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>> Yuval Peres: We're happy to welcome back Asaf Nachmias, who will tell us about
percolation on the hypercube.
>> Asaf Nachmias: Thanks for the invitation. You can sit. [laughter].
So, I'm going to tell you about a cute little problem, percolation on the hypercube. And
this is all joint work with Remco van der Hofsted from Urandom [phonetic]. And, again,
you all know me, I think, so please feel free to stop me whenever you want and ask a
question, make a comment, mockery, are all welcomed.
And here's the basic model that we'll be dealing with. We have the hypercube, it's a
graph. The vertices are vectors of 0-1 to the end and the edge set are just the vectors of
Hamming distance 1.
And we also have a number between 0 and 1. And we perform the very basic procedure
of going over every edge in the graph and independently we erase it with probability 1
minus P or keep it with probability P. And we get -- we have a random graph left and we
want to know what is this creature.
And the first people whoever looked at this model, I mean, in the '60s there was lots of
work addition [inaudible] on doing this kind of procedure on the complete graph, and we
will always keep that in mind when we do certain things.
I'll try to go from here to there. There's a problem in this audience is that some people
know too much and some people this is the first time they hear about it.
So I'm going to try to balance, but I think it will be easier if just give me some input if
there's something too quick or too slow, don't be shy.
First people who looked at this question were Erdos and Spencer in this model, and they
were interested in '79, and they were interested -- and what they showed is that they
were interested when this graph is connected, this subgraph.
They showed when this P is bigger than a half, what you get with very high probability is
a connected subgraph. When P is smaller than the half, this doesn't happen.
And the next -- the next question you can ask is, that they did manage to solve, is when
do you have a giant component? Just like you do in the Erdos-Rényi graph. And this
was solved in the '80s by the three Hungarians. So Itai Kumlash and Semerety
[phonetic]. And the few years that they had several three very influential papers.
And they showed the following result: So this is -- and I warn you up front that there's
going to be confusion because it's confusing up here, between the letter N and M. So
whenever you see an N it's going to be an M. Just because -- just saying that up front,
just because in the paper we couldn't really decide what is the best thing.
So I apologize. Always with the mention of a cube, right.
Well, no, sometimes it will be the number of vertices in the Erdos- Rényi graph. I'll try to
make this clear. But there's this confusion, notation that you can't really escape. This is
what they proved. P equals C over N.
And so you scale it like this. C is a constant. C is always going to be a constant, for
large for large and small for small, if C is smaller than 1 then the largest component C1.
So now this C is not a constant. Is at most of size, log of the volume, log of the number
of vertices, which is N. And if C is bigger than 1, the largest component is of order 2 to
the N. Linear in the number of vertices.
So this is in the Erdos- Rényi graph in GNP, this is a classical thing that goes back to the
'60s. Everybody familiar with the Erdos- Rényi -- I'm not going to -- just the same thing
on the complete graph. And this is the phase transition.
>>: [inaudible].
>> Asaf Nachmias: All right.
>>: What you have at the end there is the Erdos- Rényi graph?
>> Asaf Nachmias: No, this is on the hypercube.
>>: Oh.
>> Asaf Nachmias: Thank you.
>>: [inaudible] mention [inaudible].
>> Asaf Nachmias: Right. I'll take a note. Mental note of that.
The problem with this -- so this is a relatively easy assertion, because you can just
perform this B of S exploration process on your random graph and it's sub critical
because C is smaller than 1 and the degree is N. So this follows rather easily. This was
like a slightly harder thing, because in the Erdos- Rényi you can actually do this
comparison with a branching process, you know, to the limit.
And, you know, you get, you know, the correct thing. But here they needed this
additional trick to get this. And this it's called a sprinkling. And I'm going to show it to you
right now just because it's interesting and it's going to be useful later.
So how does the proof of two work? Proof of two? It's going to be a sketch, but it should
be detailed enough that you can make this -- you can work. So the first thing that you do
is you write it as P1 and 2 such that P1 is going to be C1 over N, with C1 still bigger than
1. Smaller than C.
And P2 is going to be some small number over N, and such a way that 1 minus P1 times
1 minus P2 equals 1 minus P. So the sprinkling, what that means, that means that we
first draw -- you first perform this edge percolation, that's when you check if an edge is
open or closed, with probability P1.
You get some subgraph. And then what you could do because of this condition is draw
another independent random graph with probability P2. And just take the union of the
edges, and that's what you get. And it's just because of this fact, right, the probability that
an edge doesn't exist is -- doesn't exist in number one. And number two it's just the
same probability, that tells you that it's distributed exactly the same.
