 Thank you very much for coming for a second lecture. So today we are going to focus on a specific lattice and study, say, for writing works on the hexagonal lattice. So basically everything I will tell you today is very specific to this lattice. And the first thing we, and you are going to understand why during the lecture. And first thing we are going to do, remember that last week we introduced a notion of connective constant, and today we are going to compute it for this specific case. So computation of mu c of the hexagonal lattice. So the story of the computation of the hexagonal lattice started by a conjecture in 1982, due to NIN house. By the way, maybe I should remind you that, so madras slade is a typical reference for safer wooding work in general, and otherwise you have lecture notes by Bauer-Schmidt, Goodman, Slade and myself, which will actually contain this computation, which is not the case of the book. Okay, so in 1982 NIN house predicted, and I will explain to you how it did it at the end of the lecture, it predicted that mu c of the hexagonal lattice should in fact be equal to square root of two plus square root of two. So that was a conjecture, which lasted for 30 years. And my goal today is to start by a proof of this conjecture. So theorem by myself in Smirnov, that indeed mu c of the hexagonal lattice is equal to square root of two plus square root of two. And if you want to do it on the square lattice, well, good luck. I do not really expect that there is a specific value for the connective constant on the square lattice. So there is something kind of special happening for the hexagonal lattice. There is actually another lattice for which you can compute explicitly. And I leave it as an exercise to you to try to compute PC for this lattice. And here what I recommend is try to deduce mu c for this lattice, from mu c for the hexagonal lattice, try to relate both. Very good. So the proof will be based on what we call a para-feminine observable. So proof based on para-feminine observable. So what I will do is I will first basically prove completely the theorem using an object which maybe doesn't sound so intuitive to you at first sight. And then during the lecture and towards the end of the lecture, I will maybe give you a hint why one can come up with such an observable. Why is it fairly natural? So first thing we are going to do is we are going to... I mean, the whole proof is based on this function, this para-feminine observable, which is really some kind of discrete version of a holomorphic function. So what we want to do is to define discrete versions of what we usually use in complex analysis. And one of the things we like to work on in complex analysis, when we work with complex analysis, is we like to work with domains, simply connected domains. So I will call first omega is a discrete, simply connected, and I will just call this domain. If it is the interior of a safe avoiding polygon on the triangular lattice, which is just the dual lattice of the hexagonal one. So what I mean by a domain is the following. You take a polygon on the triangular lattice, which is dual to the hexagonal lattice. So by dual, I mean I put a vertex at every center of a face and edges between vertices corresponding to neighboring faces. So I take a safe avoiding polygon, I cut along this polygon, I have an interior and an exterior, what is inside is my domain. So you can see it in another way, it's a connected graph whose complement in the hexagonal lattice is also connected. So just think of this basically a shape like that. So that's going to be a domain. And now what I will do is you see when you have domains like that, you have mid-edges on the boundary. So I will pick one of them, let's call it A, and I will work with safe avoiding walks starting from A and staying in the domain and ending at the middle of a vertex. So it's slightly different, sorry, middle of an edge. So our safe avoiding walks are going to be a little bit different in this class. They will go from middle of edges to middle of edges. But you agree with me that if you can count these objects, it's actually quite simple then to relate it to the counting of the original objects. That's actually what we will do. So pick A, a mid-edge on the boundary of omega, which I think you understand, we call it, for instance, the safe avoiding polygon is the boundary of omega. So you pick a mid-edge on the boundary. Pick A, okay, I realize that's written like that. It looks a little bit the same. So this is our A and the mid-edge in omega. So it can be in omega or on the boundary of omega, there is no problem with that. Okay, so the paraffin unique observable is the following. It's a function of Z. In fact, it's a function of many more things. It's a function, in fact, of omega of A of the parameter X and even of sigma. But I will just record it as a function of Z because that's what would be important for us. And this is just the sum of every walk gamma from A to Z, gamma staying in omega of exponential of minus i sigma times the winding of A to Z for the curve gamma, I will tell you what this is, times X to the length. This is the length. And this is the winding. So what is the winding? The winding is pi over 3 times the number of left minus number of right turns in gamma. So, okay, first time you see this thing, maybe you can be a little bit surprised by this definition. Forget about this term. What is it? If you forget about this term, you just end up with the partition, I mean the generating function for walks from A to Z staying in omega, right, if you ignore this one. So that looks like a much more natural object to consider if you want to count. If you want to know how many guys you have, how many walks of length n do you have between distance, between A and Z staying in omega, this is a pretty natural object to look at. The problem of this object is that it doesn't have any nice properties. If you think of trying to enumerate things using generating function, one of the first things you want to do with your generating function is you want to compute it. So usually what you do is you prove some kind of recursive relation on your objects, and this translates into some functional equation for your generating function. It allows you to compute it explicitly or at least to know sufficiently many things on the generating function to be able to extract information from it. The problem is that, I mean, we had the lecture last week, you realized pretty quickly that our safer voting walks are like the farthest away you can imagine from kind of Markov processes. Like, in order to know how you extend by one step, you need to know the whole geometry of your previous curve. So it's objects that are absolutely not well suited for recursive relations. So you have to, just to forget this hope that maybe you could explicitly compute your generating function. So the magical thing here, and it will be a little bit less magical at the end of this lecture, I hope, is that by adding this term, which is a complex number, right, by adding this term, the function is going to satisfy something completely different from what we usually see with generating function. It's going to be basically discrete holomorphic. So this term, by the way, so just to give you example of winding, let's say you start like that, so you do things like that, and let's say you end up there. Then what you did here, you can check, you did as many turns on the left and turns on the right. So the winding of your curve here is zero. So here you would just count one for this curve. Now if you imagine the same thing, but let's say you do something like that. Okay, maybe I did one more turn, but maybe not. So you end up at the same point, the perfect hexagonal lattice. Nobody can prove me otherwise. I mean, okay. When you end up like that, you did actually six turns on the left more than turns on the right. So the winding in this case is going to be 2 pi. You have six turns, which each one counts pi over 3. So you can get like that different windings. If you think about it when you arrive like that at this edge, basically any winding in 2 pi z is possible. They find up to 2 pi, but not more than that. And if you are even allowed to arrive like that, then it's even up to pi. So you have a twist like that, a complex twist to your function. So why is it a good thing to do? Why is it smart to do that? Well, let's break the suspense right away. So this was our theorem 2.1. So now theorem 