>> Eyal Lubetzky: Okay. Hi, everyone. It’s a pleasure to have Benoît Laslier from the University of Lyon,
and he’ll be talking about—it’s not in the title—but he’ll be talking about a global dynamics for … mixing
times of global dynamics for lozenge tilings.
>> Benoît Laslier: So thank you for the invitation, the … for the opportunity to talk. Yes, title is moving
around. So … I’ve … yeah … so what I will be talking about is a lozenge tiling, or equivalently, a Ising
model on Z three, or interface of the Ising model on Z three at zero temperature. So nobody …
everybod … of you has seen that this tiling can be seen as a stack of cubes, and so it’s exactly equivalent
to a … some … to one configuration of the Ising model with boundary condition that are: everything
below the curve in Z three given by the contour is, say, plus and—full cubes—and everything above is
minus and invisible. Human, we see it from a height. Because it’s convenient to parameterize such a
tiling by a height function—and here, we will choose two … for … the height function … for each
intersection of two lozenges, we will put the z coordinate of the associated—yeah—stack of cube. So
just think … we could work, also, on … with some other tilings—such as domino tilings—and this is how
we can see a domino tiling as something in R three, but—yeah—it’s much less clear in the picture, so
we’ll stick to … with lozenges.
So we will be interested in the asymptotic behavior for very large domain, when this height on the
boundary converge to some fixed curve in R three, and can consider that the size of the cubes goes as
one over n. And as you see in the picture, you have quite a rich behavior—which you know, of course,
very well—of … for the typical configuration. So it can have places like this one or …
>>: Did you take this picture?
>>: [inaudible] no.
>> Benoît Laslier: No.
>>: The one in the middle is—I think—Rick Kenyon’s.
>> Benoît Laslier: Yes, this one is Rick Kenyon’s in his paper with Okounkov. This one, I don’t remember
which paper it is.
>>: Not important. [laughter]
>> Benoît Laslier: Anyway … yeah, so you have a region where all lozenges are aligned, which we will
call frozen region—I will … it will be important afterward—and a region a region in the middle where
you see some random configuration—that will be called the liquid region. And here, we have a law of
large numbers … a typical tiling converges to some deterministic height function depending on Leon’s
boundary. And one important special case is when the boundary is contained into some plane, and
then, the asymptotic height function is planar.
So now, what we have looked at is Glauber dynamics on these tilings. So a typi … the … we make move,
adding or removing just one cube from the stack—so going from something on the left to the equivalent
on the right there. So if we … in terms of dimers, you … but … [indiscernible] it’s rotating three dimers
around one face, and the timescale we consider is that every face, or every intersection, has a rate-one
clock, and so in a … in one unit time, you have order L square moves. You know, and for this chain, our
main results are [indiscernible]. So let, say, gamma be a boundary curve such that H infinity—it’s a
deterministic limiting height function—has no frozen region. So it’s only … it’s liquid everywhere. Then,
with high probability, if we consider H … L at time t … two plus lit … rho lives here. So if we consider the
height function with a scale, L, and at a time L square, up to a small error term, then it’s, macroscopically
speaking, very close to the equilibrium shape. Note that here, this epsilon is … I mean, you still need
epsilon times L cubed to … you allow microscopic change, but … so this.
And another result is: if gamma is, let’s say, cut from an hexagonal H infinity, then the mixing time—this
time—is the … equals … so what do I mean as cut from the … an hexagonal H infinity? You consider this
big hexagon, and you draw a curve inside the liquid face, and actually you need … yeah, you don’t … you
are not allowed to go up to the boundary of the liquid face—you have to stay strictly inside—then this
gives you a curve in R three, which you can use as boundary condition, and for these boundary
conditions, we have a release … a mixing time.
>>: This is not exactly new to [indiscernible] this is some convergence of the interface, but …
>> Benoît Laslier: The first is not really a mixing time. The second is really the mixing time for lozenge
tilings. So you … with the variation distance and … yeah, so first one is much weaker. So you have no
information on local structure how …
>>: Weaker in a way, but it’s not that it is implied by the second.
>> Benoît Laslier: I don’t …
>>: I mean, there is more in the sec … in the first, you also have some concentration information about
the stationary distribution—that it’s concentrated near …
>> Benoît Laslier: Yes, but this information about the stationary distribution, it’s well known since,
actually, prop … some results, some …
>>: I have a question about the first theorem.
>> Benoît Laslier: Yes?
>>: So if there is a frozen region …
>>: [inaudible]
>>: … then what do we know? It seems like it should be all the more true.
