 Okay. Thank you for coming. Thank you for the invitation. I know it's right after lunch. I also know that it's a kind of a diverse audience, so, and it's also a big room. So, but don't feel shy about the raising your hand and asking questions because my subject is a little bit off topic. But there are some random matrices hiding in there. Let's see if you can find them. All right. This is going to be joint work with Yanda Heer from Melbourne and Sam Watson. Who's that, Brown? And it's a stat mech model. And let me just start with what you might say is the sort of granddaddy of all stat mech models, which is the granddaddy in the sense of it's one of the oldest ones, but also very simple and still not solved in full generality. So this is the six vertex model or square ice model if you like. So you consider the grid, square grid here in two dimensions. And I'm just going to direct each arrow, direct each edge by putting an arrow on it, left or right for the horizontal and up or down for the verticals. And the rule is that at each vertex, I want to have two incoming arrows and two outgoing arrows, right? It's a very simple rule. And then, so my space of configurations is a finite piece of this graph with these arrow configurations on it. It's a finite space of configurations. And I'm going to put a measure on that space. The measure is giving a configuration weight, a probability proportional to its weight, which is the product over all the different local types of vertices of some weight. For example, if you look at all, there's sort of six different local configurations at a given vertex. Four choose two, if you think about the different ways that outgoing arrows can point, right? So there's two outgoing arrows from each vertex. There's six possible configurations locally. And we give them weights a1 through a6. And when you have a big configuration like this, you look at all the vertices of type a1, all the vertices of type a2 and so on. And the weight of that configuration is a1 to the number of vertices of type a1 and so on. And then you turn that into a probability measure by dividing by some normalizing constant to get a probability measure on this finite state space. Okay, so it's a very simple model. Why do we really care about this model? That's another question. In some sense it's a very simple model and it's called the square ice model. If you put all the weights equal to each other, you can think of each vertex as having an oxygen molecule and the arrows pointing in are like two hydrogen molecules, hydrogen atoms, oxygen atom and two hydrogen atoms. So there's some version of square ice and everybody's interested in ice, especially on a hot day like this, so why not study this model? What kind of questions are we going to ask? Well, lots of questions. If I have a particular finite piece with some particular boundary conditions like I suppose I want the square and all the arrows pointing up here and down there and left there and right there. And I take a random configuration with respect to this measure, what does it look like inside? Those are the kind of questions which we're going to be considering. And another question, even simpler question is if you put this model on a torus, suppose I have periodic boundary conditions, then how many configurations are there or rather what's the total weight of all configurations? So the log of that quantity is called the free energy and this was computed actually many years ago, 50 years ago by Lieb in some special cases. The cases when these two weights are equal, these two weights are equal, those two weights are equal. There's some three dimensional sub-variety of the space of all weights. Any questions about the model? Now's your chance because this is the beginning of the lecture. Don't be shy. Another case which, oh yes, please. Oh yeah, okay. You have a finite graph and you count so the weight, so the probability of a configuration is one over a constant times a1 to the number of local vertices of type a1 times a2 to the na2 up to a6 to the na6. Right, where na1 is the number of vertices of type, the first type listed above there. Does that answer your question? And the free energy is essentially the log of this quantity z here, this normalizing constant, the sum over all configurations of their weight. Okay, another case which has been solved is this so-called free fermion case and I won't explain why we use that terminology but it's another sub-space of the set of all weights for which there are explicit solutions using determinants and linear algebra. Okay, so another way to think about this model by the way is down here. If I erase all the arrows which are pointing downward or pointing to the left, then I get the local configurations look like this and I can think of my model as a sequence of lattice paths which are on the lattice which sort of are northeast going up and to the right which are disjoint except that they can sort of bump into each other because there's a special weight a2. That's just another way to represent the model in a different point of view. Alright, so unfortunately we still can't do that for the model in full generality but we are going to consider a different five-dimensional subset, sub-space, sub-variety