2.2, I mean, lemma 2.2 is the following. Up to now, x is arbitrary, sigma is arbitrary. But the point is going to be the following. Assume that sigma is equal to 5 over 8. You are going to see where it comes from in a second. And x is equal to 1 over square root of 2 plus square root of 2. And here I really want to argue, I mean to highlight the fact that at this stage, there is no connection to the connective constant. Absolutely none. It's something completely independent. If x is equal to this, then for every v in omega, so here it's really a vertex in omega. Let's put it here. This is v. Around v, there are 3 mid-edges. So this is v and you have 3 mid-edges p, q and r that I order like that. Then for every v, I have the following relation, p minus v f of p plus q minus v f of q plus r minus v f of r is equal to 0. So I gain like that one relation, one local relation for every vertex in my lattice. So how do I interpret this thing? Notice that if I fix p as a witness, then q minus v compared to p minus v is just j. It's e to the 2i pi over 3 times p minus v. So this relation rewrite as p minus v times f of p plus e to the 2i pi over 3 f of q plus e to the 4i pi over 3 f of r equals 0. Now imagine you want to define the natural notion of contour integral in the discrete. You have a notion in the continuum. You want to integrate along a certain contour f of z d z. Well, in the continuum you have a good notion. For the discrete, how would you do that? Well, you first need to tell what is c. What is a contour in the discrete? So what I will do, what I will consider as a contour, would be simply a polygon, a safe holding polygon in the dual of my hexagonal lattice. So for instance, this would be a contour. So it passes through middle edges and goes from middle of faces to middle of faces. So if this is my thing, let's do it like that. What is the natural notion for this thing? Like let's say it's x0, x1, I mean f0, f1, f2, etc. What is the natural notion here of discrete integral? Well, for every interval like that, I need to tell what is the value of f. So what is the natural value of f for an interval like that? Well, f is defined only at the middle of the edge, right? But what you can do is just decide, it's a convention, decide that f is actually equal to this value for the whole interval, for the whole segment. If you do that and you think of the integral along this continuous contour now, then what you are going to end up with is sum for i equals 0 to k. Let's say this is fk. Of what? Well, fi plus 1 minus fi times f of zi, where zi is this thing. So here it's z0, this is z1, z2, z3, etc. Right? So this is really the notion of discrete contour when I'm thinking, well, along this whole edge from f2 to f3, where the value of my function is equal to f of z2. That basically is the only natural thing you can do. But if I define like that, if I decide, okay, this is the definition of a discrete contour when this is the polygon in omega star, like in the triangular lattice. If I define my discrete contour like that, how can I reinterpret this integral here? Well, if I take this contour like that, which is a Hones contour, and I integrate f of zizi along this contour, well, I'm going to get a certain value times f of p plus e to the 2i pi over 3 f of q plus e to the 4i pi over 3 f of r. So this thing here equals 0. Exactly tells me this is equal to 0 for every v. Now maybe you can, yes? How are we dealing with the boundary? How are we dealing with the boundary with this? At this stage, there is no problem with having a point on the boundary, right? For any vertex in my domain, even a vertex here, I do have one, two, and three guys, three mid-edge there. So actually even the contours can definitely touch this red curve, right? They cannot go outside, but they can definitely touch it and use some of the edges of the origin. Now you see why I work with domains, because if omega is a domain, then any contour can be, I mean, can be written as a sum of triangular contours when you look at the natural z2 vector space associated to contours. So in other words, so any contour can be written as a sum of triangular contours inside it. As long as there are no holes in my domain, I can do that. But then if the discrete contour integral is zero for any triangular contour, then I deduce the same for any contour. So here, that implies that the integral of c of f of z d z is equal to zero for any contour c. And that is a discrete version of a very well-known property of holomorphic function, which is that the contour integral is equal to zero. So this is kind of a discrete version of Morera theorem. It's a discrete version which is associated to Morera theorem. It's one of the definitions of holomorphic function. So really think of this as a discrete version of holomorphicity. I often have an associated question there. People ask me, well, we can also define the holomorphic functions via Cauchy-Riemann equations, right? So here, could I interpret these things as discrete Cauchy-Riemann equations? And if you think about it, what exactly corresponds to that? I mean, if you want this p minus v, this p minus v f of p is some kind of the derivative in that direction of my function. So it can also be interpreted as discrete Cauchy-Riemann equation, but we won't need it here. Okay, let me prove this lemma. So I have a big sum here, okay? And this I'm going to write it as a sum for any work gamma going from a to p, q, or r, gamma staying in omega. And here, I'm going to define something, c of gamma, this is the contribution of gamma to this sum, to the sum, and it's defined by the following relation. So c of gamma, by definition, it's what? Well, it's this weight here. So x of gamma, and there is e to the minus i sigma times the winding. Let me maybe write winding like that just to make it a little bit shorter. And there is an additional term which is this p minus v or q minus v or r minus v, which I'm going to write it as gamma, gamma minus v, right? This is the endpoint of my curve, which is p, q, or r minus v. So that's the contribution and I need to prove that this huge sum on all these walks ending at p, q, or r is zero. And in order to do that, I'm going to be extremely optimistic and just think, okay, maybe this huge sum is zero because actually the sum is zero if I do small packages of walks. Maybe there is a way to pair very small groups, the walks in such a way that their contributions cancel each other. That's very optimistic. There is no reason it should be true, but it could be. So let's try, for instance, to pair some walks together. So there is a natural type of walks we can try to pair is the following. So imagine you have a certain walk arriving at, let's say, p and doing after that, doing, okay, I want to do something non-symmetric, so let's do like that, and ending there. Okay? Then this, I can pair with the following walk. How can I associate another walk doing basically the same thing? Well, I can just, here I have a certain shape, like that and here what I did is that I did this turn here. But I could also just have done this one. So you just reverse what happens here. Take this guy and you change it to this one. Okay? These are two walks and if you think about it pairing like that, these walks completely partition all walks visiting the three mid-edges p, q and r. For any such walk there is another walk associated to it in a very natural way. Okay? Okay, so let's look at the contribution of gamma 1. Okay? It's something. But can I write the contribution of gamma 2 in terms of the one of gamma 1? So here it's, let's say it's q minus v e to the minus i sigma winding x to the gamma. If I just try to compute the same thing for gamma 2 well gamma 2 has the same length as gamma 1. So here I can write gamma 1. The p minus v here is r minus v here. So it's e to the 2i. I mean if I write both things okay maybe let's write both things instead of p minus v. Here I get e to the minus i 2pi over 3 p minus v. And then for the winding term well the winding term here is e to the minus i sigma winding of gamma 1 from a to p plus e to the minus i sigma the winding of gamma 1 to q. And this if you think about it since I started on the boundary here well whatever the end of the curve from p to q from r to q it will always be from p to q sorry it will always have a winding of minus 4pi over 3. Whatever the way you are doing and a good way of seeing it is if you take just a walk going like that there is just a computation