 Ok, donc, merci d'être venu, donc on a conclu, la semaine dernière, avec ce changement de l'EMMA, que j'ai introduit, mais je n'ai pas vraiment dit comment vous l'utilisez, j'ai juste dit que beaucoup de choses sont l'application de l'EMMA, donc je veux prendre le temps de vous dire un peu plus sur l'application, parce que ça va toujours sembler comme des applications triviales, mais c'est qu'il y avait des difficultés sur l'application de l'EMMA. Si vous avez l'EMMA, c'est très simple, mais historiquement, les gens ont des choses de plus compliquées, pas toutes les choses, mais... Donc, premièrement, je vous remercie de Fb, ici il y a un set de N, c'est-à-dire qu'il existe K smaller ou equal to N, avec sources de K equal to B. Ça ne veut pas que les sources de N soient equal to B, c'est-à-dire que vous pouvez trouver un petit current qui a ces sources. Donc, K less than equal to N, ça veut dire que... Oui, exactement, ça veut dire que l'ordering fonctionne de l'un à l'autre, ça veut dire qu'à chaque étape, le current de K est plus petit que le current de N. Donc, peut-être que l'information la plus importante que vous devez tenir en mind est que si B est juste deux vertices X et Y, puis, en trouver un petit current qui a des sources X et Y, c'est vraiment juste de dire qu'il existe un path entre X et Y. Oui. Donc, dans la définition de Fb, l'ordre H n'est pas là. Est-ce que c'est l'intention ou non ? Oh, oh, sorry, sorry. Oh, vous êtes très bien, sorry. Donc, N1 plus N2 restricté à H. Merci beaucoup. Merci beaucoup, c'est très important. Et sinon, l'ordre H n'est pas là ou dépend de H. Vous êtes bien. Donc, si je restris mon current à deux vertices, je peux trouver un subcurrent qui a des sources. Merci beaucoup pour cette correction. Donc, FxY est juste un set d'answer que X est connecté à Y en N. Ici, je vais utiliser une notation très intuitive, qui signifie qu'il existe un path de axes de X à Y avec des currents positifs. Donc, ici, vous avez X et Y, vous avez vos currents. Et ce que vous avez besoin de trouver est un pass de X à Y dans les currents. OK ? OK. Donc, on va prouver ce lemma et ensuite, on va retourner à l'application. Le proof de l'emma n'est pas si compliqué. En fait, c'est assez élémentaire, même. Donc, on va commencer avec le côté gauche, disons. Donc, ici, je ne vais pas répéter les deux choses sur le côté gauche. Je vais droiter le beta, OK ? Donc, WFN1, WFN2. Et ici, je vais juste faire un changement de variable que N1 est equal à, je veux dire, peut-être N2, ou je n'ai pas essayé de prouver. Je vois. Donc, N1 plus N2, je vais l'appeler M. Et N2, je vais l'appeler N, OK ? Donc, si je fais ce changement de variable, qu'est-ce que j'ai ? Ici, M est le current en axes de G. Et le set de sources est A, la différence symétrique avec B. Parce que je fais le summe de current avec les sources A et le current avec les sources B. Et ensuite, je summe de N, les currents de H. Donc, il y a deux choses. N est equal à M. Et les sources de N sont equal à B, OK ? Et ensuite, j'ai FM. Et WN1, WM2, je pourrais l'écrire comme WM-N times W of N. Mais je peux aussi, et c'est très facile de voir, l'écrire comme W of M times N choose M, où ici, je veux dire que c'est le produit de l'enquête de NXY chose MXY. C'est juste ces gars, let's remember what they are. Donc, indique, WN1, WN2, c'est un produit de l'enquête. Il y a cette beta JXY pour l'enquête N1XY plus N2XY. C'est parfait, ça me donnera MXY. Et ensuite, il y a un de N1XY factorial et deux de N2XY factorial. C'est le produit de ces deux choses. Maintenant, si j'include N1 plus N2XY factorial, ici, j'ai WM et ici, j'ai cette quantité, OK ? Donc, jusqu'à maintenant, je n'ai pas fait beaucoup. OK, et si j'ai essayé de l'autre côté, je vais arriver à faire le même changement de variable avec la summe. Cette fois, M est, je veux dire, encore une currente sur les g's de G. Et encore, il a des sources A, une différence symétrique avec B. Cette fois, N est un set de... Je veux dire, de la currente, c'est encore satisfait de ceci. Mais, et c'est la différence, ici, N n'a pas de sources. J'ai FM, j'ai WM, et j'ai M, choisir N, un indicateur que M est de FB, right ? Donc, ma question est pourquoi, ce que je dois répondre, c'est pourquoi j'ai une équité ici, OK ? Et vous verrez que vous aurez une équité, et en fait, parce que c'est valide pour chaque F, c'est même seulement si, vous avez la suivante identité. Donc, nous allons prouver que pour chaque M, si je prends une currente avec des sources A, une différence symétrique avec B, et puis, ici, la somme pour N, une currente sur H de N smaller que M, et des sources B de N, choisir M, est equal à cette chose. Nous allons prouver ça. Je veux dire, ici, je peux mettre F, ce qui est juste un indicateur que M est equal à une certaine M0. Donc, il n'y a pas de la génération en essayant de prouver ça. Donc, je fixe M. OK. Donc, c'est là où nous allons utiliser une cool interprétation de grandes currentes. Je veux comprendre ce type de facteur combinateur. OK ? Et je vais le faire comme suivant. Donc, associé avec M, le multigraphe, c'est-à-dire M rond, il y a des quantités, donc, vous mettez vos vertices, c'est-à-dire, ce sont les vertices de votre graph, et vous avez un current sur chaque vertex. Dis-je, 2, 1, 0, 1, 1, 3. OK ? Donc, ce que vous faites, c'est que vous associéz le graph qui a aussi de nombreux edges entre les vertices et entre les deux vertices comme le current vous donne, le nombre du current. Donc, ici, il va être 2. Ici, c'est 1, 1, 1, 3, et 0. C'est M. OK ? Donc, mettez aussi beaucoup d'edges. Je veux dire, mettez des edges de Nxy entre Mxy, pardon, entre X et Y. Donc, c'est un multigraphe, a priori, il y a plusieurs edges. Maintenant, quand je regarde à N choose M, c'est quoi ? Il devient le nombre d'edges pour chaque pair de X, Y. C'est le nombre d'edges de choisir les edges de Nxy entre les Nxy. Multiples sur tout. Donc, c'est exactement le nombre d'edges d'edges d'edges entre les Nxy et M entre les X et Y pour chaque X et Y. C'est exactement ce numéro. Mais maintenant, vous commencez à comprendre quelles sont les quantités d'edges. Cette quantité est quoi ? C'est le nombre d'edges d'edges d'edges d'edges que les sources de N sont les B où, ici, les sources d'edges sont juste des vertices avec un haut de degree. C'est la génération naturelle pour les currents. Donc, c'est juste le nombre d'edges d'edges entre les N restricatées à H ainsi que le nombre d'edges d'edges et la génération naturelle est le nombre de N inclus dans le nombre de N restricatées à H ainsi que les sources qui n'ont pas de sources. Et vous devez prouver que c'est pareil quand vous ajoutez une indication que la N est concernée par FD. C'est ce que vous voulez prouver. Si je prends une currente en FB alors, si je l'interprète comme un multigraph, le nombre d'edges de trouver un subgraph avec les sources B est le même que le nombre d'edges de trouver un subgraph n'ont pas de sources. Mais maintenant, vous êtes dans une très, très bonne forme. Pourquoi ? La première chose peut-être. Si vous regardez la multiplicité de N, vous trouverez un K qui a exactement les sources B et qui est plus petit que la N. Vous devez prouver que la N est restricatée à H. Vous devez prouver que la N est restricatée