And then the proof goes like this: The first step in P1 is that -- so here's the claim. With
high probability there will be linear meaning bigger than two to the N vertices V such that
the connected component containing V -- this is how we write it. So this is the connected
component containing that V belongs to, is bigger than 2 to the power, say, C2-N. Right?
Firstly, why is this true? So notice that this is still very -- this is still very far from 2 to the
N. This is because of this small little constant here. But this is true because -because -- well, because when you start exploring your B of S -- so let's say you start at
vector of all 0s and you look upwards and you're only going to explore, say, until the
number of 1s that you have is no more than, say, M, N over 4, right? So for each of
these vectors, no matter where you're stuck in this set of no more than N plus N over 4
vectors, you always have, and you look at your forward cluster -- you look at -- so you
always have let's say C1, for instance, say C was large enough. So you always have a
lot of -- the neighbors, the amount of neighbors that you have, instead of N, is going to be
three-quarters over N at all times.
So you always have positive drift for this branching process. And you have to run it for
very slow time. You run it about M or some constant times M times, and it's a super
critical branching process, and it has some probability of growing.
This is classical branching process things. And if you take this C2 to be very small and
this one quarter that I said to be very small, you can make everything work. And this
work -- yes?
>>: C2 ->> Asaf Nachmias: Should be more insistent and not let me ->>: Yes. C2 is delta?
>> Asaf Nachmias: No. No. It's another small constant. Soon you will see delta. So C2
depends somehow on C1, which depends on C.
All right. But is this step clear? So you could do this from every vertex and then there's a
concentration argument telling you that, you know, these things will happen, you know,
everywhere. So you get a graph, you know. So what you know about it is you have a lot
of large components. But they're not large enough.
Right? They're not this. You get a graph. There are a lot of things here. And in the next
step, what we're aiming to do is throw a lot of P2 edges that will connect a lot of these
components into like one massive component that has linear size.
All right. So step two is special in the sense that it actually -- so far we really used some
very basic properties of the geometry. But step two actually uses something about the
hypercube, and it uses the user parametric inequality. It uses the isoperimetric
inequality.
Let me know if you need me to zoom in on the font. And I will not write this the most
precise way. I'm just going to write a corollary of it, and this is what we use.
If A and B -- let A and B, A/B, they are two disjoint subsets. Two disjoint subsets, linear
size, of linear size. Yes. Size theta 2 to the N.
Then there exists a lot of disjoint paths connecting them of relatively short length, and
there exists at least 2 to the N over N to the 100th. Disjoint paths of length, what is the
length that I should put here? At most square root of N. Connecting A and B.
What are the sets that you should think of? Where does this inequality really saturated?
You take A to be -- so this is the schematic of the hypercube. This is the vectors of all 1.
This is all 0s. This is the equator.
You take A to be -- this is minus square root of N from the equator, this is square root
plus N of the equator.
You take this to be A. This to be B. And really you can show that there are a lot of
disjoint paths between them. And they have to be really of length square root of N.
That's the distance between the two sets. So this is the real thing, but the statements, of
course, much stronger. It says that anywhere you put these sets you'll get the same
thing.
How do you use this for percolation? So ->>: The hundred is ->> Asaf Nachmias: Yes. Yes, the hundred you can fix it probably to a half. But, yeah,
we don't care about these numbers. All right. So what is our situation? In our step one,
we have a lot of -- a lot of, a lot of clusters of this size, 2 to the C to the N.
And we want to exclude the possibility that after the sprinkling, these do not create a
giant component of linear size. So if they do not create a giant component, the situation
is -- so we have a linear -- that means that we will be able to split them into two. All of
these -- we'll put all the vertices into two sets. A partition of all these vertices here that
we obtain at step number one, such that all of these paths connecting them are closed.
And, of course, none of these vertices here and here are going to be in the same
component. Nothing is connected in this partition. Right? So let's figure out, you know,
how many ways can we have -- how many ways -- what is the number of partitions that
we can have? So the number of partitions is going to be in total we have 2 to the N
vertices, right? But this is the number of vertices in R. But, of course, we want to put two
connected vertices in the same partition. That's the definition. So out of the 2 to the N
ways to split these vertices together, we only get -- we have to divide this by 2 to the C2
to the N. Right? Because if you look at the component of a vertex, all of it's neighbors -sorry, all of its component has to be on the same side of the partition.
Right? So this is all the way that we can partition our components into two linear sets.
Now, we're going to use the iso parametric inequality that tells us that this is deterministic
statement. So far there's no probability that we have a lot of paths, disjoint paths of
lengths square root of N and they all have to be closed. Closed with respect to P2. This
is where we use the independence, and ->>: It's containing [inaudible].
