WEBVTT
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YUFEI ZHAO: Last
time, we started
00:00:20.330 --> 00:00:24.950
discussing the extremal
problem for bipartide graphs.
00:00:24.950 --> 00:00:28.970
And in particular, we saw
the Kovari-Sos-Turan theorem,
00:00:28.970 --> 00:00:32.810
which tells us that if you
forbid your graph from having
00:00:32.810 --> 00:00:36.320
a complete bipartide
graph, Kst, then
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you have this upper bound on the
number of edges in your graph.
00:00:42.670 --> 00:00:43.480
So we gave a proof.
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It was fairly short, used
the double counting argument,
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and it give you this bound.
00:00:49.750 --> 00:00:53.050
And the next question is
how tight is this bound?
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Is there a lower bound
that is off by, let's say,
00:00:57.760 --> 00:01:00.130
at most a constant factor?
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And that's a major open problem.
00:01:02.260 --> 00:01:07.680
It's a conjecture that
this bound is tight
00:01:07.680 --> 00:01:09.830
up to constant factors.
00:01:09.830 --> 00:01:12.960
But that conjecture is known
for only a very small number
00:01:12.960 --> 00:01:14.010
of graphs.
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And we saw a couple
of examples last time.
00:01:17.590 --> 00:01:20.400
So last time we saw
construction that
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shows that for S equals
to 2, this bound is tight.
00:01:31.670 --> 00:01:36.110
So the extremal number for k
through 2 is on the order of n
00:01:36.110 --> 00:01:37.670
to the 3/2.
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So this theta means I'm
hiding constant factors.
00:01:42.300 --> 00:01:45.090
And our construction
used this polarity graph,
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which is essentially
the point line incidence
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graph of a projective plane.
00:01:49.590 --> 00:01:53.700
And a basic algebraic or
geometric fact, if you will,
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that two lines intersecting
at most one point.
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We also sketched a
construction that
00:02:03.010 --> 00:02:08.580
showed that for S equals to
3, this bound is also tight.
00:02:11.370 --> 00:02:13.060
And this construction
here involved
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using spheres, again, in some
space over a finite field.
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So both these constructions
are in some sense
00:02:22.960 --> 00:02:24.490
algebraic geometric.
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And you can ask, is
there a way to extend
00:02:26.590 --> 00:02:31.540
these ideas to construct other
examples of Kst free graphs
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with the right number of
edges using some ingredients
00:02:36.040 --> 00:02:37.870
from algebraic geometry?
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And today, I want to show
you two different ways
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of doing that.
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So the state of the art,
which I mentioned last time,
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let me remind you what is
known about constructions
00:02:55.650 --> 00:03:00.240
that achieve the right
exponent up there.
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So in a series of two
papers by [INAUDIBLE]
00:03:07.560 --> 00:03:11.460
it is shown that if
the constants t and s,
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such that t is large
enough compared to s,
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in particular t is bigger
than s minus 1 factorial,
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then the extremal
number of Kst's is
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of the order, same
order, as given
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in the upper bound of the
Kovari-Sos-Turan theorem.
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In particular, these range of
parameters allows you to do
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(2, 2), (3, 3), which we
already know how to do.
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But the next case is 4, 7.
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And it is still
open how to do 4, 6.
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So I want to show you
this construction.
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And I will tell you
exactly what the graph is.
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I'll give you an
explicit description
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of this graph, which is Kst
free and has lots of edges.
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And as I mentioned earlier, it
is an algebraic construction.
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So as before, we
start with p prime.
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And we will take n to be
a p raised to the power s.
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And let's restrict s to
be integer at least 2.
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And, of course, that's
same as last time,
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if you have other values
of n, take a prime close
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to the desired value and
then take it from there.
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To describe the construction,
let me remind you,
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the norm map, if you have a
field extension, in this case,
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specifically I am looking
at the field extension,
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this Fp to the s, I can
define a norm map as follows.
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Sending x to be the product
of all the conjugates,
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or the Galois conjugates of
x, in this field extension.
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So explicitly written out
is just that expression,
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which I can clean and collect
and write it down like this.
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So I wrote that the
image of this norm map
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lies in the base field Fp.
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And that is because--
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well, one of the many
reasons why this is the case,
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is that if you look at--
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so I'll denote this
norm map by N--
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if you look at N of x,
it raised to power p,
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leaves this value unchanged.
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And the base field
is the field where it
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is invariant under power by p.
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So here's the graph,
which I'll denote the norm
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graph with parameters p and s.
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So the norm graph will have
as vertices just the elements
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of this field extension.
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And the edges will be the set
of all pairs of vertices, not
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equal, of course, such that a
norm of their sum equals to 1.
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So that's the graph.
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This is an explicit description
of what the vertices
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and what the edges are.
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So now, we need to verify
a couple of things.
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One is that this graph has
the desired number of edges.
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It has lots of edges.
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And two is that this
graph is Kst free.
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So let's do both
of those things.
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So the first is let's check it
has the right number of edges.
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So that's a
relatively easy task.
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What we need to do is
to count for every a
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how many choices of b are
there in this field extension
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such that a plus b
has norm exactly 1?
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And I claim that that number--
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well, so here's a
basic algebra fact,
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that the number of elements in
this field extension with norm
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exactly 1 is precisely
p to the s minus 1
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divided by p to the s.
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And this is because
really we're looking
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in the multiplicative subgroup.
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So the multiplicative
group in this Fp to the s.
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And it has a cyclical--
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so there's a generator--
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the order of the cyclic
group is p to the s minus 1.
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So you're asking,
how many elements
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when raised to this power
here ends up at the identity?
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So that's the answer.
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So that's one aspect.
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And so as a result, every
vertex is adjacent to, well,
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how many vertices?
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For every given a I
need to solve for b.
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And basically, this
many solutions,
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I have to be just
slightly careful because I
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don't want loops in my graph.
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So I may need to subtract 1.
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So it's adjacent to at least
this number up here minus 1
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to account for possible
loops, which is pretty large,
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p to the s minus 1,
which in other words
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is n raised to 1 minus 1
over s, that many vertices,
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and you see that this gives
you the right number of edges.
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So this is a graph
with lots of edges.
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So that part wasn't so hard.
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The next part,
it's much trickier,
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which is we want to check
that this graph has no Kst's.
00:10:41.060 --> 00:10:43.100
So previously in our
algebraic construction,
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we used some geometric
facts, such as no two lines
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intersect in more than one point
to show that there's no k to 2
00:10:51.230 --> 00:10:53.120
in the polarity graph.
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So there's going to be
something like that here.
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So the claim is that this
construction, this norm graph,
00:11:05.410 --> 00:11:12.880
is Ks, s factorial plus 1 free.
00:11:15.780 --> 00:11:17.890
So it's not quite
the bound I claimed.
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So it's a little bit weaker.
00:11:19.180 --> 00:11:21.430
But it is in the spirit
of what I am claiming,
00:11:21.430 --> 00:11:27.060
namely that for t large
enough, this graph is Kst free.
00:11:27.060 --> 00:11:29.570
So for t a large enough
comes constant here
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will show s factorial plus 1.
00:11:33.470 --> 00:11:41.350
And as a result, it would follow
that the extremal number 4s
00:11:41.350 --> 00:11:54.060
sub s factorial plus
1 is at least 1/2
00:11:54.060 --> 00:11:56.280
minus little 1 of the constant--
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I don't already worry
about that much,
00:11:59.010 --> 00:12:03.568
but it's on the order of
n to the 2 minus 1 over s.
00:12:03.568 --> 00:12:06.860
OK, everyone with me?
00:12:06.860 --> 00:12:10.490
So we need to verify this
graph here has no Kst.
00:12:10.490 --> 00:12:11.411
Yes, question?
00:12:11.411 --> 00:12:14.638
AUDIENCE: Should that t be an s?
00:12:14.638 --> 00:12:16.180
YUFEI ZHAO: Yes,
that should be an s.
00:12:16.180 --> 00:12:16.680
Thank you.
00:12:21.480 --> 00:12:23.152
Any more questions?
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AUDIENCE: Should that
be s minus 1 factorial?
00:12:26.960 --> 00:12:29.060
YUFEI ZHAO: So we
will show later
00:12:29.060 --> 00:12:32.450
on a better result using
s minus 1 factorial,
00:12:32.450 --> 00:12:35.420
but for now I'll show you the
slightly weaker result, which
00:12:35.420 --> 00:12:36.810
is still in the same spirit.
00:12:36.810 --> 00:12:37.310
Yep.
00:12:37.310 --> 00:12:39.953
AUDIENCE: Is the stronger
result using the same graph?
00:12:39.953 --> 00:12:41.870
YUFEI ZHAO: We'll change
to a different graph.
00:12:41.870 --> 00:12:44.412
For the stronger result, we will
change to a different graph.
00:12:50.000 --> 00:12:55.260
OK, so now let's show that
this graph here is Kst free.
00:12:55.260 --> 00:12:58.653
And for that claim, we need to
invoke an algebraic fact, which
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let me write down now.
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So suppose we have a field f.
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Any field will work
with a finite field.
00:13:14.090 --> 00:13:15.290
Any field is fine.
00:13:15.290 --> 00:13:22.520
And I have a bunch of
elements from the field,
00:13:22.520 --> 00:13:31.320
such that a sub ij is
different from a sub
00:13:31.320 --> 00:13:37.120
rj for all i not the same as r.
00:13:40.240 --> 00:13:43.200
Then the system of equations--
00:13:47.468 --> 00:13:48.760
and I'll write down the system.
00:13:48.760 --> 00:13:50.450
So x minus 1--
00:13:50.450 --> 00:13:59.460
x1 minus a 1, 1, x2
minus a 1, 2, and so on.
00:13:59.460 --> 00:14:04.286
xs minus a1x equals to b1.
00:14:04.286 --> 00:14:06.070
That's the first equation.
00:14:06.070 --> 00:14:09.240
Second equation,
x1 minus a 2, 1.
00:14:09.240 --> 00:14:13.630
It almost looks like the usual
system of linear equations.
00:14:13.630 --> 00:14:15.790
But I'm taking products.
00:14:21.170 --> 00:14:22.470
And so on.