>> Benoît Laslier: If there is a frozen region we … the best results—I think—are L si … to the six or … we
know that it’s polynomial, and … but we don’t know what happens.
>>: And it could be L squared … L to the two plus o of one, right?
>> Benoît Laslier: Yes, we think that it is L to the two plus little o of one, but as we’ll see, to get this L
two, we have to follow on an instant quite closely the relation with mean curvature motion, and if we
have frozen region, then a lot of things … I … summability could di … will diverge, surface tension
becomes singular, and we see these problem in the proof, so ...
So how does we prove this theorem? We will do a recursion on things that … we’ll define a
deterministic—yeah—a deterministic surface that will slowly move towards H infinity and iteratively
show that we will stay, with very high probability, below this interface. So it’s really the same kind of
idea as what was on first with … by Fabio Martinelli, Pietro Caputo, and François Simenhaus. And yeah,
so we show that … but the real difference between the planar case that they did and the non-planar
case is that we have to be more careful in the definition of this evolving interface, and we have to use
more complicated and fluctuation estimates. So I will sket … so more precisely, we will define—
[indiscernible] ‘kay, yeah—our moving interface …
>>: If you want to, you can just say stream, not integral rays like magic.
>> Benoît Laslier: Yes, but I have other pictures.
>>: I’m asking for perfecto.
>>: You want them at the same time? [laughter]
>> Benoît Laslier: Yes, I … can you read if I write a few things here?
>>: No.
>>: Sure.
>> Benoît Laslier: Okay, then I will try to keep it here, and you … so this Ct will be equals to the H
infinity—okay—plus we have epsilon one minus t over L two plus little o of one, and some function, psi,
which give … will give us its overall shape, and—‘kay, can give … that’s a large constant minus a xi.
Okay? So we define our shape; it’s basically something that will—when t will become of order L two
plus little o of one—it will become very close to H infinity. It—okay—it can start at distance two epsilon,
because if we know how to go from distance two epsilon to distance epsilon, then we just have to do
this argument one over epsilon time, and since epsilon is constant, then we are losing L to the little o of
one, which is still diverging—it’s not an issue. So … and we will have the following recursion. So for all t
greater than, so say, than L times L, and say t is lower than C. So the recursion will be: for all t bigger
than our nth time, we will stay below some—yeah, so cap at the nth time, and yes, t also smaller than
some very large time to avoid that we just fluctuates back up, and that’s with high probability, but …
>>: So you’re gonna prove it by induction on n.
>> Benoît Laslier: Yes, and then we work by induction on n. So here is a schematic picture of what
happens at some times. We have our Ct, which I drew concave, but actually, when we are close, it’s not
concave anymore. And we consider one point, and we want to lower by one Ct at this specific point
with high enough probability. So we define a small domain, D, of size more or less square root … one
over square root, so you have square root lozenges in … around p, and we’ll look only at what happens
inside this small domain. So now, we can try to draw everything on this … in a side …
>>: Cross-section.
>> Benoît Laslier: Yes, cross-section, in a side view. So we have these domains. C is actually defined …
something a bit bigger around D. And we—at the initial time—we know that we are somewhere below,
and so—the dynamic is monotonous—so we can replace Ht by some H tilde, which is started back at Ct,
but because we are … our recurrence hypothesis, which is forbidden to ever move above Ct. Then we
can also add some floor not too far below, and we get to study this H tilde, which li … only evolves on a
domain of size square root, and which has a floor and ceiling which are, actually, not very far from each
other, because here, you have a curvature which is of order one. When you look at a domain of size
minus one half plus some little o of one—side of D—then this distance here is … oh, I said epsilon. So if
D is of size minus one half plus epsilon, this distance here becomes some L to the epsilon over L. So one
over L is because I … and we have a first, key lemma which says this H tilde, with all its constraints, it will
mix in time of order L … of order the size of the domain on which it lives, squared, so L … of order L. And
another lemma that also said that when it’s at equilibrium, it has fluctuations on D of order …
logarithmic, so it will not fluctuate above L to the epsilon minus one in a very long time. And then, with
these trivia, we can say at time t plus L … basically, t plus L, we will stay below this H tilde at equilibrium,
and this H tilde at equilibrium will be below Ct plus L.
