 as the integral over the set A for all the particles of my density function. That's here. And this is a standard trickery of the Andreev formula to move these integrals inside the determinant. And I can write this in a compact form like this. So I can write each term of this big sum in a compact form like this. So just determinant of, and then in each entry, there's an integral q to the sum power, and then integral of these Hermite functions. OK, so now I have a big sum of determinants that look like this. It looks a little bit better. And if I introduce an operator, m, which is an integral operator whose kernel is an infinite sum of Hermite functions weighted by q. So that's a relatively simple operator. I'll take that operator, and I'll expand its Fredholm determinant, determinant of 1 plus z times m times the indicator function on A. What I see when I expand that Fredholm determinant is exactly these terms, which are written in the sum, which is written up here. So the observation is that the expansion of this Fredholm determinant in z is exactly given by 1 plus z. I need to cancel this q poc camera symbol, so I'll divide by that. But times the probability that in the model with one particle, that particle is in A. And in the next term, in the model with two particles, both of them are in A, et cetera. So OK, those are exactly what I'm looking for. Those are the gap probabilities. And so if I want to know the gap probability for n particles, I just need to pick out the nth term in that sum, which you can do by integrating around the origin. So I integrate around the origin dz divided by z to the n plus 1. That'll pick out the nth term. So that's a formula for the gap probability. That's not the one that I had on the earlier slide. So it turns out this one somehow is not so convenient for asymptotic analysis. So the formula that I had on the board involved a different kernel. It's similar. It's still an infinite sum of these Hermite functions, but weighted by something a little more complicated. So how can I go from m to k? Well, the observation that we made was that if I take, actually, this is an identity of operators. So I only care about the determinant, but this is really an identity of operators. If I take this operator, i plus z times m times a characteristic function on A, that's the same as i plus z times m on the whole real line, no characteristic function, times i minus k on the complement of A. So you multiply that out. The proof isn't too hard to see that this is an equality of operators. And then if you have an equality of operators, then the determinants are the same. Now the determinant, which is i plus z times m on the whole real line, is very simple to compute, because I'll go back to m. Yeah, here's m. So if I take i plus z times m on the whole real line, it's diagonalized by these Hermite functions, exactly. And that determinant is exactly a product of 1 plus z times q to the k, which is this Q-POP camera symbol. So this is what we do. We just take this determinant here, replace it with this one down here. And that turns out to give us a formula, which is nicer for asymptotic analysis. Now that was a relatively simple observation. But then Dong noticed that since similar formulas are showing up in these McDonald processes and other types of interacting particle systems, maybe this kind of identity of Fredholm determinants could be useful in other models. So let me generalize that identity of Fredholm determinants. So I'm going to maybe switch gears, not talk about the free fermion model and talk a little abstractly about these Fredholm determinants and operators. So OK, suppose I have, before I had Hermite functions, but suppose I have any bi-orthogonal system on some contour in the complex plane. So before my contour was just the real line. But I can take any contour, I'll call it gamma. So phi's and psi's are orthogonal on this contour, bi-orthogonal on this contour. And now let me define Kernel's analogous to what we saw before. So m is just the sum of phi times psi weighted by q. And k is the sum of phi times psi weighted by this rational expression involving q and z. OK, and there's corresponding integral operators, bold m and bold k. Then we have exactly the same result as we had before. So if I take gamma and I cut it into two pieces, gamma 1 and gamma 2, then the result is that the Fredholm, sorry, this is operator identity. So the operator i plus z times m, probably, my notation is bad here. I'm sorry. When I'm talking about operators, I should write a characteristic function there. So replace that L2 with a characteristic function, psi, kai. So i plus z times m on the first piece, gamma 1, is the same as i plus z on the big piece, gamma times i minus k on the remaining piece, gamma 2. So that's, again, an identity of operators. And if we take determinants, then I find the determinant of i plus z on L2 of the one contour is the same as i plus z times m on the big contour, i minus k on the remaining contour. And again, since I have a bi-orthogonal system, the determinant on the big contour is easy. It's exactly the same Kuhl-Pakamer infinite product. And then I'm left with a determinant on the other contour. So this gives us some kind of a way to change if I have a determinant on one contour, and maybe I don't like that contour very much, then maybe I can switch it to another one. So as an example, I can consider the Q-tasep. So that's an interacting particle system, which particles are on the integer lattice, and they can jump to the right at random. And the rate at which they jump is determined by the gap between particles. So if there's a big gap, then they tend to jump faster. If there's a small gap, they don't jump. If there's no gap, then they can't jump at all. So there's that exclusion. OK, whoops. So that's one of the models that showed up in Corwin and Borden's McDonald process paper. And they give several formulas for this certain observable in the model, which is some kind of a Q moment generating function for the position of the nth particle. So actually, this formula I read from a later paper with Sasamoto. So this is their formula. So we're interested in the location of the nth particle. That's the sum, I should say. The particles are starting all on the negative part of the real line. So they're packed. So this is step initial condition. And nth is the nth from the origin. OK, so this is their formula. So it has this two-pot camera type factor. And then there's a determinant, which I can write as i plus z times m. And the contour is any contour which encloses n positive numbers here, which are the jump rates for the first n particles. So the jump rates depend on the distance, but they also depend on some independent parameter. So we'll call those a1 through an. And this contour is supposed to enclose 0 and also all of these positive numbers, a. And the kernel for m, OK, it