>> Yuval Peres: Good afternoon, everyone. It's a pleasure to have Béla give
this talk this afternoon.
Béla Bollobás has revolutionized [inaudible] and random graphs in particular.
We really can't think of the subject without his fundamental contributions, and
often when you look at his work you think this must be a team, not a single man.
So without further ado, please.
>> Béla Bollobás: Thank you very much indeed. Actually, is the microphone
switched on? All right. I assume it's not for your benefit, but -- yeah.
So as the title says, I shall talk about percolation, but strictly speaking, it should
be essentially one pretty large theatre, but I shall not really get to proving it at all.
In fact, I shall get to stating it only after quite a while. Mostly because I think not
everybody in the room is an expert on percolation, so I shall try to give at least
the basic facts about percolation.
So if all goes well, then I shall mention these topics and some introduction,
classical results, and this should be really the new bit which is with only the
[inaudible], as I said at the beginning as well.
And just in case you think I can do a figure, I can assure you, I can't. Every good
figure was drawn by Oliver. All the clumsy ones are due to me.
So what about critical probabilities? Even if you hardly ever heard of anything in
percolation theory, you would have heard of this particular result. The basic
result, the critical probability of bond percolation on the square lattice is one half.
And it seems extremely natural. I'll return to it several times. But, still, one part
of the probabilities at most was proved by Harris, and so -- at least was proved
by Harris and at most by only 20 years later.
Now, the less obvious critical probability is bond percolation on the triangular
lattice. Actually, I shall go back and define what percolation is, but roughly I think
everybody knows it anyhow.
So here's the critical probability theorem is by John Wierman soon after Kesten
proved his famous result, although the result had been conjectured almost
20 years earlier. And here the critical probability is less obvious. It's pi over 18
appearing, which people in Microsoft research would think that I missed out a
square, but actually I didn't. But there is an sin to replace the square.
Okay. Now let's make it a little less simple. Let's introduce some complications.
So super we have got a triangular lattice, but let's focus on every second triangle.
Actually, I do have a pointer, but I have only one right hand.
Okay. And so let's define a percolation measure. Let's define some kind of -- as
follows. Put in edges originally nicely at random exactly with the same probability
as one always does, but then suppose some interaction between the bonds as
it's described there and then the obvious question is, okay, for what value of p do
we have percolation.
And another example would be this pretty lopsided setup where the probability
that none of the three sides -- so these are not connected within a block is some
probably p that these two are connected but those two are not is q. And then
they are not connected at all at zero. All connected has probability of 1r
Is the setup clear? Oh my -- I thought I took it out of my pocket and left it in the
office
>> [Inaudible].
>> Béla Bollobás: Yes. Yes. I'm sorry. P plus q plus r is 1, of course.
So is it clear? So we have a chance of having in this bond and not those two, we
have some chance of having all three connected. Whether it's using two bonds
or all three bonds, we don't care, and we have no chance of connecting those
two only.
So here are some peculiar examples, but the reason for everything I'm saying
today is some work done by Ziff, Ziff and Scullard, in several papers about four
years ago where they made several observations. The main thing was that in
several rather peculiar cases they could say that r, the critical probability, is such
and such.
But, again, these were just predictions, although they pretended a little that they
were in fact proved, but they were not, of course. And it became clear because
Wierman and Ziff proved criticality in some of the cases they predicted the value.
Okay. So now with Oliver ->> What do you mean prediction? You mean exact prediction or ->> Béla Bollobás: Yes. Exact prediction, as it was earlier. I mean, even in 1959
as soon as [inaudible] product percolation, they said, ah, and for the square
lattice, the critical probability is one half. Yeah. That was in '58 -- '57, '58. It was
obviously it was one half.
And then in 1960 Harris proved one band and 20 years later -- and, of course,
the required more and more weight, more and more note right, and when Kesten
proved that was very big result.
Similarly this way. So they could predict using -- we'll go back to it.
Okay. So now I shall define a rather general class of models and I shall say that
we have shown that these predictions are actually valid and we shall -- and what
we have done is we have determined our general result contains all the results
concerning several points.
In some sense the main thing is that we have to use new method to get these.
