>> David Wilson: So welcome to the Pacific Northwest Probability Seminar.
We're very happy to welcome back Asaf Nachmias as our first speaker. He was
a post-doc here, among other places, and he's now over the border at UVC. And
he'll talk about recurrence of planar graph limits.
>> Asaf Nachmias: Thank you. Can everyone hear me? So thanks for the
invitation, and it's nice to be back in Seattle, especially in this beautiful weather.
I'm going to tell you a story about recurrence, about planar graphs and there's
going to be circles involved and things like that.
So let me just jump ahead. All right. So we're interested in triangulations.
Triangulations, they're planar graphs embedded in the planes such that every
face is a triangle. Forgot to mention this is all joint work with Ori Gurl-Gurevich
sitting right here.
>>: Reminding you.
>> Asaf Nachmias: Reminding me. We're interested in choosing one of these
objects at random. It can be a triangulation, a quad triangulation, where every
face is a square. There are several, of course, technicalities about like several
ways you could define triangulations, but let's ignore all of them and just assume
all of our graphs are simple, they're all embedded in the plane and we never care
about the homotopic class we just care about embedding -- we've all seen planar
graphs.
We want to choose one at random and see how it looks like from -- whoops, from
afar, from far away. And I apologize for the pictures. You will see a lot of nice
pictures, none of them was made by us. The really childish pictures were the
ones that we were capable of doing, and I put the credit for the pictures, the nice
ones, like this one -- this is random quadrangulation, and ignore the lines around
it we'll get to this later, but we understand this is a finite space.
So we can always draw things at random from here. And we're interested -- and
as large number of vertices, perhaps the number of faces, edges, all relatively
immaterial right now, how does it look like?
Here's a picture of how it looks like at a large scale. And here's the credit for this
picture. And what you're seeing here is what you get when you rescale -- you
draw a random triangulation on N vertices. It turns out absolutely not a trivial
statement that the typical graph distance between two vertices is N to the
one-quarter. This is an interesting fact. And but much more exciting is the fact
when you rescale all the distances by N to the minus one-fourth you have a
random metric space. It converges in the usual Gromov-Hausdorff limits to this
random compact metric space that turns out to be homeomorphic to the sphere.
And this is a very nice heavy result by Le-Gall and Miermont. And there's
another way that you could take a limit in this for random triangulations. Instead
of scaling the distances, don't scale the distances and you want to get in your
limit an infinite connected graph. And this is the, kind of the canonical way of
doing it, the distributional limit of graphs which I will now define that was invented
by Benjamini and Schramm.
So this is how it works. You have a sequence of finite graphs GN. They could
be random but they don't necessarily have to be random. They could be fixed.
We'll see examples. You choose -- this is an important part. You choose a
random, a uniformly chose random vertex of this graph, and you look at the -you look at the ball and graph distance like here around this typical -- around this
random vertex.
And you say that the sequence of graphs GN converges to now this is a random
rooted graph. So it's a graph and it's an infinite graph and a root.
If for every radius R, when you look at -- when you look at the balls of radius R in
your finite graph, they converge? Right? Assume that the degree is bounded.
So this is all finite spaces.
This converges in distribution to the ball of radius R in your limit.
Okay. This is the important definition of this talk. So let's see a couple of
examples, just so we'll be in the right mindset. If your finite graphs are aligned, if
the path of length N or perhaps a circle of length N, then no problems. Obviously
you choose a random point it's going to be somewhere in the middle. R is -- R
you think about R is fixed and N tending to infinity.
Of course what you're going to see around you is a path of length R. So it
converges to just the infinite line.
If you start -- now here's an example where your random -- where GN is a
random graph. You start with the Erdos–Rényi random graph, complete graph
on N vertices and then you erase each edge with probability one minus C over N
and retain it with probability C over N.
It's a random graph, and you can think about it a little bit, and the truth is that this
converges to a Galton-Watson tree, because when you start exploring your
neighborhood of length, of radius R, from a typical point in your Erdos-Rényi
graph, you're not going to see any circles. It's definitely going to be a tree.
And it's going to be a Poisson tree with mean C. There's another illustrative
example. Take a binary tree of level N of height N and it converges -- well, you
would think it converges to the infinite binary tree as you would, but this is not
true.