00:14:22.470 --> 00:14:34.910
So the last one being x1 minus
a s1 x2 minus a s2, dot dot dot,
00:14:34.910 --> 00:14:40.460
xs minus a ss equals to b sub s.
00:14:40.460 --> 00:14:54.400
The system has at most s
factorial solutions, where
00:14:54.400 --> 00:14:56.320
I'm working inside this field.
00:14:59.780 --> 00:15:01.370
So that's the claim.
00:15:01.370 --> 00:15:03.782
So let me just give you some
intuition for this claim.
00:15:08.710 --> 00:15:16.840
Suppose the right side
vector is 0, all zeroes.
00:15:16.840 --> 00:15:19.730
Then I claim that
this is trivial.
00:15:19.730 --> 00:15:21.280
So what is the saying?
00:15:21.280 --> 00:15:28.740
I need to select x1 to be one
of the a's from the first row
00:15:28.740 --> 00:15:32.110
and x2 be one of the a's from
the second row, and so on.
00:15:32.110 --> 00:15:34.990
But each column
of a is distinct.
00:15:34.990 --> 00:15:36.670
So that's the hypothesis.
00:15:36.670 --> 00:15:40.660
You have all the a's in the
first column are distinct.
00:15:40.660 --> 00:15:45.940
So no two of the x's can--
00:15:49.150 --> 00:15:53.350
so I need to set one
of the xi's to be one
00:15:53.350 --> 00:15:55.330
of the a's from the first row.
00:15:55.330 --> 00:16:01.210
But you see that you
cannot set x1 to be a 1, 1,
00:16:01.210 --> 00:16:05.420
and x1 to be a s1
at the same time.
00:16:05.420 --> 00:16:08.450
So the solution just
counts permutations,
00:16:08.450 --> 00:16:10.339
which is exactly s factorial.
00:16:21.590 --> 00:16:24.910
So this algebraic
fact plays a key role
00:16:24.910 --> 00:16:29.550
in the proof of the theorem
that the lower bound
00:16:29.550 --> 00:16:34.720
that we're stating up there,
if you look at the paper,
00:16:34.720 --> 00:16:36.160
they give a proof
of this result.
00:16:36.160 --> 00:16:37.810
And it's not a long proof.
00:16:37.810 --> 00:16:41.980
But it uses some commutative
algebra and algebraic geometry.
00:16:41.980 --> 00:16:45.490
And usually in a class,
if the instructor
00:16:45.490 --> 00:16:48.790
doesn't present the proof, it's
for one of several reasons.
00:16:48.790 --> 00:16:50.950
Maybe the proof is too short.
00:16:50.950 --> 00:16:52.360
It doesn't need to be presented.
00:16:52.360 --> 00:16:54.700
Maybe it's too long or
too difficult. Maybe
00:16:54.700 --> 00:16:57.790
it's not instructive
to the class.
00:16:57.790 --> 00:16:59.920
And the last reason,
which is the case here,
00:16:59.920 --> 00:17:01.930
is that I don't actually
understand the proof.
00:17:01.930 --> 00:17:04.780
As in I can follow
it line by line,
00:17:04.780 --> 00:17:08.079
but I don't understand
why it is true.
00:17:08.079 --> 00:17:12.640
And if one of you wants to
come up with a different proof
00:17:12.640 --> 00:17:16.690
or try to explain to
me how this seemingly
00:17:16.690 --> 00:17:23.829
elementary algebraic fact is
proved, I would appreciate it.
00:17:23.829 --> 00:17:26.710
For small values of x,
you can check it by hand.
00:17:26.710 --> 00:17:29.110
So x equals to 2,
you're solving a system
00:17:29.110 --> 00:17:31.390
of two quadratic equations.
00:17:31.390 --> 00:17:32.890
And that you can check by hand.
00:17:32.890 --> 00:17:35.050
And three maybe you can
do it with some work.
00:17:35.050 --> 00:17:40.270
But even with 4 it's not
so clear how to do it.
00:17:40.270 --> 00:17:47.410
And also, one of the geometric
intuition is that if b is o0,
00:17:47.410 --> 00:17:50.320
then you have exactly
s factorial solutions.
00:17:50.320 --> 00:17:52.090
And the geometric
intuition is somehow
00:17:52.090 --> 00:17:56.290
that if you move b
around, then the fiber,
00:17:56.290 --> 00:17:59.080
the sides of the fiber,
the number of solutions x
00:17:59.080 --> 00:18:00.110
can only go down.
00:18:00.110 --> 00:18:00.940
It can not go up.
00:18:00.940 --> 00:18:05.170
And this corresponds to some
algebraic geometry phenomenon.
00:18:05.170 --> 00:18:08.440
And that's all I will say about
this algebraic fact, which
00:18:08.440 --> 00:18:09.670
we'll now use as a black box.
00:18:17.330 --> 00:18:19.520
Great.
00:18:19.520 --> 00:18:22.790
So now we have that as
our algebraic input,
00:18:22.790 --> 00:18:27.950
let us show that a
norm graph is Kst free.
00:18:27.950 --> 00:18:30.080
It's actually not
so hard once you
00:18:30.080 --> 00:18:32.310
assume that theorem up there.
00:18:36.480 --> 00:18:47.400
So let's show at a norm graph
is Ks, s factorial plus 1 free.
00:18:50.070 --> 00:18:55.170
Well, what does it
mean to have a Kst?
00:18:55.170 --> 00:18:59.160
It means that if you
have distinct vertices,
00:18:59.160 --> 00:19:05.330
which then correspond
to elements,
00:19:05.330 --> 00:19:12.580
s elements, y1 through
ys, of this field,
00:19:12.580 --> 00:19:17.580
then the common neighbors
of these elements
00:19:17.580 --> 00:19:28.800
correspond to solutions of
this system of equations
00:19:28.800 --> 00:19:32.560
where I set all of
these values to be 1.
00:19:32.560 --> 00:19:35.320
But I can write l exactly
what these guys are
00:19:35.320 --> 00:19:39.480
because I have this form,
that representation there
00:19:39.480 --> 00:19:44.690
for the norm map, so
I can write it out.
00:19:44.690 --> 00:19:48.560
And now remember
this fact that when
00:19:48.560 --> 00:19:53.080
you are in characteristic p, x
plus y raised to the power p,
00:19:53.080 --> 00:20:00.500
is the same as x to the p
plus yp and characteristic p.
00:20:00.500 --> 00:20:04.720
So I can expand the remaining
parenthesis like that.
00:20:13.340 --> 00:20:16.050
So I want the first line
to equal to 1, and so on.
00:20:16.050 --> 00:20:19.950
And each of the lines
has that equal to 1.
00:20:31.560 --> 00:20:35.790
How many solutions in x does
this system of equations have?
00:20:35.790 --> 00:20:39.510
So even if I treat each
of x and x to the p
00:20:39.510 --> 00:20:42.330
and so on as separate
variables, that theorem
00:20:42.330 --> 00:20:49.200
appear tells me that there are
at most s factorial solutions
00:20:49.200 --> 00:20:50.250
in x.
00:20:53.310 --> 00:20:57.500
Satisfies all the hypotheses
of that theorem up there.
00:20:57.500 --> 00:21:04.250
Therefore, the graph is Ks
sub s factorial plus 1 free.
00:21:04.250 --> 00:21:07.610
You do not have more than
s plus 1 different values
00:21:07.610 --> 00:21:11.360
of x satisfying this
system of equations.
00:21:11.360 --> 00:21:15.287
And that's the proof that
this norm graph is Kst free.
00:21:15.287 --> 00:21:15.870
Yes, question?
00:21:15.870 --> 00:21:22.832
AUDIENCE: Why can't
powers of like [INAUDIBLE]
00:21:22.832 --> 00:21:24.790
YUFEI ZHAO: Sorry, can
you repeat the question?
00:21:24.790 --> 00:21:29.345
AUDIENCE: Why cannot the
powers of the y's be the same?
00:21:29.345 --> 00:21:31.970
YUFEI ZHAO: The question is why
cannot the powers of the p's be
00:21:31.970 --> 00:21:33.360
the same?
00:21:33.360 --> 00:21:36.210
So you are asking,
down the second column,
00:21:36.210 --> 00:21:39.330
let's say, why are all
these y's different?
00:21:39.330 --> 00:21:42.400
Because you're working
inside a field.
00:21:42.400 --> 00:21:47.370
And raising to a p in
this field is a bijection.
00:21:51.370 --> 00:21:54.040
So think about the order
of the cyclic group.
00:21:54.040 --> 00:21:57.840
It has co-prime to p.
00:21:57.840 --> 00:22:00.040
But great question.
00:22:00.040 --> 00:22:02.640
Anything else?
00:22:02.640 --> 00:22:04.340
OK, so this gives
you a construction
00:22:04.340 --> 00:22:11.870
that gives you Kst free for t
bigger than s plus s factorial.
00:22:11.870 --> 00:22:14.630
Now, let me show you how to
improve this construction
00:22:14.630 --> 00:22:19.280
to do a little bit better,
to get s minus 1 factorial.
00:22:19.280 --> 00:22:21.080
And the idea is
to take a variant
00:22:21.080 --> 00:22:24.790
of this norm graph, which we'll
call the projective norm graph.
00:22:31.670 --> 00:22:34.430
And the projective norm
graph will define it
00:22:34.430 --> 00:22:39.100
for s at least 3
is rather similar.
00:22:39.100 --> 00:22:41.160
But there's a twist.
00:22:41.160 --> 00:22:46.410
I have as the vertex set, not
just the field extension--
00:22:46.410 --> 00:22:50.490
OK, so now I take field
extension, but one level less.
00:22:50.490 --> 00:22:54.210
And I take a second
coordinate, which consists
00:22:54.210 --> 00:22:57.060
of non-zero elements from Fp.
00:23:01.340 --> 00:23:09.650
The edges are formed by putting
an edge between these two
00:23:09.650 --> 00:23:13.730
elements, if and
only if, the norm
00:23:13.730 --> 00:23:17.070
of the sum of the
first coordinates
00:23:17.070 --> 00:23:20.340
equals to the product of
the second coordinates.
00:23:24.480 --> 00:23:27.760
So now, you can run through
a similar calculation that
00:23:27.760 --> 00:23:29.290
tells you the number of edges.
