WEBVTT
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PROFESSOR: We spent the
last few lectures developing
00:00:21.190 --> 00:00:23.530
the theory of graph limits.
00:00:23.530 --> 00:00:26.230
And one of the
motivations I gave
00:00:26.230 --> 00:00:28.720
at the beginning of the
lecture on graph limits
00:00:28.720 --> 00:00:32.830
was that there were
certain graph inequalities.
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Specifically, if I tell you that
your graph has edge density one
00:00:37.580 --> 00:00:41.840
half, what's the minimum
possible C4 density?
00:00:41.840 --> 00:00:43.810
So for those kinds of
problems, graph limits
00:00:43.810 --> 00:00:47.890
gives us a very nice
language for describing
00:00:47.890 --> 00:00:49.990
what the answer is,
and also sometimes
00:00:49.990 --> 00:00:51.830
for solving these problems.
00:00:51.830 --> 00:00:56.620
So today, I want to dive more
into these types of problems.
00:00:56.620 --> 00:00:59.470
Specifically, we're going to
be talking about homomorphism
00:00:59.470 --> 00:01:00.730
density inequalities.
00:01:06.502 --> 00:01:07.340
Homomorphism.
00:01:17.990 --> 00:01:20.450
So trying to understand
what is the relationship
00:01:20.450 --> 00:01:24.900
between possible subgraph
densities or homomorphism
00:01:24.900 --> 00:01:28.830
densities within a large graph.
00:01:28.830 --> 00:01:31.180
We've seen these kind
of problems in the past.
00:01:31.180 --> 00:01:32.640
So one of the very
first theorems
00:01:32.640 --> 00:01:35.105
that we did in this
course was Turan's theorem
00:01:35.105 --> 00:01:35.980
and Mantel's theorem.
00:01:41.710 --> 00:01:44.310
So specifically, for
Mantel's theorem,
00:01:44.310 --> 00:01:48.930
it tells us something about the
possible edge versus triangle
00:01:48.930 --> 00:01:53.730
densities in the graph,
which is something
00:01:53.730 --> 00:01:56.940
that I want to spend the
first part of today's lecture
00:01:56.940 --> 00:01:57.810
focusing on.
00:01:57.810 --> 00:02:00.130
So what is the
possible relationship?
00:02:00.130 --> 00:02:07.650
What are all the possible
edge versus triangle densities
00:02:07.650 --> 00:02:10.539
in the graph?
00:02:10.539 --> 00:02:12.630
Mantel's theorem
tells us something--
00:02:12.630 --> 00:02:18.120
namely, that if your edge
density exceeds one half,
00:02:18.120 --> 00:02:22.130
then your triangle
density cannot be zero.
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So that's what Mantel's
theorem tells us.
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And let me write
it down like this.
00:02:27.230 --> 00:02:29.150
So the statement I
just said, the one
00:02:29.150 --> 00:02:32.690
about Mantel's theorem, and more
generally for Turan's theorem
00:02:32.690 --> 00:02:42.110
tells us that if the Kr
plus 1 density in W is 0,
00:02:42.110 --> 00:02:52.930
then necessarily the edge
density is at most 1 minus 1
00:02:52.930 --> 00:02:53.430
over r.
00:02:58.210 --> 00:03:01.090
So this is what it tells us.
00:03:01.090 --> 00:03:03.130
It gives us some
information about what
00:03:03.130 --> 00:03:04.570
are the possible densities.
00:03:04.570 --> 00:03:07.490
But I would like to
know more generally,
00:03:07.490 --> 00:03:09.370
or a good complete
picture, of what
00:03:09.370 --> 00:03:13.600
is a set of edge versus
triangle density inequalities.
00:03:13.600 --> 00:03:17.340
So let me draw a
picture that captures
00:03:17.340 --> 00:03:19.870
what we're looking for.
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So here on the x-axis, I
have all the possible edge
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densities, and on
the vertical axis,
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I have the triangle density.
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And I would like to know
what is a set of feasible
00:03:40.160 --> 00:03:42.590
points in this box.
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Mantel's theorem tells
us already something--
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namely, when can
you-- so this region,
00:03:54.530 --> 00:03:58.300
the horizontal line at
zero extends at most
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until the halfway point.
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Beyond this point, it's not a
part of the feasible region.
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So far that's the
information that we know.
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Our discussion about graph
limits, and in particular--
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so let me first write
down what is the question.
00:04:16.870 --> 00:04:24.870
So if you look at the set of
possible edge versus triangle
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densities, so there
is this region here.
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What is this region?
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It's a subset of
this unit square.
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We would like to
understand what is
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the set of all possibilities.
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The compactness of
the space of graphons
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tells us that this
region is compact.
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So let me call this region
D23 for edge versus triangle.
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So D23 is compact because
the space of graphons
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is compact under the cut
metric, and densities are
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continuous under cut distance.
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So in particular, if you
have some limit point
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of some sequence of
graphs, that limit point's
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achieved by a corresponding
limit graphon.
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So you really have a nice
closed region over here.
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So we don't have to--
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I should be able to
tell you the answer.
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This is the region.
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There should not be any
additional quantifiers.
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There's no optimizer
zero, missing this point
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and missing that point.
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It's a closed region.
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So what is this closed region?
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Equivalently, we can ask
the following question.
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Suppose I give you
the edge density.
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In other words, look at a
particular horizontal place
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in this picture.
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What is the maximum and minimum
possible triangle densities?
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So I tell you that the
edge density is 0.75.
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What is the upper and lower
boundaries of this region?
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I want you to think about
why this region is--
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the vertical cross-section
is a line segment.
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You cannot have any hulls.
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So that requires an
argument, and I'll
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let you think about that.
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So I want to complete
this picture,
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and I'll show you some proofs.
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And at the end of--
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well, by the middle
of today's lecture,
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we'll see a picture of what
this region looks like.
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All right.
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First, let me do the
easier direction, which
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is to find the upper
boundary of this region.
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So what is the maximum
possible triangle density
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for a given edge density?
00:07:28.950 --> 00:07:34.680
And the answer-- so it turns
out to be what I will--
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the result I will tell
you is a special case
00:07:37.080 --> 00:07:40.012
of what's called Kruskal-Katona.
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Think about it this way.
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Suppose I give you a very
large number of vertices
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and I give you some
large number of edges,
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and I want you to put the edges
into the graph in a way that
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generates as many
triangles as possible.
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Intuitively, how should
you put the edges
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in to try to make as many
triangles as you can?
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AUDIENCE: Clique.
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PROFESSOR: In a clique.
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So you put all the edges as
closely together as possible,
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try to form a clique.
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So maximize number of
triangles by forming a clique.
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And that is indeed the answer.
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And this is what we'll prove,
at least in the graph densities
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version.
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So we will show that the upper
boundary is given by the curve
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y equals to x to the 3/2.
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So don't worry about
the specific function.
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But what's important is that
the upper bound is achieved
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by the following graphon.
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Namely, this graphon
corresponding to a clique.
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For this graphon here, the
edge density is a squared,
00:09:14.780 --> 00:09:19.980
and the triangle
density is a cubed.
00:09:19.980 --> 00:09:22.440
And it turns out this
graphon is the best that you
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can do with a given edge
density in order to generate
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as many triangles or the most
triangle density possible.
00:09:31.500 --> 00:09:34.220
In other words,
what we'll prove is
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that the triangle density
throughout W will be a graphon.
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So W is always a graphon.
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So values between 0 and 1.
00:09:50.230 --> 00:09:52.434
Then you have the
following inequality.
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So let's prove it.
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First let me draw you what
this shape looks like.
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Because of the relationship
between graphs and graph
00:10:22.760 --> 00:10:25.760
limits, any of these
inequalities about graph
00:10:25.760 --> 00:10:29.660
limits, about graphons,
it's sufficient to prove
00:10:29.660 --> 00:10:32.750
the corresponding
inequality for graphs
00:10:32.750 --> 00:10:35.840
because the set
of graphs is dense
00:10:35.840 --> 00:10:39.620
within the space of graphons
according to the topology--
00:10:39.620 --> 00:10:41.930
namely, the cut metric
that we discussed.
00:10:41.930 --> 00:10:52.100
So it suffices to show the
corresponding inequality
00:10:52.100 --> 00:10:54.350
about graphs--
00:10:54.350 --> 00:10:59.750
namely, that the K3
density in a graph
00:10:59.750 --> 00:11:08.580
is at most the K2 density in a
graph raised to the power 3/2.
00:11:08.580 --> 00:11:11.460
So let me belabor this point
just a little bit more.
00:11:11.460 --> 00:11:14.850
This inequality is a subset
of those inequalities
00:11:14.850 --> 00:11:18.210
up there because graphs sit
inside a space of graphons.
00:11:18.210 --> 00:11:22.860
But because they sit
inside in a dense subset,
00:11:22.860 --> 00:11:27.100
if you know this inequality
and everything is continuous,
00:11:27.100 --> 00:11:29.290
then you know that inequality.
00:11:29.290 --> 00:11:32.240
So these two are
equivalent to each other.
00:11:32.240 --> 00:11:36.760
Now, with graphs--
and specifically,
00:11:36.760 --> 00:11:42.850
these counts here, so triangle
densities and edge densities--
00:11:42.850 --> 00:11:46.850
they correspond to counting
closed walks in the graph.
00:11:46.850 --> 00:11:53.290
So in particular, if we're
interested in the number
00:11:53.290 --> 00:11:56.590
of K3 homomorphisms
in a graph, this
00:11:56.590 --> 00:12:01.180
is the same as counting
closed walks of length 3.
00:12:01.180 --> 00:12:04.300
And there was important
identity we used earlier
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when we were discussing the
proof of quasi-random graphs,
00:12:07.630 --> 00:12:11.080
that for counting
closed walks you should
00:12:11.080 --> 00:12:14.210
look at the spectral moment.
00:12:14.210 --> 00:12:16.030
So that's a very
important tool to look
00:12:16.030 --> 00:12:17.410
at the spectral moment--
00:12:17.410 --> 00:12:21.850
namely, the third power
of the eigenvalues
00:12:21.850 --> 00:12:24.580
of the adjacency
matrix of this graph.
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This is the eigenvalues of
the adjacency matrix of G.
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I claim that this sum
here is upper bounded
00:12:46.110 --> 00:12:52.230
by a corresponding
sum of squares raised
00:12:52.230 --> 00:12:56.307
to the power that normalizes.
00:12:56.307 --> 00:12:58.640
The first time I saw this I
was a bit confused because I
00:12:58.640 --> 00:12:59.850
remember, power
means inequality.
00:12:59.850 --> 00:13:00.990
Shouldn't go the other way.
