>> Yuval Peres: Good afternoon. us about competitive erosion. We're happy to have Shirshendu Ganguly tell >> Shirshendu Ganguly: Thank you, Yuval. So, yeah, so I'll be talking about something called competitive erosion which is a competing particle system on graphs and [inaudible] based on joint work with Lionel, Yuval, and Jim. And so yeah. Okay. So, all right, so for the model, so like I said, it's a competing particle system, so you have two kinds of particles and you have some interaction. So to define the process, you need some underlying data. So what do you need? So you need a finite graph G. It does not have to be finite. So you need some graph G, say, with vertex set V and E and edge set E. And you need two probability measures, mu 1 and mu 2, on the set of vertices. So you have a graph, you have two measures, mu 1 and mu 2 on the set of vertices, and you need an integer K between 1 and -- so suppose the graph is size N, then you need an integer between 1 and minus 1. So you have a graph, you have two measures, and you have a number K. And formally this process will be a Markov chain on all K subsets of the vertex. So you have this graph V, you have this number K, so you look at all subsets of V of size K. And formally this process is going to be a Markov chain on this state space. So to define this, so I need to define how to define how to get ST. So ST was my process, I need to define how to get ST plus 1 from ST. This is what we are going to do for the rest of the talk. So think of all the vertices as having the two colors, red and blue, and so K was my size of ST, so think of ST being the set of blue vertices. So the K vertices are going to be blue and the rest of them are going to be red. And so I have to define how to get ST plus 1 from ST. So what I do is -- so ST of size K. So what I do is I add one point, which is XT, and I subtract one point, which is YT. I'll define what those are. Okay. So yeah. So first you have to add one point, XT. So I have these two measures, mu 1 and mu 2. So what I do is I start a random walk on the graph with initial starting distribution mu 1 and then I wait till it hits this complement of ST. So ST of size K, I start my random walk and look at the first site, which is now red, which it hits, and that I include in my set ST. So ST union XT is now a set of size K plus 1. And now I [inaudible] of one more element to get the right size. So I have this other measure that I can still work with, which is mu 2, so I start a random walk with initial distribution mu 2 and like wait till it hits the existing set of size K plus 1, all the blue vertices, and then at the first site that it hits, I remove it. So I have K. I added one point. It became K plus 1, and then I removed them at their point. So I'm back to the set of size K. And so this was introduced by Jim around 2003, and so I'll come back to some of the motivations later. So the example which will make this definition clearer, but okay. So here is a very basic graph of size six and K is 3, so all the blue vertices are the set ST. And now in this example, mu 1 and mu 2, which are my starting distributions, are just point masses at these two points. So think of this as mu 2 and think of this as mu 1. So first I start a random walk from mu 1, so it does some random walk, so it picks this edge, so it's already hitting ST complement in the first step, so I make this blue. So three became four. And then after they start one more random walk from here. Suppose it picked this edge, so it's still on the red cluster, and then in the next step it hits the blue sites, so this will turn red. And so one step I went from this configuration to this configuration, so the number of blue points remain the same, but they moved around a bit. So is the definition clearer of the process? Okay, good. So this was a very general example, and now we will sort of see an example of a graph which we'll really care about during this talk. And so this is a disk of radius 30 and every vertex is colored independently red and blue with probability of one-third, and so number of reds is two-thirds fraction of the [inaudible]. So every vertex is colored blue or red with probability of one-third and two-third independently. >>: Question. There seem to be more than two colors there. >> Shirshendu Ganguly: >>: So there is some shading it shows, yeah, yeah. So -- Okay, that's just an artifact. >> Shirshendu Ganguly: Yeah, yeah. It's some pixel thing. And my mu 1 and mu 2 are, for this example, just a point north pole and the south pole. So my blue random walk will start from -- so think of this as in the complex plane and if this is the union disk, it will start from minus I and the red random walk will start from I. Okay. So let's see how this looks. So, see, there is some -- the blue vertices are getting accumulated near the blue source and the red ones are getting accumulated near the red source. And they run for some time. So let me go ahead. Okay. So yeah. So you see that they have come close to each other and, okay, of course there's some particles here which are still yet to be colored red, but now they're done. And so you have this like two different regions of blue and red. And then this is boundary or interface between them. And it's sort of like not very deviating. Like it's roughly smooth. Of course the size is just 30. So if you make it more and more, then it was even smoother pictures. So the interface is not very fluctuating. And also you see what the shape of the interface is like. So you can already guess what this looks like. There's some fluctuations, but roughly it's not a straight line. Roughly looks like an [inaudible] circular arc. Right? It might be clear, it might not be. But okay. So good. So this was a general example. Then we saw the example on the disk. And now some background and motivation. So there is some similarities at the very, very basic fundamental growth model, which is known as the internal DLA, but there are just -- there's just one particle involved, so you have a graph and you just try to see how a set of particles sort of grow with time. So we just have this one version. So IT is your cluster at time get IT plus 1 from IT you -- so you have just -- so you now have distribution, mu 1, you start a random walk with distribution mu wait till it exits the cluster IT. You have one more point that your site. So at time TT, the size of the cluster is exactly T. growing by one at every time step. T, and to just one 1 and you you add to So you keep Okay. And so this was first proposed but I think Meakin and Deutch around '86 as the model for several chemical processes like ->>: I think it's pronounced Deutch. >> Shirshendu Ganguly: Deutch? All right. Thanks. Electropolishing, corrosion and -- yeah. And later like so there's a notion of addition of sets which was invented by Diaconis and Fulton, and this turned out to be a special case of that. And this was done around '91. Okay. So this is like one particular version of that. And probably this is done on the square lattice. And mu 1 you should think of in this example as being the [inaudible]. So I'm going to always track my particle at the origin and look at what happens on the square lattice. So the first particle stays at the origin because there was nothing there. So the cluster at time 1 is just the origin. Then I start one more random box, suppose it picks this edge, so you have these two points at time 2. You keep doing this, you have size 