you do one turn on the left and five turns on the right so it's indeed four turns on the right and then if you have a larger walk you can just remove going around faces because you see that that's a move that doesn't change the winding so this thing if you take this walk you can just check that replacing by that doesn't change the winding and you do this step by step reducing the number of faces up to the point where you reach just one face so here I get minus 4pi over 3 if I do the same computation here I get on the contrary 4pi over 3 because I'm turning the other way so when I express these two guys together it is equal to p minus v e to the minus i sigma winding from a to p x to the gamma 1 and then here I have e to the i 2pi over 3 times e to the minus i sigma minus 4pi over 3 so it gives me something like that and for the other term I get the conjugate now you notice that if sigma is equal to 5 over 8 then this is minus i and this is i therefore I get 0 so if I tune sigma equal 5 over 8 then all these sums over pairs is 0 I mean r0 so that already reduces a lot my big sum upstairs because it basically tells me I only need to deal with walks visited one or two of the edges pqr although the one that visits the three are already done but now for those ones you can exactly play the same game imagine you have a walk which is visiting only one guy you can naturally associate it to walks visiting two guys by just saying well there are two ways here of extending which are these two ways you can extend because you know you don't visit the other guys so this is gamma 1 gamma 2 and gamma 3 and now if I play exactly the same game I took some times here but now I'm going to do it a little bit faster if I express everything c of gamma 1 here what is going to change is I get length which is one more so I get a multiplication by x and then the winding term is going to be the winding is pi over 3 more and the r minus v is e to the minus i 2 pi over 3 compared to the other one so I end up with i so let's do it like that e to the minus i 2 pi over 3 times e to the i 5 over 8 pi over 3 because now sigma is set and the other one is this one like that so this here is nothing more than 1 plus x cos 3 pi over 8 and with a 2 so it is equal to 0 if x is 1 over square root of 2 plus square root of 2 because this guy twice this guy is just minus square root of 2 plus square root of 2 okay well that's the end of the proof because now all the triplets are equal to 0 and these partitions all works and then get p, q, r is it okay yes the yellow one you could imagine this is a red extension of a shorter one no, no because you see the yellow one is visiting only one of the three guys so it cannot be the extension of anybody associated to this vertex here we are really partitioning guys associated to this vertex okay now proof of theorem so what do we need to do well up to now we have a relation for a certain value of x and a certain function f but this relation is not a priori related to the connective constant so what we need to prove is that this value of x which is 1 over square root of 2 plus square root of 2 the only place if you want I mean the only way all these relations are true is if x is equal to xc is equal to the critical point for our model so we need to relate this morera, I mean this critical homofficity to the I mean we need to extract juice from it to be able to say that the connective constant is what it is and notice just before I start that a priori it's not clear we can do it for the following reason is that here these relations actually there are as many relations as vertices in my lattice in my domain therefore there are way fewer relations than unknown there is one unknown per vertex one relation per vertex one unknown per edge one relation per vertex so it's a very undetermined linear system so it's absolutely unclear that we are going to be able to extract information but we will manage and the idea is to say okay there is definitely something I don't like with the para-feminine observable at first sight is that there is this winding term this winding term is actually very bad for me because if you think about it it's not even clear that our observable is not equal to 0 because you have cancellations and maybe they are so good that it exactly gives you f equals 0 so we need to kind of use it at places where we know basically that it's really not 0 and not doing anything bad and one thing which is good is that notice that what you are when you are on the boundary you take z to be this guy here well any walk going from a to z has the same winding in this case it's always 0 you can check any walk will have winding 0 in this case so f on the boundary can be very easily expressed in terms of just the generating function because the winding is deterministic so what we will do is we will just choose our domain and integrate just use the discrete contour integral on the boundary 0 so let's take let's say the upper bound so let's start with the upper bound so we are going to try to bound from above the connective constant and for this one I'm going to use exactly the domain that I drew there so I take a domain like that I have a let's say it has right t and width w ok and let's use that the discrete contour integral so this is c and this whole thing is omega and let's just use the integral around c of f of z d z is equal to 0 when the contour say goes that way ok so there are different types of contribution here the first thing is contribution due to guys at the top so in this case f i plus 1 minus f i is equal to 1 right because I'm doing this minus this this minus this and so on so I'm summing the f of z for z on the top then I have let's look at those guys they get minus 1 like that so it would be minus there are f of a and then there are minus sum of f of z for z at the bottom and z not equal to a and then there are terms on the boundary which I'm not going to write because I will leave this as an exercise for you so this whole thing is equal to 0 ok but this whole thing now I can use that f on the boundary is extremely explicit so in particular on the top f of z is equal to the sum for gamma from a to the top gamma included in omega of 1 over square root of 2 plus square root of 2 to the gamma here of course now I'm really looking at x equal x c x equal 1 over square root of 2 plus square root since I'm using that so here what I use is that the winding of the guys at the top is always 0 so I can forget the winding term ok then f of a well f of a is actually quite simple right what is the only walk going from a to a that's the empty walk therefore f of a is 1 and then what is the sum of z at the bottom but z not equal to a well it's the sum for gamma from a to bottom minus a gamma staying in omega it's 1 over square root of 2 plus square root of 2 to the gamma the only thing that remains here is that you do have a winding term right you have a winding term which is if I go like that the winding term is pi and if I go like that the winding term is minus pi and by symmetry or by taking the real part of this relation this give me of well winding pi times 5 over 8 but I have ok ok so again here you have two ways of seeing why I put cos instead of e to the i 5 over 8 times pi there are two ways of doing either you just say you take the real part here or you just use the trivial symmetry between walks and we get p and minus p that's two ways of so when I retune this thing here what do I end up with I end up with this guy I'm going to call it bt w and this guy here atw so I end up with btw the f of a is minus 1 here then I have cos of minus cos of pi over 8 which is just cos of 3pi over 8 plus cos of 3pi over 8 atw plus here terms which are due to the boundary right these terms I let you as an exercise for the whole thing to check that they are also positive so these terms here are positive actually the reason why I leave it as an exercise and I don't do the computation is because there are many ways of getting rid of these terms another one is just to let the width go to infinity because if you think about it just the walks going to infinity that way they are exponentially fewer than walks in the bulk so you can just prove that these terms just go to 0 so it's really the terms on the edges are not problematic what I really want to argue is that these terms is equal to 1 so in particular bt which is a limit when w tends to infinity of btw well bt is smaller equal to 1 for every t right simply because this