19137
>> Yuval Peres: Welcome everyone. And I'm very happy to have this talk on the self-avoiding walk and
honeycomb lattice. And I've met Hugo in several conferences earlier, and each time I'm doubly more
impressed than the previous times. I don't know how many times he can keep that up. And so please,
Hugo.
>> Hugo Duminil-Copin: Thank you very much. So thank you for the invitation. So I will be talking about
the self-avoiding walk today. And I have two main goals. My first one is to prove conjecture on the
connective constraint on this model. That will be the first part. And that will be totally basic. I hope you will
get the proof before the end.
And the second part is trying to explain somehow this proof, because you will see that it can look a bit black
magic. So I want to explain, to motivate, the introduction of the objects I will use. To do that I think it's
better to see self-avoiding walk in bigger frame. It's a part of a bigger family of models. I will explain a little
bit through these models how you can be led to this kind of proof.
Okay. So to be sure that we agree on the model, you take the honey come on the lattice, and you take
several trajectories starting at some point A. Say 0 G. So don't change, it doesn't change anything to the
model. I will start at mid edgeings and end at mid edges.
And if I did not see in the cardinality of the number -- the number of self-avoiding trajectories of length N.
There's a natural question which is to try to estimate this when N goes to infinity.
So I guess you all know that this number grows exponentially fast. That's just due to some multiplicativity.
If you take a N for self-avoiding walk, you get N step self-avoiding walk and N step self-avoiding walk.
So globally you have this inequality. And together, with other estimates, saying that you grow like an
exponential, you can deduce to C to 1 over N converges. So in other words mu C to the plus model of N.
And this mu C, this connective constraint, you don't expect it to have any nice value. But there's one
special case where there was a prediction made by physicists is the case of the hexagonal lattice. So in
1992 Nienhuis conjectured that the value was square root of two plus. How he does this is kind of even
more black magic than the proof that I would present and it's [inaudible] formalism so a lot of deep physics
behind this but it's not rigorous.
So this is the first rigorous proof of this thing. Okay. So the first thing we are going to do is not to work with
self-avoiding walk. So we're simplifying the model by looking at self avoiding bridges. So that's almost a
walk except that the first point is the lowest one and the last point is the highest one.
So this is called a bridge. And the nice property with bridges is that actually they grow exponentially fast
and at the same speed as self-services. And it's easier to work with self avoiding bridges. The reason is
that they are using the decomposition of self-services. You should really compare this to, say, a
description of random walk away from 0. That's easy to handle, usually. Why is that true, because after
you read it's not obvious. There's one [inaudible] that's straight forward -- there's less bridges than the
self-services. And what you need to prove is 0 is the bound. So how do you do that? So first thing you
can try to do is to take a self avoiding walk and send it to some bridge.
Try to see if you can associate naturally a walk to a bridge. So let's take a walk like this. And let's try to
construct a self avoiding bridge from that. So the idea is that you take the maximum of your walk, and you
just unfold your walk. See, this part I am going to refract it. And I can iterate the reasoning. I unfold and I
carry on until I get, I do the same thing below if I need to, and I will exactly get a self-avoiding bridge. So to
any walk I can actually set a bridge.
And the question is to any bridge did I associate a lot of walks or not. And in order to see that it's not that
bad, what you can try to look at is the reverse precision. And if you give me a walk in the partition of N,
say, I can give you -- sorry, a bridge and a partition of N, I can give you a walk, because the partition was
exactly correspond to points where are fold back. I start from here you tell me after two steps you fold and
so on.
So what you know is that the number of walks it's necessarily bounded by the number of bridges times the
number of possible [inaudible] which is the number of partition of N the base case. So, of course, there are
a lot of partitions that will not work, that will not send you back to self-avoiding walk. But every
self-avoiding walk will be sent, every self-avoiding walk that is sent to this bridge you can recover it for
some partition.
>>: Why is it partitions, not [inaudible] partitions.
>> Hugo Duminil-Copin: I mean other partitions in this case. Order partitions. I should say that.
>>: But then why is the number of them ->> Hugo Duminil-Copin: Because here when you unfold, the steps are smaller and smaller on necessary.
Because in the construction -- so that's a very good question. So I will be putting some stuff under the
ground over time. So if you see them that's even better.
So now what is remaining is to try to prove that there is not too many partitions. So these you know by
heart what is the equivalent of this. So you know very well [inaudible] or little word, or you don't know it and
you just notice that the generating function of order participations is this one. And this is a [inaudible] one
exactly saying that RN cannot grow exponentially fast. So you get -- so what we're going to try to do now is
work only with bridges.
