 So welcome to the the second lecture So if you weren't in the first lecture, it doesn't matter Completely independent except What you could have learned from the first lecture is that it may be a good idea to try to solve TASAP Okay So I'm Unfortunately for us or maybe fortunately for the world TASAP was solved a long time ago 1997 20 years ago by Gunter shoots So let me tell you shoots is formula for TASAP Well first if you remember what TASAP is let me just remind you what it is So TASAP you have these height function which look like this And then there's kind of a code one wants to use which you've seen in a couple of talks, which is that the The the up guys are called particles So we're on we're over Zed here, and here we just put a particle here. We put a particle No particles and here there's a particle so maybe put axis for no particles and then What's that? That's an axe and then there's a particle and then there's two axes and then there's Three particles. Anyway, you get the idea, right? So you can just encode it in particles and the dynamics of TASAP was just that the the up guys go down They jump down to down guys at rate one, but of course, that's exactly the same thing as this particle jumps here With the rule that there's only one particle per site. So if there's a particle in the way, it just doesn't jump Okay, so the entire rule of TASAP you can just think of as particles on Zed and The particles are trying to jump to the right at rate one, but they don't jump if there's no particle in the way Okay, so that's our new view of TASAP here Okay, and then there's an old convention So so here if I just decided to use this notation X's or holes and O's are particles which may seem a little odd. Okay, sorry There's a convention that we number the particles X 0 X 1 X 2 X 3 Okay, backwards. Okay, so you can also think of the positions of the particles Fair enough. Okay That's the next line. So right now. I'm gonna assume. There's a rightmost particle. Actually, I'm gonna assume. There's finitely many particles Okay, so there's seven particles Okay, seven particles So shoot solve the system for seven particles well seven could be N. So there's particles. So there's there's X Which is X N There's a Y Y N So we have n particles and we want to make the exact transition probability So the probability That X T X T is now the whole configuration of n particles as they evolve X T is equal to X given that X 0 was equal to Y Okay, so and this is going to be a formula for each n Okay, this thing is given by a determinant of A particular function so it looks like this F Of I minus J F is a function. I'll describe in a second of X N plus 1 minus I minus Y and Plus 1 minus J. It's also a function of T and This is an N by N determinant in I and J Okay, so there's an N by N determinant and Once again, I just remind you the X's and Y's are ordered backwards And the F of course is important So F and X T is given by a certain contour integral. So it's a contour integral Gamma zero one just means a contour which contains the points zero and one DW over W One minus W To the minus N To the X minus N e to the T W minus 1 That's a T Okay, so That's the formula It's not that hard to Derive this formula and I'll actually explain to you why it's true So it's not that hard to derive and it's particularly easy to prove that that this formula is correct Okay, so to prove it's correct We want to write the dynamics in the following way so this has become It's a really useful way to write the dynamics So we've got the P and we'd like to write the forward equation for P So P is these transitions, right? This is call this thing Let's call this thing P of X. So let's just fix fix Y and call it P of X And I'll just show that it satisfies the forward equation the forward equation in this language is just a minus K equals 1 to N Grad K minus Okay, so wait, this is the free evolution so grad K minus is Grad K minus just says I take the kth particle and take the minus gradient So grad minus of f of X is just f of X Minus f of X minus 1 so that's the minus gradient then there's a plus gradient I'll use sometimes to call grad plus which is obviously f of X plus 1 minus f of X This grad the K just means that it's applied to the kth particle Okay, so this is what's called the free evolution this free evolution Of course, it doesn't include the fact that the particles are not allowed to jump on top of each other Okay, so there's a free evolution, but then there's a boundary condition which stops them jumping on each other which is Grad K minus P is equal to 0 X K is equal to X K plus 1 okay So this free evolution is going to be satisfied inside this vial chamber the vial chamber just means the set of X's where X0 is bigger than X1 is bigger than X2 such a and then Couple with this boundary condition. This is actually the evolution operator Okay, okay, so why does this? Why does this thing work and where does this crazy contour integral and all that stuff? So if you differentiate Fn Well, what happens when you differentiate fn and t? There's not much that happens except the w minus 1 comes down Right The w minus 1 can be included in the other variables and so you just get this simple equation. It's minus grad minus Fn Fn is just a function of one X variable Okay, so you can see this so I've been satisfying this Now when you differentiate the determinant You just get the sum of all the time derivatives in each one of the columns right So you get a sum of the determinant with one of the columns differentiated in t Right, that's how differentiation of determinants works When you differentiate that column You get dt fn is minus grad minus of fn But in that column Is only one type of X? Right, the X is determined by the column. So all you get is grad I And that just pops out of the determinant by linearity So that's why that happened that's works. Should I say it again? Maybe I'll just say he's really that simple You differentiate the determinant if you differentiate the determinant you get the sum of the time derivatives of each of The determinant with one of the columns having a time derivative Now if you take the time derivative in the