>> Eyal Lubetzky: Right, good afternoon everyone. We’re delighted to have Van Vu who was a Post Doc
here a long time ago. We’re happy for that for many reasons particularly whenever we want to hire a
post doc we tell them well look how people here end up. It’s all because of being here.
[laughter]
Anyway and Van is now a, talking about also one of you know favorite subjects for many of us, Roots of
Random Polynomials, please.
>> Van Vu: Thank you very much, Eyal. It’s the ongoing work with Terrance Tao, Hoa, Annette’s Ohio
State, and my students Hoi and Oanh Nguyen. Yes, so let me just go right into the heart of the matter. I
will start, well it would be, there will be many other things in it. But let us start with the simplest
questions that we want to determine the number of real roots about polynomial. Then of course in the
talk there will be things about complex root as well.
But just start like this, we have a polynomial with re-coefficient in, looking at that slide. That’s like an
ancient question we want to know this. There are a lot of things for instance in high school you know
many of us learned that Descartes’ rule of signs is like four hundred years ago. That if you look at the
number of sign changes in the coefficients, right, so how many time is switched from negative to
positive or vice versa. I think that is an upper bound in the number of positive roots. You can use a
similar trick to bound the number of negative roots as well.
Okay, so for this audience we would, now we talk about the general case when we have random
polynomial. It was when, well by which we understand that the coefficient that but choosing randomly
by some law. It was first looked at by Waring in eighteen century that he proved something like this.
That the probability of all root complex divided by the probability that there is a [indiscernible] two case
is bounded by two.
Now of course in this setting in the old time it turn out that Sylvester also looked at it later. It would be
in the book of Todhunter in hundred years ago as one of the first textbook in probability. Now but of
course in [indiscernible] high probability or distribution were not determined rigorously. You can
change this number two by twenty other constant if you choose the right distribution for coefficients,
yeah but just for the sake of history that I mentioned that.
Okay, so the matter when we talk about in the modern language is that we look at this random
polynomial. When the coefficient Ci, iid is the co-piece of some real random variable C, and we denote
by N, capital N the number of real roots of this. This is sometimes referred to as a Cp polynomial.
Of course there are other models which if I have I would mention later when the things are not
necessary iid. But let’s stay with this for awhile. The first result in this direction was obtained by Bloch
and Polya in nineteen thirty-two when they look at the matter when the coefficient uniform zero plus
minus one. Then they proved that with probability tending to one. The number of real root is bounded
by square root N. Now which is much better than the root of sign if you see that we put plus minus one
there then number of sign changes would be linear in N. But they managed to prove square root N by
the following idea that you multiply the polynomial with some function which is always positive. By that
you can change the number of sign changings, so this very nice communitorial idea.
It works for other distribution. This is where if you have plus minus one it work the same way, yeah, you
don’t need the zero there. That is Polya, Bloch has no picture that I can find. Unfortunately I think he
spend most of his time in the hospital when communicated by mail. Whenever he wrote the mail he
dated April first no matter when he wrote it.
[laughter]
There is the research pick up with famous work of Littlewood and Offord as leading analyst in England in
the nineteen forty’s. This, yeah, this was mentioned as the founding work of Oxford making famous. In
the series of three papers in eight, thirty-eight to forty-three they show that if you take C to be
Bernoulli, so it’s plus minus one. Then the number of real root with high probabilities bounded between
almost log n and log square root n. They say that the same proof holds for other distributions like
Gaussian and uniform. That’d be easy to check to be the case. In most paper they focus on the plus
minus one case. It was surprise; it was sort of surprising for the math community at that time as people
guessed that the number of real root is more than log. But this number is really small to what one may
expect.
Yeah, so during the same time there’s another important work by Kac’s. He actually observed a general
formula for the expectation of the number of real roots. He can write out an explicit formula for this.
One drawback of this is that it’s all easy to evaluate in the Gaussian case, although you can write
formally for any cases even plus minus one but only in the Gaussian case, sort of easy to evaluate.
