>> Yuval Peres: Good afternoon. We are happy to have Daniel Ahlberg who always finds the
best of both worlds. He will tell us about competing growth in urns.
>> Daniel Ahlberg: Thank you very much Yuval and thanks for the invitation to give a talk here.
This is going to be a talk on competing growth and urns. This is joint work with some of my
collaborators, Simon Griffiths, Svante Janson and Rob Morris. Competing growth has been
studied in several contexts and for a variety of reasons. For example in the setting of first
passage percolation it was introduced in order to study the geodesic structure of the first
passage metric. Here I'm going to study a somewhat different model which has different
features. And then we are going to go right away and describe it. It's going to be a competition
for space between two colors on set two. And the rules for this are going to be the following.
And uncolored site can take to color either red or blue and that happens with rate one if it has
precisely one infected neighbor of that color, or it will happen immediately if it has at least two
infected neighbors of the color. An uncolored site turns red or blue at rate one if one red or
blue neighbor or at rate infinity if at least two neighbors. What we observe is if we start with
one blue site and one red site is that they are going to be growing and this effect here of instant
infections when you have more than two colors already will make it to her you are never going
to see any insight corners that will expand out immediately.
>>: [inaudible] you have one red and one blue then what do you do?
>> Daniel Ahlberg: Then it is used as rate one. It has to be at least two of the same color. If
there is one red and one blue it is still just rate one. You turn it red and then rate one you turn
blue.
>>: You never have two red neighbors and two blue neighbors?
>> Daniel Ahlberg: No. That is never going to happen. And if it happens that means that the
guy is surrounded anyway so whatever color he takes it won't matter for the remainder of the
process.
>>: But in any case you just get one thing happening according to the first row and the cascade
happens.
>> Daniel Ahlberg: Yes. Since there's only going to be one new creation at the time then an
event in that situation, unless you are already starting configuration. After sometime this will
evolve. Maybe the blue has occupied this space and the red has grown to occupy this part and
exactly how it looks here in the interface between the two doesn't really matter because what
will determine the continuation of the process is just the boundary. Typically, you will have
each monochromatic thing look like a rectangle, but there will be some competition here and
you will have these inside corners. The question we asks is what happens as time goes to
infinity. Can both colors coexist or will we have one ending up surrounding the other? In first
passage percolation if you look at some competing growth for first passage percolation what
we know already is there is a possibility that two colors can survive. And in this model we have
some bootstrap affect that somehow balances out shapes like this. And that will have some
effect on your growth of course and the question is whether the answer will be the same or will
it be a different answer and how does one prove that? What will happen is that this side will go
off in this direction and will be affected will depend on how long the sites of its neighbors is but
that is what is going to cause him to grow. At the same time this will grow but will be kind of
blocked by this guy who is trying to grow in the other direction. So we must somehow take into
consideration all these factors and see what will happen and we can right down a system of
differential equations someone says that this is what ought to happen. And trying to throw
around a the test as to what somehow governs the growth it's like the length of the perimeter
of the red and the link to the perimeter of the blue. But then in order to use that to try to say
something about what eventually will happen, one has to also somehow compare because
these areas are more somehow important than what happens around here because here is
where you really try to eat up your neighbor. So do you have a guess for the answer to your
question?
>>: Could I ask a question?
>> Daniel Ahlberg: Sure.
>>: Are there finitely many overall shapes that you can have? I mean one shape is these two
rectangles like you have drawn, yet then maybe one has to each side or something?
>> Daniel Ahlberg: You could have something like the…
>>: Something like these things that could happen?
>> Daniel Ahlberg: Like this, yeah. You'll have…
>>: Always an octagon.
>> Daniel Ahlberg: Yep, unless you somehow have this situation where you only get six wedges.
So you can have something like this and then red or the blue could also grow out in this
direction somehow and then you would have, it's too easy like that. Those that only think that
can happen. Yep. So the answer to your question will be that the answer is no, yes. You
cannot have coexistence. So theorem, the two color competing growth model has almost
surely a single survivor. Yes. Here we come already to the first open problem. I promise I
would have several open problems for you. The first open problem is what happens with three
colors? We do not know what the answer is for three colors. That may sound a little bit
ridiculous and it is a little bit ridiculous because we know that four colors cannot coexist. In
fact, an even number of colors cannot coexist but odd number of colors we do not know.