>> Asaf Nachmias: Yes. Sorry. Containing a close. Yes. So not open. The probability
that one particular path is not open is going to be 1 minus P2. P2 is delta over N to the
power square root of N. This is the probability, the one particular path is not open,
contains a closed edge. And the number of these paths is 2 to the N divided by N to the
very sharp 100.
Everybody agrees with this calculation? Now comes a tricky part of actually figuring out
this is small but this is indeed small. This is E to the -- right, this is e to the something to
the log N to square root of N. And this is -- does everybody see that this is -- this is E to
the minus C2 to the N and perhaps a 1 minus little O of 1 here. So this way I'm eating all
the polynomials logarithm, everything together. This is the essence of the Itai Kumlash
and Semerety proof, and essentially where we started our work.
Is there any questions about this proof before I go on? Okay. Great. So the question, of
course, what you can ask yourself after or perhaps you've seen other people ask, is what
happens when C equals precisely 1? What is the size of the component? And what
happens now comes, okay, you're starting to guess. And as usual this is the same
behavior as Erdos- Rényi and you would think to yourself, okay, first thing I need to know
is what happens in Erdos- Rényi. So in Erdos- Rényi this is what happens. Very briefly.
So now this is GNP.
Now I'm performing -- this is on the complete graph, Alex. No dimension, right? And P ->>: I got it.
>> Asaf Nachmias: And P is approximately 1 over N. So there's like a very intricate
phase transition, luckily none of the people who invented it is here so I don't have to be
careful about -- what I'm going to write is proved first by Bolabosh [phonetic] with
logarithmic corrections, later removed by Woodchuck, some are due to Aldos. I can see
Yuval is very unhappy. But this is what happens.
[laughter]. Very happy, glad I could be of service. So you look at 1 over N. Firstly, the
first fact is the size of the component is going to be the number of vertices to the power
two-thirds. But not only that, you can also move a little bit from 1 over N and get the
same effect. 1 plus O of N to the minus one-third divided by N, and the statement is -when this is your P, then your largest component is of order N to the two-thirds.
And moreover, I want to state this, and it is not concentrated. Not concentrated.
Meaning that if you take this largest component, divide by N to the two-thirds, and you
will find that you get a limiting random variable which is nontrivial, and it's supported on 0
infinity.
So there really is a positive variance on the normalized. Let me also comment that the
expected value of the cluster in this regime is actually of order N to the one-third.
Because of course this is the largest component. This is just a typical cluster. There's
this one-third and of course two-thirds are very much connected. So you can ask
yourself really is this really what happens? Is this -- this is the scaling window of the
random graph, et cetera, et cetera, and ->>: [inaudible].
>> Asaf Nachmias: Sorry?
>>: Average cluster, is that typical?
>> Asaf Nachmias: Okay. I'm not sure I understood this comment. But here is the ->>: Expectation of COP is much larger than typical value of COP for most V. COP is
small. Expectation ->> Asaf Nachmias: Correct. That is true. Typically, when you start exploring for a
vertex, you have probability N to the minus one-third of being in the largest one and the
rest is kind of not very important.
That's why the majority comes from here. Even that is not precise, because you see
components in all scales. Anyhow, why is this N to the minus third sharp? Because
here's the next theorem. This is your P. Epsilon is really bigger than N to the minus
third. Right? This is the condition. Then C1 divided by 2 epsilon N goes to 1. Notice
that by this condition on epsilon, this value is really larger than N to the two-thirds. And
furthermore, C2 divided by epsilon N goes to 0.
So this is -- in this situation -- in this situation actually in the first one C2 is of the same
order it's also N to the two-thirds. So here you see something like this. Number one, you
see a bunch of small components, but N to the two-thirds.
>>: [inaudible].
>> Asaf Nachmias: Yes, epsilon here is also tending to 0. But as low as you want.
Right, the fixed epsilon is covered by a theorem of this form. Right? And here is where
you actually see one significant component that looks larger than the rest of them. And
this is a smooth transition between this and this.
And the important thing also for me to state is that this quantity is concentrated. There's
also a theorem for what happens below, for P, which is 1 minus epsilon and it's in the
similar fashion, you see concentration, but it still looks like this.
I'm not going to go into that because I'm going to deal mostly with this kind of super
critical case. So this is what happens in G and P. And the first thing you would want to
say, you know, all right, and obviously because there shouldn't be any difference, we saw
one theorem where there's no difference, so why shouldn't anything behave differently.
