>> David Wilson: We're very happy for the final speaker today, Roberto
Oliviera from Brazil who will tell us on bounding the time for mixing of simple symmetric exclusion in terms of one particle mixing.
>> Roberto Oliviera: Well, I'd like to thank you all for being here, those who remain at least, and of course I'd like to thank both the organizers of this workshop for the invitation and also the group at
Microsoft Research in general I'm right in the middle of four-week visit, and like Favia having a bigger time here. Well, thanks a lot.
So today we're going to be talking about this symmetric exclusion process. But let me start with something simpler, which we all know what it is.
When you have a graph like this, an end is always going to be the number of vertices, let me say it from the outset. You can define random walk on this graph and today I want to talk about continuous time random walk. What that means is while you have a particle somewhere on your graph and each edge has its Poisson clock maybe the clocks have different rates, but for the time being we want to think all rates being one.
Say when this clock rings, if the particle is on one endpoint and moves to the other endpoint. If this clock rings, nothing happens because the particle's over here. If this one rings the particle moves again.
And it keeps going. So this is simple random walk. Simple continuous time random walk that is. And what I'm going to be talking about is a version of this with several particles. They're indistinguishable particles. And so this is the -- what I'm going to tell you now three particles symmetric exclusion process.
While symmetric means that what I already said about random walk the rate of crossing this edge doesn't depend on which endpoint you're at.
Goes in this direction as often as it goes in this direction. And now the update rule is when an edge rings, you flip the state of its two endpoints. So rings here nothing happens. If it rings here, X moves here. If it rings here, nothing happens, because there's no particle.
And it goes the length of this. Right?
And while I'm not going to give the formal definition right now we'll talk about this later. But I hope the idea is clear. I mean, all you should be able to write the genre if you so wish at this point.
So what the kind of question we're interested in is well as you all mentioned somehow mixing times it's part of the title. It should be obvious. So when you have any Markov chain, in this case we want to talk about continuous time Markov chains, you can define this parameter, the mixing time. Assuming the state space finite and the

chain is irreducible. So we know that there's a unique stationary measure. And the mixing time is the smallest T for which the transition probabilities from any point in your state space into any subset of the state space while they're close to their limiting value.
Epsilon close, I should say. So the state space for the random walk is just a vertex set of the graph. The stationary distribution in this case is uniform because of the transition rates are symmetric. For the exclusion process, the state space is subsets of size in this case size three of vertex set. And while in general going to be talking about this process here, which I'm going to call exclusion KG, which is a K particle exclusion, and we want to know what the mixing time of this is in terms of the mixing time of one particle. And now the theorem is that -- it's actually pretty simple to write down, and maybe you just move it over here so that it stays in place as I continue to write.
It's this. So while the mixing time -- it's a bit, it's a bit more than what I'm going to state here. But the mixing time of exclusion
KG, and then you have to choose the epsilon, just choose epsilon equal to a quarter which is like the standard choice. This is less than a constant. And the constant -- by constant I mean something universal.
Times log of N, N being number of vertices times the mixing time of one random walk. So RWFG is just a single random walk.
So the theorem is telling you this so that essentially well if you compare the ratio, right, of this mixing time to the mixing time exclusion process, that's at most log N up to a constant factor. And well I mean there should be a reason for why that's interesting sometimes. So let me for the time being I'm not going to say anything about the proof. I'm going to say a few things about the question that motivated this work.
>>: [indiscernible].
>> David Wilson: For any K. So the constant doesn't depend on K. So for any K between one and N. Well, with N particles the process doesn't move so it doesn't say much. Could be N over two. So this is independent of G or NK. It's universal. And now I mean there's at least in my mind there's one main motivation for this work, which is this. If you look at these Xs, at the particles, there's an alternative description of this process which is as follows: Each one of the Xs doing an independent random walk following that continuous time proof there described. Except that when this particle tries to jump over here, it's blocked. So you have this exclusion rule that forbids two particles on the same vertex. So it's a system of random walks that kind of constraint in a certain sense. They cannot jump on to each other.
And so then you can ask yourself how similar this is to a process of independent random walks. So there are obvious differences. For