>>: So we could say they cannot coexist so lease one of them will die out?
>> Daniel Ahlberg: Exactly. And the reason for this is that if you have something, we have say
alternating stretches of colors we could look something like that. But you always can do if
there are an even number of colors, you can always go along the boundary and just recolor
them alternating because this stretch of blue denotes that next to him I have some other color
on each side, but this guy over here I don't know which color he is. I just know that he's
different from that guy because there is an inside corner here, but I can't tell if he has my color
or someone else's,. So you can always recolor for everything into two colors and since we can
solve this, I said here we would start with one-on-one, but we could start with and a finite
configuration and the answer is the same for two colors. So knowing that we can also handle
the situation where we have more colors. But three colors is still an open problem.
>>: Relationship shows that there is no coexistence?
>> Daniel Ahlberg: We have no simulations, but I tell you it sounds pretty ridiculous if the parity
of the number of colors actually determined the outcome. But no. I don't have any
simulations. Before I come to talk about the proof of this, I'm going to describe something else
and this is where the urns come in and it's going to be something that I call competing urns.
We're going to have a competing urn on a finite connected graph, so we start with finite
connected graph. And each vertex we put an urn and each urn can contain either red or blue
balls that they can't contain both colors at the same time. So in this competing urns scheme
here what will happen is that discrete times null, I will first pick a ball uniformly among all balls
in the system. Second I will return the ball to where I took it but I will also place a copy of the
ball to each of its neighboring urns. And finally, I said that red and blue were not allowed to be
in the same urn at the same time, so if I attempt to place a red ball in an urn where there is
already a blue ball, they will annihilate. The red will annihilate with one of the blue and
therefore the effect is that I actually remove balls. The number of balls will decrease in that
time study. This is the competing urns scheme. Yes?
>>: Ball uniformly, not the vertex uniformly.
>> Daniel Ahlberg: No. A ball uniformly among all balls in the system, so all balls play the same
role. They are equally important independent of their colors.
>>: Can you place the neighboring -- is all the neighboring?
>> Daniel Ahlberg: Yes. You have some graph that maybe looks like this and if I pick a ball from
this urn I will look at it, what color it is, put it back. If it was blue then I put one blue in each of
the neighboring urns as well.
>>: [indiscernible] configuration of urns?
>> Daniel Ahlberg: There is one urn in each vertex and then I start with some number of red or
blue balls in each urn and I can say, for example, I start with one red here and one blue there or
something else.
>>: But they are monochromatic?
>> Daniel Ahlberg: Each ball is monochromatic.
>>: But the number of balls you need is completely arbitrary?
>> Daniel Ahlberg: Yes. Could be arbitrary, finite, but arbitrary.
>>: But you don't place a new copy in the same urn?
>> Daniel Ahlberg: No. I return the ball that I looked at and then I do that with the neighbors.
And the reason why do this will be apparent now because the observation I'm going to tell you
about next is that this competing growth model I described first is exactly this urn scheme on
the cycle of length eight.
>>: When this shape happens to have a client or…
>>: Otherwise it's out of the cycle?
>> Daniel Ahlberg: No. In this case it's not the cycle of length eight, but here it's still a cycle of
length eight and I will tell you about it.
>>: [indiscernible]
>> Daniel Ahlberg: Yes.
>>: So that initial seed it will be…
>> Daniel Ahlberg: Yes. On that initial seed we will have the cycle at one, two, three, four, five,
six, seven eight. The initial setting where we will have one blue and one red will correspond to
the case where I have one blue here, one blue here, one blue here and then I have one red
here, one red here and one red here. And these two are initially vacant. What would I have
here? I have only and coded the length of the perimeters of the different ones, so length one,
length one length one. Length one, length one length one and here I'll have two latent sides
that will only pop up when I have something growing.
>>: Can't you tell whether you have a latent side by looking at the shape?
>> Daniel Ahlberg: It can only be…
>>: [indiscernible] change colors.
>> Daniel Ahlberg: Yeah. So there's always something in between where I have go from blue to
red.
>>: Okay. But then in that picture you would have ten sides.
>> Daniel Ahlberg: Here?
>>: Yeah.
>> Daniel Ahlberg: No. It should only have eight sides.
>>: Eight real sides.
>>: So sides are alternating horizontal and vertical except [indiscernible]
>>: The question was how from looking at the shape how do you know decide that you have a
latent side? You said it's whenever you change from red to blue?