Okay. So when you want to compute this kind of thing usually you take the partition function or generating
function, depends on how you call it, and you try to find the radius of convergence. And usually you find
this radius of convergence because you have, say, modifications that allow you to compute explicitly the
partition function, because you have a functional relation that gives you exactly the partition function. So of
course here you have no hope to have an explicit formula for the partition function of bridges.
So the idea that we came with is that you can modify the partition function in order to get something where
you can compute things. So before I present exactly the modification, I need a definition. Very easy one.
The winding of a curve gamma between two points A and B is just the number of terms you did in a radian
between A and B. So in the first case, for instance here you see you have one quarter turn to the left and
one quarter turn to the right. So basically you did zero quarter turn. Winding is 0. This case you did four
quarters on the left and so on. You can have all the values, of course.
So this is the winding of a curve. And now the [inaudible] comparator, so this modification of the generating
function, it's almost generating function, this would correspond exactly to the generating function, but I add
the Y term in here. So what does it mean, this thing?
So I will be walking with a special case of self-avoiding walk. Which is the following. So I would be in this
frame. So you take finite domain, a simply connected one, but anyway. And one point on the boundary
here, A. And actually you will be working only with walks that are staying inside the domain. And so my
function my point Z. So mid edge Z, will be exactly the sum over all the walks starting at A and arriving at Z
of the [inaudible] so what should be involved in the partition function. But times this one enter. So if you
arrive with 1 and 0 you only count [inaudible] but if it's 1 and 2 pi you'll have a complex number in front of it.
And I really, I cannot infer too much that this is complex number. [laughter].
>> Hugo Duminil-Copin: For many reasons.
>>: Sigma.
>> Hugo Duminil-Copin: Sigma so far is nothing. I will explain after what it is. But that's a very good
question. Like -- oh, now we know. But here we don't know. [laughter] so that's a very good question.
>>: [inaudible].
>> Hugo Duminil-Copin: Sigma is real or not necessarily. It could be purely imaginary. So there are cases
where it's smart to look at it as an imaginary number. So let's say for now we don't know what it is. Okay.
So we have the freedom and the choice of the sigma, and the idea is that if you plug sigma equals 5 over
8, and you will see why in the next slide, and if you plug mu equals square root of 2 plus square root of 2,
you get something nontrivial, this relation, this local relation. So what does it mean?
>>: Could I make sure I understand. So summing over walks or versions?
>> Hugo Duminil-Copin: Walks.
>>: And L of gamma is the length?
>> Hugo Duminil-Copin: Exactly that's an N. Yes. That's the natural [inaudible] weight. That's the length.
Okay. So what it means is that if you look at the vertex semi your domain, vertex V, there is three mid
edges that are adjacent to this vertex, and what I mean is that some sum observable over these three
edges equals 0. And the coefficient, if you are scared by them, just think that it exactly corresponds and
that's something very important to the discrete integral of the observable on this small 2 equals 0.
>>: [inaudible] two elements?
>> Hugo Duminil-Copin: Sorry? So it's just P minus V, that's this, this and this. There's nothing
complicated in them. Okay.
>>: They're vectors and the X are numbers, complex numbers.
>> Hugo Duminil-Copin: Yes. So that's an inequality between complex numbers. Everything complex.
>>: Complex.
>> Hugo Duminil-Copin: Numbers.
>>: Vectors are ->> Hugo Duminil-Copin: I don't know -- numbers. That's complex numbers.
>>: So it's R minus V as a complex number.
>> Hugo Duminil-Copin: Yeah, yeah.
>>: And had four arguments and now the thing about this, what are --
>> Hugo Duminil-Copin: So I don't want -- so I can try to make the formula with that but you will see that
they get bigger and bigger. So I will just forget about the initial points. I fix it for N to the N, say. So this is
a real variable and mu and sigma here are fixed. So I will forget about the notations.
>>: So A is simply a starting point?
>> Hugo Duminil-Copin: Yes. Whether you should think about that as a function in Z. Okay. So this
exactly means the integral along this contour of F equals 0. So discrete integral. That's really what it
means. So that's not fancy formula. Okay. So where does it come from?
>>: Trying to simplify the sum by writing in the contour.
