 So the result of yesterday's lecture is that we have this result of boredom for our Prof. in Sasamoto, that if you take the TASAP particles, so XTNJ is just the NJF particle. Remember, there's a rightmost one in there bordered backwards, right? And so you start with some initial configuration of such things. And now if you want to know where they are at time t, just m of them, maybe m is 1, then is given by this Fredholm determinant, which is now I've written in the path integral form, which we like. Okay, and and there's a kernel. Actually, you don't need the whole kernel for this thing because this KTNM is actually with the NM and NM, right? Remember that? It's the one point. Kernel. So KTNM just means this and this are the same NM. Okay, and there's this first term which would drop out in that case. But the key thing is this part of the kernel, right? And then the size are given by some contour integrals, which I've been reminded are Charley-Pollin-Omiles, and then this Q is just the the jump probabilities of a geometric random walk jumping down with mean 2, and and that conjugation which gave you that, it changes a couple of other tiny things. There's no a 2 here. Okay, that wasn't there before. And then 2 to the minus x phi is a polynomial. Okay, but the main thing is that these phi's there's a recipe for them. You know, how do you cook the thing up? Right? So it was only known in this essentially one case. There's one or two cases one can do it. The thing was so unknown that I remember last November senior person in KPZ who I won't say who it is telling me they were going to launch a huge project to do the case particle, particle, whole, whole, particle, particle, whole, whole. Which is another flat case, right? That's right. And I think you'll see in a few minutes why that would have failed. Try to find the phi's. That was the program. Okay? Yeah. Okay, and so if you remember these things were originally written as extended kernels, and then there's this mechanics of how you turn these extended kernels into the kernel just on L2 of R. If you like, there's a paper of Borden, Corwin, and Reminec where they do this fairly generally, Reminec. So extended to path integrals. So you can look there to see the general thing. And Daniel's going to give another application of that in his lecture. Okay? So so there's a nice general thing actually this this particular one doesn't fit into the general scheme and Daniel had to work pretty hard to rewrite it to make it work. Okay, so Okay, so there's a keep this one in mind. Okay. Now also we had this thing for the area process. So remember if you took half-filled initial data, right, this step initial data, and then you rescaled it. You got this area two process and the area two process had this extended kernel, but then there was always also this path integral kernel. Right? So that was last time. And as I pointed out, you could take this path integral kernel and look on a finite box and scale the increments to be very small and then it turned into some sort of hitting problem for Brownian motions. Okay, the kernels turned into that and then you can attempt to take the box to infinity and you have a very strange limit, which I won't go over, but I'll just tell you the result of all that. So that's sort of a long computation based on the fact that for the special case of step initial data that just means fully packed to the right and empty to the left. You actually have a formula for the fies which sort of came out of topless calculations. Okay, then you calculate and calculate and calculate and you get the following thing, which will look a little bit familiar. So, okay, so first of all. So this A2 hat is just the area process minus a parabola, but it's better to write it like this because then I can rescale it. So you see the area process minus a parabola is a sort of self-similar solution of the KPZ fixed point starting from narrow wedge initial data. So if you start from narrow wedge initial data, this is the self-similar solutions in time, and we ask what's the probability that there are less than some function g everywhere? So this means for all x. Okay, and so some things in here will be familiar from the first lecture. There's i minus and then you get you get some kernel. I won't be too explicit about it. It depends on t and it looks like a kind of deformed area kernel, but let's not talk about that one too much. The important thing is that thing doesn't depend on g. That's just a funny thing for the area process you get, and we'll see what it is in a second anyway. And then you get k minus t over 2 at b. G, right now I have to remind you what this Brownian scattering operator is. So there's this operator stx, which is just e to the x d squared, the heat kernel, plus t over 3 of the third derivative. Right, and it has this beautiful kernel, t to the minus of third, e to the two-thirds x cubed over t squared minus zx over t, and then some rescaled area function. Minus t to the minus of third, z plus t to the minus fourth. x squared, okay, sorry, that's inside the area function. And the z, the z is, it's a convolution kernel with this guy, okay? So the z is the variable in the convolution kernel, okay? So stx acts on functions by convolution in z, okay? And then this, you make this at the operator, remember? at bg. Tx is the u. This expectation equals b stx minus tau u. So you send a Brownian motion starting at v, and then you wait till it hits the epigraph of g. Remember, g could have a big wall, it could be, I think, and you look at the time and the position where it hits, and you evaluate s there. Now, of course, this thing