à H et que la N est restricatée à FB. Il n'y a pas de N ou on l'a dit K qui est inclusée en MH comme la bounderie de K est equal à B. Ensuite, il n'y a pas de K comme que K est plus petit que MH et la bounderie de K est equal à B. Et vice versa. Si il existe un, il existe un K. Donc, peut-être que je devais l'écrire comme ça. Il existe un K si on n'en a pas qu'il existe un K comme ça. Simplement, en un cas, on prend le nombre de pages et en l'autre case, si il y a un K, vous pouvez trouver que ces 2 quantités sont en 0 en même temps. Pourquoi, quand elles sont en 0, elles sont en même temps? Elles sont en même temps parce que en fait, il y a un mapping entre ceci et ceci. Si il existe un K, si vous avez une différence symétrique d'un N avec un K, vous appréciez exactement ceci et ceci. Donc, si il existe un K comme ça, puis un N map à un N une différence symétrique avec un K est une évolution entre les 2 sets. Donc, ils ont la même cardinalité. Donc, si il n'y a pas qu'un set, donc les 2 quantités sont en 0, si il existe un K, donc les 2 quantités sont en même temps parce qu'il y a une bijection entre les 2 sets ceci et ceci. Je vois que certaines personnes ne sont pas convaincées. Est-ce qu'il y a une question sur ça? J'ai oublié quelque chose? Peut-être que je l'ai oublié et non. Mais oui, donc c'est assez cool et un argument short combinatériel. Ok, donc, allons-y aller aux applications de ça. Donc, comme je l'ai dit et cette fois, je l'ai déjà décrit. Donc, si je regarde remember sigma A is a product for X in A of sigma X. So imagine I look at sigma A times sigma B. So this can be written if I use the currents. Here there are going to be 2 currents with no sources and here there is going to be 1 current with sources A and 1 current with sources B. W and 1, W and 2. Ok? So if I use the switching I can put B sources on A and I end up with N1 sources A, symmetric difference with B N2 with no sources downstairs I still have no sources and here the only thing I must add is indicator that N1 plus N2 belongs to F B. But the good thing about that is that I can drop if I drop the indicator function then I'm only increasing my quantity and as soon as I did that I can drop the N2 because it appears at the top and bottom and what do I see? I see the sum of currents with sources A symmetric difference with B with no sources S. This is exactly sigma A sigma B I mean it's sigma A symmetric difference with B which is just sigma A sigma B Ok? In fact you could even not drop anything because it quantifies in some sense how much smaller sigma A sigma B is compared to the correlation imagine so define A of G so the probability of a current to be 1 over Z A G so remember this is a source I mean the sum of currents with sources A times W beta of N indicator that the boundary of N is equal to A define that it's a probability measure because if you sum on the guy on the top you get exactly this and it's just telling you it's a probability measure that is sampling a current with sources A with probability proportional to the weight of the current kind of the most natural thing you can define if I use that then for instance sigma A sigma B sigma A times sigma B I mean the correlation of sigma A times the correlation of sigma B I can rewrite them if I go there and instead of erasing I just divide and multiply by Z A symmetric difference with B what do I end up with I end up exactly with the fact that this is sigma A sigma B times the probability for one current sample according to this measure a second current sample with no sources it's N2 here I sample it with no sources and I look at the probability that N1 plus N2 belongs to FB this is just another way of putting it sigma A times sigma A the product of the spin-spin is smaller than the correlation of the product but it actually gives you by how much you have an explicit expression for that here by the way that means the product measure and because for some reason you are going to see this is going to appear more and more in fact use a simple notation when I have PGA1 PGA2 etc so if I am looking at the product measure of several of these guys I would just denote it by P A1, A2 etc Ak but you have to remember it means sampling the currents independently ok let me give you other applications like that you will become a master in the switching lemma so I am not going to erase the switching lemma so if for instance I look at sigma A H and I look at H included in G so let's say this was the first application now I look at the second one and I want to compare that to sigma A on G ok so I look this is ZAH over Z empty set H and here what I can do is I can just add on both sides the partition function on G but now here think I have my first current which is sourciless on G my second current which has sources A on H I am allowed to switch the sources from the second current to the first one so this is the same as I am going to keep this thing and here I am going to now have sources so N1 with sources A N2 on the smaller thing with no sources WN1 WN2 and here I get an indicator N1 plus N2 restricted to H belongs to FA I just switch the sources and here again I can try to make a probability appear so what type of probability can I make appear like that so here second current on H it's perfect it appears here second current so this thing is exactly the probability of the second current so the only thing I need to do is to make here a probability appear so what do I do I multiply by ZAG and divide by ZAG because now what do I see here I have sigma AG and then I have probability for a first current sample on G with sources A a second current sample on H with no sources and I want that N1 plus N2 belongs to F restricted to H so here sorry, that made no sense to FA so again I get for instance immediately that spin-spin correlations are increasing in the graph but in addition I know by how much this can be actually very useful if you want for instance to be comparing plus boundary condition to zero boundary condition because plus boundary condition can be interpreted as the easing model on the graph plus what we call a ghost vertex I will not go into detail there but then it corresponds to comparing the spin-spin with this ghost vertex to the spin-spin without this ghost vertex and you have a clean graphical representation of the difference between the two so it's very very useful when you try to prove for instance mixing properties ok, now I have a question so it looks like the sum of N1 is over N to the E of H instead of you sorry, here? yeah, it should be H yeah, sorry, here it's H and I use the switching your switching, ok so either you say ok, let me rewrite if you say here if you write what you are saying indeed it's something like 1 over Z empty set H Z empty set G and it's a sum WN1 WN2 and here indeed, you want to say N1 is on H ok, I mean N2 is on N1 is on G but we source less currents and N2 is on H and we source A but here there is no indicator and I switch the sources by introducing this term I see that I lost you, so let me try to this thing is exactly your sigma AH right? and I added artificially if you want a second current oh, if you want a first current in this case and then I switch so, I mean of course you want to be using the switching