And the reason that this is not true is that we're choosing a vertex at random.
And a vertex at random is very likely to be a leaf or a level close to the leaf, level
minus one, minus two, you can figure this out and you get what's called the
canopy tree. And this is a picture of it.
We drew this as you can see right here this is not the binary tree. It has, for
instance, it has just one line infinite line going to infinity, not more than that.
All right. Any questions about the definition of a distributional limit? All right. So
the first thing that we want to do since we're talking about random maps is we
choose one at random. The average degree is constant, at most six.
Does it have a distributional limit? And the answer turns out to be yes. And this
is a very famous paper by Angel and Schramm in the case of triangulation and
by Krikun in the case of quadrangulation, this limit exists.
And it gives you an infinite triangulation or quadrangulation of the plane. So it's
not degenerate like -- you could cook up some examples where everything
converges to a point.
And they called it this uniform infinite planar triangulation or quadrangulation
UIPT or UIPQ. So it's a random infinite triangulation of the plane. This is what
we currently know about this.
And these -- I'm not going to spend a lot of time on this issue but this is just an
issue for -- we expect so, of course, we could choose triangulation
quadrangulation. I didn't mention pentangulation or that other, whatever you call
the one with six edges.
You can continue, but we expect that all of these things behave basically the
same. We don't expect to see in the limit these small little petty matters like what
the size of faces you chose.
And this will also be evident in this work, which I will show you right now. Here's
a nice picture by Nicole [inaudible] of a quadrangulation. And this is a picture I
like more because remember it emphasizes the point that our graph is always
rooted.
So we always have a root. We're always sitting on the root and looking towards
infinity. So this is why I like this picture.
And this is a quadrangulation. It has an interesting geometry. This is a graph.
We could study its graph distance. It's a graph embedded in the plane. Also
study aspects of that.
And it has a lot of similarities with Z-2. It's one-ended. That means that if you
remove any finite set, you only have one infinite component left.
So this was proved by Angel and Schramm. The critical percolation probability
on it is a half. Just like on Z-2. So it's proved by Angel. It is [inaudible]. It's
okay. We won't -- it means that it doesn't exhibit nontrivial harmonic functions
that are bounded.
This is due to Benjamini-Curien. This happens also for Z-2, Z-3 and more. But it
is not Z-2, this graph. It has all sorts of like bizarre behavior.
It has unbounded degree, for instance. This is a main -- this is a fact. It's not
immediately clear, but when you draw a random triangulation on N vertices, your
maximal degree is going to be order log N and you'll have degrees of all
spectrum between log N and the constant.
And this is actually a major difficulty. Also has a completely different growth than
Z-2. The ball of radius R and the graph distance grows like R to the fourth
instead of R squared.
This was proved by Angel for the triangulation case and Chassaing and Shaffer
for the quadrangulation case.
And it's a fractile. It's not a nice graph. It's quite crazy. So it's, for instance, the
random walk behaves much slower on it. It's strictly sub diffusive. That's
because it has all these giant dangling ends which don't connect to infinity. It's
like a cactus. And random walk, instead of going in this straight line going
towards infinite, being efficient, wastes its time on the branches of the cactus.
And this is a behavior that we see in many examples of especially of random
nature but also of deterministic nature like Zerpenski [phonetic] gasket.
I didn't tell you in all of this whether it's recurrent, whether the random walk is
transient on this graph. And the conjecture, which by Benjamini and Schramm
and Angel and Schramm is that this graph is recurrent.
I could -- I don't really want -- I don't really have a feeling why they conjecture
this. Like you can cook up examples that all of these things hold and it's
transient. Perhaps Homer would be a better candidate to ask it. But they did get
very close in the sense that if we had, if this graph had bounded degrees you
would be able to solve this problem. And here's a theorem of Benjamini and
Schramm, the same paper where they define the concept of the distributional
limit of graphs, and they have the following statement: When you have a
distributional limit of finite graphs with the two extra assumptions, one is bounded
degree and second is planarity, very important, of course.