>> Hugo Duminil-Copin: Yes, I'm hiding it. So I want -- I want, I said, yes, I want to prove that this big sum
equals 0. So this sum, when you think about it, it's the sum over all the work that started at A and arrive
either at R, P or -- at P, Q or R, of some is weight. Depending on the curve, on the self-avoiding walk
gamma, which is a quote to mu times the length. There's this winding number, whatever, and in front of it,
if you arrive at P, you say P minus V. If you arrive at Q it will be Q minus V and so on. So I want to prove
that the sum over all this self-avoiding walks equals 0 of this contribution equals 0. The thing I will manage
to do that if I can make small groups of self avoiding walks on which at each step, so each of these groups
I can prove that the sum of the contribution equals 0. So very, very simple
Argument. And I will give you first example. If you take a walk say this one that visits the three mid edges
around the. Then you can naturally associate it to another work that we did with three mid edges that the
same start that you visit the loop that you do after in the other direction.
You can always do that. So let's try to compute C of gamma 1 for C of gamma 2. So this is C of gamma 1,
three terms involved in the sum here or so. And I see there are many, many things I can factorize actually
here. They both have the same lengths. So mu minus the length I can factor this. Q minus V and R minus
V, I can express them in terms of T minus V. That's elementary, and I factorize T minus V get two terms
that are due to these expressions.
And what is remaining is the one new term and when I look at the one between A and Q of gamma one it's
one in between A and P plus the one between P and Q. So one in between A and P that will be for the
same for gamma one and gamma two that's the same start of the curve. So the only difference remains in
the winding between P and Q is the winding between P and R for gamma one and gamma two. But when I
think about it, I know this winding. Why? Because I start on the boundary of my domain. So schematically
the only thing that I can do is is this.
I cannot do that. That's forbidden. So what I do, it's basically one term on my right and five terms on my
left. Perhaps I will do two terms on my right but in this case I will make six turns on my left. So winding will
be always minus for pi over 3 or for pi over 3 depending on which of these two curves you can see there.
But now if I put sigma equals 5 over 8, I get 0. This plus this equals 0. Trust me. You don't need to make
the computation. Okay. That's the reason why we choose 5 over 8, and I will go further, I will explain more
in more detail in the second part.
You can do the same thing with walks that visit one site or two sites. You will factorize something and you
will get this quantity, this early quantity. But if you put mu plus 2 square root of 2 you obtain 0. So sigma I
chose it to make all the pairs vanish and mu for the triplets.
>>: So mu -- there must be some trigonometric ->> Hugo Duminil-Copin: Yes, that's the co-sine of, so that's one over the co-sine of -- no that's co-sine of pi
over 8.
>>: Now I ->> Hugo Duminil-Copin: Now you see. That was obvious when you look at the model.
>>: Somehow you have to, in this last one you have to combine the ones visiting one site with the ones
visiting.
>> Hugo Duminil-Copin: Two sites because the natural thing to do is just to prolonging the walk, two ways
of prolonging walk if you start with the two sites you can remove the last site and visit only one site. So all
the walks that finish at P, Q or RI can put them into one of these groups. So my big sum equals 0.
>>: It's strange because you might -- under some circumstances there would be far more just [inaudible]
one side and then you have this [inaudible].
>> Hugo Duminil-Copin: No. That's not true. Because if you visit only one site, you can't pronounce it in
two ways to get two sites. If you visit two sites and not three, you can always remove your last site.
>>: The big sum is all the walks that end where.
>> Hugo Duminil-Copin: P, Q or R.
>>: Only -- okay.
>> Hugo Duminil-Copin: So this is the walk, that's the sum over all the walks that end at P. This one is the
ones that are at Q and R. Good question. No, but I mean it was not explained clearly enough. Apologize.
Okay. So we are happy. So we know that the integral of F on this discrete elementary contour equals 0.
So I know now it's elementary. That the integral over all close control equals 0. So the end of the proof will
be can we choose the right contour and the right domain to get something that allows me to count things on
self-avoiding walks. And it appears that this domain is a nice one, because you try to minimize the number
of computations. And this is the contour I will consider. And you will see just after why I consider this. So
saying that the integral on this contour equals 0 is saying exactly that. Nothing more. Here I replace the P
minus V and Q minus V by what it's equal to.
>>: Say again, what's alpha and beta.
>> Hugo Duminil-Copin: Alpha, beta and gamma are just a part of the boundary I'm considering, just
beta-ize and alpha-ize and epsilon the boundaries.
So why did I take something on the boundary? Because on the boundary I know the winding. So
remember that we are working with F. So okay we have a nice formula for F. I mean, relations for F. But
still there is this one term which is not positive which is a complex number, and it can -- it can make a lot of
consolation and you don't get anything on your model. So you really need to get rid of this winding term.
And the way you do that is by considering something on the boundary, because on the boundary, if you
look at walks that arrive at beta, so the winding of these walks you know it it's 0. You cannot wind around
the boundary.
>>: You're starting from the inside.