all these terms are positive and the sum is equal to 1 but if this is true not that n I mean not that bn mu c I mean bn times 1 of a square root of 2 plus square root of 2 to the n so remember bn is the number of safe avoiding bridges of length n if I'm a bridge of length n which is smaller or equal to n so this thing is going to be definitely smaller than the sum for t smaller or equal to n of bt so it's smaller or equal to n times well n smaller than n so all together I end up that bn is smaller or equal to n times square root of 2 plus square root of 2 to the n so amersley welch tells me bn is growing like the connective constant of the hexagonal lattice so amersley welch implies that mu c of the hexagonal lattice is smaller or equal to square root of 2 plus square root of 2 so here you see the only thing you have to remember is to apply your contour integral on a well chosen domain that was for the upper bound now let's look at the lower bound and we are going to play exactly the same game simply on a different domain so take this domain where this has size t look at this domain and do exactly the same use that the integrals are I mean that the windings are deterministic use that the contour integral is 0 so am going to get 1 is equal to something right and this something is a sum of different terms there are some of terms like that and there are some of terms ending there right on any of the other guys the sum on terms ending there well now I think you understood that the p minus v e to the i sigma blah blah anyway it's a complex number of modulus smaller equal to 1 so if I want an inequality here I can just get sum of a gamma going from a to this part and gamma staying in omega of x I mean of 1 of a square root of 2 plus square root of 2 to the gamma right these terms all contribute at most this quantity then there are terms ending there right but these terms what are their winding their winding is well 1 2 3 4 so get winding of minus 4 power over 3 or 4 pi over 3 when I use the p minus v and this winding what I end up with is that these guys here contribute i times the partition function of these guys and the other guy so i times let's call the partition function of this guy et and the other guy here contributes just minus i times the same quantity so the contribution of this size and this side just cancel each other therefore here you get 0 so why is it good for us it's good because it concludes the proof why because now just think of the partition function of safe avoiding works at 1 of a square root of 2 plus square root 2 well this is definitely larger we call it to the sum over t equals 0 to infinity of the sum for gamma from 0 to boundary here of size t gamma staying in this omega t of x of 1 of a square root 2 of 2 to the gamma right this is completely clear but this is 1 so this whole thing is the sum for t equals 0 to infinity of 1 so it's plus infinity and that immediately tells me that mu c is larger equal to 1 a 2 square root of 2 plus square root of 2 and that's the end of the problem so what I'm doing in this second step I'm just proving that the generating function is infinite at 1 of a square root of 2 plus square root of 2 and I'm doing this by saying at distance t when I sum even on some restricted distance t I get 1 so if I sum on t I necessarily get something infinite and that's the end of the proof so at the end you see it's I went actually fairly slowly and it fits in less than an hour it's a very short proof and it's completely elementary but really the intuition is important this intuition of using the discrete to the morphicity okay so let's try now to use the observable to go a little bit farther in enumerating the number of works so what I'm going to try to do is I'm going to try to improve how much I wish and you are going to see it's we don't improve it by much but I think more than actually the improvement it's the way we are going to do it you are going to see it's going to tell you why it's not so simple to improve it maybe and give you an idea of why it's not so simple so the theorem is the following it's saying so it's it's actually not published yet and it says that there exists epsilon positive such that Cn is smaller than e to the n to the one half minus epsilon there is a large constant in front times square root of 2 plus square root of 2 to the n the epsilon is actually explicit but it's not like a very impressive thing it's like 1 over 22 or something like that okay before we do a break let me mention like I mean first prove a lemma and this lemma is going to last week we did prove that the partition function the generating function sorry of bridges is infinite which was kind of saying that Bn was larger than something like 1 over n squared mu C to the n for infinitely many n and there was also a relation between bridges and polygons so that was also proving that polygons were larger than some 1 over n to some big power mu C to the n for infinitely many n here what I would like to have is I would like to have that for every n basically so I'm going to use the observable to improve that to say for every t we have that Bt which I already defined right is larger than C0 over t and we have also that pt which is going to be the sum for gamma for polygons let's say for gamma like that so you take lambda so you take the ball of size 8t you cut it here you cut a slit of height t and I'm summing over polygons going around like that and I sum 1 over square root of 2 per square root of 2 to the gamma and this I claim is larger than B8t which is 6 divided by t so both things together and here there is a constant C1 both things together tell you that for every t you have that Bt and pt are larger than polynomials that can look really like small gain I'm going to answer in a second it can really look like a small gain because the equations are usually not so difficult to handle but here for cephalodyne work they are it's very difficult to prove that Bn is not small for certain values of n and fairly big for other values of n so actually managing to do it here is kind of surprising and you are going to see the proof we know of this fact uses the observable and I also mention it because the result would be important for next week yes there was a question what was the second sum 50 over Bt gamma so the second sum is a sum of a polygons staying in lambda 8t and surrounding a slit like that of size t in the middle okay so let's prove that and let's do a short break remember that we just proved that Bt plus cos of 3pi over 8 8t well we prove this was equal, I mean this was smaller equal to 1 because Btw maybe it's still, no I erased it but Btw plus cos 3pi over 8 8tw was smaller equal to 1 for every omega I told you that when omega tends to infinity for acting terms these terms due to the edges we are going to 0 so trust me on this one it's true so you actually get that the limit is equal to 1 where atw at is just a limit of atw so imagine you have that well that means if you apply it to t and t plus 1 it implies that Bt minus Bt plus 1 is equal to cos 3pi over 8 times at plus 1 minus at and notice that this is nothing else than I'm summing I'm doing the partition function of objects starting at 0 ending at the bottom but I'm substriting all those so I'm starting with all those that are height smaller equal to t plus 1 all those that have height smaller equal to t so that means that these work they need to visit the top necessarily so two things first this thing is positive so it tells you that Bt is decreasing which already just to mention is not something obvious at all we have no clue how to prove this in general the second thing and it is useful it's not just the second thing is that this thing here because you see you have a work from the bottom to the top and then a work from the top to the bottom if I cut at the last point touching the top what do I end up with I end up with a bridge from bottom to top and a bridge from top to bottom so this at plus 1 minus at is smaller than Bt squared or Bt plus 1 squared and this Bt minus Bt plus 1 smaller than constant times Bt plus 1 squared if you did preparatory classes in France or if you like studying sequences this immediately tells you that Bt is larger than constant over t so that tells you Bt larger than c0 over t just this recursive inequality on Bt so that's the first claim for the second claim it's a little bit more tedious but it's the same type of manipulation and it's going to base on the following observation