YUFEI ZHAO: Last time, we started
discussing the extremal problem for bipartide graphs.
And in particular, we saw the Kovari-Sos-Turan theorem,
which tells us that if you forbid your graph from having
a complete bipartide graph, Kst, then
you have this upper bound on the number of edges in your graph.
So we gave a proof.
It was fairly short, used the double counting argument,
and it give you this bound.
And the next question is how tight is this bound?
Is there a lower bound that is off by, let's say,
at most a constant factor?
And that's a major open problem.
It's a conjecture that this bound is tight
up to constant factors.
But that conjecture is known for only a very small number
of graphs.
And we saw a couple of examples last time.
So last time we saw construction that
shows that for S equals to 2, this bound is tight.
So the extremal number for k through 2 is on the order of n
to the 3/2.
So this theta means I'm hiding constant factors.
And our construction used this polarity graph,
which is essentially the point line incidence
graph of a projective plane.
And a basic algebraic or geometric fact, if you will,
that two lines intersecting at most one point.
We also sketched a construction that
showed that for S equals to 3, this bound is also tight.
And this construction here involved
using spheres, again, in some space over a finite field.
So both these constructions are in some sense
algebraic geometric.
And you can ask, is there a way to extend
these ideas to construct other examples of Kst free graphs
with the right number of edges using some ingredients
from algebraic geometry?
And today, I want to show you two different ways
of doing that.
So the state of the art, which I mentioned last time,
let me remind you what is known about constructions
that achieve the right exponent up there.
So in a series of two papers by [INAUDIBLE]
it is shown that if the constants t and s,
such that t is large enough compared to s,
in particular t is bigger than s minus 1 factorial,
then the extremal number of Kst's is
of the order, same order, as given
in the upper bound of the Kovari-Sos-Turan theorem.
In particular, these range of parameters allows you to do
(2, 2), (3, 3), which we already know how to do.
But the next case is 4, 7.
And it is still open how to do 4, 6.
So I want to show you this construction.
And I will tell you exactly what the graph is.
I'll give you an explicit description
of this graph, which is Kst free and has lots of edges.
And as I mentioned earlier, it is an algebraic construction.
So as before, we start with p prime.
And we will take n to be a p raised to the power s.
And let's restrict s to be integer at least 2.
And, of course, that's same as last time,
if you have other values of n, take a prime close
to the desired value and then take it from there.
To describe the construction, let me remind you,
the norm map, if you have a field extension, in this case,
specifically I am looking at the field extension,
this Fp to the s, I can define a norm map as follows.
Sending x to be the product of all the conjugates,
or the Galois conjugates of x, in this field extension.
So explicitly written out is just that expression,
which I can clean and collect and write it down like this.
So I wrote that the image of this norm map
lies in the base field Fp.
And that is because--
well, one of the many reasons why this is the case,
is that if you look at--
so I'll denote this norm map by N--
if you look at N of x, it raised to power p,
leaves this value unchanged.
And the base field is the field where it
is invariant under power by p.
So here's the graph, which I'll denote the norm
graph with parameters p and s.
So the norm graph will have as vertices just the elements
of this field extension.
And the edges will be the set of all pairs of vertices, not
equal, of course, such that a norm of their sum equals to 1.
So that's the graph.
This is an explicit description of what the vertices
and what the edges are.
So now, we need to verify a couple of things.
One is that this graph has the desired number of edges.
It has lots of edges.
And two is that this graph is Kst free.
So let's do both of those things.
So the first is let's check it has the right number of edges.
So that's a relatively easy task.
What we need to do is to count for every a
how many choices of b are there in this field extension
such that a plus b has norm exactly 1?
And I claim that that number--
well, so here's a basic algebra fact,
that the number of elements in this field extension with norm
exactly 1 is precisely p to the s minus 1
divided by p to the s.
And this is because really we're looking
in the multiplicative subgroup.
So the multiplicative group in this Fp to the s.
And it has a cyclical--
so there's a generator--
the order of the cyclic group is p to the s minus 1.
So you're asking, how many elements
when raised to this power here ends up at the identity?
So that's the answer.
So that's one aspect.
And so as a result, every vertex is adjacent to, well,
how many vertices?
For every given a I need to solve for b.
And basically, this many solutions,
I have to be just slightly careful because I
don't want loops in my graph.
So I may need to subtract 1.
So it's adjacent to at least this number up here minus 1
to account for possible loops, which is pretty large,
p to the s minus 1, which in other words
is n raised to 1 minus 1 over s, that many vertices,
and you see that this gives you the right number of edges.
So this is a graph with lots of edges.