3 and size 4. So something like this. So at time T you will have exactly T vertices on the lattice that you are [inaudible]. Okay. So there's one more. Okay. So this is an IDLA cluster growing on the square lattice, and this will run for -- this is about like 3,000 particles. So you see that it's roughly isotropic. So it's almost circular. And there's some -- okay, still not circular yet, but like if you run this for long enough it will be and the boundary is something that has some fluctuation but not much. Okay. So it's roughly growing like a ball. Okay. So some comments about this process. So like I said, so IDLA at time T is exactly of size T. So if your graph is finite then -- and of size N, then IDLA is not interesting anymore after time N. So it's usually studied on infinite graphs. And questions that people ask are about what this [inaudible] cluster shape is like. And once you understand that, well, then you can ask finer questions about what the boundary is, what are the fluctuations. Yeah. So you can ask what the cluster is, whether it's like a ball and what the boundary is once you understand the first question well. Whereas erosion makes sense on a finite graph for all times. And it's a Markov chain, so you can ask -- all the basic questions about Markov chains are still relevant in this setting, so you can ask what [inaudible] measure is like, what is the mixing time, et cetera. Okay. So some references to what has been known about IDLA. So the first rigorous proof of the fact that this actually looks like a ball in some sense was by Lawler, Bramson and Griffeath around '92. And they showed that if a cluster at size N, then the right radius to look at is R so that pi R square is N because you want to conserve area, and they showed that the ball is within R1 minus [inaudible] factor of one plus minus epsilon, the ball is going to be within that cluster with high probability. But then subsequent improvements are made like first by Lawler himself and then two groups, one including Lionel [inaudible], and there were other -there was another competing group. And eventually so there's now very, very precise ->>: That's a lot of Ws. >> Shirshendu Ganguly: Yeah, yeah, yeah. >>: You said Scott, but there may be more than one Scott who works in this field. >> Shirshendu Ganguly: Yeah. Right. So Lionel, [inaudible] in one group; the other group was [inaudible]. And now they sort of really understand what the even not just order of the fluctuations but what the scaling limit is. So there's very precise understanding of this process. And there's another setting which is even more relevant to the case that we are talking about, is when you have IDLA but like there are now multiple particles. So you have -- you still have, say, the infinite graph, so you have infinite room, but there are still particles of various kinds which are trying to occupy your spot. So, the say try for example, like suppose -- so the IDLA starting my particles only from origin, but now I can say pick two points and minus one and one, and they alternately like emit red and blue particles from them. They will not to kill each other once they meet. If a red meets a blue, then it will keep working till it finds an empty site. you find an empty site. So you will have this stuff when And so here is a picture of what IDLA with two source look like. So the center of the disk are the sources and you have these particles, and the colors represent the points that the particles came from. So you have these two sources. So the purple and the green are the places where the particles only from the center of the disk occupied and then in the intersection in the outside you will have particles from both colors. So there is some here. So there are some rules about so the particles are exchangeable. So once both particles land here, you can choose which one you want to like make work again, make them work again. So the red and the sky blue portion are regions where you will have part particles from both sources coming. >>: What's the difference between the black line and the red? >> Shirshendu Ganguly: So here you can -- so potentially these are the two regions where you will have -- you can have particles from both sources occupied. >>: [inaudible]. >> Shirshendu Ganguly: Because -- so there -- so if you emit ten particles from here and particles from here, they will -- so you can first make them work, they will work like independent IDLA. So they will have these two balls. So the overlap of both sides will be exactly in the intersection, and then you can choose which one you want to like make them work again because this [inaudible] typically will have now two particles. >>: [inaudible] what's the difference between light blue and red? >> Shirshendu Ganguly: No, no, so there is no difference -- no, no, no, there is no difference. In the sense that if we run IDLA independently, they will first -- so depending on what time you occupy the sites. >>: [inaudible] generating the two source cluster in a certain order just to make a one source cluster, then you make another one source cluster, and then you take care of the overlap. >> Shirshendu Ganguly: Yeah, so depending on what time you occupy the site, this will -- yeah. Mm-hmm. But potentially because the particles are exchangeable, you can -- does not matter which particle is occupying which site. Okay. So this is still on now we will sort of switch general graphs, but now we very special graphs, which the infinite graph and you have two sources. And back to erosion. And so we've defined erosion on will, for the rest of the talk, we will talk about is sort of approximations of smooth domain. So like the disk, you discretize it. talk. And I will define what smooth means in our And at this point I should also remark that we started this process on a similar different setting with Jim and [inaudible]. The main result of this talk will be joint with Yuval, but there's a different -- but we'll have the setting which studies the same process, and that was joint with Lionel that and Jim as well. Okay. So what is the setting. So think of a smooth domain. By smooth I will mean something which I'll state precisely, and I look at the discretization of the domain using very fine mesh. So say one over NZ2. So think of this as being embedded in the complex plane and look at the intersection with one over NC2 and look at the part of the graph inside it, by which I mean all the [inaudible] completely inside it. So all the points which are in the interior of the domain and induced by them. And the class of domain that we'll work with, we'll have analytic boundaries, which means the local parses expansions and equivalently like the remain wrapping term guarantees that these domains will have maps going to the disk. Because these are simply connected domains. And, in fact, the boundary will be so smooth that the maps sort of will have analytical extensions across the boundary. And for the moment, at least, so the setting will be still like -- the sources will be like still point masses on the boundary. So you have a domain. Take two points, externally extend the boundary. They will be roughly your sources. And the graph will be the intersection of this with a fine mesh. Of course it's still like little informal because these points need not be lattice points and need not lie in the graph but like roughly this is going to be the setting. >>: When you talk about extending across the boundary, you're