WEBVTT
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YUFEI ZHAO: For the
past few lectures,
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we've been discussing the
structure of set addition,
00:00:21.310 --> 00:00:25.260
and which culminated in the
proof of Freiman's theorem.
00:00:25.260 --> 00:00:28.270
So this was a pretty
big and central result
00:00:28.270 --> 00:00:30.640
in additive combinatorics,
which gives you
00:00:30.640 --> 00:00:35.500
a complete characterization
of sets with small doubling.
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Today, I want to look at a
somewhat different issue also
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related to sets
of small doubling,
00:00:40.780 --> 00:00:44.860
but this time we want to
have a somewhat different
00:00:44.860 --> 00:00:48.760
characterization of what does
it mean for a set to have
00:00:48.760 --> 00:00:51.652
lots of additive structure.
00:00:51.652 --> 00:00:53.110
So in today's
lecture, we're always
00:00:53.110 --> 00:00:55.968
going to be working
in an Abelian group.
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Let me define the
following quantity.
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Given sets A and B, we
define the additive energy
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between A and B to be
denoted by E of A and B.
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So A and B are subgroups.
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They're subsets of this
arbitrary Abelian group.
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So E of A and B is defined to
be the number of quadruples, a1,
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a2, b1, b2, where a1, a2
are elements of A, and b1,
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b2 are elements of B,
such that a1 plus b1
00:01:41.310 --> 00:01:43.680
equals to a2 plus b2.
00:01:48.650 --> 00:01:52.610
So the additive
energy is the number
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of quadruples of these
elements where you
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have this additive relation.
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And we would like
to understand sets
00:02:02.000 --> 00:02:04.790
with large additive energy.
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So, intuitively, if you
have lots of solutions
00:02:07.460 --> 00:02:09.680
to this equation in
your sets, then the
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sets themselves should have lots
of internal additive structure.
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So it's a different way of
describing additive structure,
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and we'd like to
understand how does
00:02:19.400 --> 00:02:21.800
this way of describing
additive structure
00:02:21.800 --> 00:02:26.510
relate to things we've seen
before, namely small doubling.
00:02:30.480 --> 00:02:33.840
When you have not two
sets but just one set--
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slightly easier to think about--
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we just write E of A.
I mean E of A comma A.
00:02:42.690 --> 00:02:55.150
And these objects are analogous
to 4 cycles in graph theory.
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Because if you about this
expression here in a Cayley
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graph, let's say
over F2, then this
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is the description of a 4 cycle.
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You go around 4
steps, and you come
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back to where you started from.
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So these objects are
the analogs of 4 cycles.
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And we already saw in our
discussion of quasi-randomness,
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and also elsewhere,
that 4 cycles
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play an important
role in graph theory.
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And, likewise, these
additive energies
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are going to play an important
role in describing sets
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with additive structure.
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Consider the following quantity.
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We're going to let r sub A comma
B of x to be the number of ways
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to write x as a plus b.
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So x equals to a plus b.
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So r sub A comma B of
x is the number of ways
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I can write x as a plus b, where
a comes from big A, little b
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comes from big B. Then,
reinterpreting the formula
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up there, we see that the
additive energy between two
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sets A and B is simply the sum
of the squares of A-- r sub A
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comma B. As x ranges over
all elements of the group,
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we only need to take x
in the sumset A plus B.
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So the basic question,
like when we discussed
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additive combinatorics, in the
sense of when we discussed sets
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of small doubling,
there we asked,
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if you have a set A of a certain
size, how big can a plus a be?
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Here, let's ask the same.
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If I give you set A of a certain
size, how big or how small
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can the additive
energy of the set be?
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What's the most number
of possible number
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of additive quadruples.
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What's the least possible
number of additive quadruples?
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There's some
trivial bounds, just
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like in the case of sumsets.
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So what are some trivial bounds?
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On one hand, by taking a1 equal
to a2, and b2 equal to b2,
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we see that the energy is
always at least the square
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of the size of A.
On the other hand,
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if I fix three of
the four elements,
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then the fourth
element is determined.
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So the upper bound is
cube of the size of A.
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And you convince
yourself that, except up
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to maybe a constant
factors, this
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is the best possible general
upper and lower bound.
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Similar situation with sumsets,
where you have lower bound
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linear, upper bound quadratic.
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Which is the side with
additive structure?
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So if you have lots
of additive structure,
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you have high energy.
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So this range is when you have
lots of additive structure.
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And we would like to
understand, what can you
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say about a set with
high additive energy?
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Well, what are some examples of
sets with high additive energy?
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It turns out that if
you have a set that
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has small doubling,
then, automatically,
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it implies large
additive energy.
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So, in particular, intervals,
or GAPs, or a large subset
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of GAPs, or all these examples
that we saw-- in fact,
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these are all the examples
coming from Freiman's theorem.
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Also, arbitrary groups.
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You can have subgroups.
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And so all of these examples
have large additive energy.
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So let me-- I'll you the
proof just in a second.
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It's not hard.
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But the real question is,
what about the converse?
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So can you see much in
the reverse direction?