So that part wasn't so hard.
The next part, it's much trickier,
which is we want to check that this graph has no Kst's.
So previously in our algebraic construction,
we used some geometric facts, such as no two lines
intersect in more than one point to show that there's no k to 2
in the polarity graph.
So there's going to be something like that here.
So the claim is that this construction, this norm graph,
is Ks, s factorial plus 1 free.
So it's not quite the bound I claimed.
So it's a little bit weaker.
But it is in the spirit of what I am claiming,
namely that for t large enough, this graph is Kst free.
So for t a large enough comes constant here
will show s factorial plus 1.
And as a result, it would follow that the extremal number 4s
sub s factorial plus 1 is at least 1/2
minus little 1 of the constant--
I don't already worry about that much,
but it's on the order of n to the 2 minus 1 over s.
OK, everyone with me?
So we need to verify this graph here has no Kst.
Yes, question?
AUDIENCE: Should that t be an s?
YUFEI ZHAO: Yes, that should be an s.
Thank you.
Any more questions?
AUDIENCE: Should that be s minus 1 factorial?
YUFEI ZHAO: So we will show later
on a better result using s minus 1 factorial,
but for now I'll show you the slightly weaker result, which
is still in the same spirit.
Yep.
AUDIENCE: Is the stronger result using the same graph?
YUFEI ZHAO: We'll change to a different graph.
For the stronger result, we will change to a different graph.
OK, so now let's show that this graph here is Kst free.
And for that claim, we need to invoke an algebraic fact, which
let me write down now.
So suppose we have a field f.
Any field will work with a finite field.
Any field is fine.
And I have a bunch of elements from the field,
such that a sub ij is different from a sub
rj for all i not the same as r.
Then the system of equations--
and I'll write down the system.
So x minus 1--
x1 minus a 1, 1, x2 minus a 1, 2, and so on.
xs minus a1x equals to b1.
That's the first equation.
Second equation, x1 minus a 2, 1.
It almost looks like the usual system of linear equations.
But I'm taking products.
And so on.
So the last one being x1 minus a s1 x2 minus a s2, dot dot dot,
xs minus a ss equals to b sub s.
The system has at most s factorial solutions, where
I'm working inside this field.
So that's the claim.
So let me just give you some intuition for this claim.
Suppose the right side vector is 0, all zeroes.
Then I claim that this is trivial.
So what is the saying?
I need to select x1 to be one of the a's from the first row
and x2 be one of the a's from the second row, and so on.
But each column of a is distinct.
So that's the hypothesis.
You have all the a's in the first column are distinct.
So no two of the x's can--
so I need to set one of the xi's to be one
of the a's from the first row.
But you see that you cannot set x1 to be a 1, 1,
and x1 to be a s1 at the same time.
So the solution just counts permutations,
which is exactly s factorial.
So this algebraic fact plays a key role
in the proof of the theorem that the lower bound
that we're stating up there, if you look at the paper,
they give a proof of this result.
And it's not a long proof.
But it uses some commutative algebra and algebraic geometry.
And usually in a class, if the instructor
doesn't present the proof, it's for one of several reasons.
Maybe the proof is too short.
It doesn't need to be presented.
Maybe it's too long or too difficult. Maybe
it's not instructive to the class.
And the last reason, which is the case here,
is that I don't actually understand the proof.
As in I can follow it line by line,
but I don't understand why it is true.
And if one of you wants to come up with a different proof
or try to explain to me how this seemingly
elementary algebraic fact is proved, I would appreciate it.
For small values of x, you can check it by hand.
So x equals to 2, you're solving a system
of two quadratic equations.
And that you can check by hand.
And three maybe you can do it with some work.
But even with 4 it's not so clear how to do it.
And also, one of the geometric intuition is that if b is o0,
then you have exactly s factorial solutions.
And the geometric intuition is somehow
that if you move b around, then the fiber,
the sides of the fiber, the number of solutions x
can only go down.
It can not go up.
And this corresponds to some algebraic geometry phenomenon.
And that's all I will say about this algebraic fact, which
we'll now use as a black box.
Great.
So now we have that as our algebraic input,
let us show that a norm graph is Kst free.
It's actually not so hard once you
assume that theorem up there.
So let's show at a norm graph is Ks, s factorial plus 1 free.
Well, what does it mean to have a Kst?
It means that if you have distinct vertices,
which then correspond to elements,
s elements, y1 through ys, of this field,
then the common neighbors of these elements
correspond to solutions of this system of equations
where I set all of these values to be 1.
But I can write l exactly what these guys are
because I have this form, that representation there
for the norm map, so I can write it out.