saying you could extend to an open set that contains. >> Shirshendu Ganguly: >>: Yeah, yeah. Okay. >> Shirshendu Ganguly: So yeah. So you have this domain U and this T, so there open says U1 and D1, continuing UND so that the conformal map from U1 to D1 restricted. Mm-hmm. Okay. Good. So all right. So summary of the setup is the graph is UN which is U intersection 1 over NC2, and so now I have to give a number K, right, so for erosion I needed a graph, I needed two measures, and I needed the number K, which is the set of -- size of the blue cluster. So this is going to be a constant fraction of the number of vertices. So I fix an alpha between 0 and half, and number of blue vertices will be alpha times the size of the graph. So that's going to be the setting. And, again, like I said, mu 1 and mu 2 are roughly like [inaudible] measures. still may not be lattice points. So okay. But they So the basic question is how does a blue region look once you run this chain for long enough and with your sources X1 and X2. So on the disk we saw that the blue vertices [inaudible] the blue source and similar to the red vertices, and there was some interface. And the limiting interface, like it was roughly like an orthogonal circular arc. And the basic question is what is the truth for general domain, so what happens for general domains. And so there is some connection to this process, the Reflected Brownian motion, which I'll make like more precise later, but like it's sort of unknown that formally Reflected Brownian motion is conformal invariance. So and here we were talking about Reflected Brownian motion with normal reflection. And so this led to the conjecture by Jim which is that understanding the process on the disk sort of suffices; that if you want to understand what the process looks like on a general domain, so the final picture should be just the image of the final picture on the disk under some map. Now, one comment that I want to make here is remember that the blue was like half a fraction of the total region and conformal maps don't preserve area. So it's -- so one estimate presents what this means. So but roughly a picture like this should be true with a conjecture, and the goal of this talk is to show that this is true. So are there any questions? >>: This has to also hold for the details of the interface or [inaudible]? >> Shirshendu Ganguly: Yeah, so we will see what we -- yeah, we will see what we prove and then -- yeah. So the interface is -- so one has to even prove that there is an interface. So right. So your question is whether the properties of the interface also carries over, right? >>: Yeah. >> Shirshendu Ganguly: So first one has to prove that there is an interface. So that is not provable at this point. So we don't know how to prove that. I mean, I said we don't know how to prove that. Yeah, yeah. Yeah, yeah. Mm-hmm. >>: Sorry, you don't know how to prove what? >> Shirshendu Ganguly: That exactly a picture like this is true. That there is a -- so everything here is blue and everything -- and so there is a particular interface, and that carries over. >>: The initial state could be just like a checkerboard of when blue and there's no well-defined interface. >> Shirshendu Ganguly: Right. >>: You have to show that over time you move towards something where there does appear to be well-defined [inaudible]. >> Shirshendu Ganguly: Mm-hmm. >>: [inaudible] saying at this time in the talk or the -- >>: Yeah. >> Shirshendu Ganguly: No, so we will see -- so we will see what exactly the main result is, and then it will -- so we will see what exactly the main result is and what it exactly conforms, and then we will see ->>: [inaudible] there's more work to be done. >>: Keep going. >> Shirshendu Ganguly: Right. Okay. So the modified setting -- so till now we were talking about these two sources which are points on the boundary. But like there will be some convergence issues and we will sort of take the message, go to zero, and so it will be technically convenient. And so the main result will be sort of in the setting where instead of having point masses, the two random box, the mu 1 and mu 2, will be sort of uniform measures on small disk. So you have this point X1 and X2. We wanted to start random walk from X1 and X2, but for technical setting, like we will take two disk of radius set delta near X1 and X2 and mu 1 and mu 2 will be uniform measures on all lattice points inside the disks. So fix a small delta, take two disks and look at all that lattice points inside them, and the random walk will start uniformly from those. >>: I know you said the second clause. [inaudible]. And also the distance about >> Shirshendu Ganguly: Yeah. So this -- yeah. So this is a smooth domain. So you can fix a small delta. And then we can choose -- for any delta small enough you can choose disk which are, say, a distance delta by 2 from the boundary and also radius say delta by 4 or something. Okay. So instead of starting from points, we will start from the small disks. And then the main result will sort of involve sending the message to zero, the size of the message to zero followed by sending delta to zero so that we asymptotically sort of recover these point sources. Okay. So we need some notation. So you have the disk, union disk with these two points, IN minus I, and you have this join domain will arbitrary points X1 and X2 on the boundary. So the [inaudible] guarantees that there exists conformal maps going from the disk to disk domain so that I is sent to X2 and minus I is sent to X1. And [inaudible] are going to be inverses of each other. And just worth mentioning that there is no UNIX chip here so because a map is uniquely determined if I specify the values at three points. So here I'm just specifying the values at two points. And so this is actually a family of such maps. So I'll just do choose something arbitrarily and -- yeah. Okay. So and so now so for the disk I had this orthogonal circular arc, and these sort of have these sort of equations. So they turn out to be the level set of this function, log Z minus A over Z plus I. So every orthogonal circular arc which is symmetric with respect to minus INI will be the set of Z inside the disk so the disk is equal to beta for some beta on the real line. And now I want to define a corresponding thing for our general domain view. So just transfer with the conformal map. So serve this disk. This is the orthogonal circular arc corresponding to beta. I call this region D beta and I just look at the composition with respect to phi, which will give me corresponding regions on the general domain menu. And note that this parametrization of beta depends on this conformal map. So this was worth the value of this function on the geodesic -- on this curve is and however we're interested in area, right, so we started with alpha fraction of the area being blue, so we want to sort of find the beta which gives you alpha fraction of the area. So this we call U alpha is U beta for the value of beta which has fraction alpha of the area. As beta goes from minus infinity to the infinity, you covered the whole region. So you can find such a beta. And like I said, so this area is not preserved. So phi of D alpha is not necessarily U alpha. Of course for a particular value of alpha, you can choose your phi to be size that this is true, but that's not [inaudible] for all alpha. Yeah. So this is going to be my fraction is alpha. Okay. And I about the set of blue