>> Yuval Peres: So welcome. We're delighted to have
two lectures today, and the first is by Naomi
Feldheim on sine changes of stationary Gaussian
processes. Please, Naomi.
>> Naomi Feldheim: Thanks. So thanks for the
invitation to visit Microsoft, and indeed that's
talking about that. It's joint work with my husband,
Ohad. And so this is about a simple question that is
first defined with our Gaussian stationary processes,
then address the question.
So definition. I'll begin with the space T to be
either R or Z, and that's my time space, and I'm
looking at random function F form T to R, which is
Gaussian stationary. So Gaussian means all finite
marginals have Gaussian distribution, so for any N
and for any points T1 till TN in my time space,
looking at AF1 till FTN. This has multi-normal
distribution, mean zero. Of course, in RN.
And secondly, stationary means that my function is
invariant with respect to shifts in my time space.
So for any S in my time space and, again, any points,
if I look at the shifted values by S, this shouldn't
change the distribution.
So Gaussian processes are very natural for noise, as
we know, we like to model things by Gaussianity, and
stationary is a very natural property because it
describes invariance with time of our noise. So this
is a well known family of noises.
I'm excited in particular, there is some newer
interest in this, in the context of the analytic
Gaussian stationary function, so if we add the
assumption that almost surely our function is
analytic in the case of real time space. In this
case the zeros of the function, they form a
well-defined point process because they are
separated, and they have repulsion. So it is one of
the maybe very few processes that we know to
construct and study that have repulsion. But today
we'll talk about just general Gaussian stationary
processes.
Okay. So we'll be addressing the following question.
Look at the probability, which I denote by HN,
probability that our random function stays positive
between time zero and N.
So no sine changes, or if you're looking at the
zeros, it means that there are no zeros at all
between zero and N, or maybe it's half the
probability that there are no zeros at all. That's
why it's called the whole probability, and that's why
there's an H here.
Random matrix people and people from mathematical
physics sometimes call it the gap probability, so it
all refers to the same thing.
So I guess this is a very natural question to look
at, and one would like to understand the symptotics
of this is as N grows. So let us look at several
simple examples. So example one, let's just take -let's do some examples on Z.
So by the way, when my time space is discrete, I'll
call it sometimes a sequence, and when my time space
is continuous, I might call it just a function. So
let us look at sequences. So draw it up here.
Example one, 1Z, just FNs which are IID. Then, of
course, HN is just half to the N.
So there is exponential behavior with N. One might
guess that if correlations are weak, then this
exponential behavior should persist. So here
correlations are as weak as you can get. There are
no correlations. So what happens if correlations are
zero, say, beyond some point, maybe there is also
exponential behavior. So not exactly. Let us look
at another simple example. Let's look at the vector
at the sequence YN, which is the difference of an IID
sequence, so XN minus six, XN plus one minus 6N where
XN are IID.
So now the whole probability is just the probability
that the first difference is positive and then the
second difference is positive and so forth, so it's
the probability that they are ordered, that the XIs
are ordered in a monotonic way, and while this
probability is roughly E to the minus N again. It's
one over F plus one factorial.
So this decays much faster than exponentially. Then
after you pose for it and think, well, can this
probability decay slower than exponentially? So the
answer to this is also yes, it can. So there is very
trivial example. It says just take the constant
sequence Zed N equals Zed zero. So they're all just
equal and you randomize only the first one. Then
certainly HN is just a half. It does not decay at
all.
Okay. One might say that this is maybe degenerate
example, but it is possible to construct also
nondegenerate examples where the probability does
decay but slower than exponentially.
So we want to ask what is responsible for the
exponential decay. Looking at some sequence, how can
I know if it behaves roughly like IID sequence in
this since. So question is when does HN behave
roughly like an exponential. So we keep this
question in mind, and in order to give some
characterization of this, I need some more
terminology from the world of Gaussian processes and
Gaussian stationary processes. So this would be our
next part.
So some terminology. Okay. I would like to look at
what's called the covariance function, which gives
the covariance between the value at zero and the
value at some point T. Of course, by stationarity,
this is the same as the covariance between the value
at S and the value at S plus T for any S. So this is
a function of one variable for my time space to R.
And now since it describes covariance, this is a
positive-definite function.
In case I work over R over the wheels, I'm going to
assume that it is also continuous. I'm going to work
only with continuous correlation functions, so assume
continuous. And now harmonic analysis says that if
we have a positive-definite function which is
continuous, then it is the Fourier transform of some
positive measure.
This is served by Bockner [phonetic] at least in the
real case, so there is some measure such that this is
the Fourier transform of the measure. So let us
explain this. First of all, this is an integral over
T star, the dual space. The dual space of Z is minus
five pi. The dual space of R is just another copy of
R. You might think of minus 5 pi or the circle.
That's the same for me.
And what is raw? So raw is a positive, so it's a
measure on T star. It's nonnegative, and it is
finite, and I will also assume some small regularity
condition on raw. I will assume that for -- there is
some small epsilon such that the momentum
polynomial -- the polynomial moment of epsilon
converges, so this integral converges. This, by the
way, is enough to ensure that our processes is almost
truly continuous and I just -- I need it for some
regularity conditions.
So we have our spectral measure. Now, the covariance
function, just looking at it, is a function Fourier
variable determines everything about the process.
Why is it so? Just because Gaussian processes are
always determined by finite -- by their marginals and
the marginals, finite marginals are Gaussian, and
therefore, they are determined by the covariance
matrix. That's all you have to know to answer any
probabilistic question. So RT determines everything
about the function, and raw is a one, one bijection
with it, so if you know raw, you know everything
about the process.
There is one more property that raw must obey in
order for everything to be defined, which is
symmetry, so let me add it here. So raw is symmetric
around zero. Just because I'm in order for R to be
real, raw must be symmetric. And so it turns out
that any measure that obeys all of these conditions
defines actually covariance function and defines some
function abstractly.
Okay. So this is quite abstract. Let me tell you a
way how to construct the function with a certain
spectral measure. I will not use this directly, but
I think it is -- it helps us understand the object,
so if you're given raw, you should take psi N, which
is an orthonormal basis of L2 raw of the space, and
then take weighted Fourier transform of this.
