 Take your seats, Sylvia. All right, so thank you very much. If you attended the previous lecture by Gaétan Boreau, this lecture is going to be very, very related to what he was discussing. In fact, it's the same topic. It's about proving central limit theorems for fluctuations in these log gases or Coulomb gases. OK, so I've put back on the board the quantities that we are dealing with. So this is the energy with G, the Coulomb or log kernel. This is the Gibbs measure. And we have seen that we have a splitting of the energy in this form with the next order energy, which is just this quantity here, which is the sort of self-interaction of the fluctuations away from the diagonal. OK, so the thing we want to prove is something like this. So let's write theorem. So consider C, a regular enough function. So here, let's assume four derivatives. In fact, what we really need is that the third derivative is Lipschitz and compactly supported in R2. So I'm going to state the theorem in a two-dimensional logarithmic case, and then I'm going to comment. So you want to consider the fluctuations of C. Then this converges in law to a Gaussian random variable with a certain mean and a certain variance, which are explicit. And in particular, the variance is essentially given by some constant over beta, some constant which I forget. The gradient squared of the test function C. And so here, there is a little twist. There is this sigma here, which means that if C is supported inside sigma, which is the support of the equilibrium measure, then this is just the integral of the square of the gradient. But if C has a support which touches the boundary, which you can be interested in, then you have to take the harmonic extension of C outside of sigma. So C sigma means harmonic extension of C outside sigma. And for experts, this can be phrased in terms of the convergence to a Gaussian free field of the potential hn mu v converges to a Gaussian free field. OK, so this theorem in 2D was first proved by Ryder-Virag in the Gini-Brandsenball case, when v is quadratic. And by Ammer and Denman-Makarov, for beta equals 2, which is the determinant of the case for which you have other ways of computing. This is for any beta. And so this theorem was proven at the same time by Toma Leblay and myself, and also by the team of Bauer-Schmidt-Burgat, Nikola and Yao, with slight differences, but essentially the same. There is also a local version of this theorem. You can take C to be supported on mesoscopic scales. And the result is still true. So you remember you have to set sigma. And basically, you're testing things here in some set. And you could take C to be, let's say, some cn, which is a fixed rescaling of a fixed function c at a scale ln, which is much bigger than, so let's say ln equals n to the alpha minus alpha, with alpha strictly less than 1 half. So it means that even if you zoom down, this behavior is still true. And as I was saying yesterday, it means that the fluctuations are actually much better than order n, much better than order root n, which we obtained yesterday. But in fact, they are typically of order 1. But this is for test functions, which are quite regular. So in particular, you cannot take something for which it would be wrong. It would be to take the characteristic function of a certain set. Because if you could take the characteristic function of a set, you would be essentially able to count the number of points in a set. And the fluctuations of that are not expected to be of order 1. They are expected to be much larger. OK, so the restrictions are certainly not optimal, but there are some necessary regularity restrictions. OK, so in 1D, there is an analog result, as Gaétan was talking about, which was proved by several authors. So starting with Johansson. For the 1D logo, Johansson, Cherbina, Boroguillonnet, and Gaétan presented that. And of course, the statement is very similar. So now let us see how to prove something like that. And the method I'm going to present is based on the idea of a change of variables. So the starting point is as Gaétan described, and it's the method introduced by Johansson. You want to compute the Laplace transform of the fluctuations. So the Laplace transform is nothing else, but the expectation of t times integral of xi. So you know I call this fluked n, right? So this quantity called fluked, like fluctuation. OK, so this is the Laplace transform. If you manage to show that the Laplace transform converges to that of a Gaussian with appropriate means and variances, then you will be done. So it's about looking at the convergence of that. So why is this Laplace transform a convenient object to look at? Well, because if you compute this expectation, you're actually computing 1 over zn, integral of exponential minus beta over 2. You remember the energy, right? So you have the sum of g of xi minus xj plus n sum of v of xi. And now you're adding here plus t sum of xi of xi minus something constant, which is the integral of t mu v and with an n. OK, so the sum of xi of xi, after all, I can group it with v to make it the same thing but with a modified potential, right? So let me change things a little bit now. And let me put an n here. And this way I put an n here. And it goes with the v. So I find 1 over zn. And I find plus n sum of v plus t xi of xi. And then this term is constant. I can just take it out of the integral. So let's put it in front. Exponential minus or plus beta over 2 tn integral of xi mu v. All right, so what we find here is this integral. It's exactly the same as the partition function except for this modified potential. And so the idea is that this is something that we can hope to compute. And so I will now write it like this. So I have this constant term, which maybe is wrong. And I'm not sure. Actually, there's no