00:13:00.990 --> 00:13:03.390
But actually, no, this
is the correct direction.
00:13:03.390 --> 00:13:05.070
So let me remind you why.
00:13:05.070 --> 00:13:11.930
So if you have a positive
t, then claim that--
00:13:11.930 --> 00:13:17.550
and you have a bunch
of non-negative reals,
00:13:17.550 --> 00:13:23.430
then the claim is
that this t-th power
00:13:23.430 --> 00:13:33.200
sum is less than or equal to
the t-th power of the sum.
00:13:33.200 --> 00:13:35.635
Now, there are several ways
to see why this is true.
00:13:35.635 --> 00:13:36.510
You can do induction.
00:13:36.510 --> 00:13:40.320
But let me show you one
way which is quite neat.
00:13:40.320 --> 00:13:43.140
Because it's homogeneous
in the variables,
00:13:43.140 --> 00:13:56.480
I can assume that the
sum is 1, in which case
00:13:56.480 --> 00:14:01.280
the left-hand side is equal
to this sum of t-th powers.
00:14:05.540 --> 00:14:12.080
So because I assumed that
everything is non-negative,
00:14:12.080 --> 00:14:14.460
all these a's are
between 0 and 1.
00:14:14.460 --> 00:14:18.950
So now, this sum is less
than or equal to the same sum
00:14:18.950 --> 00:14:24.160
without the t's because
you're using it like that.
00:14:24.160 --> 00:14:27.410
And that's equal to 1, which
is the right-hand side.
00:14:36.380 --> 00:14:38.430
So this is true.
00:14:38.430 --> 00:14:41.460
And now we have the
sum of the squares
00:14:41.460 --> 00:14:44.430
of the eigenvalues,
which is also
00:14:44.430 --> 00:14:52.053
a moment of the eigenvalues--
00:14:52.053 --> 00:14:53.220
namely, corresponding to K2.
00:15:05.850 --> 00:15:08.640
So the same inequality is
true for graph homomorphisms.
00:15:08.640 --> 00:15:13.440
And to get to the
inequality for densities,
00:15:13.440 --> 00:15:18.120
we just divide by the
number of vertices raised
00:15:18.120 --> 00:15:20.880
to the third power
from both sides,
00:15:20.880 --> 00:15:23.050
and we get the inequality
that we're looking for.
00:15:27.180 --> 00:15:31.010
So that's the proof
of the upper bound.
00:15:31.010 --> 00:15:31.910
Any questions?
00:15:37.440 --> 00:15:40.350
There is something that bothers
me slightly about this proof.
00:15:40.350 --> 00:15:41.610
Look, it's a correct proof.
00:15:41.610 --> 00:15:43.360
So there is nothing
wrong with this proof.
00:15:43.360 --> 00:15:44.235
Everything is kosher.
00:15:44.235 --> 00:15:46.932
Everything is correct.
00:15:46.932 --> 00:15:48.390
You might ask, is
there a way to do
00:15:48.390 --> 00:15:51.750
this spectral
argument in graphons
00:15:51.750 --> 00:15:53.730
without passing to graphs?
00:15:53.730 --> 00:15:56.940
And yes, you can,
because for graphons you
00:15:56.940 --> 00:15:59.370
can also talk about spectrum.
00:15:59.370 --> 00:16:01.440
It turns out to be
a compact operator,
00:16:01.440 --> 00:16:03.045
so that spectrum makes sense.
00:16:03.045 --> 00:16:05.670
You have to develop a little bit
more theory about the spectrum
00:16:05.670 --> 00:16:07.800
of compact operators, but
everything, more or less,
00:16:07.800 --> 00:16:09.470
works exactly the same way.
00:16:09.470 --> 00:16:12.300
It's just easier to
talk about graphs.
00:16:12.300 --> 00:16:14.730
But what bothers
me about this proof
00:16:14.730 --> 00:16:18.000
is that we started
with what I would
00:16:18.000 --> 00:16:21.090
call a physical inequality,
meaning that it only
00:16:21.090 --> 00:16:28.290
has to do with the actual
edges and subgraph densities.
00:16:28.290 --> 00:16:32.950
But the proof involved
going to the spectrum.
00:16:32.950 --> 00:16:35.150
And that bothers
me a little bit.
00:16:35.150 --> 00:16:37.680
There's nothing incorrect about
it, but somehow in my mind
00:16:37.680 --> 00:16:40.980
a physical inequality
deserves a physical proof.
00:16:40.980 --> 00:16:43.780
So I use the word
physical in contrast
00:16:43.780 --> 00:16:47.910
to frequency, which is
coming from Fourier analysis.
00:16:47.910 --> 00:16:50.770
And that's the next thing
we'll do in this course.
00:16:50.770 --> 00:16:52.540
But this proof goes
to the spectrum.
00:16:52.540 --> 00:16:54.710
It goes to something
beyond the physical domain.
00:16:54.710 --> 00:16:55.210
OK.
00:16:55.210 --> 00:16:55.750
It's neat.
00:16:55.750 --> 00:16:58.420
But I want to show you a
different proof that stays
00:16:58.420 --> 00:17:00.540
within the physical domain.
00:17:00.540 --> 00:17:01.582
And this other proof--
00:17:01.582 --> 00:17:03.790
I mean, it's always nice to
see some different proofs
00:17:03.790 --> 00:17:07.282
because you can use it to
apply to different situations.
00:17:07.282 --> 00:17:08.740
And there are some
situations where
00:17:08.740 --> 00:17:15.730
you might not be able to use
this spectral characterization.
00:17:15.730 --> 00:17:20.190
For example, what if
your K3 is now K4?
00:17:20.190 --> 00:17:24.260
A similar inequality is true,
but this proof doesn't show it,
00:17:24.260 --> 00:17:25.420
at least not directly.
00:17:25.420 --> 00:17:27.819
You have to do a little
bit of extra work.
00:17:36.240 --> 00:17:39.420
So let me show you a different
proof of the upper bound.
00:17:42.480 --> 00:17:45.282
And we'll prove a slightly
stronger statement.
00:17:58.580 --> 00:18:03.820
Namely, that for all
not just graphons--
00:18:03.820 --> 00:18:06.380
it's not so important but
for all symmetric measurable
00:18:06.380 --> 00:18:12.135
functions, from the
unit square to r,
00:18:12.135 --> 00:18:15.330
one has the following
inequality--
00:18:15.330 --> 00:18:19.890
namely, that the
K3 density in W is
00:18:19.890 --> 00:18:26.340
upper bounded by the K2
density of W squared raised
00:18:26.340 --> 00:18:28.700
to power 3/2.
00:18:28.700 --> 00:18:31.730
Here, the square is meant
to be a point-wise square.
00:18:40.563 --> 00:18:41.480
So a couple of things.
00:18:41.480 --> 00:18:46.620
If your graph is a
graphon or a graph, then--
00:18:46.620 --> 00:18:48.910
if it's a graph, then
it's 0 comma 1 value.
00:18:48.910 --> 00:18:51.990
So taking this point by
square doesn't do anything.
00:18:51.990 --> 00:18:54.980
If you're a graphon,
you can always
00:18:54.980 --> 00:18:57.380
put one more inequality that
replaces it by the thing
00:18:57.380 --> 00:18:59.870
that we're looking for
because W is always
00:18:59.870 --> 00:19:02.000
bounded between 0 and 1.
00:19:02.000 --> 00:19:04.510
So it's a slightly
stronger inequality.
00:19:04.510 --> 00:19:10.550
Let me show it to you by writing
down a series of inequalities
00:19:10.550 --> 00:19:14.190
and applying the Cauchy-Schwarz
inequality repeatedly.
00:19:14.190 --> 00:19:18.770
So it's, again, an exercise
in using Cauchy-Schwartz.
00:19:18.770 --> 00:19:24.890
And we will apply three
applications of Cauchy-Schwarz.
00:19:29.210 --> 00:19:31.430
Essentially, three
applications-- one
00:19:31.430 --> 00:19:35.640
corresponding to every
edge of this triangle.
00:19:35.640 --> 00:19:43.510
So let me begin by writing
down the expression in graphons
00:19:43.510 --> 00:19:45.670
corresponding to the K3 density.
00:20:02.590 --> 00:20:07.690
I'm going to apply
Cauchy-Schwarz by--
00:20:07.690 --> 00:20:09.880
so I'm going to
apply Cauchy-Schwarz
00:20:09.880 --> 00:20:18.960
to the variable x, holding all
the other variables constant.
00:20:18.960 --> 00:20:22.490
So hold dy and dz constant.
00:20:22.490 --> 00:20:23.840
Going to apply to dz.
00:20:23.840 --> 00:20:27.080
You see there are two factors
that involve the variable x.
00:20:27.080 --> 00:20:29.990
So apply Cauchy-Schwarz to
them, you split each of them
00:20:29.990 --> 00:20:30.560
into an L2.
00:20:37.980 --> 00:20:39.850
So one of these
factors become that.
00:20:39.850 --> 00:20:41.850
By the way, all of these
are definite integrals.
00:20:41.850 --> 00:20:43.725
I'm just omitting the
domain of integrations.
00:20:43.725 --> 00:20:48.640
All the integrals are
integrated over from 0 to 1.
00:20:48.640 --> 00:20:50.850
So the second
application-- sorry,
00:20:50.850 --> 00:20:56.670
the second factor
becomes like that.
00:20:56.670 --> 00:20:59.315
And the third factor
is left intact.
00:21:07.700 --> 00:21:09.910
So that's the first
application of Cauchy-Schwarz.
00:21:09.910 --> 00:21:13.840
You apply it with respect
to dx to these two factors.
00:21:13.840 --> 00:21:15.806
Split them like that.
00:21:15.806 --> 00:21:18.355
AUDIENCE: There's a
normalization missing.
00:21:18.355 --> 00:21:19.230
PROFESSOR: Thank you.
00:21:19.230 --> 00:21:20.750
There is a
normalization missing.
00:21:25.550 --> 00:21:26.660
OK.
00:21:26.660 --> 00:21:28.630
Guess what the second step is?
00:21:28.630 --> 00:21:31.310
Going to apply
Cauchy-Schwarz again, but now
00:21:31.310 --> 00:21:36.430
to dy, to one more variable.
00:21:39.220 --> 00:21:43.140
Cauchy-Schwarz
with respect to dy.
00:21:43.140 --> 00:21:46.800
There are two factors now
that involve the letter y.
00:21:46.800 --> 00:21:51.570
So I apply Cauchy-Schwarz
and I get the following.
00:21:51.570 --> 00:22:01.630
The first factor now just
becomes the L2 norm of W.
00:22:01.630 --> 00:22:03.610
The second factor
does not involve y,
00:22:03.610 --> 00:22:04.780
so it is left intact.