00:23:29.290 --> 00:23:31.570
So first of all, the
number of vertices
00:23:31.570 --> 00:23:35.440
is p to the s minus
1 times p minus 1,
00:23:35.440 --> 00:23:39.600
so basically the same as p to s.
00:23:39.600 --> 00:23:49.200
And additionally, every
vertex has degree exactly
00:23:49.200 --> 00:23:54.690
p raised to s minus 1 minus 1.
00:23:54.690 --> 00:24:04.460
And the reason is that if I tell
you the values of x little x
00:24:04.460 --> 00:24:08.390
and big y, which
cannot equal minus x,
00:24:08.390 --> 00:24:10.340
or else you will
never form an edge,
00:24:10.340 --> 00:24:14.408
then they together uniquely
determine little y.
00:24:18.160 --> 00:24:20.520
So for every value of
big X and little x,
00:24:20.520 --> 00:24:23.880
I just need to run through
all the values of big Y other
00:24:23.880 --> 00:24:24.690
than minus x.
00:24:29.110 --> 00:24:34.870
So the number of edges
then equals to 1/2 times
00:24:34.870 --> 00:24:39.120
the number of vertices times the
degree of every vertex, which,
00:24:39.120 --> 00:24:48.520
as before, is the
claimed asymptotic.
00:24:48.520 --> 00:24:51.450
And the remaining thing to show
is that this projective norm
00:24:51.450 --> 00:24:52.730
graph is Kst free.
00:25:05.820 --> 00:25:11.385
So it's K sub s, s minus
1 factorial plus 1 free.
00:25:16.840 --> 00:25:19.010
It's a similar calculation
as the one before,
00:25:19.010 --> 00:25:22.240
but we need to take into
account the small variant
00:25:22.240 --> 00:25:23.520
in the construction.
00:25:27.120 --> 00:25:36.520
So suppose we fix s vertices,
labeled by this big Y's, little
00:25:36.520 --> 00:25:37.020
y's.
00:25:43.040 --> 00:25:49.100
And now we need to
solve for uppercase X,
00:25:49.100 --> 00:26:05.800
lowercase x in this
system of equations,
00:26:05.800 --> 00:26:09.230
so asking how many different
pairs, big X, little
00:26:09.230 --> 00:26:13.710
x, can appear as a solution
to this system of equations?
00:26:13.710 --> 00:26:21.030
Well, first of all, if some
pairs of the first coordinates,
00:26:21.030 --> 00:26:31.610
big I is equal to big J,
then if you have a solution,
00:26:31.610 --> 00:26:37.400
then that forces little y
to be the same as little j.
00:26:37.400 --> 00:26:40.890
And so the y's wouldn't have
been distinct to begin with.
00:26:40.890 --> 00:26:44.490
So this is not possible.
00:26:44.490 --> 00:26:48.690
So all the big Y's are distinct.
00:26:55.990 --> 00:27:03.320
Well, now, let's
divide these equations
00:27:03.320 --> 00:27:04.760
by the final equation.
00:27:10.040 --> 00:27:13.040
And we get that
the i-th equation
00:27:13.040 --> 00:27:26.970
becomes like that,
which you can rewrite
00:27:26.970 --> 00:27:37.850
by dividing by the coordinate
of the norm of big Y i
00:27:37.850 --> 00:27:39.800
minus big Y s.
00:27:39.800 --> 00:27:43.150
This is non-zero because we
just showed that all the big Y
00:27:43.150 --> 00:27:45.130
i's are distinct.
00:27:45.130 --> 00:27:49.780
If you divide by this norm here
and rearrange appropriately,
00:27:49.780 --> 00:27:53.860
we find that the equations
become like this.
00:28:15.870 --> 00:28:17.570
So after doing some
rearranging-- so
00:28:17.570 --> 00:28:20.540
this is the equation the set
of new questions that we get.
00:28:20.540 --> 00:28:29.210
And you see that if you
use new variables, x prime,
00:28:29.210 --> 00:28:33.300
do a substitution,
this being x prime,
00:28:33.300 --> 00:28:37.020
then it has basically
the same form as the one
00:28:37.020 --> 00:28:41.200
that we just saw with a
different set of constants.
00:28:41.200 --> 00:28:44.640
And in particular
from what we just saw,
00:28:44.640 --> 00:28:49.980
we see that you cannot have
more than s minus 1 factorial
00:28:49.980 --> 00:28:53.250
solutions in x.
00:28:53.250 --> 00:28:56.910
Now, they're s
minus 1 equations.
00:28:56.910 --> 00:29:00.380
And the field extension
working as the x minus 1
00:29:00.380 --> 00:29:03.880
field extension.
00:29:03.880 --> 00:29:08.690
So we saved an equation by
using this projectivization.
00:29:08.690 --> 00:29:11.060
And that's it.
00:29:11.060 --> 00:29:14.510
So this shows you the
claim of constructing
00:29:14.510 --> 00:29:20.110
a Kst free graph for t bigger
than s minus 1 factorial, which
00:29:20.110 --> 00:29:22.700
has the desired number of edges.
00:29:22.700 --> 00:29:23.583
Yes, question?
00:29:23.583 --> 00:29:26.072
AUDIENCE: Why do you
have the if some capital
00:29:26.072 --> 00:29:29.040
Y equals capital IJ?
00:29:29.040 --> 00:29:35.290
YUFEI ZHAO: OK, so the question
is why do I say this part?
00:29:35.290 --> 00:29:37.330
So I'm maybe
skipping a sentence.
00:29:37.330 --> 00:29:39.670
I'm saying, if there is
a solution to this system
00:29:39.670 --> 00:29:42.640
of equations x, if
these vertices have
00:29:42.640 --> 00:29:47.770
a common neighbor, then if
you have some x satisfying
00:29:47.770 --> 00:29:51.970
this system of equations, then
having two different big Y's
00:29:51.970 --> 00:29:56.640
being the same forces you
to have the two smaller y's
00:29:56.640 --> 00:29:57.546
being the same.
00:29:57.546 --> 00:29:58.340
AUDIENCE: OK.
00:29:58.340 --> 00:29:59.090
YUFEI ZHAO: Right.
00:29:59.090 --> 00:30:02.380
And then the Y's will
have been distinct.
00:30:02.380 --> 00:30:04.150
So for them to have
some common neighbors,
00:30:04.150 --> 00:30:08.000
you better have these
big Y's being distinct.
00:30:08.000 --> 00:30:10.620
Any more questions?
00:30:10.620 --> 00:30:12.033
Great.
00:30:12.033 --> 00:30:13.700
So as I mentioned,
it is an open problem
00:30:13.700 --> 00:30:21.090
to determine whether what is the
extremal number for K44, K45,
00:30:21.090 --> 00:30:22.880
K46.
00:30:22.880 --> 00:30:26.586
And you may ask, well, we
have this nice construction--
00:30:26.586 --> 00:30:31.600
it maybe somewhat mysterious
because of that, but explicit.
00:30:31.600 --> 00:30:33.580
And you can write
this graph down.
00:30:33.580 --> 00:30:37.180
And you can ask is
this graph K46 free?
00:30:37.180 --> 00:30:40.480
So do we gain one
extra number for free,
00:30:40.480 --> 00:30:44.410
maybe because we didn't
analyze things properly?
00:30:44.410 --> 00:30:46.580
And it turns out
that's not the case.
00:30:46.580 --> 00:30:50.200
So there was a very recent
paper just released last month
00:30:50.200 --> 00:30:54.850
showing that this graph here
for s equals to 4 actually does
00:30:54.850 --> 00:30:56.710
contain some K46's.
00:31:00.010 --> 00:31:05.140
So if you want to prove a
corresponding lower bound
00:31:05.140 --> 00:31:10.750
for K46, you better come up
with a different construction.
00:31:10.750 --> 00:31:13.030
And that's I think an
interesting direction
00:31:13.030 --> 00:31:13.680
to explore.
00:31:17.190 --> 00:31:18.300
Any questions?
00:31:18.300 --> 00:31:18.800
Yes.
00:31:18.800 --> 00:31:21.245
AUDIENCE: Do we know
of any similar results
00:31:21.245 --> 00:31:23.423
about this construction
not working for larger s?
00:31:23.423 --> 00:31:24.840
YUFEI ZHAO: The
question is, do we
00:31:24.840 --> 00:31:30.810
know any similar result of
about does this graph contain
00:31:30.810 --> 00:31:36.060
Kst for other values
of s and t less
00:31:36.060 --> 00:31:38.070
than the claimed threshold?
00:31:38.070 --> 00:31:39.250
It is unclear.
00:31:39.250 --> 00:31:42.630
So the paper that
was uploaded, it
00:31:42.630 --> 00:31:45.486
doesn't address the
issue of s bigger than 4.
00:31:45.486 --> 00:31:45.986
Yeah.
00:31:45.986 --> 00:31:50.870
AUDIENCE: Why Fp to the power s?
00:31:50.870 --> 00:31:53.530
YUFEI ZHAO: So question is
why Fp to the power of s?
00:31:53.530 --> 00:31:55.690
So let's go back to the
norm graph construction.
00:31:55.690 --> 00:31:58.660
So where do we use
Fp to the power of s?
00:31:58.660 --> 00:32:06.010
Well, certainly we needed it
to have the right edge count.
00:32:06.010 --> 00:32:08.360
So that comes up
in the edge count.
00:32:08.360 --> 00:32:12.500
And also in the
norm expression, you
00:32:12.500 --> 00:32:14.840
have the correct
number of factors.
00:32:20.960 --> 00:32:22.690
So I encourage you
to try for if you
00:32:22.690 --> 00:32:24.880
use a smaller or
bigger value of s,
00:32:24.880 --> 00:32:27.460
you either don't get
something which is Kst free
00:32:27.460 --> 00:32:31.130
or you have the wrong
number of edges.
00:32:31.130 --> 00:32:34.830
Any more questions?
00:32:34.830 --> 00:32:37.650
So later, I will show you
a different construction
00:32:37.650 --> 00:32:41.850
of Kst free graphs, again,
for t large compared to s
00:32:41.850 --> 00:32:44.790
that will not do as
well as this one.
00:32:44.790 --> 00:32:47.130
But it is a genuinely
different construction.