It turns out that the expectation is as simple technically two over by log n. Actually Kac was very
enthusiastic when he wrote his first paper and proved this for Gaussian anything that it, it should be the
same for all other random variable which have same in zero and variance one. About five years, six
years later he wrote another paper and saying okay I’m a little bit wrong. I can only do this for uniform
but maybe the result with general. Then there’s Stevens he can show this for smooth random variables.
Can even, somehow they managed they evaluate Kac’s formula.
Now if we go to the plus minus one case. It took away from Kac’s paper is the quite famous work of
Erdos-Offord in fifty-six when they proved again the number is as simple typically two over Pi log n with
the arrow term, is like log n to the two third. Actually this proof is quite fascinating. Its do the following
and it’s very hard to imagine that, why it works. What they do is they pick a bunch of [indiscernible] K
[indiscernible] on the rely. Let me just provide the lower bound. K is roughly two over Pi log n. They
look at the value of the curve of Pn at this point, right. The function at this [indiscernible] and they say
that with decent probability, say probably quite close to one. The value of the function will change from
one point to the other. The sign of the value of the function, yeah, so if you pick this two [indiscernible]
close to each other and they manage to show that this negative is positive. Sorry, this positive is
negative and so on. Say, yeah, the nice thing is that they pick this [indiscernible] in advance. They are
fixed [indiscernible] and then they say that with high probability much of the [indiscernible] pair will
have sign changes. Then you have the lower bound for the number of roots.
>>: [inaudible] so they’re not going to get local [indiscernible], zero to one.
>> Van Vu: Yeah of course they have to make some arrow term somewhere.
>>: But this gives you the lower bound up to the constant.
>> Van Vu: Yeah, the lower.
>>: It’s…
>> Van Vu: No it gives you the lower bound with right constant.
>>: Yeah.
>> Van Vu: Then they have to do some other things to do the upper bound.
>>: Right, it’s like saying that they’re really very [indiscernible].
>> Van Vu: Yeah it is. Later on Ibragimov and Maslova generalized a proof of average and Offord and
four other, well random variable would mean zero and variance one, and get asymptotic formula. If you
go into the proof it’s even more complicated in Erdos and Offord. Optimize everything probably you
would get the arrow term of all the square root look at.
For the expectation, now on the other hand from the Kac’s formula there’s something interesting from
there that you can evaluate things very exactly. I think Kac’s formula in the Gaussian case gives
something like this. The density function is equal one over Pi. It’s totally explicit square root of sum
over [indiscernible] K n, minus K square, x to the two n plus two K minus two. It is totally explicit
function. If you integrate this function from minus infinity to infinity you would get exactly two over Pi
log n. The next term is the, isn’t not literally [indiscernible] n but the explicit constant which is
asymptotically point six two five. The next arrow term is actually lower one over n.
There are people I think in paper Willis, even earlier than Edelman and Kostlan even work out all the
[indiscernible] expansion of this. Next term is one, some constant time over n. The next one is some
other constant over n square and so on. It can be computed variably [inaudible].
Right, yeah, I want to describe a little bit the argument of Edelman and Kostlan. Not for the explicit
computation but how you get the two over Pi out of this, for this formula, because yeah you can ask that
well why does Pi has anything to do here. This sort of nice [indiscernible] and worker in quite a general
case not only this case, so what they do is the following. If the C of Gaussian, this Gaussian random
variable then you normalize it would get a normalize the coding that vector. Normalize the coefficient
vector. You get a unique vector on the sphere, unit sphere. Now you look at the moment curve. We
calling it one x and up to x to the n, right, then the value of the polynomial is actually the inner product
of this [indiscernible] on the moment curve with this vector, right. Then this inner product is zero
namely that you have a root at x if and only if this vector is orthogonal to this vector, right.
The number of real roots is the number of intersection. This should be its intersection of the projection
of this curve onto the sphere and random red circle, right. This is just geometric argument. Yeah, so it
looks like this. You have some curve here which is exact and its explicit curve is the projection of the
moment curve on to the sphere. Now you look at the gray circle, the random gray circle and see that
how, how many [indiscernible] they intersect, right.
Yeah and now if you look at this curve, now this curve is complicated. But let’s look at a very simple
curve is when you have a simplest curve which is another gray circle ending up exactly two intersection,
right. Now you take a small piece of this then you will have two time the length of this in expectation,
right. Then they just break this curve into small pieces. Even what you get is just two over Pi time the
length of the curve and you can compute the curve at this formula.