00:22:12.840 --> 00:22:19.290
And the third factor is
again integrated with respect
00:22:19.290 --> 00:22:23.070
to y after taking the square.
00:22:29.080 --> 00:22:31.010
And there's now dz that remains.
00:22:33.630 --> 00:22:35.100
Last step.
00:22:35.100 --> 00:22:37.440
You can guess, you
integrate with respect
00:22:37.440 --> 00:22:42.570
to dz and apply Cauchy-Schwarz.
00:22:42.570 --> 00:22:47.420
Apply Cauchy-Schwarz to
the last two factors.
00:22:47.420 --> 00:22:50.310
And there, actually, the
outside integral goes away.
00:23:15.690 --> 00:23:16.190
OK.
00:23:16.190 --> 00:23:17.690
So you get this product.
00:23:17.690 --> 00:23:24.470
And you see every single term
is just the L2 norm of W.
00:23:24.470 --> 00:23:31.170
So you have that, which is the
same as what I wrote over here.
00:23:38.430 --> 00:23:42.002
Any questions?
00:23:42.002 --> 00:23:42.986
Yeah.
00:23:42.986 --> 00:23:46.167
AUDIENCE: Where do you use
the fact that W is symmetric?
00:23:46.167 --> 00:23:47.250
PROFESSOR: Great question.
00:23:47.250 --> 00:23:50.348
So where do I use the
fact that W is symmetric?
00:23:53.150 --> 00:23:56.065
So let's see.
00:23:56.065 --> 00:23:57.690
In some sense, we're
not using the fact
00:23:57.690 --> 00:23:59.640
that W is symmetric
because there
00:23:59.640 --> 00:24:02.940
is a slightly more general
inequality you can write down.
00:24:02.940 --> 00:24:06.810
And actually, the question
gives me a good chance
00:24:06.810 --> 00:24:11.610
to do a slight diversion
into how this inequality is
00:24:11.610 --> 00:24:13.470
related to Holder's inequality.
00:24:13.470 --> 00:24:15.690
So this is actually one of
my favorite inequalities
00:24:15.690 --> 00:24:20.940
for this kind of combinatorial
inequalities on graphons.
00:24:20.940 --> 00:24:24.690
So many of you may be familiar
with Holder's inequality
00:24:24.690 --> 00:24:25.650
in the following form.
00:24:25.650 --> 00:24:29.220
If I have three functions,
if I integrate them,
00:24:29.220 --> 00:24:33.930
then you can upper
bound this integral
00:24:33.930 --> 00:24:38.520
by the product of the L3 norms.
00:24:38.520 --> 00:24:42.400
And likewise, if you
have more functions.
00:24:42.400 --> 00:24:45.600
So if you apply just
this inequality directly,
00:24:45.600 --> 00:24:49.707
you get a weaker estimate.
00:24:49.707 --> 00:24:52.040
So you don't get anything
that's quite as strong as what
00:24:52.040 --> 00:24:54.390
you're looking for over there.
00:24:54.390 --> 00:24:58.580
So what happens is
that if you know--
00:24:58.580 --> 00:25:06.620
so if f, g, and h
each depends only
00:25:06.620 --> 00:25:18.500
on a subset of the coordinates
in the following way
00:25:18.500 --> 00:25:26.950
that f depends on only x and
y, g depends only on x and z,
00:25:26.950 --> 00:25:30.450
and h depends only
on y and z, then
00:25:30.450 --> 00:25:32.640
if you repeat that
proof verbatim
00:25:32.640 --> 00:25:34.430
with three different
functions, you
00:25:34.430 --> 00:25:38.100
will find that you
can upper bound
00:25:38.100 --> 00:25:45.060
this product, this integral,
by the product of the L2 norms.
00:25:50.280 --> 00:25:53.640
So L2 norms are in general less
than or equal to the L3 norms.
00:25:53.640 --> 00:25:57.100
So here we're inside a
probability measure space.
00:25:57.100 --> 00:26:02.140
So the entire
space has volume 1.
00:26:02.140 --> 00:26:03.713
So this is a
stronger inequality,
00:26:03.713 --> 00:26:05.880
and this is the inequality
that comes up over there.
00:26:09.550 --> 00:26:10.135
Yeah.
00:26:10.135 --> 00:26:12.718
AUDIENCE: Is there an entirely
graph theoretic proof of this--
00:26:12.718 --> 00:26:14.985
say, for graphs instead
of graphons-- that doesn't
00:26:14.985 --> 00:26:16.272
involve going to spectrum?
00:26:16.272 --> 00:26:16.980
PROFESSOR: Great.
00:26:16.980 --> 00:26:18.640
So the question is,
is there entirely
00:26:18.640 --> 00:26:20.680
graph theoretic proof of this?
00:26:20.680 --> 00:26:26.020
So the reason why I mentioned
that this result is a special
00:26:26.020 --> 00:26:27.606
case of Kruskal-Katona--
00:26:32.850 --> 00:26:35.940
so Kruskal-Katona actually
is a stronger result,
00:26:35.940 --> 00:26:43.560
which tells you precisely how
you should construct a graph.
00:26:43.560 --> 00:26:50.700
So given exactly m edges,
what's the maximum number
00:26:50.700 --> 00:26:52.200
of triangles?
00:26:56.800 --> 00:27:00.040
And the statement that
there is actually--
00:27:00.040 --> 00:27:02.380
it's a very precise
result. It tells
00:27:02.380 --> 00:27:08.760
you, for example, if you
have K choose 2 edges,
00:27:08.760 --> 00:27:11.580
you have at most K
choose 3 triangles.
00:27:14.600 --> 00:27:17.530
It's not just at the
density level but exactly.
00:27:17.530 --> 00:27:20.530
And even if the number of
edges is not in the form of K
00:27:20.530 --> 00:27:22.830
choose 2, it tells
you what to do.
00:27:22.830 --> 00:27:25.140
And actually, the answer
is pretty easy to describe.
00:27:25.140 --> 00:27:29.520
It's almost intuitive so if I
give you a bunch of matchsticks
00:27:29.520 --> 00:27:32.310
and ask you to construct a graph
with as many triangles as you
00:27:32.310 --> 00:27:33.780
can, what should you do?
00:27:33.780 --> 00:27:37.160
You start with one,
two, filling a triangle.
00:27:37.160 --> 00:27:40.680
Start filling a triangle.
00:27:40.680 --> 00:27:41.730
Another vertex.
00:27:41.730 --> 00:27:43.100
1, 2, 3, 4.
00:27:43.100 --> 00:27:44.690
You keep going.
00:27:44.690 --> 00:27:47.190
And that's the
best way to do it.
00:27:47.190 --> 00:27:49.470
And that's what
Kruskal-Katona tells you.
00:27:49.470 --> 00:27:53.410
So that's a more precise
version of this inequality.
00:27:53.410 --> 00:27:56.680
And the Kruskal-Katona,
the combinatorial version,
00:27:56.680 --> 00:28:00.100
is proved via a combinatorial
shifting argument, also known
00:28:00.100 --> 00:28:01.493
as a compression argument.
00:28:01.493 --> 00:28:03.160
Namely, if you start
with a given graph,
00:28:03.160 --> 00:28:05.500
there are some transformations
you do to that graph
00:28:05.500 --> 00:28:08.050
to push your edges
in one direction that
00:28:08.050 --> 00:28:10.840
saves the number of
edges exactly the same
00:28:10.840 --> 00:28:13.650
but increases the number
of triangles at each step.
00:28:13.650 --> 00:28:17.218
And eventually, you push
everything into a clique.
00:28:17.218 --> 00:28:18.760
So it's something
you can read about.
00:28:18.760 --> 00:28:22.033
It's a very nice
result. Other questions?
00:28:26.110 --> 00:28:30.770
So we've solved the upper bound.
00:28:30.770 --> 00:28:33.340
So from examples and from
this upper bound proof,
00:28:33.340 --> 00:28:35.560
we see that it's
the upper bound.
00:28:35.560 --> 00:28:41.050
Now let me tell you a
fairly general result that
00:28:41.050 --> 00:28:45.180
says something about graph
theoretic inequalities
00:28:45.180 --> 00:28:48.590
but for a specific kind
of linear inequalities.
00:28:48.590 --> 00:28:50.910
So here's a theorem
due to Bollobas.
00:28:55.190 --> 00:29:04.900
I'm interested in an
inequality of the form like--
00:29:04.900 --> 00:29:19.380
so I'm interested in
inequality of this form, where
00:29:19.380 --> 00:29:26.920
I have a bunch of
real coefficients,
00:29:26.920 --> 00:29:29.170
and I'm looking at
a linear combination
00:29:29.170 --> 00:29:32.120
of the clique densities.
00:29:32.120 --> 00:29:36.770
I would like to know if
this inequality is true.
00:29:36.770 --> 00:29:39.390
So somebody gives
us this inequality,
00:29:39.390 --> 00:29:41.430
whatever the numbers may be.
00:29:41.430 --> 00:29:42.910
You can also have
a constant term.
00:29:42.910 --> 00:29:46.000
The constant term
corresponds to r equals to 1.
00:29:46.000 --> 00:29:47.730
So the point density.
00:29:47.730 --> 00:29:50.100
That's the constant term.
00:29:50.100 --> 00:29:54.290
And asks you to decide
is this inequality true.
00:29:54.290 --> 00:29:55.410
And if so, prove it.
00:29:55.410 --> 00:29:57.343
If not, find a counterexample.
00:30:00.800 --> 00:30:03.260
So the theorem tells you
that this is actually not
00:30:03.260 --> 00:30:04.040
hard to do.
00:30:04.040 --> 00:30:09.230
So this inequality
holds for all G
00:30:09.230 --> 00:30:17.000
if and only if it holds
whenever G is a clique.
00:30:23.043 --> 00:30:24.710
Maybe somebody gives
you this inequality
00:30:24.710 --> 00:30:28.350
about-- it's a linear inequality
about clique densities.
00:30:28.350 --> 00:30:30.200
Then, to check this
inequality, you only
00:30:30.200 --> 00:30:35.160
have to check over
all cliques G,
00:30:35.160 --> 00:30:39.300
which is much easier than
checking for all graphs.
00:30:39.300 --> 00:30:42.927
For each clique G this is
just some specific expression
00:30:42.927 --> 00:30:44.510
you can write down,
and you can check.
00:30:51.340 --> 00:30:55.890
So I want to show you the
proof of Bollobas' theorem.
00:30:55.890 --> 00:30:59.600
It's a quite nice result.
But before that, any
00:30:59.600 --> 00:31:01.477
questions about the statement.
00:31:05.293 --> 00:31:08.640
All right.