vertices. region so I look at union of all set of blue points on my graph. U alpha. So I pick my beta so that the need my last notation. So I'm interested I want to pass on the set of vertices to the boxes around the blue points. So I have the I look at all boxes of size 1 over N, which is centered at those points. And this is [inaudible] of the main result which says that so you have the set U alpha, you have the set of blue vertices, you look at the region version of that, look at the symmetric difference and take its area, and the expectation of that under the [inaudible] sort of will go to zero if you send the mesh size to zero followed by delta going to zero. So this blue region here, take the box version which is a region on the plane, take the symmetric and you have this region U alpha, which we just defined, take the symmetric difference, look at its expectation, and then that goes to zero if you send end-to-end infinity followed by delta 20. However, there's some technical requirement. So we sort of need N to go along powers of two to infinity. And this is because some of the proofs here rely heavily on the fact that random walk on this graph's UN [inaudible] motion on the domain U. And the results that are known in the literature uses the fact that that sort of convergence is true only if N is [inaudible] part of 2. Of course, a person has spoken to the [inaudible] of this paper, and they -and they sort of believe that is true for the but the results that appear in the literature uses this assumption that N is like 2 the K. And so in words roughly if the sources are small enough, which means delta is very small, then if the message goes to zero, then the as [inaudible] the blue region, looks like the set U alpha in the sense of symmetric difference. So the area of the symmetric difference is small. So that's roughly the statement of the result. So is this clear, what the theorem is proving? Okay. So a remark which might not be pretty clear is that this chain a has well-defined stationary measure. So it's not irreducible, but still there is one recurrent class, so you can still well define the stationary measure. And also because it's a finite system, even the equilibrium, most of the time it will look like there is an interface and the interface is sort of like this image of the circular arc, which is the geodesic. There will be a positive fraction of time where it will look something very weird. So it does not have to look like this always because it's a finite system. So this has positive probability happening. So it will sort of happen like in a positive fraction of time, even though very small probability. Okay. So these are some of the initial comments. And so now let's see why one can expect this to be true, why you can expect the interface to be like a geodesic. So if there was an interface, then it should have some stability property that so red random walk would start from top and like it will stop when it hits the blue site, and similarly blue random walk would start from the bottom. And so if this interface has to stay in place, for every point roughly the push from both sides should cancel each other out. Right? And the push is nothing but the harmonic measure starting a random walk from here. So you start a random from here, you look at the chance that it hits this point, and that should sort of agree from both sides. This is one of the properties that you -- I mean, if there was some interface, it should have a property like this. And, okay, so now first we have to sort of find the candidate which has this property. So, okay, so here is where the conformal invariant nature of Reflected Brownian motion comes into play. So we don't know the geodesic will have this property. And a way to see this is so you can think of Brownian motion in the half plane. So think of Reflected Brownian motion on the half plane, which is very easy to define. And just by symmetry, you can look at the semicircles, and the semicircles will have this property that the harmonic measure [inaudible] Brownian motion from the origin, and look at the distribution of a semicircle, that will be uniform just by rotational symmetry. And, similarly, if you start Brownian motion, say, from infinity, whatever that means, so from far away, it will also sort of hit the outside harmonic measure will also be sort of uniform. So for the half plane is very easy to see just by symmetry, and then because path properties of Brownian motion don't change under conformal maps and images of semicircles, so if you take the half point of the disk, the images of the semicircle will exactly be this geodesics. And so this fact that is easy to show on the half plane will sort of imply that even on the disk the right curves which has this property will be geodesics. >>: Doesn't look like a geodesic. It should be perpendicular. >> Shirshendu Ganguly: It should be perpendicular here, yeah. is a hand-drawn picture, so yeah. Yeah. This Okay. So geodesics have this property that harmonic measures sort of cancel each other out from both sides. Okay. So this is one candidate for the interface. And now you have some self-correcting thing. So you if you have some other interface which is not like a geodesic, then you can argue that the part which is below this geodesic, this blue curve, say, will have net harmonic measure pushing it up. So the harmonic measure from the bottom will actually exceed the harmonic measure from the dock because they sort of are equal on this geodesic, so you can argue that here they will sort of be more from the bottom. So there will be net up for push in this part of the curve, and similarly the net downward push on this part of the curve. So if you didn't have the geodesic to start with, this side of heuristic tells you that it will sort of eventually sort of push yourself towards geodesic. So those are sort of the right candidates. Now, again, like I said, so interface, the notion of interface is not -- I mean, it does not exist for all configurations. So it's only if all the blues are like near the source of the blue source and if all the red points are near the red source, then only can sort of expect to define an interface. For example, like in a configuration like this, so even in each source, you will have like both red and blue particles. So it can still define an interface, maybe, but like say it's not like it will separate the two sources. So if you have a small disk here, a part of it is red, a part of it is blue, and similarly the other side. So you will not have an interface which sort of separates the two sources. It will sort of -- like it can cut to the middle of a source. Right? So it's sort not take the separation directly but like slightly different it sort of tries to quantify this idea. So that would sort of involve something called a Green function which I will define in a bit. So here's one more sort of heuristic in the same direction which is so I'll define what a Green's function is in a moment, but like let's just go with this. So it's well known that, okay, so you have this graph here. I'm still working with the setting that there's an interface, even though I said that the whole problem of making this formal was because there was no interface. And so you have these two -- you have this graph and there are two region set, the blue and the red, and you can define something called the Green function on the whole graph and on the two graphs individually. So you can look at the blue