Okay, so now the result I want to present now is the following that we can prove that for any random
variable would mean zero and variance one, and say bounded two plus epsilon moment. The
expectation is two over Pi plus a constant. Not a constant plus some amount at the big O over one
[inaudible]. This big O is, I think have to depend on the random variable. It may not be the same
constant everywhere. Well we don’t know it is a constant or not, we just bound it.
>>: It’s not just a function of epsilon on your moment bounds it’s something that can lower your
moment bounds?
>> Van Vu: No it can depend on epsilon too. It depend on Xi, yeah this depend on Xi. Epsilon is the
information of Xi.
>>: That way it only depends on your bound on a two plus epsilon moment…
>> Van Vu: Yeah, any bound on epsilon maybe it doesn’t depend on how small that is, or how big that
is.
[laughter]
>>: [indiscernible].
[laughter]
>> Van Vu: Okay, so yeah but, yeah and there is a state a few questions related to this result. But I, the
main thing is I like to use it as a bite actually to present something more significantly. The proof is
actually more interesting than the result. We see all of this is already surprising that you can get to this
preciseness. Okay…
>>: I think what James was asking was if for a fixed epsilon and a fixed bound of two plus [indiscernible]
moment. The various difference of variables Psi as your random, as your arrow term change.
>> Van Vu: Oh, I see, probably not. But I don’t, yeah I cannot reconsider it. Yeah, so let me tell you the
ideas behind the proof which actually base quite a general technique. Okay, so first of all this thing is
standard for people doing random functions. But let me go over it anyway. Either we start with a
polynom like this, right. If we look at another polynom which is Qn define like this then we just reverse
the order of the coefficient. It goes from x to one over x, so that math the spectrum from zero one to
one infinity.
That means that we can restrict our shell to the number of real roots in zero one. If we know the
number real roots here we multiple a factor of four we get everywhere. From zero one is the same and
zero one to infinity and the other side of the spectrum is by symmetry.
First we start with a small fact. This is where the big O comes from. That if you fix any small
[indiscernible] and delta is the one over a thousand then with high probability the number of real root is
bounded as a function of delta. This is known from the Littlewood-Offord paper. Where they didn’t
state this but it’s what is inside their arguments. Yeah, so the spectrum we look at is from zero to one.
Actually if I go to even go from point nine nine nine here I have only say a thousand root here that I say
is a big O. That means the bulk part of the spectrum is very close to one.
>>: It seems like this transformation was not very big?
>> Van Vu: It is too.
>>: No because now you put all the reason in one. Before they were spread out you could…
>> Van Vu: No they have spread.
>>: The symmetry…
>>: Oh, no I know I’m just saying that like you were saying all the roots are here…
>> Van Vu: Yes, all of them are there.
>>: You did the transformation the other way then.
>> Van Vu: No, no…
>>: [indiscernible]
>>: They were close to one.
>>: They were close to one the whole time?
>>: Yes.
>>: Okay, good.
>> Van Vu: No I mean this is geometric series. You think of this guy, Xi n as the plus minus one, right. It
will be dominated by one third of this random one.
>>: Is it one is a fixed point of one reverse?
>> Van Vu: Okay, so part of the proof is we say look at the bulk of the spectrum and what we prove
there is in sort of a universality result. Namely that the expectation of a number of real root doesn’t
really depend on the choice of the atom variable, so if you replace this Xi say plus from plus minus one
to the Gaussian then we lose all the small arrow which go to zero with delta, okay.
What I say that is this part actually the number of reducing expectation that exactly more or less the
same. Then we just make the arrow term in this. If you have a different approach to estimate the arrow
term here precisely then we have a precise bound. Because for Gaussian everything can be computed
very exact.
Yeah, so it’s the nice thing is that in the bulk you don’t see the influence of the distribution of Xi arrow.
Every Xi behave the same and on the other hand we cannot say the same thing for the edge, right, in
this interval. Because if we take from the interval from zero to half we know that the plus minus one
guy don’t have any root. You take random polynomial with plus minus one coefficient. Every root
should have opposite root value logic in half. On the other hand the Gaussian can have root which is
just in half. They are not supposed to behave the same way at the edge. That’s why I think that the
constant arrow term would be different from time to time.