00:31:08.640 --> 00:31:11.360
So the reason I say that
this is very easy to check
00:31:11.360 --> 00:31:15.030
if I actually give you
what the numbers are
00:31:15.030 --> 00:31:20.770
is because this
inequality for cliques--
00:31:20.770 --> 00:31:30.720
so the inequality is equivalent
to just the statement
00:31:30.720 --> 00:31:33.950
of inequality that
I'm writing down now,
00:31:33.950 --> 00:31:40.710
where I tell you precisely
what the r clique
00:31:40.710 --> 00:31:42.520
density is in an n clique.
00:31:42.520 --> 00:31:45.092
Because that's just some
combinatorial expression.
00:31:48.217 --> 00:31:49.800
So to check whether
this inequality is
00:31:49.800 --> 00:31:51.720
true for all graphs,
I just have to check
00:31:51.720 --> 00:31:54.540
the specific inequality
for all integers n, which
00:31:54.540 --> 00:31:56.103
is straightforward.
00:31:59.970 --> 00:32:00.780
All right.
00:32:00.780 --> 00:32:04.540
So let's see how to prove
that inequality up there.
00:32:04.540 --> 00:32:05.640
And here we're--
00:32:05.640 --> 00:32:07.290
I mean, we're not
going to exactly use
00:32:07.290 --> 00:32:09.668
the theorems about
graphons, but it's
00:32:09.668 --> 00:32:10.960
useful to think about graphons.
00:32:19.580 --> 00:32:25.380
So if and only if one of
the directions is trivial--
00:32:25.380 --> 00:32:28.300
so let's get that
out of the way first.
00:32:28.300 --> 00:32:32.860
But also-- so the
only if is clear.
00:32:32.860 --> 00:32:35.750
So for the if
direction, first note
00:32:35.750 --> 00:32:41.860
that this is true for
all graphs if and only
00:32:41.860 --> 00:32:48.670
if it is true for all graphons
and where I replaced G
00:32:48.670 --> 00:32:53.380
by W. By the general theory
of graph limits and whatnot,
00:32:53.380 --> 00:32:56.080
this is true.
00:32:56.080 --> 00:33:00.210
So in particular, there
is one class of class
00:33:00.210 --> 00:33:02.460
that I would like to look at--
00:33:02.460 --> 00:33:08.430
namely, I want to consider
the set of node weighted--
00:33:16.270 --> 00:33:19.900
so I want to consider the set
of node weighted simple graphs.
00:33:27.150 --> 00:33:29.020
So node weighted
simple graphs, by this
00:33:29.020 --> 00:33:35.960
I mean a graph where some
of the edges are present
00:33:35.960 --> 00:33:41.190
and I have a node weight--
00:33:41.190 --> 00:33:43.010
a weight for each node.
00:33:43.010 --> 00:33:45.210
And to normalize
things properly,
00:33:45.210 --> 00:33:50.850
I'm going to assume that the
nodes' weights add up to 1.
00:33:50.850 --> 00:33:54.690
Now, you see that each graph
like that, you can represented
00:33:54.690 --> 00:34:03.180
by a graphon where--
00:34:13.300 --> 00:34:15.800
so you can have a graphon.
00:34:15.800 --> 00:34:17.690
So they're not meant
to be the same picture,
00:34:17.690 --> 00:34:20.659
but you have some graphon
like this, which corresponds
00:34:20.659 --> 00:34:23.210
to a node weighted graph.
00:34:23.210 --> 00:34:25.699
And the set of such
node weighted graphs
00:34:25.699 --> 00:34:30.174
is dense in the
space of graphons.
00:34:37.420 --> 00:34:40.630
In particular, as far as
graph densities are concerned,
00:34:40.630 --> 00:34:44.020
they include all
the simple graphs.
00:34:44.020 --> 00:34:45.460
So it suffices--
00:34:45.460 --> 00:34:49.000
I mean, it's equivalent to--
00:34:49.000 --> 00:34:51.699
the inequality is
equivalent to it
00:34:51.699 --> 00:34:55.650
being true for all node
weighted simple graphs.
00:34:55.650 --> 00:35:01.170
But for this space
of graphs, suppose
00:35:01.170 --> 00:35:03.585
that the inequality fails.
00:35:06.900 --> 00:35:10.390
Suppose that
inequality is false.
00:35:10.390 --> 00:35:19.440
Then there exists a node
weighted simple graph.
00:35:22.750 --> 00:35:25.330
I'm going to actually drop
the word simple from now on.
00:35:25.330 --> 00:35:44.390
So node weighted graph H, such
that f of H being the above sum
00:35:44.390 --> 00:35:46.080
is less than zero.
00:35:48.840 --> 00:35:53.670
And there could be many
possibilities for such an H.
00:35:53.670 --> 00:35:54.750
But let me choose.
00:35:54.750 --> 00:35:59.490
So among all the possible
H's, let's choose
00:35:59.490 --> 00:36:04.290
one that is minimal in the
sense that it has the smallest
00:36:04.290 --> 00:36:06.090
possible number of nodes.
00:36:09.038 --> 00:36:09.955
So with this minimum--
00:36:23.060 --> 00:36:24.710
has a minimum number of nodes.
00:36:24.710 --> 00:36:40.990
And furthermore, among all
H with this number of nodes,
00:36:40.990 --> 00:36:53.230
choose the node weights, which
we'll denote by a1 through a n,
00:36:53.230 --> 00:36:54.700
summing to 1.
00:36:54.700 --> 00:37:01.030
Chooses node weights so that
this expression, the sum,
00:37:01.030 --> 00:37:02.753
is minimized.
00:37:08.680 --> 00:37:11.117
And by compactness--
and now we're
00:37:11.117 --> 00:37:13.450
not even talking about compact
in the space of graphons.
00:37:13.450 --> 00:37:15.075
You have a finite
number of parameters.
00:37:15.075 --> 00:37:16.340
It's a continuous function.
00:37:16.340 --> 00:37:19.000
So just by compactness,
there exists
00:37:19.000 --> 00:37:21.340
such an H for which the
minimum is achieved.
00:37:24.530 --> 00:37:26.155
This is minimizing
over integers.
00:37:26.155 --> 00:37:28.840
And here, minimizing
over a finite set
00:37:28.840 --> 00:37:32.300
of bounded real numbers.
00:37:32.300 --> 00:37:34.750
So the name of the
game now is we have
00:37:34.750 --> 00:37:37.870
this H, which is minimizing.
00:37:37.870 --> 00:37:41.620
And I want to show that
H has certain properties.
00:37:41.620 --> 00:37:43.120
If it doesn't have
these properties,
00:37:43.120 --> 00:37:45.340
I can decrease those values.
00:38:01.450 --> 00:38:06.250
So let's see what
properties this H must
00:38:06.250 --> 00:38:11.150
have if it has the minimum
number of nodes and f of H
00:38:11.150 --> 00:38:12.310
is minimum possible.
00:38:16.180 --> 00:38:25.620
So first I claim that all the
node weights are positive.
00:38:25.620 --> 00:38:27.480
If not, I can delete
that node and decrease
00:38:27.480 --> 00:38:28.400
the number of nodes.
00:38:38.880 --> 00:38:41.970
I would like to
claim that H must
00:38:41.970 --> 00:38:54.010
be a complete graph because
if some ij is not edge of H--
00:38:54.010 --> 00:38:56.715
so here i is different from j.
00:38:56.715 --> 00:38:57.590
I do not allow loops.
00:38:57.590 --> 00:38:59.410
It's just simple.
00:38:59.410 --> 00:39:05.412
Then let's think about what this
expression f of H should be.
00:39:05.412 --> 00:39:06.870
So I don't want to
write this down,
00:39:06.870 --> 00:39:08.520
but I want you to
imagine in your head.
00:39:08.520 --> 00:39:15.990
So you have this graphon H. I'm
Looking at the clique density.
00:39:15.990 --> 00:39:19.600
It's some polynomial.
00:39:19.600 --> 00:39:23.990
In fact, it's some multilinear--
00:39:23.990 --> 00:39:27.380
it's some polynomial
in these node weights.
00:39:30.700 --> 00:39:34.490
So I want to understand
what is the shape
00:39:34.490 --> 00:39:39.010
of this polynomial as a
function of the node weights.
00:39:42.130 --> 00:39:53.950
And I observe that it has
to be multilinear in--
00:39:53.950 --> 00:39:58.130
has to be multilinear
in particular in
00:39:58.130 --> 00:40:01.470
alpha i and alpha j.
00:40:06.090 --> 00:40:06.915
It's a polynomial.
00:40:06.915 --> 00:40:07.790
That should be clear.
00:40:07.790 --> 00:40:10.940
It is multilinear
because, well, you have--
00:40:16.810 --> 00:40:20.070
why is it multilinear?
00:40:20.070 --> 00:40:24.180
Why do I not have
alpha i squared?
00:40:33.910 --> 00:40:35.266
Either of you.
00:40:35.266 --> 00:40:38.580
AUDIENCE: It says the
0 is not [INAUDIBLE]..
00:40:38.580 --> 00:40:41.270
PROFESSOR: So we're forbidding--
00:40:41.270 --> 00:40:45.350
so here's alpha 1, alpha 2,
alpha 3, alpha 4, alpha 1,
00:40:45.350 --> 00:40:48.630
alpha 2, alpha 3, alpha 4.
00:40:48.630 --> 00:40:52.877
So understand what
the triangle density--
00:40:52.877 --> 00:40:54.460
if you write down
the triangle density
00:40:54.460 --> 00:40:58.990
as an expression in
terms of the parameters,
00:40:58.990 --> 00:41:01.420
think about what comes
out, what it looks like.
00:41:01.420 --> 00:41:06.340
And they essentially consist
of you choosing a subgraph,
00:41:06.340 --> 00:41:10.260
which you cannot have repeats.
00:41:10.260 --> 00:41:11.460
So it's multilinear.
00:41:11.460 --> 00:41:14.690
So it's multilinear
in particular in alpha
00:41:14.690 --> 00:41:17.440
i and alpha j.
00:41:17.440 --> 00:41:31.010
So no term has the product
alpha i alpha j in it
00:41:31.010 --> 00:41:33.950
because ij is not an edge.
00:41:36.720 --> 00:41:41.820
So here's where we're
really using that we're only
00:41:41.820 --> 00:41:44.580
considering clique densities.
00:41:52.380 --> 00:41:56.670
So the theorem is completely
false without the assumption
00:41:56.670 --> 00:41:57.660
of clique densities.
00:41:57.660 --> 00:42:02.573
If we have a general inequality,
general linear inequality, then
00:42:02.573 --> 00:42:03.990
the statement is
completely false.
00:42:06.800 --> 00:42:11.240
So it's multilinear.