And now remember this fact that when
you are in characteristic p, x plus y raised to the power p,
is the same as x to the p plus yp and characteristic p.
So I can expand the remaining parenthesis like that.
So I want the first line to equal to 1, and so on.
And each of the lines has that equal to 1.
How many solutions in x does this system of equations have?
So even if I treat each of x and x to the p
and so on as separate variables, that theorem
appear tells me that there are at most s factorial solutions
in x.
Satisfies all the hypotheses of that theorem up there.
Therefore, the graph is Ks sub s factorial plus 1 free.
You do not have more than s plus 1 different values
of x satisfying this system of equations.
And that's the proof that this norm graph is Kst free.
Yes, question?
AUDIENCE: Why can't powers of like [INAUDIBLE]
YUFEI ZHAO: Sorry, can you repeat the question?
AUDIENCE: Why cannot the powers of the y's be the same?
YUFEI ZHAO: The question is why cannot the powers of the p's be
the same?
So you are asking, down the second column,
let's say, why are all these y's different?
Because you're working inside a field.
And raising to a p in this field is a bijection.
So think about the order of the cyclic group.
It has co-prime to p.
But great question.
Anything else?
OK, so this gives you a construction
that gives you Kst free for t bigger than s plus s factorial.
Now, let me show you how to improve this construction
to do a little bit better, to get s minus 1 factorial.
And the idea is to take a variant
of this norm graph, which we'll call the projective norm graph.
And the projective norm graph will define it
for s at least 3 is rather similar.
But there's a twist.
I have as the vertex set, not just the field extension--
OK, so now I take field extension, but one level less.
And I take a second coordinate, which consists
of non-zero elements from Fp.
The edges are formed by putting an edge between these two
elements, if and only if, the norm
of the sum of the first coordinates
equals to the product of the second coordinates.
So now, you can run through a similar calculation that
tells you the number of edges.
So first of all, the number of vertices
is p to the s minus 1 times p minus 1,
so basically the same as p to s.
And additionally, every vertex has degree exactly
p raised to s minus 1 minus 1.
And the reason is that if I tell you the values of x little x
and big y, which cannot equal minus x,
or else you will never form an edge,
then they together uniquely determine little y.
So for every value of big X and little x,
I just need to run through all the values of big Y other
than minus x.
So the number of edges then equals to 1/2 times
the number of vertices times the degree of every vertex, which,
as before, is the claimed asymptotic.
And the remaining thing to show is that this projective norm
graph is Kst free.
So it's K sub s, s minus 1 factorial plus 1 free.
It's a similar calculation as the one before,
but we need to take into account the small variant
in the construction.
So suppose we fix s vertices, labeled by this big Y's, little
y's.
And now we need to solve for uppercase X,
lowercase x in this system of equations,
so asking how many different pairs, big X, little
x, can appear as a solution to this system of equations?
Well, first of all, if some pairs of the first coordinates,
big I is equal to big J, then if you have a solution,
then that forces little y to be the same as little j.
And so the y's wouldn't have been distinct to begin with.
So this is not possible.
So all the big Y's are distinct.
Well, now, let's divide these equations
by the final equation.
And we get that the i-th equation
becomes like that, which you can rewrite
by dividing by the coordinate of the norm of big Y i
minus big Y s.
This is non-zero because we just showed that all the big Y
i's are distinct.
If you divide by this norm here and rearrange appropriately,
we find that the equations become like this.
So after doing some rearranging-- so
this is the equation the set of new questions that we get.
And you see that if you use new variables, x prime,
do a substitution, this being x prime,
then it has basically the same form as the one
that we just saw with a different set of constants.
And in particular from what we just saw,
we see that you cannot have more than s minus 1 factorial
solutions in x.
Now, they're s minus 1 equations.
And the field extension working as the x minus 1
field extension.
So we saved an equation by using this projectivization.
And that's it.
So this shows you the claim of constructing
a Kst free graph for t bigger than s minus 1 factorial, which
has the desired number of edges.
Yes, question?
AUDIENCE: Why do you have the if some capital
Y equals capital IJ?
YUFEI ZHAO: OK, so the question is why do I say this part?
So I'm maybe skipping a sentence.
I'm saying, if there is a solution to this system
of equations x, if these vertices have
a common neighbor, then if you have some x satisfying
this system of equations, then having two different big Y's
being the same forces you to have the two smaller y's
being the same.
AUDIENCE: OK.
YUFEI ZHAO: Right.
And then the Y's will have been distinct.
So for them to have some common neighbors,
you better have these big Y's being distinct.
Any more questions?
Great.