beta. And there's a minus. Yeah, so I did something completely. OK, my factors beta are wrong. So instead of changing it into tn, I'm also going to put a minus beta over 2. It will not change anything. And then everything goes into the beta over 2. So it's good. And so here I find the ratio of the partition function associated to vt. So let me write it like this with the partition function associated to v0. And I write vt equals v plus tx. All right, so I have changed things a little bit here. So instead of computing the Laplace transform, I'm computing the Laplace transform up to a factor. So later, what I will want to do is I will want to take tau equals minus beta over 2 tn. And this way, I will have the expectation of exponential tau times the fluctuation. So this means that at the end of the computation, I want to be able to take t equals minus 2 over beta n times tau, which means I will be interested in very, very small perturbations. You see, t will be very small. It will be of order 1 over n. But for notational purposes, it's nicer to keep it as t. OK, so this is why now we're interested at looking at the ratio of partition functions for a modified gas with respect to the original gas. OK, so what do we do with that? Well, we have this splitting formula here, which allowed us, if you remember, to remove the dominant terms of order n squared from all the quantities, and in particular, from the partition functions. So I can write a reduced partition function, let's say, which is k n v, which is just z n v exponential beta over 2 n squared i v of mu v for any potential. And this thing sort of comes out of the Gibbs measure. So I can rewrite p n beta as equal to 1 over k n v exponential minus beta over 2. And what I have here is my next order energy, f n mu v, plus the confining effective potential, dx n. OK, so when you want to compute this ratio of partition functions, you can compute the ratio of the k ns, because the other part is something that's explicit. And you can compute the ratio of these guys properly. And so if you put that together, you find that this expectation that you're looking at, minus beta over 2 t n flux c. OK, I can rewrite it after a little bit of not much, as the ratio of the k n v t, k n v, and a term that's explicit, and which is essentially going to give the variance. So it's going to give you something like this, exponential over t squared, some constant here. OK, so this thing in the end, you remember I take tau to be this. This is the variable for my Laplace transform in the end. So this term is going to give exactly the variance, because when you take the log of this, it's going to give a quadratic function. So this is already settled. Now what we have to study is the ratio of these reduced partition functions. OK, so if you have to look at a modified log gas with a modified potential, then of course there is a modified equilibrium measure. So now I'm going to denote mu 0, the original equilibrium measure, and mu t, the one associated to v t. And the first question is, what does mu t look like? So it turns out that if you're in a Coulomb case, it's not too difficult to study mu t. So let's say in the 2D log case, you may remember that mu v had a density, which was given by Laplacian v over 4 pi times the characteristic function of the set sigma. So if you're in the 2D log case, if c is supported inside sigma, then perturbing the potential there like this does not change the support. And for t small, you can check it's not difficult. That mu t will simply be mu 0 plus the appropriate thing. So plus t Laplacian c over 4 pi. And the support doesn't change. So this is this. And since Laplacian c is inside the support, I don't have to do anything. So this is the new equilibrium measure. And now if c is not supported in sigma, then this is not true. Because the support of the equilibrium measure will move a little bit. So you have a support like this. And it will move typically by an order t. So things will move a little bit like this. OK, this case is harder to treat for that reason. You have to understand the new equilibrium measure. However, that can be done. And I will not go into the details of that case. Just know that it can be done. And I will continue with the setting where it's supported inside the sigma. And so now the next step is to build a transport map, a map phi t such that phi t push forward of mu 0 is approximately equal to mu t. This will create a transport map for which we can make good change of variables. And of course, since t is very small, we are doing something asymptotic. So we can expect that phi t, we can build it as a perturbation of identity. It will be an identity plus t psi. And the question is to build the psi that will do this. So you will not transport exactly to mu t, but to leading order in t, it will be the same. So it will be approximately mu t if you want plus little o of t, in some sense. And so if you're in 2d, and mu t is, as I said, mu 0 plus t plus n psi, then it's actually not difficult to see that psi equals gradient psi over mu 0 does the job. So you have an explicit transport. And if psi is regular enough, this thing is regular. So why is this equation the right one? Why is this psi the right one? Well, I don't know if you know the definition of transport and push forward. But let me write it now. Phi t push forward mu 0 is a measure that you can call mu tilde t. So it's an approximation of mu t, which is such that if everything has a density, the determinant of d phi t is equal to mu 0 divided by mu t of phi t. This is the formula for change of variables. So you're saying that your measure is pushed forward by phi t. And now if phi t is identity plus t psi, I can linearize. And I find 1 plus t times the divergence of psi equals to mu 0. And then I can also linearize