00:42:11.240 --> 00:42:14.360
So if we now fix all
the other variables
00:42:14.360 --> 00:42:17.480
and just think about
how to optimize,
00:42:17.480 --> 00:42:23.530
how to minimize f of H by
tweaking alpha i and alpha j,
00:42:23.530 --> 00:42:27.350
well, it's linear, so
you should minimize it
00:42:27.350 --> 00:42:29.620
by setting one of
them to be zero.
00:42:32.190 --> 00:42:36.100
And that would then decrease
the number of nodes.
00:42:36.100 --> 00:42:49.410
So can shift alpha i and alpha
j while preserving alpha i
00:42:49.410 --> 00:42:52.755
plus alpha j and not changing--
00:42:55.545 --> 00:43:05.670
so not increasing
f of H. And then
00:43:05.670 --> 00:43:13.310
we get either alpha
i to go to zero
00:43:13.310 --> 00:43:16.340
or alpha j to go to
zero, in which case
00:43:16.340 --> 00:43:23.030
we decrease the number of
nodes, thereby contradicting
00:43:23.030 --> 00:43:24.808
the minimality assumption.
00:43:30.180 --> 00:43:36.290
So this argument here then tells
you that H must be a clique.
00:43:36.290 --> 00:43:42.110
So hence, H is complete.
00:43:46.470 --> 00:43:53.020
And if H is complete, then as
a polynomial in these alphas,
00:43:53.020 --> 00:43:55.660
what should f look like?
00:43:55.660 --> 00:43:58.810
Well, it has to be
symmetric with respect
00:43:58.810 --> 00:44:01.270
to all these alphas.
00:44:01.270 --> 00:44:05.750
So in particular, it has to be--
00:44:05.750 --> 00:44:13.440
so since H is complete, we
see that, in fact, now you
00:44:13.440 --> 00:44:15.720
can write down
exactly what f of H
00:44:15.720 --> 00:44:20.000
is in terms of the parameters
described in the problem.
00:44:20.000 --> 00:44:26.400
Namely, it's Cr times
r factorial times Sr,
00:44:26.400 --> 00:44:34.150
where Sr is a symmetric
polynomial where you look at--
00:44:34.150 --> 00:44:40.290
you choose r of the
terms, r of these alphas
00:44:40.290 --> 00:44:42.790
for each term in this sum.
00:44:42.790 --> 00:44:46.980
It's just elementary
symmetric polynomial.
00:44:46.980 --> 00:44:50.080
And I would like to know,
given such a polynomial,
00:44:50.080 --> 00:44:54.810
how to minimize this number
by choosing the alphas.
00:44:54.810 --> 00:44:56.940
But if you think
about what happens
00:44:56.940 --> 00:45:01.350
if you fix again everything
but two of the alphas--
00:45:01.350 --> 00:45:11.310
so by fixing all of, let's
say, alpha 3 to alpha n,
00:45:11.310 --> 00:45:12.750
we find that--
00:45:12.750 --> 00:45:20.170
so as a function in just
alpha 1 and alpha 2, f of H
00:45:20.170 --> 00:45:21.550
has the following form.
00:45:32.860 --> 00:45:35.475
And because it's
symmetric, these two B's
00:45:35.475 --> 00:45:36.750
are actually the same.
00:45:40.770 --> 00:45:47.390
So if we now vary alpha 1 and
alpha 2 but fixing everything
00:45:47.390 --> 00:45:53.870
else, because alpha 1
plus alpha 2 is constant,
00:45:53.870 --> 00:45:57.230
I can even get rid
of this linear part.
00:46:00.850 --> 00:46:03.050
So that linear part is
fixed as a constant.
00:46:07.020 --> 00:46:09.810
I want to minimize this
expression with alpha 1 plus
00:46:09.810 --> 00:46:11.790
alpha 2, how it's fixed.
00:46:11.790 --> 00:46:13.770
So there are two
possibilities depending on
00:46:13.770 --> 00:46:19.170
whether C is positive or
negative or, I guess, 0.
00:46:19.170 --> 00:46:20.370
So now you're here.
00:46:20.370 --> 00:46:35.250
So depending if C is
positive or negative,
00:46:35.250 --> 00:46:43.400
it's minimized by either the
two alphas equal to each other
00:46:43.400 --> 00:46:52.340
or one of the two of
alphas should be zero.
00:46:52.340 --> 00:46:55.520
The latter cannot occur
because we assume minimality.
00:46:55.520 --> 00:46:58.150
So the first must occur.
00:46:58.150 --> 00:47:01.880
And hence, by symmetry, if
you apply the same argument
00:47:01.880 --> 00:47:04.550
to all the other
alphas, all the alphas
00:47:04.550 --> 00:47:09.190
are equal to each
other, which means
00:47:09.190 --> 00:47:11.190
that H is a simple clique.
00:47:15.490 --> 00:47:17.150
It's basically an
unweighted clique.
00:47:20.060 --> 00:47:22.780
So in other words,
if this inequality
00:47:22.780 --> 00:47:28.360
fails for some H,
some node weighted H,
00:47:28.360 --> 00:47:33.740
then it must fail for
a simple clique H.
00:47:33.740 --> 00:47:36.630
And that's the claim above.
00:47:36.630 --> 00:47:37.130
Yeah?
00:47:37.130 --> 00:47:38.630
AUDIENCE: So in the
statement, there
00:47:38.630 --> 00:47:41.137
are two n's, are those two
n's different n's then?
00:47:41.137 --> 00:47:41.720
PROFESSOR: OK.
00:47:41.720 --> 00:47:42.230
Question.
00:47:42.230 --> 00:47:43.210
There are two n's.
00:47:43.210 --> 00:47:44.670
Yeah.
00:47:44.670 --> 00:47:45.380
Thank you.
00:47:45.380 --> 00:47:47.450
So these are two--
00:47:50.460 --> 00:47:50.960
yeah.
00:47:50.960 --> 00:47:53.122
So these are two different n's.
00:48:01.554 --> 00:48:03.042
Great.
00:48:03.042 --> 00:48:04.034
yeah.
00:48:04.034 --> 00:48:05.540
AUDIENCE: I have a question.
00:48:05.540 --> 00:48:10.130
Which is the node weight
such that f of H [INAUDIBLE]??
00:48:10.130 --> 00:48:11.590
PROFESSOR: Question
is, why can we
00:48:11.590 --> 00:48:17.590
assume that you can choose H
so that f of H is minimized?
00:48:17.590 --> 00:48:18.590
Its because once--
00:48:18.590 --> 00:48:19.090
OK.
00:48:19.090 --> 00:48:20.830
So you agreed the
first thing you
00:48:20.830 --> 00:48:24.187
can minimize because the number
of nodes is a positive integer.
00:48:24.187 --> 00:48:25.770
So if there's a
counterexample, choose
00:48:25.770 --> 00:48:27.590
the minimum counterexample.
00:48:27.590 --> 00:48:32.215
Now, you fixed that number of
vertices, and the number of--
00:48:32.215 --> 00:48:33.970
then this is an
optimization problem.
00:48:33.970 --> 00:48:36.040
It's minimizing
continuous function
00:48:36.040 --> 00:48:38.200
with a finite
number of variables.
00:48:38.200 --> 00:48:42.600
So it has a minimum
just by compactness
00:48:42.600 --> 00:48:44.095
of a continuous function.
00:48:44.095 --> 00:48:45.220
So I choose that minimizer.
00:48:50.370 --> 00:48:53.380
Any more questions?
00:48:53.380 --> 00:48:56.950
So we have this rather
general looking theorem.
00:48:56.950 --> 00:48:59.283
So in the second part
of today's lecture,
00:48:59.283 --> 00:49:00.700
after taking a
short break, I want
00:49:00.700 --> 00:49:02.800
to discuss what are
some of the consequences
00:49:02.800 --> 00:49:06.960
and also variations of
that statement up there.
00:49:06.960 --> 00:49:10.240
And I want to also show you
what the rest of this picture
00:49:10.240 --> 00:49:10.750
looks like.
00:49:15.030 --> 00:49:18.390
So let's continue to deduce some
consequences of this theorem
00:49:18.390 --> 00:49:20.190
up there that
tells us that it is
00:49:20.190 --> 00:49:22.650
pretty easy to decide
linear inequalities
00:49:22.650 --> 00:49:25.830
between clique densities.
00:49:25.830 --> 00:49:31.380
Namely, to decide it, just check
the inequalities on cliques.
00:49:31.380 --> 00:49:44.590
So as a corollary for each n--
00:49:44.590 --> 00:49:47.890
yes, for each n, the
extremal points--
00:49:55.200 --> 00:50:09.440
so the extremal points of
the convex hull of this set
00:50:09.440 --> 00:50:29.140
where I record the clique
densities overall graphons W.
00:50:29.140 --> 00:50:32.610
So think about this set
as the higher dimensional
00:50:32.610 --> 00:50:35.910
generalization of that
picture I drew up there.
00:50:35.910 --> 00:50:39.810
But no previously we
had n equals to 3,
00:50:39.810 --> 00:50:42.480
and we're still interested
in n equals to 3.
00:50:42.480 --> 00:50:51.330
But in general, you have
this set sitting in this box.
00:50:51.330 --> 00:50:55.220
And so it's some set.
00:50:55.220 --> 00:50:59.240
And if I take the convex
hull of the set, what
00:50:59.240 --> 00:51:03.620
that theorem tells us-- and
it requires maybe one bit
00:51:03.620 --> 00:51:04.890
of extra computation.
00:51:04.890 --> 00:51:09.980
But what it tells us is that the
extremal points are precisely
00:51:09.980 --> 00:51:23.600
the points given by W equals
to Km for all m equal to 1.
00:51:23.600 --> 00:51:27.950
So evaluate, find what
this point is for each m,
00:51:27.950 --> 00:51:31.250
and you have a bunch of points.
00:51:31.250 --> 00:51:34.520
And those are the convex hull.
00:51:34.520 --> 00:51:37.250
So I'll illustrate by
drawing what the points are
00:51:37.250 --> 00:51:38.840
for the picture over there.
00:51:38.840 --> 00:51:42.170
But it essentially follows
from Bollobas' theorem
00:51:42.170 --> 00:51:44.390
with one extra
bit of computation
00:51:44.390 --> 00:51:47.340
to make sure that all
of these are actually
00:51:47.340 --> 00:51:49.100
extremal points of
the convex hull.
00:51:49.100 --> 00:51:51.770
None of them is contained
in the convex hull
00:51:51.770 --> 00:51:54.750
of the other points.
00:51:54.750 --> 00:51:59.998
So for example, we can
also deduce very easily
00:51:59.998 --> 00:52:00.665
Turan's theorem.