After you have this Fourier transform basis, you can
define a function, a random function F, random
sequence, which is sum AN phi NT where any N are IID
normal random variables.
So it is not very difficult to check that this random
sound indeed converges and converges to some Gaussian
process under the margin of the Gaussian to a
stationary process and the spectral measure is indeed
raw. So this is a quick way how to construct such
functions.
Okay. As I see some examples, so first if I take,
say, a sequence XN which is IID, what would be the
spectral measure? So in order to compute this, I
first need to compute R and the covariance, so R of N
is just delta 0N. The Fourier transform of this is
just an indicator or a constant function indicator of
minus 5 pi. So that's my measure in this case.
Good.
Now, this measure can be viewed also as part of the
real line. So now suppose I want a function whose
measure is the indicator of minus 5 pi. I regard it
as part of the real line. What function would that
be? So what I need to do is begin with raw, which is
the indicator, and then construct R, which is now the
real Fourier transform of this. So this is a sync
function. So this would be the covariance, the
correlation between two points in distance T of such
a function.
And now if I really want to understand what it is, I
can follow the recipe here of constructing it and
construct it as a random sound. So let me tell you
the answer, write it here. So the answer is FT being
sum AN sine pi T minus N divide by T minus N where AN
are again IID [indiscernible].
So what's going on here? You have a sync function.
So that's function which looks like this. It
vanishes at all other integers, and these least one
and zero, and you shift it to any integer, and at
every integer you take your random normal amplitude.
Then you sum this all up, different things, and you
get your random function.
So in this form here you can see that FN for a
certain integer is exactly AN. So this is indeed an
IID sequence. So this function is rather interesting
because it is some kind of real stationary
interpolation of the IID sequence on the lattice.
Right. So this is the function it has spectral
measure of the indicator.
>>: In your construction of the Gaussian process
from the measure?
>> Naomi Feldheim:
Yeah.
>>: I guess is there any easy way to see that F of T
is stationary?
>> Naomi Feldheim: I would compute the covariance
and just it's for, yeah, probably part of the -yeah, and by the way, I did not say, but you probably
have to worry about the realness of phi N because
this has complex numbers, but it is always possible
to do this construction so the phi would be real.
Just choose psi N to be odd and even functions, but
yeah. Right.
So that was our first example. Second example. The
second example that we had before, so YN, the
sequence which is XN minus XN, XN plus one minus XN.
Differences of an IID sequence.
So what's the spectral measure here? So first we
have to compute the covariance function. So at zero
it's two. That's the variance at the point. At
distance one or minus one it's -- well, just minus
one. And from there on it's just zeros.
And now we have to Fourier transform this to get your
measure, and this turns out to be two, one minus
cosine lambda. So that's like the density, which
looks like this. Okay. So this is our second
example. Not much to say here.
Third example which might -- we might keep in mind,
similar to number three there, but write something a
bit more general. If we have raw, which is an atomic
measure, so sum of sum weights, and then some delta
minus lambda N and delta at lambda N. I must keep it
symmetric because I want the Fourier transform to be
gree [phonetic].
So suppose I have something like this. Then by the
construction we have, we Fourier transform we get
that our function or sequence is sum with weights of
cosines and sines. So it's where AN and B now are
both IID norms. So its sum of random waves we've
given frequencies, lambda N and some given weights.
Now, taking delta measure just at zero will give you
the constant function, the examples that we had
before. So this is some generalization. Okay. So I
think this gives us some understanding between the
spectral measure and the process, itself. So at last
we are ready to state the result.
So on there, so say that the Feldheim theorem. I
want to indeed give some condition for our process to
have exponentially decay. I'll give separately the
conditions for upper bound and lower bound. Even so,
they're quite similar. So first let's just give the
condition for upper bound.
So suppose there is some small alpha and some
positive number M, and positive number big M such
that if I compare my spectral measure to the Lebesgue
measure and it is between those two constants, for
any interval inside a small neighborhood of the
origin, then there is an upper bound.
So, of course, I mean for large enough N, I can
borrow this from above my exponential. Lower bound
very similar, but I need only the bound from below.
So I compare my measure to Lebesgue measure. I want
it to be bounded from below for any I in some small
neighborhood of the origin, then it is bounded from
below.
So a few remarks here. First, this comparison to
Lebesgue measure, I think the intuitive way to
understand it is that Lebesgue measure or the
indicator we said is somehow related to an IID
sequence. So the indicator of minus pi pi is exactly
the IID sequence on the integers, and so an indicator
of some smaller interval, the spectral measure of
such a thing would be independent on some different
lattice, a more space lattice. And so we are
comparing it either from both above and below or just
from below.
Second remark is that indeed for the lower bound we
need only the lower bound on the comparison. Here we
need both, but maybe it's just part of our method and
not truth. Maybe the truth is that you need only
comparison from above to give this above.
Actually, we can refine this a bit. We can ask for
an upper bound near zero in some neighborhood of the
origin and the lower bound somewhere else. Just give
me it anyplace where this lower bound holds and not
necessarily in minus alpha alpha. So the proof still
works. So this hints that the lower -- this can be
improved. Moreover -- yeah.
>>:
The lower bound is somewhere else?
>> Naomi Feldheim:
means that ->>:
So lower bound somewhere else
The interval or --
>> Naomi Feldheim: Yeah, that this bound is the same
and this bound for any I in some other interval may
be very far from zero. So -- yeah. I need
comparison to Lebesgue measure somewhere from below,
but not necessarily in the same interval. Right.
And the second thing I want to remark that constants
here are very different and it would be interesting
to understand if there is really exponential behavior
asymptotically.
So now let me tell you a bit about the proof in the
few minutes I have. And how does this show up. Oh,
by the way, before as I am erasing the examples, you
can see that in the examples we had, number two,
which was faster than exponential, indeed has
spectral measure that does not obey this bound from
below. It is zero, near zero. So it obeys the upper
bound but not the lower bound. It's faster.
Okay.
So idea.
