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>> Yuval Peres: Today Fabio Toninelli will tell us about metastability and logarithmic energy
barriers.
>> Fabio Toninelli: So, first of all, thanks for all the opportunity to give this talk here. I'm going to
tell you about some work we did in Rome here with Pietro Fabio and Barry and Francois who
were post-docs there.
So it's about the stochastic dynamics for model of -- so it's a model of coming from statistical
physics. It's one plus one dimensional poly model. I'll introduce it.
And let's say if you look at the invariant distribution, you will see that it sort of has two phases.
You will see a picture in a moment. And we want to understand the dynamical signature of this
phase 2 phenomenon, double well phenomenon. So I'll introduce the global dynamics which
most of you know anyway, but just go through it quickly.
And then since we have this double word structure, we expect -- and actually prove results of the
type that the system jumps from one phase to the other in sort of -- metastability result about the
way the system jumps from one space to the other and back.
So somehow very specific model that is maybe -- okay. It's very specific, you will see. But the
phenomenon which arise are, we think, more general than the specific one. Some interests with
respect to all the results about metastability for the system. I'll come to that discussion towards
the end.
I'll come to the model. It's about like that. The generic configuration of the system is nearest
neighbor path, start from minus L ends at plus L and at each step goes up or down by one.
Right? So and then there's this horizontal line, and here have marked the point where the path
intersects the horizontal line. This you can see is a polymer configuration if you want in two
dimensional space, and it is a directed path. So that's why I mention one plus one dimension.
So anyway the configuration is just one of these paths. And we put a measure on it by weight in
the path. So the dimension, group of measure will be P depending on the size of the system and
on some interaction parameter lambda, which is non-negative, and we just weight the path by
taking the lambda to the times the number of red dots.
Okay. So lambda to the number of intersection between the polymer and this horizontal line.
Good. So this is end of eta. And of course you normalize it to be a probability.
So if lambda is larger than 1, then you are giving a reward to the polymer, to this path for touching
the interface if lambda equal one you have just a uniform measure. If lambda is smaller than 1,
you are penalizing those sections. So actually we will mostly intersect lambda smaller than 1.
Penalizing if you want this line is repulsive.
Good. So just at equilibrium, this invariant measure doesn't hide anything -- I mean, nothing -I'm not going to say anything new about this equilibrium measure. The equilibrium of the system
is very easy to analyze. For instance, you can compute exactly the free energy which means
computing the large [inaudible] symptotics of 1 over L times the log of the partition function. This
is exactly computable. You can compute the correlation function in the larger limit.
This is all explicit. You can do it by [inaudible] theorem. Let me just tell you a couple of facts
about this equilibrium measure. So as I said, there is a special value of lambda, which is 1.
Lambda equal 1 all configurations are equiprobable. So we just have essentially a random walk
condition to come back to 0 at the end. So sort of Brownian and the walk breach if you want.
And for lambda -- lambda larger than 1, the configuration is typically localized. Meaning if you
look at the average of the number of contacts and divide by the size of the system, this converges
for large size to a constant, which is positive. Depends on lambda. If lambda is smaller than 1,
then this constant is 0, but you can say more than that. You can say that the probability that you
have at least certain number of K of contacts, the case exponentially with K. So you have really
the number of constants is really tight. You can say more if lambda is smaller than 1 not only the
number of contacts is typically order one with exponential tail. You can also say they are close to
the boundaries.
So you can say that -- so take some parameter small L which will be growing with the system size
but much less than L. And look at the configurations which are positive. Well, I have a picture -no. Picture is up here.
Well, it doesn't matter. Look at the configurations which are positive at distance small L away
from the boundaries. And call omega plus the set of configurations. Omega minus will be minus
the configurations which are omega plus.
And something one can prove easily is the sets omega plus minus have probability essentially
one-half for N large. Meaning the probability of having an intersection or a 0 of the random walk
say in the middle, very far away from the end point, is going to 0 with L.
