 Brownian initial data, that says you take a Brownian motion on one side and zero on the other side for some bizarre reason you can do the integration. So that case we can do now. So there's some magic formulas still to come, that's for sure. Okay, another thing which is definitely not in this picture. Yeah, in fact, it's sort of the same thing, which makes you, it's just the old trick happens to work here. In fact, it works this, it's the same thing, yeah. It's because the way that that trick worked, use other solvable models, which we can also do. The Brownian initial data, two-time formulas. Well, you're gonna hear, this is an advertisement for Gepeng's talk this afternoon. Two-time formulas, well, as you can imagine, to get a two-time formula out of this, you have to take the Chapman-Komogorov equation and integrate over almost all internal variables, over this enormous space of internal variables. So of course, the formula you get is useless. And yet, they get simple closed-form two-time formulas. So it's not really something in this picture, okay. But I'll tell you something else, which, oops, which is in the picture, can you hear? Here's another important process, which is that you just take the KBZ fixed point and you differentiate the an-axe. Okay, now, okay, maybe you don't like that because it's not differentiable, but you can take a distributional derivative, there's no problem, it's linear operation. So you get a UTX, which is a distribution-valued Markov process. It's not completely obvious it's Markov because you may have lost some information by differentiating, but it is. So this is a distribution-valued Markov process, which is the stochastic burger fixed point. And just like H is dying to be the solution of DTH equals, well, it's actually a quarter of DXH squared. This is sort of the equation for H, you can't prove it. U is supposed to be the solution of DTU equals DXH squared. They are some sort of weak solution of that, though nobody can prove it. But on the other hand, one amazing thing is that we can compute U, we meaning the community, not particularly our group. This is actually known in equilibrium. So U, H has almost invariant Brownian motions. They're not, you could take a two-sided Brownian motion and then later you see a two-sided Brownian motion, but with some shift, which is supposed to be this by-grains distribution, but U should have invariant white noise. Okay, so let's take this and calculate the spacetime covariance in the equilibrium process. So you start this U with white noise and that's supposed to be the invariant process for this stochastic burgers fixed point. It exists, it's a thing. Now of course, it is far from obvious that this formula makes any sense whatsoever because these are distributions, but in fact it does and it's equal to TN minus 2 thirds. This thing called the KPZ scaling function, T to the minus 2 thirds, X. And this just comes from an old calculation of a Ferrari and Spahn. So this FKPZ scaling function, I won't write it down. It's a function which is a lot like the formulas for the polymer endpoint distributions, which Daniel was writing yesterday. So just some crazy formula full of airy functions. So this was computed by a Ferrari and Spahn, maybe also a little bit in a paper of Praha from Spahn. But, and this just comes out because they had computed this for the space-time covariance of TASAP and it just scales. So that's their scaling limit of space-time covariance. So the thing existed in the literature before and it was just waiting for the process to exist to have that covariance. So the last thing I wanna do is talk about the time, how it changes in time. Okay, so we talked about the space regularity. So the space regularity, it basically looks like a Brownian motion in space, at least locally. So if you start from some initial data, it's gonna produce a Brownian motion version of that initial data. But now we're interested in the time regularity. Yeah, that's coming. You keep anticipating, just wait five minutes. Okay, so time regularity. Oh, I erased the, so it's invariant under this KPZ one, two, three scaling. And from that, it's not too hard to guess what the time regularity is. So the time regularity is you should see that HT plus SX minus H. I'm probably gonna change my notation later. SX should be about T to the one third. That's about how big it should be. So you expect it to be, you expect it to be a little bit less than how they're one third in time, okay? So let's try to prove that. But to prove that, you need some variational formulas. And so this goes back to what Pierre was talking about in his lecture a few days ago. So variational formulas, I have 15 minutes and I know you guys are hungry for lunch and I keep stopping you. So I'm gonna skip a lot of the, I'm gonna give you a simple derivation. At the level of TSAF, one has the following property. Suppose you start TSAF with two different initial datas. So now I'm writing the initial data here. So F1 and now I just take the max of F1 and F2. So I have two functions when I start TSAF, well they have to be fair functions for TSAF, okay? Of course, we know what functions you can have for the TSAF, high function. Suppose I take two functions and I start TSAF with the max of two functions. Turns out that that's equal to, and I'm gonna have to qualify this statement a lot. So don't start asking me questions too fast. That's equal to the max to the initial data. So TSAF has this wonderful property. It preserves max. Now, now I have to qualify this. Okay, so what does it