So the main observation is that if
you take a spectral measure which is actually the sum
of two positive finite measures raw one and two, then
the corresponding functions, the corresponding
function F can be written as the sum of two
independent copies, F1, and F2. F1 has spectral
measure raw 1 and F2 has spectral measure raw two.
So this is very easy to see this linearity from the
definitions, but it is very useful here. How so?
Our conditions, so here both conditions our lower
bound gives that you can write raw as maybe M times
indicator of minus alpha alpha plus something.
Now, the indicator corresponds to something that has
independence on the lattice. So our function F is
sum of F1 plus an independent copy of F2 where F1 -so if we suppose -- so F1 has this spectral measure,
so if we suppose alpha, for example, is pi over K for
some integer K, then F1KN is IID.
Then we want to use this IID sequence in order to
bound from below and from above. So let me assume
just for simplicity that K equal one, so just F1 on
the lattice is IID. Okay. So for the upper bound.
So both for function and for sequence. So this was
my original whole probability. I can always bound it
by some lattice or some sublattice, so this is the
sequel then being positive for every N on the lattice
I chose, so in this case I choose just the integers
one, two to the N. Oh, so that's not F1. Sorry.
That's F. Still not F1. So my original function on
the lattice.
Now, how would I bound this? So as we recall, FN is
just F1 plus F2 where F1's are actually just IIDs and
F2 is some other Gaussian sequence.
So I want to spread it into sum of two events, sum of
two probabilities. So the first probability is just
F1 being bigger than minus F2N for every N, but given
that the sum of F2N is small, smaller than, say, one.
The reverse. Sorry.
Okay. So the rest I will just bound by the
probability that this did not happen. So if this did
not happen, the sum of F2Ns is bigger than one.
Now I want to show that they are both square. So
this is the event I want, but given something that
happened to F2s, and this is just probability that
this did not happen. I want to show that each is
exponential. So for the first parts for -- so
that's, say, probability, the first probability you
just have -- I will write it. You just have an IID
sequence being bigger than some number minus CN where
CNs you can think of as given numbers that their
average is less than one. That's what you want to
compute to bound from above for any N.
So of course, this is just multiplication of the
probabilities that -- of the standard random variable
to be bigger than minus CN, so that's three of CN.
And now we use the fees, the local K function and
also a monotonic function in order to bound this by E
to the N fee of one, something like that.
So this is just some Gaussian computations. So
that's for the first part. For the second part, this
probability, so you notice that what you have here is
also a Gaussian random variable. It is the sum of
Gaussian random variables, and so you just have to
compute its variance and understand this bound. So
this is again a Gaussian bound where the hint is that
the variance of the sum is roughly the spectral
measure at zero. So this is by doing some harmonic
analysis. Okay.
Lastly, about the lower bound. So lower bound. I
rewrite this probability I'm interested in, so this
is the probability, and now I want to again use this
decomposition into F1 plus F2, in order to -- yeah.
>>:
So in part two is where --
>> Naomi Feldheim:
>>:
So this --
-- use the -- part of two you're --
>> Naomi Feldheim: This is -- so, yeah. So there is
a sum of two probabilities. I want to show that each
is exponential, so this will be ->>: But in part one you just use the lower bound on
the --
>> Naomi Feldheim:
here, so --
But maybe there is a one over
>>: But the upper bound spectral measure is used in
part two?
>> Naomi Feldheim: Yeah, it's used in part two.
It's used in part two. I'm saying that this variance
is roughly what happens with the spectral measure at
zero and I want to bound this from above, exactly.
So this is roughly spectral measure at zero, which is
less than some constant big M divide by N. Exactly.
Then once you have the variance, it's just a Gaussian
computation. Right.
Now for the lower bound. So I want to construct some
event using my decomposition that will make my
function positive on the whole interval. So for
instance, I can ask for F1 to be bigger than one on
the interval, and simultaneously F2 should be less
than a half, say.
>>:
That's a derivative of a spectral [inaudible]?
>> Naomi Feldheim:
>>:
Right.
Yeah.
[inaudible].
>> Naomi Feldheim: Yeah. Just M. In case of this
condition, actually the spectral measure is
absolutely continuous in this small neighborhood, so
there is meaning for this.
Okay. So this is just the product of two
probabilities because F1 and F2 are independent, and
now I have to study each of them separately. Now,
the first -- the second one has been studied. It's
just probability that some Gaussian process is in a
small ball as in ran use half for a long time, and so
it's called the small ball probability. There are
many bounds on that. So we have to do some work. We
can apply some bound by telegrand [phonetic] and
we're able to bound this by an exponential bound
below. Actually, there are bounds from above.
As for the first one, what's written here is a very
concrete question. We have the concrete function F1
of T, which is actually written here. That's F1 of T
because it has spectral measure of an indicator, and
I want to understand what's the probability that this
is bigger than one for a long time.
And so this was done or similar work was done by a
few guys a few years ago, so it's Antazena Bachme,
Malzo and Torson [phonetic], so actually we have some
improvement on their proof, but the main idea is to
look at this special series and to construct an event
on the series.
So for instance, the event will be more or less
asking N of the LANs to be bigger than two and show
that the rest of the things do not ruin your event
and that this event is exponential and yields this.
>>:
To lower bound --
>> Naomi Feldheim:
So there is some construction.
>>: -- small [indiscernible] for F2 you could just
use the fact that you can borrow a little bit of the
measure from -- I mean, from the upper bound.
>> Naomi Feldheim:
>>:
For this, you mean?
Yeah.
>> Naomi Feldheim: The small probability? Yes. So
there is some raw truth that is left and there is the
function F2, yeah, if you apply -- you have to apply
some small [indiscernible]. But -- so this is more
or less the idea.
So I'm at end just with some interesting open
questions here, so I think this is just the beginning
of understanding asymptotics, as you see that we have
only bound but they are very far from each other and
be interesting to understand the real asymptotics.
Also would be interesting to understand what happens
if the spectral measure blows up at zero or if the
spectral measure is indeed vanishes at zero, or even
in some interval around zero. How does this affect
the whole probability?