looks complicated, but of course, this thing is just an integral of the hitting density of Brownian motion against this, at this function, g against this function. So I could write the whole thing out, and it's just an explicit integral against hitting probabilities of Brownian motion for the, for the g. Fair enough? And then k, kt epi g, if you remember, is just i minus stx minus s bar epi gx minus tx star st minus x minus st minus x epi tx plus, okay? Okay, so this is the Brownian scattering operator, which you're supposed to think of as some sort of hitting operator. It's supposed to go somehow from infinity at its, it's got, you know, it's a kernel. It has variables u and v, and you start at u and infinity, and you come in and you hit g somewhere, and then you go out to infinity v. Okay, and then it's messed up with all these backward heat kernels. All right, so, so that's the, sorry, that's the formula for the area process to be less than a problem. And as I said, so the, so the story is you just, you start with the extended kernel formula for the area process, make a path integral version of it, take a fine mesh limit, then take the box to infinity, and you get this, this formula for the area process, for the rescaled area process. Okay, good. So that's, that's actually where things stand, and so now we solved it for some special initial data, right? We solved it for starting from a narrow wedge. So this is the solution of the KPZ fixed point starting from a narrow wedge. The narrow wedge, if you remember, is this function. Okay, it's like the delta functions for this, in this business. Okay, the problem is, of course, you just get it for that one solution, it's the one thing, right? And how on earth are you supposed to get the general solution out of this? Because it's not like the thing's linear. It's not like if you solve for some delta function, the thing's just a sum of solutions or something like that. But actually everything we need is on the board already. So I can tell you how to solve it. Okay, so suppose you had this formula. There's one little thing that we're not using, and that's time reversibility. So the model has built in a time reversal invariance, which is exactly true in TSEP, and so it has to be true in the limit. And that's that if I start with function g and look at the height function and ask whether it's less than or equal to f at some time t later, I could just start with minus f and run the thing just because backwards in time, TSEP looks like minus itself. Particles and holes just get flipped. Particles, holes, flipped for the height function just means the height function itself just gets flipped. Okay, so this is the time reversal invariance. What does that mean? We got a formula for the probability that this area process is less than or equal to any function at time t. That's starting from a narrow wedge. You have a formula for any t that you're less than that. So now flip it using the time reversal. That means now I have a formula for the probability starting from any function g to be less at one point than stuff. Because the flip of a narrow wedge is just the probability to be less than a narrow wedge just means to be less than something at one point. Does everyone see that? If you flip a narrow wedge, it's just an upside down narrow wedge. If you ask if you're less than that, that's just asking a one point distribution of the thing. Okay? So now we have a formula starting from anything for the area process at one point. Starting from anything at the fixed point level to be less at one point. But now look back here. Remember that x and the height function are almost the same thing. So asking if x is bigger than aj is sort of the same thing as asking the height function somewhere is less than something. Okay? So this is basically a formula for the height function being less than something at m points. Right? So what this formula tells you is that if you know the one point kernel, you know the m point kernel. Does everyone see that? Because the m point kernel just involves things you know. So if you know the one point kernel, you know the m point kernel. But we know the one point kernel for anything now. So we can just guess. So the point is you can actually keep going with this thing and really figure out what the kernel is. But it's messy and complicated. But now the thing just gives you a chance to guess because the whole thing was just set up perfectly for us. As long as you can guess that the kernel must really be some hitting probability, you can just start guessing what are the fies. And all you have to do is guess and check because it was all set up perfectly. It's a recipe. If you know the right things, you can just guess them and check them. So you don't even have to go through with this horrible calculation. By the way, you didn't need all of this. This thing already tells you everything you needed to know. You knew the probability starting from a not narrow wedge, one of these guys, right? To get anything. So it's right there and you knew the fees in that case. Now you flip it. That means you know the probability starting from a bunch of narrow wedges to be less than something at one point. But that tells you what's the KT, right? Because you only knew the one point KT and then it extends it for you. Once you see that, you can just guess that this guy here is already this K-hypo f. Oh, sorry. The general K-PZ fixed point formula just means that this guy would be the K-hypo f. So you can guess the K-PZ fixed point formula, which I wrote to you before. So you guess that formula and now you guess what's the fees, check that they