take this is t take this domain this smaller domain you can do exactly the same manipulation in this domain you integrate along the contour integral and what you are going to end up is well this guy here let's call it Et you are going to end up with Et plus and maybe times a certain constant which here is cos pi over 8 plus At but in the triangle which I call it like that is equal to 1 exactly this is integrating along the contour integral of this triangle but notice that this is a triangle so this is just the partition function of this type of object well At triangle is clearly smaller equal to At of the strip that's completely clear so At triangle is smaller than A or maybe this is now this is height say well it's not height 2t but height 2t it's smaller than the strip A of the strip for 2t so that immediately tells you that Et is larger equal to 1 I mean to a certain constant C1 times B 2t if this guy is larger than this guy this guy must be larger than B2t simply because we have B2t plus A2t basically equal here there is sorry a cos of 3pi over okay so you know that you can turn like that by not paying more than constant over t I mean not more than Bt so now the game is going to be simply to combine this to turn several times and do the same at the bottom with a polygon so I let you think about it but Pt from this thing you can deduce easily that Pt is larger than B8t divided by t and I leave it as an exercise first because I'm a little bit late and second because it's an interesting small exercise to combine this kind of uniform estimate plus the fact that Bt is decreasing to create polygons you can easily see that you can just combine 6 times the thing the problem is the difficulty is to end up at the same place so here there is a small game with Cauchy Schwartz to manage to try to say that at a certain distance squared has like you use Cauchy Schwartz in a smart way and you end up with exactly the estimate there so I let you think about it we kind of already used this Cauchy Schwartz type trick last week with a Pm we did exactly the same thing where we had an end point and we sum over end point and use Cauchy Schwartz too it's the same trick here so I let you think about it short break maybe and then we prove our theorem 2.3 let's start again with the proof of our theorem 2.3 and then after that I will give you a little bit more informal discussion on all of this proof of theorem 2.3 so the first thing that I want three steps are three steps are going to be actually quite simple simply because well you see step one for instance is it's sufficient to prove the bound for half space walks right? so by lemma so what was the number of the lemma 1.4 we focus on hn the step 2 is refinement so my goal is to prove that we can also restrict ourselves to walks which are not too big and in order to do that simply do the following so start with your half space walk okay and so do the Hammersley-Wesch decomposition up to bridges of height smaller or equal to n to the 1.5 plus epsilon so you fix epsilon and you start unfolding up to the first time when you unfold something of height smaller than n to the 1.5 plus epsilon okay when you did that the first time you do it you just cut so you end up with a walk a half space walk and a bridge and this bridge you know that every single time you unfolded the height was at least n to the 1.5 plus epsilon okay notice that these walks here the reconstruction cost for these walks I need to locate where did I unfold right so for each one of the unfolding I pay a n but how many unfolding can there be well not so many because each one of this length is n to the 1.5 plus epsilon so I can have at most n to the 1.5 minus epsilon such unfoldings so here why before we were having a reconstruction cost which was e to the square root n if I just restrict myself to things up to size n to the 1.5 plus epsilon I pay a reconstruction cost which is of this order so here actually you notice maybe here I could have written it like that here I can put if you want this is still an improvement n to the n to the 1.5 minus epsilon so here if I cut there the reconstruction cost is only that so what I only need to do is repeat half space walk with height smaller or equal to n to the 1.5 plus epsilon if I can bound this I will be done but I can actually improve that because any half space walk if I take a half space walk of size smaller than n to the 1.5 plus epsilon maybe this walk is extremely wide well cut it at the right most point think of rotating everything by pi over 2 and what do I end up with I end up with 2 half space walks that way which have width smaller than n to the 1.5 plus epsilon and height which can be arbitrarily large well do exactly the same as here start unfolding as long as the unfolding is larger than n to the 1.5 plus epsilon and cut after the first step where the unfolding is smaller than n to the 1.5 plus epsilon the reconstruction cost is going to be of this order so what I just argued is that I can restrict myself to hn and I'm going to use this notation which means number of half space safe avoiding walk of size m m is smaller than n a priori included in lambda n to the 1.5 plus epsilon I can really look at walks for which the width and the height is smaller than n to the 1.5 plus epsilon so I confined like that my walks these walks the number of such walks is actually really the type of guys that are problematic in Amerslie-Welsch the bad guys in Amerslie-Welsch are kind of guys that are kind of space feeling and going back and forth at height square root n that's the bad things so here I'm going to take myself a little bit more freedom go to 1.5 plus epsilon and say for free I can confine like that once walks that are doing that so that's step 2 I'm going a little bit faster because these are less important theorems somehow next week we will go back to a theorem that I will prove really with care but this one I want just to give you a glimpse at the proof step 3 is going to be to say well so now I'm working with confined half space walks I want also to work with walks that actually do a lot of back and forth so let's call how did I call it I didn't call it, very good you go let's say bad confined half space walks let's call the number of such walks h m tilde and what does it mean to be bad it means that you actually when you do the Hammersley wedge decomposition it contains at least n to the one half plus epsilon bridges I mean back and forth unfoldings of height larger we call to n to the one half and minus epsilon so I want to restrict myself to walks so they are half space walk in the box of size n to the one half plus epsilon and n to n to the one half plus epsilon and I want to restrict myself to walks that are going to actually do a lot of back and forth like that it's at least n to the one half minus epsilon okay yes why is it sufficient to count these walks exactly for the same reason as before if I take a walk which is not bad what does it mean it means it does less that n to the one half minus epsilon less in the Hammersley-Wesh decomposition but if larger than n to the one half minus epsilon but if it does less than that I mean the decomposition after for those that are smaller than n to the one half minus epsilon well remember it's strictly decreasing the unfoldings in height so overall it will do less than twice half minus epsilon unfoldings therefore the reconstruction cost would be at most n to the 2 n to the one half minus epsilon so these walks I can just bound them very easily so the problematic ones are those that indeed do a lot of back and forth of big heights and that's my step 3 and that's the difficult part so here I let you think more about all these steps like try it's a very good exercise try to really write them down following the Hammersley-Wesh decomposition and try to see that the reconstruction cost are small every single time so now let's bound h and t to the square and really I think if you ask anybody who tried to improve Hammersley-Wesh then you will see that all these first steps they are trivial like clearly everybody starts by doing that and then you end up with this kind of space feeling walks at scale n to the one half plus epsilon and you want to bound these guys so how do you do so the idea is going to be the following I'm going to first so pick delta very small let's really think of delta extremely small and this delta in morally is 2 epsilon so think delta equal to epsilon I choose epsilon tiny really tiny