>> Daniel Ahlberg: Exactly.
>>: Without [indiscernible]
>> Daniel Ahlberg: It's horizontal here.
>>: Okay. Without changing direction?
>> Daniel Ahlberg: Yes. There is always horizontal, vertical, horizontal, vertical and so on. So if
you don't see that then something is wrong and that means that it's a latent side. And why is
this the correct way, the correct model? What happened here if this side would advance one
step? It does proportion to its length, so if the length is the number of balls in there then you
will jump one step. His length will remain the same, but it will increase the length of his
neighbors. And it's the same if this one would advance one side, he will increase this neighbor
by one, but it will also eat up his blue neighbor. It's rather straightforward to convince oneself
that this actually is just the same process.
>>: So you are using the fact that the same color corners are convex and so the opposite color
corners are concave? If in the top left it was red, then red with the red, but you don't get that.
>> Daniel Ahlberg: No. You never get that because somehow you always have these corners
where there's…
>>: [indiscernible]
>>: And so is this equivalent for any number of colors?
>> Daniel Ahlberg: For any number of colors, yes. You can do the same, yes. But the graph will
change. For example here it depends on some multiple of whatever you have. Here you have
one, two, three, four, five, six different colors. I guess it would be eighteen or something. But
you just have to do the calculating.
>>: What colors couldn't you end up with a number of convex shape? I mean couldn't you
have like a C shape?
>> Daniel Ahlberg: Yes. It doesn't necessarily have to be convex, but what you will see is this is
convex to begin with.
>>: You never let it…
>> Daniel Ahlberg: Well you could close in and then everything collapses. The evolution
collapses, but that can only happen finite number of times and then you restart. That is
something that can happen, but it only happens a finite number of times and when it happens
you will just have, be in the situation like this and here it cannot happen. Here what can
happen is that what happens from here is you can go to a situation where it looks like this and
red is here in between, and then eventually blue could close in on the red and of course then
everything collapses. But we don't care because we know that red doesn't exist anymore.
Theorem two is that for any finite connector graph the competing urn scheme will have only
one winner.
>>: That's with two colors?
>> Daniel Ahlberg: With two colors. I should emphasize that, for two colors. Another
observation is that there is a reason why we can't prove this theorem for three colors and are
proof goes through this theorem which is more general because it considers any finite
connector graph. But theorem two is in general false for three colors. And I will give you a
counter example. We take the following graph and if you start with these two guys, red, these
blue and these two green and we may assume they have a large number in each of these
components and that they are also very equal work what will happen if red is trying to conquer
the others he will first have to get the good advantage here in the middle, but while doing so it
has to compete against to others of his opponents which are equally strong. So for him to grow
large in the middle he has to have like a large expression of a random walk with a very negative
[indiscernible] and that's unlikely to happen.
>>: So the center keeps coming back zero.
>> Daniel Ahlberg: Yes, exactly. So theorem two is false with three colors and then the obvious
open question but we still expect for some graphs three colors to be able to coexist. And in
particular, what happens for the cycle? The cycle is interesting because it's really what
corresponds to the first model. Open question number two, may three colors coexist on the
cycle, cycle of arbitrary length now? And even better it could even give us like a criteria for
graphs for when you can have three colors coexisting and when you cannot.
>>: So the relationship between these two, this would involve…
>> Daniel Ahlberg: If you can do that problem for the cycle then you do this problem.
>>: If the answer was no [indiscernible]
>> Daniel Ahlberg: Or the other way around.
>>: If the other answer is no then the answer to that is no. Is that right?
>> Daniel Ahlberg: Yes.
>>: Any to annihilate here, you couldn't have, also some graph on the others and have some
color? Here you wrote red into annihilate, but three colors any two annihilate?
>> Daniel Ahlberg: Any two annihilate, exactly. Yes.
>>: You could think of red, you know, blue and green coexist peacefully but they both
[indiscernible]. So with red that would kind of reduce.
>>: [indiscernible] red and green would combine?
>> Daniel Ahlberg: Yes. You could say almost the same.
>>: Yeah, but from, you could have a situation where only three balls of the different colors
forming a relation curve.