Okay, so, yeah so let me go over the main ideas of this, it’s actually the proof about something, some
more general universality phenomenon, not only the expectation but other things as well. Yeah, so the
step one in this work is that we proved the universality holds but for small delta which is not, no longer
one over a thousand or one over a million. But say a small power of n, small negative power of n. We
go even to here. It’s like one minus n to the minus a thousand, yeah.
Then we prove that. That is actually where eighty percent of the work is. Then we extend this, use a
different trick to extend the first result to O small delta. Yeah, so now me, because in order to prove
this universality I need to understand the distribution of O roots, not only the real one, and here is a few
experiment. Some picture that I copy from John Baez work page. If you look at a set of O roots of
complex and real together there wouldn’t be n of the more [indiscernible]. It is ground know that the
limiting distribution they go to the inner circle, it’s start of the picture of that.
I don’t know he make a lot of experiments with the highly real polynomial and get this. This is some
[indiscernible] you see here, there’s some hole you see near the point one and minus one, so it’s one
and minus one. There’s a [indiscernible] of the boundary that I don’t know if anyone understand it yet.
There’s also some black hole on the circle, yeah. This convergence was proved a long time ago that we
know that it’s converged to this by these people.
Now let me zoom in the real axis when we really care about. We see a bunch of roots on the real axis.
But we also see something very nice or something very remarkable here. This is some repulsion
between the real root and the complex. When you have the real roots actually we see some hole there
that the complex root did not get in.
>>: I’m sorry was this for Gaussians or…
>> Van Vu: No you can do for any. They give you the same picture.
>>: Okay.
>> Van Vu: Actually we prove that they would do the same picture.
>>: Not at the level of the same fractural features, right?
>> Van Vu: Actually its correlation is a level of one over n, it’s the right level. It’s universality of the
local. Yeah, so yeah that’s what I just mentioned to [indiscernible] that we look at the local limit now.
Yeah, so we have this picture. We have a huge circle. Now we look at just a small region, right. Near
this circle we expect to see end roots. Now we look at same region, D which is really small when we
expect, in expectation we see say one root, right. That’s what I just, okay.
We are going to prove that if in the Gaussian case we expect to see indeed one root then you’ll replace
Gaussian by say Bernoulli we would have more letters, same expectation, right. This is going to be true
not only for the first correlation function. It’s true for any fixed correlation function. But I would not
mention that.
Now, but what is the point? If, now this take care of the O kind of root, real and complex alike, but if
you go to the like D to be close to the real axis. Then we know that then by repulsion it we have a real
root there’s no complex root. Actually if in this guide we expect to see one real root to the Gaussian
case we also expect to see one real root in the other case as well, because there’s just no complex root
by repulsion. Well we have to prove that there’s a repulsion, the same repulsion in the Gaussian case
and the plus minus one case. That’s also part of the universality, but.
Yeah, so basically the thing we really prove with theories that we prove local universality for all
correlation function. That gives you the density, the two [indiscernible] and so and so forth. That will
take care of step one and two. I am not going to define correlation function because it’s sort of
technical. But I would be glad to discuss more later.
Yeah, okay, so there’s some technical requirement here. That’s why we need this condition is that we
have to be close to one. Namely that we need the center of D must be reasonable to the unit circle. It
cannot be like a distant one over a thousand from one. It have to be a negative polynomial distance.
Okay, so the idea behind this. Let me give you the outline through very simple example. Then we will
go through the key observation or the key new idea. Let me introduce this Lindenberg’s Switching idea
which is very useful and for proving limiting distribution or universality of the limiting distribution. The
way I understand universality is something like this. You heard the word a few times. But let me define
what I think about. Let me have a black box here and then there’s a sequence of inputs, say Xi one up to
Xi n. They are iid random variable and the black box give you some function, this guy have some limiting
distribution.
Now there’s some other sequence Xi one prime of Xi n prime coming in. You have another sequence
coming out, right. By universality I mean that if we cannot differentiate the Xi one and Xi one primer
looking at the output they wouldn’t have the same limited distribution, right.