of the space of all weights which we called quite unimaginatively the five vertex model but it's a special case of the six vertex model and it's a generalization of another model which is an example which is a special case of the free fermion model but a model which has been talked about already the Lawson-Scheiling model. So what is our model? You just take here the six possible configurations again and we're just going to throw this one away by giving it weight zero. Okay, so the paths are not allowed to bump into each other anymore but other than that, that's the only restriction so I have a five-dimensional space of weights but if I scale all the weights by a constant the measure doesn't change so I can be free to set the first weight here to be one without loss of generality and there's another hidden symmetry here which is these two weights, well if you think about it in terms of the lattice paths every time you turn right you're going to have to turn left eventually so you can pair up the left turns and the right turns and so it doesn't hurt to give these last two configurations the same weight. So even though it looks like it has five different parameters in reality there's only three parameters which I call and I chose this weight to be e to the x that one e to the y and the last one to be while r times e to the x plus y over two the last two. So there's three parameters x, y and r left over and right so I'm just repeating the definition the configuration has a certain probability one over z times the r to the c where c is the number of corners but now e to the vx plus h y where v is the number of vertical edges so if you notice how I defined it this configuration gets an e to the x and that one gets an e to the x over two so the net number of x's there are, well I get one e to the x for every vertical edge in the model and one e to the y for every horizontal edge. And so here's a simulation of the case, well when x and y are zero and r equals one in this case all five configurations are equally, all five local configurations are equally likely. And that's with the toroidal boundary conditions. Alright, well what's the Lawson's Tiling Model? It's this model of random tilings with these three types of tiles, the red, the green and the blue and here's a little simulation of the Lawson's Tiling Model just think about the, for example the uniform weight on all possible tilings of a torus with these three kinds of tiles. And you can see that actually these models are quite similar in fact they're exactly the same when that quantity r is one. You can think about these chains formed from blue and green Lawson's and they look just like these lattice paths here, the northeast lattice paths. They don't intersect each other and so on. And well, that's the r equals one case. Remember r is the weight per corner here, r is kind of counting the number of corners. And so if you're familiar with the Lawson's Tiling Model then you can think of the r not equal to one case as sort of an interacting Lawson's Tiling Model where every blue Lawson's, every time you have a blue and a green Lawson's adjacent to each other you get an extra factor of r. So if r is less than one then you want to have fewer corners so that these paths want to go straight. If r is bigger than one you want to have more corners. Okay, well these two parameters x and y they, well if you're a physicist they play what you call the role of what you call a magnetic field. Right, if you, x is in some sense telling you that your preference for going vertically and y is your preference for going horizontally. If x is large then configurations with lots of vertical edges will have large weight and therefore large probability, similarly for y. If x is small then you're sort of, you know, you don't want to have lots of vertical edges. So as you vary x and y, what's going on in the model is you're varying the number of these lattice lines. Right, if x and y are both large then you want to have lots of lines going both vertically and horizontally. If they're both small then you want to have very few lines. If you go back to this case here, you know, most of the picture here is red because I took x and y to be small, I mean or negative I should say. So that, you know, these blue-green lines they're not, they're not as frequent as they would be in the uniform model. Okay, so there's this somewhat interesting relationship between these two parameters x and y and the density of lines which I'm calling, well the density is determined by two real numbers, s and t. S is like the horizontal density of lines, it's the number of lines per unit segment horizontally and t is the number of lines per unit segment vertically. Right, if you increase x, you increase s, if you increase y, you increase t. Right, does that make sense? But this, it's only a, there is sort of one-to-one correspondence between x, y and s, t but the actual, you know, quantitative relationship between x, y and s, t is not at all obvious. Right, if I double x, you know, if I add one to x, what happens to s? You know, it depends on everything else. So that is the sort of fundamental relationship which we need to find because that will actually tell us the quantity, the first, the most