>> Daniel Ahlberg: Yeah, you could also do something like that. Okay. Let's have a look at
what we do to prove this. Or if we can prove. And they would become more evident why we
do, our proof works for two colors, but it doesn't work for three colors. Our first proof of this
theorem included, we used this language of stochastic approximation algorithms, if you know
what that is, which essentially like random sequences you can split into a [indiscernible] part
and a drift part and the [indiscernible] part gets weaker and weaker over time, so it will
disappear in the end and so when that happens you somehow will, the drift term will tell you
where you will end up based on some vector field which leads to some local attractor.
However, there is a problem with that approach because it requires you to know the number of
balls in your system is growing linearly. And when the underlying graph is a regular graph we
managed to prove that, but when the graph is not regular it was more complicated. So after
why we ended up with this other proof where we instead are going to embed our system again
in continuous time and used techniques on branching process theory. First step of the proof
embed into continuous time and let y of t be the vector that counts number of reds minus
number of blue at each site. And here already see why it works for two colors because we can
count red positively and blue negatively. How to do it for three colors you would have to have
some, another dimension.
>>: So here you are telling me that every ball has a poisson clock which means [indiscernible]
>> Daniel Ahlberg: Exactly. So assign a poisson clock and now, of course everything will be
growing exponentially fast, but it's just like embedding in continuous time and results will be
the same. It's just a technicality. And this at a time t. What we show that if I take a limit as
time goes to infinity of e to the minus lambda t, whatever we explain soon what lambda is, we
will have a limit which will be some random variable times pi. And surely where lambda and pi
are, lambda is [indiscernible] eigenvector of the adjacent [indiscernible]
>>: Eigenvalue.
>> Daniel Ahlberg: Eigenvalues, sorry and pi is the eigenvector, the corresponding eigenvector.
And w random variable and the probability that w equals zero, that happens with probability
zero. And this is precisely what tells you then this is condition that says that really the number
of balls in the system grows linearly that this random variable here cannot be zero. So it tells
you that on the long-term you will have, the number of balls will be some random number
which would be either positive or negative. Positive if red is winner, then red will distribute
according to some proportionate to the [indiscernible] eigenvector and if w instead is negative
it means that blue one and we can make this conclusion precisely because this eigenvector is
known to have all entries strictly positive. So proving this proves the theorem, theorem two.
Let me first comment on that we know that this holds if we only have one color. If we only
have one color then there is no annihilation. And the process here is just a multi-type
branching process in continuous time. And then from the classical theorem of Markov
branching processes we know that this limit exists and this is a random variable which is
nonzero almost surely. So what we need to do is somehow argue why this also holds in our
case when we have an annihilation in the system. It's the annihilation which is a little bit
complicated and how we handle it is by actually converting the system into a conservative
system where we don't have annihilation anymore. I will do that by instead of annihilating red
and blue points, we will just merge them into forming a purple ball instead and then I will take, I
will continue looking after the purple balls to help me out in the process. So by doing this, what
I do is I somehow couple the balls. I have a red and a blue, like red nucleated eats a blue, so I
keep both of them but I somehow couple them so that in the future they will always have
descendents at the same time. So we look at a conservative system, so red plus blue equals
purple.
>>: Is it two purple?
>> Daniel Ahlberg: No, one purple. It doesn't matter. One purple. It does matter a little bit,
but one purple. [laughter]. But I'm going to do -- one purple.
>>: It's not too late to change your mind.
>> Daniel Ahlberg: No. One purple. Okay let's r of t, e of t and p of t, the vectors counting
these balls.
>>: What says something is conserved, but yes.
>> Daniel Ahlberg: Yes. Instead of annihilating the red and blue, I still keep them in the system
but is purple balls. And purple cannot interact with red and blue. So wants to balls merge to
put become a purple ball, they will not interact more with the remaining balls in the system.
They will continue to give descendents according to their now, and poisson clock but they will
not interact with the red and blue balls in the system.
>>: What is conservative?
>> Daniel Ahlberg: It means that I don't annihilate balls. I will keep them in the system.
>>: But you annihilate one of them.
>> Daniel Ahlberg: No. I merged them.
>>: But the number of balls…
>> Daniel Ahlberg: The number is not conservative.
>>: Nothing. He's got this thing going, for example, you could just increase the number, you
could have two red balls.
>> Daniel Ahlberg: So think of each purple ball as two.
>>: Another thing is you could just add two red ball to the system.
>> Daniel Ahlberg: Yes.