So here I have a picture which said that essentially -- just what I wrote here. Configurations are
either mostly plus, with few 0s at the boundaries or mostly minus.
And I want to refer to these omega plus, omega minus as the two wells, the two potential wells,
but okay. So for pictorial reasons.
So we have these two wells on which the probability measure concentrates in the larger limit
would like to understand what is the effect on the dynamics which I'm going to introduce in a
second. So we introduce the stochastic dynamics, which is just the heat bath dynamics for those
who know it.
It is stochastic dynamics, which -- well, okay, so it's defined like this. You have your polymer -your path goes from minus L to L. So to each point X from minus L to L integer, I mean. You
attach an independent mean 1 exponential clock, rings to exponential times average 1. And
when -- let me draw a picture. Okay. There is a picture here. So suppose that -- so this X here,
the plot rings. And at that time your configuration is the blue one. At that time you flip -- well, you
put to equilibrium this portion of the path. Condition all of the rest.
So just to be more explicit, if your configuration is something like this, and suppose that X is here
and the clock rings, if you are at tight -- if the path at that moment is a type two, then you flip the
height from like this to this with a probability which is lambda over 1 plus lambda. And otherwise
if your distance -- you are away from the interface for this integer here, then when the clock rings
at this X, you flip on with probability one-half. I mean, this is the more explicit way to describe
this.
So whenever the clock rings you equilibrate that part of the path given conditioning over all the
rest. Of course if the clock rings somewhere here where you cannot flip any value amount and
you do nothing.
Okay. This is the usual heat bath dynamics. So it's well known the equilibrium measure is just -I mean, this measure I introduced is invariant, reversible. You have continuous time Markov
process, which converges to it. As long as you are in finite volume, no problem you converge
exponentially.
And one object -- one quantity one looks at is the exponential -- sorry, the relaxation time which is
just given by the inverse of the smallest eigenvalue of minus generator.
It just gives you the speed of convergence to equilibrium to L-2 norm. So we call this relaxation
time. Good. So as I said, so this is a very pictorial image. Just very rough.
Essentially, you have -- you imagine that the system, the configuration of the system is just a
point. And the at equilibrium, it is either in the -- this omega minus set or in the omega plus set.
So in one of these two wells. So what will happen is that if you let the system evolve starting from
somewhere, suppose in the omega minus phase, for some time it will stay there. Sort of moving
there but not exiting the well. And on a much -- and then losing memory inside this phase. And
then on a much longer time scale given by the relaxation time we'll jump to the opposite phase
and so forth.
So this is a very rough picture, because somehow in this picture the system is one dimensional.
So if you really had the particle in one dimensional potential with two wells, or even finite
dimensional space, you could apply the [inaudible] theory, and it would be, it would be -- it's
standard sort of to justify what I've just said, that you spend a lot of time in a well then after
exponential time you jump to another well and essentially these jumps are independent, I mean,
the time between jumps are independent, exponentially distributed random variables.
Now, of course in our case, as in many other cases like easy model, examples that I will make
later, the problem is that the system lives in of course a very large dimensional space which
grows with L. So it's really hard to make sense of this picture.
But this is I mean common to most nontrivial statistical mechanic systems. In addition, the
difficulty in our case, which I'll discuss a bit later, so this picture of metastability requires
somehow that there is a very large separation between the time scale, which the system takes to
equilibrate inside pure phase. I'll call it pure phase, won't define it precisely. Time scales which
is needed to jump to the other phase.
We need inspiration to lose memory inside phase. We'll see in our case the separation is not so
big. So that is why this makes things a bit subtle in our model. Good.
So I will just --
>>: [inaudible].
>> Fabio Toninelli: Oh, you put a wall?
>>: Yes.
>> Fabio Toninelli: Then the relaxation -- well, so what is the question?
>>: What's the relaxation?
