 Some formulas from yesterday so if you remember this was the the energy we are looking at and It was pointed out to me by a good year last night. I used them The same letter for two different things yesterday, which was very bad And so I'm going to fix this by defining the energy with a curly H here Okay, and we have a splitting of this That we proved yesterday of this form Mu V is the equilibrium measure Zeta is this sort of effective confining potential and then yesterday I introduced the potential generated by the fluctuation measure and So this one should be a H with straight straight H So it's the integral of G of X minus Y It depends on X and of course it depends on the configuration So for each configuration you have a given a given function okay, so this is the This is the potential and if you remember it says minus Laplace H n equals sum of Dirac minus n mu V and the the point was to try to prove that this Next-order energy can be expressed with this electric potential Let me Stop here for two to talk about what's happening in the one dimensional logarithmic situation because you remember that in 1d the logarithm is not the coolant kernel So I cannot write this however. It's easy to generalize these formulas by thinking that The real line are Can be embedded in the plane So usually people do it in the complex plane, but let's embed it in R2 and so each point X i you can view it as X i comma 0 And then this function H n you can view it as defined over R2 So now X is in R2 It's going to be integral of minus log X minus Y the sum of the Dirac masses at these X i 0 Minus n and mu V is a function of the real line only so Let's write it X sorry y1 for the force coordinate And here d y right but of course mu V of y1 I can also Think of this as a measure in the plane with density mu V of y1 times Delta R of y So I write here Delta R to denote the sort of Dirac on the real line right, it's the it's the uniform measure on the real line and it takes you When you integrate a function Against Delta R Say X1 X2 It just integrates With zero for the second coordinate okay, so Once you view H n as an as a function on R2 this way It's simply the potential generated by the configuration the distribution of charges at X i 0 minus n mu V times the Dirac mass on the real line and then you can write That this holds true In R2 okay, so this is a sort of trick That allows to see The logarithm interaction in 1d has a sort of restriction of a 2d interaction For measures supported on the real line and then you can continue the calculations in the same way as if you were in R2 For specialists and when you do When you look at log gases or random matrices on the real line you compute the silt still ts transform It's it's exactly the same procedure. So you view things in the upper complex plane It's the same thing as expanding this and computing the gradient of age So it's essentially the same as the still ts transform Okay, so now I want to give a rigorous meaning to the computations. I was doing yesterday about computing f n mu V of Xn and relating it to integral of gradient H n Squared and so in order to do that I'm going to have to introduce a truncation procedure For these Potentials H n which are singular at each charge right so H n you remember blows up like the Coulomb kernel Near every point of the configuration So I define I'm given Give myself some eta which is just a vector of numbers It's a one eta n and think of them as small numbers and I define H n comma eta To be H n It's the same as the function H n minus the sum of G of X minus X I Minus G of eta I Positive part, okay So you see if you remember H n near X Near each X I is going to blow up exactly like G of X minus X I So this thing is removing the singular parts and Replacing it by G of eta I which is a constant right, so effectively it's the same as Saying you have a you have a function whose graph is going to look like this and you replace this graph by the truncated version where you truncate at The value G of eta I so eta I is this little radius and you're just going to chop off all the peaks You're all the points you chop them off at distance eta I Okay, so this is this is a formula that does this All right, and now You can try to compute the laplacian of this guy So what are you going to find? Well, it's going to be the same as the laplacian of H n except in the ball Centered of at X I and radius eta and there instead of having a Dirac mass at X I you're replacing it by the laplacian of the function whose Graph is like this So let's say this is the function f eta of X. It's just G of X Minus G of eta positive part Okay, so what this what does this function do? It's zero Outside of a ball of radius eta it has this infinite peak And then it stops here When you reach G of eta Okay, what is the laplacian of this guy? It's the Dirac at the origin There's the CD everywhere So it's the Dirac at the origin minus What I will call? Dirac zero eta Or Dirac zero eta or Dirac p eta is a measure of mass one on The sphere of radius eta Normalized to have total mass one. Okay, so why is this true? It's a computation, but the other way to see it is This thing is a radial so the laplacian of f eta has to be something radial and so it's natural then that would be a uniform measure well because This laplacian is just simply the jump of The normal derivatives when you reach the boundary of the ball right so it's supported on the boundary of the ball and Then by symmetry it has to be uniform and the fact that it's total mass one It's easy to check because the total mass here has to integrate to zero All right, so another way of saying all this is that this truncation procedure What it does it removes the Dirac mass and replace it by replaces it by a smeared Dirac mass