So can you show that when you percolate on the hypercube at 1 over N, you get a
cluster. This time should be of cluster of size 2 to the N over times two-thirds, right. The
number of vertices to the power of two-thirds.
But there is a problem. So the problem is -- many people call this value or one particular
value of it the critical probability or PC or whatever, whatever you want. But the problem
is that in the hypercube, the critical probability PC is really not 1 over N. It really is -- this
is undefined. Now I can see Yuval is a little less happy by his smile.
But what I mean here is that -- and this could be a formal thing. If you percolate. Here's
a fact. If you percolate exactly at P which is 1 over N on the hypercube. The hypercube
has very short cycles and a relatively large degree -- relatively small degree. That's the
difference between G and P and the hypercube.
You will find that the largest component is at most N to some power, with high probability,
whatever.
All right. So this is definitely not -- and what we're looking for is volume to the power
two-thirds. If we believe -- so here comes, there's this belief always in this background,
you know, a lot of physical words that these models are high dimensional enough. So
everything should behave the same because it's a critical statistical physics model. But
I'm just repeating words that I read and heard. Can't really explain this.
But the truth is that you do see -- so the point of this talk is to tell you do see this
phenomenon here, this phased transition, but the problem isn't -- the problem with this
whole project, that's why it's also taking so long, is it doesn't happen at this point that you
can write down very nicely as 1 over N, and in fact it happens in some other point, which
is 1 over N plus 1 over N squared, second term is -- third term is seven over two over N
cubed, plus some O to the minus 1. No one knows what the fourth coefficient. And I
don't think no one really cares.
>>: What is the one node?
>> Asaf Nachmias: The window will be this. I'm coming to this statement. I want to
explain the first key idea in this.
So essentially, I forgot to mention one theorem. So we were here we were at the '80s.
Now we're at the '90s, and in the '90s this group Bolabush, Kawakawa and Muchuk
[phonetic], they decided, okay, let's take a step forward and they showed what you kind
of want to show.
They showed that when P is 1 plus epsilon over N, this time epsilon has to be very large.
Log cube of N divided by N, then the largest component divided by two epsilon two to the
N goes to one.
So this is what they proved. But notice this is weak. Because we want this condition.
Right? In the hypercube this condition translates -- you want to write a theorem that
holds when epsilon is larger than 2 to the minus N over 3. So this is very weak. In fact,
in my notes I plan to show this first. But of course you can't improve their theorem
because we know -- well, we don't know yet. But we know that it doesn't really happen at
this location. And the correct place -- it's one plus epsilon times something, but this
something is not going to be 1 over N. The question is what is this something. And this
was first integrated by Borgs, Chayes, van der Hofsted, Slade and Spencer. And they
came up with the following very neat definition.
And they said the following. Borgs, Chayes, van der Hofsted, Slade and Spencer, in '05,
and they said here's a definition, define PC as -- here it is, the expected size of a cluster
and vertex is precisely 2 to the N over 3 volume to the one-third. Just like you have here.
Note that you can always do this because on the right-hand side you have a monotone
polynomial that goes from 0 to 2 to the end and you're just choosing the end that it equals
at.
And what did they show? They showed that P equals P C1 plus 0 of -- now 2 to the
minus N over 3. So the same -- so they showed a statement close to number one. You
look at this. And then the largest component is aborted, two to the, volume to the power
two-thirds, with high probability; and, in fact, it's also Hagen Reich [phonetic] van der
Hofsted so that it's not concentrated.
But what they lacked is, you know, a statement of this form. That when you go a little bit
above this volume to the one minus one-third, you actually get, you see a giant
component.
And theoretically if you just have this statement, you don't know if you put in here
something much bigger than this correction, it could be staying still in this same situation.
Camera man probably hates me right now. So here's what they conjectured.
So their conjecture was take this PC and write down P as PC, one plus epsilon. And
epsilon as before is little 0 of 1. And bigger than 2 to the minus N over 3. .
Then C1 divided by 2 epsilon goes to 1.
>>: Does that mean bigger, bigger?
>> Asaf Nachmias: It's the opposite of the little 0. The ratio goes to infinity.
>>: [inaudible].
>> Asaf Nachmias: Yes.
>>: You write it [inaudible].
>> Asaf Nachmias: You write it little. So this is their conjecture, this is what we proved.
So the theory which is still in the process of final touches, is, yes. All right? Is there any
question about the statement of the theorem?
>>: [inaudible].
>> Asaf Nachmias: Yes, they did have -- sorry, what is it, funny?
>>: So why does this expansion for PC, what does that prove?
>> Asaf Nachmias: Okay. Thank you. This is a Harrah Slade van der Hofsted.