>> Fabio Toninelli: It's L squared. I have a slide on that. But, no, maybe not exactly on that
question. So relaxation time should be S squared. We can prove the mixing time is of order at
most L squared log L and I guess for the lower bound I'm not sure we have L squared or S
squared log L.
>>: Depends on lambda. For large lambda, one expects even better.
>> Fabio Toninelli: Yeah, for large lambda one expects mixing time in that case to be L. Sorry, S
squared.
>>: Less than 1. The gap ->> Fabio Toninelli: I have a slide just a couple of slides. Okay. So before stating the results,
given a set S, let me call tau S, the hitting time of this set. So the first time where you enter this
set.
And let us look at this scale process, which is the following. Omega of S. Omega of S is one, if
the system, if the configuration at time S times the relaxation time is inside omega plus. It is
minus 1 if it is inside omega minus. And it is 0 otherwise. It just tells you which well you are and
gives you 0 if you're outside of both.
So the result we prove here, let's say, let me give it in three parts. The first one says that justify
this meta stable picture. So we can find a set S plus inside the well omega plus, which is not
exactly omega plus itself, but has an equilibrium measure, which is one-half apart from small
correction.
So it almost feels the omega plus I've just introduced. And whenever you start in this set, then
the time to jump to S minus or the opposite of this set is exponential. And it is exponential with
this rate. Twice the relaxation time. 1 over twice the relaxation time.
And so in the order of 1 it's important that it's uniform in time. So otherwise this claim could be a
bit empty. So whenever -- we can build two sets. S plus and S minus. Altogether we have
almost full measure. When you start from one of them you jump to the second set in an
exponential time like this.
So the second thing is that if you start inside S plus and you look at this scale process omega S,
then it just converges to a Markov chain on two states plus or minus with unijump rates.
So if you look at the system on this scaled time, so you just multiply by relaxation time, you're
either in plus or minus and you make jump between the two at exponential times.
And you never sort of for fixed time you are always inside one of the two wells. So in principle
this omega also takes the value 0 which means you're outside of both things. But in practice you
will never see it. So words.
So thirdly, we have some estimate on the relaxation time on which I will give some heuristics,
because it also explains some order of the results which is between two powers of the system
size. Power 5 over 2 and power 9 over 2.
So somehow it's perhaps a bit surprising that we have such a sort of precise picture without
knowing, having really a good control on the relaxation time, maybe. Good. So this is the picture
that comes out. I would like to give some -- no, before doing that, maybe I come back to the
question before. This is all about lambda smaller than 1. Maybe I just repeated.
So what happens for lambda -- so the system exhibits a dynamic on positioning in the sense for
lambda larger or equal to 1 the relaxation is quicker. For instance, for lambda equal 1 where
equilibrium is equal on all the path the mixing time is known to be of order from above and below
by L squared log L. So this is proven by David Wilson by coupling argument. And relaxation time
is of order CL squared with total expediency. Something like 1 over 2 pi squared, something like
that.
For lambda larger than 1, so you see here the relaxation occurs on time L squared, more or less
apart from logs. Where in the other case for lambda smaller than 1, we have at least L to the 5
over 2.
For lambda larger than one, we have Bounds, in an earlier paper, which form between L squared
and S squared log L and relaxation between L and L squared. As a conjecture, we think that this
bound, the L squared bound on the mixing time and the L bound on the relaxation time are the
optimal ones.
These we can prove with a limit where the direction parameter grows with the lambda size. We
believe it to be true in this whole region.
Anyway, for the system with the wall, if it's sufficient attractive to keep the, to have localization,
we also conjecture these things to be true.
Anyway, but this was mostly to show that for lambda in the localized or in the neutral, we could
call it phase or lambda equal 1 or larger, there's really a different scaling of the relaxation time
with respect to lambda equal to 1.
>>: Random and [inaudible].
>> Fabio Toninelli: Yes, this showed the -- actually.
>>: The wall.