So as I mentioned, it is an open problem
to determine whether what is the extremal number for K44, K45,
K46.
And you may ask, well, we have this nice construction--
it maybe somewhat mysterious because of that, but explicit.
And you can write this graph down.
And you can ask is this graph K46 free?
So do we gain one extra number for free,
maybe because we didn't analyze things properly?
And it turns out that's not the case.
So there was a very recent paper just released last month
showing that this graph here for s equals to 4 actually does
contain some K46's.
So if you want to prove a corresponding lower bound
for K46, you better come up with a different construction.
And that's I think an interesting direction
to explore.
Any questions?
Yes.
AUDIENCE: Do we know of any similar results
about this construction not working for larger s?
YUFEI ZHAO: The question is, do we
know any similar result of about does this graph contain
Kst for other values of s and t less
than the claimed threshold?
It is unclear.
So the paper that was uploaded, it
doesn't address the issue of s bigger than 4.
Yeah.
AUDIENCE: Why Fp to the power s?
YUFEI ZHAO: So question is why Fp to the power of s?
So let's go back to the norm graph construction.
So where do we use Fp to the power of s?
Well, certainly we needed it to have the right edge count.
So that comes up in the edge count.
And also in the norm expression, you
have the correct number of factors.
So I encourage you to try for if you
use a smaller or bigger value of s,
you either don't get something which is Kst free
or you have the wrong number of edges.
Any more questions?
So later, I will show you a different construction
of Kst free graphs, again, for t large compared to s
that will not do as well as this one.
But it is a genuinely different construction.
And it uses the idea of randomized algebraic
construction, which is something that actually was only
developed a few years ago.
It's a very recent development.
And it's quite nice.
So it combines some of the things we've talked about
with constructing using random graphs to construct
H free graphs on one hand, and on the other hand,
some of the algebraic ideas.
In particular, we're not going to use that theorem up there,
but we'll use some other algebraic geometry fact.
OK, so let's take a quick break.
So what I want to discuss now is a relatively new idea
called a randomized algebraic construction, which
combines some ideas from both the randomized construction
and also the algebraic construction that we just saw.
So this idea is due to Boris Bukh just a few years ago.
And the goal is to give an alternative construction
of a Kst free graph with lots of edges
provided that as before, t is much larger compared to s.
So this band here will not be as good as the one
that we just saw.
And I will even tell you what it is.
But it's some constant.
So for every s there is some t, such
that this construction works.
As before, we working inside some finite field geometry.
So let's start with a, a prime power.
You can think of prime if you like.
It doesn't make so much difference.
So we're working inside a finite field.
And let's assume s is fixed and at least 4.
Let me write down some parameters.
Don't worry about them for now.
Just think of them as sufficiently large constants.
So d is this quantity here.
So we'll come back to it later when it comes up.
OK, so what's the idea?
When we looked at the randomized construction,
we took a random graph.
We took an Erdos Renyi graph.
Every edge appeared independently.
And saw that it has lots of edges,
if you choose p property and not so many copies of h.
So you can remove all the copies of h
to get a graph with lots of edges that is h free.
What we're going to do now is instead
of taking the edges randomly, we're
going to take a random polynomial.
F will be a random polynomial chosen uniformly
among all polynomials with--
so I wrote uppercase Y and uppercase X and Y. Actually, X
and Y, they're not single variants.
They are-- so each of them is a vector of s variables.
So in other words, x1 through xs are the variables
in the polynomial.
And then y1 through ys.
So it's a 2s variable polynomial.
So among all polynomials with degree, at most d--
d being the number up here--
in each of x and y sets of variables.
So you look at it as xs variables,
each monominal has their exponents sum
to our most d, and likewise with each monomial for the y
variables.
So this is the random object.
It's a random polynomial in 2s variables.
And the degree is bounded.
So you only have a finite number of possibilities,
and I choose one of them uniformly at random.
And now, what's my graph?
We're going to construct a bipartide graph
G. The bipartideness, it's not so crucial.
But it'll make our life somewhat easier.
So it's a bipartide graph.
So it hs two vertex parts, which I will label left and right,
L and R. And they are both the s dimensional vector
space over f cube.
And we'll put an edge between two vertices,
if and only if that polynomial up there
f evaluates to 0 on these values.
That's the graph.
So I give you a random polynomial f.
And then you put in edges according to when f vanishes.
So f, if you view the bipartide graph as a subset of Fq
to s cross F to the s, then this is the zero set.
The edge set is the zero set.
Just like in random graphs, with the construction
with random graphs, we'll need to show a couple of things.
One is that has lots of edges, which will not be hard to show.