00:32:47.130 --> 00:32:50.340
And it uses the idea
of randomized algebraic
00:32:50.340 --> 00:32:54.210
construction, which is
something that actually was only
00:32:54.210 --> 00:32:55.600
developed a few years ago.
00:32:55.600 --> 00:32:57.947
It's a very recent development.
00:32:57.947 --> 00:32:58.780
And it's quite nice.
00:32:58.780 --> 00:33:01.260
So it combines some of the
things we've talked about
00:33:01.260 --> 00:33:03.860
with constructing using
random graphs to construct
00:33:03.860 --> 00:33:06.730
H free graphs on one hand,
and on the other hand,
00:33:06.730 --> 00:33:08.250
some of the algebraic ideas.
00:33:08.250 --> 00:33:11.250
In particular, we're not going
to use that theorem up there,
00:33:11.250 --> 00:33:14.777
but we'll use some other
algebraic geometry fact.
00:33:14.777 --> 00:33:16.110
OK, so let's take a quick break.
00:33:19.500 --> 00:33:23.820
So what I want to discuss
now is a relatively new idea
00:33:23.820 --> 00:33:28.740
called a randomized
algebraic construction, which
00:33:28.740 --> 00:33:33.930
combines some ideas from both
the randomized construction
00:33:33.930 --> 00:33:36.170
and also the algebraic
construction that we just saw.
00:33:39.970 --> 00:33:43.620
So this idea is due to Boris
Bukh just a few years ago.
00:33:47.660 --> 00:33:53.900
And the goal is to give an
alternative construction
00:33:53.900 --> 00:34:09.620
of a Kst free graph
with lots of edges
00:34:09.620 --> 00:34:19.610
provided that as before, t
is much larger compared to s.
00:34:23.659 --> 00:34:26.050
So this band here will
not be as good as the one
00:34:26.050 --> 00:34:27.159
that we just saw.
00:34:27.159 --> 00:34:29.280
And I will even
tell you what it is.
00:34:29.280 --> 00:34:30.520
But it's some constant.
00:34:30.520 --> 00:34:32.650
So for every s there
is some t, such
00:34:32.650 --> 00:34:33.870
that this construction works.
00:34:37.630 --> 00:34:44.239
As before, we working inside
some finite field geometry.
00:34:44.239 --> 00:34:47.062
So let's start with
a, a prime power.
00:34:47.062 --> 00:34:48.520
You can think of
prime if you like.
00:34:48.520 --> 00:34:49.810
It doesn't make so
much difference.
00:34:49.810 --> 00:34:51.550
So we're working
inside a finite field.
00:34:54.389 --> 00:34:59.180
And let's assume s is
fixed and at least 4.
00:35:03.513 --> 00:35:04.930
Let me write down
some parameters.
00:35:04.930 --> 00:35:06.320
Don't worry about them for now.
00:35:06.320 --> 00:35:08.730
Just think of them as
sufficiently large constants.
00:35:08.730 --> 00:35:11.470
So d is this quantity here.
00:35:11.470 --> 00:35:15.080
So we'll come back to it
later when it comes up.
00:35:15.080 --> 00:35:19.000
OK, so what's the idea?
00:35:19.000 --> 00:35:21.430
When we looked at the
randomized construction,
00:35:21.430 --> 00:35:22.690
we took a random graph.
00:35:22.690 --> 00:35:24.970
We took an Erdos Renyi graph.
00:35:24.970 --> 00:35:27.280
Every edge appeared
independently.
00:35:27.280 --> 00:35:30.010
And saw that it
has lots of edges,
00:35:30.010 --> 00:35:33.730
if you choose p property
and not so many copies of h.
00:35:33.730 --> 00:35:36.160
So you can remove
all the copies of h
00:35:36.160 --> 00:35:41.410
to get a graph with lots
of edges that is h free.
00:35:41.410 --> 00:35:43.310
What we're going to
do now is instead
00:35:43.310 --> 00:35:45.380
of taking the edges
randomly, we're
00:35:45.380 --> 00:35:48.205
going to take a
random polynomial.
00:35:54.030 --> 00:36:07.690
F will be a random
polynomial chosen uniformly
00:36:07.690 --> 00:36:19.720
among all polynomials with--
00:36:19.720 --> 00:36:24.510
so I wrote uppercase Y and
uppercase X and Y. Actually, X
00:36:24.510 --> 00:36:26.340
and Y, they're not
single variants.
00:36:26.340 --> 00:36:39.280
They are-- so each of them
is a vector of s variables.
00:36:41.870 --> 00:36:46.550
So in other words, x1
through xs are the variables
00:36:46.550 --> 00:36:47.570
in the polynomial.
00:36:47.570 --> 00:36:51.800
And then y1 through ys.
00:36:51.800 --> 00:36:54.058
So it's a 2s
variable polynomial.
00:36:58.180 --> 00:37:06.050
So among all polynomials
with degree, at most d--
00:37:06.050 --> 00:37:07.790
d being the number up here--
00:37:07.790 --> 00:37:13.840
in each of x and y
sets of variables.
00:37:13.840 --> 00:37:16.700
So you look at it
as xs variables,
00:37:16.700 --> 00:37:20.750
each monominal has
their exponents sum
00:37:20.750 --> 00:37:24.170
to our most d, and likewise
with each monomial for the y
00:37:24.170 --> 00:37:25.790
variables.
00:37:25.790 --> 00:37:29.260
So this is the random object.
00:37:29.260 --> 00:37:32.300
It's a random polynomial
in 2s variables.
00:37:32.300 --> 00:37:33.450
And the degree is bounded.
00:37:33.450 --> 00:37:35.990
So you only have a finite
number of possibilities,
00:37:35.990 --> 00:37:38.120
and I choose one of them
uniformly at random.
00:37:44.750 --> 00:37:47.030
And now, what's my graph?
00:37:47.030 --> 00:37:49.120
We're going to construct
a bipartide graph
00:37:49.120 --> 00:37:52.810
G. The bipartideness,
it's not so crucial.
00:37:52.810 --> 00:37:54.690
But it'll make our
life somewhat easier.
00:37:58.870 --> 00:38:00.330
So it's a bipartide graph.
00:38:00.330 --> 00:38:03.820
So it hs two vertex parts, which
I will label left and right,
00:38:03.820 --> 00:38:09.320
L and R. And they are both
the s dimensional vector
00:38:09.320 --> 00:38:11.600
space over f cube.
00:38:11.600 --> 00:38:18.530
And we'll put an edge
between two vertices,
00:38:18.530 --> 00:38:21.200
if and only if that
polynomial up there
00:38:21.200 --> 00:38:27.960
f evaluates to 0
on these values.
00:38:27.960 --> 00:38:29.200
That's the graph.
00:38:29.200 --> 00:38:32.090
So I give you a
random polynomial f.
00:38:32.090 --> 00:38:38.720
And then you put in edges
according to when f vanishes.
00:38:38.720 --> 00:38:44.420
So f, if you view the bipartide
graph as a subset of Fq
00:38:44.420 --> 00:38:49.210
to s cross F to the s,
then this is the zero set.
00:38:49.210 --> 00:38:52.970
The edge set is the zero set.
00:38:52.970 --> 00:38:56.480
Just like in random graphs,
with the construction
00:38:56.480 --> 00:38:58.950
with random graphs, we'll need
to show a couple of things.
00:38:58.950 --> 00:39:02.330
One is that has lots of edges,
which will not be hard to show.
00:39:02.330 --> 00:39:06.050
And second, that it will
have typically a small number
00:39:06.050 --> 00:39:09.650
of copies of Kst.
00:39:09.650 --> 00:39:13.000
And that will have
some ingredients which
00:39:13.000 --> 00:39:16.660
are similar to the random
graphs case we saw before,
00:39:16.660 --> 00:39:22.820
but it will have some new ideas
coming from algebraic geometry.
00:39:22.820 --> 00:39:27.360
First, let's show that this
graph has lots of edges.
00:39:27.360 --> 00:39:34.530
And that's a simple calculation,
because for every pair
00:39:34.530 --> 00:39:44.620
of points of vertices, I
claim that the probability--
00:39:44.620 --> 00:39:46.940
so here f is the random object--
00:39:46.940 --> 00:39:53.790
the probability that f evaluates
to 0 on this pair is exactly q.
00:39:53.790 --> 00:39:58.970
So exactly 1 over q, 1
over the size of the field.
00:39:58.970 --> 00:40:01.930
So this is not too hard.
00:40:01.930 --> 00:40:06.850
And the reason is that
the distribution of f
00:40:06.850 --> 00:40:32.210
is identical to if you are an
extra random constant on f,
00:40:32.210 --> 00:40:34.070
chosen uniformly at random.
00:40:34.070 --> 00:40:35.840
So I took a random polynomial.
00:40:35.840 --> 00:40:37.430
I shifted by a random constant.
00:40:37.430 --> 00:40:39.320
It's still uniformly
random polynomial,
00:40:39.320 --> 00:40:41.180
according to that distribution.
00:40:41.180 --> 00:40:46.670
But now, you see
that this, whatever f
00:40:46.670 --> 00:40:49.810
evaluated to, if I shift
by a random constant,
00:40:49.810 --> 00:40:53.700
you will end up with a
uniform distribution.
00:40:53.700 --> 00:41:03.640
So it tells you that that guy
up there is uniform distribution
00:41:03.640 --> 00:41:12.600
on every fixed point
u, v. So in particular,
00:41:12.600 --> 00:41:18.220
it hits 0 with probability
exactly 1 over q.
00:41:18.220 --> 00:41:25.310
And as a result, the number of
edges of g in the expectation
00:41:25.310 --> 00:41:31.620
is exactly n squared
over q, where
00:41:31.620 --> 00:41:34.270
n is the number of vertices.
00:41:34.270 --> 00:41:37.410
So n is actually not
the number of vertices,
00:41:37.410 --> 00:41:43.655
but the size of each vertex
part, namely q to the s.
00:41:43.655 --> 00:41:48.770
So you see that it gives you
the right number of edges, so n
00:41:48.770 --> 00:41:51.290
to the 2 minus 1 over s.
00:41:51.290 --> 00:41:53.280
So we have the right
number of edges.
00:41:53.280 --> 00:41:58.280
And now, we want to show that
this graph here typically does
00:41:58.280 --> 00:42:01.690
not have too many
copies of Kst's.