>> Fabio Toninelli: Yeah, it was a system with a hard-walled constraint. But probably you can
extend that to get L2 something.
So let me for a moment abandon this -- I mean, I would like to explain first of all why this 5 over 2
appears and also a bit why what are these exponential -- logarithmic barriers I mentioned in the
title.
So you have a system -- so suppose you start the system inside the phase omega minus. So we
have configuration which is mostly negative, and eventually you know it will jump to omega plus.
So what's a reasonable way for the system to do that?
And you have to remember that this horizontal line is repulsive, so you don't want to touch it too
often or too much.
But a simple way is the system creates a bubble here one or two sides of the system let's say on
the left. The end point of the bubble, let me call it C. Then this bubble will fluctuate very large.
So this C will move fluctuate back and forth. In the end if we move to the right boundary so the
bubble has expanded to everything. You have jumped to the plus phase.
So this is a reasonable mechanism. Of course, if you -- somehow it will not exactly happen like
that. You can create more than one bubble. For the purpose of this heuristic let's disregard that.
The reason being that at equilibrium it's very unlikely to have many intersections. This is the case
exponentially with number of zeros. So it's reasonable that even during the dynamics, you have
no reason to create many of them. So let's just suppose there's one of them. One of those.
Good. So then the second observation is that if you look at the equilibrium, what condition on
having exactly one bubble. So having one crossing, what is the probability that at equilibrium it
occurs at some point X? If you do that simple computation you'll see that probability behaves like
some normalization factor times 1 over the distance of this from the left boundary to the power of
3 over 2 and by symmetry it's also 1 over the distance from the right-hand side to the power
minus 3 over 2.
So you can see -- you can write -- as being a particle in exponential V which is in this logarithmic
form. It grows logarithmically from both end points and there exists these sort of shifting just to
normalize the measure. So the potential is done like this.
And in the middle, the potential is three halves log of L. So then, again, arguing heuristically, one
imagines that this end point of the bubble will -- okay, will fluctuate back and forth. But it will force
the minimum of the potential. It will feel the drift driven by the gradient of this potential. The
gradient will be if the X is in the first half of the system it will be something like minus over 3 the
distance from the left-hand side and otherwise -- and symmetrically in the second half of the
system.
Somehow in the whole heuristics there's a hidden conjecture that along the evolution of these two
bubbles, they stay at equilibrium, conditionally on the position of C, which is something we cannot
at all justify.
But it's sort of reasonable. C should give you sort of the slow mode and then the two bubbles
should follow and equilibrate along the way. It's very hard to justify.
Good. So let me -- let's see what comes out from this kind of heuristic. So, first of all, you have
this particle [inaudible] which feels the drift. But the drift is almost 0, most of the system, the case
like 1 over the distance from the end point. So if you're at the microscopic distance from the end
point, the drift is almost here in a sense.
The second is that the barrier, the potential barrier which is in the middle is logarithmic in the site
of the system. So like 3 over 2 log L.
And, okay, there is some connection between this bubble dynamics and the 0 range process
introduced by Lamdean and Beltran [phonetic] one year ago. But I will not define the system.
The last observation is there's a nontrivial competition between a diffusion effect and activation
effect, and so what is that? Consider two, these two pictures. First one you have a particle which
moves in the potential which grows -- it's still between L minus L and the potential grows linearly
in the first half of the system and then decreases linearly symmetric. It will feel a drift, which is
just the slope of the potential, which is V.
Here it goes to the left. Here it goes to the right. In the second picture, it's the one we are
looking at. The potential is logarithmic, and the drift is what I had just before. And the barrier -so barrier here is linear in the size of the system and here it's logarithmic.
So what is the time this particle will take suppose you started here to jump to -- to go to the other
end point?
So, first of all, suppose there's no potential, then you have just a simple random walk between
minus L and L and it will take L squared, just diffusion.
Secondly, in the case of uniform drift, the minus V plus V. So the time to jump a barrier of height
V-0, which is V times L, is sort of essentially the exponential, exponential of this barrier height.