And if you are not accustomed to the language of
spectral measures, I'll rephrase it. This means that
the integral of LX, meaning the correlation function
is infinity and this means that the correlation
function integrated is zero. So in a sense, we show
that if it's finite but not -- but somewhere in
between, then the behavior should be exponential, but
what exactly would happen if it goes to infinity or
to zero. Our guess is that it should depend on the
weight that it goes to infinity or zero, but we don't
have any concrete work on that.
Okay.
So that's all.
Thank you very much.
[applause]
>> Ohad Feldheim Ohad Feldheim:
>> Yuval Peres: Any other questions?
>>: So seems like the obstruction imposed by example
number two does not exist if you assume that the
correlations are positive. Do you have any idea how
to apply the fact that correlations are nonnegative
and ->> Naomi Feldheim: Yes, that's a good point. So if
correlations are positive, indeed, it's very easy to
show that there is existence of the limit. So if all
correlations are positive, so there's existence of
the limit log of this probability divide by N,
because if all correlations are positive, you can
show that log of this probability are -- there's some
sub [indiscernible] and ->>:
[indiscernible].
>> Naomi Feldheim: Yeah, and there is -- so exactly.
So existence of limit is a one line proof, but still
it would be interesting also to say what is this
limit and it does not apply to correlations like the
sine [[indiscernible] correlations.
>>: That limit is positive, can you infer anything
about the process?
>> Naomi Feldheim:
>>:
Again?
Suppose it's positive.
If limits is -Can you infer anything
about the process?
>> Naomi Feldheim: I guess that for many processes
it would be positive, so ->>:
Yeah.
It rules out certain processes.
>> Naomi Feldheim:
Oh.
>>: If you say anything causes about the process
that can be ruled out by this [inaudible].
>> Naomi Feldheim: Yes, so again, my guess would be
that the information you get is precisely that you do
not have either blowup or zero in -- at zero. So I
guess that it would be exactly in correspondence with
saying that there was finiteness of the spectral
measure at zero or finite integration of the
correlations on the whole line, I guess, but I think
we do not know something.
>>: My compliments to [indiscernible] to say
something [indiscernible] Direction.
>> Naomi Feldheim:
happens.
>> Yuval Peres:
Yes.
Okay.
>> Naomi Feldheim:
Yes.
I agree.
Yeah.
That
Thank Naomi, once again.
Thank you.
[applause]
>> Yuval Peres: So for the second part Ohad Feldheim
will tell us about 3-coloring the discrete torus.
>> Ohad Feldheim: Yes. So thanks Yuval, for the
invitation and the hospitality and, well, it is going
to be completely different now with the presentation
and not a blackboard talk, and the topic is
completely different, so I hope we'll make this
context switch easily.
And I'll tell you today about 3-coloring discrete
torus and the there is also a subtitle for people
interested in statistical mechanics which said the
rigidity of zero temperature 3-states
anti-ferromagnetic Potts model, and I will explain it
as we progress. And this is all joint work with the
Ron Peled, also from Tel Aviv University.
So the object we're going to deal with is a proper
3-coloring of the discrete lattice ZD. So we take
the lattice with nearest neighbor adjacency, so for
example, in two dimensions every vertex is adjacent
to four other neighbors, and we would like to take
uniformly chosen 3-coloring, prophecy coloring of
this lattice, but of course, since it's infinite, we
can't. So we must limit ourselves to some boundary
conditions.
There are two classical ways of doing so. So one
way's what's called 0-boundary conditions, so we take
some bounded set, and this set, we choose it to have
an even boundary. So what do I mean? The lattice in
every dimension is a bipartite graph, so we want all
the vertices on the boundary to be of the same
bipartition class, let's say from now on the even,
even by position class.
We color all of those vertices by the color zero, and
then we pick uniformly chosen 3-coloring. And the
periodic boundary condition is when we will actually
just coloring the torus, so we will have N to denote
the side of the torus in each direction. And what we
do is we look only on the set from zero to N minus
one in each coordinate and we wrap around, so this
one is also adjacent to this zero. So this is a
discrete torus.
We're looking with 3-colorings of this. The benefit
of this boundary condition is that it didn't force
many vertices to be colored by particular color. So
in a way it might be more natural boundary condition
to reflect our thought of the infinite, of the
infinite lattice.
So we will be interested in uniformly chosen
purposely colorings of those objects, and we're only
going to discuss high dimensions. So imagine that
the lattice is of dimension, say, 10,000, and what my
pictures depict is just a two-dimensional section of
it. Otherwise, things would not be -- I mean, they
would not be representing it correctly.
And, well, we could say that there is a natural
combinatorial motivation to look on this object
because while the lattice is natural graph and the
3-colorings are natural here because there are only
two 2-colorings of those sets, so the coloring is the
minimum number of colors where this would make sense.
But there's also some motivation from statistical
mechanics, so there is a phenomenon called
anti-ferromagnetism, and in the narrow sense it
reflect the fact that on a solid matter the electrons
of adjacent items or molecules tend to take opposing
spins, but more generally it is a situation wherein a
solid matter in lattice, a adjacent sites tend to
have a different state.
This is modeled by the q-states anti-ferromagnetic
Potts model, which is the model of q-coloring of the
lattice where every configuration is chosen with the
proportion which is inverse exponential of the number
of forbidden edges. Those are edges that connect the
color to the same -- two vertices colored by the same
color.
And when we -- so there is a parameter here bitter
which says how much we dislike those edges, and when
bitter is taken to infinity, which corresponds to
temperature being taken to zero, then we are left
with just q-colorings of the discrete torus.
So this is how those two problems relate and, well,
we'll get back to this idea mainly in our open
questions because we would very much like to
generalize what I'll tell you about today on the
3-colorings of discrete torus to the real Potts
model, and with bitter, which is not infinity.
So what could be asked about this model? So I think
very natural question is what is the relative
frequencies of the color in a typical coloring. So
for example, is it one set, one set, one set. So I
pick a coloring at random. Will I see one of set of
the vertices colored by zero, one set colored by one,
one set colored by two.
And we can ask more sophisticated questions, such as
does the typical coloring follow some pattern. So if
we see a coloring, does it look like a random one.
And in particular, there's a question which
physicists think interesting, and this is there is a
long-range correlation or what is the nature of
long-range correlations, so what does this mean?
