Alexander Holroyd: Welcome everybody. It's a great pleasure to introduce Tom Hutchcroft. Tom
claims he's still a Ph.D. student, although, if you look at the astonishing list of his achievements so far,
it's quite difficult to believe this. So he's a student at University of British Columbia and he's currently
here at MSR as an intern with about a week to go. And during that time he's added several more
impressive proofs to this list. So today he's going to tell us about circle packing.
Tom Hutchcroft: Thank you [indiscernible]. Thank, everyone, for coming.
So I'm going to be talking about circle packing. And specifically I want to talk about circle packing in
relation to random triangulations at the plane. And this is joint work with Omer, Angel and Asaf
Nachmias, and [indiscernible] Ray.
Okay. So let me start by telling you what I mean by circle packing. Suppose we take a finite simple
planar graphs. Simple just means that for any two vertices, there is at most one edge connecting them,
and there are no loops that start one vortex and return to the same vortex. So suppose we have such a
graph. So planar obviously means we can join in the plane. But what's the best way to draw it in the
plane? Well, one quite convincing answer to that question is given by the circle packing theorem.
So the circle packing theorem really gives us a canonical way to draw a planar graph. So the circle
packing theorem says that every finite simple graph is the tangency graph of the circle packing.
So what does this mean? A circle pack is a collection of discs in the plane with disc joint interior. So
they cannot overlap but they can touch or be tangent. And if I have such a circle packing, I can use it to
draw a planar graph by just drawing straight lines between the centers of circles if they touch.
So clearly I can get planar graphs from circle packings in this way. The hard and surprising fact is that
every planar graph can be drawn in this way, and moreover if the graph is a triangulation, which means
that every face is a triangle, then in fact this representation as a circle packing is unique as much as it
possibly could be. So it's unique up to Möbius transformations, which if you don't know just the
functions that send circles to circles and reflections.
So let me go over those definitions again quickly. So a circle packing is just a set of circles in the plane
with disjointing interiors. The tangency graph of the packing is the graph which has the circles as its
vertices and two circles are adjacent in the graph if they touch. And something I didn't already say is
the carrier of the packing, which is the union of the circles, or really discs, together with the regions
that are bounded by a path of circles representing a face in the graph. So this is the area of the plane
that is taken up by the packing.
And when we have an infinite triangulation, as we will do shortly, we call it a packing in the disc if it's
carries the unit disc and in the plane if it's carrier is the plane.
So the original circle packing theorem, which is due to, it has quite a complicated and interesting
history so it's accredited to Koebe, Andreev, and Thurston, says that every finite planar graph has a
circle packing. This was extended to infinite triangulations of the plane by He-Schramm in '95.
>>: The people that rediscovered this theorem get credit for it?
Tom Hutchcroft: Well, first, Andreev had not so much to do with the theorem. Thurston rediscovered
this theorem as a [indiscernible] to a theorem of Andreev, and popularized it.
>>: [Indiscernible].
Tom Hutchcroft: It wasn't realized until some time afterwards that it had already been proven fifty
years earlier.
>>: Did they encounter an alien civilization that also proved it? Or did they get priority?
Tom Hutchcroft: Right. We'll just have the names together in a long list.
>>: [Indiscernible].
Tom Hutchcroft: So He and Schramm proves an extension of the circle packing theorem to infinite
triangulation. So if you take a triangulation of the plane, it can always be circle packed in the plane or
in the disc, but not both. Exactly one of the two. So we call a triangulation circle packing parabolic if
it gets packed in the plane, and circle packing hyperbolic if it gets packed in the disc. And also this
packing, again, is unique up to, you know, it's unique much as it could be. So packing in the plane if
it's parabolic is unique up to translation and scaling and things that obviously preserve the circle
packing of the graph and in the disc, it's unique up to Möbius transformations that fix the disc. Or in
other words, the isometries of the hyperbolic plane if you think of the unit disc as being the hyperbolic
plane.
And together this really gives you a discreet version of the uniformization theorem for Riemann
surfaces. So here you can see this picture is from Ken Stephenson's book. At back you can see the
circle packing at the triangular lattice, which takes up the whole plane. Here we have the circle
packing of the five regular triangulation, which is finite, so it lives on a sphere. And on the right we
have the circle packing of the seven regular triangulation, which is in the disc.
And you can see how the different types, these are all very different graphs. The circle packing type is
telling us something about what the graph is like.
And when we have this circle packing, we can also draw the graph, as I said before, in the Euclidean
case, it makes sense, in the parabolic case it makes sense to draw the graph with straight lines between
the centers, wherein the hyperbolic case we can draw the graph with hyperbolic geodesics between the
hyperbolic centers of the circles. And these drawings are actually determined by the graph. That's
what rigidity tells us. That we take this graph, which it's just this commentatorial description of a
triangulation, and we get all this geometry for free. But it's really a function of the graph.