Then the resulting graph, the limit is almost surely recurrent. Remember, these
graphs could be deterministic but the limit, since the limit -- you always are
choosing in your finite graphs you're choosing a random root then the limit is a
random graph.
So hence the almost surely should be there even if we're choosing deterministic
math, which I'll show you a sketch of this proof later.
An example is Z and Z-2 that we saw that we got. The canopy tree that we also
saw.
And you really do need -- there's no hope of proving that the UIPT is recurrent
using this theorem because this theorem really does require that the bounded
degree condition, because okay here's a graph similar to the canopy tree. It has
weights on it. But you can always mimic the effects of -- so the random walk is
going to walk according to the weights when it's here it's going to walk with
probability, what, two-thirds upwards and one-third downwards.
This graph is transient.
>>: [inaudible].
>> Asaf Nachmias: Sorry. Yes. Sorry? I got it. Yes, there's another one.
Thank you. This graph you can convince yourself is transient. It's a weighted
graph. We didn't talk about weights at all, I agree, but you can always mimic
these weights by having multiple edges. This is an easy thing. You can
convince yourself that like we're taking a binary tree with these multiple edges,
you can get this as a distributional limit.
So indeed the theorem does require the fact that you have bounded degrees. So
here's our main result. Instead of bounded degrees require that the root has an
exponential tail which means that the probability, that the degree of the root, the
vertex degree of the root, the probability that it's bigger than K, the K is
exponentially in K.
So we have that assumption replacing the bounded degree plus planarity gives
you almost sure recurrence. And it's good because we do know these facts, the
exponential, the exponential tale of the degree of the root for the UIPT/UIPQ/you
PIQ, result of Angel and Schramm and Benjamini and Curien.
And we really do need a comment that I won't explain much about, we really do
need the exponential tail anything, any tail that's a little fatter we have an
example for that that is transient.
Any questions about the result?
>>: How much more slightly ->> Asaf Nachmias: Sorry?
>>: How much is slightly fatter, can you quantify that in any way?
>> Asaf Nachmias: If it decays like E to the minus K to the alpha for any alpha
smaller than 1, then there's an example. We could push it more to anything, but
it's a necessary thing.
So if there are no questions, let me try to tell you a few words about the proof in
the little time I have. The proof is we like it, it's cute because it has circle packing
in it.
So let me tell you -- give you a brief introduction into circle packing, a very
interesting theory and it's very elementary, and it's like easy to get into. So I
highly recommend it. There's a really nice paper by Stephon Odachom
[phonetic] from circle packing from SLE that's very nice. Here's the basic
concept.
You have a finite simple planar graph. I'm giving you an embedding in the plane.
Here it is right here. But we're looking for a canonical way of embedding it. What
does it mean canonical? I'm actually not that sure. Another way that people
describe it this way to represent the graph, it's conformally natural. I read it in
your paper, Stephon, I'm still trying to figure out -- but I can convince you perhaps
that these are correct terms to use but let's not get into the there's a lot of
connections to conformal maps but I want to keep this very simple.
But the main theorem that we use is this Koebe Andree Thurston theorem.
Koebe discovered it in '36, was forgotten, and Thurston realized and
rediscovered it and found out there was a corollary of theory by Andree on
polyhedras, and it says that any finite simple planar graph can be presented as a
tangency graph of a circle packing drawn on the right. So circle packing is just a
bunch of circles in the plane with disjoint interior, some of them may be tangent,
some may not. We're drawing them in the plane.
If two of them are tangent we put an edge between them. This is the tangency
graph. Obviously, if you have a bunch of circles satisfying this and you draw this
tangency graph, it's going to be planar. That's easy to see. But the strength of
this theorem is the fact that the converse holds.
Any simple planar graph can be, you can find a packing of the circles in the plane
that realizes it.
In fact, when your finite graph is a trangulation, then this drawing is unique up to
trivialities and maybe use transformation that keeps -- sends circles to circles so
nothing really changes.
All right. This is the theorem. What does it have to do with probability? This is a
geometric theorem. It's nice. But it does have like a lot to do with probability.
And this is why it's so very fascinating and we're just starting to discover all sorts
of things that Odachom knew of course many years ago.