And second, that it will have typically a small number
of copies of Kst.
And that will have some ingredients which
are similar to the random graphs case we saw before,
but it will have some new ideas coming from algebraic geometry.
First, let's show that this graph has lots of edges.
And that's a simple calculation, because for every pair
of points of vertices, I claim that the probability--
so here f is the random object--
the probability that f evaluates to 0 on this pair is exactly q.
So exactly 1 over q, 1 over the size of the field.
So this is not too hard.
And the reason is that the distribution of f
is identical to if you are an extra random constant on f,
chosen uniformly at random.
So I took a random polynomial.
I shifted by a random constant.
It's still uniformly random polynomial,
according to that distribution.
But now, you see that this, whatever f
evaluated to, if I shift by a random constant,
you will end up with a uniform distribution.
So it tells you that that guy up there is uniform distribution
on every fixed point u, v. So in particular,
it hits 0 with probability exactly 1 over q.
And as a result, the number of edges of g in the expectation
is exactly n squared over q, where
n is the number of vertices.
So n is actually not the number of vertices,
but the size of each vertex part, namely q to the s.
So you see that it gives you the right number of edges, so n
to the 2 minus 1 over s.
So we have the right number of edges.
And now, we want to show that this graph here typically does
not have too many copies of Kst's.
It might have some copies of Kst's.
Somehow that is unavoidable.
Just as in the random case, you do have some copies of Kst's.
But if they are not too many copies,
I can remove them and obtain a Kst free graph.
OK, so what is the intuition?
How does it compare to the case when
you have a genuine Erdos-Renyi random graph?
Well, what is the expected number of common neighbors?
So if you fix some u with let's say
on the left side with exactly s vertices,
I want to understand, how many common neighbors does u have?
But because the common neighbors,
if he has too many common neighbors, then that's a Kst.
It is not hard to calculate the expectation
of this quantity, both in the random graph case
as well as in this case.
And you can calculate if you pretend every edge occurs
independently, the expected number of common neighbors
is exactly n to the q to the minus s.
So there are s elements of u, which is exactly 1.
And you know that for a binomial distribution with expectation 1
and a large number of variables, the distribution
is approximately Poissonian.
Ah, but that's in the case when it's independently distributed,
which is the case in the case of GMP.
But it turns out for the algebraic setting
we're doing here, things don't behave independently.
It's not that you're doing coin flops for every possible edge.
We're doing some randomized algebraic construction.
And for algebraic geometry reasons,
you will see that the distribution is very much not
like Poissonian.
It will turn out that either the number of common neighbors
is bounded or it is very large.
And that means that we can show using some Markov inequality
that the probability that it is very large is quite small.
So typically, it will not have many common neighbors?
And that's the intuition, and so let's work out this intuition.
Any questions so far?
So how do we do this calculation?
So first, let's start with something that's
actually fairly elementary.
So suppose you have some parameters, r and s.
So I think of r and s--
so I have some parameters r and s.
And thick of them as constants.
Have some restrictions, but don't worry too much
about them.
Suppose I have two bounded subsets of the finite field,
where you have size s and v has size r.
Then the claim is that the probability
that f vanishes on the Cartesian product of u and v. OK,
so what do you expect it to be?
So I have s r elements, and I want f, this random polynomial,
to vanish on the entire product.
Well, if s, its value behaved independently for every point,
you should expect that the probability is exactly
q to the power minus s r.
And it turns out that is the case.
So this is true.
This is an exact statement.
OK, so why is this true?
So this is in some sense a generalization
of this claim over here.
And you have to do a little bit more work.
But it's not too difficult.
So let's first consider this lemma
in a somewhat simpler case, where all the first coordinates
of x are distinct--
of u are distinct.
And all the first coordinates of v are distinct.
Suppose u and v have that form.
So I write down the list of points for u
and first coordinates are all distinct.
What I want to do is to give you a random shift,
to do a uniform random shift.
And I will shift it by a polynomial g, which
is a bivariate polynomial.
So these are not vectors.
They are just single variables.
And I look at all possible sum of monomials,
where the degree in i is less than s, and the degree in j
is less than r.
And these a's are chosen uniformly independently
at random from the ground field fq.
And as before, we see that f and f plus g
have the same distribution, the same probability distribution.
And so all it remains to show that is whatever f comes out
to be, if I tack on this extra random g,
it creates a uniform distribution on the values
on the entire u cross v. But, actually, see,
I have sr choices exactly for these coefficients.
And I have sr values that I'm trying to control.
So really it's a counting problem.