And this allows us to apply geometric techniques to study planar graphs. So I want to just quickly
mention a few classical, or not so classical applications of this, that I won't be able to get into further.
So first, circle packing can be used to approximate conformal mappings. So this is a very interesting
thing that was conjectured by Thurston and eventually proven by Rodin and Sullivan.
Something I really like is that you can use circle packing to prove the planar separator theorem, which
is a statement about the isoperimetry of planar graphs. And there's a very nice proof using circle
packing due to Miller, Teng, and Thurston.
And a new thing that I've unfortunately won't be able to talk about today, but in upcoming work with
Asaf Nachmias, we used circle packing to prove that the free uniform spanning forests of every
bounded degree planar graph is connected. So it's okay if you don't know what that means because I'm
not going to talk about it any further.
So suppose we have a bounded degree triangulation. Then the type of the packing and the geometry of
the packing encapsulates a lot of probabilistic information about the triangulation. So I guess I forgot
to mention, and my slides got mixed up or something, but another part of the He-Schramm theorem is
that when you have a bounded degree triangulation, the circle packing type, parabolic or hyperbolic, is
equivalent to whether the simple random walk on the triangulation is recurrent or transient. So if the
triangulation is recurrent, then the circle packing type is parabolic, it gets packed in the whole plane. If
it's transient it gets packed in the disc.
And it also tells us about things like harmonic functions on the graph. We can also do lots of things
like resistance estimates using geometric techniques instead of forests and things from complex
analysis.
So this is where that came up. So this is what I just told you. Another part of the He-Schramm
theorem is that the circle packing type in the bounded degree case is equivalent to recurrence in the
simple random walk. The circle packing type being parabolic is equivalent to recurrence. And actually
when you have a CP hyperbolic triangulation, it's always transient. No conditions. However, in the
unbounded degree case, things can go wrong. So you can get a CP parabolic triangulation of
unbounded degree that is transient. And you can construct one as follows. So you start with just the
circle packing of the triangular lattice, and then you're going to add some circles in to create a drift. So
if I put just one circle in each of these two gaps here, and then the next one along I put two circles in
the gaps, and then the next one I put four and so on. And I can create a drift to the right that makes the
graph transient. So probably everyone in this room knows this, but transient means that the random
walk on the graph escapes to infinite. It does not revisit every vertex infinitely often. And recurrent is
the opposite of that.
So there's more that you can say though in the bounded degree case. So Benjamini and Schramm
looked at circle packs of triangulations, and they proved that if you're in this hyperbolic case, so you
have your triangulation circle packed in the disc, and you run the random walk on it, and you look at
where it is, then it converges to a point in the boundary. And they use this to deduce this kind of
dichotomy that holds for all bounded degree triangulations. And in fact you can extend it to all graphs
that you find that essentially transient planar graphs of bounded degree have to be very transient in
various senses. So you get this dichotomy that breaks up parabolic, hyperbolic, recurrent, or transient.
So either the random walk on the graph is recurrent, the graph is circle packing parabolic, and all the
bounded harmonic functions on the graph are constant.
So I should just remind you what a harmonic function on a graph is. So a function on a graph on the
vertex set of a graph is harmonic if for every vertex U, H of U is equal to the average of H of V over
the neighbors. So the value of the function is that every vortex is the average value of the function of
the neighbors. Or equivalently, function is harmonic if when I plug in a simple random walk into the
function, I get a martingale.
So you'll remember from complex analysis that there are no bounded non-constant harmonic functions
on the plane. And similarly if you have a recurrent graph, it can never have bounded harmonic
functions. However, it's not true that transient graphs admit non-constant bounded harmonic functions
in general. So if you take any Euclidean lattice set to the D, even, so when D is bigger than 3, the
random walk is transient; but for no D are the bounded non-constant harmonic functions.
However, for the planar graphs, this dichotomy does hold. So whenever I have a bounded degree
transient planar graph, then I get bounded non-constant harmonic functions. And why is that? Well, I
can circle pack, if it's a triangulation, say, I can circle pack it in the disc, and my random walk
converges to a point in the boundary. Moreover, Benjamini and Schramm also proved that the law of
this limit point where the random walk converges to, has full support. So any interval in the boundary
has a positive probability for the random walk to converge into it. And I can use this to define
harmonic functions by I just take some function on the boundary, and I just take the harmonic function
of V to be the expected value of the limit, the function applied to the limit at the random walk.
>>: So do we continue this environment?
Tom Hutchcroft: No, if I take, I mean it has to be like measurable. So whenever I have function on the
boundary, I can take ->>: [Indiscernible].