00:42:01.690 --> 00:42:03.500
It might have some
copies of Kst's.
00:42:03.500 --> 00:42:05.620
Somehow that is unavoidable.
00:42:05.620 --> 00:42:09.650
Just as in the random case, you
do have some copies of Kst's.
00:42:09.650 --> 00:42:11.360
But if they are not
too many copies,
00:42:11.360 --> 00:42:14.300
I can remove them and
obtain a Kst free graph.
00:42:17.830 --> 00:42:20.070
OK, so what is the intuition?
00:42:20.070 --> 00:42:22.590
How does it compare
to the case when
00:42:22.590 --> 00:42:27.260
you have a genuine
Erdos-Renyi random graph?
00:42:27.260 --> 00:42:32.603
Well, what is the expected
number of common neighbors?
00:42:36.050 --> 00:42:43.120
So if you fix some
u with let's say
00:42:43.120 --> 00:42:49.600
on the left side with
exactly s vertices,
00:42:49.600 --> 00:43:02.300
I want to understand, how many
common neighbors does u have?
00:43:09.845 --> 00:43:11.220
But because the
common neighbors,
00:43:11.220 --> 00:43:14.070
if he has too many common
neighbors, then that's a Kst.
00:43:16.820 --> 00:43:20.070
It is not hard to
calculate the expectation
00:43:20.070 --> 00:43:22.760
of this quantity, both
in the random graph case
00:43:22.760 --> 00:43:24.780
as well as in this case.
00:43:24.780 --> 00:43:27.780
And you can calculate if you
pretend every edge occurs
00:43:27.780 --> 00:43:32.570
independently, the expected
number of common neighbors
00:43:32.570 --> 00:43:39.210
is exactly n to the
q to the minus s.
00:43:39.210 --> 00:43:42.870
So there are s elements
of u, which is exactly 1.
00:43:45.980 --> 00:43:50.600
And you know that for a binomial
distribution with expectation 1
00:43:50.600 --> 00:43:53.720
and a large number of
variables, the distribution
00:43:53.720 --> 00:43:55.574
is approximately Poissonian.
00:44:03.802 --> 00:44:11.640
Ah, but that's in the case when
it's independently distributed,
00:44:11.640 --> 00:44:14.760
which is the case
in the case of GMP.
00:44:14.760 --> 00:44:17.760
But it turns out for
the algebraic setting
00:44:17.760 --> 00:44:22.240
we're doing here, things
don't behave independently.
00:44:22.240 --> 00:44:26.170
It's not that you're doing coin
flops for every possible edge.
00:44:26.170 --> 00:44:28.800
We're doing some randomized
algebraic construction.
00:44:28.800 --> 00:44:31.110
And for algebraic
geometry reasons,
00:44:31.110 --> 00:44:33.670
you will see that the
distribution is very much not
00:44:33.670 --> 00:44:35.660
like Poissonian.
00:44:35.660 --> 00:44:40.040
It will turn out that either
the number of common neighbors
00:44:40.040 --> 00:44:43.700
is bounded or it is very large.
00:44:46.810 --> 00:44:53.620
And that means that we can show
using some Markov inequality
00:44:53.620 --> 00:44:57.130
that the probability that it
is very large is quite small.
00:44:57.130 --> 00:45:02.300
So typically, it will not
have many common neighbors?
00:45:02.300 --> 00:45:06.740
And that's the intuition, and so
let's work out this intuition.
00:45:06.740 --> 00:45:07.620
Any questions so far?
00:45:12.220 --> 00:45:13.650
So how do we do
this calculation?
00:45:13.650 --> 00:45:16.070
So first, let's start
with something that's
00:45:16.070 --> 00:45:17.240
actually fairly elementary.
00:45:17.240 --> 00:45:25.060
So suppose you have some
parameters, r and s.
00:45:25.060 --> 00:45:27.160
So I think of r and s--
00:45:27.160 --> 00:45:29.980
so I have some
parameters r and s.
00:45:29.980 --> 00:45:34.160
And thick of them as constants.
00:45:34.160 --> 00:45:36.160
Have some restrictions,
but don't worry too much
00:45:36.160 --> 00:45:38.630
about them.
00:45:38.630 --> 00:45:47.600
Suppose I have two bounded
subsets of the finite field,
00:45:47.600 --> 00:45:54.590
where you have size
s and v has size r.
00:45:54.590 --> 00:45:58.310
Then the claim is
that the probability
00:45:58.310 --> 00:46:09.240
that f vanishes on the Cartesian
product of u and v. OK,
00:46:09.240 --> 00:46:12.840
so what do you expect it to be?
00:46:12.840 --> 00:46:17.310
So I have s r elements, and I
want f, this random polynomial,
00:46:17.310 --> 00:46:19.240
to vanish on the entire product.
00:46:19.240 --> 00:46:23.370
Well, if s, its value behaved
independently for every point,
00:46:23.370 --> 00:46:26.030
you should expect that
the probability is exactly
00:46:26.030 --> 00:46:30.060
q to the power minus s r.
00:46:30.060 --> 00:46:32.710
And it turns out
that is the case.
00:46:32.710 --> 00:46:33.480
So this is true.
00:46:33.480 --> 00:46:35.297
This is an exact statement.
00:46:38.570 --> 00:46:39.730
OK, so why is this true?
00:46:39.730 --> 00:46:42.100
So this is in some
sense a generalization
00:46:42.100 --> 00:46:44.250
of this claim over here.
00:46:44.250 --> 00:46:46.040
And you have to do a
little bit more work.
00:46:46.040 --> 00:46:48.520
But it's not too difficult.
00:46:48.520 --> 00:46:55.080
So let's first
consider this lemma
00:46:55.080 --> 00:47:11.950
in a somewhat simpler case,
where all the first coordinates
00:47:11.950 --> 00:47:14.080
of x are distinct--
00:47:14.080 --> 00:47:16.370
of u are distinct.
00:47:19.440 --> 00:47:23.720
And all the first coordinates
of v are distinct.
00:47:23.720 --> 00:47:25.660
Suppose u and v have that form.
00:47:25.660 --> 00:47:29.140
So I write down the
list of points for u
00:47:29.140 --> 00:47:31.270
and first coordinates
are all distinct.
00:47:33.990 --> 00:47:39.000
What I want to do is to
give you a random shift,
00:47:39.000 --> 00:47:43.470
to do a uniform random shift.
00:47:43.470 --> 00:47:47.280
And I will shift it by
a polynomial g, which
00:47:47.280 --> 00:47:48.960
is a bivariate polynomial.
00:47:48.960 --> 00:47:50.600
So these are not vectors.
00:47:50.600 --> 00:47:55.180
They are just single variables.
00:47:55.180 --> 00:48:03.770
And I look at all
possible sum of monomials,
00:48:03.770 --> 00:48:12.560
where the degree in i is less
than s, and the degree in j
00:48:12.560 --> 00:48:13.880
is less than r.
00:48:16.640 --> 00:48:24.360
And these a's are chosen
uniformly independently
00:48:24.360 --> 00:48:32.330
at random from the
ground field fq.
00:48:34.910 --> 00:48:39.800
And as before, we see
that f and f plus g
00:48:39.800 --> 00:48:45.890
have the same distribution, the
same probability distribution.
00:48:45.890 --> 00:48:49.660
And so all it remains to show
that is whatever f comes out
00:48:49.660 --> 00:48:53.530
to be, if I tack on
this extra random g,
00:48:53.530 --> 00:48:56.680
it creates a uniform
distribution on the values
00:48:56.680 --> 00:49:04.270
on the entire u cross
v. But, actually, see,
00:49:04.270 --> 00:49:09.350
I have sr choices exactly
for these coefficients.
00:49:09.350 --> 00:49:14.970
And I have sr values that
I'm trying to control.
00:49:14.970 --> 00:49:17.420
So really it's a
counting problem.
00:49:17.420 --> 00:49:27.210
And it suffices to show
a bijection, namely
00:49:27.210 --> 00:49:37.400
that for every possible
vector of values,
00:49:37.400 --> 00:49:40.630
there exists a choice
of coefficients,
00:49:40.630 --> 00:49:50.230
as above, such that g evaluates
to the prescribed values
00:49:50.230 --> 00:49:52.544
with the given coefficients.
00:50:02.710 --> 00:50:07.390
And that uniformity will follow
just because you have the exact
00:50:07.390 --> 00:50:10.410
if you just do a counting.
00:50:10.410 --> 00:50:13.663
And the one-dimensional
version of this claim,
00:50:13.663 --> 00:50:15.080
so let's think
about what that is.
00:50:15.080 --> 00:50:19.000
So if I have, let's say,
three points on the line
00:50:19.000 --> 00:50:22.330
and a degree 2 polynomial,
what I am saying
00:50:22.330 --> 00:50:26.860
is that if you give me the
values you want on these three
00:50:26.860 --> 00:50:31.090
points, I can produce for
you a unique polynomial that
00:50:31.090 --> 00:50:35.460
evaluates to the prescribed
values on these three points.
00:50:35.460 --> 00:50:39.250
And that you should all know
as Lagrange interpolation.
00:50:39.250 --> 00:50:42.720
So it tells you
exactly how to do that.
00:50:42.720 --> 00:50:45.390
And that works for many reasons.
00:50:45.390 --> 00:50:51.240
One of them is that the random
indeterminate is invertible.
00:50:51.240 --> 00:50:54.250
Here, we have multi-variables.
00:50:54.250 --> 00:50:58.290
So let's do Lagrange
interpolation twice, once
00:50:58.290 --> 00:51:01.920
for each variable.
00:51:01.920 --> 00:51:16.030
So we'll apply Lagrange
interpolation twice.
00:51:16.030 --> 00:51:23.360
So the first time, we'll see
that for all values of u,
00:51:23.360 --> 00:51:29.890
there exists a single
variate polynomial
00:51:29.890 --> 00:51:38.850
in the y variable with
degree at most r minus 1 that
00:51:38.850 --> 00:51:46.940
evaluates to the correct
values on the fixed little u.
00:51:54.280 --> 00:51:56.090
So do it for one
variable at a time.
00:51:56.090 --> 00:51:59.600
For fixed u, do
Lagrange interpolation
00:51:59.600 --> 00:52:02.200
on the y variable.