Imagine that our torus is huge. So I said that D is
large, but imagine that N is much larger now, and
with we've examined it in a small region, and we want
to know what can we tell from its behavior in this
region very far away from them.
And if all centralized observables on those two
regions, their correlation in the case exponentially
with their distance, we would say that there is no
long-range correlation.
Now, after studying this object, one can make the
following educated guess. So maybe a coloring looks
roughly like this. So what is this coloring? All
the even sites here except for one are colored by
zero, and the odd lattice points are colored
uniformly by -- uniformly independently by zero, by
one and two.
So I guess that coloring makes sense. So indeed,
we've limited ourselves very much in respect to the
even sublattice, but now on the old sublattice we
have complete freedom. So if all the vertices were
colored -- all the even vertices were colored by
zero, we would have a two to the number of vertices
over two configurations.
Now, of course, it pays to have some vertices on the
even sublattice not colored by zero because, well,
for example, the first one that we had gives us
another degree of freedom equal to half the number of
vertices and all that we lose is that it's forced its
neighbors, but we can expect that as the dimension
becomes higher, the percentage of those vertices will
decay dramatically.
So is this really how it looks? So the conjecture is
that on two dimension, not at all. So my pictures
here, if you thought of them as two dimensions, it's
not the case. But for all dimensions higher than
two -- and higher than two because there are some
senses of fractional dimensions and in those senses
really you can see the two should be the critical
one, the critical dimension. So for all dimensions
higher than two there is some rigidity. So there is
some behavior like this that on the one sublattice
the vertices tend to take the same color.
Okay. So this conjecture has been established for
0-boundary conditions in high dimensions. So what
Peled shows in 2010 that the typical 3-coloring with
0-boundary conditions has nearly all the even
vertices taking the color 0. So formally we say that
if d is large enough and we have uniformly chosen
3-coloring with 0-boundary conditions has the
expected number of even vertices not colored by zero
out of all even vertices, decays roughly
exponentially with the dimension. Yes. And whether
this log factor is really necessary is open.
So this result didn't work for periodic boundary
conditions, and the reasons will become clearer as I
advance along this talk. And of course, it's opening
low dimensions, and as I said, it's widely open and
when we have no clue how to advance towards it, it's
not clear that combinatorics is the right approach to
get to low dimensions. Could be that one should do
completely different things.
Now, towards showing this on bounded tori, we have
the following remarkable result by Galvin and
Engbers. So what they show in our context is that if
you fix N and you take high enough dimension, so this
is not what we wanted. We wanted to take large D but
to take N to infinity. Then the result holds.
Typical 3-coloring with periodic boundary conditions
is nearly constant on either even/odd sublattice. So
of course, this is the counterpart of the result I've
shown on the lattice slide because, well, when we're
coloring with periodic boundary conditions, we have
no bias towards the even sublattice or the odd
sublattice or towards one of the colors. So this is
what we would expect.
So why is this so remarkable? Mainly because it
works also for q-colorings, and there are very few
results that work for general q-colorings, and as
you'll see when I'll explain a bit about the proof of
a 3-coloring, you will see that we really use
something special about them. It's really not easy
to generalize it.
But it is not our purpose. So what I will talk about
today is the parallel phenomenon for periodic
boundary conditions. So what we show is that a
typical 3-coloring with periodic boundary conditions
is nearly constant in either the even or the odd
sublattice, so this was our goal.
And formally, so what is the formal counterpart if d
is large enough and we uniformly choose a periodic
boundary conditions 3-coloring, and now we're looking
on all colors and both sublattices and we look on
where there is a color that is so dominant that there
is as few as possible vertices not colored by this
color. So the number of vertices not colored by this
color of all the vertices on this sublattice will
decay exponentially with the dimension, roughly.
So this is more or less natural. Natural
generalization.
Okay. Observe that when I said that there is even
and odd sublattices, I have implicitly said that N
must be even. This is not a weakness of the proof.
It's how things really are. Odd tori do not resemble
the entire lattice in the sense that they are not
bipartite, and so conjecture like this that I've
presented here cannot hold for them, and something
else happens. We have a very good idea about what
happens in odd tori. Nothing of this is written, but
maybe it will help for other results. We're not sure
it's interesting enough on its own, on its own
accord.
And the proof that I'll tell you about today
introduces topological techniques to the problem, so
it's not that we're going to do again what was done
for the periodic boundary conditions. Actually,
we're going to use that result, but it's going to be
not so simple to transfer it to the torus, and this
is what I will explain, how we do this
transformation.
So what I'll tell you about now is what is the real
challenge. I mean, well, where is the effort in
doing this. So in order to tell you anything about
proofs in 3-colorings, I absolutely must tell you
about homomorphism height functions. Those are the
objects that allowed us to understand so much better
3-colorings that has the numbers of close.
So what are homomorphism height functions? Those are
simply homomorphisms form a graph G, in our case the
ZD2Z, and this means just a mapping from this graph
ZD2Z that satisfies that adjacent vertices have their
images deferred by exactly one.
And you should think about it like a topographical
map on our set. So this reflects the height and the
requirement of difference one represents some kind of
continuity. And like we did for 3-colorings, we can
pick a uniformly chosen such height function with
certain boundary conditions. So for 0-boundary
conditions everything is just the same. Pick a set
with even boundary, color its boundary, so pick the
height of its boundary vertices to be zero, and then
pick the rest uniformly along such -- among such a
height functions.
For periodic boundary conditions we must add at least
one vertex whose height we fix, because if we will
not do so, we will have a degree of freedom of
lifting the entire height function by any constant.
So we pick one such vertex and fix it and we call
this usually pointed periodic boundary condition
uniformly chosen height function.
So why do I tell you about this object? So actually,
on ZD, this object in 3-colorings are just the same
object. This is the same. So what do I mean? There
is a bijection between those two objects, and it goes
as follows: So if I give you a height function and I
take a modulo three of it, you will get a proper
coloring.
Well, of course, if two vertices their height differs
by one, then modulus three, their height will also
differ by one, so they will not be -- will not get
the same color.