One case we know very well is a case when you put in the side of the black box you just take the sum of
the Xi i, normal i by square root n which give you the central limit theorem, right. No matter what the
random variable would mean zero in vary once you take, you take the sum and normalize it by square
root n and you get the same limiting distribution.
Let me prove it here for you by, well there are several proof essential. Let me tell them but the
Lindenberg ones is what we try to imitate later. It goes like this, yeah, so we have to care about the
limited distribution of this guy x which is just the sum. If we want to control the distribution of this it’s
small s enough to control the distribution of the expectation of a test function on axis. We can take the
test function to be a [indiscernible] function like this.
Okay, so now in order to consider the X prime we do the following or Lindenberg did the following trick
that he switch Xi to Xi prime, or Xi tilde one at a time. Sorry this guy should be Xi as well, right. First you
switch the first one to Xi tilde in the second step you switch the second one, and so on. After N step you
would get from X to X tilde, right. We just try bound this thing by trying to go in the quality one at a
time we see how much is the effect in one switching. Even one switching we condition and everything
expect the pair that we are switching.
Then you look at the difference between the two guy and you can use [indiscernible] extension here.
You condition on everything else which give you this Xi bar…
>>: The Xi bar is not what you wrote…
>> Van Vu: Yeah, I realize but yeah, right. Xi what we get out there we subtract which we need to
subtract, yes sir, you are right. But, yeah with the right definition of Xi bar then [indiscernible] extension
you have this. But now if we take the expectation and we know that’s Xi i and Xi i but I have the same
mean and the same variance. The first two term disappear so we wouldn’t have the third term which
quite magically have the n to the three, what? Okay it have n three half here and that would queue off
the fact that something n terms. You would have an arrow term which goes to zero. You can even have,
yeah you can have an arrow term which is negative power of n. Okay so that’s the main idea behind the
Lindenberg Switching.
Yeah, now of course we in order to do this we cannot take F to be the [indiscernible] function. We have
to smoothen it to take the derivative. The method works very well for explicit variable X. If we have the
way to write now X as some explicit function in the Xi. For instance if a, not the linear form but the
higher polynomial in Xi then for instance there’s a work of Mossel that’ll deal with such a case. I have
application in computer science too.
Okay but what we care about here is some function which is fairly implicit, right. Because what we care
is this function X which is the number of root in region D. Some region D is given near the circle. Now
sure these are function in terms of the atom variable Xi one and two Xi n. But it is not an explicit
function like some of the polynomial or any way you cannot write down in any way that’s in order to.
Yeah, so here’s an idea actually which I really like. Took us awhile to find it, but once you find it, it
worked quite simply, in a simple way that you convert the, the conversion from implicit to explicit.
Yeah, so what you do is that you look at the root of the polynomial and let F be the function you care
about. It is the indicator of this region, right. The number of root we care about it’s just the sum of the
indicator, right. The indicator will give us one if there’s a root and then zero otherwise.
Now of course you have to smooth this function a little bit to take derivative. The key is that this is
green formula that says that if you have a nice function F then F can be.
[laughter]
No, yeah it’s the same green in calculus. F can be F zero can be written in this way, right. This is
something that actually in calculus we learned. But now we just plug it in and shift everything, right. Fzi
should be log of z minus zi time the Laplacian. When we sum up the right hand side then basically we
sum up the log. Namely we take the log of the product. But the product of Z minus Zi, Zi is the root of
the variable, of the polynomial. Actually it is the polynomial itself normalized by the first coefficients
say. This thing we really care about the number of root is now an explicit function. It’s completely
explicit it’s just the interval.
That’s the first part of the conversion. The second part is that now although we have this integration we
can prove it. We can estimate it by sampling. We can sample, well even if we have integration we can
take m random [indiscernible] inside the region and take the mean that in principle have to give us a
good approximation of this number.
Now, okay so now after we take the sample [indiscernible] y one up to ym and fix them. Then now this
function X which count the number of roots is an explicit function of [indiscernible] G y one up to G ym.
Now you can use the [indiscernible] key to prove the universality of the specter vary the same way, yes.