>>: This is this definition.
>> Asaf Nachmias: Yes. So what I wrote for this, I meant this is the value in which you
get this. And they suppose there are like infinitely more coefficients. It's asymptotic. It
doesn't necessarily converge to the right place.
>>: [inaudible].
>> Asaf Nachmias: The fourth term, I think, is still unknown. Perhaps the fifth term is still
unknown.
I think they kind of gave up. Also, you know, when you don't really suspect that it
converges, then who cares. All right. So before I'll tell you some things about the proof,
unless there are any questions, let me tell you what is more or less a general statement.
So our method kind of -- really uses some random walks estimates and their connection
with percolation, and eventually we write this down you discover what you really use, and
this is what we use. So there's the general statement is the following:
So the general statement instance will allow this theorem with exactly as written, with this
PC and every time that you have Ns involved, every time you can switch, you can
translate by taking V, which is going to be the number of vertices, to be 2 to the N.
Right? And then the statement that I'm going to write is going to be exactly this. And the
general -- this is not the right page. So the number of vertices is V, as we said. Large V.
The mixing time, the uniform mixing time of the simple random walk or nonvector tracking
random walk, this is a bit of a technical issue, is just going to be the first T.
So everybody knows what a random walk is. This connection between random walk and
percolation, this is what I aim to spend a little bit of time on.
The first time that I say P, T, 0-0, your return probability, is smaller than 1 over the vertex
set or V, plus the O of 1. This is some -- not a very standard definition. But it's
essentially the strongest mixing time that you can have. It means that the probability,
essentially it means that the probability of being at any particular vertex, 0-0 always
maximizes this quantity, if I replace this by another -- on a transitive graph. Even that is
not completely true.
But anyhow, the time that it takes the distribution of the random walk after T steps to be
close to uniform, what exactly the statement that we'll need for that you will see in what
I'm going to show you. And then do we need? So the conditions are the assumptions.
Assumptions. Is there a question?
The assumptions are 1. The degree goes to infinity. So we have a sequence of graphs.
Degrees going to infinity. Yes?
>>: [inaudible].
>> Asaf Nachmias: 0 is any vertex. We're only going to deal with transitive graphs. So it
doesn't matter what vertex you start from.
>>: [inaudible].
>> Asaf Nachmias: Sorry? [laughter].
>>: The same -- [inaudible] the whole.
>> Asaf Nachmias: I thought it was a joke. So the rest of them. Two, so this is -- 2 is the
only significant condition. 1 and 3 are technical, not very important. Here is the
important condition. PC that we defined in this way, V to the one-third, times the degree
minus 1 to the power of the mixing time is no more -- it's essentially 1 plus little 0 of 1. It's
equal. Easy to show, by the way, that PC is always bigger than 1 over the degree minus
1 with this definition. So this really tells you something. I will show you exactly where
this comes from. So a bit mysterious. And 3 is another technical condition that I don't
want to tell you what it is. Something technical.
But also it's also a random walk technical. So it's usually something that it's easy to
check. Some -- I can write this down for you if you're very much interested. Technical
random walk condition. But somehow it's a local condition. So this condition, in terms of
random walk, looks at how many cycles of length three, of length four, of length five do
you have and things like that.
>>: [inaudible].
>> Asaf Nachmias: All right. Here it is. It's not going to enlighten you in any way. For
every XY these are vertices. [laughter] sum over UV in the vertex set. Sum over
numbers S1, S2, S3 from 1 to the mixing time of the return probability S1 of XU, return
probability of U -- not the return. The visiting probability. These are all random walks.
And S3 from V to Y is smaller or equals two the delta function at XY plus little o of 1.
All right. So now we can all blame Yuval for seeing this completely pointless line. I can
explain this relates to something that's called the triangle condition, it's a random walk
version of it. And it will be too deep inside to go.
But, sorry.
>>: Like [inaudible].
>> Asaf Nachmias: It's not. But that's why Yuval asked the question. It kind of controls
how your random walk behaves before the mixing time. Sorry? If you look at this it
essentially counts triangles and squares and all sorts of things like that. And they're all
very close. It's squares -- everything that is a square that is length at most the mixing
time.
So you don't get very long paths here because all these numbers go up to the mixing
time. So it's some sort of a local condition. And you can verify these assumptions, for
instance, here's a class of graphs that I like. Take expanders. Not necessarily -- so,
okay, expanders, say, with finite degree.
So, okay, this theorem doesn't work, but for technical thing -- never mind. Expander with
degree log M, log of the volume. And which have logarithmic girth, together with the girth
you can easily verify this condition, the mixing time.
