 over 2 and over d log n if I'm in the logarithmic cases. So let me write this like this, one log. And then I add minus beta. So c beta n to the 2 minus 2 over d. OK, so everything comes out of the integral, except what's left in my integral. I need to keep something in the integral, otherwise it will not converge. The term that makes it converge is this effective confinement. So I find in my integral exponential minus 2 minus n beta sum of zeta of xi dx1 dxn. Yes? OK, so here I think it's OK. This one comes with a plus because there's a minus. It's the other one that should come with a plus, yes. Thank you. OK, so if you look at what's going on here, if you're in the Coulomb cases with d bigger than 3, you should scale temperature differently, in fact. So instead of looking at this exponential minus beta over 2, you should look at this with the power n to the 2 over d minus 1 positive part times hn. So you see 2 over d minus 1 in dimension 1 or, sorry, it's positive part or negative part. 2 over d minus 1, negative part. So in dimensions 1 or 2, this will just not be active because the negative part is 0. And in dimension 3 and higher, it will start to kick in. And it will make these terms be more interesting for the sequel. So OK, I prefer not to go into these details because it's not so important. But if you want to get an interesting expansion in dimensions 3 and higher, you should scale temperature a little bit differently. And so from now on, let's be in the simple case. Let's say d equals 1, 2. This way I don't have to deal with that term. All right, so if I continue here, I have obtained an upper bound log zn is bounded from above by this plus an n long n term, which is fixed. And then I have the log of this. And now observe that this is an integral of n variables, but you can separate the variables. The exponential of a sum is the product of the exponentials. And in fact, it's exponential minus n zeta of x dx to the power n. And this, you can compute. When n goes to infinity, it's going to behave like. So this is what makes the integral converge. So you have to assume that v grows sufficiently fast at infinity so that zeta grows sufficiently fast so that these integrals converge. And then this, just by monotone convergence or whatever you want, this converges to the volume of the set, zeta equals 0. So you have to remember here that zeta was this function which is obtained from the equilibrium measure. It's always non-negative, and it's 0 in the support of the coincidence set, let's say. So when zeta is positive, this goes to 0 point-wise. And when zeta is 0, you recover the volume of this set. So let's say by monotone convergence. And so when you take the log to the power n of this, this is going to give you a term of order n. Of course, I had forgotten some terms here. So I have a c beta n, essentially. OK, so this goes into the order n term. And we have concluded a bound from above of the partition function, n beta over 2d log n plus c, which possibly depends on beta, times n. And in fact, this is good. This is the right behavior. So what is much harder is to obtain a lower bound. You want to obtain a corresponding lower bound with maybe a different constant. So if you think of how you obtain an upper bound, well, you bounded the energy from below. So to obtain a lower bound, you will have to bound the energy from above. But OK, the energy is not always bounded from above. If you think of the energy, it's this Coulomb interaction. If you have configurations of points where two points are very close to each other, just two points, they can make the energy go to infinity. So what you have to show is that there is enough configurations with bounded energy, enough in the sense of the volume. When you integrate here over configurations, you want to show that good configurations create enough volume where you can construct a large enough volume of configurations for which you have a corresponding upper bound on the energy. And you can actually do that. It's a little, there's some work to do. And in the last lecture, I will try to explain how to do that. And in fact, what we will obtain in the last lecture, it's not only that you have a lower bound with another constant, it's that you have an explicit value for this constant, or not explicit, but you have the existence of an asymptotic in n. So there exists a constant C such that log of the n is equal to all these blah, blah, blah, all these leading order terms, which are actually not so interesting, plus a constant C beta plus little o of n. So there exists such an expansion with a constant C beta that depends on beta, and it can be characterized. And in fact, this expansion will come together with a large deviations principle. So there will be a large deviations rate function for which this constant C beta is the minimum of the rate function. When you get a large deviation principle, you actually get an expansion of the partition function with it. So as I said, this thing is actually quite optimal. And you can work at getting the constant and how it depends on v, et cetera. These results, of course, as I said before, in 1D, it was well known by other techniques in 2D and higher for general beta. This works well, and this works in these Coulomb cases. With larger dimensions provided, you normalize things properly. All right, so let's now assume that we have such an expansion, and I will try to see how the consequences we can get from that. So another fact that is interesting is that this energy Fn, you see this next-order energy, controls the fluctuations. So the idea is that this energy, and you see it with the electric formulation, this energy has good coercivity properties. So we have a proposition, which is quite easy to prove, which is that if phi is a test function, lip sheets, let's say, supported in some set, let's say u, and you integrate phi against the fluctuation associated to some configuration, then you can control this in terms of the energy plus follows. So you have the lip sheets norm of phi, and then you have the square root of the volume of the support, and then you have the energy of the truncated fields. So we don't have enough space here, but this would be l2 of u plus n to the power 1 minus 1 over d. So this is true for every choice of eta, such that eta i is less than n to the minus 1 over d, which you remember as this typical lens scale. And so you see this proof, as I was saying, that the fluctuations are well-controlled if you control these electric energy terms. And not only that, but if you want to understand the fluctuation in a set u, it suffices to understand the part of the energy that lives in u. And this will be important when you want to localize things. If you want to look at the small set, you don't have to understand what's going on far from it. This energy expressed in terms of the electric potential has