00:52:05.840 --> 00:52:08.350
So what does Turan's
theorem tell us?
00:52:08.350 --> 00:52:20.180
It tells us that if the r
plus 1 clique density is zero,
00:52:20.180 --> 00:52:32.930
then the K2 density
is at most 1 minus r.
00:52:32.930 --> 00:52:36.245
So why does Turan's theorem
follow from the above claims?
00:52:41.350 --> 00:52:44.740
It should follow because
all the data here has
00:52:44.740 --> 00:52:46.360
to do with clique densities.
00:52:46.360 --> 00:52:47.860
And everything we
saw so far says
00:52:47.860 --> 00:52:50.110
that if you just
want to understand
00:52:50.110 --> 00:52:52.630
linear inequalities
between clique densities,
00:52:52.630 --> 00:52:53.320
it's super easy.
00:53:01.600 --> 00:53:04.372
Maybe I'll draw the
picture for triangles,
00:53:04.372 --> 00:53:05.830
and then you'll
see what it's like.
00:53:08.700 --> 00:53:12.060
So the corollary tells
us for this picture,
00:53:12.060 --> 00:53:15.540
corresponding to n
equals to 3, what
00:53:15.540 --> 00:53:19.670
the points, the extreme
point of the convex hull are.
00:53:19.670 --> 00:53:22.100
So let me let me draw
these points for you.
00:53:22.100 --> 00:53:26.340
So one of these points
is this 1/2 comma 0.
00:53:26.340 --> 00:53:29.180
So that corresponds
to Mantel's theorem.
00:53:32.400 --> 00:53:36.680
Now, if you go to the
other values of m,
00:53:36.680 --> 00:53:39.290
you find that those points--
00:53:39.290 --> 00:53:41.930
so the extreme points--
00:53:41.930 --> 00:53:50.200
they are of the form m minus
1 divided by m, m minus 1 m
00:53:50.200 --> 00:53:57.370
minus 2 divided by m squared
for positive integers m.
00:54:00.396 --> 00:54:03.760
So for m equals to 2, that's
the point that we just drew.
00:54:03.760 --> 00:54:04.630
And the next point--
00:54:09.270 --> 00:54:12.110
so next two points, one
third and one fourth,
00:54:12.110 --> 00:54:14.690
they are at, if you plug it in--
00:54:17.850 --> 00:54:18.350
thank you.
00:54:18.350 --> 00:54:21.760
2/3 and 3/4.
00:54:21.760 --> 00:54:25.720
They correspond to 2/9 and 3/8.
00:54:28.780 --> 00:54:30.940
So let me show you
where these points are.
00:54:30.940 --> 00:54:38.740
So they are at here
and over there.
00:54:38.740 --> 00:54:41.860
And you have this sequence
of points going up.
00:54:51.170 --> 00:54:53.110
So this is the convex hull.
00:54:53.110 --> 00:54:55.900
And from that information,
you should already
00:54:55.900 --> 00:54:58.710
be able to deduce
Mantel's theorem
00:54:58.710 --> 00:55:03.870
because this right half is
not part of this convex hull.
00:55:03.870 --> 00:55:05.670
So that's what Mantel's theorem.
00:55:05.670 --> 00:55:09.900
And similarly, the
deduction to Turan's theorem
00:55:09.900 --> 00:55:11.490
also follows by a similar logic.
00:55:13.810 --> 00:55:14.310
OK.
00:55:14.310 --> 00:55:17.110
So you have this
sequence of points.
00:55:17.110 --> 00:55:21.430
Now, it happens that all of
these points lie on a curve.
00:55:21.430 --> 00:55:26.190
So let me try to draw
what this extra curve is.
00:55:30.140 --> 00:55:34.966
So there is some
curve, like that.
00:55:38.438 --> 00:55:40.230
So there's some curve like that.
00:55:42.780 --> 00:55:51.750
The equation of this curve
happens to be x 2x minus 1.
00:55:51.750 --> 00:55:53.940
And because the
regions is contained
00:55:53.940 --> 00:55:56.010
in the convex hull,
the yellow points,
00:55:56.010 --> 00:56:01.720
it certainly lies above
this convex red curve.
00:56:01.720 --> 00:56:03.160
You've seen this
red curve before.
00:56:06.470 --> 00:56:06.970
From where?
00:56:19.010 --> 00:56:20.010
So what is that saying?
00:56:20.010 --> 00:56:23.740
It's saying that if your
edge density is beyond
00:56:23.740 --> 00:56:27.010
above one half, then you
have some lower bound
00:56:27.010 --> 00:56:29.730
on the triangle density.
00:56:29.730 --> 00:56:31.560
Where have we seen this before?
00:56:31.560 --> 00:56:32.658
Problem set one.
00:56:32.658 --> 00:56:34.450
There was a problem on
problem set one that
00:56:34.450 --> 00:56:36.890
says exactly this inequality.
00:56:36.890 --> 00:56:38.660
So go back and
compare what it did.
00:56:43.580 --> 00:56:46.710
But of course, the
convex hull result
00:56:46.710 --> 00:56:48.410
tells you even a
little bit more--
00:56:48.410 --> 00:56:51.950
namely, that you can
draw line segments
00:56:51.950 --> 00:56:54.770
between these
convex hull points.
00:56:59.660 --> 00:57:01.850
So you have some
polygonal reason that
00:57:01.850 --> 00:57:03.230
lower bounds the actual region.
00:57:06.312 --> 00:57:07.520
So what is the actual region?
00:57:07.520 --> 00:57:09.390
So leaving you in suspense.
00:57:09.390 --> 00:57:11.990
So let me tell you what
the actual region is now.
00:57:11.990 --> 00:57:14.680
So it turns out to be
actually-- it's beautiful
00:57:14.680 --> 00:57:19.043
and it's quite deep,
that the region is now
00:57:19.043 --> 00:57:19.960
completely understood.
00:57:19.960 --> 00:57:22.600
And it's a fairly recent result.
It's only about 10 years ago
00:57:22.600 --> 00:57:31.250
roughly that there are
some concave curves.
00:57:31.250 --> 00:57:39.120
The sequence of scallops going
up to the top right corner.
00:57:39.120 --> 00:57:42.930
And this is now understood
to be the complete region
00:57:42.930 --> 00:57:44.580
between these lower
and upper curves.
00:57:47.510 --> 00:57:50.680
So this is the complete set
of feasible regions for edge
00:57:50.680 --> 00:57:54.310
versus triangle densities.
00:57:54.310 --> 00:57:57.100
So this lower curve
is a difficult result
00:57:57.100 --> 00:57:57.870
due to Razborov.
00:58:02.730 --> 00:58:07.600
And I want to give you a
statement what this curve is.
00:58:07.600 --> 00:58:12.760
And Razborov came up with this
machinery, this technique,
00:58:12.760 --> 00:58:14.770
known by the name
of flag algebra.
00:58:14.770 --> 00:58:16.930
So actually, he came
up with this name.
00:58:22.390 --> 00:58:24.390
So I won't really tell
you what flag algebra is,
00:58:24.390 --> 00:58:26.670
but it's kind of a
computerized way of doing
00:58:26.670 --> 00:58:28.440
Cauchy-Schwarz inequalities.
00:58:28.440 --> 00:58:33.930
So many of our proofs for this
graph through inequalities,
00:58:33.930 --> 00:58:35.850
they go through some
kind of Cauchy-Schwarz
00:58:35.850 --> 00:58:38.130
or sum of squares equivalently.
00:58:38.130 --> 00:58:42.033
But there are some very large
or difficult inequalities
00:58:42.033 --> 00:58:43.200
you can also prove this way.
00:58:43.200 --> 00:58:45.840
But it may be difficult
to find exactly what is
00:58:45.840 --> 00:58:48.630
the actual inequality--
the chain of Cauchy-Schwarz
00:58:48.630 --> 00:58:50.970
or the sum of squares that
you should write down.
00:58:50.970 --> 00:58:56.550
So this machinery, flag
algebra, is a language,
00:58:56.550 --> 00:58:59.910
is a framework for setting
up those sum of squares
00:58:59.910 --> 00:59:02.220
inequalities in the
context of proving
00:59:02.220 --> 00:59:04.710
graph theoretic inequalities.
00:59:04.710 --> 00:59:07.120
So it can be used in
many different ways.
00:59:07.120 --> 00:59:11.760
And notably, a lot of people
have used serious computer
00:59:11.760 --> 00:59:13.180
computations.
00:59:13.180 --> 00:59:14.820
If I want to prove
something is true,
00:59:14.820 --> 00:59:19.050
I plug it into what's called
a semidefinite program that
00:59:19.050 --> 00:59:23.190
allows me to decide what kinds
of Cauchy-Schwarz inequalities
00:59:23.190 --> 00:59:26.680
I should be applying to derive
the result I want to prove.
00:59:26.680 --> 00:59:28.310
So that's what flag
algebra roughly is.
00:59:32.130 --> 00:59:35.948
So what Razborov proved
is the following.
00:59:42.430 --> 00:59:46.030
So Razborov's theorem,
which is drawn up there--
00:59:46.030 --> 00:59:47.830
that's the lower curve--
00:59:47.830 --> 00:59:50.820
is that for fixed--
00:59:50.820 --> 01:00:02.410
so for a fixed value
of edge densities,
01:00:02.410 --> 01:00:15.670
if it lies between two
specific points, drawn above,
01:00:15.670 --> 01:00:23.820
the minimum value
of triangle density
01:00:23.820 --> 01:00:26.530
with a fixed value
of edge density
01:00:26.530 --> 01:00:30.164
is attained via the
following construction.
01:00:33.960 --> 01:00:37.920
It's attained by the step
function of the graphon
01:00:37.920 --> 01:00:48.340
corresponding to a K clique.
01:00:48.340 --> 01:00:51.160
So a complete
graph on K vertices
01:00:51.160 --> 01:01:03.450
with node weights alpha 1
through alpha K summing to 1,
01:01:03.450 --> 01:01:07.680
and such that the first K
minus 1 of the node weights
01:01:07.680 --> 01:01:09.300
are equal.
01:01:09.300 --> 01:01:18.970
And the last one is Smaller
01:01:18.970 --> 01:01:19.690
All right.
01:01:19.690 --> 01:01:24.610
And the point here is that if
you are given a specific edge
01:01:24.610 --> 01:01:27.490
weight, edge density,
then there is
01:01:27.490 --> 01:01:33.580
a unique choice of these alphas
that achieve that edge density.
01:01:33.580 --> 01:01:37.420
And that is the graphon you
should use that minimizes
01:01:37.420 --> 01:01:38.800
the triangle density--
01:01:38.800 --> 01:01:40.998
describes the lower curve.