work, and then we'll rescale and get proved that the K-PZ fixed point formula is what we guessed it was. So it's the structure of the fixed point formula, which you guess from the Bartholome ensembles, and then you go back, use that to guess the fees, prove that they work, and then rescale to get the actual fixed point formula. Okay, so now let me just tell you what's the fees. So let's look in here. Okay, so to figure out what's the fees, we should actually figure out what's the size. Okay? Figure out what's the size. I need to know one little thing and that's that Q actually is invertible, sort of. So Q has a left inverse. I'll call it Q inverse, which is just 1 plus 2 grad plus. So you can check very quickly that 1 plus 2 grad plus on that Q gives you 0. Okay? So it's not a right inverse because the function Q inverse of 2 to the x, 2 to the x or 2 to the minus x equals 0. Okay, so it's not exactly a right inverse, but actually that's the whole kernel. So if you define things right, it's both a left and a right inverse because that function 2 to the x isn't actually an L2. Okay? So we want to compute psi k, n. Okay? Of course, we know it's a Charli polynomial, but I'm a barbarian. I'm just going to compute it. Okay. So I'm going to take Q inverse. Now you take Q inverse, you just take this 1 plus 2 grad plus inside and you get 1 over 2 pi i, integral dW, 1 minus W is k. But then the Q inverse acts on the 2W to the x, right? So you get 2 times 1 over 2W minus 1 e to the 2W minus 1 over 2W to the x minus x0. That's, so that's equal to psi n plus 1. The main thing is all these contour integrals, they're just codes for solutions of certain difference equations. Okay, which of course is the same as the Charli polynomial. But that tells you that psi nk is nothing but Q to the minus k of psi 0 and minus k. So you go all the way down to 0. Now psi 0, well you can look at what's psi 0. And you can check the following thing. DT, so I won't keep checking these things by hand. You can almost do them in your head. DT of e to the t over 2 psi 0 n is minus a half grad minus of e to the t over 2 psi 0 n. You can probably check that in your head. Which tells you that psi 0 n is nothing but e to the minus t over 2 grad minus plus 1 of delta x0 n. So it's actually something pretty simple. And this operator I'm going to be using it a lot. So I'm just going to name it RT, okay? Alright, so the upshot of all this is all these counter integrals. Well, a lot of them you can just compute things because all they are is solutions of difference equations. So what does that tell you? So I want to think about that as Q is the generator, the one step transition probabilities for a geometric random walk, right? So that means I can think of this guy here as the probability starting at xi to be at the boundary x0 at time k. Fair enough? That's all it means. Sorry. Oh, modulus these ends. Okay? And the ends you don't worry about so much. So there's lots of backwards, there's a Q to the minus n, but you can just throw that outside and do the calculation inside. Okay? Now remember what we believe. We believe the whole thing is just a hitting probability of the curve x0. You're supposed to go from infinity to come in, hit the curve and then go out. Okay? So k must be the hitting time. But this was a free evolution. So what that means is that the fees, let me just draw a picture here. There's some curve here which is x0. I'm drawing it very schematically. Actually the x0s are decreasing, but don't worry about that too much. So this is, so here's xi and you come in and there's k. That's what it told you to do. Well if it was supposed to be the hitting time, it must mean that on the other side, xj, the fee is the probability you went to xj without hitting the curve. In other words, looking backwards in time, so if you run the geometric random walk backwards, you should start at xj. Your first hitting time should be k and then you should have a free evolution to xi and that's what this guy should be. So here it is. That's a, where hk is the solution of the following problem. It's with q stars because the geometric random walk is going backwards. Geometric random walk going backwards is just geometric walk in the other direction. It's q star, so instead of jumping down, it jumps up. That's all. But of course I've got an inverse. So here's our problem. I want to solve, you start at k and then you go backwards. At k, you're given this function, which is one at the boundary x0 and set up for geometric distributions elsewhere. You solve the heat equation, but we're in discreet. Backwards, back to time zero. With a zero condition along the curve x0. So I just want to emphasize that this is a backwards heat equation. You're solving the thing backwards with a boundary condition of zero along your curve. Now first of all, you might ask, well, we were told long ago never to try to solve backwards heat equations. But q's, you can write q inverse. But q inverse is a kernel of one dimension, exactly one dimensional kernel. And that means every time I walk backwards trying to solve this thing, I have one degree of freedom which I can adjust and make the thing zero on the boundary. So the thing's perfectly solvable. There's no problem. And now I have to show you why it satisfies one and two. That's all we need. Once it satisfies one and two, it's over. So it's solvable. It's uniquely solvable. So let's do two, so this is zero. Now let's do two. We want to prove that it's a polynomial of degree k. Well, if I define h tilde, k and lx to be 2 to the minus x, h, k and lx, okay, well then I get an equation for it. Just write that.