Tom Hutchcroft: Yeah, bounded functions. To X infinity is the limit point. Then I can always get a
harmonic function this way. So I just take the harmonic function of V to be the expectation of the
function on the boundary applied to the limit point, if the random walk started at V.
So a natural question to ask is, well, are there any other bounded, it should say, harmonic functions on
the graph? And in fact the answer is no. So this was proven by Angel, Barlow, Gurel-Gurevich and
Nachmias. And so they proved that every bounded harmonic function on the graph can be represented
as a function on the boundary in this way.
And probabilistically what this means is if you know where the random walk converges to in the
boundary, then you know all of the tail information about the random walk trajectory. So these are
equivalent statements.
And so the techniques here are things like resistance estimates, Harnack inequalities. And the bounded
degree assumption is really essential in all of this.
So we wanted to rebuild this theory for random triangulations without the bounded degree assumption.
And as I said, these techniques that existed really very heavily required the bounded degree. There was
no way to adapt them. So it required ->>: Are there other examples [indiscernible].
Tom Hutchcroft: Yes, they are. Sorry, this is where I'm supposed to show you those.
So you can make a, it's the same thing if I start with some circle packing, I can keep adding circles
inside the gaps to create drifts. I can basically, if I have no conditions, I can make the random walk do
absolutely anything I so desire. So I can make random walk just go around in a spiral or something so
that it always has to go close to the boundary, but doesn't have to converge at a particular point. Or
another thing you can do is make two very strong drifts that go into the same point. And then you get
some tail information, namely, did I take this drift or this drift, which is not given to me by just
knowing where I converged to in the boundary.
So we asked the following questions. First, is there an analogue of the He-Schramm theorem to
characterize the circle packing type of a random triangulation by some kind of probabilistic property?
Secondly, if I give you a random triangulation, can you tell me whether it's CP parabolic or hyperbolic?
So if the degrees are bounded, we know it's just recurrence or transience. But if it's unbounded, what
do we do? Then in the hyperbolic case can we recover this boundary theory? So does the random walk
converge to a point in the boundary? And does the limit have full support; i.e., every interval has a
positive chance the random walk converging into there and no atoms, i.e., no single point has a positive
probability for the walk to hit it at a point in the boundary. And lastly, is the unit circle a realization of
the Poisson boundary? I.e., does every bounded harmonic function on the graph arise as an extension
of a function on the boundary?
So here's an example of a random hyperbolic triangulation. This is the Poisson-Dalonie, Poisson
[indiscernible] complex in the hyperbolic plane. So it looks pretty hyperbolic; right? But is it circle
packing hyperbolic? Well, it's not so obvious, right? It has unbounded degrees. Of course it is. But
we don't know that yet.
Okay. So before I talk about the answers to these questions, let me tell you a bit about what I mean by
random triangulation. So Benjamini-Schramm convergence of graphs was introduced to study
questions of the following form. What does a typical triangulation of the sphere with a large number of
vertices look like microscopically near a typical point? So in order to study this question, Benjamini
and Schramm introduced the following limiting operations. Take can sequence of finite graphs, Gn,
and for each n, choose a root vertex, Rho, of Gn uniformly at random. We say that the sequence of
graphs Gn, Benjamini-Schramm converged to a random rooted graph, G Rhu, if for each fixed radius
R, the balls of radius R in the finite graphs around the roots converge in distribution to the balls of
radius R around the roots in the infinite graph.
So let me show you some examples. If we take a big Taurus, or a big box and these Benjamini
Schramm converge to the whole lattice, well, the Taurus it's pretty obvious, the box is still pretty
obvious. If I pick a uniform point, it's very likely to be near the center of the box. So locally it doesn't
see the boundary, it just sees the whole lattice.
Very cryptical of the Schrammian random graph converges to a Poisson Galton-Watson tree. People
are very familiar with these things I think. And the height and binary tree converges to what's called
the canopy tree. So you might think it converges to the binary tree, but it doesn't. Because if you
choose a uniform point in this truncated binary tree, it's very likely to be a leaf ordinarily. So you get
this thing that looks like an infinite binary tree viewed from a leaf.
So if you think about this example a bit, you'll soon realize that there is no way to get the infinite
binary tree as a limit to finite trees. For example, because the average degree of a finite tree is always
at most 2. But the degree of an infinite binary tree is 3. So you obviously can't possibly get it as a limit
of finite trees.
>>: And by the infinite binary tree, you mean the 3-regular.
Tom Hutchcroft: The 3-regular, yes.
And in fact much more is true. So Benjamini and Schramm proved that every Benjamini Schramm
limit of finite, well, they proved that every Benjamini Schramm limit of finite simple triangulations is
CP parabolic almost surely. And they deduced that every Benjamini Schramm limit of finite planar
graphs 6 bounded degree is recurrence almost surely. So not only can you not get the binary tree as a
limit of finite trees, you can't get it as a limit of any sequence of finite planar graphs.