important quantity about the model which is the free energy. In fact, there's a relationship between the free energy, I mean the derivative, the free energy is a function of the weights x, y and r, r, you should think of a subscript r here and the gradient with respect to these x and y variables is in fact the s and t variables. So if I happen to know what s and t corresponds to a particular value of x and y for a fixed r then I would be able to integrate that to get f. There's an interesting sort of, well, this is sort of a general relation for models of this type. Once I know x and y, if I know the relationship between x, y and s, t, I can get f. Conversely, if I know f then I can get s, t from x and y by this equation. Yes, well you can integrate up to a, you can get the, that's right, you need to find at least one value somewhere, you're right. But in some sense we don't care about any, any single value, we care about the functional relationship as x and y varies. Okay, so let me just tell you the answer in this case, the easy case, which is r equals 1. Right, that's just the uniform, well it's the Lawson's Tying Case with these weights x and y. Right, and here there's a very simple relationship between x, y and s and s, t. But it's, I said simple and mysterious because it is mysterious. And the relationship is the following. You take a triangle with edge lengths 1, e to the x and e to the y, there's three positive numbers. And you look at these two angles and the two angles are s and t, pi times s and pi times t. Right, that, well, like I said it's mysterious. I don't have any, I can't prove that to you in the, in the five minutes or even 15 minutes. Right, it would take some time. Y is so simple, I don't know. But, you know, you can notice certain things like if, if x is very large and y is not very large, then there is no such triangle. Right, you have to satisfy the triangle inequality. If x and y are too small, then what happens is this triangle sort of flattens out and s and t go to zero, which means you have no lattice pass at all. Okay, so you can, you can figure out various interesting things just from this relationship. But, you know, once you have that relationship, you know, it's just a matter of trigonometry or something to, to figure out the relationship between x, y and s and t, and you can compute f that way. f is this free energy. And in fact, while I'm at it, s and t naturally live in a triangle. Right, the horizontal density goes from zero to one, and the vertical density goes from zero to one, the sum of the two densities can't be larger than one because, you know, you can't, you can't fill up more than the lattice. They have to be disjoint. The two, the lattice pass all have to be disjoint from each other, which makes that the s and t varies over this unit triangle. And in fact, the x and y can be any real numbers in the whole plane. But if x is too big, then this triangle sort of flattens out completely and you get a degenerate measure over here where s is one and t is zero. So this entire region over here maps under this gradient to that vertex and this region maps to there and this region maps to there. But there's an interesting region in here in this amoeba shape that maps to sort of injectively onto the interior of the triangle. So that's this nice relationship. That's the case r equals one. Right, and part of the goal of the talk is to at least explain what the relationship is when r is not equal to one, is some generalization of this fact. Are we good so far? While I'm at it, this is still in the case r equals one. There was this other two. Let me go back one slide. Given x and y, I can tell you s and t, it's an invertible map and the inverse is also the gradient of some function sigma called the surface tension that is essentially measuring how much the, you know, if you think of this Lawson's Childing Model as a surface in r3, it has a certain, the growth rate of the number of surfaces as a function of the slope, s and t is like the slope of that surface in r3. The growth rate as a function of the slope is this number sigma, is this quantity sigma. Okay, and you know, if you do that integral, which Gerard asked about, and you happen to know the value somewhere like on the boundary, then you can compute it everywhere and well, there it is. The funny thing is, or the interesting thing is that that integral does not give you some elementary function. It involves a dialog and some variant of the dialog rhythm function if you know what that is. But anyway, there's a plot of that function on the, so it's a function, here it's a function of s and t, which gives you the surface tension as a function of s and t. It's a nice concave function. And you can see that it's sort of minimized at 1 third comma 1 third, that's the uniform measure. The uniform measure is the one where the growth rate is largest. And therefore the surface tension is minimal because that's where the system is happiest is being in the 1 third, 1 third, 1 third slope. Okay, so let, but now let's go to the general case, the case of general R. And, right, so the question, the fundamental question is how do we find the free energy, right? And unlike the Lawson's case, we don't have nice determinant