You know the degree. You kind of need to estimate PC, which is a little bit tricky, but you
can show. So, for instance ->>: Grouped?
>> Asaf Nachmias: Logarithmic say logarithmic degree, right. And girth, which is
something that you can attain. Large girth. Right.
>>: [inaudible].
>> Asaf Nachmias: Yes. It also works if you want for constant, for fixed degree. But
then this 2 is not going to be correct here, right? If you have constant degree, then
survival probability of a branching process poison 1 plus epsilon is not 2 epsilon but
something else.
All right. But these are all technical, not very interesting things. How does the proof
work?
>>: [inaudible] for expander, if it's transitive and the degree is just slightly turning to
infinity.
>> Asaf Nachmias: Yes, it works but under some girth assumptions. Oh, yes, it should
work for any expander. This statement should be true for any expander. But with the
different constant. We don't know that. How does the proof work? I like this talk,
because I can give you the last line of the proof. So here's the last line and I can give it
to you and it actually makes sense. And it's not QED. Preparing for the joke.
So here's the last line of the proof. I shouldn't have erased that. So the last line of the
proof is going to look very similar to the last line of the proof of step two that we saw in
the [inaudible] it will tell us something interesting.
So the last line -- this doesn't work. -- the last line is going to be this. So, remember, we
are in this case that epsilon is bigger than 2 to the minus N over 3. Now I'm back just for
the sake of to be clear I'm in the back in the hypercube setting.
So here's the last line. It's going to be the last estimate is going to be 2 to the power
epsilon V, where epsilon 2 to the N divided by KN epsilon to the minus 2. KN is going to
be -- K is going to grow to infinity very, very slowly.
This will correspond to this number of partitions. Times 1 minus some fixed constant
delta epsilon N to the power of epsilon squared, N to the 2 N.
And you can -- this one I can estimate. This will be exponential in epsilon cube 2 to the
N, which we know goes to infinity. And this will be slightly smaller, right? So all of this is
still XE to the constant epsilon cube 2 to the N.
So, again, this will correspond to our partition. So just by this -- let's reverse engineer the
proof. So we will have -- so we will have, as you remember in step one of the sprinkling
argument, we find a lot of components, a lot of vertices which their components are of
size, of big. So we know -- so we know the number of vertices.
So the large would be, large component in step one will be, large will be bigger than this
KN over epsilon to the minus 2 and the number of components is going to be -- and there
are numbers of these vertices in large is going to be bigger than epsilon 2 to the N.
So this is the first step. What does the second step tell us? The second step is more
informative, or at least more innovative.
You remember that previously we had this to the power square root of N, because we
can only find -- the isoparametric inquality only allowed us to find paths of square root N.
This is in deed tight. The main thing we do is get rid of these parametric inequality, and
saying a statement of the form percolation clusters cannot look like these extremal sets
they're closer to random sets so they have to be closer to each other, have to clump
together.
And we will find is that these two such percolation clusters will have a lot of paths of
length one, which are closed. In other words, edges which are closed. So if we -- in the
splitting, once we find a lot of large components and we try to split them together, what
we will find out is that we have a lot of closed edges.
This is long, but it's just one edge. And the number of these closed edges just by this is
going to be epsilon squared N 2 to the N.
All right? So now I need to explain to you -- I'm not going to explain to you this part. This
is -- every part here requires some work in technology, some from combinatorics, some
from statistical physics. And this is less interesting. This is where you need to find this
randomness built inside the percolation.
Let me -- so this gives you -- this is essentially what we prove. And I won't be able to
show you the whole proof, of course, but I want to give you a hint how do you find, where
is this randomness hiding? And how do you use the simple random walk?
So I'm going to show you a very small simple inequality and then try to give you a hint on
how it's used to prove something like this.
Is there any question about this very high level proof? All right. So here's the inequality
lemma. Before the lemma a little thing, a little exposé. So what is special about GNP.
What is a very convenient thing about it? Is that if you look at this function, if you look at
the connective function, the probability that the two vertices are connected.
Okay. Then this, as long as these are two different vertices, is the same number. It's the
constant function. Right? It's the same number for any two vertices, just because of
symmetry, for any P also. For hypercube we can't expect anything to hold. Right? If
you're looking at your immediate neighbor, right? The probability that you are connected
to it is something like, if we're looking at PC, which is close to 1 over N, then it's going to
be about 1 over N. But the probability for anything on the equator to be connected, 2 is
going to be 2 to the N over 3 divided by the size of the equator, because of symmetry,
and so we're working here at PC, not in general P.