instance you're never going to have two particles on the same vertex.
And on the other hand there are similar processes, at least from kind of for a high level perspective, perhaps. So it seems natural to ask, if you can say anything in general about the relationship between the two mixing times. It's one way to ask me whether the way exclusion converges to equilibrium somehow similar related to the one single random walk versus equilibrium. Right now I'm going to prove the theorem starting with an assumption or conjecture or whatever that's, well, far from being proven and I think very hard to prove.
And this is the conjecture. Well, okay, think of this as heuristic but
I'm willing to bet that something like this is true.
And maybe I have to explain what this means. Whilst I'm going to use the symbol here RWKG for K independent random walks. And one quarter here as well. And maybe want to put a constant here. So what I'm saying is this: That K particle is doing exclusion shouldn't take more, much more to mix than K particles doing independent random walks.
And while this might seem a bit weird in some sense because exclusion has obstacles. One particle blocks the other. But because of the symmetry we have many properties that sort of suggest it's true. I'll even given a stronger form of this later. But at least if you believe me by now you have this. And this, of course, I mean this is -- yes.
>>: [indiscernible] you said just if you can tell us something about the proof of the real theorem without that assumption.
>> Roberto Oliviera: Sure. Yes. But let me just say that if you have this, while since I got started I'm going to finish this and then I'll jump to the proof.
So if you have this, then you can prove this. With a different constant perhaps. And this is less than -- and this by general fact, this is not hard to prove that -- I mean, K random walks to achieve distance recorded as equilibrium for one K-1 random walk is enough.
And you can get this inequality comes from general theory of Markov chains and this is essentially the same result that I have over there except for having an end here. So in some cases we know that you can get this with a K. So, for instance, for walks on the cycle or on the discrete Taurus any dimension you can get this with a K, I can only get it with an N. But that's the way it comes out of the proof. So in terms of applications, you know, there are many previous bounds about symmetric exclusion. Maybe what I should say is that this matches or improves upon all previous bounds except in this specific case that I just mentioned which is when you have a discrete Taurus and you have K particles, you can get log K instead of log N. Following Yuval's request I'm going to try to move as quickly as possible to the proof.
And okay so there's kind of a very rough outline of the proof is that you first consider mixing for two labeled particles. And what I mean by this is well you look at this exclusion process over here and with

two particles but you give them names. So let's say A and B. Right?
And you have the same rules. So when you have an edge update you flip the labels. And then well so this is the first step and I'll say something about it. It's not so hard to understand. But then you need to go from two to K. Of course, it's not meant to be obvious, right.
And let me just mention that always suffice consider that K is less than half of all the vertices because in the exclusion process, if you look at the complement of the set of particles, right, you get another exclusion process. So if you have K particles the complement has N minus K. You can always assume that the number of particles is at most as half of the total.
>>: [indiscernible].
>> Roberto Oliviera: Yes, I can actually do this part for labeled particles. So that's what's called the interchange process. And I can get this when K over N is less than -- well, is bounded away from one.
So anything that's bounded away from one can do with a constant that depends on this ratio here. So that's actually the way the proof works. Wouldn't know how to do it any other way. So need to consider labels. For exclusion suffice to go this far. And right so well how for two particles for two particles it's actually, there's two kinds of couplings that you can consider and they're sort of complementary, and that's the goal as part of the proof is to show that.
So here's coupling number one, say you have A and B, and you have A and
B like this. Right? So it's two different validations starting from different states. So if you're just trying to couple single random walks one way to do it is to let them evolve independently until they meet and coalesce from there. That's how you -- one way to prove the usual Markov chain theorem. The label chain or Post Office particles you do the same thing what you do you meet and A and A and continue from there until B and B meet. So there's one case of part one let me erase part two for the time being, which I called the easy case and the easy case, this is coupling number one, is for all vertices in the graph, the expected time -- the expected meeting time -- and I hope what that means is clear. For random walks from VW. So meeting time means you let them evolve independently until they find one another.
So the easy case is when the expectation is less than, I don't know,
8,000 times T mix of a single random walk. So I mean I should say what's our goal here, to say that mixing time for two labeled particles is at most a constant times the mixing time for one single particle.
Right? So the easy case is when these expected meeting times like this. In this case you can use this strategy I just described and it's going to show that the mixing time in this case C mix of -- well, interchange process to use the kind of two labeled particles. And when
I write C mix without an epsilon here means epsilon equals quarter. So in this case you can prove it's something like maybe I guess the actual constant they have four and then it's 40,000 [indiscernible] enough.
Just want a constant.