To see that this is a bijection, we have to explain
how to calculate the height of a particular vertex,
so how do we do it? So we can think of the coloring
as encoding height differences on this lattice. So
from zero to one we go up by one, and from one to two
we go up by one, from two to zero we go up by one,
and in the other direction we go down.
And now to calculate the height of a vertex, what I
can do is to integrate those height differences along
a path. So for example, if I want to know what was
the height of this vertex, I am saying, okay, from
zero to one I go up by one, so it's one. From one to
two I go up by one, so it's two. From two to zero I
go up by one, so it's three. And now from zero to
two I go down by one, so it's two again. And from
two to zero I go up by one and it's supposed to be
three, and indeed, this vertex is called by height -is the height [indiscernible].
So how can I convince you this is really a mapping?
I should show you that on different passes you get
the same height. So this is the same as showing that
the longest cycle you don't accumulate any height.
To show this, it's enough to show that on the cycles
which generate all the cycles, you don't get any
height.
In all ZD, in all -- no matter what is the dimension,
the cycles are generated from small four cycles. And
now to see that you can't gain height on a four
cycle, you can, for example, say that it is even, the
height that you gain, because you go over four
vertices, and the resulting height might be the same
modulo three as the starting height, so it would be
the same modulo six, and so in four steps moving from
one value modulo six to another value modulo six, so
it must be the same value.
Okay. Now, this bijection of -- sorry -- does not
extend to the torus. Well, actually, this is an
example, right? So this coloring matches only this
height function. And this seven, say, and this zero,
they do not fulfill the requirement to be a height
function.
And this is main obstruction.
Why so?
Because we
know very much about height functions. So in that
paper of Peled from 2010, he shows that no matter if
it's on the lattice or on the torus, height functions
have what we want. They tend to be a nearly constant
on either even or odd sublattice.
And for 3-colorings, the way that the result is
obtained is just saying, okay, these are the same
objects, so they also satisfy this property. And the
challenge here is that because we don't have the
bijection, and we wish to transfer this to -- the
result to 3-colorings, then we would have to do
something else. We will have to somehow go around
this problem that we don't have the bijection.
Okay. So in order to try and understand 3-colorings
on the torus, now that we don't have bijection with
height functions, let's see what we can put them in
bijection with.
So we take the coloring on the torus, and one thing
that we can do with it is to pull it back to the
plane. All right? So the plane is ZD, the lattice
ZD is universal color of this torus, and so every
function here could be pulled here. And what we do
is just put copies of the same coloring all over the
plane.
Now, on the plane we have bijection, so we can apply
it and get the height function. And we can ask
ourselves, what kind of height functions will we get?
So they will not be periodic like the coloring, not
necessarily, but at least because the coloring is
periodic, then the height differences would be
periodic. So if here we have some height difference,
then in the next copy we'll have the same height
difference.
So functions whose slope, whose gradient is periodic,
we call them quasi-periodic functions, and they are
defined by the heights on one copy, on one
quasi-period, and by how much height do you
accumulate when you wrap around the torus in each
standard basis direction. Okay?
So here, for example, you have a slope six when you
go along E1, and you have slope zero when you go
along E2. So these are the slopes that they write
here on this site.
And it is not so hard to convince ourselves that,
first of all, we'll always get slopes which are zero
modulo six. The argument is similar to what I
explained before. A vertex in its translation must
be the same modulo three in height and must be the
same modulo two, and that every such quasi-periodic
height function defines a periodic 3-coloring.
So we have this bijection. Now, if we were lucky,
let's say, and our coloring turn out to be a
quasi-periodic function with slope zero, then it's
excellent. We can pull it back to the -- we can push
it back to the torus and we will have exactly the
function which corresponds to this coloring.
So in fact, 3-coloring, flat 3-colorings correspond
homomorphism height functions. So what is the plan
halfly [phonetic]? We have 3-colorings. We want
to -- we know that they're the same as periodic
3-colorings on ZD, which are the same as
quasi-periodic functions on ZD, and we know that the
periodic functions on ZD are the same as the
homomorphism height function on the torus, and for
them we have the result that we want.
So the entire challenge, the entire purpose of this
work is to show that those two things are roughly the
same, that is, that if we pick a quasi-periodic
function on the ZD, usually it will be periodic.
And how are we going to do this? So first abstractly
we know that flat 3-colorings are in one-to-one
correspondence with height functions on the torus.
Now we have those slope 3-colorings. What you would
like to do is we'd like to flatten them. We are
going to map them to flat 3-colorings and then use
this mapping to get height functions on the torus.
Now, we would like to show that, first of all, our
mapping is few to one. So not -- there are not many
sources to a particular height function; and second,
that the resulting height functions are very unusual,
and because they are very unusual and we know that
this is a small subset and this is few to one, then
small times few is not so much, and we are maintain
with the notion that most height functions are flat
3-coloring -- most 3-colorings are flat and we can
grow results from height functions to flat
3-colorings. This is the entire purpose of our work.
Now, let's be more precise about what we do. So
let's look only on quasi-periodic functions with a
particular slope M. So we're restricting ourselves
to one slope.
>>:
Is it a vector or --
>> Ohad Feldheim: It's a vector. It's a vector. It
has in every direction different. What we're going
to do is we're going to map all of those functions
with this particular slope to functions with zero
slope, and this we're going to do in a one-to-one
way. Yes.
So given the slope, the mapping is one to one, and
this is the sense in which I said it's few to one.
One image can have -- if it has different sources,
they must have different slopes.
And now after doing this, we will find in the image a
very long level set. So think of level set as in -so this is an approximation of a continuous function,
so think of it as a level set in a continuous
function, and if time will permit, we'll also give a
definition later on.
And the fact that long level sets are rare was
exactly the instrument that was used in this Peled
2010 paper to show that things are flat. He showed
that long level sets are extremely uncommon. This
was the main instrument.
And this would tell us that those images are very
rare. And so we would know that if the image of
those psi M is just a small subset of each. Psi M is
just a small sum. So we would be left only with the
periodic function, or mainly with the periodic
function.