>>: So in some sense nothing is conserved. You can change the number.
>>: The mass of the purple ball is twice the mass of the ordinary ball so mass is conserved.
>>: You can increase mass by just adding two red balls.
>>: The process has a lot of increase of mass.
>> Daniel Ahlberg: But when the red ball, if he nucleates and produces two red balls them
maybe one has to merge with the blue one to form a purple one but it's still there.
>>: I guess the reaction is to the word conservative. [indiscernible]
>> Daniel Ahlberg: Everything is going to be increasing, but I mean as soon as you have a
nucleation everything is increasing. So conservative maybe is the wrong word, but it just means
that no annihilation. Okay, start. But the observation here is that if I count all the red plus the
purple that is just the number of balls we would have this there were no blue in the world. In
particular, since I know that I am monochromatic system that I have some conversion,
convergence, multiplying each by lambda t this converts some random variable wr times this
eigenvector. But at the same time if I just count blue plus purple that's the number of blue I
would have if there were no red from the beginning. This will also have to converge to some
variable wb then this happens almost surely. The good thing about this is that what I'm trying
to track is the number of red minus the number of blue, which I can just take by taking
differences. So this implies that et minus lambda ty of t is just e to the minus lambda t red
minus blue, which then will have to converge to wr minus wb times pi. So we have
convergence.
>>: And it also shows it's a martingale. It's the difference of two normal martingale, which is a
martingale just changing sides.
>> Daniel Ahlberg: Yeah, I guess.
>>: But wr and wb are not independent?
>> Daniel Ahlberg: No. They are not independent. They are certainly dependent, which means
that I cannot directly conclude that wr minus wb cannot be zero.
>>: Precisely.
>> Daniel Ahlberg: And that's something that's missing still and that's going to be the third step
of the proof, essentially, and the last step of the proof because once we know that we are
done. In the third step we need to show that this limit is nonzero. And how would we do that?
Now we need to include another ingredient. What we're going to do is essentially condition on
what happens in the first nucleation. The first nucleation pick a red or a blue. Let's assume it's
a blue ball and the ball will nucleate. I will mark each of its descendents and his descendents
will have marks on them and their descendents will in the future also have marks on them. I
will keep track of them separately.
>>: [indiscernible] blues that was created in the first nucleation?
>> Daniel Ahlberg: Do. I pick a blue ball. I will not mark him, but he will produce offspring and I
will put a mark on each of his offspring.
>>: [indiscernible] go away?
>> Daniel Ahlberg: What?
>>: Also the purple ball stay?
>> Daniel Ahlberg: What did you say?
>>: If you mark the blue ball for the red ball?
>> Daniel Ahlberg: The purple will also keep the mark, yes. And purple mark will also have
descendents which are marked. So the marks are kept for all descendents. Why will I do that?
Because in one nucleation I can only give rise to one color. So the first is blue then this will be
blue so keeping track of them separately will guarantee that we will have convergence with
something like this. We know this is nonzero because it's monochromatic. So mark balls of first
nucleation and all their descendents. So for marked balls what will happen? Let's call them m
of t the mark balls. If I multiply by this they will converge again to some random variable wb if
we assume that they are blue. And then if I suppress the first nucleation, so I marked
everything here. And if I just ignore them and keep track of the system without the first
nucleation, it will just evolve like the system where the first nucleation was suppressed. And it
will converge again to this random variable here just with a time delay because I am
suppressing for some random time and then we'll just differ by a positive constant.
>>: Is this the conceptive system or the original system, the statements that you are making
now?
>> Daniel Ahlberg: I'm making this system in the conservative system. We are working in this
system for the remainder of the proof. Yes. If I look at the system with first suppression, first
nucleation suppressed then I will have something e to the minus lambda ty put a tilde on it.
This will converge to some, put a prime here because it's a copy of that w there times some e to
the minus or yeah, to the minus lambda tall, some random time for the first nucleation to occur
times this vector. But if I also somehow keep track of everything at once, I will still have the
regular system, competing system, conservative system…
>>: Mark [indiscernible] marked ball, the one that you returned is not going to get the mark,
only the -- so why do you get the exact? The mark balls would be smaller than all the blue balls.
Why would you get the same [indiscernible]?
>> Daniel Ahlberg: It's not necessarily the same, but for any initial configuration. It depends on
the initial configuration. It certainly depends on where those balls landed and depends on
which ball that nucleated and so that's not the…
>>: It's not the same wb that you have there? If you can…
>>: It's wn.