Now there's the non-easy case and the non-easy case, the idea that you have to make rigorous is this. So what does it mean for a graph not to be easy. It means that random walks take a long time to meet. So what does this mean in terms of the interchange process? Right? It means that if you start A and B from two states where random walks are unlikely to meet then the interchange looks like two independent random walks. That's good, right, because we know how to bind the mixing time at two in terms of one. Now, it's what -- this is not quite the proof, because here you are asking for all V and W. When you negate that, you have there exists a pair V and W from which random walks take a long time to meet.
But then what you show is that actually you have the same property for most pairs of initial states. That's the idea. If you have one single pair, what random walks take a long time to meet, well, you can imagine like this, right, you have this pair here VW and random walk for a long time they take a long time to meet. Well, if they walk for a time that's like the mixing time, these two places are more or less random.
Of course you can say that the pair is random but you can say something to that effect. Right? Using one of the things you're going to use here is what's called the negative correlation property of symmetric exclusion essentially says what's the probability of the two particles are inside a set is at most the product of the probability.
So if it's non-easy it's similar. After a burn-in period. It's similar to two independent random walks and then you're done as well.
So this part of the proof is very clean except for the constant. But it's conceptually very simple. Now let's move on to the hard bit. And the hard bit, I should say, that comes from -- well, the ideas come from a paper by Ben Morris, who already mentioned it's a paper about symmetric exclusion on discrete Tauri. Say you take two dimensional
Taurus with large side length, you want to know what the mixing time is. In order to determine the time he came chameleon process. Let me try to show you what it is.
Okay. Let's see. Part two. From two to K. And what you want to do is this. So let's say I want to move from 2 to 3 particles. That's not terribly exciting. Could actually use the same ideas for any bounded number of walkers. That should be clear.
But I mean so the idea I'm going to use here doesn't move from 2 to 3 to 4. That would give you a [indiscernible] over hat and K which is bad. So now I'm going to show you something that allows you to move straight to any K you wish. Any K mod N over 2. So let's say I want to consider three particles and one of them is going to be red. And actually let me paint this all red here. And what I want is not the mixing time of the whole thing but I want to know how close is this to uniform given the other two particles. It should be clear it should be somehow equivalent. You have two particles you have N minus 2 vacant

vertices and how close is this to uniform conditionally in the position of the other two? And so that's the idea. And now you're going to do a series of tricks. So the first one is this, is well you remember that each edge had its Poisson clock, which we're saying that rings at rate one. Let's make the clocks ring at rate two. When clock of edge
E rings you flip a coin to decide whether to use it.
And the effect of this is really doing nothing. Right? Because we know if you take the Poisson process with intensity two or rate two, and you erase half of the points as random fashion you get Poisson process with rate one. This really doesn't do anything. But what we're going to do is this: So, well, so you can imagine what happens here. So while this edge rings and you look at the coin, to decide whether you're going to move or not. The edge rings and you look at the coin to decide whether or not. Now, trick number two involves what happens when an edge rings that is adjacent to this. So you could look at the coin to determine whether this is going to move here or not.
But you're going to force yourself not to look at the coin. And what's the idea here? If you don't look at the coin, the conditional probability of this being here is equal to condition probability of this being here. And now what you do is you pretend you had one unit of red ink here and, well, I don't really have pink here. The right way to do this according to Morris would be to paint this pink and this pink. Meaning that these are half red and half white. And so this by definition for today is going to be pink. And the trick number two is do not look at edges -- don't look at coins of edges that touch some red particle and for the time being we only had one particle. Now it's disappeared. And eventually get more red particles because we have trick number three, which is when -- so I should say well we had one red vertex which was the location of the stack particle and we had wide vertices which have no ink and we have pink which are sort of halfway between. So the rule now is that when the number of pink is greater than the number of white or number of red, well, I'm going to write it here like all pink, flip the same coin. And what I mean by this is while the pink particles are going to all decide whether they're going to turn red or turn white, altogether, and they're going to flip the same coin. So here the outcome could be that either these two particles become white and now all red ink disappears from your system or they both become red. Let me pretend they became red. Continue this otherwise it ends or just disappears. And now you have extra moves. So maybe this edge rings and again this becomes half red and half white. So have two pink now. And I guess here they really need to be bigger. Well, eventually this particle will move somewhere and it will share ink with some other white vertex. And you have something like this. And now they're going to all flip a coin together to decide whether they all turn red or all turn white. And the main point of all of this sort of gymnastics here is this. So the probability that this tagged particle, let me call it red here, is that a vertex X given all the rest, it can be described by the expected amount of ink as site X given all the rest. So you can see that the blue stuff is going to