01:01:40.998 --> 01:01:43.540
So you can write down specific
equations for the lower curve,
01:01:43.540 --> 01:01:44.510
but it's not so important.
01:01:44.510 --> 01:01:46.052
This is a more
important description.
01:01:46.052 --> 01:01:48.070
These are the graphs
that come out.
01:01:48.070 --> 01:01:51.430
And what is something
that is actually quite--
01:01:51.430 --> 01:01:53.680
I mean, why you should
suspect this theorem is
01:01:53.680 --> 01:01:57.220
difficult is that unlike Turan's
theorem-- so Turan's theorem,
01:01:57.220 --> 01:02:00.580
which corresponds to all
those discrete points.
01:02:00.580 --> 01:02:04.370
In Turan's theorem, the
minimizer is unique.
01:02:04.370 --> 01:02:07.950
I tell you the number--
01:02:07.950 --> 01:02:10.680
I tell you that the
edge density is 2/3,
01:02:10.680 --> 01:02:13.220
and I want you to minimize
the number of triangles.
01:02:17.190 --> 01:02:19.250
Not from Turan's
theorem, but it turns out
01:02:19.250 --> 01:02:22.790
that this extremal
point is unique.
01:02:22.790 --> 01:02:26.750
Essentially corresponds to a
complete three partite graph.
01:02:26.750 --> 01:02:30.080
But for the intermediate
values, the constructions
01:02:30.080 --> 01:02:31.920
are not unique.
01:02:31.920 --> 01:02:43.560
So unless the K2 density
is exactly of this form,
01:02:43.560 --> 01:02:49.760
the minimizer is not unique.
01:02:52.770 --> 01:02:54.680
And the reason why
it is not unique
01:02:54.680 --> 01:03:03.130
is that you can replace--
01:03:03.130 --> 01:03:04.790
so what's going on here?
01:03:04.790 --> 01:03:06.400
So you have this graphon.
01:03:11.040 --> 01:03:14.270
Alpha 1, alpha 2, alpha 3.
01:03:14.270 --> 01:03:34.600
I can replace this graphon here
by any triangle free graphon
01:03:34.600 --> 01:03:35.830
of the same edge density.
01:03:41.240 --> 01:03:43.670
And there are lots
and lots of them.
01:03:43.670 --> 01:03:46.550
And the non-uniqueness
of the minimizer
01:03:46.550 --> 01:03:54.370
makes this minimization
problem much more difficult.
01:03:54.370 --> 01:03:56.580
So Razborov proved
this result for edge
01:03:56.580 --> 01:03:58.320
versus triangle densities.
01:03:58.320 --> 01:04:04.290
And this program was
later completed to K4,
01:04:04.290 --> 01:04:07.800
and more generally,
to Kr So K4 is
01:04:07.800 --> 01:04:17.060
due to a result of Nikiforov,
and the Kr result of Reiher
01:04:17.060 --> 01:04:18.200
So a similar picture.
01:04:18.200 --> 01:04:19.850
It's more or less
that picture up
01:04:19.850 --> 01:04:22.840
there but with the
actual numbers shifted.
01:04:22.840 --> 01:04:26.380
Instead of edge versus triangle,
it is now edge versus Kr.
01:04:29.740 --> 01:04:32.818
I should say that it's worth--
so this is a picture that I
01:04:32.818 --> 01:04:35.110
drew up there, and this is
roughly the picture that you
01:04:35.110 --> 01:04:36.520
see in textbooks--
01:04:36.520 --> 01:04:38.730
how they draw these scallops.
01:04:38.730 --> 01:04:42.970
I once plotted what this picture
looks like in Mathematica,
01:04:42.970 --> 01:04:46.090
just to see for myself
where the actual graph is.
01:04:46.090 --> 01:04:48.340
And it doesn't actually
look like that.
01:04:48.340 --> 01:04:50.500
The concaveness is very subtle.
01:04:50.500 --> 01:04:53.170
If you draw it on a computer,
they look like straight lines.
01:04:53.170 --> 01:04:55.320
So in some sense,
that's a cartoon.
01:04:55.320 --> 01:04:59.170
So the concaveness
is caricatured.
01:04:59.170 --> 01:05:02.563
So it's not actually as
concave as it is drawn,
01:05:02.563 --> 01:05:04.480
but I think it's a good
illustration of what's
01:05:04.480 --> 01:05:05.355
happening in reality.
01:05:09.700 --> 01:05:12.830
Questions?
01:05:12.830 --> 01:05:18.040
So on one hand, every
polynomial graph inequality--
01:05:18.040 --> 01:05:20.840
so what do I mean by a
polynomial graph inequality?
01:05:20.840 --> 01:05:27.270
So something like--
suppose I have
01:05:27.270 --> 01:05:28.920
some inequality of this form.
01:05:38.890 --> 01:05:40.890
And I want to
know, is this true?
01:05:43.420 --> 01:05:44.850
It turns out that
I don't actually
01:05:44.850 --> 01:05:49.380
need these squares in some
sense because I can always
01:05:49.380 --> 01:05:58.600
replace them by what happens
if you take disjoint unions.
01:05:58.600 --> 01:06:03.100
So all I'm trying to say is
that every polynomial graph
01:06:03.100 --> 01:06:09.670
inequality can be
written as a linear graph
01:06:09.670 --> 01:06:11.980
inequality of densities.
01:06:18.710 --> 01:06:21.170
But nevertheless, this still
captures a very large class
01:06:21.170 --> 01:06:22.890
of graph inequalities.
01:06:22.890 --> 01:06:25.850
And if I just give you some
arbitrary one that is not
01:06:25.850 --> 01:06:27.380
of that form, it
can be often very
01:06:27.380 --> 01:06:30.000
difficult to decide
whether it is true or not.
01:06:30.000 --> 01:06:31.250
So over here it's not so hard.
01:06:31.250 --> 01:06:32.625
You just plug it
in, and then you
01:06:32.625 --> 01:06:35.690
can decide whether it is true.
01:06:35.690 --> 01:06:39.080
I mean, it turns out to decide
whether this inequality is
01:06:39.080 --> 01:06:42.110
true, it's really a polynomial.
01:06:42.110 --> 01:06:44.600
And then you just check.
01:06:44.600 --> 01:06:46.200
It's not too hard to do.
01:06:46.200 --> 01:06:50.890
But in general, suppose I give
you an inequality of this form.
01:06:50.890 --> 01:06:54.610
So some generalized version of
a linear inequality, like that.
01:06:54.610 --> 01:07:00.000
It's even decidable if
the inequality holds.
01:07:00.000 --> 01:07:04.110
Decidable in the sense of
Turing halting problem.
01:07:04.110 --> 01:07:07.350
So is there some
computer program
01:07:07.350 --> 01:07:10.110
give you this
inequality is true?
01:07:10.110 --> 01:07:12.120
I wonder, can you write
a computer program
01:07:12.120 --> 01:07:16.140
that decides the truthfulness?
01:07:16.140 --> 01:07:17.072
It turns out-- OK.
01:07:17.072 --> 01:07:18.780
So before telling you
what the answer is,
01:07:18.780 --> 01:07:22.500
let me just put it
in some context.
01:07:22.500 --> 01:07:24.452
What about more classical
questions before we
01:07:24.452 --> 01:07:25.410
jump into graph theory?
01:07:25.410 --> 01:07:37.790
If I give you some polynomial
p over the real numbers
01:07:37.790 --> 01:07:46.500
and I want to check
is that true--
01:07:46.500 --> 01:07:49.900
so this is not too hard.
01:07:49.900 --> 01:07:51.010
So this is not too hard.
01:07:51.010 --> 01:08:02.220
But what if you have
multivariate for all real?
01:08:06.440 --> 01:08:08.360
Does anyone know the answer?
01:08:08.360 --> 01:08:09.320
Is this decidable?
01:08:14.250 --> 01:08:15.720
So as you can
imagine, these things
01:08:15.720 --> 01:08:18.810
were studied pretty classically.
01:08:18.810 --> 01:08:22.950
And so it turns out
that every first word
01:08:22.950 --> 01:08:25.500
or theory over the real
numbers is decidable.
01:08:25.500 --> 01:08:27.939
So this is a result of Tarski.
01:08:27.939 --> 01:08:31.979
In particular, such
questions are decidable.
01:08:31.979 --> 01:08:34.946
And in fact, there is a very
nice characterization of--
01:08:34.946 --> 01:08:39.569
so there's a result
called Artin's theorem
01:08:39.569 --> 01:08:41.850
that tells you that
every such polynomial,
01:08:41.850 --> 01:08:45.000
if it is non-negative,
then if and only
01:08:45.000 --> 01:08:47.729
if, it can be written
as a sum of squares
01:08:47.729 --> 01:08:50.250
of rational functions.
01:08:50.250 --> 01:08:52.170
So there's a very
nice characterization
01:08:52.170 --> 01:08:57.080
of positiveness of
polynomials over the reals.
01:09:00.109 --> 01:09:03.410
But now I change the
question and I ask,
01:09:03.410 --> 01:09:04.970
what about over the integers?
01:09:09.350 --> 01:09:11.950
So if I give you a polynomial,
is it always non-negative
01:09:11.950 --> 01:09:16.760
if I have integer entries?
01:09:16.760 --> 01:09:18.078
Is this decidable?
01:09:21.214 --> 01:09:24.910
So turns out, this
is not decidable.
01:09:24.910 --> 01:09:25.859
And this is related.
01:09:25.859 --> 01:09:29.560
So it's more or less the
same as the undecidability
01:09:29.560 --> 01:09:42.670
of Diophantine equations,
which is also known
01:09:42.670 --> 01:09:44.080
as Hilbert's tenth problem.
01:09:50.660 --> 01:09:54.770
So there is no computer program
where we give you a Diophantine
01:09:54.770 --> 01:09:58.370
equation and solves the
question or even tells you
01:09:58.370 --> 01:10:02.180
whether the equation
has a solution.
01:10:02.180 --> 01:10:05.223
And this is part of what
makes number theory,
01:10:05.223 --> 01:10:06.890
makes Diophantine
equations interesting.
01:10:06.890 --> 01:10:11.630
So it's undecidable,
but we talk about it.
01:10:15.360 --> 01:10:19.620
So undecidability is a famous
result due to Matiyasevich.
01:10:24.790 --> 01:10:28.870
So what about graph
theoretic inequalities?
01:10:28.870 --> 01:10:45.555
So is a graph homomorphism
inequality decidable?
01:10:51.477 --> 01:10:53.310
I mean, the question
you should ask yourself
01:10:53.310 --> 01:10:55.290
is, which one is it closer to?
01:10:55.290 --> 01:10:59.280
Is it closer to deciding the
positiveness of polynomials
01:10:59.280 --> 01:11:03.430
over reals or over integers?