tiny I want a tiny improvement on Hammersley-Wesh and let's use you can use lemma that I just proved before the break the 2.4 to choose k such that if I look at lambda of 8n to the delta the number of bridges the number of polygons of size exactly k here right so let's call it pk theta is larger than 1 over n to say 10 delta uc to the k so the partition function is larger than n to the 7 or maybe I should take yeah ok so how do I choose this k I know that the partition function of the polygons with non-fixed length surrounding this by the lemma is larger than 1 over 8n to some power which is 7 delta then that means there is a length k for which it's larger than here I guess I can take 9 ok there is a length between n to the delta and 2 I mean n to the 2 delta for which with length exactly k I have more than this otherwise my pt that I just defined before for t equal n to the delta so let's fix this k and now the idea is going to be the following we are going to take our bridge our half-space walk and we are going to unfold it exactly like you would do for a musley wedge so that seems like a terrible thing to do right you take your half-space walk and you unfold so you have your guy like that so this is a very bad map because remember this map a priori well is up to n to the n to the 1 half plus epsilon 2 1 if you do a musley wedge brutally the reconstruction is very expensive you can have up to n to the 1 half plus epsilon reconstruction height ok I could improve by using the result of Ramanujanardi but at this stage it's not going to be relevant so basically it's a very bad map and if I just do that of course I'm not improving on a musley wedge because I'm actually doing something even more stupid a musley wedge was not stupid in the first place I'm doing something stupid ok but the idea is going to be the following the idea is that we are going to prove that in fact rather than sending this map to only one bridge I'm going to send it to many bridges because what I can observe is that here it's a fairly long and high bridge so I can try to glue polygons of the form there I can try to glue them a little bit like we glued two polygons together I pick a polygon and I try to glue it here to my work and I try to glue it here to my work ok so of course if I just do that I didn't progress much for the following reasons that I need to know where I glued things if I give you some bridge where I glued these polygons then you know where I glued them if you don't know then probably I didn't get I didn't get much entropy by doing that so the trick is going to be to glue them in such a way that if I give you the bridge with the polygons you are actually able to tell me where I glued the polygons more precisely what I want is to glue polygons in such a way that if I give you the work and the places where I unfolded you are able to tell me where I glued the polygons so how will I do that well I will glue the polygons in such a way that when I refold they are actually intersecting the work they are self-intersection of the work and this you know is forbidden so you know something wrong happened at this place therefore I necessarily glued the polygons that's the trick so let's try here to do it so let's call a good place so I have my polygon my half-space work and let's call it so remember it's doing a lot of back and forth between a certain g-mean and g-max and here the distance between the two is n to the one half minus epsilon because I'm a bad confined work so I'm going to call a place a good place for gluing if when I put a polygon like that I actually so I put it like that here I have an excursion here I have another excursion I'm going to call this thing good if when I glue this polygon I intersect the next guy okay so let's call it a good I mean let's call good of g-max a set of places like that and also I'm going to ask that this red polygon when I glue it it doesn't exceed this line or doesn't go below this line okay exercise because it's actually a cool exercise I liked it there are many good places so the claim is the following if delta is indeed equal to 2 epsilon which I basically assume then the set of good gamma is always larger than the small constant times n to the one minus 2 epsilon and this thing from now on I will call it capital N this number there are many places where you can glue and the reason for it is that imagine you are not a good place if you are not a good place basically it means that if I glue the polygon I do not intersect anything on my right but that means in particular that the whole work needs to avoid a segment like that n to the delta n to the 2 epsilon so you have n to the one half minus epsilon back and forth the width of these guys is most they are all confined in an area which is n to the one half plus epsilon so if for many of them you have a whole segment which is forbidden of size n to the 2 epsilon then there is just a pigeon hole there so to turn it to a proof it's a little bit more tedious but I let you try to do it so this is an exercise but once I have that let me just conclude because then it's not so complicated so if I have n places where I can glue then I can choose as long as I can choose smaller than say small constant times n to the one minus 2 epsilon minus delta so I take a small proportion so this is basically capital N divided by n to the delta basically as long as I choose r smaller than that I can actually find r places among the n places in such a way that in addition you see if I put one polygon then I cannot put another polygon next to it because they will intersect each other even when unfolded but they are at most n to the delta places forbidden for each one of my guys so if I choose I'm realizing I'm not very clear so let me do it on a drawing here if I choose to glue a polygon here I clearly don't want to glue a polygon here because when I'm going to unfold these two polygons are going to intersect which is forbidden so every single time I choose a polygon like that it forbids me a certain number of places where I could fold another polygon glue another polygon and if you think about it it forbids n to the delta places so that means here as long as I take r smaller than basically that say c1 times that then I have basically capital N over 2 to the r choices for r coherent places to glue consistent consistent places so not only I can glue one polygon but what I'm saying is that if r is like that I can even glue r polygons so now I have a map let's call it glue which is going to be a map from my bad confined half space work times polygons like that of length k to the r times 1 capital N over 2 to the r into the safe avoiding bridges of length m plus k r and that consists in doing what? I take my work gamma I take polygons r and I take a subset s of places to glue and I associate to it the gluing of pi 1 pi k to unfold gamma so I unfold gamma and I glue the polygons at these places okay? the claim is still you could think it's actually even more to one map but my claim is that it is actually basically m to the n to the 1 plus epsilon to one map basically so this is m to n to the 1 half plus epsilon to one map so what does it give me overall is this fact that it's n to the 1 to I mean n to the n to the 1 half plus epsilon to one map where it gives me that h n tilde confined times here I get 1 over n to the 10 delta mu c to the k to the r times n over 2 to the r it's more than m to the n to the 1 half plus epsilon times the number of bridges of this length which is like that so here there are many things that can sell these things can sell with this one right? and here if capital N is much larger than n to the 10 delta this term here r is basically of order n to the 1 minus 2 epsilon minus delta so if this is true and r is much larger than log m times n to the 1 half plus epsilon then these terms is going to completely kill this one this term is going to be much, much larger than this other term what did I get what did I get from that I basically got that h m tilde is smaller than mu c to the m which is exactly my claim now I need just to check that I can choose these things so I chose epsilon to be 2 delta so this here n is this guy clearly if I take delta small enough I can get that this 20 epsilon is smaller than 1 minus 2 epsilon ok and clearly here r is this one I can pick it to be much, much larger than this so by plugging epsilon small enough I get my improvement ok you see I mean it's not very simple to get an improvement on the range