So this is the grand scheme, okay? So what we have
done so far is that our challenge is taking
quasi-periodic functions and mapping them one-to-one
to periodic functions. And this is the call of this
work. So how can we do this?
>>: Kind of
it cannot be
for every -every N, but
could get --
like decay with slope, right? I mean,
that, right? I mean, you showed that
you know, you get some small subset for
presumably if you summed over M, you
>> Ohad Feldheim: No, so actually we'll show that
all the images are very, very rare. So the entire
set of images for all Ms together, its percentage out
of colorings will be just a -- so it is something
like some constant to minus N to the power D minus
one. So this is very, very, very, very rare. So
even though we have a polynomial number of slopes,
it's insignificant in comparison to how rare those
sets are.
Okay. So how do we flatten a function? Well,
fortunately, in one dimension it's really easy. What
you do is you say I take this function, I know that
it gains a certain slope. In this case it gains
slope six. Let's find the first time that it
gains -- that it gained a three height. This is this
green line. First time that we gain three.
Now, let's reflect the slope of the function from
this point and until we finish one period, okay? So
if we gain that to here half the slope and we were
supposed to gain another half here, so instead we
will lose this half and we will come up with
something flat, and if we do this periodically, we
will get a flat function. So we reflect immediately
after height M1 over 2, but what can we do if we have
several Ms? So it doesn't generalize very easily.
And to see how it generalizes, we must know something
about how multi-valid functions look, how do they -continuous functions look on the torus. And topology
will give us this answer. So this is how a typical
height function looks from very far away, let's say,
on the torus. I mean, this is how such functions
look. And this is a continuous on the loop, of
course.
So the black lines here, they are level lines, okay?
And you can see that the height when they heat
this -- the X-axis, and they have a very interesting
structure. So first of all, there are two kinds of
them. One kind is those that we call trivial. One
way to think about those is that they are the level
lines that when we project them to the torus, they
split it in two parts. Okay?
So, and the other way to think about
you start inside one of them and you
wrap around, you will definitely end
it, a translation of it. Okay? And
going to be important in the proof.
them is that if
make an entire
inside a copy of
this notion is
And there are nontrivial ones that when you start
from here and you start traveling around, you might
cross them just one. You'll never meet them, a copy,
I mean, you will not meet another copy until you get
to the other side.
So those nontrivial level contours, at least in the
continuous setting we would like to think of them as
some kind of co-dimension one manifolds. They are
very important in understanding how to do this
reflection, so they have the following structure.
They are some sort of interlaced. So you see this is
one copy of such a level line. This is another
translation of it. If I translate this level line by
one and by N in this direction, I'll see here a copy.
If I translate by another one, I'll see another copy.
And the same goes for another level line. So if this
is the next level line of height six, then there will
be here another level line which is just of the same
height difference, and they have this ordered
structure which we -- I'm not sure that I will get to
describe it more accurately, but it is the parallel
of what happens one dimension. So you can think of
it that in this direction things look very similar to
how the one dimension and case looked.
So what is going to be my tactics? I'm going to,
first of all, mark for myself only increasing level
lines. So first I go to height three, then go to
height six. I know those three and six. I already
reached height six. Then I reach height nine and
height 12, et cetera. And only the nontrivial ones.
So I don't mark this and I don't mark this.
Now I will have some way of finding the first level
line which is nontrivial. Let's say it's this one
for the moment. And I can now find also its
translation. So its translation is this level line.
And I can go back and look where did I gain half the
height, so I go from this 15 and half the height
between 15 and three is nine, so it's this line.
And I'm going to reflect everything that goes on
between those two lines. So it's going to be this
region. And so I will get a flat function. But how
will I be able to recover it? So even after the
flattening, if I'll take these three, go to the right
to the next translation, then go back until I see a
height difference of six, which I can calculate
because I know the slope, I will know that between
this line and this line is the domain that I'm going
to have to flip again to get back to my original
function.
Okay. So this is roughly the idea. This idea is
very topological. So if I want to do this for
continuous functions, I know to do it only for
polynomials or for very nice functions. And I will
have to take those notions of what is nice function,
what is a -- what are what are those co-dimension one
manifolds and somehow formalize it in the discrete
terms.
This is going to be the main challenge in taking this
topological concept and creating a discrete
algorithm.
So first of all, let's discuss a bit about what are
those level lines. So far they were very abstract.
Let's make them a bit more concrete.
So originally in the work of Peled, level lines were
defined in the following way. So he defined, first
of all, a sublevel set of a vertex, say this is
vertex V and the sublevel set of a certain height,
height K, is just taking the connected component of
this vertex in the graph without the vertices of this
height. So I'm traveling without crossing vertices
of height one.
This was the notion of a sublevel set, and then the
level set would be the contours bounding this set.
This was the original definition. So apparently this
is not good enough for our purposes, so we had to
improve upon it a bit. And the reason is that this
notion of a level lines that I described to you of
those co-dimension one manifolds, it turns out that
the corresponding term in the discrete settings are
sets such that they're connected and the complements
are connected.
If a set is connected and its complement is
connected, then we call the boundary separating them
a co-dimension one discrete manifold.
So in order to turn this set which didn't have this
property into a set whose boundary satisfies this,
what we do, instead of defining a level set for one
vertex, we define a level set for two vertices. So
we want to look at the level set between this vertex
and this vertex. This zero and this three of height
two.
And what we do, we add to this. So we look at the
connected comparance [phonetic] of these three, and
we add to the connected component of zero everything
that is not in this connected component. So we pick
this blue regions and add them into the set, and so
now we will have a set which is connected and also
its complement is connected. So this is the
definition of what is a sublevel set separating two
vertices in our context.
And the boundary of this is what we call a level set,
and now, that may be the most interesting topological
object that we have in our proof, the important is
COM is the following: If we have a set of such level
sets that they are co-connected, that is, that they
are connected and their complement is connected, and
that we know that their translates do not intersect,
so we can prove this also for our level sets, then
they will satisfy one of three of three properties.
So it goes like this. We look at such a level set
and we look at all of its translations in the world.
This is the set, the set -- do I have it here? No, I
didn't write its name, but all the U plus T where T
is some translation from the set in time ZD.
have some set and all of its translations.