So you can also look at the uniform infinite plane triangulation. So this question that they started with
was about you look at all triangulations of the sphere, you choose one uniformly at random, and you
take one of these limits. And this limit was proven to exist by Angel and Schramm in 2003.
And this limit, the UIPT, it has a natural markup property, which is that if you explore the UIPT, so you
have some region of this triangulation that you've revealed already, and you just reveal more of it bit by
bit by peeling away at the boundary, then the law of the part that you haven't yet uncovered depends
only on the length of the boundary, of the piece that you have revealed already. Which is it obviously
has to depend on this. This is the minimum that it could depend on. So you might ask, well, is the
UIPT the only triangulation with this property? And in fact the answer is no. [laughter] so the question
I phrased differently is yes. [laughter] so there is a 1-parameter family of random planar triangulations
with this markup property. So a half planar version of these was constructed by Angel and Ray, and
Nicholi Curien constructed the full planar ones. So they have this one parameter family where with
index by this parameter capital, which goes from 0 to a mysterious number to 2/27, and when I plug in
this end point value to 2/27, you get the UIPT. But all the others are somehow hyperbolic in flavor.
And conjecturally, these other values of kappa, you can get them as local limits where instead of
looking at uniform triangulations at this sphere you look at a uniform triangulation of a high genus
surface where the genus is growing linearly with the number vertices.
And so here's a picture of a circle packing of one of these Markovian triangulations with kappa that's
quite close to the critical way of 2/27.
>>: [Indiscernible].
Tom Hutchcroft: No.
>>: [Indiscernible].
Tom Hutchcroft: No.
>>: [Indiscernible].
Tom Hutchcroft: Yes, the circles get smaller as you go closer.
Okay.
>>: [Indiscernible].
Tom Hutchcroft: No, this one is [indiscernible]. The other one ->>: [Indiscernible].
>>: Is it possible to say what kappa represents? Like why is it not [indiscernible]?
Tom Hutchcroft: The way he constructs it, it's a parameter that comes up in some distribution. It's not
a natural parameter like the curvature or something. It's just something weird.
Okay. So everything that you obtain as a Benjamini-Schramm limit to finite graphs always has the
property of unimodularity. So what is unimodularity? We saw that a random rooted graph, G Rho, is
unimodular if it satisfies the mass transport principle. So this says that if I take a function, whose input
is a graph with an ordered pair of distinguished vertices, and the output is a positive real number, or a
non-negative number. So I think that this is a rule of how to send mass from some vertices in the graph
to some other vertices. So I'll say f(G,U,V) is the amount of mass that U sends to V in the graph G.
So the mass transport principle says that for every such function f, the expected amount of mass sent by
the root, which is the sum f(G, Rho, V) summing over OV, is equal to the expected mass received by
the root. So if you have a finite graph and you choose the root uniformly, this is kind of a funny way of
writing in exchanges of the order of summation. But it's inherited by these Benjamini-Schramm limits
to finite graphs. And it turns out to be an extremely powerful thing. Really, if you want to prove
something about unimodular random graphs, the trick is almost always to find some mass transport that
does all the work for you.
>>: Why is it called unimodular? [indiscernible].
Tom Hutchcroft: It comes from group theory.
>>: Does it is have something to do with unimodular [indiscernible]?
Tom Hutchcroft: It comes from the modular function in group theory being what is one. A group is
unimodular when it's modular function is equal to 1.
>>: The math principle is used along in transient graphs? Or extension and it doesn't hold in all
transient graphs, but holds in all [indiscernible] graphs and generally it holds in other planar graphs
which means that the left par measure equals right par pleasure on the [indiscernible].
Tom Hutchcroft: Okay. So as you've all mentioned, this property holds on every Cayley graph, it
holds on finite graphs, and by extension it holds on limits to finite graphs.
So this is our answer to the first two questions. Well, second question. So suppose you have a
unimodular random planar triangulation. And I need to put in this ergodistic condition here, but this
just rules out something done like, you know, you take the triangular lattice with probability of half and
the seven regular hyperbolic classes with probability of half.
And what this theorem says is that I can determine the circle packing type just by calculating the
expected degree at the root. So if the expected degree is 6, then the triangulation is almost surely CP
parabolic. And if the expected degree is bigger than 6, then the triangulation is almost surely CP
hyperbolic. In fact, without this theorem, it's not. We don't know an elementary proof of the fact that
the expected degree is always bigger than equal to 6. There should be an infinite in the hypothesis. We
don't know an elementary proof that for a unimodular random triangulation, an infinite one, the
expected degree is always bigger than regular 6. So this is a problem you might want to think about.