but that's the way you do it ok so that was kind of a parenthesis in the whole thing because the style even of the proof was a little bit more informal and now let me describe in the last 20 minutes a little bit more conjecture and heuristic on 2D safer wedding group so the first thing this one I will not prove and it says the following it says that cn so there exists a constant a such that cn is smaller than n to the a mu c to the n so this would be absolutely fantastic if we would get that that would be a huge improvement on Hamas-Lewelch because you will get to a linear correction instead of stretch exponential except here it's going to be mu c bar where mu c bar is the following it's a limit when n tends to infinity of the number of safer wedding group on you I'm going to tell you what it is starting at z and I'm going to put the max on z to the power of 1 of n and what is u u is a universal cover of hexagonal lattice minus a face like minus 1 face so what do I mean in this theorem I mean the following you take your lattice and you remove a face this face you remove think you puncture the plane okay now take the universal cover of the plane so it's going to go as a parking lot like that around the singularity well clearly this hexagonal lattice has a pre-image by the canonical projection on from the universal cover to the plane okay so the hexagonal lattice itself is going around and going up one layer every times you go around the singularity so it's a lattice I mean it's not a lattice it's a graph sorry and I can define the connective constant on this graph let's be a little bit careful to define the connective constants what did I need I needed the sub multiplicativity of the number of works the problem is that here I'm not transitive so in order to recover the transitivity I mean to recover the sub multiplicativity I just take the max over any starting point of the number of works of length m you can check that it's because it's bounded degree it's well defined and this is sub multiplicative if I take the maximal guy I look at works of length n plus m I will get works of length n times works of length m starting at another point but that I can bound by the max for this so this connective constant is well defined this theorem would say that if you can prove so conjecture if mu c bar is square root of 2 plus square root of 2 then you get the polynomial bound and really think of it as it's not surprising globally this graph really looks like the hexagonal lattice so you really expect actually the connective constant to be equal to square root of 2 plus square root of 2 so we did some simulations and it's really I mean I didn't but some people did and since they are not extremely conclusive maybe I would not tell who they are because if that it should actually be square root of 2 plus square root of 2 so that's a decent conjecture and one exercise you can prove try to do is the following if you want to try I mean basically you know as much as I do on this problem now so you are as well armed as I am but you can do the following here mu c you could actually do the same if instead of taking the universal cover of just one face you take the universal cover of the hexagonal lattice minus a box you remove more than a face you remove a lot of faces if you think about it you could actually exactly get the same result so if for one of this if like removing a sufficient amount of boxes of faces to start with but finite number if you get that mu c bar is equal to mu c then you are done so exercise for you try to check that as k goes to infinity when you remove k boxes k faces sorry when k goes to infinity you can prove that mu c bar goes to mu c it converges to mu c the question is is equal to mu c for some finite value of k so is this thing converges to mu c because at some point it just drops to mu c and start being equal to mu c or is it because it just goes gently to mu c without ever reaching it by the way really I mean it should be true already for this guy yes you could also disconnect them but maybe indeed the best thing to do to first do is try really to remove a box of size k define mu c bar k say like that so now you see you need to go around the enemy is what is that if you remove one face you could think that maybe going around like that is actually quite good for you you never want to go in the bulk you want to stay near but the bigger the box you remove the harder it gets to actually profit from this singularity and question can you prove that for some k this mu c of k is equal to that was the conjecture moment actually there will be another conjecture okay connection to conform all invariance now so we saw that our parafamilic observable looks like a discrete version of a holomorphic map well there is a good reason for that it's a following so imagine you pick your domain omega but now it's a domain in the continuum okay so omega is a simply connected domain and now pick a very thin mesh size lattice of size delta so exactly no lattice but of mesh size delta okay let's call the finite graphs that you obtain by taking a safe holding polygon in the triangular lattice which is dual and goes around like that cut it and see what is inside let's call it omega delta okay so what you can do in this case is pick a point a delta on the boundary another point b delta on the boundary and you can look at the safe avoiding walk from a delta to b delta but not really with fixed length let's create gamma delta so pick gamma delta among walks I mean safe avoiding walks in omega delta going from a delta to b delta and let's pick this walk in such a way that the probability of gamma delta is proportional to xc to the gamma xc being 1 over square root of 2 plus square root of 2 okay pick a walk so you don't fix the length now you just penalize the length so it's a very good exercise for you to try to see that if you penalize the length by something x here which is smaller than xc then the walk will typically be extremely straight will want to be as straight as possible it's a much harder exercise but it's a cool thing to try or to try to read the paper that if you penalize here by some x which is larger than xc so you don't penalize enough by the walk then what you are going to obtain is a walk which is basically space filling typically it's going to want to be extremely long if you do it exactly at xc what happens we don't know mathematically but we have a good conjecture for that so conjecture is that which is due to Lohler, Schramm and Werner and this conjecture is saying that as omega delta A delta B delta tends to omega AB and here I'm going to tell you it's called Karateodorif sense but just think like if you look more and more like the origin like omega AB so you approximate here by a polygon going roughly along the boundary and you take the limits and you take A delta which tends to some A and B delta tends to some B basically when you do that let's call it gamma delta let's call it gamma delta gamma delta converges in Loh to a process which is called the Schramm-Lohler evolution of parameter h0 so it's called SLE h0 it's a process in the continuum which is kind of a continuous safe avoiding walk here the convergence of Loh here so it's convergence in Loh and here the Loh the convergence in Loh is defined from the topology corresponding to the metric, the following metric right, to have the convergence in Loh I need to tell you what is the metric first one to the metric so the distance between two curves and gamma and gamma prime is going to be the mean over phi and psi of max over T between 0 and 1 of gamma of phi of T gamma on psi so you take and gamma and psi are increasing so you take two parametrization it's a little bit of mystery to make people come here live so you can read I should really rewrite it I'm late, I'm not rewriting too bad so here for two parametrizations you optimize over parametrizations of gamma and gamma prime and then you compare the distance that you obtain so it's the standard I mean that's the most standard thing you could try okay let me just tell you so I mean this is a hard conjecture I mean quite hard I would say but let me just make a remark is that a way maybe toward this conjecture would be to prove the convergence of the function f that I defined before so it's a second conjecture which is due to it's not a remark, it's rather a conjecture which is due to to Schramm to Smirnoff so conjecture 2.9 is that if I mean as omega delta a delta b delta converges to