Okay. So, still, back to the simple example, what we shall have is, for example,
some kind of triangular lattice. It will be infinitely more complicated in a moment
but in some rule telling us in which way to join the various sides in these shaded
regions. So we have some connecting mechanism drawing the same picture as
curved lines, so we have got this shaded triangle, and in each triangle there are
some black box which tells us that the probability that, let's say, within this
particular triangle you can go from a to c, but neither from a to b nor to b to c. In
other words, this edge is [inaudible] has this probability. And the other black box
will tell you that the probability that they are all n has this probability which can
[inaudible] in four different ways, either all bonds r in or just two of the bonds are
in.
We don't have many possibilities because the triangle has only three sides.
So often this black box is attained by putting in some generator in a shaded
triangle. So, for example, here the generator put in has this particular shape.
And then the natural thing would be this is your lattice. Okay. Let's say all of
them have this -- all bonds have the same probability. What is the critical
probability? And what I'm going to say tells us instantly what this probability is.
Okay. Now, to make sure that we have the right notation, so, for example, p, a,
b, or p, a, b bar c is the probability that the bonds are in such a fashion that within
the black box, a and b are in the same component and c is in a different
component.
And an immediate consequence and one of the easiest consequences you'll find
main result is that if you have a triangular -- if in your setup every black box is a
triangle, every polygon in the model is a triangle, then all that matters is whether
the probability that they are all connected is bigger than the probability that no
two are connected. The rest is irrelevant.
So the probability of connecting all of them is strictly greater than the probability
of connecting no two -- some of us are busy, but -- yeah. No, poor [inaudible],
I'm sorry. But he has three phones so he's at least three times as likely to get a
phone call as anyone of us.
Okay. And the other way ->> [Inaudible].
>> Béla Bollobás: Yeah, that's what he says. Occasionally one phone calls the
other and then he doesn't have to do anything.
Okay. So that's where it will come, back to the very beginning '57 was the
beginning of everything, Broadbent and Hammersley --
>> On the previous slide [inaudible].
>> Béla Bollobás: Oh, this result has nothing about it. And in fact we don't know
why it percolates or not. I mean, there's no result about that. And I shall return
to it. Much of the future work should center on exactly what happens at the
critical probability. So I'll go -- return to it several times.
Wait. I think I am going backwards. Ah. Good.
Okay. So just a historical peculiarity was introduced at almost exactly the same
time when the theory of random graphs was started by Edos and Renyi, and still
for 30 years or so there was no connection between the two and now there is a
lot of cross-fertilization. It really is one subject at long last.
So what's the setup? We have -- in percolation, rather than just having a nice
finite graph which is often a complete graph, in percolation we have an infinity
graph which is almost always lattice-like or an actual lattice, and [inaudible] was
introduced by Hammersley. He used different technology, site and bond and so
on.
So the point is exactly as in the theory of random graphs. Every bond, for
example, can be in one of two states, open or closed. Open really counts that it's
in the random graph we are taking. And there is some measure which tells us in
what way these selections happen.
And, of course, you can read everything quicker than I should say so let me just
pass onto it.
Now, the easiest -- by far the easiest percolation measure is one in which
everything inside is independent. So you have got the probabilities of various
bonds or sites and everything is independent.
This is really the classical case studied by hundreds of people.
Okay. For any site let cx be the component of the -- formed by the bonds or site
you have selected and theta is the probability that this component is infinite.
And then the critical probability is the place where this theta goes from zero to
non-zero. So that seems to be very meek transition, but zero [inaudible] instantly
as we can make it look much more dramatic like that. So below the critical
probability almost surely there is no infinite cluster [inaudible] there is an infinite
cluster. The two are trivially equivalent.
Now, it's not that -- this one looks somewhat similar to the triviality but it is far
from a triviality. Just to mention a result, the basic result of Menshikov which
says that below the critical probability it's not only that you don't have an infinite
open cluster but the component or the origin or any site you care to name shows
exponential decay. So the probability it's large is exponentially small.
And one could define the same for selecting sites rather than bonds. Irrelevant.
Okay. What about examples? Even I could draw a square lattice, triangle lattice,
hexagonal, Kagome and even the Archimedean lattices are getting a little more
complicated. And there is, for example, this one, Martini lattice of Scullard. I
think a little imagination does tell you what do you see here.