>>: [indiscernible]…
>> Van Vu: Well, well it [indiscernible].
>>: But everything looks very natural except the last step. Surprise this open sampling here just add
[indiscernible].
>> Van Vu: [indiscernible].
>>: Can’t you just work with, do the same method but do a replacement on the [indiscernible] ones
instead of on the second?
>> Van Vu: What?
>>: Why are you sampling these?
>> Van Vu: Because I need this guy to be fixed. Here m would be a small power of m.
>>: Yeah, but can’t you just move directly to the replacement method on the [indiscernible] and show
the [indiscernible] change much [indiscernible]?
>> Van Vu: I don’t know how to do that. Maybe there’s a better way to do that. Yeah and…
>>: [indiscernible]
>> Van Vu: If there is a simpler way to do that that would be nice. But yes, so that’s the [inaudible].
Okay and so in order to prove that the sampling is give you the right thing then we have to bound the
variance, and so on. That’s this part is, it is the technical tooling we need to bound the variance here.
That say because we want the sample to make the right estimate so we have to say okay the variance of
this guy is small, and basically [indiscernible] to say that the log of the function is not too large for most
[indiscernible] z.
That part you have to deal with. Why this thing is not entirely trivial. We want the log is not to be large.
Now the value of Pn itself is very easy to bound from [indiscernible], right, because we sum up n term
and they are random variable with bounded variance. From a [indiscernible] it is not too bad. But one
thing is the, which is less trivial that you need to [indiscernible] from below that it is not too small. That
is not entirely trivial.
That, how you use that because it can be sometimes small. Yeah, if you look at the value of, some
special value of z then the value of the polynomial evaluated at z can be really zero. For instance if you
look at z to be say the cubic root of unity, right. Then now you look at this guy a w and w square. This
guy would be one and then now you’ll w to the fourth would be w and so on, right.
Actually if you have to sum this guy up then they are periodic. Then you can [indiscernible] your
polynomial like this like I did earlier the periodic, with [indiscernible] K. But, yeah but now this guy can
be zero with some decent probability. Yeah, so if we split our things it would be, right it would be write
it as a polynomial in w so the free coefficient is this and is a sum of n over k terms. The next coefficient
of cells is sum of n over K term.
Each of these guys have a decent chance to be zero. This have a chance of square root, one over square
root n chance to be zero. The probability equals zero is not small. It can be like n to the minus k over
two. It can be n to the minus one half or n to the minus two, three halves, or.
This cannot be ignored totally. The Lemma we really have to prove is that actually this k root of unity is
the only troublesome points. If we can avoid getting a little bit farther away from the k root of unity
then the probability that your, our polynomial is small, is very small.
It use something that I built; we built in the last few years of so called the Theory of Inverse, LittlewoodOfford, which deal with the following question. This can be seen an as an antique concentration results
that the main question is that we have a linear form like this. We want to know that how much mass
can it pick up at zero, or any point for that reason?
Right, so the classical result, Littlewood and Offord say that if the ai at non-zero then this probability
called zero cannot be more than one over square root n. Now let’s see that if we are not satisfied with
one over square root n we ask what happen if I want the probability to be as small as one over n cube?
What can you do there?
Okay, so now we look for reason that makes this concentration probability to be large, say and the trivial
one is the following. Let’s imagine that O the ai are integer. They are integer in the small interval of
sum like n to the three say, right. Then by the central limit term we know that with a fair chance say at
least point nine s should be in the square root and [indiscernible] up of this interval, right.
The probability that s equals zero by pigeon hole principle is the reciprocal of the length of the interval,
right. You have a short interval and the mass is in there. There would be the [indiscernible] when you
pick up more mass and just the less reciprocal of the length. You take the length to be polynomial.
Then the probability can be polynomially large, okay.
You can generalize this to higher dimension if you take ai to be [indiscernible] in [indiscernible] sum of
the automatic progression, and with volume N to the C. Then you would have similar [indiscernible]
which is right here. But this is enough to keep it this is the first case in mind.
What we really prove in this direction is that if we assume the inverse. Namely that we assume this
bound in advance that the probability it have a mass [indiscernible] with mass at least this much can
actually most of the coefficient ai has to be in the arithmetic progression of the right length.