01:11:03.430 --> 01:11:07.000
On one hand, you
might think that it
01:11:07.000 --> 01:11:10.238
is more similar to the question
of polynomials over real.
01:11:10.238 --> 01:11:12.280
So first of all, why it's
similar to polynomials,
01:11:12.280 --> 01:11:16.090
I hope that's at
least intuitively--
01:11:16.090 --> 01:11:17.830
nothing's a proof,
but intuitively it
01:11:17.830 --> 01:11:19.800
feels somewhat similar
to polynomials.
01:11:19.800 --> 01:11:22.140
And all of these guys
you can write down
01:11:22.140 --> 01:11:23.630
as polynomial-like quantities.
01:11:23.630 --> 01:11:28.633
And we saw this earlier in the
proof of Bollobas' theorem.
01:11:28.633 --> 01:11:30.300
So you might think
it's similar to reals
01:11:30.300 --> 01:11:32.610
because, well, for
graphons, you can
01:11:32.610 --> 01:11:34.380
take arbitrary real weights.
01:11:34.380 --> 01:11:37.150
So it feels like the reals.
01:11:37.150 --> 01:11:46.530
So it turns out, due to a
theorem of Hatami and Norine,
01:11:46.530 --> 01:11:50.250
that the answer is no.
01:11:50.250 --> 01:11:53.940
It is not decidable.
01:11:53.940 --> 01:11:59.380
And roughly the reason has
to do with this picture.
01:11:59.380 --> 01:12:04.720
Even though the space of
graphons is not discrete,
01:12:04.720 --> 01:12:07.840
it's a very continuous
object, even if you just
01:12:07.840 --> 01:12:10.120
look at this picture
here, you have
01:12:10.120 --> 01:12:14.750
a bunch of discrete
points along this scallop.
01:12:14.750 --> 01:12:18.580
So here's a potential
strategy for proving
01:12:18.580 --> 01:12:24.190
the undecidability of graph
homomorphism inequalities.
01:12:24.190 --> 01:12:29.310
I start by just restricting
myself to this curve.
01:12:29.310 --> 01:12:32.892
I restrict myself
to the red curve.
01:12:32.892 --> 01:12:34.850
If you restrict yourself
to the red curve, than
01:12:34.850 --> 01:12:36.620
the set of possibilities--
01:12:36.620 --> 01:12:39.845
it's now a discrete set, which
is like the positive integers.
01:12:43.670 --> 01:12:46.600
And now I start with--
01:12:46.600 --> 01:12:51.070
I can reduce the problem to
the problem of decidability
01:12:51.070 --> 01:12:54.510
of integer inequalities.
01:12:54.510 --> 01:12:57.220
I start with an
integer inequality.
01:12:57.220 --> 01:13:03.480
I convert it to an inequality
about points on this red curve.
01:13:05.990 --> 01:13:14.170
And that turns into a
corresponding graph inequality,
01:13:14.170 --> 01:13:16.231
which must then be undecidable.
01:13:19.020 --> 01:13:22.970
So this undecidability result
is related to the discreteness
01:13:22.970 --> 01:13:25.270
of points on this red curve.
01:13:29.990 --> 01:13:32.370
So general undecidability
results are interesting.
01:13:32.370 --> 01:13:35.220
But often, we're interested
in specific problems.
01:13:35.220 --> 01:13:38.970
So I give you some specific
inequality and ask, is it true?
01:13:38.970 --> 01:13:42.720
And there are a lot of
interesting open problems
01:13:42.720 --> 01:13:43.680
of that type.
01:13:43.680 --> 01:13:45.990
My favorite one, and also
a very important problem
01:13:45.990 --> 01:13:48.170
in extremal graph
theory, is known
01:13:48.170 --> 01:13:50.026
as Sidorenko's conjecture.
01:13:58.657 --> 01:14:00.740
So the main cause conjecture--
it's a conjecture--
01:14:00.740 --> 01:14:11.500
says that if H is bipartite,
then the H density in G or W
01:14:11.500 --> 01:14:14.890
is at least the
edge density raised
01:14:14.890 --> 01:14:21.060
to the power of the
number of edges of H.
01:14:21.060 --> 01:14:27.820
So we saw one example
of this inequality
01:14:27.820 --> 01:14:29.990
when H is the fourth cycle.
01:14:29.990 --> 01:14:31.630
So when we discussed
quasi-randomness
01:14:31.630 --> 01:14:33.945
we saw that this is true.
01:14:33.945 --> 01:14:35.820
And in the homework,
you'll have a few more--
01:14:35.820 --> 01:14:37.445
so in the next problem
homework, you'll
01:14:37.445 --> 01:14:39.040
have a few more
examples where you're
01:14:39.040 --> 01:14:41.790
asked to show this inequality.
01:14:41.790 --> 01:14:42.940
It is open.
01:14:42.940 --> 01:14:44.660
We don't know any
counterexamples.
01:14:44.660 --> 01:14:49.240
And the first open example,
it's known as something
01:14:49.240 --> 01:14:50.260
called a Mobius strip.
01:14:55.240 --> 01:14:59.920
So the Mobius strip
graph, which is
01:14:59.920 --> 01:15:04.150
a fancy name for the graph
consisting of taking a K55
01:15:04.150 --> 01:15:05.500
and removing a 10 cycle.
01:15:16.390 --> 01:15:17.570
So that's the graph.
01:15:17.570 --> 01:15:20.560
It is open whether this
inequality holds for that graph
01:15:20.560 --> 01:15:22.010
there.
01:15:22.010 --> 01:15:24.187
And this is something
of great interest.
01:15:24.187 --> 01:15:26.020
So if you can make
progress on this problem,
01:15:26.020 --> 01:15:27.415
people will be very excited.
01:15:31.750 --> 01:15:37.072
Now, why is this
called a Mobius strip?
01:15:37.072 --> 01:15:38.530
This took me a
while to figure out.
01:15:38.530 --> 01:15:40.363
So there are many
different interpretations.
01:15:40.363 --> 01:15:42.610
I think the reason why it's
called a Mobius strip is
01:15:42.610 --> 01:15:47.170
that if you think about the
usual simplicial complex
01:15:47.170 --> 01:15:47.980
for a Mobius strip.
01:15:50.950 --> 01:15:56.110
And then this is the face vertex
incidence bipartite graph.
01:15:56.110 --> 01:15:58.570
So five vertices,
one for each face.
01:15:58.570 --> 01:16:02.530
Five vertices, one
for each vertex.
01:16:02.530 --> 01:16:05.890
And if you draw the incident
structure, that's the graph.
01:16:05.890 --> 01:16:10.150
I'm not sure if this
topological formulation will
01:16:10.150 --> 01:16:15.305
help you improving Sidorenko's
conjecture or disprove it,
01:16:15.305 --> 01:16:17.680
but certainly that that's why
it's called a Mobius strip.
01:16:17.680 --> 01:16:19.222
And there are some
people believe who
01:16:19.222 --> 01:16:22.170
believe that it may be false.
01:16:22.170 --> 01:16:23.920
So it's still open.
01:16:23.920 --> 01:16:25.068
It's still open.
01:16:27.820 --> 01:16:30.130
The one last thing
I want to mention
01:16:30.130 --> 01:16:35.330
is that even though the
inequality written up there
01:16:35.330 --> 01:16:38.630
in general is
undecidable, if you only
01:16:38.630 --> 01:16:41.180
want to know whether
this inequality is true
01:16:41.180 --> 01:16:45.830
up to an epsilon error,
then it has decidable.
01:16:45.830 --> 01:16:49.850
In fact, there is an
algorithm that I can tell you.
01:16:49.850 --> 01:16:58.010
So there exists an
algorithm that decides,
01:16:58.010 --> 01:17:04.568
for every epsilon,
that resides--
01:17:04.568 --> 01:17:06.860
so I just want to know whether
that inequality is true.
01:17:06.860 --> 01:17:10.310
But I allow an
epsilon error, meaning
01:17:10.310 --> 01:17:27.860
it decides correctly
this inequality is true
01:17:27.860 --> 01:17:36.530
up to an epsilon error for
all G or outputs a G such
01:17:36.530 --> 01:17:45.616
that the sum here is negative.
01:17:49.430 --> 01:17:52.930
So up to an epsilon of error,
I can give you an algorithm.
01:17:52.930 --> 01:17:56.230
And the algorithm follows--
01:17:56.230 --> 01:17:59.800
I mean, it's not too
hard to describe.
01:17:59.800 --> 01:18:02.500
Basically, the idea
is that if I take
01:18:02.500 --> 01:18:07.010
an epsilon regular
partition, then all the data
01:18:07.010 --> 01:18:10.090
about edge densities
can be encoded
01:18:10.090 --> 01:18:12.340
in the epsilon
regular partition.
01:18:12.340 --> 01:18:17.590
So apply even the weak
regularity lemma is enough.
01:18:22.190 --> 01:18:32.150
And then we can test the
bounded number of possibilities
01:18:32.150 --> 01:18:37.820
with some fixed number of parts.
01:18:41.600 --> 01:18:47.270
And by the counting lemma,
you lose some epsilon error
01:18:47.270 --> 01:18:51.270
if I check over
all weighted graphs
01:18:51.270 --> 01:18:54.860
on some bounded number of
parts whose edge weights are
01:18:54.860 --> 01:18:59.330
multiples of epsilon, let's
say, whether this is true.
01:18:59.330 --> 01:19:02.090
If it's true, then it is
true with this epsilon.
01:19:02.090 --> 01:19:03.560
If it is false,
then I can already
01:19:03.560 --> 01:19:06.860
output a counterexample.
01:19:06.860 --> 01:19:08.870
So there is only finitely
many possibilities
01:19:08.870 --> 01:19:11.090
as a result of weak
regularity lemma.
01:19:11.090 --> 01:19:15.870
And therefore, this
version here is decidable.
01:19:15.870 --> 01:19:20.190
So today, we saw many different
graph theoretic inequalities
01:19:20.190 --> 01:19:21.520
and some general results.
01:19:21.520 --> 01:19:24.780
And there are lots of
open problems about graph
01:19:24.780 --> 01:19:27.210
homomorphism inequalities.
01:19:27.210 --> 01:19:31.055
So this concludes roughly
the extremal graph theory
01:19:31.055 --> 01:19:32.020
section of this course.
01:19:32.020 --> 01:19:33.810
So starting from
next lecture, we'll
01:19:33.810 --> 01:19:36.130
be looking at Roth's theorem.
01:19:36.130 --> 01:19:37.710
So looking at the
Fourier analytic
01:19:37.710 --> 01:19:40.550
proof of Roth's theorem.