>> Hugo Duminil-Copin: Yes. You start from here and if you wind -- if you wind around it it means you get
out of your domain. So you can express same thing for alpha, epsilon and epsilon bar, you can get what is
winding in this case and so you get relation that only involves both mean weights now.
>>: Is this the reason you need the bridge so you have the winding on it.
>> Hugo Duminil-Copin: Exactly. Why do you need a bridge? Because the bridge is typically something
that goes from one side of a domain to another boundary of the domain. A walk is something that ends in
the middle. And in the middle you don't control anything. So that's exactly that. So I was going to say that.
So you have nice formula, 1 equals the co-sine. You have some terms that are now just the generating
function, and here, for instance, if you let the width going to infinity, you let L goes to infinity, you're walking
the strip now.
And this term is exactly the generating function of the self-avoiding bridges of height say so if you walk with
a height T of height T. So here I'm counting self-avoiding bridges of height T.
>>: What's the angle.
>> Hugo Duminil-Copin: So the -- don't matter. Really to minimize the computations. And actually you let
L goes to infinity. So even this appears this term. You only get these two terms. This one disappears.
So really what I want to emphasize is that here I have the generating function of bridges of height T. So,
for instance, for free what I get is that the number -- so generating function of bridges of height T is smaller
than 1, because this term is positive. So I will not do it because I guess you can do it by yourself, but so
now, for instance, you can deduce that mu is not smaller than mu C because if it was smaller because
there would be this mu that times the minus the length, it will not be the entropy. There would be too much
entropy and this here would explode exponentially fast, and it's not the case. It's smaller than one. You
get one inequality making this reasoning rigorous and the other one is a little bit tricky but nothing
completely crazy.
So finally you finish the proof just by using this combinatorial equality. So I'm afraid not have time to finish
so I will not finish the proof. But if you want after the talk you can ask me and I will finish it.
So if you want by the way the proof is not on our caveat. But it's some Internet site and the one of stars.
So you can go the proof is completely written there.
>>: [inaudible].
>> Hugo Duminil-Copin: Because we are waiting to prove more. We hope at least to be able to do more.
Okay. So what I did so far is I did the negative constraint. So I sketch -- of course I skip some steps but
easy steps. And I did that by introducing this weird function, this [inaudible] observable. We don't know
exactly where it come from, but it works.
And the important thing on this discrete observable is that it's discrete holomorphic in some way, the
integral over all the close parts equals 0. So here you see the first connection with [inaudible] variance.
You have something homomorphic in the discrete lever.
So what I'm going to do is now I'm going to explain what we are going to do, what we would like to do
further, and just to avoid one annoying question every time I give, I explain this kind of thing is what you
should not do next.
Okay. So perhaps I should have started by this, but it's a good place also. What are the real conjectures
for the self-avoiding walk. So complete one. So the first one is so this conjectures are due to [inaudible].
And so it was a guy who guesses this. But you will see had the argument to show that the correction in
front of this mu to the N was a polynomial correction. And with very special value, gamma, which is 43
over 32. And this gamma is universal in the sense that if you take self-avoiding walks on, say, the square
lattice, obviously you don't have as many self-avoiding walks. You have much more self-avoiding walks in
the square lattice, the mu constraint is not the same.
But the correction in front of it is exactly the same. It's N to the gamma minus 1. That's this kind of weird
things you can get from [inaudible] invariance. Just geometric questions because we're probablists.
>>: Maybe it's worth putting in the word why this sounds completely crazy if you don't ->> Hugo Duminil-Copin: So I would explain it just after.
>>: [inaudible].
>> Hugo Duminil-Copin: I will explain it just after. And I will explain it for this one, because that's one
sentence explanation. For if you look at the geometric question, like what is the mean square displacement
of your walk, you will also get something which is equivalent to power or something, and it will be the mu
will be universal as well.
So this conjecture, mu is gotten then by looking at [inaudible]. So continuum limit of fancy objects. And I
guess it's not -- it's hard for us at least to understand where it comes from.
>>: Three-quarters [inaudible].
>> Hugo Duminil-Copin: Three-quarters, but it was ordered but only numerical.
>>: [inaudible].
>> Hugo Duminil-Copin: There is this thing, yes, but I mean that's just saying I have two things that should
balance each other. So I look exactly at the time where they are exactly the same power and gives me the
right quantity. So that's true. But it was really, really heuristic.