So here's a very nice theorem of [inaudible]. It's a completely -- there's no
randomness in the underlying graph. So if you have -- so this theorem states
that if you have an infinite circle packing in the plane such that it covers
everything, of course you can imagine circle packing that don't cover everything
and they're infinite.
Here, you have this is the disk and you put a circle here and a little circle here.
And like more and more and they converge like little points here.
And they become smaller and smaller, and don't worry the entire plane. And the
theorem of Hand Schramm tells you when it covers the entire plane, it is
recurrent. I'll show you a proof of this.
>>: Cover the entire plane exactly?
>> Asaf Nachmias: Going to be an annoying question. It means if you look at
the union of all the circles and all the space between them, the inter ->>: Bounded space.
>> Asaf Nachmias: All the bounded space between them then it covers
everything.
Okay. So here an accumulation point, I mean exactly so you don't have a
sequence of circles converging to a point. That does not happen.
When this happen, the graph is recurred. In fact, they also have the converse:
When you're in a situation like this, when the graph is packed in a bounded area,
it is transient.
So you have a dichotomy. This is very nice. Also bounded degree here is
necessary because here's the circle packing of the usual triangular lattice, and
now I'm going to change it. I'm going to turn it into -- I'm going to rule -- I will not
have bounded degree here. I want to create a drift to the right.
So I'm going to add a bunch of little circles. I hope that you can see them here
and here. So whenever you're in a big circle, just because you're performing a
simple random walk you have way more chances of going to the right. And here
I put like twice as much circles than I put here so you still have more chances of
going to the right.
So once, like your random walker travels around here, once he gets to here he's
sucked into infinity.
So you do need this bounded degree situation. And let's see how you would
prove the Benjamini Schramm theorem. Let me remind you what it is. Any
distribution of planar bounded graph is almost surely recurrent. Let's go over the
strategy. So GN is our finite planar graphs.
We apply Koebe's theorem and find a circle packing, whatever circle packing we
want, doesn't really matter. We're going to normalize it by choosing a random
circle. This will also be a random root.
So we chose a random circle. We have a picture. We can always shrink it.
Nothing changes really and move it around, the same graph.
So we're going to normalize we'll put the root to be the unit circle at the origin.
And we're taking a limit of this thing, right?
We have a sequence of graphs. We chose the random root. We drew it in the
plane. We took a limit. So what we need to prove here is of course what we get
is something that covers the entire plane. And then we can appeal to the Hand
Schramm theorem, and this is absolutely not trivial to show. There's a lemma
there, it's short but it's fascinating. It's really interesting.
And the lemma is it's not that it covers the entire plane. There are situations that
you can have an accumulation point but it's only going to be one. It's going to be
an isolated accumulation point. So, of course, I'm not applying immediately Hand
Schramm, but one accumulation point, in fact any isolated set of accumulation
point is easy to handle.
By inversions -- I won't go into this, but it's easy.
We apply the Hand Schramm theorem or perhaps slightly stronger version of it
that allows some accumulation point and we get that the graph is the recurrent.
We use the bounded degree assumption because we use the Hand Schramm
theorem.
There's a nice picture by Maxiem Krikun [phonetic] of how a random triangulation
may look like.
Okay. So in order to prove that the UIPT is recurrent, we needed two new
ingredients to add to the strategy. One is a quantitative result of the Hand
Schramm theorem. The Hand Schramm theorem tells you the graph is recurrent.
We'll study the effective resistance between the origin and balls of radius R in
your circle packing, and the fact it's recurrent tells you the effective the
resistance, the electrical resistance goes to infinity, we'll see how fast -- I already
told you it grows logarithmically, in the radius of your packing, if you look at the
effective resistance -- we'll go through that.
The second ingredient is to take these like, like these UIPT large degrees.
Think about them as stars and we'll replace them with a tree that will have
degree 3, and we'll kind of embed it and we'll need to study how this changes the
effect of resistance. We'll use the quantitative version of the Hand Schramm. I
won't be able to show you the entire thing, but I'm hoping to convince you. I'm
not sure about that. But I'm hoping to convince you that it's correct and that you
can easily read this. Our paper is short. It's elementary.
So I guess I'm considering this talk as a sales pitch. All right. So let's tackle the
first ingredient. We have a circle packing of a bounded degree trangulation.