Which is just smeared on the sphere of radius eta All right, so when I go to compute the laplacian here Every time I had a Dirac mass I replace it by a Dirac mass smears smeared radius eta I And then nothing changes the rest is the same, okay so it's one way of regularizing these Potentials that are singular and what what's convenient with this way is that it's completely explicit and it's it's fairly straightforward. It's just these truncations Okay, so with that I can proceed to computing And what I'm going to prove is the following proposition that f n Mu v so let's say for any new mu This next order energy which if you remember was defined by I'm going to write it here Fn of xn Respect to mu Is equal to the double integral on the complement of the diagonal of this fluctuation against itself okay, so the Proposition is that this is always greater Than 1 over cd the integral of gradient h and eta squared minus The sum of g of eta i plus error terms Which are small essentially so I Will control these error terms by the sum of eta i squared Let me check c times n times this So the idea is that later. I will choose this eta i properly this eta i is properly In such a way to make these error terms Small enough Okay, and There is this inequality With equality that's important if all the balls Bxi eta i or disjoint So it will be sometimes useful to choose the eta i so that the balls are disjoint. Okay, so this Formula is doing what I was promising. It's it's making rigorous this this relation By saying well instead of having this integral that's divergent we're going to compute the integral of the truncated The quantity for the truncated potentials this one now is finite and convergent, but we have to remove a certain divergent Quantity which is exactly the the energy that the tip carries so you If you look at what happens when eta goes to zero this thing Tends to infinity again, but this thing subtracts off exactly the divergent behavior And in effect you are computing things in finite parts right Okay, so let me explain roughly how This can be proved so the first step the first step consists in studying these energy of this truncated potential and showing that this is roughly It's actually equal to CD the double integral of G of X minus Y so this the into the Coulomb interaction of the smeared Dirac's with themselves Okay, so this This is what this is the computation I was doing yesterday Except now for the truncated guys. It's it's legal. You don't have to worry about the diagonal anymore The integral is convergent So basically you look at this formula you multiply the formula by You integrate it against the same right-hand side and Then you integrate by parts and you find this okay, so this is the the legal version of it Okay, and now what we have to do is we have to compare this and this Okay, so these two Quantities sort of look the same except I have the Dirac's here and I have the truncated Dirac's there and here of course I can remove the diagonal by writing that I Have diagonal terms first So what is the diagonal? I'm integrated G of X minus Y the Dirac on Smirred on the ball radius eta. This can be computed. It's going to give you just G of eta I for each For each point so this is the diagonal term and so I'm writing that this is roughly equal to this plus CD The double integral on the complements now of the diagonal And I do the same. I have the smeared Dirac's and the smeared Dirac's here Okay, and so now when I make when I subtract this Minus that I have this term Which is not formally very much equivalent to this term. Yeah When I equals J here Right so when X equals Y or when I correspond to Y equals J So it's the interaction of the point with itself so if it was a True Dirac the interaction of the Dirac with itself. It's infinite. You don't want to see that But when it's a smeared Dirac, it's okay. It's a finite quantity. You can compute it And it gives you this right so this is the integral of G of X minus Y Delta X I eta I of X Delta X I eta I of Y All right, so this is an explicit computation. It's all radial. It's easy to compute So it's the self interaction of a smear charge and it blows up of course when eta goes to zero You should expect that it behaves like this All right, so now my my task is to prove that this quantity is bigger than this quantity up to error terms And that there is equality if the balls are disjoint and so that It's based on the following argument So I'm just going to look at the terms that correspond to the interaction of the charges So these these types of terms X I eta I X J It's I J and now I is different from J Because I've taken into account the terms for which I is equal to I okay so now let's look at a function like this G of X minus Y Delta X I of Y For example, this is the potential generated by a Dirac at X I We know what it looks like. It's just G of X I minus X but if you think of it as You can view it as the solution to minus Laplace H equals CD Delta X I So it's a function which is harmonic away from X I And otherwise, it's always super harmonic So H is super harmonic So here we use the Coulomb nature of the interaction we're this is really where it comes to play a role and If I do the same with eta I The same is true, so it's super harmonic and it's harmonic Away from B X I eta I Okay, not too hard Now, you know something that's called the maximum principle right maximum principle tells you that if you take a harmonic function and if you Average it over a sphere It's the same as taking the value at the center of the sphere Right the average is equal to the value at the center so if I take this function if it's harmonic and Now I average it on some other Sphere, so I take this whole thing and I average it It's the same as if I take The same