And you can see that this implies the Benjamini-Schramm theorem that I mentioned earlier, because
any limit of finite triangulations have to have an expected degree of at most 6 by [indiscernible]
formula degree the average degree of a finite triangulation is less than 6, and the limit, the expected
degree is at most 6.
So how do we prove this? Well, the proof is actually extremely easy, once you've seen it. So let me
remind you of the slide that I showed you earlier which says that the circle packing actually gives us a
drawing of the triangulation, and the drawing is determined by the graph. And the important thing here
is this fact of the drawing is determined by the graph means that we can use the drawing to define mass
transport.
So what do we do? Well, suppose I'm in the parabolic case. So I have my drawing. And for each
corner of the triangle, I send the angle at the corner to each of the three vertices of the triangle. Right?
So remember, the mass transport principle says that the mass into the root, the expected mass into the
root is equal to the expected mass out of the root. Well, what's the mass out? Well, I just have all these
angles. Sorry?
>>: From X?
Tom Hutchcroft: From X to each of X, Y, Z.
So the mass out of X, well, I just have each corner around me and I'm sending three copies of each of
them. So the mass out is just 6 Pi. Well, what's the mass in? Well, each triangle that I'm adjacent to is
sending me the sum of its internal angles, which is Pi. How many of them are adjacent to me, well,
that's the degree. So I apply the mass transport principle and I find that 6 Pi is equal to Pi times the
expected degree. And, well, cancel the Pis, and you're done. And you can see that in the hyperbolic
case, exactly the same thing is going to work; right? So we just do the same transport. Now the mass
out is still 6 Pi. But now the sum at the internal angles of a hyperbolic triangle is strictly less than Pi.
So now I get a strict inequality here instead of the equality that I had before, and I find that the
expected degree is strictly larger than 6. And that's it. And this gives you a completely trivial proof of
the once difficult Benjamini-Schramm theorem.
So we were also able to recover the boundary theory with the triangulations. So a theorem states that if
you have a random circle packing, a unimodular random circle packing -- a unimodular random circle
packing hyperbolic triangulation, and the second moment of the degree is finite, and you circle packet
in the disc, then each of the following hold almost surely. One, the random walk converges to a point
in the boundary, almost surely. Two, the law of the limit point has full support and no atoms. And
three, it's a realization of the Poisson, also adjoint to Furstenberg boundary.
And I should stress again in a while the results exactly the same as the ones for the bounded degree
case, the proofs are completely different. And so the key techniques here were we're going to see a lot
of mass transports and another technique is this notion, or not a technique so much as a concert, which
is what we called invariant non-amenability, which was a notion of non-amenability introduced by
Russ and David Aldous.
Okay. So let me remind you what normal standard non-amenability is. So suppose we have an infinite
graph G. Then we define it's Cheeger constants to be the infimum of the size of the boundary of a finite
set divided by its volume. So here the boundary as a set of edges that have one point endpoint in the
set and one point endpoint out of the set. And the volume is the sum of the degrees in the set. So and a
graph is said to be amenable if the Cheeger constant is zero, and non-amenable if it's positive.
So for example, this lattice set is amenable, because if I take a big box, right, then there are, it has the
volume is n to the D. But the size of the boundary is n to the D minus 1. So as the box gets larger, this
ratio gets smaller and the infimum is zero. However, if you take say a 3-regular tree, then it's nonamenable. Any set in the 3-regular tree has the size of its boundaries comparable to its volume.
And probabilistically, you can think of non-amenable this way. So it's actually equivalent to the
exponential decay of return probabilities for the simple random walk. So an infinite graph is nonamenable if and only there exists a constant A less than 1 such that the probability if you start a random
walker time X, and you say what's the probability that it's add X again at time n, then this is decaying
like A to the n. So it's decaying exponentially.
And so if we have, say, a hyperbolic lattice or something, these are non-amenable. But are they
unimodular CP hyperbolic triangulations non-amenable? Well, no, the condition is too strong to be
satisfied by random triangulations. Because in the triangulations we have, we'll have these large
regions that look like the triangular lattice, and that's going to kill the non-amenability. Because there's
this very strong global condition that these rare bad sets are going to run.
But we want to come up with a different notion of non-amenability that captures the same kind of
properties but applies in our setting. So this was introduced by Russ and David Aldous. So we start
with some terminology. A percolation on a unimodular random graph G Rho is a random subgraph of
G Omega, such that this triplet G Rho Omega is unimodular. So what does that mean? It means I can
define mass transport that depend first on the graph and on this random subgraph and the mass
transport principle continues to hold. And we write K Omega Rho for the component of the percolation
that contains the root. And we say that a percolation is finite if all of its connected components are
finite almost surely.
So now here's the trick. Instead of taking this infimum of all sets, we look at finite percolations, and
we take the infimum over the expected ratio of the size of the boundary of the component containing
the argent to its volume.