00:52:02.200 --> 00:52:06.490
And now, once we have
those things there,
00:52:06.490 --> 00:52:16.580
viewing the g that we want
to find as a polynomial whose
00:52:16.580 --> 00:52:22.840
coefficients are polynomials in
the x variables, but is itself
00:52:22.840 --> 00:52:30.570
is a polynomial
in the y variable,
00:52:30.570 --> 00:52:34.500
we find that again using
Lagrange interpolation,
00:52:34.500 --> 00:52:40.310
there exists these values
for these coefficients
00:52:40.310 --> 00:52:51.510
here, such that each
coefficient of if you plug
00:52:51.510 --> 00:53:04.650
in the first entry into little
u agrees with the coefficients
00:53:04.650 --> 00:53:09.460
of the g that we just found.
00:53:09.460 --> 00:53:11.540
And that should be the
case for every little u.
00:53:14.910 --> 00:53:17.320
So once you find
these polynomials
00:53:17.320 --> 00:53:20.280
and now you have a
bonafide polynomial in g.
00:53:20.280 --> 00:53:23.610
And that's the claim above.
00:53:23.610 --> 00:53:26.990
So using Lagrange interpolation
twice, once for each variable.
00:53:26.990 --> 00:53:31.370
So this is-- if you're
confused, just think about it.
00:53:31.370 --> 00:53:34.590
There is nothing deep here.
00:53:34.590 --> 00:53:37.890
So that finishes the
claim in the case
00:53:37.890 --> 00:53:42.160
when the first coordinates
are all distinct.
00:53:42.160 --> 00:53:45.180
So we use that fact crucially
in doing this Lagrange
00:53:45.180 --> 00:53:47.200
interpolation.
00:53:47.200 --> 00:53:52.350
Now, for general
u and v, where we
00:53:52.350 --> 00:53:53.910
don't have this
assumption of having
00:53:53.910 --> 00:53:57.090
distinct first
coordinates, well,
00:53:57.090 --> 00:54:00.840
let's make them to have
distinct first coordinates
00:54:00.840 --> 00:54:04.350
by considering a random
linear transformation,
00:54:04.350 --> 00:54:07.980
so using a probabilistic method.
00:54:07.980 --> 00:54:20.880
So we suffice as to find
invertible linear maps, p
00:54:20.880 --> 00:54:34.170
and s, on this vector space,
such that TU and SV have
00:54:34.170 --> 00:54:35.712
the above properties.
00:54:40.910 --> 00:54:43.150
So let me show you
how to do it for u.
00:54:43.150 --> 00:54:45.170
So I need to find you
a invertible linear
00:54:45.170 --> 00:54:48.060
transformation t.
00:54:48.060 --> 00:54:50.170
Well, it's just the first
coordinate that matters.
00:54:50.170 --> 00:54:55.910
So it suffices to
find just a linear map
00:54:55.910 --> 00:55:03.600
corresponding to the
first coordinate that
00:55:03.600 --> 00:55:05.886
is injective of u.
00:55:08.840 --> 00:55:12.020
Whatever linear map you have,
even if it's zero, that's fine,
00:55:12.020 --> 00:55:18.845
I can extend it to the
remaining coordinates.
00:55:18.845 --> 00:55:20.220
Actually, if it's
zero, then it's
00:55:20.220 --> 00:55:21.585
not going to be injective of u.
00:55:21.585 --> 00:55:24.990
So it better not be zero.
00:55:24.990 --> 00:55:27.720
OK, well, let's find
this map randomly.
00:55:27.720 --> 00:55:40.520
So pick t uniform via random
among all linear maps.
00:55:46.310 --> 00:55:48.310
And I want to understand
what is the probability
00:55:48.310 --> 00:55:53.440
of collision, bad event,
if two elements of u
00:55:53.440 --> 00:55:56.460
end up getting mapped
to the same point.
00:55:56.460 --> 00:55:58.710
Well, that's not too hard.
00:55:58.710 --> 00:56:07.150
So for every distinct pair
of points in fq to the s
00:56:07.150 --> 00:56:23.100
the probability
that they collide,
00:56:23.100 --> 00:56:24.600
think about why this is true.
00:56:24.600 --> 00:56:27.212
It's exactly 1 over q.
00:56:30.990 --> 00:56:33.750
If x and x prime, they're
differing at least one
00:56:33.750 --> 00:56:35.920
coordinate, then even just
along that coordinate,
00:56:35.920 --> 00:56:38.650
I can make them distinct.
00:56:38.650 --> 00:56:41.470
So this is the case
for every pair.
00:56:41.470 --> 00:56:47.170
So now by union
bound, the probability
00:56:47.170 --> 00:56:55.900
that t1 is injective on u
is at least 1 minus the size
00:56:55.900 --> 00:57:01.370
if u choose 2 times 1 over q.
00:57:01.370 --> 00:57:07.150
And that's why we chose
q to be large enough.
00:57:07.150 --> 00:57:13.540
So q is at least r squared, s
So this number here is positive.
00:57:13.540 --> 00:57:16.730
So such a t exists.
00:57:16.730 --> 00:57:21.350
And so we can transform this u
and v, two configurations where
00:57:21.350 --> 00:57:23.120
the first coordinates
are all distinct,
00:57:23.120 --> 00:57:26.790
and then run the
argument as before.
00:57:26.790 --> 00:57:27.892
OK, great.
00:57:30.660 --> 00:57:36.780
So what we've shown so far is
that if you look at these Ksr
00:57:36.780 --> 00:57:41.620
structures, they appear
with probability exactly--
00:57:41.620 --> 00:57:44.580
well, with expectation
exactly what
00:57:44.580 --> 00:57:49.320
you might expect as in
independent random case.
00:57:49.320 --> 00:57:51.210
But what we really
want to understand
00:57:51.210 --> 00:57:55.610
is the distribution of the
number of common neighbors.
00:57:55.610 --> 00:58:00.050
In particular, we want to
upper bound the probability
00:58:00.050 --> 00:58:02.360
that there are too
many common neighbors.
00:58:02.360 --> 00:58:08.160
We want to understand some
kind of tail probabilities.
00:58:08.160 --> 00:58:11.600
And to do that, one way
to do tail probabilities
00:58:11.600 --> 00:58:13.155
is to consider moments.
00:58:13.155 --> 00:58:15.622
Yes, question?
00:58:15.622 --> 00:58:21.920
AUDIENCE: [INAUDIBLE]
00:58:21.920 --> 00:58:25.485
YUFEI ZHAO: Sorry, can
you repeat the question?
00:58:25.485 --> 00:58:27.900
AUDIENCE: How do you
have the equality right
00:58:27.900 --> 00:58:29.832
before the lemma?
00:58:29.832 --> 00:58:32.790
YUFEI ZHAO: Question, how
do I have the equality right
00:58:32.790 --> 00:58:34.800
before the lemma?
00:58:34.800 --> 00:58:43.470
So there, I'm actually saying
for the Erdos-Renyi random
00:58:43.470 --> 00:58:44.860
graph case--
00:58:44.860 --> 00:58:47.820
that's the case-- in
Erdos-Renyi random graph,
00:58:47.820 --> 00:58:52.680
each edge, if you have the same
edge probability, is 1 over q.
00:58:52.680 --> 00:58:55.932
And then that's the
number of common neighbors
00:58:55.932 --> 00:58:56.640
you would expect.
00:59:00.200 --> 00:59:05.150
So that's a heuristic for the
Erdos-Renyi random graph case.
00:59:05.150 --> 00:59:07.220
OK, so now let's
try to understand
00:59:07.220 --> 00:59:10.680
the distribution of the
number of common neighbors.
00:59:10.680 --> 00:59:19.850
So let's fix a u subset of Fp to
the s with exactly s elements.
00:59:19.850 --> 00:59:22.790
And I want to understand how
many common neighbors it has.
00:59:22.790 --> 00:59:36.330
So let's consider the number
of common neighbors of u
00:59:36.330 --> 00:59:40.140
and the d-th moment of
this random variable.
00:59:40.140 --> 00:59:44.470
So this is a common way
to do upper tail bounds.
00:59:44.470 --> 00:59:47.590
And one way to
analyze such moments
00:59:47.590 --> 00:59:51.670
is to decompose this count
as a sum of indicator
00:59:51.670 --> 00:59:53.000
random variables.
00:59:53.000 --> 01:00:03.670
So let me write I of v to be the
quantity which is 1 if f of uv
01:00:03.670 --> 01:00:08.770
is 0 for all with ou
and big U. In the words,
01:00:08.770 --> 01:00:10.730
it's a common neighbor.
01:00:10.730 --> 01:00:16.737
It's 1 if v is a common
neighbor for u, and 0 otherwise.
01:00:20.220 --> 01:00:22.410
So then the number
of common neighbors
01:00:22.410 --> 01:00:31.200
would simply be the sum of
this indicator as v ranges
01:00:31.200 --> 01:00:33.080
over the entire vertex set.
01:00:36.590 --> 01:00:39.170
And I can expand the sum.
01:00:39.170 --> 01:00:41.780
Then all of these
are standard things
01:00:41.780 --> 01:00:43.925
to do when you're trying
to calculate moments.
01:00:56.150 --> 01:01:06.240
OK, so I can bring
this expectation inside
01:01:06.240 --> 01:01:10.170
and try to understand what is
the expectation of this object
01:01:10.170 --> 01:01:10.670
inside.
01:01:14.100 --> 01:01:17.420
Well, if all the
v's are distinct,
01:01:17.420 --> 01:01:22.045
then this is simply the
expected number of Kst's.
01:01:30.090 --> 01:01:32.090
But the v's might
not be distinct.
01:01:32.090 --> 01:01:34.220
So we need to be a
little bit careful.
01:01:34.220 --> 01:01:36.290
But that's not too
hard to handle.
01:01:36.290 --> 01:01:39.440
So let me write M sub
r to be the number
01:01:39.440 --> 01:01:45.640
of subjective functions from
d element set to an r elements
01:01:45.640 --> 01:01:53.100
set, an M to be sum of these
M sub r's for r up to d.
01:01:53.100 --> 01:01:58.390
Then, let's consider how many
distinct v values are there.