this mu t of phi t. So I'm going to find mu 0 minus. So plus, OK, maybe I should write it here. Well, you can linearize everything. So we have mu t plus t over 4 pi Laplace-Sensenk psi. I'm keeping only terms of order t. And I will have a grade mu 0 dot t times gradient mu 0 dot psi. And if you look at this and you identify terms of order t, you find that you must solve this equation. Divergence of mu 0 gradient psi equals Laplace-Sensenk psi. And now look at this. This psi obviously solves this. OK, so this is the linearization of the Mon-Jean-Père equation, if you want. OK, so this is nice in the Coulomb case because in the Coulomb case, we're dealing with local operators, the Laplacian. If you were in 1D, you could still build a good psi. And in 1D, a good psi can also be built. And building this good psi is actually very similar to inverting the operators that Gaëton was talking about earlier. So in some sense, this transport business is containing some of the ideas of the Schringer-Dyson equations. But OK, you can view it in a very naive way, just trying to build a transport. All right, so now we have the good psi. And we want to compute the ratio of partition functions. So I go back to my partition function. So I have to compute this. And I remember my splitting in the formula. So I have fn. And now it's with respect to mu t of xn plus 2n zeta t dxn. And let me make the change of variables. That xn is phi t of yn, where phi t is the transport I just defined. And so if I do that, I find myself with exponential minus beta over 2 fn mu t of phi t of xn. I have the zeta terms of phi t. So everything that continues here is in the exponent, so 2n. And then I have to include the Jacobian of the change of variables. So if you remember, the Jacobian of a change of variables is the determinant of d phi t. I can put the determinant inside the exponent. And it will not be very clear, but I will continue. So plus, which is still in the exponent. So whatever, there is exponent of sum of log that d phi t of xi. And finally, I close my integral. OK. So the zeta t terms, don't worry about them. They don't really play a role. Now I find myself with this modified, the energy with the modified measure, the points that are transported. And I have this Jacobian term. So the idea is I want to compare them with kn v0. So I can certainly write this as the integral of exponential minus beta over 2 fn mu t of phi t minus fn mu 0 of xn. So I take it, and I put it back. So let me drop the zetas, because as I said, they're not so important. Plus, so I have exponential sum of log that d phi t. And then I have exponential fn minus beta over 2 fn xn. So I have here the terms that I would have for the partition function of the original gas. So if I rewrite this, I am just computing the expectation under the Gibbs measure associated to v0 of exponential minus beta over 2, the modified thing minus the original one, plus this sum log that d phi t. All right. So this is just a computation. I started with the ratio of two partition functions. And I compute this ratio by making a change of variable. So sorry, I should write here. This is the ratio kn vt by kn v0. And if you remember, if I know this ratio, I will have my Laplace transform at the end. So this is what we need to compute. And so now it's about comparing energies, these next-order energies, when you modify the charges by pushing them by phi t. And now I'm going to cheat and go to one dimension. So I'm going to do this comparison in 1d. And at the end, I will tell you a little bit of what you have to do in 2d. Because in 1d, this computation can be made completely explicitly, and it's quite easy. So let's go. So I want to compute this thing, f and mu t of phi t of a configuration minus f and mu 0 of the same configuration. I just have to use the definition here. So I have the double integral of minus log of x minus y. And here, I have x minus y, sum of the Dirac's at phi t of xi minus mu t and the same minus the same thing with mu 0 to plus the complement of the diagonal. And then I have the sum of the Dirac's. OK. So now I remember that mu t is approximately, and for the purpose of this talk, I will assume it's equal, to the push forward of mu 0 by phi t. So by definition of a push forward, when I'm integrating something against the push forward of a measure, it's the same as changing the variables in the integrand. So this is the same as looking at minus log of phi t of x minus phi t of y and the original fluctuation against itself. And then, of course, I add the same with log of x minus y. But now this thing I can simplify. I can group things together. And this becomes equal to minus log. So I have simply the ratio of phi t of x minus phi t of y divided by x minus y. And this thing, this fluctuation. So let's write it deflux of x, deflux of y. Next thing, I remember that I'm expanding in t. 1t goes to 0. And phi t is identity plus t psi. So I can expand the logarithm. And if I do that, in 1d, I'm going to get this thing. I'm going to get t times psi of x minus psi of y divided by x minus y, deflux of x, deflux of y. OK, so I have this term. And you see it's t times something. I will write this term A. So if you add the patience to follow this whole calculation, what we have done, yeah? No, so there is no log because you see I'm expanding. So phi t of x, I write it as x plus t psi of x. This I write as x plus t psi of y. And so I have x minus y divided by x minus y. And then I use the expansion that log of 1 plus t h is approximately t h. So log disappears by linearization. You find a term that's, if you compute, you will find a term exactly like this. OK, so I was computing the ratio of partition functions. And I arrived at the expectation of something which I can approximately write as exponential minus beta over 2.