Raj, what do you see? Pardon?
>>: [inaudible].
>> Béla Bollobás: Really? I see martini glasses.
>>: [inaudible]
>> Béla Bollobás: Really? Oh, okay. For me these are martini glasses as they
are usually stacked. I mean, here there is the glass and the stand. Yeah. Okay.
But, again -- and that was a genuine question. Here is a very simple kind of
lattice. Suppose every bond has the same probability. What is the critical
probability?
Okay. So these are the basic questions. But the [inaudible] critical probabilities
was really fashionable at the beginning. Now we would like to do much, much
more. We are really interested in what happens at the critical probability. It will
be wonderful to know what happens.
In particular, here is this big, big question do we have conformal invariance in
various cases.
And, of course, it seems that -- okay, officially this question is independent of the
critical probability, but in fact when it -- the only case in which it has been
determined, the critical probability was as simple as possible. So there is some
point in starting it, trying to prove that the critical probability is such and such and
then maybe one can get conformal invariance in a different way.
Okay. So bond percolation, again, there is some origin. Some bonds are
selected. We are interested in the component structure.
Okay. Why is the critical probability of the square lattice one half? Now, if you
ask essentially anybody, especially not in percolation, then he will tell you, of
course, that he knows it. It's because it's self-dual, and this statement is meant
as a proof of the fact that the critical probability is the one half.
But this is complete rubbish. I mean, I'm afraid Hammersley knew in 1957 that
the square lattice was self-dual. That's not exactly a deep, deep fact. And still,
no chance of proving the actual fact.
>>: [inaudible]
>> Béla Bollobás: He knew it. That's it.
>>: [inaudible]
>> Béla Bollobás: Absolutely, yeah. Exactly. Which is why it took Kesten only
20 years to prove it. Yeah.
So, now, actually looking at this, you do see that if you take an n plus 1 by n grid
and you take it's dual the way I have taken so the dual will be an n by n plus 1
grid, then it is a coffee-time problem, but not entirely trivial coffee-time problem.
No matter how you put in various bonds and in what way you assign various
states to the bonds in the original grid, you transfer it to the dual by putting in a
bond when the other one is not in and vice-versa so at every crossing exactly
one of the two bonds should be in. So no matter how you assign states to the
original lattice, there is either a crossing, a left-right crossing, in the original or a
top-bottom crossing in the dual.
Which tells us instantly that the probability, if you put in bonds with probability
one half, then the probability that you find the crossing from left to right, you find
a left-right crossing, is exactly one half.
Okay. It's 101 by 100, but it's the same as a square. So essentially it tells us
that in a square the critical probability is 1 -- sorry, the probability of crossing s
square left to right if the bonds are selected, the probability one half is exactly
one half. So that's what it tells us. But it's very far from a proof, and that's,
again, back to history.
Now, we do have a really essentially trivial proof with Oliver which has rather
annoyed lots of people, but if you look at [inaudible] book then he proves this
result on page 175 or whatever, and in fact you can do it very quickly and without
any dirt flying around using results which were not available to Kesten at the
time.
Okay. Triangular hexagonal, again, there are these guesses, but proved only
later.
Okay. Let's go closer to what we really are interested in.
Weighted plane lattices. So let's start with a nice graph, discrete vertex set, and
let's let it have a lattice structure in the sense that we have two independent
translations under which the lattice is invariant.
And it's weighted because every edge carries its own weight, its own probability.
And let's call is symmetric if the reflection in the origin gives you an isomorphism.
Now, what's dual, again, we get the dual, but you always have dual. Dual edges
are crossing and they come in pairs, and to get the dual probability distribution
assign 1 minus the probability of the origin edge to the dual edge.
Okay. So here is a quintessential plane lattice which was obviously drawn by
me. Actually let's go -- this really is out of Kesten, but he didn't quite -- no, he
didn't dot the i's at all and Grimmett had plenty of i's to dot to get the result.
So if the horizontal and vertical bonds are put in with two different probabilities
rather than p and p but you use p and q, then what is the condition on p and q to
get percolation? And the result is that p plus q is greater than 1.
Okay. What about symmetric, the symmetric plane lattices? So a little deeper
results. Let me just state these two results with Oliver.