move as it moves before, right? And what I mean by ink is that, well, this is one for red. This is half for pink. And this is 0 for white.
So this is what I mean by ink. And this is just sort of Martingale property. Just need to show it's preserved at every single step.
Certainly when they share ink, means if they a particle has probability of P being here, has probability of P over 2 being here, probability of
P over 2 being here. And you have to show it's continuous at all times. But now comes something else. This by itself doesn't do much.
What it does a lot is this thing here. So, by the way, this is a chameleon process. And so now let's look at the amount of ink in the system. At these times, there are times when pin particles completely disappear. They all become white or red. So the amount of ink in the system, it doesn't change. You can see turning one red into two pink doesn't change the total amount of ink, right. So its evolution happens of these so-called depinking times when pink become red or white. What you have start from one.
Well, if you have K particles you have N minus K plus 1 possible vertices that might be occupied by red. There's also possibility that you end up at 0. This evolves randomly. It evolves by taking ever bigger steps depending how high you are. And I mean they're unbiased so they might bring you down as well. So you could decide or you could somehow end up here. So I guess it's drawn logarithmic scale. But anyway you could do both things. And while the probability is that well let's say fill is the event you have N minus K plus one red particles. That means that every single vertex that could be occupied by a red particle is going to be occupied. And because of this
Martingale property, each time step the amount of ink goes up or down by the same and equally likely to do so. The probability of this event, given you start for one is just one over N minus K plus one.
And Morris proved this that if you're looking at the total variation distance between the law of this red particle at time T, given all else and you want to know how close it is to uniform over the appropriate set. So vertex, vertices minus what else is occupied by else, this -- while you have to take an expectation here you're taking conditional, this total variation, this have to take an expectation. So what Morris showed is that this thing here is less than one minus the total amount of ink in your system divided by N minus K was one. You have to take an expectation and you have to condition on this event that it fills up the whole thing.
So I'm not saying this is obvious. I'm just throwing lots of things at you right now. But that's the main step in this chameleon process argument. Is just saying that while you have this ink describes the evolution of the continuous, the conditional density of this guy but of course at some point you want to get rid of the conditioning, this is exactly what happened. Except you have to condition this fill event here.

But this is nice because you don't really have to think too much anymore about the condition distribution of this given the rest.
That's more or less the idea. And for people who know this evolving set technology that Morris and Yuval developed, this is similar to the idea they used there, except it's different. It's similar but it's not the same thing. But, right, so now -- let me make a remark here. So I mean now that your problem is to analyze the evolution of this ink, random variable condition of this, and well the way Morris does it is just well he knows everything he needs to know about his graph, which is as I said it's a discrete Taurus. You have heat kernel estimates.
Many things that allow you to show to control the evolution of this.
Here we don't have any of this. So what we're going to do is we'll add some quiet periods of length, some constant, let's say T times the T mix of I P2 G. Remember, this is two labeled particles. So this is sort of the same thing as a mixing time. And the thing about these periods is that in these periods where we look at all coins, meaning no color changes. Well, the colors might move around, right? What I mean by this is that you apply the same rule that you have in the usual process. So if this edge rings, the blue particle over here and the pink particle over here they switch places. So particles are simply going to move around with no color changes. And, of course, that means that we don't apply this trick number two here for those times. So that's why I say we look at all coins. Right. And we need this because we essentially don't know what a heat kernel looks like. So, right, so what happens here? So I should -- well basically I drew this evolution here. These are so-called depinking times. Which means the moments at which the pink decide to turn all red or white and when you condition this event fill, all you're doing is you're biassing the coin flips by a certain amount that can explicit compute towards up. Right because you can't go down. You can't go to zero. So you're biassing well this is no bias random walk which sort of the size of the steps increases as you go up now you're biassing it towards N minus K plus 1.
So and what we need to show is essentially, so if you want the amount, total amount of ink to be large, which is what we want here, you want to show that, well, I mean this is telling you that essentially you're moving very quickly. Discrete time walk moving up in the sense they can make precise. What you want to show the durations of these periods they're not constant as I'm drawing here. They're random variables.
The time it takes between one depinking and another and you want to show that those times they are of the order of the mixing time, essentially. So in a very strong sense. So I mean that's not completely obvious. But the main goal of the remainder of the proof is this. So show something like the, I'm going to be a bit technical here. But that if you take any condition, anything on -- you look at what happens up to depinking time K minus one and you want to know that this thing here, well, divide by some constant T mix of I P2 G which already said is the order in their mixing time. I show this less than some constant here. Meaning that it's very likely that this ends in a time of this order. And so the way to do it is really start with any configuration of red and white and you want to know what happens,