Okay. But now I guess we have much more precise intuition why it should be true and that's due to this
conjecture by [inaudible] that the scaling limit of the self-avoiding walks is SLE. So I guess here everybody
knows about SLE so I'm not going to give too much, too many explanations but I would like to say what is
converging to the SLE. So you take a domain, your favorite domain. And two points A and B on the
boundary and you consider hexagonal lattice of very small mesh size. And you look at all the self-avoiding
path that start at A and arrive at B. So there is a very natural measure to put on them is like always you
want to penalize by the energy. So you will put a measure where the probability of some curve gamma
would be proportional to some mu minus the length of the gamma. Sorry, you don't see. You want
something like this. You can see if you take mu very, very large, too large. You will penalize a lot by the
length. So basically your work they would like to be straight, to go straight from A to B. It's a fluctuation but
it will disappear at the limit. On the other hand, if you take mu very small you don't penalize enough and
you get something which becomes space fitting and you get something which is not an [inaudible] that's not
a limit. That's not a curve that is nice. I will explain to you what it is later. But if you take mu exactly equal
to the critical value in some way and this critical value its square root of 2 square root of 2, that's the
connective constant. Then you expect to have a scaling limit for this object.
So when the mesh size goes to 0 you should have a random [inaudible] curve and this curve should be a
[inaudible] curve.
So just two words about SL 8 serve. That's a Lobner chain [phonetic], and so it has a drift -- a driven
function, driving function, sorry. Which is called of [inaudible]. Two only things that I want to explain on
that. Because it will be useful for the next point.
So before I can explain to you where these things come from. So you can do the same thing for the square
lattice and the right mu will be the connective constraint for the square lattice. Will you expect to have the
exact same scaling limit. And somehow now you start to understand that what will be universal is exactly
what you can compute via scaling limit. And that exactly the spirit of universality.
This thing, so I don't remember which is easiest to explain. So this one, this mu here it corresponds, it's
related to [inaudible] dimension of the curve. So that's an example why this mu should be universal,
because it's connected really directly to the [inaudible] and there are things that you can -- you have to be
smarter but you can get 0 exponents [inaudible] so everything that is universal should be, you should see it
at the scaling limit and vice versa, in the sense what you can compute at the scaling limit is universal.
>>: I had one thing if you take two self avoiding walks, then concatenate them as [inaudible] the
concatenation is still self-avoiding. The mu cancels and you can see ->> Hugo Duminil-Copin: Exactly.
>>: That's one way to see ->> Hugo Duminil-Copin: That's completely true.
>>: So this [inaudible] is assumed to be universal.
>> Hugo Duminil-Copin: Yes.
>>: That's something globally.
>>: The self-avoiding ->>: [inaudible].
>>: Things that you can compute.
>> Hugo Duminil-Copin: So, for instance, in this case ->>: Self avoiding.
>> Hugo Duminil-Copin: The quantity you will talk about, it will be the limit will be exactly the exponent,
universal exponent for the SL 8 [inaudible]. Okay. So what we want to do now so of course what we'd
really like to do is to prove con [inaudible] invariance, to prove this convergence. So I'm not going to do
that today. Unfortunately, I would say. But I can explain to you what should be the strategy.
If you want to prove something like this, usually you need two ingredients. So first one is a
precompactness argument. You have your discrete curves here. You let the mesh size go to 0. Can I
extract some [inaudible] that are converging? And actually -- so converging to a continuous curve. That
would be the right thing. And even a little bit more than that you should have something that makes you
converge to a Lobner chain. So just think that it's slightly bigger. You are slightly more than just
continuous curve as a limit. You want a Lobner chain. So precompactness is a strong one, which gives
you a Lobner chain. So once you know that that's a Lobner chain, that's not necessarily a conformant
invariant object. I'm just saying you want a Lobner chain at the limit.
The second step is to say okay now what I need to do is to compute the driving, the driving function of this
lobner chain. And prove that that's a boolean motion. And that technique that you can use here is the
following. If you know nightingales of your curve that you can compute. Actually, the square is nightingale
from my curve. What you can do is you write the formula and you say the a priori drift, it actually equals 0.
It vanishes, since that's Martingale. But hidden in this drift there is your driving process somewhere. It's
perhaps not very explicit. But it is involved in as a driving, in the drift term.
So if you have enough Martingales then you hope to have enough information on your driving process to
get at that boolean motion. And that's exactly the technique. So easily what you manage to prove is
something like your driving process is a Martingale and your driving process squared minus 8 [inaudible] of
T is a Martingale and you use the theorem to say okay data mine my process. So that will be the
technique.
What you need is a Martingale which you can compute. And usually necessary because you are working
with a conformant invariant object you'll get Martingale that comes from the invariance. But a priori you
would not need that. This conformant invariant Martingale that should become observable. This is a
Martingale, small computation and it should converge to something holomorphic that you can compute.
We know the boundary conditions.
You know already there's discrete integrals that vanishes, the discrete vanishes. Okay.