part, you can look at the red part, so the two Green function that you can define are done at the Laplacian, whatever, so I'll -- so this is like an informal thing. So [inaudible] the Laplacian of the Green function, here is the harmonic measure at this point. And similarly for the red one it's harmonic measure at this point. And we said that this interface should have the property of the harmonic measure to cancel each other out. So you'll have some condition like this, that the green -- the Blue Green function that's a red Green function should have a Laplacian zero near on the boundary, and of course they're also harmonic in the blue part and the red part individually except for the sources, so ignore the sources for the moment. So what you end up with, this difference of the Green function is sort of harmonic everywhere except these two sources. And the actual Green function on the whole graph also satisfies the same property. That is also harmonic everywhere except the two sources. So this difference of Green function has the same Laplacian condition as this Green function. So you have this -- so this is the Green function of the whole graph, this is the difference of the Green function. They have the same Laplacian condition which means that the difference is the harmonic function. And harmonic functional bounded graphs are constant. So these are same of the constant, which means -- and notice that -- so I didn't mention this, but like these two green function are also sort of vanishing. So the Blue Green function sort of vanishes on the red region, and the Red Green function sort of vanishes on the blue region. So on the interface, both of them are sort of like zero. So you have this GN. This function is roughly like zero on the interface, so GN is roughly like a constant on the interface. And the blue region was one side of the interface. It's reasonable to believe that the set of blue sites is all points where GNX is bigger than some constant. So GN was constant on the interface. On one side it was bigger than the constant. On the other side it was smaller than the constant. So roughly the blue region should be occupying a level set of the Green function. Okay, good. So these are heuristics sort of involving the harmonic measure, but this one is a Green function and the fact that derivatives of the Green function are like the harmonic measure near the boundary. Okay. Good. So this sort of gives you an idea of what the blue region is. So it's a level set of certain function. Okay. All right. So now we'll state a quantitative version of the main result which will imply this. So we had this result which said expectation of some area was going to zero if I took these limits. And now we'll have a quantity version of that result. So I need some notation. So okay. So look at all configurations which is almost like what we want it to look like. So there are only a few blue particles in this region where we want it to be all red, right? And this quantifier is -- so you only allow at most epsilon N square vertices of this type, which is in this region, you have a complement and has blue color or is in U alpha and has red color. So at most epsilon N square particles are colored opposite to what they should be. So they're separate from the geodesic and has the wrong color. Because they're falling in the wrong region. Okay. So okay. Good. And so for any such configuration, the blue region and U alpha, the symmetric difference is at most has area at most epsilon because there are just epsilon N squared particles on the other side. Okay? So if I show that the measure of this set is large, then I'm done. Right? Because I want to show the expectation of the area goes to zero N probability. Expectation of the area goes to zero, it suffice that you show that this thing goes to zero N probability. Area of the symmetric difference goes to zero. Mm-hmm. Okay. So this is the formal statement. So, okay, [inaudible] quantifiers, so take a small delta and large N, depending on epsilon, then this measure of this set of epsilon is 1 minus E power of some constant times N square. Okay. So have this epsilon fixed. Then for a small delta and N large enough, the [inaudible] measure of the set is 1 minus exponentially smaller N, square actually. And then these are automatically impressed because this is epsilon [inaudible] arbitrary, I will set epsilon to zero. Okay? So proving this suffices. Okay. So okay. So the key to using the proof is identification of a certain function. So this is two properties that I want to show that omega epsilon is a large measure. That is what I want to show. So I identify a function W on the space of colorings which has -- which is maximizing omega epsilon. And when it's outside of omega epsilon, it sort of increases on average. So you have the safe space of all colorings, you have the set omega epsilon that you want to show has large measure and identify a function which is maximized inside that set and whenever it's outside it sort of increases on average, which is what this means. So if you start from something which is outside omega epsilon, then in one step the value of the function increases. And this implies that -- this should imply that no matter where you start from, you should hit the set epsilon quickly. Because the maximum is inside this set and you cannot continue increasing forever, which you will do if you are outside the site for all times. Okay. And to motivate the construction of discussed says that the blue region So all the blue particles should be K, right? Which means, okay, which candidate for what this W should be this W, recall that the heuristic that I is roughly GNX bigger than K for some K. in this region GNX bigger than K for some then sort of gives you a natural on the space of colorings. Okay. Before I'll define formally what -- or like semiformally what Green function is. So it's harmonic everywhere except blue and red sources. We have these two disks, it's harmonic everywhere else. And the Laplacian is sort of one and minus one on these two sources. Only I'm suppressing some constant, so I actually don't want one and minus one, I want it to be something else. But after constant, this will be the function. And there are formulas you sort of [inaudible] the random walk on the graph, and then you basically look at the time that a random walk spends in one source minus the time it spends on the other source. Now, this integral goes from zero to infinite. So individually these two integrals are infinite because both of them have positive measures, so random [inaudible] spend infinite time in both of the them, but the difference is sort of integral because these two disks had the same area. So what I'm saying is that you look at the expected local time in one source minus the local time in the other source. And that will have these properties. So the Green function that I defined in -- I was talking about in the heuristic is going to be this function. And then the heuristic set that the blue region was roughly a level set of this function, okay, so which means that a natural candidate for W should be this function; that is, some GN to be -- some GN over all the blue sides. So you have some coloring, you look at the sum of GN over all have these two properties, the data increases in average when have some function on the graph, and then you the blue sides. So I said that the W should W sort of maximizing omega epsilon, and the you're setting outside. So