become local somehow. So proving that is straightforward. Let me remind you that we had minus Laplacian hn eta equals cd times the sum of delta xi eta i minus n mu v. So this is almost the fluctuation. So when I integrate phi against this guy, which is almost the fluctuation, except I replace the Dirac by the Smirred Dirac, I can write this as 1 over cd with a minus sign, the integral against the Laplacian. And then I can integrate by parts. If phi is compactly supported, this becomes the grow of grad phi, grad h. And then use Cauchy-Schwarz. You bound by the L infinity norm, constant. And then you get the L2 norm. And then you get the square root of the volume of the support. So this is almost the answer. Now you have to correct for one mistake, which is that you replace the Dirac by the Smirred Dirac. OK, what error does this create? Well, it's equal to this. How can I bound this? Well, these are two measures of supports that are distant by eta i from each other. And they have the same mass. On the other hand, phi is Lipschitz. So this is bounded by the Lipschitz norm times eta i. And now if I use that eta i is bounded by n to the minus 1 over d. And I multiply by the number of such terms, which is n. I get n to the 1 minus 1 over d. This is actually a small term. So the dominant term is this one. The idea is that these terms now, this gradient h n eta, over a set u, we expect this to be proportional to n times the volume of u. So this is what I was saying. The energy scales like the volume. And so this gives you an estimate for the fluctuation if you are able to control these energies. But these energies, you control them because they're controlled by f n. You have to remember that the previous proposition told you that, at least if I look in order d, this is bounded above by f n plus the sum of g of eta i. So I have a 1 over c d. I had plus errors, but the previous proposition, if I rewrite it slightly differently, it bounds this by this plus that. And so it's all about controlling f n. But the idea is that we do control f n. Since f n is up to constant factors, it is the energy. So we control its exponential moments. And so we control the exponential moments of the fluctuations. So we'll show you this in what second. All right, so maybe I can say something before that. I was saying that this thing, you expect it to behave like f n. We expect it to be like n, typically. Because this is what we have seen when we have expanded and obtained these relations of the partition function. So the typical f n is of order n. Of course, f n could be much larger, but this would have a small probability. So if typically you have an order n, it means that for the L2 norm here, you have square root of n. So this becomes typically of order square root of n. Because this is the L2 norm. And over there, I have the square of the L2 norm. Of course, I take f n to be of order n. And I take eta i to be n to the minus 1 over d. And so I put my terms together. So in fact, f n is not only n. It's c n plus or minus n over d log n. But this term always gets subtracted from every quantity. So what comes out here is square root of n. And so I want to point out that this is already an improvement compared to what we started with. I told you the idea is that we want to obtain stronger results on the bounds and fluctuations. We know the convergence of the empirical measures to the equilibrium measure. So we know that this thing is much less than n. But now we find that if we test it against Lipschitz functions, the norm of this thing in the dual of Lipschitz functions becomes less than square root of n with large probability. And so this is already an improvement, of course, from n to square root of n. What we will see tomorrow is that if you look at even smoother test functions, you can improve to c. But OK, this is already a nice bound. And so now I was going to talk about controlling exponential moments. So it's to make what I just said a little more rigorous. So what we have is this. If you take the expectation under the probability pn beta, of course, of exponential beta over 4 fn plus n over d log n. So this is the quantities that should always go together in the logarithmic case, at least. Then you do control this by n. So this is the correct meaning to give to what I was saying before, which is that we expect this thing to be typically of order this minus this. Well, we have it in exponential moments. And then you can apply Markov's inequality to say that once you have this, the probability that this thing is much larger than n will decay at a certain rate. OK, so how do we control these exponential moments? We remember that we are trying to compute here. This expectation is equal to 1 over z exponential beta over 4 fn plus n over d log n. OK, so this is what we're trying to compute. And so we have to multiply by exponential minus beta over 2 hn. But this guy is nothing else but constants. So this n squared i v of mu v plus n over 2 d log n plus fn plus 2n sum of zeta xi. OK, so if we put everything together, the leading order terms here will come out of the zn. So we'll have exponential minus beta. This will go with zn. And then what you find is, essentially, you have a beta over 4 fn. And here you have a minus beta over 2 fn. So it looks very much like minus beta over 4 fn plus these other terms that are meant to do what they should do. So there's going to be a minus beta over 4 d n log n plus 2 n. So minus n beta sum of zeta of xi dxn. OK, so this whole computation, to try to show simply that when you compute this expectation, you're just computing effectively the partition function for a different temperature. So for a temperature beta over 2 instead of beta. So you have just divided beta by 2. But otherwise, you get all the same terms. So this is zn beta. And this is roughly zn beta over 2 multiplied by some factors which are involving these n log n. And so if you believe in the expansion that I gave you here, if you believe in an expansion like this, if you plug it into here, you're going to see that all the leading order terms cancel out. And what comes out is an order n. And this gives this. So you control exponential moments. And it gives you controls on exponential moments of the fluctuations as well. So for example, you get log in the same way by applying this bound, the bound that controls the fluctuations by fn, essentially. And so this gives you an estimate on deviations, what's called moderate deviations. The probability that the fluctuation will be much larger than expected can be quantified like this in bounded above. So tomorrow I will show you how to get the improvement to see here in a central limit theorem for the fluctuations. And in the last lecture, I will talk about obtaining an expansion like this and the large deviations principle for the microscopic configurations. Thank you. Nothing to be tiring.