>>: What is the difficulty.
>> Hugo Duminil-Copin: Sorry.
>>: What is the difficulty.
>> Hugo Duminil-Copin: Okay. So what is the difficulty? Everything. That's very simple.
Precompactness in this case we don't know how to do. Because you don't have -- so usually for statistical
model you use like an FKG [inaudible] estimate. So crossing of right angles. Here you don't have this kind
of thing, because it's strongly nonMarkovian and it depends on the whole curve. You don't know anything.
So this is already a big open question to have precompactness. The second question is okay let proof that
this observable property with scale converges. I don't know. I don't know how to do that. Why? Because
this discrete holomorphosity should not be compared with any mean you don't have the same strong ridgity
that you have when you work with continuous holomorphic conditions. If you fix the domain, the boundary,
the value of the boundary on the domain for holomorphic function, you data mine your holomorphic
function. Here it's not the case..
When you count the number of information you have, you have one unknown variable by edge. And how
many functions, how many relations do you get? One per vertex. So that's twice less relation for -- twice
less relation than variables.
>>: That argument for [inaudible] is that the same problem.
>> Hugo Duminil-Copin: No, because so how it's a different problem because what they prove is
something stronger than that. And in the case -- sorry. In the case of [inaudible] you have something
which is even nicer that you know precompactness of your curve. Of the observable and everything which
is not the case here.
>>: Referring to the last part, the discrete.
>> Hugo Duminil-Copin: But they have precompactness. If you give me precompactness, it converges to
a continuing function, then I will say [inaudible] theorem, I have integral over the contours are equal to 0
and holomorphic.
>>: If you know 1-2 follows.
>> Hugo Duminil-Copin: So that's very good question again. No, that's true. It's very rare I can talk about
that. So that's good. So that's a very good question. No, because the winding term is something so ugly
that precompactness of the curve doesn't even guarantee that this winding term behaves nicely. So
that's -- and that's a problem. It makes a lot of problems.
Okay. But what you can do -- let's try to prove something. So what you can try to do is to have a bound on
the correction term. So after you have one which is totally for free from [inaudible] you get that it's bigger
than mu to the N. It's the number of self-avoiding walks is larger than square root of 2 plus square root of 2
to the N. Here's a bound. You would like to have polynomial bound. And the best one you have is exactly
the one I described to you which is square root of N. So that's far from being the truth.
So a little bit like you want to have good estimates on mixing time here you will have you want to have good
estimate on the correction term. So polynomial bound would be good, very, very good. And here you -- it's
a work in progress and it's going well. So I mean you can play with the geometry and using more involved
combinatorial arguments and it should be working.
So you can still do a little bit more than just the connective constraint. So let's try to explain where this
parafermion comes from. In order to do that I will need to consider a new model. This new model is a
model on loops on my lattice. So I have still a discrete domain. And I consider a model where you have
loops like this. And the priority of a configuration is proportional to the number of X to the number of edges
you use to make your loops. Times N to the number of loops. Okay.
So that's just for people who know interest in model that's very close to this model, that you don't have the
same constraints on the configuration. So that's a model which is a representation for the spin model
which is even more famous. So the spin modern model, you put here it will be on the faces of your graph.
You put a spin, which is an element of the N dimensional sphere. So O of 1 is the Ising model. So that's O
of 1 is the Ising model. And O of 2 for people who know that is the X-Y model. And so on. So exactly like
the random crystal model as a geometric representation random of [inaudible] model, here you get
geometric representation of OM spin model. So already there's a large strong parallel of these two models.
So [inaudible] actually studied the model for every N between 0 and 2. And it obtains the following
diagram. So there is a critical line. [Inaudible] when you take X smaller than this you get the sub critical
model in the sense that there is a correction length. You see very small loops logarithmic loops. There is
this line which is exactly corresponding, you see it's a generalization of the formula I presented to you
before. And here that's critical. There's no correction length. And there is a second phase and for us it's
not common to see that. But there is a second phase where you see again critical phase. But it's not the
same critical phase in the sense that it's critical exponents and everything that is depending on your model
are not the same.
I will -- so O of 0 is self-avoiding walks. This line corresponds to self-avoiding walks. X is 1 over mu. The
line here is it's the Ising model. Okay. So for this model now I can try to create the same observable, see if
there is a common frame for everything. So let's start with the Ising model. So you look at 0 of 1. So you
have your lattice like this. And I'm going to put an Ising model on this mid edges. Just because I like it.
So there are plus or minuses. So when you look at the Ising model there's a natural thing you want to
compute on it is the correlation between two points. So I mean I don't need motivation to introduce this.