Divided by 2 to the N divided times square root of N, divided by square root. So this
number is much smaller than this. So, of course, you can't expect such a thing to hold.
But lemma tells you you still can retrieve the symmetry you see in the Erdos-Rényi graph
in the following sense. So we're not going to look at the event that X and Y are
connected to each other, we're going to consider the event that they are connected to
each other in a long path.
So the probability that they are connected in a path -- and here comes perhaps the
surprise -- this length needs to be at least a mixing time. Probability that the two vertices
are connected is going to be at most the expected cluster size divided by the number of
vertices. So the power of N with a small correction.
>>: There's the longest haul to the shortest path.
>> Asaf Nachmias: This is for every X and Y. So the stronger statement, so the
probability that there exists N path doesn't necessarily have to be the longest path, the
shortest path.
>>: It could be connected ->> Asaf Nachmias: They could also be connected by that. Uh-huh. And the advantage
of this statement is that it's true uniformly for all the vertices. So this kind of thing -- so I
believe that this should be something general that should hold in two dimensional lattices
and all of these reasonable graphs that you perform percolation.
Of course, what does it mean? Of course, in the two dimension tours you don't expect to
put in here the mixing time to get this. You expect to put some other quantity. So the
conjecture is that there's always a quantity you could put here that you get this. Of
course, you could put here something so small, so large that you will get zero, and
everything. But you also want that the sum over Y of this event, whatever you put here,
in this case it's T mix, is going to be larger than 1 little O of 1 of the expected cluster size.
>>: Speculation that the T mix.
>> Asaf Nachmias: Sorry? What is the stipulation?
>>: What is ->> Asaf Nachmias: I forgot to say T mix in the hypercube. I think this answers your
question. T mix is M log M. Or a constant times M log N. Was this your question?
>>: The question was so X and Y are connected by a path of [inaudible] at least T mix?
>> Asaf Nachmias: Yes. This is the notation. This is somewhat strange because here
there's a random walk estimate and this is a percolation event. But the proof is -- it's
complete triviality so let me show it to you. Switch pens. So if this event occurs, this
event occurs, then there exists a vertex V and then open simple path and an open and
open simple path Amaga [phonetic] of X to V of length T mix.
And V is connected to Y disjointly. In other words, there's another open path of
unbounded size on that connecting V to Y. Right? And now we just sum over everything
and use the fact that these paths are disjoint. So familiar thing, inequality disjoint so the
probability is smaller than -- can you see here? Claire? The sum over V. The sum of
these paths W, the probability that this path, this path is open. So open simple path is P
to the power M log M. Right?
And here I write the probability that V is connected to Y. So far this is clear? All right.
And now I'm going to do the first -- the first thing I'm going to do is instead of considering
all the open simple paths, I'm going to consider all the non-backtracking, not necessarily
simple paths.
So this is the first relaxation, inequality is in the right direction, instead of simple,
self-avoiding I'm going to write non-backtracking.
All right. The P to the probability M log MI put the other way. And now I can enumerate.
So here is where the mixing time comes in. How many paths do I have of length M, of
length T mix which are non-backtracking? It's D to the minus 1 to the power T mix. D is
the degree. So let's write N. N minus 1 to the power -- okay. So it's N times N minus 1
to the power T mix minus 1 if you want to be precise. Doesn't really matter. This is the
number of paths.
How many of them end up in V? These are the ones we are enumerating on. And this is
exactly what this is telling you. This tells you that this is at most the number of such
paths is at most this to the power 2 to the N times 1 plus little o of 1. Right? Now we can
put everything outside.
There's a P to the power T mix, and here you see exactly where condition 2 and the
assumption comes in. P to the power of T mix. This is that. And the sum over V here
gives you, because of here I'm using the fact that the graph is transitive, I get the
expected cluster size. At any vertex.
All right? And because of this assumption, this is all 1 plus over 1 and we get this. Any
questions about the lemma?
All right.
>>: [inaudible] on the right-hand side you put max of X and Y [inaudible] it's not
necessary.
>> Asaf Nachmias: Did I say there's something -- I said that I believe that this should
hold, that this condition doesn't -- I can only prove it when I have condition number two.
But this should hold I think perhaps with another value in say two dimensional
percolation, which all of these -- and any lattice, all of these assumptions don't work.
So I don't really know. It could be just a useful thing. So let me try to convince you ->>: [inaudible].
>> Asaf Nachmias: What? Sorry?
>>: The definition of T mix.
>> Asaf Nachmias: Okay. How do I use this thing? I'll try to tell you in three minutes.
So why is this a useful thing? It's useful in two important ways.
