part yet. But that, as I said, is something that banishes in the limit. Okay, so the question is how do you choose trees? And so over here, when I sort of wrote out this PDE in some sense, I'm making a choice on what sort of trees I'm taking the scaling limit of. And there's, you know, to do the CRT, you actually have to do some conditioning. So let me actually sort of state that a little bit more carefully. So there's a pretty well-known theorem of Aldous that basically says that if you choose the trees to be distributed as critical binary Galton Watson trees with exponential lifetimes of mean one over two square root of k, and that these trees are conditioned to have k edges, then these trees converge and distribution to CRT as k goes to infinity. Okay, so I'll call this the conditioned case. And so the sort of natural question is that if we choose the driving measures to be determined in this way, and if the trees are distributed as critical binary Galton Watson trees, does this process and do the hulls converge to anything? So this is kind of standard stuff. So you want to prove that you've got a scaling limit. So you need to establish tightness, and then you need to establish uniqueness of the limit point. And so, you know, the way you set this up. So what you've really got is a sequence of measure-valued processes, and what you want to do is you want to say that the sequence of measured-valued processes is tight. So somehow you have to look at the score cut space of, you know, paths into the space of measures, and the limit that you would get is something called a super process because it's a measure-valued Markov process. Okay, so it turns out, and, you know, somehow I guess by hindsight this was clear, but, you know, we're not really super process people. And it turns out that with hindsight, to actually establish tightness for this problem is actually remarkably straightforward. The fact that sort of we are doing Dyson Brownian motion, or, you know, the tightness works for Dyson Brownian motion, or we're just doing Kulubink repulsion, that we're doing interacting particles, just doesn't matter that much because it turns out that, you know, there's a very neat estimate, very simple estimate that basically gives you sort of the fact that the spatial motion is not too bad. So the tightness proof actually is very much in line with just sort of classical tightness for the branching process itself. And these are theorems that are worked out, so, you know, let me first state the results. So the scaling limit, the first theorem on the tightness of the scaling limit is that let me choose the trees to be distributed as critical Gault and Watson trees, and let me choose the corresponding sequence of measures, and if I choose the scaling constants in this way, then the sequence is tight. So the reason I'm choosing the constants in this manner is I want to build trees where I have some control over the geometry, and ideally you'd like something like isometry, but what we're doing is we're building trees where we can control the branching angles. So it doesn't matter how large the tree is, the branching angle is always controlled, okay, so which fits very nicely with sort of conformal mappings. And it turns out that somehow this rescaling that you choose over here, you know, this goes back to Aldous's theorem, is that this is the rescaling for which the total population process has a scaling limit, and this is well understood. It's the local time of the normalized Brownian excursion. So, you know, let me just give it to you in a picture. So what I've got is a Gault and Watson tree, a binary Gault and Watson tree in continuous time, and you can think about sort of the population process of this Gault and Watson tree. So what's going on in this picture over here is that I'm, that's the time axis going up, and at every time over here I'm just counting the number of individuals that I have. Okay, so if you flip that around, that's time, and here are the number of individuals that you have. So what's going on in the limit is that both of these guys are converging. So this guy's converging to something that's known, the Brownian excursion. This guy's converging to something which is known, which is the local time of the normalized Brownian excursion. So it turns out, so somehow when I'm writing, so this goes back to what I was saying about conditioned and unconditioned, it turns out that somehow there's one limit where it's easier to write out the SPD. There's SPDs in both cases, there's stikeness in both cases, but in one case the SPD is particularly clean. It's this one. In the other case it involves an additional term. The additional case corresponds to the CRT, this corresponds to the fellow diffusion. So there are sort of different scalings that you can do, depending whether you're conditioning or unconditioning or not conditioning, and so there's a result of Bitman which basically says that the total population process converges to the local time. So that's this picture over here, in the case when you have critical binary Galton Watson trees. And there's another case where if you just take a discrete critical Galton Watson process and you say that the offspring have finite variance, then this is sort of standard, okay it's like textbook, so you can find this in the book by Ethier and Kurtz, for example, a very efficient proof of this that tells you that somehow the Galton Watson process is converging to the fellow diffusion. So the SPD I'm actually describing is in this setting, and in this setting too there's an SPD which I'm not actually telling you what it is. So the characterization of the limit is basically the following, for example, in the unconditioned case, each sub-sequential limit solves the following martingale problem, and so this is really somehow the term that you expect, okay, this is like a weak form of the Coulomb repulsion, or a weak form of what you expect in the free probability limit, but now there's a fluctuation term, and the fluctuation term has a quadratic variation which is pretty much standard in, you know, for example, it's pretty much exactly the same thing as you would get in the case of Dawson Watanabe. Okay, so there's a problem, the problem is that we don't yet know if the solution to the martingale problem is unique, and there's a kind of stochastic calculus framework for it which I don't understand yet, but I believe that actually this problem, I mean the uniqueness is actually not that big a deal, somehow Dawson has, the repulsion is just right. Okay, so let me now sort of in the last few minutes just give you a bunch of pictures to tell you why we did what we did, and to sort of connect with sort of a couple of other things that people have done, which I think are very interesting. So as I mentioned at the start, somehow, you know, I mean Stefan Roder was quite helpful to us, and Stefan has been interested in the following problem for a long time, which is how do you embed a tree into the upper half plane? Sorry, how do you embed a tree into all of the complex plane in a particularly natural way? So I want a natural embedding of trees, and some are natural to truly be isometric because the tree is a metric space, and there's a beautiful fact, so if