right? And I mean this really does not depend on this conditioning, as
I said, because this conditioning only affects what happens when they decide to depink. Right. So what do you do here? It's actually not very complicated. So you start with any configuration of black. Well, blue, sorry. So blue are the other particles, the ones K minus 1 remaining particle. It's A and B for this case. So you have a set of blue particles here. You have a set of red particles here. And you have a set of white particles. And I'm going to assume that you have more white than red. And because of what I said, I'm going to assume they go through a quiet period where no depinking takes place, but which is sufficient because of it's of the order of mixing time of two particles. It's sufficient to have the following thing, I mean, essentially if you look at any two particles over here, they're going to be mixed up. So you can pretend they're as likely as any other to be in this state. So that's the important thing that you only need pairwise attraction here. So you have this thing here. And now you want to know what's the probability -- so you need the number of depinkings that's well erase the rule but you want to know how long you need to wait until the number of pinks becomes larger than the number of white or the number of red. Since I have less red than white here in the beginning, the other three are symmetrical, I need to know how long does it take until the number of red goes down by one-half. So essentially the main thing is you can check for yourself you need to bound the number of times, the amount of times you have to wait. So this is quiet. This is color changing. You need to know how many of these windows you need to see until you see many color changes. And essentially what you do, the basic idea is I want to bound the number of collisions that I have in each one of these here between a white particle and the red particle, because collisions are good. Collisions lead to things becoming pink. Except I mean if you just count collisions you're not going to succeed because there might be double collisions. Sometimes you're going to have two red particles that bump into the same white thing. And so then you're bound to get into trouble. Maybe this one traveled a lot came here, bumped into white and then it's bad because then you're counting the same thing twice.
So what you need to do is besides -- and I mean bounding collisions, as long as this is of the order of mixing time is really no big deal, because this white part here, it's a positive fraction say more than
25 percent of graph, more than 25 percent of edges. So there are lots of particles in here. So proving that there's at least one collision for each one of these is very easy. As long as this is the order of mixing time. But for bounding double collisions you need this extra idea, which comes from these Poisson processes, the coins and everything else.
So in the end you end up looking at events like this. So you define this notion here for to take a vertex V here, you call F of V the first collision time, well, the particle, the colored particle. So sorry the red -- first red or white particle that whoever started from V collided with. There might be none, but it's very likely to have some particle.

So the bad event and this you compute for all these guys, the bad event would be something like this. You have red here and you have -- and you move this way. At this point it collided with some white. So this means that here you have V. Here you have W, which equals F of V. And here you have some D prime that also collided with this one so this guy's also F of V prime. How do you control this? Right? So the basic idea is this. If you look -- so here you had a collision between
V prime and W. At this point you don't look at the coin and this will become pink, right? Well, maybe there is an even earlier collision but let's say this is the first one. But you can pretend you just evolved the process as usual. Look at the coins those two particles flip and they continue. And then you can ask something like this. Well, okay, given that these particles collided, we can say something like this, given that there's a collision here, this guy is equally likely to have
W equal F of V and to have V prime equal to F of V. Meaning I'm looking at the first collision. So when they get here, because of the exclusion property and the way I set up the rates and everything else, there's a coin. So maybe things continue in a way that view hits W.
Maybe things continue in a way that V hits V prime and those three events are equivalent likely. So using this is somehow saying that the event that the two particles meet the same guy is essentially the number of bad collisions this word is essentially the number equal in expectation to the number of collisions within the red set. So the number of bad events, well the expected number of bad events, it's less than you expect the number of collisions within red. And why is this good? Because as we started out, we said we have less red than white.
So you have the expected number of collisions with white minus the expected number of collisions with red that's typically going to be positive. And then you are good to go. So the idea essentially is that once you go through this mixing stage, because these guys are all mixed up here, you can prove that a white particle will probably collide with, sorry a red particle will probably collide with a white one and the probability of double collisions is smaller. And that's it. And that's essentially how you -- so you show with a constant probability the number of pink grows by a certain factor in this phase, and then repeat it and you're done. So I guess that's it. Thanks a lot.
[applause]
>> David Wilson: Any more questions or comments?
>>: What's the best lower bound; it's just a random walk?
>> Roberto Oliviera: Actually not even that, if you have the process with labeled particles, sure, one random walk gives you the lower bound. If you have K unlabeled particles, it's not the case that one random walk is the projection of K -- you essentially have no lower bound, no general lower bound that I know of. And that's kind of you know the most annoying probability you can think of in regards to some of this. It surely is the case that exclusion N over two particles

mixing lower than one random walk as far as I know nobody knows how to prove that.
>>: In general true with log K?
>> Roberto Oliviera: I think it should be true with log K in this generality. Actually my own feeling is if you take the process which
David called the interchange process with any number of particles, and you put any epsilon here, this should be less than T mix of the corresponding system of K independent random walks. Maybe the constant here. And with that. And then you have, you get log K automatically.
But this is very, very far from being something proven or provable even. I don't want to push this so far. But it really seems to be beyond our current knowledge of things. And I actually -- I think that actually this holds both ways. But you can upper or lower bound these two points.
>> David Wilson: Okay. Let's thank all of today's speakers and Roberto. [applause]