One can see that this is -- it's not difficult.
So usually I guess this is called something like activated dynamics, to jump across an integer
barrier E you need exponential of E and then there should be [inaudible] temperature, but here
there's no temperature.
Okay. So what in our case. If you believe in this kind of picture, you would guess that the time to
jump from one end point to the other should be exponential of the barrier, which would give you L
to the 3 over 2, which clearly is wrong. Because already without potential, we know that the time
is L to L squared.
So there's sort of competition between this diffusion effect and the activation effect. And for this
one dimensional dynamics, with this drift, the relaxation time can be computed exactly apart from
constant.
For instance, using a meta by [inaudible] law, using Harding's inquality, or by using conical path,
the way you compute this. Relaxation time turns out to be L 5 over 2. Personally I don't have
clear heuristics on just looking at the system why the 5 over 2 should come out.
Anyway, so this is about the simplified dynamics. One bubble approximation, local equilibrium
blah, blah, blah.
>>: Is the 5 over 2 some fluctuation over 2?
>> Fabio Toninelli: No, no ->>: [inaudible].
>> Fabio Toninelli: No. No, it's not something -- I guess it's -- I think if the potential were just 1
over X -- I mean, alpha log X, it's alpha plus 1 -- so it's really ->>: You solve ->> Fabio Toninelli: No, it's a simple function. It's just that it's not so easy to guess, I think. So
anyway, these heuristics something we find a bit rather convincing. We conjecture this to be true.
For the true dynamics, let's forget this one dimensional one.
>>: Let me check -- the exponential would have broken down to sufficiently small constants. In
other words constants that aren't constants, sort of what you're looking at here.
Anyway, you have conviction three-halves would have broken down wherever you had the
exponent going to 0. Why was 3 -- want to go back a little bit.
>> Fabio Toninelli: Yeah, sure.
>>: The V 0 is 0, your formula is wrong in 1 and 2. So it's certainly misleading, too.
>> Fabio Toninelli: No, I mean ->>: Predicts three-halves.
>> Fabio Toninelli: I agree. No, I mean -- I agree. I mean but ->>: So if you set the constant such that -- I see you get three-halves just by plugging in
pretending it's a straight line.
>> Fabio Toninelli: Yeah, it was just ->>: Yeah.
>> Fabio Toninelli: I agree. It's misleading argument.
>>: Talking about the shape ->>: The shape is important.
>>: Just for -- talk in the case where the belief is constant, it's the minus B plus B. So the picture
is really that -- so the particle spends a lot of time to the left. And then at some random time
exponential time decides to go to the other side, but not slowly. Not with random walk, takes
essentially a drift that is plus V to the guy goes to the top of the mountain then falls down. So
here's totally the system is doing something immediate.
>>: So what altitude can you make that statement?
>>: I see.
>>: How high does V 0 have to be before you can make ->>: Instead of 2 over log L it will be hundred log L and then it works.
>>: But you said this is not the hope, right, because many, many times it climbs halfway and ->>: Not many times.
>>: Very few times.
>>: Exponential [inaudible] too.
>>: [inaudible].
>>: It does exponentially two times before it does the whole thing.
>>: Right. The last ->>: [inaudible] is very quick. Of course, the like with the positive velocity. Because it's so
costly ->>: But average distance goes [inaudible] stays bound before it needs to be ->> Fabio Toninelli: Okay. So as I said, for the true dynamics we prove these two bounds. We
believe the lower one to be correct based on the heuristics.
>>: Both of them to be correct, the lower bound to be ->> Fabio Toninelli: Thank you. [laughter].
We hope both will be correct. [laughter].