So that’s the basic way the recursion works, and so like I said—yes—we have lemma one for the
dynamics constrained between … so the first lemma, it says that this H tilde, with its constraint, mixing
time’s the size of the domain squared. And lemma two: inside, we have that we can write in a very
informal way, like that. So the proof of this lemma for lozenge tilings, it’s actually deduced from the
work of on David with Anosov dynamics, where mixing time L squared was proved for … without any
constraints, and the constraints allows you to reasonably accurately compare the two mixing times,
and—yes—or … and you have also another argument for domino tilings. For lozenge, you can get a little
bit better in the power of H and logs and that, but it doesn’t matter. The second lemma was one of …
shows a big improvement with respect to the planar case, because fluctuation estimates are quite
difficult to …
>>: I don’t understand …
>> Benoît Laslier: [indiscernible]
>>: … what you mean by L to the minus one half plus epsilon.
>>: His basics tile is size one over L.
>> Benoît Laslier: Yes. So the idea was to use fluctuation estimates on large hexagons. So yeah, it’s the
same kind of picture with … I showed in the beginning—that were obtained by Petrov, so like, two years
ago—where you do have logarithmic fluctuation for something which is not planar, but it’s only for
hexagons. Fortunately, when you look at something of radius square root of L, then you can do a Taylor
development of any shape, and you see that the third order would give a contribution, the L to the
minus the three half, which is smaller than the one over L, which is the smallest significant distance. So
you can really ignore all higher orders than two, and then you compute the thing, and you see that in …
you can find somewhere in the hexagon … any second-order-zero development of the macroscopic
shape. So … and then you—okay, for any small domain—you find a big hexagon; you fit it so that it has
the proper boundary—yes—it has the proper boundary; and since you know the fluctuation inside the
big domains, you know, in particular, that you have logarithmic fluctuation inside the smaller one. And
that’s how we get this lemma two.
Yes, this is the reason why we have this discrepancy between the general boundary and really boundary
cut from an hexagon, ‘cause if we are cut from an hexagons, we can use the hexagonal … you can adapt
lemma two with bigger domains than L to the minus one half, ‘cause we are always with domains—
yeah, hexagonal types. Why, for general boundaries, we cannot … we are stuck in our recursion? When
… I mean, in our recursion, we are only allowed to look at very small domains, and that means we
cannot … we have to stop when the curvature of Ct becomes too small. And the curve … so … and that
means we cannot really approach more than at a fixed distance our H infinity. I know how much time I
have left … I …
>>: Twenty-five minutes.
>> Benoît Laslier: Okay, then …
>>: That’s just an upper bound.
>> Benoît Laslier: Hah. I can say just a word about … I can either speak about lemma one or give a bit
more details on the recursion to explain why this particular shape for the domination—Ct. I don’t know
what you would prefer.
>>: I mean, you could start by maybe give … discussing lemma one, because it … this is a … or what was
known without the constraints, how you … of the ceiling and how you were … how you used the ceiling
in the proof.
>> Benoît Laslier: Okay. So I can … think I will present the proof of lemma one that works exactly in the
same way for lozenge or for dominoes, because for dominoes, it wasn’t known, and for that, I think—
yes—I have one picture. So the general idea is to define an auxiliary dynamics for which it will be easy—
or for which we will manage—to compute the mixing time, and then make a comparison between these
two. So for … to define the auxiliary dynamics, we consider another representation of tilings or
dominoes, which is—kind of—bead model. Think bead model is finally for the continuous part of … the
continuous analogue, but that’s really it. So definition goes like this: forget about any dimer or that …
do not … which is not horizontal. Then you have a configuration, and it’s actually enough to recover the
whole dimer, because—yeah—if you have this configuration, if you look at these two vertices, you know
they have to be—and more generally, all the vertices in this region—they have to be matched with each
other, or else because—yeah—you know that they are not matched, and there is an even number of
hits, so you have only one way to have sent it. And furthermore, the interaction between these beads is
quite simple: they have to stay around or organized like in this picture, so between two beads on the
same thread, you have exactly one bead on the right and one bead on the left, and …
>>: What are the beads again?
>> Benoît Laslier: Beads are horizontal dimers. Okay? This works for some hexagonal lattice, but you
can also find threads and … for domino tilings, so these diesmer on the square lattice. And then, the
dynamics—the auxiliary dynamics—we use is just—at an update—is: pick either even or odd, and
resample all the beads which lie on even or odd threads—because threads are or … labelled in order—
conditionally on everything else. So if you take … if this is zero and you pick odd, you will put these two
back—this one inside this region; this one inside this region. You will resample, also, these four, et
cetera, et cetera. So this is an also monotonous dynamics, and we can check that for this dynamics, if
you look at the volume between two interfaces—so the volume betwe … being the sum of the value of
the height function on all sides, so really in the … the volume in R three—then it is a martingale, or it
tends to decrease, so it’s over martingale. I’m not …
>>: Super.