looks a little complicated, maybe, but it can be written as some function of the first variable, psi, divided by psi minus q times the second variable, eta. And it doesn't look much like the m that we had on a previous slide. But if I assume that the ratio of the psi and the eta is smaller than 1 over q, which is true for certain contours, then you can expand this as a geometric series. And I can write, if I expand as a geometric series, I get an expansion like this. So there's a q to the k. There's a phi of psi. There's a psi of eta. And I just have to pick these phi and psi to be appropriate things. So I picked the phi to be exactly that f times an appropriate power of psi. And the psi function is just a power of eta. So the contour that we're talking about is the big one here, the big solid one on the outside, which encloses all of the a's and also 0. Then it's not so hard to see. This function, f, satisfies f of 0 is 1, which means that the psi and the phi functions actually do form a birethogonal system on that contour. Just the Cauchy theorem tells you that. So we have a kernel m, which can be written in the form that we like, expansion of birethogonal system, weighted by q. So we can try to apply our theorem and see if it helps. So I need to think about this big contour as part of an even bigger contour on which the phi and the psi are orthogonal. So what I can do is take the big one and add to it so I can think about the dotted contour, I guess. So the dotted contour has a big contour, and then it has another dotted contour on the inside, which is the opposite orientation. So what do those do? They enclose the a's, they will just enclose 0. And that's exactly what I'll call gamma. So gamma is the dotted contour here. And then gamma decomposes into two parts. Gamma 1 is the, sorry, this is the notation I use. Gamma 0 comma a is the big one that encloses 0 and a. And gamma just a is the one that just encloses the a's, these positive numbers. So now we just apply this identity of Fredholm determinants. The original determinant was this one, i plus z times m on the contour enclosing everything. Now, all right, so that should be the same as the determinant of i plus z times m on the union of these two contours, which are oriented in opposite directions, times the determinant of i minus k on the smaller contour that only encloses the a's. And once again, I get my pock hammer. I get then, OK, I reverse the orientation, so i minus k becomes i plus k. And I have a different kernel now and a different contour. And so this gives us an alternative formulation of this moment generating function. And what's this k now? Well, I just use my formula for the k. So it can be expanded in the phi and the psi with these pre-factors. And you can notice, if you look at some books on hypergeometric functions, that this is actually a basic hypergeometric function. So there's a explicit pre-factor times basic hypergeometric function with certain values for the parameters. So that's nice. But what's really nice is that there's a contour integral formula for that hypergeometric function. So in the end, we have a kernel that looks something like this. There's a very similar kernel for the same object, which appears in the Borden-Corwin McDonald Processes paper. But it's not exactly the same. And it's not clear to me if this might be easier for asymptotic analysis. Now, in this particular example, the asymptotic analysis has been done, probably by people sitting in this room. But the point was just to show how this simple identity can give different formulas for these determinant type formulas, which in some cases may be more useful than what's out there already. So the process is very similar for at least what's called the Q Whitaker processes and possibly beyond. So I'm not sure. OK, so what now? So one thing that we're asking is, is this Fred Holm determinant identity useful for asymptotics in any other models? I think the answer is yes, but I have to look and see. Back to the Free Fermion model, we gave results in one dimension. You can ask about higher dimensions, which was done in the physical setting by the DDMS group last year. But I don't know of any other mathematical results. We do, I think, lose contact with random matrix theory in higher dimensions. I don't think there's any random matrix interpretation of, say, two-dimensional free fermions. What else? In these canonical ensemble and the grand canonical ensemble of this MNS model, we saw that the local statistics are the same. So somehow in the larger limit, the local statistics are the same, which is to be expected. But it's another question. What about non-local statistics? So for the grand canonical ensemble, there's recent work by Johansson and Lambert. And they found, so they considered temperature as a parameter, but also the scaling for the linear statistics. So in some mesoscopic scaling, they found that there's some regime. You can tune those two parameters, temperature and the mesoscopic scaling, in which there is non-Gaussian fluctuations for the linear statistics. So at low temperature, they find just random matrix-type linear statistics. At high temperature, they find the usual central limit theorem. But in between, there seems to be something non-Gaussian. And as I understand, it doesn't seem to be a universal transition between those two. But you could still ask, maybe it's not universal distribution, but is there anything universal about it? Are there any universal properties to it? So they did linear statistics in the grand canonical ensemble and actually a little more general. What about in the canonical ensemble? So there's less results that I know of. There's a recent paper of Yasek Greila, Majumdar and Cher, in which they considered some specific type of linear statistics. But for the canonical ensemble, there's still lots of open questions. So in particular, should we expect that the canonical and grand canonical ensembles will agree in the linear statistics? And then finally, just a very recent paper, which was on the archive two weeks ago by Kundin, Mazzadri, and O'Connell, they gave some interpretation of free fermions on an interval with various symmetries. They interpreted it as finite temperature version of the random matrices associated with classical compact groups. So that's another direction. I'm not sure what to do with that, but it's very interesting, and I just wanted to point it out. So that's all. Thank you very much. Questions? So I have a question. For this crossover distribution that you got, for the largest one, what sort of tail estimates do you know, or can you use the matrix models to establish some? Oh, good question. I don't know, maybe Satya here. He probably knows better. Do you know the tails of this crossover distribution? Yeah. Other questions? OK, so let's thank Carl again. And we'll start in a few minutes, three minutes.