And it suffices to show a bijection, namely
that for every possible vector of values,
there exists a choice of coefficients,
as above, such that g evaluates to the prescribed values
with the given coefficients.
And that uniformity will follow just because you have the exact
if you just do a counting.
And the one-dimensional version of this claim,
so let's think about what that is.
So if I have, let's say, three points on the line
and a degree 2 polynomial, what I am saying
is that if you give me the values you want on these three
points, I can produce for you a unique polynomial that
evaluates to the prescribed values on these three points.
And that you should all know as Lagrange interpolation.
So it tells you exactly how to do that.
And that works for many reasons.
One of them is that the random indeterminate is invertible.
Here, we have multi-variables.
So let's do Lagrange interpolation twice, once
for each variable.
So we'll apply Lagrange interpolation twice.
So the first time, we'll see that for all values of u,
there exists a single variate polynomial
in the y variable with degree at most r minus 1 that
evaluates to the correct values on the fixed little u.
So do it for one variable at a time.
For fixed u, do Lagrange interpolation
on the y variable.
And now, once we have those things there,
viewing the g that we want to find as a polynomial whose
coefficients are polynomials in the x variables, but is itself
is a polynomial in the y variable,
we find that again using Lagrange interpolation,
there exists these values for these coefficients
here, such that each coefficient of if you plug
in the first entry into little u agrees with the coefficients
of the g that we just found.
And that should be the case for every little u.
So once you find these polynomials
and now you have a bonafide polynomial in g.
And that's the claim above.
So using Lagrange interpolation twice, once for each variable.
So this is-- if you're confused, just think about it.
There is nothing deep here.
So that finishes the claim in the case
when the first coordinates are all distinct.
So we use that fact crucially in doing this Lagrange
interpolation.
Now, for general u and v, where we
don't have this assumption of having
distinct first coordinates, well,
let's make them to have distinct first coordinates
by considering a random linear transformation,
so using a probabilistic method.
So we suffice as to find invertible linear maps, p
and s, on this vector space, such that TU and SV have
the above properties.
So let me show you how to do it for u.
So I need to find you a invertible linear
transformation t.
Well, it's just the first coordinate that matters.
So it suffices to find just a linear map
corresponding to the first coordinate that
is injective of u.
Whatever linear map you have, even if it's zero, that's fine,
I can extend it to the remaining coordinates.
Actually, if it's zero, then it's
not going to be injective of u.
So it better not be zero.
OK, well, let's find this map randomly.
So pick t uniform via random among all linear maps.
And I want to understand what is the probability
of collision, bad event, if two elements of u
end up getting mapped to the same point.
Well, that's not too hard.
So for every distinct pair of points in fq to the s
the probability that they collide,
think about why this is true.
It's exactly 1 over q.
If x and x prime, they're differing at least one
coordinate, then even just along that coordinate,
I can make them distinct.
So this is the case for every pair.
So now by union bound, the probability
that t1 is injective on u is at least 1 minus the size
if u choose 2 times 1 over q.
And that's why we chose q to be large enough.
So q is at least r squared, s So this number here is positive.
So such a t exists.
And so we can transform this u and v, two configurations where
the first coordinates are all distinct,
and then run the argument as before.
OK, great.
So what we've shown so far is that if you look at these Ksr
structures, they appear with probability exactly--
well, with expectation exactly what
you might expect as in independent random case.
But what we really want to understand
is the distribution of the number of common neighbors.
In particular, we want to upper bound the probability
that there are too many common neighbors.
We want to understand some kind of tail probabilities.
And to do that, one way to do tail probabilities
is to consider moments.
Yes, question?
AUDIENCE: [INAUDIBLE]
YUFEI ZHAO: Sorry, can you repeat the question?
AUDIENCE: How do you have the equality right
before the lemma?
YUFEI ZHAO: Question, how do I have the equality right
before the lemma?
So there, I'm actually saying for the Erdos-Renyi random
graph case--
that's the case-- in Erdos-Renyi random graph,
each edge, if you have the same edge probability, is 1 over q.
And then that's the number of common neighbors
you would expect.
So that's a heuristic for the Erdos-Renyi random graph case.
OK, so now let's try to understand
the distribution of the number of common neighbors.
So let's fix a u subset of Fp to the s with exactly s elements.
And I want to understand how many common neighbors it has.
So let's consider the number of common neighbors of u
and the d-th moment of this random variable.
So this is a common way to do upper tail bounds.
And one way to analyze such moments
is to decompose this count as a sum of indicator
random variables.
So let me write I of v to be the quantity which is 1 if f of uv
is 0 for all with ou and big U. In the words,
it's a common neighbor.