>> Benoît Laslier: Super-martin … so it tends to decrease. Then, if we … we can also know the variance
of this process. Why? Because—oh, I don’t have this picture here; I can cheat and take … so to bound
the variance, we have to know: where are we allowed to make movements? And one place where we
are sure we can make movements is when we have … when we could just add the cube or remove one.
And how do we find someplace where we can add a cube? We start from any vert … anywhere, and we
try to go away from the board at the maximal possible speed. Then we follow, for example, this path—
we have lot more choices: here, you could just keep on, but [indiscernible]. And if we … if at some point,
we are stuck—we are in a local minima—we could add a cube. Now, if we are constrained between a
floor and a ceiling which are quite close—some H over L—then this process of moving at maximal
possible speed, it will … it has to stop quite quickly. And actually, it has to stop in at most a constant
time H steps, because we are moving towards the floor and ceiling at the positive speed.
Yeah, so we have around L squared divided by some power of H sides where we can make a move—so
the variance should be of order L squared. And we also have … so the volume, it’s a martingale or a
submartingale, which has variance L squared, which is positive, and has a maximum L to the three. So,
putting it all together, we get a mixing time for this auxiliary dynamics, and notice the comparison is also
easy, because when we moved one bead, the set of allowed position was actually made of one path
going up and one path going down—so it was also of order H. And this means that we can simulate one
update of the auxiliary dynamics by freezing everything in, say, the odd threads, and letting the Glauber
dynamics runs for H square steps, because in H square steps, each bead will make—yeah—H square
move inside—do a simple random walk—inside something of size at most H square. And—hey—if we
can mimic, then again by this Ferris-Winkler censoring, you compare the two … so to … mixing time, so
that’s … and that’s also gives us a lower bound—yes.
We can find some conf … spatial boundary condition for which the volume drift is lower-bounded by
something like L. And then, we can … additionally, these conditions look like big pyramids. Truly, for the
lozenge case, it’s just you take hexagonal boundary conditions and … and now, if you want to say that an
initial configuration cannot move down too much, you fit below some pyramid, and you know that the
pyramid can erode volume only at speed L. So at times some small constant times L squared, it only has
removed a small part, and it has not moved too much down, so it … you cannot have converged yet. So
it’s … gives a long … lower bound constants time L squared—a bit better than what was known before.
And I think this should be it. I don’t … much more time. [applause]
>> Eyal Lubetzky: Question? Could you say a few words about showing the fluctuations … this lemma
two?
>> Benoît Laslier: Yes, the Petrov computations rely on the exact solvability of the model, and you can
express exactly with integral the probability that you see one dimer at some point, and—yes—then you
approximate this integral, and you get the fluctuation in the … inside the big hexagon. And that’s
really—yeah—not my work, but to get the—yes—the fluctuation in a small domain, the key point is this
bijectivity, actually, between points inside hexagonal boundary conditions and order-two development
of some curve or of some interface. And then, you have an exact formula for the limiting height
function inside hexagons. You can derive it, and compute the differential of the application that—to
some point, on some parameter of the hexagon—gives the local … the Taylor development at order two
at this point. That’s some application; you compute its derivative; you compute its determinant; you see
that it’s nonzero, so at least locally, you have some bijectivity, and then you say that actually, the
boundary of the set of points inside some hexagons is sent by this application to the boundary of the set
of Taylor development—order two, which is basically infinity—and if you have some function which—
yeah—maps bijectively small open sets to open sets and maps the boundary to the boundary, then it’s
globally bijective, and that’s it. So …
>>: I noticed one of your slides showed one of Rick Kenyon’s kind of pictures. And I’m just curious what
that had to do with your work.
>> Benoît Laslier: Okay, that’s … that was another tentative to … yes?
>>: Without having [indiscernible]
>> Benoît Laslier: Sam … to compute fluctuation estimates in domains with non-planar boundaries. So
to … Rick Kenyon has a paper where he showed how to relate those random or harmonic function on
these graphs, where each half-edge is oriented away from the point where it … so a harmonic function
on these graphs are related to the kastonane metrics and ultimately, to the fluctuations of the dimer
models, but to get, really, a great fluctuation estimate, you have to have precise estimates on: how to
harmonic function on this converge to true, continuous harmonic functions? And yeah, something that I
tried to do it, and I proved that the random walk converged to Brownian motion on this graph, but not
with accurate enough … or to … not with any speed of convergence that could be used for later.
>> Eyal Lubetzky: Okay. Thanks again. [applause]