omega a b well f here I'm gonna maybe put all the so it's omega delta a delta xc and 5 over 8 of z delta divided by f of omega delta a delta xc 5 over 8 of b delta so I do what I do the ratio of two parametrization observables one is just the observable value that z the other one is the observable value that b well this should converge as delta tends to 0 to the following psi prime of z over psi prime of b to the 5 over 8 where psi is a conformal map from omega a b to the upper half plane where this is phi of b psi of b sorry and psi of a is infinity so we have a clear candidate for the limit of our observable think of this as a renormalization this is our unrenormalized observable that we work with to prove that the connective constant was equal to square root of 2 plus square root of 2 I just renormalize it by this quantity because in fact it's not very hard to see that it's going to 0 so you renormalize it by something it should converge to something very explicit when delta tends to 0 and observe in particular so this is a conformal map in particular this thing is holomorphic so it's not that surprising that this thing should be discrete holomorphic in the discrete world okay if you prove that you did a very big step that's the conjecture some people say you did everything I will not go as far as I would say that but you did a big step let me I still have 5 minutes so let me tell very briefly two things actually maybe one and then the other one I promise it at the beginning of next class like that you maybe want to come next time so let me just tell you how now we can guess the 3 fourths and the 11 of us 32 okay assuming you are specialists which I'm sure you are all specialists of SLE of course nobody is but you are going to have to trust me a little bit on this one so first thing so really think if you want to study random work many things are much simpler to study once you know brony and motion brony and motion gives you a lot of process you have a lot of tools in the continuum that you don't have in the discrete but it's exactly the same for the SLE the SLE themselves are very well understood so in particular one thing which is known due to BFARA is that the Hausdorff dimension of SLE 8 third is 4 third and that you can really it has nothing to do with safe forwarding work it's a computation in the continuum but that gives us a pretty good idea of how to guess the 3 fourths why because it tells you okay I have my SLE when it's going to go to distance 1 basically how many balls of size epsilon do I need of size delta sorry do I need to cover it well I basically need delta to the fourth third boxes to cover it right if I take balls of size delta I need basically delta to the fourth third to the minus fourth third because this is not a big number delta to the minus fourth third boxes to cover it but think here imagine this gamma delta is basically your SLE 8 third so this is our safe forwarding work SLE 8 third here if you have length if you want to go to distance I mean sorry then each step of this safe forwarding work is basically like a small step of size delta here so how many steps do I need here for my safe forwarding work to go to distance 1 well if I do the analogy I need delta to the minus fourth third okay so for my safe forwarding work on the lattice of mesh size delta to go to distance 1 the convergence to SLE is basically telling me I need delta to the minus fourth third steps if I rescale back this to scale 1 that means to go I mean if I do n step which is delta to the minus fourth third I would go to distance n to the 3 quarter okay go to distance that's the first step the second thing we wanted so in particular this is a way of predicting of course it's not a proof far from it but it's a way of predicting the n to the 3 fourth in a slightly hopefully more rigorous way than Flory's computation remember Flory predicted this n to the 3 fourth out of two mistakes now there was another weird exponents that we were getting which was this n to the 11 over 32 for the cn right? so how can we do to guess this 11 over 32 well take c to n and divide it by cn squared right this should behave like what basically it should behave basically like constants over n to the 11 over 32 right? each one is behaving like mu c to the n times n to the 11 over 32 the mu c to the n is exactly cancelling between these guys and the n the 2n to the 11 over 32 is going to simplify with one of the n to the 11 over 32 leaving one alone okay? this is basically the probability for two safe avoiding work one starting at 0 and one starting just next to it at an edge next to it the probability that these two independent walks do not intersect right? I picked two independent walks at random starting at 0 and 1 well I have cn squared choices so I get probability for each one of the cn squared and exactly the guys that succeed are guys for which I get a safe avoiding walk of length maybe 2n plus 1 instead of 2n basically here if you put 2n plus 1 you are also fine okay? so the 1 over n to the 11 over 32 seems to be the probability that two safe avoiding walk of length n starting near each other do not intersect but now we know I mean these walks they go to distance n to the 3 fourths right? on both sides so if I rescale this and work on a lattice on mesh size delta equal 1 over n to the 3 fourths this is basically the probability that our two rescale walks starting at neighbors distance delta of each other do not intersect right? replace that by the SLE computation this c 2n plus 1 over cn squared should approximately be as a probability for 2 SLE 8 third of gamma 1 intersecting with gamma 2 is empty so 2 SLE 8 third starting at distance 1 over n to the 3 fourths going to distance 1 of not intersecting but it's actually very well known what is the probability for 2 SLE starting at distance epsilon of each other not to intersect it's known to be epsilon to the 11 over 24 so this should be approximately epsilon which here is 1 over n to the 3 fourths to the 11 over 24 and this is 1 over n to the 11 over 32 and that's how you predict the n to the 11 over 32 so again the same idea you need to reinterpret you need to reinterpret geometrically your exponents and then you try to show that you can compute them in terms of the scaling limits meaning in terms of SLE and there in the continuum you know everything so in the continuum you know this exponent at the same thing as here you know this exponent and these two exponents give you the exponent for several reading books again there are many many steps which are unclear in what I'm doing not just the conjecture they are also here I'm cheating because it's not clear that I can exchange all these limits there are many things that are not very clear but at least physically you think that if you are in a good world where everything goes fine that should be the exponent so what I didn't have time to do and for the people who are courageous enough to come back on Friday I didn't tell you how Ninhaus predicted the thing that's a pretty cool cliffhanger right so on Friday I will tell you how Ninhaus predicted that then we will leave the two-dimensional world and try to prove things on the geometry of safe forwarding work and there I promise I will go back to a slower pace and really prove things completely here I went a little bit further because I wanted to give you ideas of what is happening in this two-dimensional world but next week on Friday it will be more reversed so you can sort of relate these exponents to the continuum object it means that they are also sort of implied that they are universal yeah so that's actually a very good way exactly like safe forwarding, random works on different graphs converge to the same object which is Brodyian motion here this SLE is expecting to be the limit for many objects as well I mean for the safe forwarding work on many graphs so if you take the square lattice at criticality should also converge to the same thing and therefore because the exponents I mean the two exponents are only expressed in terms of the scaling limit you should get the same for different lattices yeah sorry we don't have the convergence we don't know even some pre-compactness yeah this is actually a very big question I will maybe come back a little bit to that on Friday because it's a very good question yes if you say the inverse the gamma one and gamma two do not intersect but you wrote that they do intersect yes that's blackbird dyslexia right thank you very much