It's 1 if v is a common neighbor for u, and 0 otherwise.
So then the number of common neighbors
would simply be the sum of this indicator as v ranges
over the entire vertex set.
And I can expand the sum.
Then all of these are standard things
to do when you're trying to calculate moments.
OK, so I can bring this expectation inside
and try to understand what is the expectation of this object
inside.
Well, if all the v's are distinct,
then this is simply the expected number of Kst's.
But the v's might not be distinct.
So we need to be a little bit careful.
But that's not too hard to handle.
So let me write M sub r to be the number
of subjective functions from d element set to an r elements
set, an M to be sum of these M sub r's for r up to d.
Then, let's consider how many distinct v values are there.
If there are distinct v values, then they
take on that many possible values.
And Mr for the number of subjections,
and for each possible r, the exact number
for the exact value of this expectation
is q to the minus rs.
And that's exactly what we showed.
So this comes from the lemma just now.
But we chose-- I mean, look, this
is that binomial coefficient.
And you have this number here.
So they multiply it to at most 1.
So we have this number, this quantity there,
which I think of as a constant.
So this is a constant.
So the d-th moment is bounded.
And one way to get tail bounds once you have the moments
is that we can use Markov's inequality.
It tells us that the number of common neighbors of u,
the probability that u has too many common neighbors, more
than lambda common neighbors.
I can rewrite this inequality here
by raising both sides to the power d.
And then, using Markov inequality, take expectation.
So all of these are standard techniques
for upper tail estimation.
You want to understand the upper tails on random variable,
understand its moments and use Markov on its moments.
But now, we know we have some bound for the d-th moment,
right?
Which is M as we just showed.
So there is this bound here.
So far you can run the same argument
in the random graphs case.
And you wouldn't really do much different.
I mean everything is more or less the same
what I've said so far, although we
had to do a special calculation algebraically that
didn't really make sense--
I mean, that you have to show some kind of near independence.
Question?
AUDIENCE: Is that less than or equal to, right?
The Markov inequality.
YUFEI ZHAO: Ah, thank you.
So this is less than or equal to.
Thank you.
But now is where the algebra comes in,
so the algebraic geometry nature of this argument comes in.
It turns out that this quantity here--
previously, we said, at least heuristically,
in the random graphs case, it behaves like a qua Poisson
random variable.
So it's fairly uniform in a Poisson sense.
It turns out because of the algebraic nature
of the construction, this random variable
behaves nothing like a Poisson.
So it turns out it's highly constrained due to reasons
using algebraic geometry.
And I'll tell you exactly why.
So that the number of common neighbors
is either very small or very large.
And here is the claim that for every s and d,
there exists some c, such that if I
have a bunch of polynomials on fq to the s of degree
o at most d, then if you look at the number of zeros,
common zeros, of the f's, how many common zeros can you have?
It turns out it cannot just be some arbitrary number.
So this set has size either bounded at most c.
Or it is at least q minus something very small.
And I'll explain just a little bit
why this is the case, although I will not give a proof.
So either somehow you are working
in a zero dimensional case, or if this algebraic variety that
comes with it has positive dimension,
then you should have a lot more points.
And the reason for this dichotomy
has to do with how many points are there
on the algebraic variety over a finite field?
So I will not give a proof, although if you
look on the course website for a link to a reference,
that does have a proof.
But I will tell you what the key algebraic geometric
input is to that claim up here.
And this is a important and famous theorem
called a Lang Weil bound.
So the Lang Weil bound tells you that if you
have an algebraic variety, v. And for now, it is important
in order to say this properly to work in the algebraic closure,
so Fq bar as the algebraic closure,
it's the smallest field extension where I can
solve all polynomial equations.
Then, the variety cut out by a set of polynomials--
so v is this variety.
So if it is irreducible, it cannot be written as a union
of finite number of smaller varieties,
irreducible over Fq bar.
And all of these polynomials have degree bounded.
Then the question is, if I take these polynomials
and I look at how many Fq points does it have?
So in other words, now I leave the field.
I come back down to Earth, to the base field and ask,
what's the number of solutions where
the coordinates are in Fq?
OK, so how many points do we expect?
Well, the simplest example of an algebraic variety
is that of a subspace.
If you have a d-dimensional subspace over Fq,
you have q to the d points exactly.
So you expect something like q raised to the dimension
of the variety.
Now the dimensions is actually a somewhat subtle concept.
I won't define.
But there are many definitions in algebraic geometry.
It turns out it's not always exactly as nice
as in the case of linear subspaces.
But the Lang Weil bound tells us that is not too far off.