>> Fabio Toninelli: The reason why we get this 9 over 2 it's better to see it's 5 over 2 plus 2. The
reason we use some decomposition for Markov's chains. So the point is the slow mode which
should give you the correct mixing time is the motion of this bubble end points. It would be more
than one. So that should give you the 5 over 2. Then the motion of the excursions given the
bubble end points should be diffusive, like simple excursion with the wall. That should give you L
to the 2. So by this composition method instead of finding maximum between them you find the
product. So that's why we lose at 2. Already giving that is quite a technical pain, but this explains
the 9 over 2. So let me come to this question about the not so big separation of time scale. So
the difficulty which appears is that times to relax inside of single phase is of order L squared,
which is not much smaller than N to the 5 over 2, which we believe to be the time to jump from
one side to the other. It's much larger, but only by square root of L. So once -- for instance, one
thing we can prove is that if you look at the evolution starting from the maximum configuration,
which I denote by this sort of 10, the maximum configuration would be a path like that. Cannot
have anything higher than that. You look at the evolution at time T, this is the semi group. Then
look what is the difference from just looking at the plus phase. Plus phase meaning you take
equilibrium condition to be positive. Then if you look at this at time much smaller than L squared
you're in variation distance essentially 1. You've not moved variation distance. If you take times
which are sufficiently larger than 2 and sufficiently smaller than L to the 2, and L to the 5 over 2,
then this variation distance is smaller of 1. So in this time range, starting from the maximum
configuration, you have relaxed one of the two phases. The one [inaudible] stuff. So this is sort
of -- so we are able to prove it starting from the maximum configurations because we use a lot of
monotonicity arguments. We do not work for a given arbitrary configuration inside omega plus.
But it's smaller -- it should be true for most configurations omega plus, just that we cannot prove
it.
Anyway, so this proves that this deep phase I've not defined is of order more or less L squared,
and as I said, the global time should be over the L to the 5 over 2, which is not so distance. So in
the beginning we were not even sure that this separation would be enough to have at least be
able to prove metastability.
So let me just, as contrast, let me recall that in another case, well known case where metastability
has been proven is 2-D. 2-D is the model with free bound conditioning, you're below the critical
temperature, so you have two phases, plus minus. Also there you start from let's say the old plus
configuration in some time which is of order -- this is not really proven, but let's say more or less,
in time of order a power of the system size you will converge the plus phase, which is the one
close to where you started from. And then on a much larger time scale, you will jump to the
minus phase. And the time scale which is of the order of exponential of the size of the system,
where the constant here is related to the surface tension.
So very quickly, you converge to one of the two phases. It takes enormously much, enormously
longer to jump to the other phase and having this big separation it allows -- allows, might have
Fabian Machelli [phonetic] to prove this is the meta stable picture.
So here we don't have a lot of time to lose memory. This is the difficulty. So essentially -- well,
okay. We say 30, 40 minutes. I can just flash one slide that will give a few ingredients of what
we do, and then if anyone is interested.
So one ingredient is proving that from the maximum configuration in time L squared more or less
you converge to half of the equilibrium. So it will be the plus phase. The second one is to -okay. The mixing -- as I said the relaxation time is the lowest eigenvalue of the universe. There
must be some function which solves generator times G equal minus 1 over relaxation time G.
When one step is proving there exists, we cannot prove that you have a simple solution. But
there exists a solution to this, which is close to indicator of the plus phase, minus indicator of the
minus phase.
And from this two ingredients, it is almost easy to prove that starting from the maximum
configuration, looking at things at time, of time of order TREL then this distribution looks like this.
You are in the plus phase with this probability and the minus phase with this other probability.
And then this is, say, the first step to prove stability, is definitely more messy. So maybe I stop
here. [applause].
>> Yuval Peres: Questions or comments?
>>: You have more slides for those of --
>> Fabio Toninelli: No. [laughter].
I pretended like I had more. But I don't. [laughter].
>>: You need to stop pretending.
>> Fabio Toninelli: We can discuss on the white board for those who are interested in details.
>> Yuval Peres: Okay. So any other personal questions? Let's thank Fabio again.
[applause]