lemma so you systematically in some sense add a second current artificially there are two cases so either you already have a product and when you have a product in some sense you get very naturally two currents sometimes you want to be comparing quantities that have only one one quantity which in some sense correspond to one current so then in this case you add artificially a second one I want to be taking time on this I know it's, I mean first it's not really pushing us away from the ultimate goal because I'm actually giving you quantities that are not irrelevant for the full proof of the of the triviality but also I mean this is such a powerful tool that I think it's useful to be discussing so let me give you a third application which is called the Simon Libby's Equality and which is maybe the first kind of clear connection between easing and random works and people forget it quite often so it says the following take a set S and a point zero in it and take X which is not in S I claim that Sigma zero Sigma X in G is probably equal to the sum for Y on the boundary of S meaning really here of Sigma zero Sigma Y but when I look at the easing model on the graph S only times Sigma Y Sigma X on G why do I say this is a connection to random work if you think of the green function of the random work and you decompose the random work so green function is expected number of visits here for a random work starting from here if I decompose the random work on the first exit time so I stop it the first time then what do I get I get that Sigma zero Sigma X is an average on the probability of exiting S somewhere at a given point times now where the expected number of visits here starting from this point so this is if you interpret this as a green function you recover exactly the green function starting from Y here and this is well that should be interpreted as a green function S but this is clearly like the expected number of visits of Y starting from zero when you look at random work in S is clearly larger than the probability that you exit S through Y so it's really what you get it's exactly what you get for random work it's something that you get from the Markov property in random works there is another model that has this property percolation for people who know there is exactly using the BK inequality or actually doing it by hand some things saying the probability that zero is connected to X is smaller than the sum of the points on the boundary probability that zero is connected to Y in S times probability that Y is connected to X and this so easing percolation and random works are really actually self-auditing work also satisfies these things there are really processes for which they are special proof for instance of exponential decay and things like that that's what they share it's this kind of random work inequality that you have so how do we get it out of the random current by the way this is maybe not that easy to get without the random current but with the random current it's quite straightforward so yeah can you see a little why it's less or equal not equal so for random work it's equal if you replace this by exit probability already if you replace it by a expected number of visits you lose something so it's only an inequality anymore it's a question basically it's an inequality because here it's kind of doing a union bound like instead of saying there exist a place where you exit you bound it by then it's a sum over the points of probability of exiting through this point and because it's not exactly probability of exiting but number of visits you lose something sorry yeah exactly when you look at the picture it looks good because there is a unique exit point and for random work you can cook up a slightly modified version of this by looking I mean by replacing this by probability of touching the boundary at y and there it's an equality but of course this is only a caricature of what happens for easing easing doesn't have this mark of property so it's going to be an inequality and we will see actually in the formula you will see where you get an inequality ok yes of course I don't know if it's relevant but could you get an equality by doing some kind of inclusion exclusion principle I mean we will get an equality with a certain probability thing and then we will go back to a union bond but I mean if you don't do the union bond you will get something where you get an equality so here so the idea is going to be so again it's the same trick as here we have only one guy and we would like to cook up two especially that here we have product of spin-spin so here it's this thing right this is exactly our spin-spin so I'm going to add artificially the empty set S so it's one of the empty set G the empty set S and here I get some so N1 is on G and has sources 0x and N2 is on H is on S sorry and has sources empty set ok up to now it's an equality but here is inequality and indeed you could do something smarter if you want to keep something sharper but by the way I don't know any application where somebody is doing something sharper well ok no let me rephrase there is a very important application but I don't want to talk about you about it here because it's some ongoing project but in high dimension you can improve on that and it's important so I'm going to do an inequality by just observing that if there is a source from 0 to X then necessarily there is Y on the boundary of S such that N1 plus N2 restricted to S I mean such that if you want 0 is connected to Y in N1 plus N2 restricted to S it's just trivial because even in the first current there is already a pass from 0 to X so there is already a Y on the boundary which is connected so I force you in the sum of the two currents so this I mean maybe this is a better way to write it is maybe saying that this is N1 plus N2 restricted to S belongs to F0Y but that's perfect this is exactly the thing that you are allowing I mean this is exactly the thing you need to be allowed to switch the sources this time in the other way usually we add an indicator here we are going to remove it so if we remove it let's me invert just the two sums so now I have WM1 WM2 indicator of N1 plus N2 belongs to F0Y so I'm allowed to switch 0Y on both sides so that gives me N1 with sources 0X, symmetric difference with 0Y which is XY and N2 with sources 0Y if I take this and I switch 0Y I exactly end up with this top ok well now what do I see this gives me sigma X, sigma Y in G and this give me sigma 0, sigma Y in S so once you know the switching lemma you really get Simon Lib in I mean I didn't lie in two lines ok just for your character I want to mention one thing one thing which is true for easing and wrong for many many quantities is that if you take beta so define phi beta of S to be the sum for X on the boundary of S of sigma 0, sigma X S ok I mean no I could have let me write Y just to be more transparent to what is above so it's exactly this quantity here ok notice that if phi beta of S is smaller than one strictly then sigma 0, sigma X G is smaller equal to phi beta I mean actually it's always true so maybe I should not but to the X over radius of S simply because what I can do is I can for every X not in S why because so radius is just right I mean the max