column, there's only one of all the X's Being differentiated. So you just get in the if column you just get from that formula Grad minus of f in that column, but only in that column But that grad minus can just come out now by linearity So that proves that the P satisfies that You know, why does it satisfy that? now So again if you If you take grad k minus Then you get a grad k then you get a grad minus but grad just in the in the in the kth column right, so just the kth column that is Acting on by this grad minus because this is grad k minus, which just means it's only acting on the kth variable Okay So you get a you get the matrix But with just the kth column with a grad k minus on it Now so you get grad minus K of fn of x k say Okay, with with the y's too. I'm just being the word with minus y of course Okay, but that we could rewrite it as grad k plus of Fn of x k minus one So let's erase that just so it's clear No, it's just the same thing Like talking you get minus. I'm sorry minus grad came up grad k plus of k minus one But now if x k is x k plus one plus one This is just grad k plus Fn of x k plus one, but now what happens when you take grad k plus of fn Well grad well, okay Grad plus of fn Grad plus of fn just acts on the x So all it changes is the end The grad k plus fn, which you could just think of as grad plus of fn Is actually just fn plus one Of x. I don't think that I think I really made a typo there. I think it's that is that correct? Constantly, I think it's a typo in here. Okay. Anyway But what that means is two columns are exactly the same That's where you get zero from the then the determinant is equal to zero. So these two properties this Property of the function and and this property Show that this determinant actually satisfies the equation that it's boundary condition. Okay, it's really that easy and This crazy contour integral is just some weird construction to get something to satisfy those two things Okay, it's just a representation of solution of this equation Sorry, I got to talk a little louder and I can't even see. Oh, yeah Yeah, so of course This is the evolution of the P and then you would really need to show initial condition Which is completely correct, right? If you have the initial condition and then this this is actually just a representation of the forward equation Okay, so I showed it satisfy the quarter equation. You can also check it satisfies the initial condition I just don't feel like doing it right now. We got to get somewhere today. Yeah. Oh, okay So you take a grad. Oh, I hope it's true. You take grad plus in then you're taking a grad of x, right? So that means I have a w x plus 1 on the bottom, right? So there's one thing so it's 1 over w Minus 1 times that thing but whenever w minus 1 is w minus 1 over w. So you just get an extra n It's nothing more than that. You see how the thing works. It's amazing Okay, of course, this thing is just constructed out of these these objects Are just some general solution of these equations and then this construction is a special thing to get you the initial condition corrected It's in the notes. So I won't do it Well, we all know that if you have a P and it satisfies the initial condition and satisfy the forward equation And then it's your P Right. This is just the forward equation. It's just not the way you're used to writing it Okay, okay Well, there's several steps here. Okay, so so that's it. So that's how you solve TASAP That would be the end of the talk Except Except I don't like this very much because we're trying to go you see what if you remember from last time We're trying to go to a huge time and space scales So if I have seven particles that kind of gets washed out in the limit, right? So of course, we need a formula which goes to n equals infinity nicely and this formula doesn't that's the problem there's another thing which is that It turns out that the thing doesn't really want to ask you what's the probability to go from one point to another if we saw last time The right question to ask is if you start with some function What's the probability you're less than something at some later time? And if I want to look at the KpZ or the say sub height function I want to look at some later time and I just want to say well at this point here Is it less than that or maybe at two or three points? That's the kind of question I want to ask but that means I want to start with 1,000 particles and Then end up with three Points later. I want to know what's what's less than or equal to y at three points later But here you only have 1,000 to 1,000 or what was our 1,000 to 1,000? So that would mean I would have to sum over the other 997 variables and so I'd have to write a sum of this thing, but that's some is intractable Fair enough So the moral of all this is there's formulas and then there's formulas Sometimes you have exact formula, but they're useless. You need a formula which is useful Which can pass the limit, okay? so actually The person who discovered what to do is is sasamoto Yeah, this amazing idea in 2005 Okay so You want to try to create out of this a formula Which is a formula for starting with a lot of particles and then you're just less than or equal to some X's at maybe one or two or three points Fair enough Here's what he realized Okay, you're the following thing This is just from this this guy here It means fn plus one x T Is that correct? The problem is all I have is all I have my notes The following thing is true and and if I have to change that I think I made a typo here Daniel or Constantine is there a plus one or a minus one there? What Daniel? Oh, thanks Of course, we could do it in a second. Maybe someone can just check in this. Well, I'm talking okay But but anyway, if you just sum the thing you get this Sorry, why? Okay, you get a summation like that But now this is wrong This this is this is correct The other one I think there might have you a one plus one. This is a minus minus one here and This is mine and minus one and this is minus one No Okay, I'm very bad at getting these things exact. I'm sorry Okay, let's look at the last column. So the last column of the thing looks like this All right, of