Actually, it's not quite clear what the point is of having a projector where you can
see it but just about. I mean, surely if you look at the screen and you see one
which is half hidden, which one do you read? The one you should read or the
one -- of course you read the one which is half hidden. So it is a bit silly.
>>: [inaudible]
>> Béla Bollobás: That's the aim, to emphasize it. Yeah, I didn't think of it. But
then it should be the other way around. Yeah.
So first, no matter what distribution we have either in the original or in the dual,
we do not have percolation.
And as a consequence -- sorry. The only condition needed was the symmetric,
and it's important -- the more symmetric we have in the structure, the easier our
proofs are. Being symmetric is just about the weakest you can get away and one
has to work quite a bit harder to prove it than if you have, as they used to have,
two x's of [inaudible], for example.
>>: What does the symmetric ->> Béla Bollobás: Oh, essentially symmetric. Symmetric in the origin. That's all.
So just one.
And there used to be results that said if it's at least six-fold symmetry or whatnot,
then it goes. If it's four, then it goes. And that's about the weakest.
Okay. There's another ingredient which looks like the kind of things we have
used, and it has been used a lot to give various predictions and in fact to get
results later.
So super in our graph there is a triangle and then instead of taking the triangle,
let's put in a new vertex and join this new vertex to these three vertices, three
edges.
Now, the question is under what conditions -- so you have some probability. In a
moment I think we have it. No, that's going back.
So with some probability p -- I'm sorry, it does say it. So these are in it with
probability p, those are in with probability r. Is there a choice of p and r such that
as far as the outside world is concerned, these three sides behave exactly the
same way as those three sides.
So when you take this unit, this little gadget, connecting them in a triangle, let's
say in one of these shaded triangles, then the two behave exactly the same way.
So is there such a path?
Why doesn't it go?
So what is the extremely simple question -- so what we want is that the
probability that the within a region, within a triangle, all are connected in one
should be the same as in the other. So all connected in one will be whatever.
Either all are connected or at least two -- or two are connected and the third is
not connected. That is the first expression. Or they are all connected from the
start.
Okay. And so on. So what conditions do we want? We want whether these are
the same for some choice of p and r.
We do have three equations, and of course there's another condition. So it's not
entirely obvious that there is such a choice of p and r, but if we set r to be 1
minus p, then two become the same, we are left with a single equation, namely,
this one.
So for this particular value of p, does the star and the triangle give exactly the
same black box.
And using exactly that, they made this guess that this was a critical probability of
the triangular lattice in the bond case. And this was the one that was proved by
Wierman quite a bit later.
Now, let's see how it can generally be applied to connect various lattices. Let's
take the triangular lattice and view it as a union of these units where all the
connections happen within a unit.
And now from what we know is if you choose the probabilities of these edges
appropriately and you replace all three edges by a star with a different
probability, then globally nothing has changed. If you have a long path, then of
course you'll have exactly the same long path. Okay, it may be twice as long or
whatever. Irrelevant. It goes roughly from one place to roughly the other place.
Okay. So if we choose them, then one goes into the other, and what you get if
you go from the triangle, you get the hexagon. So it tells you these two behave
exactly the same way.
So here is a proof of Wierman theorem, and the coupling provided the probability
of a triangle with edges p naught and then the hexagon we choose the dual one,
1 minus p naught [inaudible]. The two percolations are equivalent. Whenever
you find a long path into one, you find essentially the same long path in the other.
So if that is the case, then of course it has to be true. Either both are
supercritical or both are subcritical. Yes. But what I told you a little while ago,
that we know that they go against each other. So if one is supercritical, the other
one is subcritical, then the dual is subcritical and vice-versa.
But simply applying it to this triangle -- star triangle transaction formation, we
know instantly that they are both critical.
Somehow it doesn't want to go.
Okay. Now we're getting to the actual topic. And here really my aim is just to get
it across what the whole setup is and I shall say very little about the rest.
So, again, we have got lots of the sites, vertices, and rather than edges we have
got polygons put down. We have got these hyperedges, set of polygons.
All the time so far I have been harping at these triangle arrangements. We can
view those that that was exactly a hypergraph in which every phase was a -- or
every edge was a triangle.