distance between a point in S and 0 why because whatever the point I take when I can apply my Simon Libe inequality on S centered at 0 but now for sigma Y, sigma X I can apply the Simon Libe inequality centered I mean by looking at the translate of S around Y and I can keep going like that let's say I exit here and because I can apply it at least X over radius of S times before I reach X I'm certain to have this inequality but if phi beta of S is smaller than 1 what does it tell me it gives me automatically exponential decay of correlations for sigma 0, sigma X so if phi beta of S is smaller than 1 then you have that and in particular correlations decay exponentially fast so in particular what do I learn on beta well beta then must be smaller than beta critical here I mean smaller equal at least because above criticality they don't even decay when you take G to be ZD but in fact even add criticality why because if this is strictly smaller than 1 you are on a finite set S you can increase a little bit beta and still get something smaller than 1 so this will be contradictory with the fact that you are exactly at beta C because you will get exponential decay a little bit above it so this is an open condition if you want there exists S finite such that you have this is an open condition so it cannot be and it's included in 0 beta C so it cannot be more than the open set 0 beta C but that tells you one interesting thing is that phi beta C of S is largely equal to 1 for every S finite containing 0 it's a very strong inequality and in fact it's a characterization of the critical point and it's a cool characterization you can use this to prove for instance sharpness meaning exponential decay as soon as beta is smaller than beta C and another thing you can do for the physicists in the room you can look at the susceptibility of your model which is the sum of the correlations well, chi of beta C is larger equal to the sum for N equal 1 to infinity of phi beta C of lambda N if I take boxes like that the correlation between 0 and X in the full plane is larger than the correlation in the box lambda N so you can write this by monotonicity but I just told you that phi beta C of S was always larger equal to 1 so what it gives me automatically is that this is always larger equal to 1 so it tells me that the susceptibility of the easing model is always infinite at the critical point which is a strong indication but actually one has to be very careful because that's not equivalent in fact you can look at the easing model with long range interaction 1 over X minus Y squared in 1D which exactly satisfies this because here there is nothing special about nearest neighbor so it has infinite susceptibility but still positive spontaneous magnetisation but at least physically let's say that if the correlations explode it's a very strong indication but it tends to be passive from below it's a strong indication that you have a continuous phase transition when the model is not too weird so that's why I wanted to mention that does it give a good estimate of the susceptibility on the box or how it goes so it gives you that sigma 0 sigma X is roughly at least like 1 over X to the D minus 1 because you have roughly 1 over N to the D minus 1 points at distance N so it gives you this bond we will see that you get the infrared bond on the other side that gives you smaller or equal to 1 over N to the D minus 2 so you can sandwich the spin-spin correlation between these two values and let's face it this is not a very good lower bound because in dimension 4 and more you expect actually exactly to have behavior D minus 2 so you are actually equal to the top one and in dimension 3 the current estimates are that you get like exponent 1 point 0 3 or 0 1 I never remember but it's extremely close actually to 1 over N to the D minus 2 so it doesn't give a good lower bound but it still gives a polynomial lower bound which is as a probabilities with this type of things ok before I get actually it will be a good we are going to make a smaller break which is a little bit earlier but let me mention 1 so actually I'm going to mention what you just ask so in particular the max of the sigma 0 sigma x beta c for x on the boundary of lambda n this is larger than 1 of a volume of lambda n this is a polynomial lower bound let me mention and this is a cool exercise but I will not do it here but for your culture yet another inequality of the using model so this is the fourth application something like that it's called the messager miracle solide and it says the following if you take 0 and you take either a vertical line like that or a diagonal one like so either it's x equal y plus constant or y equal constant or actually x equal constant also works but I mean this is pretty ok and it says the following let's say I take a point and I draw it with this but then it will be obvious what I mean for the others let's take a point x on this side of the line and define tau x to be the reflection with respect to this line well sigma 0 sigma x is always larger equal to sigma 0 sigma tau x ok, why is it good we can check that so this is true for vertical lines so if you want in higher dimension hyper planes along one coordinate it's also true for diagonal hyper planes so in particular it implies something that is extremely natural from the point of view of physics but which is not at all obvious when you look at a lattice model it implies something like sigma 0 sigma x is always larger than sigma 0 sigma y if y is larger than d tan x something like that if the distance between y and the origin is at least d times the distance between x and the origin so it allows you to tell something completely obvious that if I take a point constant distance farther than you the spin spin is smaller well I mean there is a scaling hypothesis or something like that so then it's completely obvious but I mean when you are not you don't have the scaling hypothesis at your disposal it's not that clear that things are decreasing in fact this type of questions are completely open for instance for I mean basically for every other model like Bernoulli percolation or FK percolation it's not something obvious for this inequality are the horizontal and vertical line arbitrary or yeah you can take any horizontal line, any vertical line that's your question I mean you will need you will need that it has this type I mean either like that or diagonal it's going to be clear in a minute so how do you prove that just I mention it and I don't do it what you do is that again you try to have N1 and N2 2 currents but here the trick is not to take 2 independent currents is to take 1 current in the plane and to fold it so N2 so you are going to have N1 which is the part of the current here and N2 is going to be the reflected version this is tau of N2 in some sense it's the reflected version of this current you fold like that like you fold your page into 2 why is it not that stupid to do that is that notice that if you have 0x then the sources of N1 2 is 0 and x but if you have 0 and tau x then the sources of N1 2 are still 0 and x because I folded the second current and then what happens is that in fact you will notice that this sigma 0 sigma tau x for this folded current it corresponds