>> Daniel Ahlberg: What wn?
>>: Now I'm not sure.
>> Daniel Ahlberg: Yes. So it's a different random variable, but the important thing is
probability of this random variable being zero is zero. And this random variable here is a copy
of this random variable and have the same distribution. But if I would keep track of everything
at once, then all this system would also evolve as, and have this limit. So if I manage you that
what will happen is that I have w pi. That's the limit of e to the minus lambda ty of t, but this is
the same as just a superposition of the suppressed configuration plus the marked balls, which
will converge to wm plus w prime e to the minus lambda tall pi. So somehow I did a relation
between this guy and this guy. I am not been a precise enough for all of this to make sense
now. I'm going to make that clear. Because what can happen here is that I need to make the
marks jump around a little. The reason is what could happen is, imagine that here I have a blue,
it's nearest two neighbors. Its neighbors at least one of his neighbors is red. What the other is
doesn't matter. Imagine that he is the first to nucleate, so he will produce marked offspring.
But I will also want to keep track of these marked balls. I will keep track of the suppressed
system and I will keep track of the whole system. So if I want to keep track of the whole system
I need the offspring he places here to annihilate with one of the red, the former purple, which
will affect what happens in the suppressed system. For example, at a later time comes another
blue which is not from a descendent of the first nucleation and wants to annihilate with that
red which isn't there anymore. I have to somehow allow the two blues that are there, one that
is marked and one that is not marked, they change so that the unmarked blue actually pairs
with the red instead. So the marks, if there's a purple mark ball in the same urn as a blue ball
which is not marked, the mark has to change to the blue ball.
>>: The blue grabs the marks from the purple?
>> Daniel Ahlberg: Yes. So that's a [indiscernible] I need to make this correct. But the point is
if I have this, since I know that this guy is nonzero, so w prime and w cannot be zero at the
same time. But since they have the same distribution, it importantly means that we probably
that most 1/2 they can be zero. So this implies that probability w equal zero is less than or
equal to 1/2.
>>: Is w here equal to your m plus w prime e to the minus lambda tau sigma, [indiscernible]
lambda tau, okay. And tau itself has a continuous distribution. So does not already tell you
that w has a continuous? The w is the sum of these two terms and tau is independent of
anything. If you condition on w [indiscernible]
>>: Wm depends on it.
>> Daniel Ahlberg: Yeah, so I think these guys are not independent and I don't think I can make
that.
>>: But tau is independent of everything, right? Tau is the time of the first nucleation.
>>: So the time in which wm starts to…
>>: [indiscernible] somehow conditioning on the nucleation being on blue.
>>: What are the same? What did you say have the same distribution, w prime and…
>> Daniel Ahlberg: Yes, w and w prime.
>>: That's all promising. I mean conditioning on tau. Then you can look at the some of the
second nucleation which is… It seems like you still have some continuous round of this.
>>: [indiscernible] the existing proof [laughter].
>> Daniel Ahlberg: Maybe it's possible that it is like that. But anyways, from this it's very easy
to also reach the conclusion, because what we will do now is this here holds and it has to hold
independent of the starting configuration.
>>: Why did you say that this is less than half?
>> Daniel Ahlberg: Because this thing has to equal to this over here and that guy and that guy
are the same distribution then neither of them can be…
>>: Because wm is never zero.
>> Daniel Ahlberg: Never zero. The result of them is monochromatic evolution. And this
happens independently of the initial configuration. So in particular, we will apply now
[indiscernible] law which says that the more and more you observe of your system the more
certain you are going to be of your outcome, so really you would converge to like the indicator
of this event. But if that indicator can't be more than 1/2 then only on a null set that you really
have zero as the limit. So result follows from that.
>>: [indiscernible] was positive then you could condition on some initial evolution where the
probability would be .9 [indiscernible].
>>: So this is true no matter what [indiscernible]?
>> Daniel Ahlberg: Maybe I would need to give some more details in order to precisely
convince you that this is a proof, but I will not do that because I also want to end with a few
more open problems. What is really interesting we think is that first the connection between
growth model and the competing urns scheme. But then also this model is quite interesting in
its own right because you can ask many questions for the competing urns scheme. There two
questions about open problem when it comes to the existence of or the possible coexistence of
three or more colors. But other problems, other open problems concern what happens for the
competing urns scheme now, for example if you study it on [indiscernible].