Okay. And when you look at the height and [inaudible] expansion of the Ising model you get loops and
strings and this is exactly the O of 1 as I describe it. That's exactly the loops and exactly the loops of the
height and [inaudible] expansion of the Ising model. In this case, this quantity, when I compute it in terms
of the O 1 model is exactly the partition function of my model times the partition function of my model but I
add a string between X and Y. I must have trajectory which is X and Y. So let's say here I mean the
domain with a lot of loops, and here I'm in the same domain with a lot of loops but I must have a string
between X and Y.
Okay. So that's -- I mean, very natural to do. Now what physicists like to do, and I will not be able to tell
you why, is to add what they call a disorder comparator. What does it do, this disorder comparator it takes
the two points X and Y and say, okay, I take one trajectory between these points and I switch the
[inaudible] on all this, along all this trajectory. So I force the spins to be frustrated in some way. I change
this.
Here I had, I don't know, between these two J now I will have minus J. When you look at this and you try to
see what is equal to in the loop representation, you can still do the same thing that before, but the thing is
that you need to add a term in front of every of your configuration and it will be one. So here this was just a
sum of X to on the configuration gamma of X of the number of edges times N to the number of loops. So
here I will have also this.
But not ready to count the thing with disorder [inaudible] I need to add a term and this term will be called to
one if the curve that was here has even number of terms around Y. And minus 1 if it does an odd number
of terms. So here I must add something which is 1 or minus 1 regarding the number of terms I do my curve
does before reaching X.
And this if you think about it, you can rewrite this, so these two things. You can rewrite this like this. If you
do an even number of terms, we will have winding which is multiple of 4 pi. So when you divide by 2 it
gives you a multiply of 2 pi, which is 1. You get 1 when you put the exponential. If you do an even
number -- an odd number, you will get minus 1. So what I found finally is something that looks very close
to what I considered before. There is this winding term which is naturally happening here. And I'm not
taking a fancy quantity. And this one-half, it is spin in the case of the Ising model. And if you want, you can
replace for all the models, you can look at the natural spin somehow.
>>: 5, eight.
>> Hugo Duminil-Copin: Exactly. You change the sigma. Here you will say I will not take one-half. I will
take some sigma and in terms of N there is a natural value of the sigma which is this one. And you can
check that for 0 it exactly gives you 5 over 8. So, of course, perhaps the best motivation for the sigma is to
say it works. Like when I do for all in models I can do the same combinatorial trick and I'll get the right thing
by choosing this sigma. But where it comes from deeply is from the fact that in the Ising model that's really
a very natural quantity to look at and it's a natural generalization.
So this sigma is connected to the [inaudible] charge and the spin of the model. So it's [inaudible] one-half
was a fermion. It was really the spin of the particles in the fermionism Ising model. Of course, here sigma
not equal to half integer. It doesn't give you the spin in the sense of physics. That's why it's called a
parafermion. [inaudible].
>>: Isn't N equals 0 self-avoiding.
>> Hugo Duminil-Copin: Yes.
>>: But if you put five-eighths there.
>> Hugo Duminil-Copin: Five-eighths and three-eighth are the same spin, actually. It's symmetrical. So
when you try to introduce the observable it's more natural to get this one, but when you want to work with,
it's more natural to get the other one. And the dualities is too hard and you will get an holomorphic if you
get this sigma, that's why we don't like it we prefer discrete holomorphic.
So it's where it comes from. And so with that you can try to do the same thing as in the self-avoiding walk.
You'll have the same problem. F is discrete -- so [inaudible] so first F is discrete holomorphic, exactly as
before. And that's basically the same proof.
But as before the discrete holomorphicity does not determine your function so you have the same problem.
>>: [inaudible] holomorphic function [inaudible].
>> Hugo Duminil-Copin: Yes. But ->>: In that situation. One in the graph and one in the [inaudible].
>> Hugo Duminil-Copin: Yeah, but actually they are -- so in the case of the random crystal model, which in
some way it appears a little bit more natural, you have a function which is directly defined on the middle
graph. So somehow in the primal and the dual graph already and it's not sufficiently 0. There's something
said about that. But there's missing information. And if anybody has an idea of where to find, to look for it,
very good. But somehow you've already had holomorphicity which is behind you already have discrete
holomorphicity. The conjecture, just to finish, so there's this diagram. So conjecture is that on this line
when you look at the curve here, so let's say that you put this point here and this point here on the
boundary, this curve should converge to an SLE, with a certain kappa that you can compute.