thing against the Dirac mass Right, it's this is the maximum principle the average is equal to the value at the point But that's going to be true only if I'm In the zone where the function is harmonic, so this is okay If I'm away from B X I eta I so if B X J eta J Does not see B X I eta I Then this is true because I'm in the region where it's completely harmonic And so I can replace the function the average by the value Okay, but the maximum principle also holds for super harmonic functions with an inequality It tells you that if you have a super harmonic function, and if you take averages Over a sphere the answer is going to be smaller than the value at the point So in any case I always have an inequality Which is true? Okay, and so when I put all that together I find that The integral of G of X minus Y Delta X I eta Delta X J eta That's less Then the integral Sorry, so this is eta I eta J Eta I so here I can replace by the value at the center and Then I can exchange the roles of I and J and I can do the same again Okay, so you find that Basically, this is telling you that the interaction energy between two smeared Charges is smaller Than the original interaction energies that you had before smearing Okay, so this is an effect of The Coulomb nature of the interaction moreover All the inequalities here are equalities if the balls are disjoint And so if I put all that together It proves my proposition So there is just some error terms that come from dealing with the mu V here What I see when I have mu V and I like I change delta X I by the smeared by the smear charge, but since mu V is a nice Measure with the density it will create error terms that are proportional to the Radius of the of the spheres So this is how you get these these types of terms Okay, so now With this proposition I Want to show you that we've already obtained some interesting information so we have a way of re-expressing the this total pair wise Coulomb energy of the system That's composed of the neutral system that's composed of positive charges and a negative background We have a way of re-expressing it in terms of these integrals Right and the idea is that these integrals are going to be extensive in space and scale like Essentially like the volume of the area in which you integrate them So as a corollary we have a Lower bound for Fn Which is valid for any configuration of points Take it up So you take it Equals for every eye Okay, you want to find the right value of eta i that you should plug in there The idea is you want eta i to be small so that these error terms are small But you also don't want it to be Too large too small because otherwise these terms will become very negative and you're looking for a lower bound So if you optimize those two things you find out the right eta i that you should take Is n to the minus one over D And that should not come really as a surprise because this is the natural Length scale Which we expect to be the distance between two points right the distance between a point and it's nearest neighbor We expect it to be of that order and so we are truncating here at an order that's comparable to the Distance between the points I could take here any small constant in front. It would not make any difference Okay, so if I plug that in what do I get? The idea is I'm going to take Fn of Xn. I have a Lower bound here and this term is obviously positive so I can just discard it And so I find minus So I'm going to have n identical terms Equal to this. This is the choice of eta i and the error terms I Bound them below by Cn The eta i squared will give me n to the minus 2 over D and I have n terms so I have a 2 Okay, and so now let's compute what this is So you have to distinguish two cases in the log cases Then G is minus log right so what you find here And I have Sorry, there is a plus. It's a plus No, it's a minus okay, so it's G is minus log so it's a plus and then there's a minus So I get minus n over D log n minus C and To the 2 minus 2 over D This is good and then in the other cases Because I have to assume that the error term has the bad The wrong sign right I have to assume that it's bad for me. I Want a bound from below. I don't know the sign of the error term So if I want a bound from below I put it with a negative a Negative constant. This is the worst case scenario My lower bound is very low Okay, so in the other cases G is n to the power sorry G is R to the power 2 minus D if you remember And so I put that in factor Get n to the minus 1 over D to the power 2 minus D and if you believe me It's going to give you minus C and 2 minus 2 over D okay, so the logarithmic cases are a little bit different because you get these terms that Pop out which you don't get in the other Coulomb cases Okay, I think I made a mistake somewhere Okay, so for example in dimension 2 You see that this lower bound is of order n and is this is actually the idea is that this lower bound is actually optimal it gives you the right order and Proving that of course is It's a more difficult endeavor and so once you have a lower bound for the energy You can obtain easily an upper bound for the partition function. So now let's think We're in the situation with temperature and we're looking at the probability density of this form Now we have a lower bound for hn this curly hn because we have this splitting formula And we have a lower bound for fn And so naturally this gives us an upper bound For Z because Z Okay, it's the integral of this The xn So let's input The lower bound that we just found Into the the integral Okay, so Sort of obvious computations, but everything is going to come out of the integral and You're going to get Say n squared IV of mu V if you remember well then these terms are going to come out so I add a plus beta