That’s the, okay, but if we look at, go back to the problem that we had before is that we have the sum of
the ai, sum of the Ci, w to the i, right. We care about this probability zero say that this probabilities
large, right. This is only the Tao case because we really care about is that this probability is small, not
only zero. But let me take this Tao case then my ai could be this, right.
If I say that this probability is large that mean my ai have to be on a short arithmetic progression. An ai
here’s equal to z to the i, say. That would imply that the z should be an algebraic number with low
degree. This somehow give you a, well now you have a feeling that why they say the cubic root of unity
would be the only math example. That this come for the fact that you really work with a [indiscernible]
of low degree and somehow you can go to that from here.
Okay, so that part proves the universality project sort of take care of the first part of the argument when
we can have universality from one to one minus some polynomial function of n. By the way the reason
that we, this has a long history that why we come to this idea. Because it started a long time ago when
we started to work on universality of random matrices, so the key problem there is to prove universality
of the correlation function.
There’s two group working on that Terry and I, and [indiscernible]. We work a lot on random and
Hermitian matrices, right. It’s either complex or real asymmetric, complex, and Hermitian, and real
asymmetric. In this case we know that the Eigenvalues are in number. The problem here was solved
quite successfully for the complex guys and partially solved for the root guys. We had good result there.
On the other hand another model popular than Matrix Theory just take iid with no symmetry. You have
n square random bit in the matrices. Now the Eigenvalue will not be real they are mostly complex and
what the correlation function question makes them at the same way as before.
The original, the first methods that Terry and I and also the method that [indiscernible] and Ed develop
it doesn’t work for these guys at all. Because it rely very much on the fact that Hermitian matrices are
not. They do not react to small perturbation. On the other hand we know that a small perturbation to a
known Hermitian matrices can change the spectrum quite a bit. We have to find new method to deal
with this. Then actually this method was sort of invented to deal with the iid curse. Then we realized
that it worked for polynomial as well. You can also use it for the original setting. Of course it doesn’t
lead to any new result. But it’s sort of, gave a common approach to deal with the random processes.
Now the last part of the puzzle is that we have to go from this, we have to go from the universality very
close to one to universality not so close to one. We know that we get result from here for Pn. But I will
go exactly one to go to one minus delta for some small delta. We need to extend that range.
This boiled down to understanding something from numerical analysis. This is another question which is
interesting but it’s old. Namely that how often does a random polynomial have a double root or even
near a double root in some sense? For instance if you want to run like the most popular method to
compute roots is the Newton algorithm and the running time is very crucially depend on the existence
of double roots. Yeah, so you don’t have double root in the running time improve exponentially.
Okay, so what we can show is that we can have, well this is actually not zero. We can have, yeah if you
look at the interval from one minus delta to one minus say like log n square over n. Then you are not
going to have a double root. Actually not even going to have a near double root namely that the root
and the derivative is not zero but very small. Then we don’t even have this case. Yeah, it’s not zero it’s
like point nine nine.
Let me tell you why this is helping our course of extending this universality. The thing is that I try to
cheat here a little bit. Yeah, so even Pm I can go from one to one minus and to the minus delta. Now
we have to go from here to a next one, say one minus, from L to the minus delta, right. Now I can look
at the for some L which is say some small power of m. Is that a Pm? I look at Pm, so it’s a polynomial
degree m. I’m saying that in this interval actually Pn and Pm the roots in this interval are more like the
same. Because if you look at Pm is like this, C to the m and Pm is exactly this. Now you add term like
this in there, right.
>>: Can you take…
>> Van Vu: What?
>>: What is the assumption of the [indiscernible] this statement of the [indiscernible] is not true if you
want double roots.
>> Van Vu: Yeah of course that’s why. That’s why we need to put the non-existent double roots, yes.
>>: Right, we’re putting…
>> Van Vu: Yeah, that’s the beginning. That’s the beginning of how I try to extend it, right. If I can have
something like this then actually I can replace Pn to Pm by some smaller m. Then I say that instead of n
to the minus delta I can take m to the minus delta and I iterate that. Now these only work if F and G
each cell they don’t have double root. That is exactly the [inaudible].