So we need the obvious conditions. The vertices should form a discrete set, the
interiors should be disjoint connected, and there should be no accumulation
point. All degrees are finite, and any time two hyperedges meet each other, they
meet each other, of course, in a vertex.
Okay. So what's an example? Here is an example of a hypergraph, a plane
hypergraph. This is an edge and that is an edge and that is an edge, but rather
than just drawing it as a usual edge, let's make it a little thicker and so on.
Okay. So, once again, the setup will be that in each edge we have got a black
box which tells us which way the various sides are joined or with what probability
they are joined.
And the question is under what distribution do we have percolation in the whole
lot. And how should percolation -- how do we think of percolation, that within a
black box -- well, there are no edges. It need not be realizable. It need not be
realizable by a plane graph at all. It's just a black box that tells you that with
probability 10 to the minus 10 you can go from -- you can go all across, or with
probability 1 over 10 you can go from here to there, but not there. Whatever.
Another picture would be more of the way we would think about it. We would
replace every Kagome by 2 Kagome and would replace every vertex of degree
incident with k with d hyperedges by 2 [inaudible].
And not even Paul is supposed to sleep. His jet lag should be over [laughter].
No, that's all right. I am used to waking people up. And it's much harder in a
class where I don't know the names [laughter]. Then I have to rely on T-shirts
and whatnot. So even the person in the pink T-shirt should know the answer to
this question. And it usually works.
Yes. Okay. I'll return to this picture a little bit -- in a second.
So here is the setup, and these are the hyperedges and then that's the area
shaded and vertices is black and there is some region in between. The regions
in between are the phases of the whole setup, and then we would like to define
the dual. How?
Of course, exactly as we always do. So we put a vertex in each phase and then
for every edge, every hyperedge, define a dual hyperedge. What will it be? The
dual of this particular edge will be the hyperedge which has a vertex here, a
vertex there, there, there. These four. So they would look like that.
Okay. And then, again, what's a plane hyperlattice? Exactly what we had plus
two independent translations under which it's invariant.
So here is the picture of what I have been talking a minute ago. So for a phase,
the dual would be, for example, like that.
And if you look at this picture, you see immediately that this particular
arrangement of phases is in fact self-dual. Look at the phases in the other case,
it looks exactly the same translated by 1.
Okay. But we don't really know what dual another define. It was extremely easy
in the case of edges because we just said there is an edge, the probability in the
dual that the edge will be 1 minus this probability.
Here it's not likely at all because we have lots of distributions. So a k phase, a k
hyperedge, really has officially 2 to the k states, and we will have to say what the
translation is.
Yeah, but actually, we should not take on all of them because we can't deal with
them, but we shall take only plane partitions.
So it says so that given a polygon, the vertex and the partition, the partition is a
plane partition if essentially it can be realized by an internal tree. That's exactly
what it says [inaudible].
So no two of its parts interlace. Again, what's a partition? When we have -- the
black box will tell us that if you can join a to c and b to e and f, g, and k all
together and so on. I mean, this would be a state. That's what the black box will
tell you. You can go from this, you can run around to this group and that group
and that group.
Okay. But we take only the plane partitions. So those -- is it fairly -- Paul, now
that you are awake, is it clear what a plane partition is? No. No. Not as such.
Okay. So suppose this is one part and this is another part and this is the third
part. Is this a plane partition?
One part, another class, third class. Is it a plane partition? No. Yeah, you are
absolutely correct, because these two, this one and that one, are interlaced, and
that's what I mean. Which is exactly saying that there's no way of putting a forest
into the phase which tells us which bits are connected because you would have
to put in something which connects these two and you would have to put in an
edge which connects those two, and it contradicts planarity.
Okay. So that's what the plane partitions are.
Now, what about the duals? That's extremely easy. Again, you think, of course,
in the dual you have a vertex in each phase next to it, and you just ask yourself if
you have the partition inside as described, are, again, these two -- are tease two
in the same part? And what's the verdict? Of course, there's no way of getting
from there to there, so these two are in different parts and these two are in
different parts.
Yeah, but let's take a more normal example. For example, so one, two, three
and then these two are the same and this is the third class. Now, that is
definitely a plane partition. And what is the dual partition?