formulas. I didn't tell you about how to do the Lawson's Tiling Case, but it's much easier than this one. There's no determinant formulas, but we need, but we can go back to Lieb's method, which is, in fact, older than his work, but it goes back to, I guess, Hans-Beta, the beta ansatz. And so what we're going to do is put the model on a cylinder, maybe I'll draw a picture, think about a square grid on a cylinder, on an infinite cylinder like that, circumference, you know, n, and sort of bi-infinite cylinder. And we're just interested in counting. We're summing the weights of all configurations. So we're going to define a matrix which tells us, given a configuration on a given row, how many configurations does it give rise to on the next row? So our matrix is going to be indexed by configurations on a given row. So that's this matrix T called the transfer matrix. And if our circumference is n, oops, we got too hot. If our circumference is n, then, you know, if you think about what a configuration looks like, it has some paths coming in, and at the next row up here, it's going to have some paths coming out, which are connected like this. So the state of the transfer matrix is a configuration on this row. So it's a subset of vertical edges on a given row. So this matrix T is a 2 to the n by 2 to the n matrix indexed by all possible configurations on a given row. And the entries are just the, if I have a given configuration on this row and a given configuration on this row, I need to know if they are connected or not, and if they are connected, what's the internal weight of the vertices involving that connection? Those are the entries in the matrix T. So what's the point of T? If you take powers of T, this counts the number of configurations starting at a given configuration. If I start at some initial configurations, E, EI, and EJ, this is the number of configurations. There's a total weight of configurations starting in state I here and ending n layers higher in state J. So the goal to compute the free energy, it suffices to understand the powers of T and in particular, all we need is the leading eigenvector of T. So we've reduced the problem, yes? We've reduced the problem to a linear algebra problem. We have this matrix, find the largest eigenvector and eigenvalue. Find the, well, we need both eigenvector and eigenvalue, but the free energy is just given by the leading eigenvector, sorry, eigenvalue of this matrix. Okay. How are we doing? It's coming up. Well, we're, you should not be happy yet. You shouldn't be jumping up and down yet because it's a 2 to the n by 2 to the n matrix, right? That's pretty large, right? Even when n is 10. And we're interested in the case when n is large. Well, there's a sort of a partial diagonalization of this matrix into blocks, right? T is a big matrix, but there are in fact blocks according to the number of particles, the number of paths, right? In this case, if there are three paths coming in on one layer, then there has to be three paths coming out on the next layer. So there's this, for each k, where k is the number of particles, there's a sub-matrix, an invariant subspace, and let's call Tk the matrix on that subspace. So Tk is the transfer matrix for k particles, and it's only n choose k by n choose k, okay? So that's a slight improvement. But the key observation of beta is that, well, there's an explicit form for the eigenvectors for all these little matrices. And so let me just run through this briefly in some small cases, like T1. What is T1? T1 means there's one particle. So if you have one particle coming in and one particle coming out, well, the graph has this, the matrix has this circular symmetry because the cylinder has this circular symmetry, and which means that the eigenvector, it's a circulant matrix. The eigenvectors have to be, you know, exponentials, e to the i 2 pi j over n, where n is the circumference. And if you know the eigenvectors, it's easy to compute the eigenvalue, okay? So for the one-by-one matrix, it's completely trivial to diagonalize that matrix. But for the two-by-two matrix, it's already much harder. But the kind of amazing fact is that the eigenvectors, but this formula kind of generalizes, the eigenvectors have this nice form that has a sum of two exponentials. Here, this is my eigenvector, which has two parameters, zeta 1 and zeta 2, and x1 and x2 are the positions of the particles coming on a given row. So this is x1, that's x2, and this is, you know, the output. And as a function of x1 and x2, the eigenvector looks like this. It's some constant, zeta 1 to the x1, zeta 2 to the x2, plus another constant, zeta 1 to the x2, zeta 2 to the x1. You just switch x1 and x2. Well, maybe I will skip the reason that we should expect something of that form. Well, it's basically the fact that the state space is like a vial chamber. You know, you have this set of pairs, x1, x2, where x1 is less than x2, and you can think of it as a random walk on some higher-dimensional, some higher-dimensional triangle. Anyway, but the fact is that for Tk, when you have k particles, this formula generalizes, and the eigenvectors always have this nice form, well, nice in quotes, right? It's still kind of complicated. It's a sum of a bunch of wave of