to 0 connected to the boundary of your half plane so it has sources 0 and x but there must be a path so now I am in your half plane if you want I have N1 plus N2 I have 0 and x and in some sense in this case I must have a path because in the original thing they must have a path from 0 to tau x so when I fold this path doesn't disappear and that is the difference that this is smaller than this because there is this additional condition and there is no reason a priori to have a path from 0 to the boundary it could just be that 0 is connected to x without being connected to the boundary so I mean I let you look at the details on a l'exercice and again it quantifies also a little bit this so you can for instance try to interpret the gradient of the spin-spin correlation in this way to try to understand that it becomes like x is just next to the boundary it's connected I mean the difference the gradient will be 0 is connected to x but not to the boundary so it makes a half plane ok let's make a short break I mean a 10 minutes break and then we restart with this time trying to quantify like how close you are from vix rule so here I gave you a lot of examples of manipulations we can make now we are going to focus on the one that we are interested in which is trying to quantify how good vix rule is for the easing model and we are going to see some nice quantities appearing ok ok, so new part of the class let's try now to look at two end point correlations and see how close they are from the vix product so 2.3 quantifying maybe I should put quantifying vix rule for easing meaning quantifying how close it is to be true and we are going to start so the question is how close how close sigma x1 sigma x2n and g2n x1 x2n which by definition is the sum of a pairings pi of the product for i equal to 1 to n of sigma x pi 2i minus 1 sigma x pi 2i so you take a pairing how close this quantity are when mutual distances go to infinity ok, that's going to be the game let me keep that and as a starter so I'm going to do it completely usually I only do the 4 point function but actually I think we have time to do the 2 end point function then next week to really do a full proof of the triviality under a certain assumption so step 1 let's do the 4 point function the 4 point case 2 4 beta of x1 x4 which is sigma x1 sigma x4 and we subtract g4 of x1x4 and my goal is to know how small I mean to get how small this is ok, so let me start not a fail attempt but like something which is good but not optimal so let's look at so it's proposition something like 2.9 I think it's going to be a backbone representation so I'm going to try to express I mean to get a bound on that using the backbone ok, and the bound is saying the following 0 is larger than u4 which is larger ok, maybe let me put 0 is less or equal to minus u4 which is so u4 is always negative and this is smaller than v4 which is the sum of the following so it's the sum over k and k which is just 2, 3, 4 ok of sx1 sxi sx xj sxk so this is exactly if I stop here this will be exactly the g4 but I'm adding probability or maybe yeah, maybe let me write it like no, I'm going to write it in a non normalised way otherwise it's going to be the end sum on gamma1 which goes from x1 to xi and gamma2 which goes from xj to xk so I have x1 x2 x3 x4 and I'm going to look at one pass from x1 to xi and one from xj to excel and here I'm going to put rho of gamma1 rho of gamma2 so this is exactly taking two backbones and looking at indicator that gamma1 is not intersecting gamma2 it's intersecting gamma2 sorry so I can bound u4 using in some sense something like the probability that if I sample a backbone from x1 to xi and a backbone from xj to excel as they intersect ok ok, don't worry if you don't quite understand too much what I'm doing anyway this will not be the object we will work with we will have a better expression using the double random current so next try still to prove this so remember this is a sum of a gamma backbone connecting x1 x4 of rho of gamma expression sigmaa is the sum of the weights of a backbones pairing the things and here if I look at a backbone that is pairing the four points automatically let's say the backbones are started from x1 they are going to connect to one of the three guys and then there will be a second pass connecting to the second guy so I'm going to write this as sum of a gamma1 that goes from x1 to sum xi and gamma2 that goes from xj to xk ok except now rho of gamma this is rho of gamma1 times rho gamma2 gamma1 bar of gamma2 this was the chain rule right the weight if gamma is the concatenation of two backbones except that for the second backbone I'm looking at the weight in the depleted graph where I remove the first backbone and actually a little bit more around it all these points for which I know it's even ok first observation is that here so left inequality if I just sum I fix gamma1 and I sum on the gamma2 then I get from gamma1 from x1 to xi rho of gamma1 and then I get sigmaxj sigmaxk in the graph g-gamma1 bar it's exactly what it corresponds to I'm summing all the backbone from xj to xk just in the smaller graph so I get the spin-spin in the smaller graph but we saw earlier actually brilliant pedagogical presentation I know there is monosonicity in the graph therefore I can remove this and as soon as I remove this here this becomes the correlated from this and this is just sigmax1, sigma xi so I get sum over the possible i of the product so the 4-point function is smaller equal to the g4 sigmax1 sigmax4 is smaller equal to g4 of x1, x4 by the way therefore u4 is negative this is called Le Bovitz inequality now let's look at the right inequality for the right inequality when things are not much worse if gamma1 and gamma2 do not intersect then we saw that rho gamma1 bar of gamma2 was in fact larger equal to rho gamma2 so that implies that sigmax1 sigmax4 is in fact larger equal to the sum of gamma1 connecting x1 to xi gamma2 connected x2x of rho of gamma1 rho of gamma2 so here you feel oh well I got exactly what I wanted I got g4 so I mean what is wrong well it's just that before when I was writing rho gamma2 of gamma1 bar in some sense there was something implicit here which was that gamma2 and gamma1 should not intersect because otherwise in some sense you set the weight to be 0 if you prefer so here I need to add gamma1 intersected gamma2 is the intersection is empty because otherwise I lost too much if the two guys do not intersect I can put the inequality otherwise I get 0 so I need to put this but now you exactly saw that I mean this is g4x1 x4 minus the same sum with the intersection voilà so at this stage already there is something very clear that should appear in your brain so there is a random walk interpretation of u4 if you don't have this indicator function you get so now if I imagine that I sample at random rho1 and rho2 with probabilities that are proportional to their backbone weight then what am I doing I am sampling two independent backbone if you want one pairing x1 and xi one pairing hj and xk and I am looking at whether they intersect with good probability now if I ask the same question for random walk I take two points and I sample a random walk from there condition on ending there and a random