01:01:58.390 --> 01:02:03.820
If there are distinct
v values, then they
01:02:03.820 --> 01:02:07.570
take on that many
possible values.
01:02:07.570 --> 01:02:10.540
And Mr for the number
of subjections,
01:02:10.540 --> 01:02:16.240
and for each possible
r, the exact number
01:02:16.240 --> 01:02:18.640
for the exact value
of this expectation
01:02:18.640 --> 01:02:20.380
is q to the minus rs.
01:02:20.380 --> 01:02:22.210
And that's exactly
what we showed.
01:02:22.210 --> 01:02:25.970
So this comes from
the lemma just now.
01:02:29.770 --> 01:02:31.710
But we chose-- I
mean, look, this
01:02:31.710 --> 01:02:34.420
is that binomial coefficient.
01:02:34.420 --> 01:02:35.670
And you have this number here.
01:02:35.670 --> 01:02:37.530
So they multiply
it to at most 1.
01:02:42.280 --> 01:02:45.580
So we have this number,
this quantity there,
01:02:45.580 --> 01:02:47.090
which I think of as a constant.
01:02:47.090 --> 01:02:49.520
So this is a constant.
01:02:49.520 --> 01:02:51.590
So the d-th moment is bounded.
01:02:57.900 --> 01:03:02.260
And one way to get tail bounds
once you have the moments
01:03:02.260 --> 01:03:07.286
is that we can use
Markov's inequality.
01:03:07.286 --> 01:03:12.850
It tells us that the number
of common neighbors of u,
01:03:12.850 --> 01:03:18.460
the probability that u has too
many common neighbors, more
01:03:18.460 --> 01:03:22.590
than lambda common neighbors.
01:03:22.590 --> 01:03:29.690
I can rewrite this
inequality here
01:03:29.690 --> 01:03:34.730
by raising both
sides to the power d.
01:03:34.730 --> 01:03:42.350
And then, using Markov
inequality, take expectation.
01:03:51.950 --> 01:03:54.010
So all of these are
standard techniques
01:03:54.010 --> 01:03:56.740
for upper tail estimation.
01:03:56.740 --> 01:03:59.800
You want to understand the
upper tails on random variable,
01:03:59.800 --> 01:04:03.670
understand its moments and
use Markov on its moments.
01:04:06.710 --> 01:04:12.920
But now, we know we have some
bound for the d-th moment,
01:04:12.920 --> 01:04:14.600
right?
01:04:14.600 --> 01:04:16.880
Which is M as we just showed.
01:04:16.880 --> 01:04:18.560
So there is this bound here.
01:04:21.320 --> 01:04:23.420
So far you can run
the same argument
01:04:23.420 --> 01:04:25.100
in the random graphs case.
01:04:25.100 --> 01:04:27.550
And you wouldn't really
do much different.
01:04:27.550 --> 01:04:29.300
I mean everything is
more or less the same
01:04:29.300 --> 01:04:31.010
what I've said so
far, although we
01:04:31.010 --> 01:04:33.140
had to do a special
calculation algebraically that
01:04:33.140 --> 01:04:34.290
didn't really make sense--
01:04:34.290 --> 01:04:37.910
I mean, that you have to show
some kind of near independence.
01:04:37.910 --> 01:04:38.705
Question?
01:04:38.705 --> 01:04:43.455
AUDIENCE: Is that less
than or equal to, right?
01:04:43.455 --> 01:04:44.393
The Markov inequality.
01:04:44.393 --> 01:04:45.850
YUFEI ZHAO: Ah, thank you.
01:04:45.850 --> 01:04:47.470
So this is less
than or equal to.
01:04:47.470 --> 01:04:50.030
Thank you.
01:04:50.030 --> 01:04:52.950
But now is where the
algebra comes in,
01:04:52.950 --> 01:04:56.180
so the algebraic geometry nature
of this argument comes in.
01:04:56.180 --> 01:04:58.940
It turns out that
this quantity here--
01:05:02.990 --> 01:05:05.630
previously, we said,
at least heuristically,
01:05:05.630 --> 01:05:08.720
in the random graphs case,
it behaves like a qua Poisson
01:05:08.720 --> 01:05:10.140
random variable.
01:05:10.140 --> 01:05:13.900
So it's fairly uniform
in a Poisson sense.
01:05:13.900 --> 01:05:15.940
It turns out because
of the algebraic nature
01:05:15.940 --> 01:05:17.860
of the construction,
this random variable
01:05:17.860 --> 01:05:20.020
behaves nothing like a Poisson.
01:05:20.020 --> 01:05:31.590
So it turns out it's highly
constrained due to reasons
01:05:31.590 --> 01:05:32.790
using algebraic geometry.
01:05:32.790 --> 01:05:35.400
And I'll tell you exactly why.
01:05:35.400 --> 01:05:37.320
So that the number
of common neighbors
01:05:37.320 --> 01:05:40.620
is either very
small or very large.
01:05:40.620 --> 01:05:50.550
And here is the claim
that for every s and d,
01:05:50.550 --> 01:05:54.660
there exists some
c, such that if I
01:05:54.660 --> 01:06:07.040
have a bunch of polynomials
on fq to the s of degree
01:06:07.040 --> 01:06:14.260
o at most d, then if you
look at the number of zeros,
01:06:14.260 --> 01:06:29.860
common zeros, of the f's, how
many common zeros can you have?
01:06:29.860 --> 01:06:32.230
It turns out it cannot just
be some arbitrary number.
01:06:32.230 --> 01:06:42.400
So this set has size
either bounded at most c.
01:06:42.400 --> 01:06:48.600
Or it is at least q minus
something very small.
01:06:51.520 --> 01:06:52.990
And I'll explain
just a little bit
01:06:52.990 --> 01:06:55.790
why this is the case, although
I will not give a proof.
01:06:55.790 --> 01:06:58.930
So either somehow
you are working
01:06:58.930 --> 01:07:06.890
in a zero dimensional case, or
if this algebraic variety that
01:07:06.890 --> 01:07:10.670
comes with it has
positive dimension,
01:07:10.670 --> 01:07:12.600
then you should have
a lot more points.
01:07:26.360 --> 01:07:29.300
And the reason
for this dichotomy
01:07:29.300 --> 01:07:33.740
has to do with how
many points are there
01:07:33.740 --> 01:07:39.060
on the algebraic variety
over a finite field?
01:07:39.060 --> 01:07:41.362
So I will not give a
proof, although if you
01:07:41.362 --> 01:07:43.570
look on the course website
for a link to a reference,
01:07:43.570 --> 01:07:45.180
that does have a proof.
01:07:45.180 --> 01:07:47.670
But I will tell you what
the key algebraic geometric
01:07:47.670 --> 01:07:50.950
input is to that claim up here.
01:07:50.950 --> 01:07:54.910
And this is a important
and famous theorem
01:07:54.910 --> 01:07:56.330
called a Lang Weil bound.
01:08:01.180 --> 01:08:04.970
So the Lang Weil bound
tells you that if you
01:08:04.970 --> 01:08:13.250
have an algebraic variety, v.
And for now, it is important
01:08:13.250 --> 01:08:19.760
in order to say this properly to
work in the algebraic closure,
01:08:19.760 --> 01:08:27.060
so Fq bar as the
algebraic closure,
01:08:27.060 --> 01:08:29.560
it's the smallest field
extension where I can
01:08:29.560 --> 01:08:33.510
solve all polynomial equations.
01:08:33.510 --> 01:08:41.211
Then, the variety cut out
by a set of polynomials--
01:08:45.450 --> 01:08:46.710
so v is this variety.
01:08:46.710 --> 01:08:53.420
So if it is irreducible, it
cannot be written as a union
01:08:53.420 --> 01:08:57.300
of finite number of
smaller varieties,
01:08:57.300 --> 01:09:02.490
irreducible over Fq bar.
01:09:02.490 --> 01:09:09.570
And all of these polynomials
have degree bounded.
01:09:12.240 --> 01:09:18.220
Then the question is, if
I take these polynomials
01:09:18.220 --> 01:09:22.830
and I look at how many
Fq points does it have?
01:09:22.830 --> 01:09:25.390
So in other words,
now I leave the field.
01:09:25.390 --> 01:09:29.649
I come back down to Earth,
to the base field and ask,
01:09:29.649 --> 01:09:33.220
what's the number
of solutions where
01:09:33.220 --> 01:09:34.460
the coordinates are in Fq?
01:09:39.040 --> 01:09:40.790
OK, so how many
points do we expect?
01:09:40.790 --> 01:09:43.020
Well, the simplest example
of an algebraic variety
01:09:43.020 --> 01:09:45.189
is that of a subspace.
01:09:45.189 --> 01:09:48.640
If you have a d-dimensional
subspace over Fq,
01:09:48.640 --> 01:09:51.850
you have q to the
d points exactly.
01:09:51.850 --> 01:09:57.460
So you expect something like
q raised to the dimension
01:09:57.460 --> 01:09:59.010
of the variety.
01:09:59.010 --> 01:10:01.930
Now the dimensions is actually
a somewhat subtle concept.
01:10:01.930 --> 01:10:02.750
I won't define.
01:10:02.750 --> 01:10:06.070
But there are many definitions
in algebraic geometry.
01:10:06.070 --> 01:10:08.380
It turns out it's not
always exactly as nice
01:10:08.380 --> 01:10:10.360
as in the case of
linear subspaces.
01:10:10.360 --> 01:10:13.630
But the Lang Weil bound tells
us that is not too far off.
01:10:13.630 --> 01:10:19.330
You have a deviation
that is at most
01:10:19.330 --> 01:10:22.360
on the order of q 1
over root q, where
01:10:22.360 --> 01:10:25.060
there are some hidden
constants depending
01:10:25.060 --> 01:10:28.690
on the description of
your variety in terms
01:10:28.690 --> 01:10:30.940
of the degrees of the
polynomial with the dimension
01:10:30.940 --> 01:10:32.410
and the number of polynomials.
01:10:32.410 --> 01:10:36.240
But the point is that the
number of points on this variety
01:10:36.240 --> 01:10:40.300
is basically should be
around the same as the model
01:10:40.300 --> 01:10:41.665
case, namely that of a subspace.
01:10:44.540 --> 01:10:46.640
And that brings
us some intuition
01:10:46.640 --> 01:10:48.230
to why this lemma is true.