>>: This is how you prove that you are doing pure math, right? [laughter]. [indiscernible]
>> Daniel Ahlberg: Yeah, but it's interesting I think because you can ask many questions which
all relate quite a late to other known models which, for example if you start with just one ball at
origin on set D, how will evolution look like? You will have some exponential growth in the
volume but how will the spread look like and somehow it's an interesting question to try to
determine what happens.
>>: It's even on set, right?
>> Daniel Ahlberg: Even on one dimension.
>>: And that's more relevant to the cycle itself as you say.
>> Daniel Ahlberg: Yep. So even in one dimension there are nontrivial questions to ask. To
focus here on the competing aspect is that what happens if you start with two balls, one red
and one blue? It could even be on set. And is there a possible for coexistence or is it not? Let's
start writing the problem. For one initial red and one blue on set D is there positive probability
for coexistence? In one dimension you will have something like this. You will have some
interface in between the two at some point and then you would have something that looks like
that perhaps. You have, what?
>>: You have to color it.
>> Daniel Ahlberg: Yes. One blue and one red. What should happen here is the volume should
grow exponentially because we have this in continuous time. But the growth horizontally
should still be linear. It seems to be, it's not clear what will happen and if there is a possibility
for both colors to coexist in this model even on the line because it's not as in first pass
percolation where once they separated there is no interaction anymore. But here the one
that's stronger could, gradually eat up the other. That would be a very slow process in the
beginning if they are rather equal but the question is what happens over time. And we don't
really know what will happen with this interface here. Maybe it's the case that there is positive
probability for both to survive indefinitely but that somehow one should maybe have an
advantage and maybe somehow have an interface pushing in that direction. But it's really not
an easy question because it depends very much how the interface looks here.
>>: [indiscernible] is it possible to have two of them survive and the interface moving to the
left?
>>: Actually, the simulations seem to suggest that it is.
>>: But isn't that impossible? If you have positive probability of…
>>: I don't see where that should be impossible.
>> Daniel Ahlberg: In order to try to answer these questions somehow one would have to
understand what happens here on the boundary in between the two. Is this something that is
very slow, low and one rather quickly each of the other or would it be rather high and somehow
-- it seems hard to actually come up with an answer. To begin with is there a possibility for
coexistence and if so what can be said about the interface? Instead of starting here with two
balls or a finite number of balls in the beginning, why don't we start with an infinite number of
balls? For example, we could start at each site we could toss a coin to determine whether to
place one red ball or one blue ball. We start with a random configuration in the whole lattice
and as time evolves this looks more similar to maybe a glabo [phonetic] dynamics or something.
And in glabo dynamics there's a big open question whether you will stabilize in all pluses or
minuses and you can ask a similar question here. This recording you use in the beginning is
biased, so for p larger than 1/2, p would probably be red, is it the case that each, or eventually
turns out to be all red or all future, or will it be that you fluctuate between red and blue in
between times? And natural conjecture would be that for p larger than 1/2 you would stabilize
at blue, but to be equal to 1/2 you should fluctuate in between being red for blue.
>>: That's a setting where [indiscernible] something, right? [indiscernible] everything in
translation barrier?
>> Daniel Ahlberg: Yes. [indiscernible]
>>: So maybe does that tell you in then that it has fluctuate?
>>: It's not stationary [indiscernible]
>> Daniel Ahlberg: No, because you have the volume growth. You don't have just one guy at
each position, but like somehow the law of everything changes. So I don't think you directly
could get like a cheap answer to the question.
>>: To which question? Could you have some of the lattice coming, settling on red and another
point settling on blue? Suppose they [indiscernible] then they would keep somehow, one
would have to [indiscernible] the other.
>>: [indiscernible]
>>: [indiscernible] [laughter].
>>: You eliminate in C [indiscernible]
>>: So if you what?
>>: I mean if you run timeframe, have had a large time you will either see a big red window or
a really big blue window around you. It doesn't tell you what happens when you fix the time
and then look at a really big scale.
>> Daniel Ahlberg: Yes. And typically with larger and larger components, supposedly, but what
would happen with the boundary in between those components? They can grow maybe slower
and slower and slower or move slower and slower and slower.
>>: But in the limit with some probability everything is red and in some probability everything
is blue.