walk from here condition on ending there and I look at whether they intersect or not so this is now random walk probabilities these things are going to be very different depending on the dimension in dimension d smaller I mean strictly smaller than 4 this remains larger than constant in 2d it's kind of clear you have your first kind of path from x1 to x2 and the second one it's kind of non degenerate so they have a positive proportion to cross in 2d it's clear actually true it's a little bit less clear but actually it's not a difficult theorem at all and it is steep true in dimension d larger than 4 in fact it decays if I take points at mutual distance l of each other it decays like 1 over l to the d minus 4 so random walks in dimension 5 and more and in fact if you take long range random walks in any effective dimension larger than 4 they will actually intersect with probability tending to 0 when the distance goes to infinity and in fact in dimension 4 it's a little bit more subtle but it's also true it should decay like 1 over log l so now what does it tell me it tells me that when I'm comparing u4 to g4 well in dimension 4 and more if I believe that my backbones behave like random walk then I should get a factor tending to 0 in front so u4 should be much smaller than g4 or equivalently sigma 1 sigma 4 of course the very big cave art here is that these backbones they do not behave like random walk a priori dimension 3 they do not behave at all they are much more like safe avoiding walk type objects so I mean we are going to have to do something to kind of compare the behavior of these backbones to random walk but if you believe that in dimension 4 and more they behave like random walk then you have a very good heuristic singling out 4 as a critical dimension ok let me mention now a slight improvement on this it's going to be important because that's the one that works in dimension 4 and also it will lead it will help getting the higher like the what you get in for 2 endpoints ok so proposition 2.10 so here it was a backbone representation so let's now get a double random current representation and it says the following u4 beta or if you want minus u4 or u4 is minus 2 sigma x1 sigma x4 times probability for one current with sources x1 to x4 and one source less current of x1 x4 all connected in n1% meaning that I can find for any 2 points I can find that all these points are connected by path of positive current and actually another way of writing it just I put it here I can also just keep sources a little bit more balance all the sources on the first guy I could also keep them on both sides and ask the same event so the advantage of this representation with respect to the previous one is that as you noticed it's inequality and the proof as often with random current is going to be straight to a round if I look at sigma x1 sigma x4 and I subtract sigma x1 sigma x2 sigma x3 sigma x4 minus minus so this is zx1 to x4 over z empty set minus zx1 x2 zx3 x4 over z empty set z empty set minus minus and as usual we are going to try to minimize a little bit I'm going to put the empty set here and the empty set here to have two currents also for the first one and now notice what I'm not even going to write these big sums anymore you are specialist now so here I have one current with sources x1 x2 and one current with sources x3 x4 if I switch if I say this is my second current and I switch the sources x1 x2 or let me write it below so here I get the same thing I'm going to get zx1 x4 z empty or I'm going to get let me put it like that sorry sum over n1 with sources x1 x4 and here I switch sources x1 x2 to the first current so I'm going to get indicator connected to x2 in n1 plus n2 and I can do the same for the others so eventually what do I get here I can make the sigma x1 sigma x4 appear as usual look at the probability so this is always like adding and subtracting zx1 x4 I get probability x1 x4 empty set for the first term I get 1 there is no condition for the second one I get indicator function that x1 is connected to x2 in n1 plus n2 for the third one it's going to be switching x1 and x3 with the thing so I get x1 connected to x3 and for the fourth one it will be x1 and x4 that I switch so I get this but notice now you have four sources you are at least pairing the sources so what are the possibilities either x1 is connected to one person or to everybody if it's connected to one person then one of this indicator function exactly pops up and is equal to one the two others are equal to zero so I get zero they are connected to everybody in this case it's three indicator functions that are equal to one and I get minus two so this is a very complicated way under the source constraint that you need to pair x1 to x4 it's a complicated way of just writing minus two indicator function that everybody is connected ok as soon as everybody is connected in fact you could also switch back sources x1, x2 to the second guy and you will get a second expression because this is x1 and if they are all connected then you can switch any source you want so this is just minus two indicator function of x1, x4 all connected sorry if x1 is equal to x2 this equality becomes trivial or you are right sorry I should take this disjoint point it's actually very good I am so much having in mind that we are going to let the distance go to infinity that I didn't even mention that but you are entirely right very good no question there for the product launch sorry P with x1, x4 and MTC is the product launch yeah, yeah, it's this P here I wrote it like that I should, because here I wanted to be careful that it's not the same graphs and I didn't introduce the notation for that that's all, but when they are the same graph I will always use the comma but you are right, it's a product measure it's just that very soon we are going to have four currents so if this is three crosses and three additional piece and you know my I'm getting old so I need to be careful with my yes anyway just I'm going to keep the three diagram bond for next week I think and just a small remark let's write a small remark after this you have to respect switching lemma nobody will erase the switching lemma you should repeat after me anyway okay just a remark how bad was the estimate with v4 how bad is it it's not a bad bound so let me mention that in fact minus u4 is also larger or equal to 2 third of v4 so up to constant v4 is actually a good estimate and in order to see that it's kind of a little bit we know now what this is right this is one half I mean this is twice sigma x1 sigma x2 sigma x3 sigma x4 times probability x1 x2 x3 x4 off x1 everybody x4 all connected right this is u4 it's one and this is true for any permutation of 2 3 4 but notice that this in fact is clearly larger sorry, this whole thing here is larger equal to sum of a gamma 1 from x1 to x2 and gamma 2 from x3 to x4 of rho of gamma 1 rho of gamma 2 indicator than gamma 1 and gamma 2 intersect why is it true this is just interesting my x4 times the probability that the backbone intersect it's actually exactly what this is if the backbone intersect obviously x1 and x4 are all connected in the sum of the 2 currents it's not an equality because you could imagine that the backbone