 this term a of xn plus sum of log of that d phi t of Xi. All right, and so now what I'm claiming is that what's in the inside this exponent? Is it essentially some fluctuations? Because you see this a term is the integral of a certain Okay, let me retrace my steps a little bit. I want to modify this term a little bit So I look at this term this a and I observe that what I'm integrating here is a function so I have x minus psi of y over x minus y which is actually regular if psi is say c1 or c2 because When x goes to y this will go to the derivative of psi So this function is actually regular But I'm integrating in the complement of the diagonal So what I can do is I can put back the diagonal and write it as let me call it c hat of xy the fluct of x fluct of y plus the Diagonal terms that or minus the diagonal terms that I put in there and you can check It's not too difficult that those diagonal terms are going to be related to the derivatives At at the point so with a plus or a minus I don't remember Yeah, so it should be this Okay, and that psi hat is just this function Psi of x minus psi of y Divided by x minus y All right, so what's what's happening to this term? This term is actually going to go together with the Jacobian because if I Linerize again that of d phi t phi t is identity plus C prime Identity plus tip C prime So when I linearize the log that when 1d the data I Just get some of psi primes so This can again be rewritten As the expectation of something where I'm going to get 1 minus beta over 2 The sum of the C prime I with a t So this is what the term should give If I'm not mistaken with the signs and then I get this guy Plus T So it comes with a minus beta over 2 So still in the exponent times T Times this term. Let me call it B double integral of psi hat D-flug D-flug Okay, so now I'm claiming that I can evaluate those terms The first term you see here. I'm summing a certain function against Dirac masses right, so I can write this first term as Exponential 1 minus beta over 2 Times T times the integral of C prime Against the equilibrium measure and that would come with an N Because that's my approximation of Some of the Dirac's So I can take this out of the integral And then inside the expectation I can put back the fluctuation With a minus so this one will have a minus I think Okay, so you see when I compute C prime against the fluctuation I regain this term minus the term I took out of the integral so this is computable and The other one is here. It's also computable. All right, so my claim is that Everything that's in the expectation is actually negligible or it will be like one And that the leading term is this one that just came out So we have to look at what's inside the expectation it's the exponential moments of some integrals of fluctuations and Now I remind you what we proved yesterday we proved that if I have a test function and I integrate it against the fluctuation I can control this essentially by the integral of fn plus n over d log n to the power one-half and in addition the exponential moments of Fn plus n log n. I always control them By n so this is exactly what I need here because I'm dealing with Something against the fluctuation So something regular enough so that's if phi is lip sheets And here I have a t. All right, so what is going to be the size of this guy? well Let's say I can bound it by root n typically in exponential moments, so the exponential of This t, c prime fluct I will be able to bound by Tn, is that what I want? No, sorry, so I would say square root of n so this thing I essentially control it by Something that's like square root of n and when t is of order Like order tau over n if you remember I had to take t of order tau over n This goes to zero So this term actually does not contribute Once you take its expectation And then we have the second term a second term. I'm trying to look at Exponential moments of something of the form t times some constant times this double integral of psi hat of xy Deflect of x Deflect of y But now I can use again this This this control I can use it twice, right? I'm integrating twice So I can expect that this is going to be controlled by c Fn plus n over d log n Because here I have a square root and I use it twice So it's going to make it appear with a square and the exponential moments of that I control By n, but here I have a t So the the log of this sorry, I should take the logs This is a log There's a log So the log of this expectation I finally bound by Tn But he is of order 1 over n So this is of order 1 okay, so This is the end of the computation now We retrace our steps and we look at what we've done. We've computed the ratio of the partition functions We've obtained some term here that is explicit that comes out of the integral Multiplied by so this term by the way since Tn if of order 1 this is of order 1 multiplied by something whose Expectation is essentially of order 1 Okay, so we plug that back into the computation of the Laplace transform And we remember that the ratio of these partition functions up to Some constant factor is the Laplace transform of the fluctuations So we have obtained the first step that the Laplace transform is of order 1 independent of n is bounded independently of n but so That means I control the exponential moments of any fluctuation Instead of controlling it here by n If the if the test function is regular enough, I have upgraded