>> Daniel Ahlberg: Yes. I think.
>>: Because this event that everything is eventually red, because there is an event that every
[indiscernible], that event is shifting back.
>>: And so it has to fluctuate.
>>: But it could weakly converge to the mixture of that [indiscernible]
>> Daniel Ahlberg: So what indicates that if p is larger than 1/2 we actually know the answer
and Ohn's [phonetic] answer is that you fixate on all red eventually on each site with probability
one fixate. Let me just finish by writing this down. Another theorem, [indiscernible] p
[indiscernible] to determine whether to place red or blue at each site independently if p is
strictly larger than 1/2 then each site fixates at red almost surely. So question, what about p
equals 1/2? I'm going to end there and we can continue to discuss whether these problems can
be solved or not. [applause].
>>: What to the possibilities for people that [indiscernible] on z?
>> Daniel Ahlberg: Yeah. Maybe I didn't emphasize, but this is in set z for any dimension. The
question here is also for set z in another dimension and for z, well it's unclear so you should
have something that your monochromatic components should be growing and growing and
growing and become bigger.
>>: But every site should go back and forth so many times?
>> Daniel Ahlberg: Yeah, but somehow it's hard to -- you need to say something about the drift
of the interfaces. [indiscernible] they should behave some random walk or something like that
and coming back and forth. But it could be somehow that the randomness in this somehow
stabilizes and you only have like a finite [indiscernible] variation or something so you somehow
get the [indiscernible].
>>: But that, sorry, doesn't seem like it's a theorem that sites kind of fixate on C. They can't
fixate. [multiple speakers] [indiscernible].
>>: So what's the proof?
>>: Because there is a probability that red fixates with either zero or one.
>>: [indiscernible]
>>: Everything fixates.
>>: But maybe everything fixates to different colors.
>>: No. Because if you have an interval that fixates on red and the next one interval fixates on
blue, so the growth will be determined, the growth numbers will be determined by the length
of [indiscernible]. There will be a brief, at least a brief [indiscernible] for a longer interval
towards [indiscernible].
>>: This doesn't sound like a proof. It sounds like a heuristic algorithm and I agree. But…
>>: [indiscernible].
>>: If you have just these two neighbors that are, they would have to, if one is bigger than the
other he kills, the bigger one kills the small one.
>>: But they are both growing and maybe they are just kind of keeping pace with each other on
some scale. Everything's growing.
>>: So if you have an interval that fixates on blue then the growth rate within the interval is like
the lambda would depend on the length of the interval as well as on what happens on the
adjacent intervals.
>> Daniel Ahlberg: It's not like in [indiscernible] dynamics where you…
>>: I don't believe there is a thing where you…
>> Daniel Ahlberg: [indiscernible] dynamics when you, at each time point you have like when
you update you have the probability of updating to either thing. But now like on the time
interval in which this happens with this model since the number of balls is growing then it
doesn't stay stationary over time. I very much agree the arguments but I do have a proof.
>>: [indiscernible] I was working on a related problem [indiscernible] stood in you can see and
so I am not sure if you are meeting next week but it's essentially equivalent to what we were
doing as well. One thing that we can more or less show is if you look at the interface with one
red in one blue I can show that the interface essentially moves it [indiscernible]. We don't
know that it's possible to have coexistence. Simulation suggests that it is.
>> Daniel Ahlberg: That's interesting.
>>: But it's hard to simulate it properly with extended lengths of time.
>>: So you could if the interface exists?
>>: In a certain sense we interface this to be [indiscernible] possibly random seemingly also
random.
>> Daniel Ahlberg: But you know that there is some convergence to like some constant, or do
you just have a lower bound?
>>: It's a slightly weaker statement, but yes.
>> Daniel Ahlberg: Yeah. So then one would have to compare that velocity to the velocity
which the front moves which…
>>: Yes. It's smaller than the velocity.
>>: I think we should vacate the room because it's…
>>: So for three colors, I mean even if it's on a triangle?
>> Daniel Ahlberg: Well in a triangle you have to kill one because once you nucleate you will kill
your neighbors, so the number of balls will decrease until you…
>>: But you might stop at that moment.
>> Daniel Ahlberg: Yeah but you still have to decrease until you -- have the total number of
balls will be decreasing until you have your neighbors are the same colors.
>>: There are not three colors.