exponentials. But in this case, there's k-factorial different exponentials, which it's maybe convenient to write it in this form. There's this matrix, this sort of van der Maan-like matrix, where the entries are zeta i to the xj. And then we have this thing, which is not a determinant, even though it looks like it's a determinant sub a. Well, because these coefficients here, when you sum over the symmetric group, the coefficients are not just minus 1 to the signature. They're not just the signature of the permutation. They're some other function on the permutation group. They may pi as some arbitrary function on sk. And then when you make this sum, let's define that to be determinant sub a. So that's the form of the general eigenvectors for Tk, for any k. Of course, if a pi happened to be something nice, like minus 1 to the pi, then this would be an actual determinant, and we'd be in business. But it's not. We'll get there. It's not, but there's some manipulation you can do to make it look like a determinant. Okay, what about these constants? So I have to tell you what a's are, and also the zetas. And it turns out that the zetas satisfy some nice equations. While nice is in the eye of the beholder here, these zetas are called beta roots, and they satisfy a certain system of polynomial equations. Here we're on the n by n space, little n. And so there's little n of these zeta i's, and they satisfy this nasty actually looking system of polynomial equations, rational equations. But here's where the six vertex model and the five vertex model differ, because in the five vertex model, here I wrote down the equation for the five vertex model, and you see in the denominator here, there's no j. This is a product over j, there's no j in the denominator, so I can take this denominator and just scoot it over to the other side, and then I get a function of i over here and a product over all the j's of some polynomial there. And probably this is not important, this is going too fast, but let me just redefine these new numbers, w sub j. These are complex numbers, which are just scaled versions of the zetas, the beta roots. And then the equation looks like this. So when you move this denominator over, it looks as simplified. So here we get an interesting polynomial, w i to some power, 1 minus w i to some power equals this thing. So here's the interesting observation. This is a product over all the roots that we care about. And in particular, it's a symmetric function of all the w j's. So we might as well think of that as a constant, which depends on our parameters, x, y, and r. For example, you know, I suppose my two constants were 12 and 4, then I essentially, to get the roots, these beta roots, I just need to solve some single polynomial equation. w to some power, 1 minus w to some power equals constant. So these roots turn out to lie on some nice friendly curves in the plane called Cassini ovals. Cassini ovals are the locus of points such that the sum of the logs of the distances to 1 and 0 is a constant, like 1. So it's kind of like an ellipse. An ellipse has a property that you sum the distances to two points and is a constant. Here you sum the logs of the distance to two points and you get a Cassini oval. And they look like that. The red, blue, and green here are three different Cassini ovals with the same ratio of alpha over beta, the coefficients here. And so, you know, you should think of these two numbers, 12 and 4, as being very large, because we want large in the limit of large system size. These two things, these two exponents are going to go to infinity, but their ratio is going to be constant. The ratio depends on the density, the horizontal density or something like that. Okay, so, you know, think of, as you think at large, the curves are not going to change. It's just that the roots are going to become more and more dense on these curves. And this constant is a function of our input parameters, x, y, and r. And you can see something interesting happening, that, you know, when this constant is small, there's two ovals. And when it's large, there's one oval. They sort of merge. And in fact, there's one case, which is that curve. Who knows what that curve is called? Well, it's a Lemnis gate. The Lemnis gate is actually the symmetric version, but I guess you can call it a generalized Lemnis gate. It's one of those classical algebraic geometry curves. There's no clock in here. How much time is there? Okay. Oh yeah, there's an orange line. What is that orange line? Well, when these two numbers, 12 and 4, get large, the sum of these two numbers is capital N, the system size, and the smaller number is the little n. It's the number of paths. And that orange line separates the little n roots from the rest of the roots. So the roots we care about are the ones that are near one here. And notice that when this constant is small, there's exactly four roots, the little n roots on this guy and the other roots are on that guy. But when it gets larger, so this orange line separates the little n roots from all the big n roots there. The roots we really care about are the ones on that side of the line. Okay. And I might as well just show it up, put it up here. The leading eigenvalue is some function of these...