You have a deviation that is at most
on the order of q 1 over root q, where
there are some hidden constants depending
on the description of your variety in terms
of the degrees of the polynomial with the dimension
and the number of polynomials.
But the point is that the number of points on this variety
is basically should be around the same as the model
case, namely that of a subspace.
And that brings us some intuition
to why this lemma is true.
So you have those polynomials up there.
So there are some subtle points one
needs to verify about your disability,
but the punchline is that either you
are in the zero-dimensional case, in which case
you have something like [INAUDIBLE] theorem,
and that tells you that a number of solutions is bounded.
Or you're in the positive dimensional case, in which case
the Lang Weil term tells you, you
must have lots of solutions.
And there is no middle ground.
And now, we're ready to finish off Boris Bukh's construction.
So we see that applying this lemma up there with this, what
should be my polynomials be?
I'm going to use my polynomials f sub u of y to be--
well, I have a random polynomial up there.
So I'm trying to find common neighbors.
I'm trying to find common solutions.
So these are my polynomial as u ranges over big U.
So for q large enough, we find that the number
of common neighbors of u, the probability that it is bigger
than c, where c is supplied by that lemma,
is equal to the probability that a number of common neighbors
of u exceeds q over 2, where q over w is this quantity
rounded--
smaller than that quantity up there.
So if it has more than c solutions,
that automatically has a lot of solutions.
And now, we can apply the Markov inequality up there,
the tail bound on the moments to deduce that this probability is
at most M divided by q over 2 raised to the power d.
And the moral of it is that we should
have a very small number--
I mean, this should occur with very small probability.
So let's call u being a subset of v bad if u has r elements, u
is contained entirely on the left side or the right side
of the original bipartition.
And most importantly, u has a lot, namely more than c,
common neighbors in j.
So how many bad sets do we expect?
Basically, a very small number.
So the number of bad sets, bad use,
is upper bounded by-- well, for each choice of s elements,
the probability that something is bad
is this quantity up here, which we chose d to be a large
enough constant depending on s.
If you look at the choice of d up there,
see that this quantity here is quite a bit smaller
than the number of vertices.
And now, the last step is the same as our randomized
construction using Erdos-Renyi random graphs,
where we remove--
well, it's almost the same, but now
we remove one vertex from every bad set.
And we get some graph g prime.
And we just need to check that g prime has lots of edges.
We know got rid of all the bad sets.
So g prime is now K sub sc plus 1 free.
We got rid of all possibilities for s points having
more than c common neighbors.
Now, we just need to check that g prime has lots of edges.
Well, the expected number of edges in g prime
is at least the x--
what we removed one vertex for every bad u.
And each bad u carries with it at most n edges,
because there are only n edge on each side of the bipartition.
And, well, the number of edges of g
has expectations exactly n squared over q.
And the number of bad u's we saw up there is not very large.
So in particular, this quantity, the second term
is dominated by the first term.
And so we obtain the claimed number of edges.
Also, the graph has at most 2n vertices.
So we may have gotten got rid of some.
But actually, fewer vertices, the better.
At most, 2n vertices, and it is a case of s--
So it's Kst free for t large enough.
So this gives you another construction
of Kst free graphs.
And so today, we saw two different constructions
of Kst free graphs for constants s and t, but in both cases,
t is substantially larger than s.
But the most important thing is that they both match
the Kovari-Sos-Turan bound.
So it gives you some evidence that maybe
the Kovari-Sos-Turan conjecture, the theorem,
is tight up to at most a constant factor, although that
is a major open problem.
And it remains a very difficult it seems open problem, but one
that is of central importance in extremal graph theory
to try to come up perhaps with other constructions
that can do better.
Maybe they will have some algebraic input.
But maybe they will have some input from other ideas.
We do not know.
Question?
AUDIENCE: So is this q defined?
Because I remember q as a prime power,
but it doesn't say there.
YUFEI ZHAO: So question is, is q defined?
So just like in the proofs of polarity graphs and what not--
so you have some n.
You rounded down to the nearest prime powers.
So s is a constant.
So n is basically q to the s.
So take large n, round it down to the nearest prime power.
q could be a prime, for instance.
It could be a prime.
It could be a prime power.
Think q to be a prime.
So I'm saying for every q, there is a construction.
And for every n, you can round down to the nearest q to the s
and then run this construction.
Any more questions?
Great.
So next time I will begin by telling you a few more things
about why people really like this construction
and some conjectures that were solved using this idea
and some conjectures that still remain open
along the same lines.
And we'll also go beyond Kst, so other bipartide graphs
and show you how to do upper bounds
for those bipartide graphs.