Yeah and so if they don’t have double root then the number of roots defer by very little and I don’t
make any mistake when I do this trick. Yeah, so that’s why we need this near double root. Not only
double root but near double root zero. Yeah, so the root has separated and the root of F and the root of
G are close to each other. They have the same number of root.
Okay, so that is the, about the main idea. Yeah, so there is one [indiscernible] of questions that I…
>>: You didn’t say how you argue about the non-existence of double roots.
>> Van Vu: Yeah, so that’s the separate [inaudible]. That’s another paper to. Yeah, so we do this and
iterate. Yeah, so that’s basically the main ideas. Each of, and here’s some other things that we can do
with this technique that we mentioned, right.
Yeah, so we can prove a correlation function of this. Yeah, we can look at more other model of random
polynomials when the variance of the Xi i are different. There are classes which are popular in the
literature like we can look at polynomial of this type. Like this code of very polynomial and polynomial
of this type, and it would be square root and I don’t remember. It’s called a binomial polynomial.
[indiscernible] had a book about the case when we don’t look at polynomial but look at infinity
extension the series. There’s a lot of interesting properties of this series, random series. Any way we
can extend the random series too. As I mentioned random polynomial with dependent coefficients can
also be considered. The only case we consider here was the Characteristic function of random matrices.
Basically we do this correlation for Eigenvalues. Yeah, so there’s many other interesting case that may
come to mind.
One interesting question that I don’t know how to solve is what is the constant in the big O? We know
the very first result I mention which I started with is that E of n C n equals two over Pi log n. Now we
know the adult term which is [indiscernible].
Well there are two questions actually. First of all does it tend to limit? That’s a question I mentioned in,
at lunch last week. We know this for the plus minus one case. Of course it was no for the Gaussian
case. Otherwise we don’t know and even when it tended the limit we don’t know that the limit is. We
have no reason to go [indiscernible] for this. We know that it’s less than ten to the million. But that’s
about it. It is intimately connected to the existent of double root from say zero to point nine nine. The
question I mentioned last time. Thank you very much.
[applause]
>> Eyal Lubetzky: Is there any comments or questions?
>>: So what of the guess, so it’s all going to be the same concept, right?
>> Van Vu: No.
>>: You have to explain…
>> Van Vu: Yeah and then I have like no medical experiment that it shows off by two, three or
something.
>>: It’s bigger?
>> Van Vu: Smaller.
>>: Smaller, that’s for the [indiscernible]?
>> Van Vu: Yeah, for the [indiscernible]. But I don’t know it is bounded by a million [indiscernible] or,
now if you prove the bounds less than a million, so maybe new.
[laughter]
>>: You mean the constant for plus [indiscernible], plus…
>> Van Vu: Yeah, plus minus one. I don’t know if it is less than a million. I know it is there.
[laughter]
About the experiment it’s like point four…
>>: Farther distributions there’s still a problem to show double zero root?
>> Van Vu: Yeah, that we haven’t try very much. Actually there are recent development here that for
other distribution like in [indiscernible] case. Not for the very polynomial now we manage to actually
prove universality closer to the origin. You don’t need to be polynomial rebounded away actually is
enough to be at a constant [indiscernible]. Then that already gives the big O zero arrow term in
constant [inaudible]. Yeah, but the big O is bigger.
>>: On the interval next to the [indiscernible] one you have very good control, right, is that what you’re
saying?
>> Van Vu: Yes, yes...
>>: [indiscernible]. There the big O is like one over n.
>> Van Vu: Yes.
>>: But then with this delta…
>> Van Vu: From zero…
>>: What you get is the old term eventually?
[laughter]
Okay.
>> Van Vu: Yeah, well from there…
[laughter]
What the G [indiscernible] if I fix the delta I can tell you the arrow term here depending on delta quite
nicely. I don’t know what delta is. If my delta is one over ten to the million, that’s a bracket. The
number of root would be the log of that which is a million which is [indiscernible]. We don’t have a root
[indiscernible] in delta. If you have a root [indiscernible] in delta they are by root estimate for the
number of root from zero to one minus delta.
>> Eyal Lubetzky: Okay, let’s thank Van again.
[applause]