these are the two things that are actually like considered the main body of the work. So these are not -- these are technically challenging things to prove. >> [inaudible]. >> Shirshendu Ganguly: >>: Sorry? First one is challenging? >> Shirshendu Ganguly: Yeah, because so I defined my set, so omega epsilon was defined in terms of U alpha, right, and U alpha was defined in terms of this Lyapunov function, right, and here I have a discrete function, so you have to show some convergence. Okay. So like I said, so you have to show something like this is true, that roughly GN, if N is large enough, then roughly looks like this function. Okay. So okay. So I'll -- so I will -- all right. So do you remember what site was? So you have these domains D and U, and site was the average then from U to D. And this was exactly -- this bigger than beta was exactly the region U beta. And if GN is -- okay. So and the proof sort of uses the fact that random walk on this graph UN converges to Reflected Brownian motion on U, and you have to sort of use the fact that Reflected Brownian motion has some nice differential properties. So it is a solution for [inaudible] with some [inaudible] boundary conditions. So you use the fact that the Reflective Brownian motion satisfies some PD and the fact that the random walk heat kernel converges the Reflective Brownian motion heat kernel. And this set of convergence uses -the things that we use are actually very recent results about local convergence. So the fact that random walking Reflective Brownian motion is not very old, so it's pretty new, and the fact ->>: Wait. The fact the random walking convergence [inaudible]. >> Shirshendu Ganguly: >>: Is new? In this reflective setting. >> Shirshendu Ganguly: Yeah, yeah. So if you have a domain and do random walk on that, that it will converge to Reflective Brownian motion. >>: I thought it was classical. >> Shirshendu Ganguly: No, no, no. Random walking Reflective Brownian motion -- depends on what your time length for classical is. >>: For my intuition back in the -- >> Shirshendu Ganguly: >>: I thought it was conjecture -- >> Shirshendu Ganguly: Mm-hmm. >>: Yeah, yeah. But -- yeah. Proving thing is -- yeah. [inaudible] but in special cases -- >> Shirshendu Ganguly: >>: I mean, of course -- For example, on the half plane, it's easy to show. Okay. >> Shirshendu Ganguly: Okay. And so there's a recent paper which sort of proves quantitative bounds for this convergence. So they show local CTL estimates of this convergence. And this is what one can use to show that this [inaudible] function which is used -- which is defined in terms of the random walk actually converges to this continuing function on the whole domain. And then you can use the fact that this heat kernel for the Brownian motion is a solution to some PD. Okay. So what does that imply? So we had this W occurring which is [inaudible] of GNX over the blue vertices. You said that GNX is roughly like this function. So which means that this region was roughly the region where this function was less than beta and this function was bigger than beta, and GN is now -like these functions of GN bigger than GN less than beta roughly the same regions. Right? So this region, U alpha, was defined in terms of this function being bigger than some beta. But now I show that GNX is also roughly like that same function. So U alpha is roughly the region where GN is bigger than beta. And so to maximize W, you want to back in all your blue vertices in this region. Summing GNN over the blue vertices, I want to maximize it. This is a place where GN is bigger than beta. It has alpha fraction of the whole area. So I should put in all my blue vertices in this region to make W maximum. Okay. Which means that the maximum realized in the set of omega epsilon because you cannot allow too many blue vertices in this region for it to do maximum. Okay. So this was first one that this sort of is maximizing omega epsilon. I still have to show this positive drift condition that if you're outside, then you increase on average. Okay. So this sort of uses an energy argument. So look at this graph. Look at this blue region on this side. Look at the red region on this side. So now you have three graphs, one is the whole graph, one of these two colored graphs. And so now look at the effective resistance, which I'll define in a moment, for this whole graph and these two other graphs. So you have this three effective resistances for these three graphs, and it turns out [inaudible] it turns out that this difference is exactly the effective resistance of the whole graph minus the sum of the effective resistance of these two small graphs. So of this blue region -- sorry? >>: Exactly that. >> Shirshendu Ganguly: >>: Okay. [inaudible]. >> Shirshendu Ganguly: I mean, it's not because -- okay, so the interface is not like a line. So you have these two regions of the red. Yeah. So it's not exact. So there is some small error here, but like roughly -- and the small error is polynomial is small NN, so it's 1 over N to the some beta or something. But roughly this is true. Okay? Okay. So let me -- okay. So let me define what refractive resistance is in the setting. So these are not blind sources. You have these two small disks, and you start your random walk from the disk and you stop it -- okay. So you have these two small sources, which are not points, and effectively in the whole graph in this setting is energy of the unit current to the blue source to the red source so that the current through each starting point is one over the total number of sides. So roughly you have some current flow going from the blue source to the red source. That will give you some energy on the whole graph. And effectively the whole graph is that energy. And similarly for the blue and the red regions, the energy of the unit going from the blue source to the interface will be your resistance to the blue part and similarly to the red part. So you have these three graphs. One of them you have these two sources, so you send unit current from one source to the other source, compute its energy. For the other two you have a source and an interface, so send a unit current, compute its energy. And Thomson's principle sort of says that if you look at the graph, look at any flows, then the energy of the flow which is -- the minimum energy of the flow would be only when the flow is occurring itself. So current [inaudible] all flows. And so then you can see why this thing should be non-negative at least. Because if you take the flow which is on the whole graph, going from one source to the other source, and restrict it to the blue part and the red part, they will give you honest flows on them which has the same divergence as this current flow. So they have strictly more energy. I mean, they have at least as much energy as these ones. So RF should be at least as big as the sum of these two, because RF restricted to blue is bigger than this. RF restricted to red is bigger than this. And these are sort of these joint sets. And but then you can actually prove some quantitative lower bound if you started from something which was away from -- so if you had this initial thing that there were at least epsilon N square vertices on the wrong region, then you can actually prove a quantity of lower bound. This is just non-negative but it's bigger than some constant A depending on epsilon. So if I start from omega epsilon complement, you will have some positive [inaudible] which is lower bounded by a function of epsilon. Okay? >>: So I guess I thought the effective resistance will be smaller than the sum of those two. >> Shirshendu Ganguly: Why? Is that these joint graphs? So yes. So I look at the energy of the flow from going from here to here, and I look at the sum of these two, which is the same as shorting this curve, so I should think of this as one point. So what is the sum? The sum is the effective -- the energy of the current going from here to here, if you glue all these points along the boundary. >>: [inaudible] series, you're connecting things in a series. >> Shirshendu Ganguly: >>: You short out the boundary, you make the resistance less. >> Shirshendu Ganguly: >>: But so -- and even so -- It's same potential along the whole boundary -- >> Shirshendu Ganguly: interface. >>: And because this -- No, no, no, so this is some -- this is some arbitrary Oh. >> Shirshendu Ganguly: If this was the same [inaudible] then this would be zero. But the fact that you are starting from omega epsilon -- outside omega epsilon tells you that this interface is not of the same potential, should we actually make some quantitative gain, which is this epsilon. Okay. And so now -- so I'll quickly say how these things will imply the result that we wanted to prove. So you have this function W which increases on average when you're outside. And the maximum is inside this set of epsilon. So this should imply that hitting time of that set should be small. So the first result is that hitting [inaudible] omega epsilon, no matter where you start from, is going to be like order N square. So it's being bigger than CN square is exponential, has exponentially small property. So suppose this is [inaudible] of all colorings and omega epsilon is a small set inside, you start from anymore, in order N square time you hit the set omega epsilon. Right? So this just involves what I just said. But be sure that decision measure, this is close to one. You also have to prove that once you enter the set you don't escape quickly. So this is the next step which is that once you reach -- so you first hit omega epsilon, then you carry on and hit something smaller on epsilon prime. And once you're in epsilon prime, then you will actually take exponential time to get out of omega epsilon. Okay. So these are just submartingale [inaudible] bounds. So the function W of sigma T as T runs is a submartingale, and so you have [inaudible] bounds that will sort of imply this. And this is the same that you -- same argument that you have for a biased random walk. Okay. So now we have two -- so we have the sitting time results, and then we have to transmit that stationary measure. Okay. So this is not very hard. So stationary measure is the amount of time that is spent asymptotically in the set. So you start from anywhere and you look at -- and you want to measure the stationary -- you want to measure the stationary measure of the set, and you look at the proportion of time that it spends in that set. And the hitting time results show that no matter where you start from, in order in square time, it will hit the set omega epsilon and it will stay there for an exponentially long time, right? So you typically spend only order N square time outside omega epsilon or in time interval of exponentially large length. So this tells you that something like this is true; that the mass of omega epsilon is bigger than one minus some -- this is exactly the proportion that you spend outside. Okay. So the proof of the quantitative the main result just because -- just by now I'll quickly wrap up just by making result was in terms of this area of the result is complete which then implies sending epsilon to zero. Okay. So some final comments that the main symmetric difference, right? And in the picture, in the simulation we saw that exactly there was like no red particles in this region UL, so there was some interface, but there was absolutely no red particles in this region. So in general with Lionel Levine, Yuval and Jim, we manage to prove this result in a slightly similar setting, but like which was the simpler geometry. So the setting was we started this process on the cylinder, so you have the cylinder and now the sources are the top and the bottom. So the random box starts uniformly from the top, the red random walk and the blue random walk starts uniformly from the bottom. And there we would actually show that, on equilibrium, and because of symmetry in one coordinate, this should be the interface because there is nothing going on in the other coordinate. So the blue should -- so if the blue is like alpha fraction, then the right height should be alpha, if this height is one. And then you could show that if you look at the region which is outside alpha plus minus epsilon, then with high probability there would be no blue particles in this region above alpha plus epsilon, and there would have been no red particles in this region alpha minus epsilon. But [inaudible] is the fact that some things were very, very precisely known on this geometry, which is not true for general domains. So, yeah, we use the fact that IDLA was very well understood on cylinder. So first we prove the similar [inaudible] for the cylinder that there would be some dust particles but like very small, and then we use some precise estimates to show that those will be wiped out very quickly. Okay. So a relevant question in this setting would be to prove such a term for this general context. So having a general domain and show that even a picture like that says that is true even in this setting. And once you do that, then the next question would be to understand the interface better like people did for IDLA. So once they show that this was roughly like a ball, what was the interface going to be like. And because you have forces from both sides, it's believable that the fluctuation should be even smaller than what happens for IDLA. So for IDLA we know that this -- the fluctuations or the boundaries are like logarithmic, so it should be even at most that much. And then you can ask the question in high dimensions for general domains. And the current condition would be again this Green function intuition which says that the level -- the separating hypersurface should be still a level set of the green -- of a suitable dipole potential function on this region. And then you can ask a question, so I'll start working with two measures which were separate from each other, but then you can ask the other extreme, which is what happens if both the blue and the red random box starts from the same source. And now suppose you have the infinite lattice so you have Z2 but the model is use alternatively released red and blue particles from the origin and they will keep working if it finds a site which is either empty or of the opposite color, so blue work can hit a red guy or and empty site. In either case it will turn it into blue. It so it will treat white and red as particles of opposite color. And it turns out that even though you have like ten to the ten particles, only like a few thousand [inaudible] have been occupied. So most of the activity would be like red trying to eat up blue, blue trying to eat up red. So it would really like fill up these empty sites. But nothing is conjectured in the setting, what is the picture like. So this is one. And then you can ask the question about what happens if you have multiple colors. So this is a simulation where