Before I tell you how to prove that the effective resistance grows logarithmically
in the radius. And let me tell you about this important ingredient called the ring
lemma by Ruden and Sullivan. And it tells you that if you have a circle here in
the middle, which is completely surrounded by other circles, as you can see here,
then the ratio of the radii is bounded by any two circles, the inside one and one of
its neighbors is bounded from above and below in a number that depends only
on the bound you have on the degree.
So this is a very now, once you think about it it's relatively easy. Let's say you
have a circle here and a giant circle here.
You will need to take -- you'll need to put a lot of little circles here, if this one was
in the middle you'd need a lot of circles here in order to make it completely
surrounded.
This is the basic idea. I guess perhaps you could put another large one here, but
then you will have a problem. I realize I can see by Russ's smile that I'm giving
you a very shady proof of this fact. And okay ->>: Can't you just have about four of the same sizes they sent to us?
>> Asaf Nachmias: Not if you make this as big as you want.
If you want to come draw it.
>>: No. But lexical would be very, very small. Then the one adjacent could also
be very smart.
>>: Also we need all the others more or less the same size as the central one.
>>: You have a small circle that one of the two [inaudible] cycle left is also
smaller, at least one of the two.
>> Asaf Nachmias: This is not [inaudible] mistake that's one minimum radius.
>>: The central one has radius one. So the others have radius about one. Why
can't he have that?
>> Asaf Nachmias: I'm not going to try to draw it.
>>: So what happens?
>>: That circle, that type of two neighbors no, no, this is not a directive.
[inaudible].
>>: You can get lots of circles around the big circles.
>> Asaf Nachmias: Yeah. Right. Here you would have to have a lot. So I
guess I was wrong, actually.
>>: This is correct ->> Asaf Nachmias: The ratio of the tangent -- yeah.
>>: Imagine ->>: Goes into that further.
>>: There's --
>> Asaf Nachmias: What's written here is correct but I made it sound as if like
this picture, but you think about you have an infinite trangulation so everything is
around it.
This is not a triangulation because the other, because the outer face has more
than three edges in it if you want.
Okay. Let's move on from this point. So here's the lemma that I want to prove to
you. The effect of resistance if you have triangulation, bounded degree, the
effective distance across an anulus is at least a constant.
There's the picture. I'm simply looking at the effect of the electrical resistance
between all the points here in the middle and all the points here outside.
And just like you have in Z-2, the effect of resistance here is at least a constant
and also at most a constant. Why is that? There's the proof.
The effect of resistance is a minimizing problem, right? You look at all the
functions that are 0 on one side and one on the other side.
So functions that are going to be 0 here and 1 here.
You compute their energy. The energy is just the sum over all the edges of the
difference square of the function. And you take -- and you take the [inaudible] of
that and that gives you the inverse of the effective resistance.
To show this lemma I need to exhibit a function that has a finite energy for any
circle packing.
And this is just going to be a very simple function. It's just going to interpolate
between 0 and 1. So on the circle packing. So if this is your anuli and you have
a point here, the function is simply going to give the value of this point simply
going to give it the distance from the smaller, from the smaller circle.
And this is a Lipschitz function. And let's look at the contribution to an edge of, to
the energy. We're looking at FX minus FY square. So if this was an edge right
here, right, then FX minus FY is simply, is at most the distance between the two
points. Right? This is just the triangle inequality. But now we're coming to the
useful fact that we have a circle packing.
This is the distance square right here are the two circles, can you see this?
The distance squared, because of the ring lemma, all of the radii here are
comparable, the distance square is simply the area of this circle. You have
bounded degree. You sum over all the edges. What I'll get is an approximate
constant of the area of the entire anulus. And that is R square. We'll also had,
this is normalized so the function is 0 here and 1 here.
So we get exactly the area is 1 over R. We divide by 1 over R square. And this
little computation here gives you the result.
Any questions about this? It's very simple. And now, of course, to get the sum is
bounded, the energy is bounded so the effective resistance is at least a constant.