What is this one connected to? It's single, correct? It can't be connected to
anything. It would be interlaced.
What about this one? Which is it connected to?
>>: [inaudible]
>> Béla Bollobás: At the top. What about these? This one is, again, single, and
these two are connected. So the dual, that partition is one in which you have two
singles and two pairs, as it happens.
Anyhow, so that's -- that's the setup. And in the case of a triangle, it's extremely
easy. It's exactly invariant.
Okay. What about this martini lattice? Take p1 for the star bond -- this has
probability p1 and those have probability p2. If you view it like that, then, of
course, this gives you a interpolation between the hexagonal and the triangle
lattices.
And if you consider this as a generator, then, of course, that's exactly what we'll
have. And if you view the whole thing as one triangle, then the dual of that
triangle, of course, is this triangle. Okay, we have to supply symmetry to go from
one to the other, but you have a chance of being self-dual.
And when they are self-dual, then we have got exactly the critical probability.
Okay. Of course, to get anywhere with percolation we need measures. We have
some percolation measure on a set of configurations which are produced by
some random things, but the main thing is the different hyperedges produce -whatever you produce here in hyperedge is independent of whatever you see in
every other hyperedge. So hyperedges are independent of each other and
percolates if there's infinite components. And then what we are interested in is
this hyperlattice percolation model.
Okay. And the dual is exactly as I already described. Once again, every
partition gives you a dual partition and the probability will be just the same as
before.
And that was exactly what happened in the original case. I hope it's clear that we
haven't changed anything.
In the case of normal bond percolation, when one edge is in the probability p
then the dual edge is in with probability 1 minus p. But that is exactly the same,
that it's not in with probability p, and dual of the partition and this ridiculously
simple two-point space, the dual, is exactly that.
So it's the exact correspondent. We haven't changed anything whatsoever.
Now -- yes, but now this day we have lots of states and lots of measures. A
measure is one that assigns to each state at probability. So unlike in the simple
case where we had just one up and down thing, here we have measures on
these posets, measures on partitions. Partitions form a nice poset, and then the
obvious thing is that percolation measure dominates one if on every up-set it's
bigger provided this up-set is non-trivial.
And here is the main theorem I sort of mentioned already, but now we should
more or less understand what it says.
So if we have a self-dual hyperlattice percolation model with probability p wherein
the p is a distribution, p is a distribution on all planer partitions of the edges. If
you have another probability measure on these which strictly dominates p, in that
case you get percolation. If it's strictly below, then we have exponential decay.
But in particular, Hp is critical.
Notice that here much, much more is missing than in the other cases because
we carefully require that it's a strict domination. For every up-set we have strict
domination, and that's not what we should have. We should have that if there
is -- it's always at least and there is one poset in which it's strictly bigger, then it
should be true. But we can't prove it.
Okay. And there's a corollary. We get exactly this -- a very simple case, a
triangular case, but all that matters is whether it's cheaper to connect all three
vertices, then are we more likely to connect all three vertices and none or the
other way around.
And it implies absolutely everything I mentioned earlier. For the martini lattice it's
a tiny bit of calculation. You just write down your equation that everybody in the
past could write down -- oh, except Dominick. Oh, my God. Oh, horrible.
Yeah, almost everybody could write down. And the result is 1 over square root 2,
which is a very nice result.
Okay. What about tools? Because in many ways this is the main thing that we
want more tools to get things, to prove things.
Here are four of the tools, four bits we have used. Some correlation inequality
ones. Everybody knows -- in fact, usually people know as the SKG inequality,
which really is just Harris's lemma that in, let's say, in a cube up-sets are
positively correlated. This is the lemma Harris put into the paper and which he
proved the first result about critical probability being one half, and very nicely he
put it in with apologies that, of course, this is such a trivial fact that it must be
known to everybody, but he couldn't find it in the literature so he puts it in.
And, ridiculously, that was the first occurrence, which I still find it amazing. I
completely agree with him that it's inconceivable that in 1960 such a basic thing
was not used by probabilists.
Okay. The heart of everything is this rectangle-crossing lemma about which I
shall say essentially nothing, which was how did the proof go in the original
Harris-Kesten theorem? We know that crossing a square, if the probability of a
bond is one half, is about one half. So far so good.