this into a control by C So this is this is not good, but what I have upgraded and I should find my eraser Upgraded things into saying that exponential moments of Something like this Thanks to the computation of this ratio of partition functions is bounded by a constant if C is regular enough Okay, but now I can bootstrap this Information I can go back to my calculation This term I already knew it was small it was contributing one over root n anyway But the other term that was troubling me. I had a control by an order of a constant, but now I can upgrade this and since I have Essentially something of the form of exponential moments of a fluctuation. It's essentially of that form But he has a T in front so the log of that it's going to be bounded by CT and This is like C over n and so it goes to zero and that finishes my proof because Once I put everything all the pieces together. I find that this term I can treat it as one and The Laplace transform of the fluctuations. I have computed them explicitly the The the the log of them is going to converge. So this is going to be exponential mc sorry exponential tau times mc. This is my Original term and so I have found that the log of the expectation Flux times C Is behaving like so if you remember there was a variance term Which was like that? And there is this tau mc Plus little o of 1 and this is exactly what we needed to prove convergence to a Gaussian To a Gaussian So at this point if you have infinite regularity on C you can actually bootstrap this procedure And you can obtain a rate of convergence. So all these convergences You can make them You can sort of input the information you already got on the convergence of the exponential moments of the fluctuation to explicitly describe what happens and obtain an expansion in all orders of 1 over n and The other thing is that as Gaétan was mentioning in his in his talk in fact computing Laplace transforms of fluctuations or computing ratios of partition functions. We've seen it's essentially the same and so if you want you can also obtain Expansions of the ratio of partition functions for two different potentials By applying this method and if you have as Much regularity as you want you'll be able to expand in powers of 1 over n. So this is what Gaétan Borot and Alice Guionnet have done with an approach which is Philosophically similar but technically a little bit different Okay, so I said that I have cheated because I have presented the proof to you in 1d In fact, if you're in if you're in 2d It's not It's not going as smoothly and the place where it's difficult is when you're trying to compute the difference of the of the Energies So I don't know if you remember at one point. I was having to compute something like this right you want to compute evaluate the difference between the energy of of a system I have particles of Charges and the transported system where you move you move everything by phi t so in one day I have used the explicit form of F in this form to to evaluate this This difference and to linearize it in t obtaining these terms that involve this psi of x minus psi of y over x minus y But if you do that in 2d the equivalent terms is Basically is going to be a psi of x minus psi of y scalar with x minus y Divided by x minus y squared and this term is not continuous. So There's rotations that happen near the diagonal so it's not It's not convenient to do it this way instead what we do is we resort to the formulation in terms of the electric Potentials that I presented yesterday. So you express things in terms of a difference of something like this Minus the original one And you evaluate this difference and in the end to prove that the corresponding term that comes here is negligible you You call the expansion of the partition function So you have two ways of computing the expansion of the partition function one way is to To compute the ratio in this way, but we have another way of computing it So if you want the KN Vt divided by KN V naught has an expansion The log of it has an expansion up to CN That I mentioned in the first lecture with a constant here that is a little bit explicit and This is obtained by our other result with Toma Loble that I will present tomorrow So borrowing from that result and this expansion and the explicit dependence of C In the equilibrium measure you have another way of computing this and it allows you to show that this This harmful term is actually well controlled Okay, so I think it's it would be a good time to stop if I remember I may have some additional comments, but I Think I quoted everybody It's a good time for lunch. Thank you Any questions before lunch? Yes, I Think it I don't think it works, but we can we can try to talk about it I mean typically the terms that you're obtained you're going to obtain in they're going to look like Signs and cosines near when you look near the diagonal. Yeah Yeah, yeah, that's that's an interesting thought, but okay I'm not sure it would work, but we can we can talk about it. Okay, so tomorrow. Yeah, maybe I'll say tomorrow I will completely switch From this topic and and the goal will be to just describe a little bit the the large deviation principle that you can get for which the As a by-product you get this expansion of partition functions in in any D Okay, so if you didn't attend to today, which is not your case obviously You can still come tomorrow Okay, let's make Sylvia good