So we
And now this set will -- all of those translations
will satisfy one of three states. Either they are
all disjoint, or all of the complements are disjoint,
or they are all ordered, totally ordered by
inclusion.
This idea that they are all totally ordered by
inclusion is the counterpart of this topological
observation that we have those translations of a
level set containing one another. So those are
nontrivial level sets.
So since the description was a bit hard, let's look
at some pictures. So here are -- three level sets
and their translation. So one is those blue level
sets. You can see that they are the sets that
separate this minus one and this zero. And indeed,
this reflects like the peak of a mountain. It is a
trivial level set. And you can see that when you
translate it -- sorry -- when you translate it, you
get disjoint copies.
The counterpart of those are those yellow sets where
it's like we're trying to get from a vertex outside
the peak of a mountain to the mountain, itself. So
if those were valleys -- maybe this is a better
description -- those are valleys and those are
mountains, and they satisfy that their translations
are the complements of their translation of disjoint.
So the complement of the set here is just this
mountain peak, and the translation of it is this one,
is this next mountain peak.
And finally, we have those red level sets, and for
those red level sets you can see that they are really
ordered. So this is one level set, and the next
translation is this one, and all of the translations
will be ordered.
Okay. So I think this is more or less enough about
those objects. I can just tell you that what we
should do now, if we would posit, is to show that if
we have a slope, we must have some nontrivial level
set, and then we should explain how to recover -recover our reflections, how to define it, and those
things are very difficult technically. I mean, if I
would really like to give you a concrete proof,
unfortunately, this turned out to be -- so only the
technical elements of showing the topological
counterparts took more than 20 pictures.
Okay. So let me conclude with some open problems.
So odd tori, we know very well what's going on. It's
not really -- I mean, I wouldn't bet my right arm on
it, but it seems very plausible what is the structure
that we're going to see, and I think it would be
interesting in the context of measures where you have
a slope, you are forced to have some particular
slope.
But the real challenge starts showing this for four
colors and more. So as soon as you get to four
colors and more, you no longer have this nice
bijection with height functions, but you still can
define different phases and their interfaces. So it
doesn't look completely hopeless. There are some
preliminary ideas, but I think it would still be
remarkable if someone would know how to solve it and
there is -- there are some people who believe that if
you'll solve it for four colors, you probably know
how to solve it for a number of colors. I haven't
made up my mind if I share this view.
Another challenge is nonzero temperature. So this
says we allow some edges which are not respecting the
coloring. Now, we don't have the real correspondence
with height functions, but some structure of height
function still remains, so we still have some
abstract cover space where the height function is
defined. It still is related to height functions in
some way, and there are very interesting topological
phenomenon that happens here, but it still seems a
bit far. And of course, low dimension, which I don't
know. I wouldn't recommend trying to tackling it
using those methods. I mean, it would be interesting
to push the dimension down, but if you are trying to
reach to three dimensions, you'll have to have a very
brilliant idea.
I would like just to comment a bit about some small
relation between this thing and small -- and low
dimensions. So I discard my -- our results here for
tori with all the sides of the tori of length N. But
they would work also when there are non-even sides of
the torus.
And in particular, they would work for tori of the
form, so N to the two times two to the D. So those
are tori that have many dimensions, but many of them
are very compact. So here you would still see this
phenomena, all -- everything I said here would hold,
while if the torus was just N times two to the D,
then it is not. It doesn't hold. So at least this
is one way to see that there is something special
about dimensions two and more. But of course, this
is not really the low dimension case that we're
interested in.
Thank you.
[applause]
>> Yuval Peres:
Questions or comments for Ohad?
>>: Yeah. Do you have result for, say, bitter
depending on N?
>> Ohad Feldheim: For bitter depending on N? See,
so what you are asking me is if I can -- so we were
interested always in result that are independent of
N, right, right.
>>:
[inaudible].
>> Ohad Feldheim: But you're asking if Galvin -- so
I don't know. I would start at looking at the result
of Galvin Engbers, because they are less sensitive to
those things probably, but I don't know.
>>: But do you expect that -- I mean, it seems that
there's a fundamental problem to move from real
3-colorings to almost 3-colorings.
>> Ohad Feldheim: Yes, but also, I mean, this line
of salt also made people think that it's hard to
prove it on the torus, and we were able to do it
using some topological tricks. So it's possible that
using more sophisticated topological tricks you can
still overcome the fact that you would have some bad
edges.
Actually, when you look on configurations with bad
edges, you can classify them according also to
certain topological structures, and you can -- or
even then say that certain topological structures are
not so likely to happen.
>>: But I mean, like, it really becomes some kind of
a analytic property if you have only few bad edges.
I mean, topology, you know, presumably, you know,
doesn't it, you know, detect so well whether, like
the density of the bad edge. I mean, you can -- even
if there is one bad edge, you'll have some serious
topological flaw [inaudible].
>> Ohad Feldheim: So you might and you might not.
Right? So when you have a bad edge, you have now
this kind of a four cycle, and if you have a four
cycle of this site, of this type, this is very bad.
Right? Because it means that you have -- that you
are increasing in height when you go around it and
you are no longer correspond any height function.
You don't have to have cycle like this.
So for example, 0011 is okay for us. I mean, it
wouldn't cause any trouble. There would be trouble
around, okay? But -- and when you look on those bad
cycles, then those co-dimension two point -co-dimension two manifolds they wrap around, they
create some kind of structure, and it might be
possible to say something about the structure, and if
the structure you can show that it's really small, so
you will never have big loops when you wrap around
something bad happens, it might be possible still to
obtain a combinatorial result.
So I'm telling things that are really half-baked, but
at least you can see how I think on approaching
topological -- I mean, approaching topologically
those faulty 3-colorings.
>>: Okay. [inaudible]. Can you just briefly
explain the heuristics, why the change of behavior
between the dimension two and three, and not, say,
three and four? What's the ->> Ohad Feldheim:
Briefly explain.
So I think not
briefly enough for this situation, but I can explain
it not briefly, if you'll have time later on.
>> Yuval Peres: Okay. So we'll continue in private.
Let's thank both our speakers.