do like that and that it's a loop or even let's say the backbones these are the backbones but you could imagine that there is a loop of the first current that connects to a loop of the second current that connects to a loop of the third current that connects to the loop of the first current to end up with everybody connected so it's not an equality but clearly there is an inequality between the two ok ok so that means that minus u4 is larger than twice this but this is true for any mutation of x2, x3, x4 so 3 times u4 is larger than twice the sum of gamma 1 from x1 to xi gamma 2 from xj to xk of rho of gamma 1 rho of gamma 2 indicator of gamma 1 and gamma 2 do not intersect which is exactly v4 so 3 times u4 is larger than twice v4 ok so true for permutation for the 3 permutation of x2, x3, x4 and then you get that minus 3 u4 is larger than 2v4 ok so at least you are up to constant v4 was actually a good approximation but with a double random current you get an equality ok ok well we are getting closer to the end of this class to the juice so let's just mention step 2 of two endpoints this is going to be important because when we are going to try to prove gauchanity in particular when we are going to express the characteristic function of our random variable I mean how we we will express it, we will just expand and then we will have two endpoints correlation that appears so there the proposition 2.11 is the following is that g2n is not always larger than s2n so s2n is just sigma x1 sigma let me call it s2n so s2n x1 x2n sigma x1 sigma x2n ok, it's just the schringer I mean I'm using the schringer notation actually I didn't need it I don't know why I did that ok let's not do it sorry I thought I needed it but I don't so I have a bound like that which sometimes is referred to as the gauchan bound so the spin-spin correlation are always smaller than the sum of of the pairing so what is given by thick law and again, here you want to be expressing it in the other direction so you would lack exactly equal to g2n but you are going to get something a little bit weaker and this something is going to be the following I need to take 4 points out and what I get is g2n-4 of x1 xi when I remove xi xj xk xl I think the notation is pretty self-explanatory you just remove the 4 points and you look at the correct, I mean at the weak law and here I get u4 of xi xj xk xl so why is this proposition nice well it's quite nice because it tells you you only need in some sense to treat u4 and then you will get an estimate for everybody so at least next week I mean tomorrow we will just focus on the 4 points function ok let's prove this, it's not that complicated ok, so first thing is that g2n, I mean a way of writing g2n is just to say well it's a collection of n-path pairing x1 to xpi1 to xpi2 xpi3 to xpi3 of rho of gamma1 rho of gamman it's just a way I mean so pairing xi this is g2n sigma x1 sigma x2n is what it's just a sum on gamma1 gamman pairing the xi but this time it's rho of gamma1 rho gamma gamma1 bar of gamma2 rho gamma1 concatenitide with gamma2 bar of gamma3 etc I remember now this thing I forgot about it but now I remember you know how my n will come one day I will just I didn't remember how maddening this noise was but then ok this is sigma1 sigma2n it's complicated a priori this is ok g2n larger equal to sigma x1 sigma x2n follows well exactly from the same argument as for 4 points what did we do we were just summing I mean fixing gamma1 to gamman-1 summing over the gamman we were getting sigma x1 sigma x2n in the smaller graph where you remove these guys and by griffith I mean by the inequality increasing in the graph it's smaller than sigma x2 pi pi2n-1 sigma x pi2n so this is follows from the same proof as for n equal 4 so what is truly interesting is the other one and the other one is going to be again let me write it like that the same trick if I look at something like that it's larger equal to what I get in the in the guy without the thing so here I can I mean where I'm not looking at it so I can sum over gamma1 gamman rho of gamma1 rho of gamman and the only thing is that there is an indicator function that the gamma1 gamman do not intersect well again write this as 1 minus sum over let's say r and s of rho of gamma1 rho of gamman indicator than gamma r intersect gamma s it's clearly a lower bound right? I should have indicator than nobody intersects where it's 1 minus probability that they intersect this is 1 minus the expected number of intersection if you want pairs of guys that intersect ok but here now what do you see if you look at gamma r and gamma s you see rho of gamma r, rho of gamma s times the indicator function that intersects this was exactly before so here upstairs so this is now sum over i smaller than j smaller than k smaller than l, these are going to be exactly the points that are paired by gamma r and gamma s and then I will get sum over gamma1 gamma r hat gamma s hat gamman rho of gamma1 rho of gamman and again there is no I think you understood what I mean and here there is sum of rho of gamma r rho of gamma s indicator than gamma r is intersecting with gamma s this is v4 so here if I should be careful I should sum I mean ok maybe let me add well yeah I mean ok it's v4 of the points that are paired like that and this is exactly g2n-4 of the points that are not paired by gamma r and gamma s ok and the only thing that remains is to just see that this is smaller so we said u4 is larger than 2 thirds of v4 so v4 is smaller than 3 half of v4 of u4 ok so that was the end of this part now let me prepare the field maybe I mean I still have 7 minutes so I'm just a little bit afraid of being lacking some time tomorrow so let me try to start a little bit the third part of the talk which is a conditional proof of triviality can I ask a question yes of course so in the continuum limit so this these bounds are for the for the lattice model but in the continuum limit we have to rescale the spin field yeah we will rescale the spin field I think for the triviality of the continuum limit one has to divide all these quantities by g2n then yeah but everything is homogeneous here so yeah you divide both sides by g2 you could define a small g2n which is a guy rescaled by whatever you want and then everything is homogeneous but can we still show that the remainder takes to zero at the left distance yeah yeah yeah u4 is expressed in terms of u4 is kind of g4 times a probability ok so if you my point is that here you are g2n is kind of expressed in terms of sum of g2n-4 times something that looks like g4 when you rescale it's perfectly homogeneous so the only thing you need to prove is that this is much smaller than g4 but now if you rescale g4 by what you want you also have to rescale u4 so the ratio of the two will not change the only I mean the only problem is kind of when you don't know what renormalization you should put or if you want to prove that any renormalization is going to give you the same result like that you don't have a way to save yourself with a smart renormalization is to always work with ratios so u4 here for instance if you prove that u4 is much smaller than sigma x1 sigma x4 then you can rescale as you want sigma x1 sigma x4 you will have to rescale u4 also