>>: Where finite percolation means [indiscernible].
Tom Hutchcroft: Yes, right. That's what this is.
>>: [Indiscernible].
>>: [Indiscernible].
Tom Hutchcroft: So it's a random subgraph such that this graph plus this extra, this 0, 1 labeling of the
edges, is unimodular. So I can define mass transport on the graph with this extra information so they
can depend on both things, and the mass transport principle still holds. So the expected mass into the
root X equals the expected mass out of the root.
>>: So it's not like you're doing [indiscernible].
Tom Hutchcroft: Right, exactly. [indiscernible].
And we say that a unimodular random rooted graph is invariantly non-amenable if this guy is positive.
So this is saying not only if it's conversive it's invariantly amenable, so this is zero, it's saying not only
do sets of bad expansion exist, but I can actually title the whole graph in an invariantly defined way so
that the origin is in a set of bad expansion with high probability at the root. And you can phrase this as
the maximum density of a finite percolation. So this invariant Cheeger constant is the expected degree
of the graph minus the densist you can make a finite percolation without it becoming infinite. So D is
still invariantly amenable as far as being amenable. And in fact the Cayley graph, it turns out that this
notion is equivalent to the classical one.
However, here's an example of a random graph which is amenable but nevertheless is invariantly nonamenable. Take a 3-regular tree and randomly stretch all the edges, so replace each edge with some
path of an unbounded length but with finite mean. And if you do it with finite mean, then you can play
around with the root and move the root so that it still is unimodular. And what you get is amenable,
obviously, because you have these long things that just look like parts. So they have their boundaries,
just 2, and they have some big volume. So it's definitely amenable. However, it is invariantly nonamenable. And critical Galton-Watson tree conditioned to survive is invariantly amenable. And any
recurrence of unimodular random graph is always invariantly amenable.
So our next direct ingredients of this classification is in fact not only does the expected degree
determine whether the triangulation is circle packing parabolic or hyperbolic, it also determines
whether it's invariantly amenable or not. So you get this really nice dichotomy that either the expected
degree is 6, it's parabolic, everything is amenable, or conversely all the opposite things hold, so it's
parabolic, the expected degree is bigger than 6, and it's invariantly non-amenable.
And this is good news, because ->>: [Indiscernible].
Tom Hutchcroft: It is the same.
So we've defined this weird thing in variant non-amenability, and we know that it holds for hyperbolic
triangulations. But what can we actually do with it? Well, it turns out that you can do most of the same
things that you can do with usual non-amenability. Because there's this result of Benjamini-LyonsSchramm that says that if we have a graph that's invariantly non-amenable, then we can take a random
subgraph of it, which is really non-amenable. And this let's us carry through a lot of the same
arguments that we would use in the really non-amenable sets.
>>: So there's an implication of one way, which is invariant non-amenable.
Tom Hutchcroft: Yes, non-amenable is stronger than invariantly non-amenable.
>>: You just said if you can think of a [indiscernible] that's legitimately non-amenable?
Tom Hutchcroft: Yes.
>>: Doesn't that make sense if we were to stretch the tree?
Tom Hutchcroft: You can't. Because it's not obvious, but this tree does have a random subgraph,
which is not --
>>: [Indiscernible].
Tom Hutchcroft: You have to do something a bit cleverer than that. Because once you remove those
you end up with some parts that have formed out of the formerly short paths. But you can do it.
[indiscernible].
So now I can tell you how we use all this machinery to prove convergence. So let's suppose we were in
this really non-amenable setting and that we have bounded degrees. Then there's some constant A less
than 1 such that the probability for n, if I start the random walk at Rho, and I look at probability that
I'm at some other vertex V at time n, it decays like a constant depending on the maximum degree times
A to the n. But now the total area of all the circles is at most Pi. So that means that there are most 1 of
A to the n over 2 circles of radius at least A to the n over 4. And this means that just take a union bound
and I find that the probability that the radius at random walker time n is bigger than A to the n over 4 is
at most this exponentially decaying thing. Right? And this means that the -- what does this mean?
Sorry. It means that the expected radius of time n is also decaying exponentially.
And that means that we can sum over all the radii and the expectation of this sum is finite; right? And
that means that the sum is almost surely finite. But this means that the sequence of centers a
[indiscernible] sequence, right? For obvious reasons, right? Because if I just take the paths between
XN and XM, it's bounded by the sum at the radii from XN2. So we deduced that the random walk
converges almost surely.
So I've been cheating by assuming that we have this real non-amenability. But we can use this theorem
that I showed you on the previous slide to push this through to the invariantly non-amenable setting.
So we perform the same argument looking at the time the random walk expands in this really nonamenable graph, and then there we get that the radii are decaying exponentially, and then, you know,
you just have to control things on the time you spent outside and show that the radii are still decaying
exponentially over the whole random walk. And in fact more is true. So in fact the radii of the circles
at the random walker decay exponentially; in fact it's equal to the speed of the random walk as
measured in the hyperbolic measurement on the circle packing.