What about crossing 200-by-100 square grid the long way? It should be banded
away from zero, and the RSW, Russo-Seymour-Welsh, lemma tells us that is
exactly the case, that you name your favorite aspect ratio, 25, and then if you
have a rectangle of height 100 and side length two and a half thousand, then the
probability of crossing it the long way is at least 10 to the minus 10.
And once you have it, you are in business. Depends whether you finish is
quickly or not and depends whether you prove is RSW quickly or not or correctly
or not, as it happens, because one of the two proofs was incorrect. I mean,
Russo-Seymour-Welsh is two companies. It's Russo and Seymour-Welsh.
Yeah. And then one uses the sharp-threshold results, as I've heard here
mentioned several times, and the very simple fact that K independent
percolations also have non-trivial critical probabilities. I shall not go into this, but
let me just say what kind of thing we need instead of Harris's lemma, this up-sets
our poset correlated [inaudible] infinitely weaker will do, but we can do much
better.
So you takes a finite poset in which the greatest element has some given -- I'm
sorry, which has a greatest element and has a strictly positive probability, fix it
[phonetic].
Now, if you take whatever power of this poset, p to the n, and n can be anything,
then two up-sets are correlated according to some silly horrible function.
Irrelevant. The main thing is that if you tell me the probability of one of the sets is
one quarter and the probability of the other set is 1 over 10, then I can say, ah,
but the intersection is at least 10 to the minus 10. Some function.
Okay. And the way one tends to use it, that if the union of lots of up-sets is big,
then at least one of the up-sets is also big.
Okay. And sharp thresholds, so similar thing, is a huge literature, but it's always
on grids and cubes and solid cubes and whatever you like. Here, again, we need
quite a bit worse situation. But the extension itself is, in spite of that, is nothing to
write home about, although it's not terribly simple.
As I said, the main thing by far is this rectangle crossing lemma, and it's very
much like you're asking somebody to swim across -- whatever you like to name -with your hands tied your back. I mean, and that really is the whole -- proving all
these rectangle-crossing lemmas, people always use symmetry and a lot of local
knowledge.
Here we have to use it essentially, I mean, with infinitely little local knowledge,
and that's what makes it non-trivial.
Okay. Are there open problems? Yes, of course. There are infinitely many.
Of course, the one for which you will get a Fields Medal is if you manage to
extend Smirnov's theorem for -- if you just do it for everything I mentioned here,
you extend Smirnov, then I'm sure that's enough for a Fields Medal, which is
conformal invariance. At the moment it's known only in by far the simplest case, I
mean, for the triangle lattice and not even for the square lattice is it known.
But, also, lots of the things I mentioned are not in the final form. At least I hope
they are not in the final form.
Also, as sort of a pie in the sky, I would love to see this method used to genuinely
approximate critical probabilities of essentially whatever by superimposing or
whatnot some of these lattices about which we do know quite a bit.
So, for example, back to Smirnov and conformal invariance, it would be
fascinating to see whether this percolation which we know is critical, we
discussed it is critical, which is totally lopsided. So we have some probability of
joining all three, we have some probability, tiny probability, of joining none of
them, and we have a fairly large probability of putting in this drift.
So all these edges are in almost certainly, and otherwise there's a little bit that
they're all joined. So it looks not -- it doesn't look like the kind of thing that should
be where connections should be conformally invariant.
Of course, you can have transaction formations and everything looks in a
different way, so who knows.
Okay. Really, my aim was not to tell you anything about the proof, because that
would take actually quite a few lectures, but just to draw your attention to the fact
that there are these fairly general models for which one can actually say fairly
easily what the critical probabilities are.
Thank you very much.
[applause].
>> Yuval Peres: Questions or comments?
>>: [inaudible]
>> Béla Bollobás: No, I don't think so. No.
>>: [inaudible]
>> Béla Bollobás: No. No, no, no. No, there the two are the same. Here they
are not. Here one of them is shorter, if you like, and the other one stays big.
Actually, there's an obvious extension of net result which is incredibly cheap.
You take the squares and then you don't do it at the top and the bottom have the
same probability and left and right, but the top and the left have the same and the
right and the bottom have the same. And the same result applies, that p plus q
has to be bigger than 1 and there is percolation.