you're working in the whole plane, somehow there's a beautiful characterization of these embeddings. So a planar tree is said to be confably balanced if each edge, so this is a picture taken from Joel Barnes, he says this is an embedding of what I'm calling a true tree. So the important thing about this tree is that every single edge has equal harmonic measure from infinity, okay, and edge subsets have the same measure from either side. So there's kind of an amazing result due to Bishop and sort of a short note by Bjorn which says that this set of trees is actually in one-to-one correspondence with a family of polynomials called Shabbat polynomials. So these are polynomials which have just two critical values, and you know algebraic geometries came up with it somehow from their own, through their own considerations, but these polynomials are poorly understood. So what Stefan's student did was to basically try to understand the asymptotics of these true trees as n gets larger and larger and then he sort of got stuck. So the way he was trying to do it is basically in line with, so I'm going to show you another couple of pictures, this is from a paper of Bjorn, so somehow when you're trying to build up trees, so you know there's a purely combinatorial picture. So up here what I have is non-crossing partitions. So what I'm doing is I'm taking a circle with an even number of endpoints and then I join edges in pairs. Okay so that's a purely sort of, you know this is a classic sort of combinatorial bijection that says that non-crossing partitions of this nature are in one-to-one correspondence with planar trees. But now what you do is you try to make this somehow conformal. So you add another piece to it where you basically try to build a conformal mapping which takes exactly this guy and pairs these edges. So you build a bunch of conformal mappings which are pairing these guys. So somehow what happens when you're working in the entire complex plane is that you can prove that the set of solutions to this is rigid. So there's a polynomial solution then you can prove that you make a conformal mapping by some continuity arguments you can say that you've got an entire function and then by Liouville's theorem this has to be the Shevat polynomial. So there's this natural setting in which you're working on the whole complex plane and you truly get these isometric embeddings of these trees and these are very rigid. I mean they're just a set of solutions to some polynomials. So this is not the case in the upper half plane so you don't have a Liouville theorem and somehow, you know, Lebanon revolution has an extension to this setting but we don't quite know what it is just yet. So that's one remark that I wanted to make in terms of just the embeddings of trees. Okay, so the really interesting thing is not just having the trees but what you actually want is labelling on the trees. So this is sort of this amazing work of Lagal and Mermon which was sort of one of the things that got us excited. So there's a fundamental bijection. So if you've got a label plane tree then there's fundamental bijection that takes you from the label tree to the quadrangulation. So all that I've been telling you so far relates to just having the tree. Now I'm adding on this additional structure which is the labelling. Okay, so my claim is the following. So if you know how to embed the tree putting on the labelling will actually give you a very clean way a sort of conformal way of doing this bijection. Okay, so it's just sort of a cute two-minute thing. So here's what I'm going to do. I'm just going to take exactly the same tree which is in Lagal's lecture and I'm going to show you how to sort of conformalize it. So as you can see there were slides in this talk that were made by the graduate student and the slides in the talk made by the advisor. So I spent like three hours yesterday trying to get this right on my computer and finally abandoned hope I drew it on paper and took photos with my camera. Okay, so you see there's the tree. Okay, and now let me just explain to you how the bijection works. So what you're going to do is sort of the CVS bijection actually takes you. It's a bijection from corners of the tree. Okay, so you know you have to sort of... Okay, let me explain it to you in this picture. So the numbers over here refer to the corners of the tree. So corners are where two directed edges are meeting. Okay, and what I'm going to do is I'm going to sort of flatten this out. Okay, so I'm going to flatten the edges in cyclic order and label the corners. So those are the corners and then each of the corners also corresponds to a vertex. So I take the labeling that goes with each of these vertices. Okay, so now I'm going to just show you what the CVS bijection means in this setting. So what it means is that... So what I'm going to do is the following. I take... Let me go back a second. Yikes, this guy goes... Okay, this is... Yeah, okay, so here we are. So what I'm going to do is I'm going to imagine that each of these points over here is on the boundary of the half per half plane. And what I'm going to do is I'm going to join these points in the way that you do it for the CVS bijection. So the way it works is that you take... So all I've done over here is to put down the labels. So the first thing I do is I look at points which are labeled one. I connect them to the point at infinity. Then I look at the points which are labeled two and I join them to the next point with sort of label which is one below in cyclic order. And then I go to three, the points which are labeled three and I do the same thing. So the CVS bijection what it gives you is actually gives you a family of nested geodesics in the upper half plane. Okay. And so what I'm doing if I undo my conformal mapping what I'm really getting is an interval... is an interval which is on the endpoint of the upper half plane. So if I just take my... So what I've given you is a method for constructing trees or conformal mappings which will have the combinatorics of the tree. Now if I put on the labeling what I'm really doing is I'm looking at geodesics and I apply my conformal mapping to these geodesics and that automatically gives me a quadrangulation. Okay. So in some sense the work that we're doing is actually sort of very closely tied to actually the problem of building up quadrangulations. It doesn't fully work because our mappings have the right sort of combinatorics but they're not isometric so we don't have isometric embeddings of trees by this method but somehow I'm quite optimistic that we should be able to do something interesting. Okay. So I'm sorry for going a couple of minutes over but let me stop here. Thank you. So that's exactly so that's exactly where we stuck. So here's the thing. So what beyond actually is doing is subordination which is actually just laminar evolution. And so somehow I want to use beyond's argument but somehow to do that I actually need to know my solution to the SPD to begin with. So I've not been able to close that loop. So that's exactly right. So somehow this fits beautifully with what he's doing. I mean we're really just doing subordination. What's happening is that the measure is sort of fluctuating at exactly the same time.