And this is a nice thing. Because so far we've been recovering a lot of things that we already knew
about bounded degree triangulations. But this is something that is actually new to the random setup.
So let me tell you a bit about why the exit measure. So this is the law of the limit point. Why does it
not contain any atoms? So when we have a unimodular random graph, we can always bias it in a
certain way. So just change the measure a bit to get an equivalent measure which is stationary. So that
just means that if I take G Rho, which is this rooted graph, then it has the same distribution as GX1,
where CP parabolic 1 is the first step at the random walk.
Okay. So I'm first going to claim that almost surely there are either no atoms in the boundary, or there
is just one big atom that has all the mass. And this is a pretty straightforward argument. You just look
at the atom of biggest way and get a contradiction if it's not 1. So for each atom, let's define this
function. So HCV to be the probability, so this should be a V. The probability starting from V that the
limit is equal to this point. So this gives us a harmonic function. And Levy's 0-1 law, which is just a
special case of the Martingale convergence theorem, tells us that when I plug random walk in to this
function, it just converges to the indicator that I actually converge at that point. So if I look at the
probability from XN, then I'm going to converge at that point, it just converges to the indicator of
whether I actually do converge at that point or not.
So if we define MV to be the maximum at these HCV over all the atoms, then this has a limit in 0-1 as
well. Right? Because, you know, it's obvious.
But since the graph is stationary, see this thing M Rho actually it doesn't depend on the circle packing
rate. Because it's just saying look at the biggest atom in the boundary, what's it's weight. So if I apply
some Möbius transformation to the circle packing, this does not change. So this M Rho is just a
function of the rooted graph G Rho. But we've just seen that it converges to something in 0-1. But if
you have a stationary sequence of random variables and they almost surely converge to either 0 or 1,
then they all have to be 0-1, almost surely.
>>: Can you explain what [indiscernible]?
Tom Hutchcroft: Yes, so you can be a bit more careful with this. So I request also look at a different
harmonic function, which is the probability that I converge to an atom at all, or all atoms. It also
converges to, and this is just the sum of these guys. But this also converges to the indicator that I've
converged to an atom. So what does that mean? Well, it means that if I converge to an atom, then that
atom has weight tending to 1, and the total weight of all the atoms is also tending to 1. That certainly
means that this biggest atom is also converging to 1. It's just that one. Conversely, if I don't converge
to an atom, then the sum of all of them converges to 0 and that certainly the maximum [indiscernible].
And so yes, so you have this stationary sequence of random variables, it converges something in 0-1, it
must just be 0-1 almost surely to start with.
So now we're in this setting so we want to get a contradiction from the assumption that atoms exist. So
now we know that either there are no atoms, which is what we want, or there's just one big atom that
the random walk always converges to. So what do you do? So let's assume that we're in this bad setup,
there's just this one big atom that the random walk always converges to. And in some sense, the sets of
that graph is really not hyperbolic. So what does this mean? Well, take the disc with the circle
packing, and apply a Möbius transformation sending this one atom to infinity. So I get the circle
packing of the graph in the upper half plane now. And because this atom is unique, I can make it stay
infinity and stay out of infinity, and this circle pack in the upper half plane is unique up to Möbius
transformations that fix infinity; right? But these are just translations in scaling.
>>: You said before, so many Rho equals 1.
Tom Hutchcroft: Yes.
>>: Why is there a single atom in this case?
Tom Hutchcroft: Well, because the maximum weight of an atom is 1.
>>: And the sum is a maximum.
Tom Hutchcroft: We use the [indiscernible]. Okay?
So now we have this packing in the upper half plane, and because this atom is fixed to infinity, it's
determined to the translation of scaling. So now we can draw the graph using straight lines between
the center, right, and this drawing is also determined up to translation in scaling. So now we can do
exactly what we did before and deduce that the expected degree has to be equal to 6, right, because
we're basically in this Euclidean setting again. But we knew from our assumption that it's hyperbolic
that the expected root is bigger than 6, so we get a contradiction. And this shows more generally that
you cannot have any kind of distinguished point in the boundary. Here's a picture of a circle packing.
>>: So it's a different drawing; right? The angles are not the same curves that you would get from the
[indiscernible].
Tom Hutchcroft: Right, exactly. They're the same circles, but now we're drawing it with straight lines
instead of the hyperbolic [indiscernible].
So I think I have time to show you this last thing which is that the exit measure has full support. I.e.,
for every interval in the boundary, there's a positive probability of the random walk converges to.
Okay?