mean to start TSAF with two different initial data? We never really talked about that. So the problem is the right-hand side of this thing actually in principle doesn't make any sense at all. Because you start from some initial data and you start flipping guys down. And the two things are flipping fights. The flips have nothing to do with each other, right? Okay. But of course, you could start TSAF. So TSAF at every site, there's a Poisson process, PX. And the thing flips down at rate. The indicator function that you see that there, okay? And makes a flip of size two. And so it's a minus. That's actually the rate of TSAF flipping down, right? You start with a Poisson process at your site and that Poisson process rings. And whenever you see one of these guys, that's what the one, this guy, you flip it down. Okay, that's the rate of TSAF. So DH and TSAF is just that. Okay? But what it means is that you can take the same Poisson process for TSAF starting from two initial data and run the two on the same Poisson process. Does that make sense? Okay, so just take the same P's. Start with two initial data and run the thing. And now it should be completely clear if you think about it for 10 seconds that if you start with one profile above another, that's preserved for all time. Because there's no way one can flip down below the other one. It just can't happen. That never happens, okay? So if one's above, it stays above. Okay? So that's called the basic coupling of TSAF. You need a coupling. You need a coupling in order to be able to even say the right-hand side. So the basic coupling of TSAF is just use the same Poisson processes. And then this is just the property I just described, actually, okay? Just the fact that if you start above, you stay above is just that property, okay? That's true in ASAP. Yeah, yeah, yeah, that's true. Yeah, it's true even in finite range exclusion. It's a general fact about exclusion processes. It's true in KPZ, yes. Well, proof it's true in ASAP, so it's true in KPZ. Because KPZ is just ASAP, right? Okay. Now, is it true in the KPZ fixed point? Well, yeah, because it's the limit of TSAF. But the problem is I had to use the coupling, right? Now once you couple the two guys, so now we take two initial data for TSAF and now try to take the limit, okay? You can take the limit of the coupled process because you know that they're both tight. Because we have enough information here to show tightness. We even showed that everything's, you know, I didn't say it, but we know everything's uniformly held as you take the limit. So there's no problem. So there's tightness. So you get two copies of the KPZ fixed point with this property, okay? The problem is you don't have uniqueness because of tightness. All you could use is tightness, so you don't know uniqueness. So I'm gonna keep talking, but just keep in mind we don't know uniqueness at the level of trying a couple things. So this is because Markov processes, when you talk about them in terms of their transition probabilities, are missing something for when you write things like stochastic equations for them. Stochastic equations basically couple all initial data. And so it's a stronger thing than just the Markov process using the transition probabilities, okay? So we have the Markov process using the transition probabilities for the KPZ fixed point, but the coupled things come to us in a soft way where we don't know uniqueness anymore, okay? All right, so the fixed point has this property too. So, and the fixed point. But with the caveat that I just talked about, modulo uniqueness problem, okay? So that tells you, let's get a formula from this. So of course, if you can couple two points, you can couple three points, you can couple four points, so I can couple a billion points and I can put a fine mesh of little narrow wedges everywhere and then you get the following formula. This is for the fixed point now. HTXH0, is equal to the soup over Y of HTX starting from a narrow wedge at Y. Okay, is the notation clear? There's a narrow wedge at Y there, plus H. This thing here, for every Y, it's a perfectly processed in X, but in X it's a rather inga-puchket process in Y because what you had to do, if you had 1,000 Ys, you had to do that coupling 1,000 times. I'll call this thing T to the 1 third A of T to the minus 2 thirds, because of the scaling, AXT to the minus 2 thirds Y minus a parabola. If you put these scalings of Xs in that, you would get the same thing, so you can just think of it as A hat, if you like. So this AXY is a two-parameter process. In the first variable, it's a nice stochastic process. In the second, it's a little bit mysterious. It's not unique, we don't know it's unique. This is called the area sheet and this variational formula tells you in a sense how to solve for any initial data as a variational problem using this sheet. Now, the variational, it looks bad because in Y it's so horrible, but on the other hand, because of time reversibility, AXY and AYX are the same thing in distribution. So it's actually just as good in Y as it is in X. And so the only problem isn't the fact that it's badly constructed in Y, the problem is you just don't know it's unique. So modular uniqueness is a perfect formula. And the statement says something bizarre, which is that even if you don't have uniqueness, any area sheet gives you this variational formula, which gives you the KBZ fixed point. So the only thing is I shouldn't use the word the. It's not the area sheet, it's an area sheet. So the problem with this, you might say, ah, well, why