course, I wrote it sideways Okay, so the last column looks like that now What it means is that you could rewrite this as the summation Over some variable I'll call z to two and you'll see why in a second greater than x one of fn minus two of Z to two minus yn f minus one Set to two This is just this is just that same formula that written in for the cold column. Okay, or you could do it again So it said to three Greater than said to two X one now I've got fn minus three I've said three three because of that fair enough. You just did it twice. Ah Okay, so this guy you could just think of as the sum of these guys. Okay. Well, okay That's because there's a difference operator, right? But also, of course, you need to know that as y goes to infinity this thing goes to zero But actually that's not too hard because you have any any contour around zero one So if you make the contour size two or something then you let x go to infinity that goes to zero. So that's fine okay, now Now this thing is just a sum of those so this guy just the sum of these guys It's just being written in column notation Okay, it's just the same formula that would be this guy is equal to the sum of these guys Something bigger than I did then then that thing I just write it some all the guys bigger than him. That's it There's nothing deep here Okay Or if I took the second last column So second-last column and you take said 23 again, you'll see why in a second Fn minus three You do the same thing you get that okay, but now You see if you change z two three and Z three Three if you change them then there's a change of sign So you're always looking at fn minus three in these two columns Okay, but now if you change those variables those two very that very those variables only exist in that last two columns So if you change the variables, then you've changed two columns right This is writing the thing as the sum of the determinants okay, and That gives you a minus sign That means that if you look at the whole thing the only part that contributes is the sum So only Z three three greater than equal to z two two Three Contributes well actually z two two Itself has to be greater equal to x one Okay, so this is the only part that contributes There's there's just a cancellation in the other part where the two overlap Okay, what does that mean that means that the probability can be rewritten So of course you do this again and again and again and again You do it over all the columns and the whole probability can be rewritten as The sum over these Gulf and Zetlin patterns So you've pulled the I down to a one By doing this all those times Z one and minus y and plus one minus j and Now the sum is over all okay, so we've got z one three Z two three Three and these things all have to have these inequalities Can you guys see the inequalities like hey? It's just rewriting what's up there, and of course it be this pattern keeps going up all the way to end And down here is your initial data x one x two x three So the sum is over all such patterns. Okay, that must be an eye sorry This is an eye minus j z I this is what it is Sorry, we've seen this before Alright, okay Sorry, what should be here F? Oh, this this is a one. Yeah, sorry. This is an eye This is correct now, right? Yes That's actually the key thing I wanted to show you because because it's a really wonderful idea and Now the thing of course is related to things like random matrices That thing can actually be rewritten You can write this the summation over pattern Z. I am now no longer with the inequalities with z one and is equal to x Z I and That one and is equal to x n. Sorry of W and of Z where W and of Z is equal to the product and equals one Two and that's my number of particles determinant Q or these things have to be defined now So Q Q x y is just the indicator function that x is greater than y and Psi k n of x is almost f It's minus one to the k f minus k X minus y and Minus k t. Okay Now this is the sum over all patterns. They have no order, but You can check that this particular thing. It's an exercise to check that that Because of that Q. This is just the color McGregor formula in a funny form Just tells you this is the indicator function of this thing being a Gelfand's Evelyn pattern. Okay, and Then this is just a different way of writing this. So this is really just the same formula Okay now The reason for all this rewriting is that this kind of thing is Recognizable as some kind of Determinant a point process. It's not exactly These guys are not positive things. They're not probabilities They could actually be signed. So it's come kind of signed Determinant a point process So you can write the whole thing as one Type of determinant on some slightly extended space and Then you know from this language of determinant a point processes That if you want to compute things then you should compute the correlation kernel of this determinant a point process which I won't do and This kernel is basically obtained by inverting Some matrix involving these guys Which I won't say explicitly what it is. Okay? What I'll do is I'll tell you the result of that computation because that would take two or three lectures just to go through it So the result of that computation, which is actually our starting point after we have 20 minutes So I probably won't finish today The result of that computation. I'll leave that up is The following so here's the actual recipe for solving pace up. This is a 2007 Warden was I want to write Corwin. Sorry. We're in Ferrari and Sasamodo Okay So the probability starting with so now now x sub zero is Is the name of the taste up initial condition? Okay It's it's got a last particle and if you want just assume that there's a finite number of particles But it definitely has a last particle to the right So that's the initial condition and now if you want to know Xt and j greater than a j J equals 1 m and here M. M is just one or two or three This is given by the determinant of I minus the Correlation kernel with these kais so I'll remind you these kai bar Has kai bar a at the i-th coordinate is the indicator function that x is greater than a I It's that vector Oh, sorry less