So that's not a conjecture for N equal 1 because here you have enough information. So spin one-half it's a
very nice spin when you think about it, because you know the argument of your function in this case. So
somehow you have twice more variables but there are real variables, not complex variables. Because you
know the argument. And so in this case you have enough information and you can start to work really hard
on that. If you are very smart, you get this observable converges and everything works fine. So you get
convergence.
>>: You get the precompactness.
>> Hugo Duminil-Copin: Yes. You get it for free -- not for free but from the fact that -- no, no, for free.
Erase everything. So -- no, you get it from the discrete holomorphicity which is an interesting fact. Which
is very natural as well. When you look at harmonic functions, they have naturally smoothness which is very
strong. You have very strong rigidity theorem for how many functions. And in this case you can prove
discrete holomorphicity. But somehow a stronger form of discrete holomorphicity, has the same rigidity
theorem. Converges for free. Meaning you get it from the observable I describe. So observable itself I'm
converging because I decided to.
>>: Not really free, try one get one free.
>> Hugo Duminil-Copin: Yes, yes that's exactly right. You really have to catch -- you get for free but here
you see you try to catch it and it's very hard. The proof is very, very smart.
Just for the conjecture, in this area you get this SLE. And that's not the same. So it's where I'm telling that
that's not the same critical phase. And there are two cases that are very famous for self avoiding walk. So
here I said I'll get SLE 8 served. That's what I said before. Here I get SLE 8. And I mean that's the natural
guess. You will say -- so as I said mu is not large enough in this case or X is too large so I get space filling
that's exactly SLE 8. For N equal 1, SLE 3 when you look at the criticizing model.
>>: No matter what your penalty is as long as it's small enough to get SLE 8 it doesn't change.
>> Hugo Duminil-Copin: Yes. But you will understand it in the similar example. I will not manage to
explain, but for 01 you get SLE 3 which corresponds to the critic [inaudible] model. If you get above it, you
look high extension in super critical phase what you get is percolation.
And somehow it's natural because you are the connection is not strong enough so at the scaling limit you
are just dependent. And no matter which super critical comparator you take, it's up to equal so [inaudible]
you take. Super critical.
>>: The same relation you see in the random cluster model.
>> Hugo Duminil-Copin: No. So in the random crystal model you don't see that, because you don't have -you really have critical value where you have something nonintegral, and above it and below it you get
super critical phase and sub critical phase so that's treasure limits. You don't get SLEs there. That's
what's spanning these models.
Okay. And so just to finish, a few slides. What you cannot do is to walk with the square lattice, because II
know I will get this question. And so no, no, no. Please, I'm begging you. Why is you can -- okay, let's
take even more latitude. I'll say much more things. To put fancy weights everywhere. And the thing that is
only two families of solution where you can find a parafermionic observable and one is trivial or at least it's
not interesting us in the scaling, I mean in our work and second one is hexagonal in model which is
included so it's not necessary easy to see that you can create hexagonal ON models in it but if you put the
right weights you'll get the exact hexagonal model.
So we fall back on the one we were working with. And these points where you can find parafermionic
observable they're the same where in physics you can find what we call the young Baxter equality. And
this is something very, very strong in physics. Basically physicists know how to handle a model only if they
know that, this equality.
I guess I will finish here. Thank you. [applause].
>>: If you want to compute the connectivity constant like in the similar lattice in 3D, didn't use anything ->> Hugo Duminil-Copin: Yes, but somehow the reason why there's a square root of 2 plus square root of 2
is the peer related to the exact what we call exact integralability of the model and this is very typical from
two dimensional lattices.
>>: [inaudible] did you find same lattice will tell you.
>> Hugo Duminil-Copin: I don't think so. I think it's too much related to this Young Baxter equality and
things like this. You never know, but this kind of miracle somehow, they hold in two dimensions.
>>: The miracle of 5 over 8 cancels, gives you --
>> Hugo Duminil-Copin: No, that's not the real miracle. It was a miracle that was at [inaudible] takes this
model and innerves it on this model and modify it again and for another model and finally he can describe
this model. But all this projection between models that's really something you know in two dimension. So
it's a very, very [inaudible] on exactly solvable model so other work by Baxter and everything.
>>: Is there an analog for the winding number of 3D.
>> Hugo Duminil-Copin: Even this. Even the interface, the notion of interface is not obvious in 3D. And
usually my bet will be that if you have an exact value for a model is that there is self-duality some where.
That's the case for random crystal model. If you find a nice value. Take decoration. If it's one-half for the
triangle lattice because have you a subduality property, and I will guess in most cases that's why. Perhaps
it's very well hided, but that's the idea. So here -- and self-duality is something typically in two dimension.
>> Yuval Peres: We will thank the speaker again.
[applause]