So the strategy, in order to prove step number two, is, you know, given that you know
step number one is that you start with two clusters. You start with two, sorry, vertices,
XY, and you want to grow them. And you want to say if they have grown enough, then
I'm starting to see edges connecting them.
So first you grow them a little bit and you condition on what you see. So you kind of have
to have a conditional statement. So you grow them separately a little bit. They've grown
to here, and they are big. They contain about epsilon to the minus 2 vertices. Their
diameter is epsilon to the minus 1. And now you are conditioning exactly on what you
see, and you want to say that -- now you have a vertex here, and you want to look at it at
the future and how much does it have expectation, how many vertices does it see in the
future.
And you want to say that -- so first thing you want to say is that this is large, that this did
not consume a lot. Epsilon to the minus 2 is very small compared to epsilon to the 2 to
the N because of this condition.
And now -- so now you want to perform some second moment argument, counting, you
know, counting the number of edges that you see that this one is connected to. This one
is connected to from here. But this one is not connected.
>>: The foundation [inaudible].
>> Asaf Nachmias: What is the condition?
>>: You mean the triangle?
>> Asaf Nachmias: No, this is not the triangle. This would be a very large -- this would
be a very large path or cycle, and the thing is the fact that you have -- if you just wanted
to compute this probability that X is connected to this and Y is connected to this, then
there is no problem, because you know the probability -- and these paths have to be
disjoint, because you know the probability of both of these being connected, you know
them by the lemma.
But now once you condition on like this crazy configuration that holds here, you can -what you see here can be completely thwarted. So the probability -- so the expectation
of what you see, you want to say that expectation of your future cluster is still of the size,
hasn't changed by much. It's this size. But of course now that you've conditioned it, it
could be diverted to this side of the hypercube. And from this point it could be converted
to this side of the hypercube, and then they don't really have a chance to meet. But this
lemma tells you that that can't happen, because all of the vertices and what's left have to
receive more or less the same mass.
If you have this -- if the sum of what you see in the future is this, and the lemma holds,
even with the conditioning, because everything is monotone, then you know that it has to
spread equally in all the remaining vertices so you can perform -- you just count the
number of edges here. You perform a first moment. You perform a second moment, and
you show that what you get here is enough edges. Everything works out so you can get
exactly the number of edges you want, and that's it.
And I think I'll stop here. [applause].
>>: [inaudible].
>> Asaf Nachmias: No. We replaced it. If you want I can tell you things we don't know
how to do. But I'll take that as a no.
One thing we don't know how to do, for instance, is we think that the largest component
has a COT. We don't know how to prove it. So we have a law of large number for it but
we haven't obtained a central lemma theorem. The second thing we cannot do -- so we
can show this picture that the second largest cluster is really smaller. We don't know how
to show that its distribution or that it looks like the sub critical case, which we believe is
true.
The same thing happens in Erdos-Rényi. And there are other related questions we want
to study.
>>: [inaudible] an annoying question but I was curious if this has any applications?
[laughter].
>> Asaf Nachmias: Of course ->>: [inaudible].
>> Asaf Nachmias: The study of glasses in quantum physics.
>>: The study of what?
>> Asaf Nachmias: The study of glasses in quantum physics.
>>: I mean algorithms anything that doesn't -- it doesn't have to be like real world
edifications, you know what I mean, it's like how can this be used?
>> Asaf Nachmias: To get me a job. [laughter].
There is no like -- if you want a serious answer there is no serious applications that I
know of. The method -- [inaudible] -- the sprinkling this has been used all over the place,
and a lot of computer science, the same ideas. It's not even questions about
percolations that Jan used it recently. The methods are sometimes useful. I'm not even
claiming that these methods will be useful to anyone or anywhere.
>>: The focus of moving on beyond [inaudible] equalities in general set using structure of
the specific case, how is some of that earlier work, I think that's an important point.
>>: I'm not trying to say this is not important or ->> Asaf Nachmias: But that would be right. [laughter].
>>: By no means, but I was wondering where I could use the results. If I wanted to, for
hypercube or something, you never know. I'll figure it out myself.
>>: I thought it was one thing -- in other applications where iso parametric inequalities are
used, they're used a lot, can you so -- so one lesson from this is in your specific
application maybe your sets have more structure or in such case such randomness which
means they have better for properties generated by the general ->> Asaf Nachmias: I should write this down. [laughter].
>>: That's really important. And in fact it's -- that's been used in other cases, but this is
maybe the most spectacular example.
>> Asaf Nachmias: Oh boy.
>>: Sounds like it.
>> Asaf Nachmias: All right.
>> Yuval Peres: Let's thank the speaker.
[applause]