01:10:48.230 --> 01:10:50.822
So you have those
polynomials up there.
01:10:50.822 --> 01:10:52.280
So there are some
subtle points one
01:10:52.280 --> 01:10:54.200
needs to verify about
your disability,
01:10:54.200 --> 01:10:56.990
but the punchline
is that either you
01:10:56.990 --> 01:10:59.467
are in the zero-dimensional
case, in which case
01:10:59.467 --> 01:11:01.300
you have something like
[INAUDIBLE] theorem,
01:11:01.300 --> 01:11:05.590
and that tells you that a
number of solutions is bounded.
01:11:05.590 --> 01:11:08.085
Or you're in the positive
dimensional case, in which case
01:11:08.085 --> 01:11:09.460
the Lang Weil term
tells you, you
01:11:09.460 --> 01:11:11.440
must have lots of solutions.
01:11:11.440 --> 01:11:13.180
And there is no middle ground.
01:11:16.170 --> 01:11:20.636
And now, we're ready to finish
off Boris Bukh's construction.
01:11:26.100 --> 01:11:35.580
So we see that applying this
lemma up there with this, what
01:11:35.580 --> 01:11:37.690
should be my polynomials be?
01:11:37.690 --> 01:11:46.080
I'm going to use my polynomials
f sub u of y to be--
01:11:46.080 --> 01:11:49.300
well, I have a random
polynomial up there.
01:11:49.300 --> 01:11:53.640
So I'm trying to find
common neighbors.
01:11:53.640 --> 01:11:56.010
I'm trying to find
common solutions.
01:11:59.970 --> 01:12:08.450
So these are my polynomial
as u ranges over big U.
01:12:08.450 --> 01:12:17.840
So for q large enough,
we find that the number
01:12:17.840 --> 01:12:24.320
of common neighbors of u, the
probability that it is bigger
01:12:24.320 --> 01:12:30.600
than c, where c is
supplied by that lemma,
01:12:30.600 --> 01:12:35.000
is equal to the probability that
a number of common neighbors
01:12:35.000 --> 01:12:41.430
of u exceeds q over 2, where
q over w is this quantity
01:12:41.430 --> 01:12:42.780
rounded--
01:12:42.780 --> 01:12:45.410
smaller than that
quantity up there.
01:12:45.410 --> 01:12:47.220
So if it has more
than c solutions,
01:12:47.220 --> 01:12:49.170
that automatically has
a lot of solutions.
01:12:52.230 --> 01:12:55.890
And now, we can apply the
Markov inequality up there,
01:12:55.890 --> 01:13:00.330
the tail bound on the moments to
deduce that this probability is
01:13:00.330 --> 01:13:05.400
at most M divided by q over
2 raised to the power d.
01:13:08.580 --> 01:13:10.560
And the moral of it
is that we should
01:13:10.560 --> 01:13:12.900
have a very small number--
01:13:12.900 --> 01:13:16.030
I mean, this should occur
with very small probability.
01:13:16.030 --> 01:13:29.140
So let's call u being a subset
of v bad if u has r elements, u
01:13:29.140 --> 01:13:32.780
is contained entirely on the
left side or the right side
01:13:32.780 --> 01:13:35.480
of the original bipartition.
01:13:35.480 --> 01:13:41.240
And most importantly, u has
a lot, namely more than c,
01:13:41.240 --> 01:13:49.110
common neighbors in j.
01:13:53.086 --> 01:13:56.650
So how many bad
sets do we expect?
01:13:56.650 --> 01:13:58.940
Basically, a very small number.
01:13:58.940 --> 01:14:07.320
So the number of
bad sets, bad use,
01:14:07.320 --> 01:14:12.180
is upper bounded by-- well,
for each choice of s elements,
01:14:12.180 --> 01:14:14.130
the probability that
something is bad
01:14:14.130 --> 01:14:22.810
is this quantity up here,
which we chose d to be a large
01:14:22.810 --> 01:14:24.700
enough constant depending on s.
01:14:24.700 --> 01:14:27.880
If you look at the
choice of d up there,
01:14:27.880 --> 01:14:32.350
see that this quantity
here is quite a bit smaller
01:14:32.350 --> 01:14:34.540
than the number of vertices.
01:14:40.730 --> 01:14:52.070
And now, the last step is
the same as our randomized
01:14:52.070 --> 01:14:55.970
construction using
Erdos-Renyi random graphs,
01:14:55.970 --> 01:14:59.180
where we remove--
01:14:59.180 --> 01:15:00.880
well, it's almost
the same, but now
01:15:00.880 --> 01:15:06.640
we remove one vertex
from every bad set.
01:15:17.360 --> 01:15:22.520
And we get some graph g prime.
01:15:22.520 --> 01:15:26.540
And we just need to check that
g prime has lots of edges.
01:15:26.540 --> 01:15:29.000
We know got rid of
all the bad sets.
01:15:29.000 --> 01:15:38.180
So g prime is now K
sub sc plus 1 free.
01:15:42.390 --> 01:15:46.110
We got rid of all possibilities
for s points having
01:15:46.110 --> 01:15:49.040
more than c common neighbors.
01:15:49.040 --> 01:15:52.720
Now, we just need to check
that g prime has lots of edges.
01:15:52.720 --> 01:15:56.870
Well, the expected number
of edges in g prime
01:15:56.870 --> 01:16:01.180
is at least the x--
01:16:01.180 --> 01:16:09.750
what we removed one
vertex for every bad u.
01:16:09.750 --> 01:16:15.210
And each bad u carries
with it at most n edges,
01:16:15.210 --> 01:16:19.034
because there are only n edge
on each side of the bipartition.
01:16:26.950 --> 01:16:29.810
And, well, the
number of edges of g
01:16:29.810 --> 01:16:33.530
has expectations exactly
n squared over q.
01:16:36.360 --> 01:16:46.780
And the number of bad u's we
saw up there is not very large.
01:16:46.780 --> 01:16:51.910
So in particular, this
quantity, the second term
01:16:51.910 --> 01:16:54.890
is dominated by the first term.
01:16:54.890 --> 01:17:01.640
And so we obtain the
claimed number of edges.
01:17:09.300 --> 01:17:16.550
Also, the graph has
at most 2n vertices.
01:17:16.550 --> 01:17:18.630
So we may have gotten
got rid of some.
01:17:18.630 --> 01:17:21.300
But actually, fewer
vertices, the better.
01:17:21.300 --> 01:17:28.790
At most, 2n vertices,
and it is a case of s--
01:17:28.790 --> 01:17:31.190
So it's Kst free
for t large enough.
01:17:34.320 --> 01:17:36.510
So this gives you
another construction
01:17:36.510 --> 01:17:39.180
of Kst free graphs.
01:17:39.180 --> 01:17:41.180
And so today, we saw two
different constructions
01:17:41.180 --> 01:17:45.020
of Kst free graphs for constants
s and t, but in both cases,
01:17:45.020 --> 01:17:47.690
t is substantially
larger than s.
01:17:47.690 --> 01:17:50.270
But the most important thing
is that they both match
01:17:50.270 --> 01:17:54.380
the Kovari-Sos-Turan bound.
01:17:54.380 --> 01:17:57.840
So it gives you some
evidence that maybe
01:17:57.840 --> 01:18:01.290
the Kovari-Sos-Turan
conjecture, the theorem,
01:18:01.290 --> 01:18:04.710
is tight up to at most a
constant factor, although that
01:18:04.710 --> 01:18:06.570
is a major open problem.
01:18:06.570 --> 01:18:11.430
And it remains a very difficult
it seems open problem, but one
01:18:11.430 --> 01:18:15.360
that is of central importance
in extremal graph theory
01:18:15.360 --> 01:18:20.500
to try to come up perhaps
with other constructions
01:18:20.500 --> 01:18:22.200
that can do better.
01:18:22.200 --> 01:18:25.050
Maybe they will have
some algebraic input.
01:18:25.050 --> 01:18:28.120
But maybe they will have
some input from other ideas.
01:18:28.120 --> 01:18:29.220
We do not know.
01:18:29.220 --> 01:18:30.026
Question?
01:18:30.026 --> 01:18:32.406
AUDIENCE: So is this q defined?
01:18:32.406 --> 01:18:34.429
Because I remember
q as a prime power,
01:18:34.429 --> 01:18:35.750
but it doesn't say there.
01:18:35.750 --> 01:18:37.510
YUFEI ZHAO: So question
is, is q defined?
01:18:37.510 --> 01:18:40.900
So just like in the proofs of
polarity graphs and what not--
01:18:40.900 --> 01:18:43.750
so you have some n.
01:18:43.750 --> 01:18:47.140
You rounded down to the
nearest prime powers.
01:18:47.140 --> 01:18:48.370
So s is a constant.
01:18:48.370 --> 01:18:52.150
So n is basically q to the s.
01:18:52.150 --> 01:18:56.200
So take large n, round it down
to the nearest prime power.
01:18:56.200 --> 01:18:57.712
q could be a prime,
for instance.
01:18:57.712 --> 01:18:58.545
It could be a prime.
01:18:58.545 --> 01:19:00.140
It could be a prime power.
01:19:00.140 --> 01:19:01.300
Think q to be a prime.
01:19:04.050 --> 01:19:07.420
So I'm saying for every q,
there is a construction.
01:19:07.420 --> 01:19:12.790
And for every n, you can round
down to the nearest q to the s
01:19:12.790 --> 01:19:17.400
and then run this construction.
01:19:17.400 --> 01:19:19.950
Any more questions?
01:19:19.950 --> 01:19:20.450
Great.
01:19:20.450 --> 01:19:24.230
So next time I will begin by
telling you a few more things
01:19:24.230 --> 01:19:27.770
about why people really
like this construction
01:19:27.770 --> 01:19:31.198
and some conjectures that
were solved using this idea
01:19:31.198 --> 01:19:32.990
and some conjectures
that still remain open
01:19:32.990 --> 01:19:34.650
along the same lines.
01:19:34.650 --> 01:19:39.530
And we'll also go beyond Kst,
so other bipartide graphs
01:19:39.530 --> 01:19:41.060
and show you how
to do upper bounds
01:19:41.060 --> 01:19:43.390
for those bipartide graphs.