you have like three colors, each of mass one-third and the sources of the roots of unity and then you run it and this is what it looks like. But if you did not start from the setting where every color was equal and you have -- so remember that in the term we had these two sources, we took a conformal map. So those map to the -- so two sources on one domain would map to two sources on the other domain. And then we had this one more degree of freedom which said that you could map this region D alpha to actually U alpha using -- but here, once you fix these three sources on two different domains, you have no control on what the areas are going to be like. So you don't exactly know what [inaudible] set of true here and what exactly is the -- what exactly is the truth in this setting. Okay. So I'll think I'll stop here. Thanks. [applause] >>: You said the -- in IDLA the fluctuations logarithmic. >> Shirshendu Ganguly: Um-hmm. >>: It's just bounding [inaudible] or is there any understanding of what the limit is with log scale? >> Shirshendu Ganguly: Yeah, so it's very well known, and so you should probably asks Lionel because he was one of the -- yeah. >>: So two dimensions. So the largest fluctuation is logarithmic, but typically you think they are square root of log. And there's a limit which is a close relative of gaussian tree field. >>: That's known? >>: Yeah. I mean, you can always start a strength in the results. known for certain class of test functions. You're >>: Yes. So the answer is you're getting gaussian [inaudible]. >>: It's not quite the gaussian free field because the origin is special. But the origin is a special point. So we called it augmented Gaussian. But it's very, very close to Gaussian in the space. >>: [inaudible] normalization. >>: Right. >>: Normalization. Any point is tight. Normalization. The fluctuations themselves are tight. Only the maximum gives you the [inaudible]. >>: Two dimensions you could [inaudible]. >>: So there is a root log normalization. >>: I think this might be the difference between Gaussian and augmented Gaussian free field. So... >> Shirshendu Ganguly: So you had a question? >>: Higher diminishes, fixed dimensions, this type. >>: Is there anything you had the sources [inaudible]. >> Shirshendu Ganguly: Right. So, first of all, the Green function order is more. So if a point was -- so if you first start random from a point, it will be hit like more that than -- so like roughly like log N times. So if you're stick random walk, starting from a point and kill it on the other point. But and so -- and this kind of [inaudible] that's now how because Brownian motion does not hit this point, so we could not make that work that if you take this discrete function and then work with the limiting behavior of the -- how to show the direct convergence so to something like this log Z minus A over C plus A. But one approach would be just without bypassing all this Brownian motion convergence, so you can start with this function log Z minus A over Z plus I, which you know is harmonic and has a singular disk and has a normal boundary condition and the continuum and like restrict it to the lattice. So it's not going to be exactly harmonic. So it can have some error. And then you can show that even with -- so I had this weight function which I was summing this GNX over blue sites, right, so instead of summing GNX over blue sites, you can take this actual continuum harmonic function and then just sum that over blue sites and show that that works. So I have this function GN, I show that GN convergence is some function log of something over log -- log of something over something, right? But you can start with this log of something over something directly instead of having the discrete function. Then it will not be exactly harmonic. You will have some errors. And then but just show that you will still have this positive drift condition even in that case. >>: But that still incurs some technical -- >> Shirshendu Ganguly: Yeah, yeah, of course. Yeah. So this will be a robust way because this set of convergence of random walking Brownian motion is only known for square lattice. So... >>: But this still incurs some technical -- >> Shirshendu Ganguly: sure. Mm-hmm. Oh, yeah, of course. Yeah, yeah. But like, right, >>: I just want to mention some of the motivation behind what I was doing in 2003, which was I was very interested in [inaudible] router and what they might be good for, so I was trying to come up with probabilistic questions from this kind of derandomization would give a close match. And, in fact, this is an example of that. So you can do this kind of competitive erosion and derandomize things. And if you do it the right way, you get the same interfaces, level sets of these functions, conjectural. >>: Right, right, but you mean just doing [inaudible]? something else? So you have to do >>: Here's the thing I always forget. I could look it up in my notes and fine out which is which. There's two obvious candidates for how to do [inaudible]. Depending upon whether the red particles and blue particles use the same rotors [inaudible]. And one of them gives you the interface you expect, creates microscopic disturbances of the interfaces. I never understood it and I can't remember which one was which. anyone is interested, it's kind of -- But if >> Shirshendu Ganguly: I would imagine the independent [inaudible] would sort of give you -- if you use two different set of verticals for the two walks, probably it should make the random walk closer. And it's not clear. >>: [inaudible]. >> Shirshendu Ganguly: Oh, opposite? >>: The red should spin the rotors backwards from the blues. Never tried that. That would be the way to do it. Same direction for both of them. Basically one of them you can definitely see there was a kind of a skewing, a clockwise skewing of the interface in a very systematic way. >>: Do you is still have pictures? >>: I could dig them up. >>: It be nice to [inaudible] there are several natural versions, and only some of them work. >>: The other thing I'll mention is I did an experiment with three colors using the kind of the same thing that you did on the last slide, but where it was lopsided, or not symmetrical, and there still appears to be a median clock where the angles go around in 20 degrees, which is a sort of a natural [inaudible] ->>: But how did you arrange to keep the math equal? >>: The areas were unequal. >>: No, but how did you arrange to keep it unequal? >> Shirshendu Ganguly: So what was the rule? Did you -- So you -- >>: What was the rule? Each time did you -- >>: I believe the rules just cyclically alternate between them. >>: That should be equal to the [inaudible]. >>: Sorry? >>: That should equal like the mass. >>: If you introduce a red particle and then the blue particle and then the yellow particle ->>: Because each time it captures something of another color. reds are in the minority. Then when -- so when the ->>: I remember now, yes. >>: That's the right. Suppose the It's the particle that just captured -- >>: So when red steals from yellow, then yellow gets to go next. yellow steals from -- I'm pretty sure that was it. >> Shirshendu Ganguly: Yeah, yeah. >>: Yeah, that's the way to keep it -- >>: Yeah. Then when >>: Otherwise it's -- >>: Anyway, the pictures are part of a Mathematica notebook, so we can see what the code does, but I'm pretty sure that was it. >>: We just went through the same realization. >>: Thanks. [applause]