>>: So is there any reasonable way of [inaudible]
>> Béla Bollobás: That would be nice, but then, I mean, we're into the problem
of using duality. That's a wonderful task, but the answer is, of course, I have no
idea.
Almost certain there isn't a simple way of relaxing it.
Yes?
>>: How straightforward or how difficult is it to construct the black boxes from
[inaudible]
>> Béla Bollobás: Oh, you don't construct the black box. That's the whole point.
I mean, God gives you these connections. You can't look into it how the
connections come about. And they need not be -- there is no mechanism
connecting. That's the whole point.
If it were a planer graph then it would say, okay, here's a planer graph. But in
that sense they are not planer graphs at all because you can't have this which is
realized and that thing which is realized and so on.
>>: [inaudible]
>> Béla Bollobás: No. But different things need to have different things. No.
So, for example, it's -- the measure is strange on it. If the measure were nice,
okay, you can realize it. And in fact that's exactly the way we think of it originally,
because there is a -- there is a black box telling us which way various things are
used and then we put in various rings. So that's another set of phases and so
on.
And then you put them into -- these into various states, which is essentially what
you are saying. And now one says that when are two connected, if there is a
black path from one vertex to another vertex to another vertex, and here is a
black path going around.
And normally what black box, the black box tells you, you can realize it by finitely
many rings, three rings, and then you get it all as a coloring. Yes. But the
measure on the coloring is hopeless.
So, for example, up-sets here are not correlated. So there is no chance of
defining it properly because this very basic thing gets -- actually, I was hoping to
announce a very big result, but I'm afraid I've been let down. And, also, my
stupidity let me down, of course.
I had a wonderful email from [inaudible] some years ago on the 1st of April which
he sent to Martin Grechel [phonetic] as well, and it was -- and I know I had it on
my damned laptop somewhere and I couldn't find it. It was actually a brilliantly
written piece saying that two physicists using the cavity method combined with
the latest progress in nuclear physics or whatever and [inaudible] managed to
prove the Remand [phonetic] hypothesis.
Now, those of you who are old enough, like you, Val, did you get an email from
Martin Grechel at the time? Martin Grechel was the main organizer of the
congress, and [inaudible] he got this email he sent an email to everybody he
knew, including me, that essentially breaking news, great news. The Remand
theory has been solved.
And he was of a bit of a fool, because they would give away signs in the
message. So I was desperate trying to find it and say, actually, I talked about
various results, but there is an even more important, a much more important
result. I would have liked to have seen your reaction, but it was not to be.
You didn't ->>: [inaudible]
>> Béla Bollobás: It's actually -- it really is a masterpiece [laughter].
>>: So just one more question. At the end you said because of the lopsided
nature and problems with performing things.
>> Béla Bollobás: Okay, it's not -- it's not --
>>: This is just local [inaudible]
>> Béla Bollobás: Yes, true. I know. So you can -- by squashing, you can
connect it. But even then, it's so -- it really is very different. It may be just my -just my feeling that if it goes for this, and it's a very, very easy distribution -- I
mean, the point of it is partly it's lopsided, but more that it's an extremely easy
variation.
Maybe one can prove it for that, and that I would find as a very encouraging sign
that it's true always. In fact, some years ago Oliver and I worked a lot to prove
that the critical probability in [inaudible] percolation is one half. And it took us
ages to do it, but we did it. And that was supposed to be the first step towards
conformal invariance of a [inaudible] because if there's any percolation which
should be conformally invariant, it's [inaudible] percolation, which is the one you
get by dropping some points in the plane and take for each the [inaudible] cell
and then color the [inaudible] cell black or white or put it in or out with the same
probability. Again, looking at it, it's obvious that critical probability is one half and
70 pages later you actually conclude that it is indeed -- well, maybe it was longer
[laughter] -- that it is one half.
It was supposed to be the warmup exercise. Let's prove it today that it's one half
and then tomorrow we start on the real project, which is that it's conformally
invariant. Unfortunately by the time we proved it we were dead tired and we still
haven't managed to get back to it. But we shall. It's certainly one of the things
we would love to do.
>> Yuval Peres: Any other questions? Thank you.
[applause]