don't you just compute the distribution of the area sheet from the KBZ fixed point formula. But what does the KBZ fixed point formula tell you? It's going to tell you exactly the following thing. Suppose I wanted to know, suppose I had two X's and two Y's. So I had the area sheet at XI, YJ, IJ equals one to two. So I have four points of the area sheet. And I wanted to know the joint distribution. So this should be a four-dimensional distribution. The KBZ fixed point formula tells you exactly the following thing. And if you think about it, it's just precisely this. It says that area XI, YJ. Actually, let me put hats on them. So the hat just means that you subtracted the parabola. That's a common notation here. Okay. So it tells you exactly that that thing is less than... Oh, okay. In general, it tells you exactly this. You can compute that. For any F and G, you can compute this. This is exactly what the fixed point formula tells you about the area sheet for any F and G. So if I put... Get that. So you get that information. But here's something a little bit surprising. If you check the span of that in four dimensions, it's only a three-dimensional space. It's not obvious. So it's a three-dimensional linear subspace of four dimensions. And so you actually do not get the information of the area sheet. Okay. So you don't know uniqueness of area sheets. Okay. So how is all of this useful? Okay. All this sheet stuff. I said I wanted to prove this. Let me show you that the proof of this is very easy using this area sheet technology. So one thing you do know, and it's easy, is that the area sheet is an area process in each coordinate. Okay. So in x, it's the area two process, and in y, it's the area two process. That's not enough to characterize it. So it's the area sheet in analogy to the Brownian sheet being Brownian in both coordinates. But you could write this formula. Equals suit. Oh, sorry, with T's. Okay. There's a variational formula. That's a variational formula, which Pierre wrote down. So I want to actually discuss it for just a second. That's the variational formula, but I'm missing the second coordinate. Okay. So that variational formula is true for each x, but it's not true as a function of x. So this formula gives you one point distributions of the resulting object. But if you want this thing as a process in x, you need the whole sheet. Okay. That's the missing information. But I'll just use this formula to prove the time regularity. Okay. So we want h of T plus s minus h of s. Okay. T is positive, and let's suppose that s is positive. Okay. I can't prove the time regularity for any uc function for s equals 0 because initially it's actually false. Okay. Especially if you start with a narrow wedge, you can see immediately that's going to be false. The time regularity is the difference that can be infinite. So because of the Markov property, I could assume that I'm starting with hs. So I might as well just take hT minus h0, but h0 is the kind of function which I get at time s. But at time s I already know that I have a function which is held there beta less than a half. Sorry, there's an x here, of course. And so h0 is held there. But my shift invariance of the whole process, of course the whole process is shift invariance, so we might as well just take out x equals 0. So now let me just write down the formula. We've got hT0, hT0, yep, is equal to sup over y. T to the 1 third of an area process, t to the minus 2 thirds x, minus x squared over t, plus h0 of x of y. This is sup over x, sorry. So x minus h0, 0. That's the variational formula. Now this guy here is locally a Brownian motion, particularly it's held there beta, beta something less than a half. So this thing here is less or equal to c, t to the 1 third, t to the minus 2 thirds x to the beta, because it was held there. And this thing here is held there 1 beta, so this is x to the beta. And you can center the a. Sorry, I can write minus a0. I should just, yeah, thanks. I sort of realized that myself as I was writing it. It doesn't matter in the calculation because you can take that out of both. Okay, anyway, so there's that, right? And now you just do the variational problem. And you immediately see that the sup appears that, the sup is that x is about t to the minus 1 over 2 minus beta, sorry, 2 minus beta. And the sup and gives t to the beta over 2 minus beta. And as beta goes up to a half, that thing just goes to t to the 1 third, right? So beta equals 1 half minus delta just gives, but that proves the time regularity and hits questions for Jeremy. So the problem is, as far as I can tell, and I would love to be told otherwise, if someone has any idea, there's no way to prove uniqueness for such objects except to compute their distributions because we don't have an equation for them, okay? So the only reason we know uniqueness of the KPZ fixed point is because I can give you a formula for it. Since I can't give you a formula for two coupled KPZ fixed points, I can't prove they're unique. That's just that. Any other questions? Okay, one more. Devar, could you talk a little louder? No. No, there's no dual variational formula. I wish. Okay, last question. Yeah. Well, you know, for the stochastic heat equation, it's a quarter in time and a half in space because it has this one, two, four scaling invariance. It's the same thing, and this has this one, two, three scaling invariance, so it's a half in space and a third in time. It's kind of clear from the scaling that it should be like that, okay? Okay, so let's all thank Jeremy again.