And what do we get? Of course when we have an infinite triangulation and now,
of course, we can put more and more anuli growing dyadically, and each one we
have effective resistance at least a constant. We're using again the ring lemma
to say there are no large edges between far away anuli and we just add the
whole thing up, and we get that it grows logarithmically. And in particular it's
recurrent.
So this gives you also a simple proof if you want of the Hand Schramm theorem.
All right. So in the last five minutes I'm going to try to convince you that this
already gives you enough information to be able to control the resistances once
you replace the degrees in your random planar map from when you replace the
large degrees, which is a star, you replace them by a tree. So this is what we do.
This is really, you know, a fix for this problem. We do it in two steps just to be
consistent. First we put on every edge we put a point, a vertex in the middle. It
has degree 2 right here. And then all the vertices marked in black here, we are
going -- they look like a star, and after the first step, and we're going to replace it
by a balanced -- as balanced as we can, binary tree.
Right in such a way that we maintain planarity we can always think about it
already embedded in the plane instead of this vertex we put a tree and we
connect it to the previous neighbors that were here before.
So we have transformed the graph, let's call it G star. And it's important -- this
graph has degree at most 3, as you can see by the picture.
So of course we are here we're ready to end another important fact that you can
like easily convince yourself that if you perform this transformation on the
distributional limit on the UIPT, you could also perform it on the finite graph, and
you'll get the same -- and you'll get the same limit.
So there's a continued issue that's relatively technical. So we know that this
graph is recurrent, this G star. But we know more than that, we know a
quantitative version of how recurrent it is in a given circle packing.
So we also need to -- we have that, and we need to relate the effective
resistance back to the old graph. This is the key lemma and it's the fact that the
effective resistance every time we change -- of course if you look at the effective
resistance, say this was your entire graph, the effective resistance between these
two vertices, two, and now it is more than two. So the effective resistance in the
new graph is bigger, as you would expect, it's recurring.
So there is -- we are losing something here, but we want a quantitative estimate
of how much we're losing and the fact that every time we do such a thing, we're
losing a factor of the degree to the effective resistance. Because what's exactly
here's a precise statement. If we put smaller edges distances on our graph by
how small, we put on this edge we will put resistance which is 1 over 7, 1 over
the degree of the original vertex.
Right. Then -- so now we decrease the degrees -- we decrease the resistance of
G star but we're claiming that the resistance.
In this new thing the resistance between big circles, between the origin say and
big circles remain the same. Okay. So here this is a statement that I'm not going
to prove to you, it's a calculation with resistances and this transformation.
It's relatively straightforward. So what we need, the last thing that we need is
that in this UIPT or any distributional limit, the degree grows practically
logarithmically. What do I mean by that? Right? Because we take this G star of
ours, we pack it in the plane, and we want to say or -- and we want to -- so this
has bounded degrees. Sorry, let me rephrase. We look at our, say, UIPT. We
pack it in the plane. And we want to say that the -- so we already know, right,
that the degrees of the -- in the graph distance, you look at a ball of radius R in
the UIPT in the graph distance. The degree there is at most logarithmic. We
need a different statement. We need a statement that says once you pack it in
the plane and you look at a ball of radius R, the degree there is also at most
logarithmic. It's not the same metric. Now we have two vertices, and we're
looking at the Euclidian distance between them in the packing versus the metric
of the graph distance.
All right. So this is -- so my whole point of this was to say that if you know that,
say, in a ball of radius R, the degrees at most logarithmic, by the previous, when
you do this, when you do this transformation, G star, right? On G star we know
by the quantitative version of Hand Schramm, the degrees grow logarithmically.
If you believe that I've only lost a factor of the degree, then the new resistance in
G between, say, the ball at the center and the rate -- and the circle of radius R,
the outside of it, is at most log R -- sorry. It's by the quantitative Hand Schramm,
you have here this log R and you divide by the maximal degree.
So the resistance is at least a constant. Of course, we need to prove that this is
the resistance goes to infinity. But the point is that the resistance is at least a
constant, not only when you look at the vertex at the center, but from any finite
set, the same calculation works. And it's easy to see that if for any finite set the
resistance to infinity is at least a constant, that means the resistance, that the
graph is recurring, the resistance is actually infinity. This is easy. Thank you for
your attention.
[applause]