So what we're going to do is we're going to define a mass transport on the graph in which every vertex
sends a mass of at most one, but some vertices are going to receive infinite mass. And this contradicts
the mass transport principle. So this is not quite immediate if you're not used to working with these
things. But if some vertex receives infinite mass, then the root receives infinite mass with positive
probability. So this is not too hard to show. And obviously if the root receives infinite mass with the
positive probability, then the expected mass it receives is infinite. But everything sends a bounded
amount of mass. So you would get a contradiction.
And we're going to define the transport in terms of the hyperbolic geometry of the packing and of the
support of the measure on the boundary. And so by rigidity, it's going to really be a function of the
graph.
Okay. So suppose the support is not the whole boundary. Then we can write the compliment as a
union of disjoint open intervals in the circle. So let's take this set and let's draw the hyperbolic
geodesic between the endpoints of each such interval. And we'll write so these intervals are indexed
[indiscernible]. Let's wry AI to be the set of circles which are contained in the region between the
boundary interval Theta I Phi, and the geodesic between the two points.
Okay. Now suppose I have some circle here that's inside AI. So what do I do to define my mass
transport? I look at the left-hand endpoints of the interval. So I'm inside this region. And I take the
hyperbolic geodesic from the hyperbolic center of this circle and I follow it along until I hit the circle
which also intersects this geodesic. And so I'll define a mass transport by just sending mass 1 to this
circle, to this vortex corresponding to that circle. And it could be that I never encounter such a circle
and then I just don't send any [indiscernible] back.
Now, let's look at the circle that's on the line V and look at the set of circles that send mass to V.
>>: How much sent from U [indiscernible]?
Tom Hutchcroft: Every vertex sends mass by the 0-1 in total.
So we can for each vertex V that intersects this line, we have this boundary interval, BV, which is just
the interval between this geodesics that starts here and just grazes V, this circle corresponding to V, and
this one that comes along and then just gazes something else. It's the first one to do that. So I get one
such interval for each vortex V.
>>: [Indiscernible].
Tom Hutchcroft: Sorry?
>>: [Indiscernible].
Tom Hutchcroft: This guy here as an MV.
However, aside from some stuff here that might avoid everything, there is an interval of positive length
from some point inside here all the way up to Phi I, such that which is equal to the union of all the BVs.
Right? But there are only countably many circles, and they each have an interval associated to them
and the union to the intervals has a positive length. That means there's some interval BV that has
positive length. But now this vertex has to receive infinite mass, because, you know, it has this open
neighborhood of a boundary interval in which every circle n there sends mass to it. And the circles
accumulate everywhere. So this such a vertex V receives infinite mass and this contradicts the mass
transport at this point.
>>: Where do you see your assumption?
Tom Hutchcroft: That there were, [indiscernible] well, that's how these arcs were defined; right?
>>: The arcs -Tom Hutchcroft: The arcs as in the arcs over the compliment of the support.
>>: So BV is not -Tom Hutchcroft: Right, exactly. So this is not in the support, this is not in the ->>: You're only saying some things that are not? [indiscernible].
>>: [Indiscernible] it's thing to say that the circles accumulate even though the hypothesis is that the
support is not there.
Tom Hutchcroft: Right.
>>: But then where are you using the other hypotheses?
>>: This game was played on the gap.
>>: But where is the fact that [indiscernible].
Tom Hutchcroft: Because otherwise there would be nothing, none of these arcs would exist.
>>: [Indiscernible].
Tom Hutchcroft: Yes. Exactly.
>>: So it doesn't really use [indiscernible].
Tom Hutchcroft: And there's nothing really about the random walk.
>>: It's just you cannot define a set of arcs of such.
Tom Hutchcroft: It's similar to with the atom, right. We just said there's no distinguished point.
>>: You're saying there's no distinguished end point.
Tom Hutchcroft: Yes.
>>: [Indiscernible].
Tom Hutchcroft: Oh, well, so that's part of the He-Schramm theorem. It's part of the definition of what
a circle packing is.
>>: We're waiting for the [indiscernible].
Tom Hutchcroft: Just this first sentence.
>>: [Indiscernible].
>>: There's no distinguishing. Because that's the strongest.
Tom Hutchcroft: Although I think it's implied by the thing about, you could also prove that a stronger
statement, just knowing the statement about random walkers.
>>: But neither of your proof [indiscernible].
>>: [indiscernible].
Tom Hutchcroft: Right. Sure.
Okay, well, thank you very much. [applause]
Alexander Holroyd: Any more questions?
>>: So I'm going to take this opportunity to mention a question. [Indiscernible] we used an argument
very similar to the one of VM to prove that every invariance forced in the hyperbolic plane has a
[indiscernible]. But it's all been in higher dimensions, hyperbolics.
Tom Hutchcroft: So if you have an invariant forest inside.
Alexander Holroyd: Any more questions?
[Applause]