by homogeneity so the ratio of the two will always be the same so if I understand correctly your conclusion is that u4 is equal to u4 times something and that's something exactly yeah right I mean that's what you get basically there imagine you prove the probability is small it tells you that u4 is much smaller than sigma x1 sigma x4 u4 is actually sigma x1 sigma x4 minus something so if you have the difference of two things one is actually the guy on the right I mean the difference of two things is much smaller than this guy but this is this one it tells you that both guys are of the same order just the difference is much smaller right and then the ratio is so there is no problem of renormalization everything is homogeneous from the point of view of ok ok let me actually not start this let me just mention one last property I was planning to put it next week but maybe tomorrow but let me try to so actually there is one case where things are not that complicated and this is something probably snow very well is that when you are trying to estimate a probability of something sometimes you can replace by expectation so here we are trying to prove that there is an intersection if in some way I can prove that the expected number of intersection is small it's a very simple way of proving that there is no intersection ok so here it's called the tree diagram bound and it's a way to bound u4 in some sense saying that you bound the probability of intersection by the expected number of intersection so I just want to mention it here because we will use it tomorrow so this is smaller than the sum for y belonging to zd of sigma x1 sigma y sigma x4 sigma y it's called the tree diagram bound because in some sense it looks like you have your points and you sum over y spin-spin correlation so you have this and there is a 2 here so let me finish this class by giving you an idea of this estimate ok so in fact it's not going to be difficult once you admit one bound ok so u4 remember is 2 sigma x1 x2 sigma x3 sigma x4 times probability x1 x2 x3 x4 of x1 x4 all connected or maybe another way to put is that x3 belongs to the cluster of n1 percent 2 of x1 it's a stupid way of saying I look at the connected component so this means the connected component sorry the connected component of x1 in n1 percent 2 meaning all the points I can go to by only walking on edges with positive current ok so I have x1 I have x2 this is all the set of edges that have positive current and I'm checking whether x3 is on it or not and because I have sources x3 x4 remember that x3 and x4 are necessarily connected so this is another way of putting so here is what I will not prove I will maybe prove it at the end tomorrow if really I have time but this is not priority is that in fact here I can get a smaller equal I can send the intersection easier by adding a current here and I think that the cluster in n3 of x3 x4 intersect the cluster of x1 in n1 percent 2 it's not that clear like that why this is true but it's a monotonicity so now there are 3 currents the first one has sources x1 x2 the second has no sources and the third has x3 x4 and you can check it's a kind of natural exercise in some sense you can check that this is just improving your probability of intersection maybe if you make a drawing maybe I can make a drawing and then after we have that you are going to see things are easier so let's look at the probability of not intersecting ok so in one case I have a backbone from x1 to x2 so if I have only 2 currents so this is n1 and I have a second current but this current is not connecting to x3 so I mean the sum of the 2 currents is not connected to x3 so it looks like something like that if you want if I explore the whole connected component in n1 plus n2 in some sense when I have only 2 currents and the second current must have sources from x3 to x4 I still need x3 to be connected to x4 in the complement in this picture with 3 currents the exact equivalent is that this current here is allowed to do whatever it wants it's not in the complement of this first current and in fact the smaller the graph the bigger the weight you remember they get bigger and bigger when you take smaller and smaller graph and in fact when you plug here what that means it exactly gives you the right inequality so it's not something super deep it's fairly simple to get but once you have that you are done and this is I take you just 2 more minutes and then we can all go to to T and you can blame me for this 2 minutes delay or you can blame the clock for being 2 minutes early it depends so u4 so I can bound it so I can do exactly the same thing I can rewrite the same thing except now I'm going to do an inequality which is even more obvious I'm going to so for now I had these sources let me add even 1 more current this becomes like crazy but let's add a 4th current and ask that the cluster of n1%2 intersects the cluster of n3%4 this is clearly bigger probability because I'm just making this state bigger by adding 1 more current but why is it nice now because notice that's this random variable and this one they are independent one depends only on the first 2 current and the other one on the first on the 2 others so if I now say bound and that's what I wanted to do from the beginning bound this by the expected number of intersection probability of intersection smaller than expected number of intersections I mean this is either 0 if it does not intersect or it's larger equal to 1 if it does so it's clearly larger than this random variable so here I can just rewrite this a sum for y in zd of the probability that y belongs to the first cluster which you can write as y is connected 2x1 in n1%2 times probability x3x4 empty sets of y is connected 2x3 in n1%2 or if you want in n3%4 ok so what did I do probability of intersection there it's fine what is the expected number of intersection sum over every point of the probability that the point is in the intersection so I write sum over every point of the probability that y is in this cluster meaning y is connected to x1 and probability that y is connected to x3 but because things are independent I can just split it and now ok but the switching NEMA gives you an exact formula for that we did so many of them that now I don't have to present it anymore but this is actually sigma x1 sigma y sigma y sigma x2 over sigma x1 sigma x2 and the other one here there are sources x3x4 so it's sigma x3 sigma y over sigma x3 sigma x4 I had sigma x1 sigma x2 and sigma x3 sigma x4 here so things cancel and I get just the sum I wanted ok so like take home message of this 3 diagram bound it's a bound that you obtain when you bound the probability of intersection by the expected number of intersection and in dimension 5 and more it will actually be sufficient to prove triviality in dimension 4 it won't be because this intersection probabilities for random walks are such that expected number of intersection in dimension 5 and more is tending to 0 as well while in dimension 4 it's not tending to 0 so that's we will need to do something more tomorrow to treat the case where we are in dimension 4 but before that we will use a 3 diagram bound to get something in dimension 5 and more ok, thank you very much and see you tomorrow