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But, first, let me show you
this claim that small doubling
00:07:20.670 --> 00:07:23.690
implies large additive energy.
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Well, if you have small
doubling, if a plus A is size,
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at most, k times
the size of A, then
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it turns out the
additive energy of A
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is at least the
maximum possible,
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which is A cubed divided by k.
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So that's within a constant
factor of the maximum.
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It's pretty large.
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If you have small doubling,
then large additive energy.
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So let's see the proof.
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So you can often
tell how hard a proof
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is by how simple the statement
is, although that's not always
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the case, as we've seen
with some of our theorems,
00:08:05.080 --> 00:08:07.660
like Plunnecke's inequality.
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But in this case, it turns
out to be fairly simple.
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So we see that r sub A comma
A is supported on A plus A.
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So we use Cauchy-Schwarz
to write--
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so, first, we write additive
energy in terms of the sum
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of the squares of these r's.
00:08:38.260 --> 00:08:46.670
And now, by Cauchy-Schwarz,
we find that you can replace
00:08:46.670 --> 00:08:50.510
the sum of the squared
r's by the sum of the r's.
00:08:50.510 --> 00:08:55.370
But now the key point
here is that we take out
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this factor coming from
Cauchy-Schwarz, which is only
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A plus A. So if the support size
is small, we gain in this step.
00:09:06.310 --> 00:09:11.110
But what is the sum of r's?
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I mean, r of x is
just number of ways
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to write x as little
a1 plus little ab--
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little a2.
00:09:17.800 --> 00:09:23.500
So if I sum over all x, I'm just
looking at different two ways--
00:09:23.500 --> 00:09:28.810
we're just looking at ways of
picking an ordered pair from A.
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So this last expression
is equal to the size of A
00:09:34.210 --> 00:09:39.980
to power 4 divided by
A plus A. And now we
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use that A has small
doubling to conclude
00:09:43.430 --> 00:09:47.700
that the final quantity is at
least A cubed divided by k.
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So we see small doubling
implies large additive energy.
00:09:58.512 --> 00:09:59.720
And this kind of makes sense.
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If your set doesn't
expand, then there
00:10:03.830 --> 00:10:08.180
are many collisions of sums.
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And so you must have lots of
solutions to that equation
00:10:11.110 --> 00:10:13.220
up there.
00:10:13.220 --> 00:10:15.230
But what about the converse?
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If I give you a set with
large additive energy,
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must it necessarily
have small doubling?
00:10:24.922 --> 00:10:27.015
Oh.
00:10:27.015 --> 00:10:28.140
Let me show you an example.
00:10:30.760 --> 00:10:38.320
So, well-- so a large
additive energy,
00:10:38.320 --> 00:10:45.070
does it imply small doubling?
00:10:47.680 --> 00:10:50.730
So consider the
following example, where
00:10:50.730 --> 00:10:53.610
you take a set A which
is a combination,
00:10:53.610 --> 00:10:56.970
is a union of a set
with small doubling
00:10:56.970 --> 00:11:03.874
plus a bunch of elements
without additive structure.
00:11:10.170 --> 00:11:12.230
So I take a set
with small doubling
00:11:12.230 --> 00:11:16.940
plus a bunch of elements
without additive structure.
00:11:16.940 --> 00:11:21.190
Then it has large additive
energy, just coming
00:11:21.190 --> 00:11:25.120
from this interval itself.
00:11:25.120 --> 00:11:31.990
So the energy of A
is order N cubed.
00:11:31.990 --> 00:11:34.630
N is the number of elements.
00:11:34.630 --> 00:11:38.320
What about A plus A?
00:11:38.320 --> 00:11:41.650
Well, for A plus A,
this part doesn't--
00:11:41.650 --> 00:11:43.120
that's the part
that contributes,
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or the part of this A
without additive structure.
00:11:48.270 --> 00:11:52.250
And we see that the
size of A plus A
00:11:52.250 --> 00:11:56.830
is quadratic in the size of A.
00:11:56.830 --> 00:12:00.470
So, unfortunately,
the converse fails.
00:12:00.470 --> 00:12:05.480
So you can have sets that have
large additive energy and also
00:12:05.480 --> 00:12:07.590
large doubling.
00:12:07.590 --> 00:12:11.100
But, you see, the reason why
this has large additive energy
00:12:11.100 --> 00:12:14.110
is because there is a very
highly structured additively
00:12:14.110 --> 00:12:15.990
structured piece of it.
00:12:15.990 --> 00:12:20.290
And, somehow, we want to forget
about this extra garbage.
00:12:20.290 --> 00:12:24.790
And that's part of the reason
why the converse is not true.
00:12:24.790 --> 00:12:26.820
So we would like
a statement that
00:12:26.820 --> 00:12:30.150
says that if you have
large additive energy, then
00:12:30.150 --> 00:12:33.750
it must come from some
highly structured piece that
00:12:33.750 --> 00:12:36.380
has small doubling.
00:12:36.380 --> 00:12:38.260
And that is true, and
that's the content
00:12:38.260 --> 00:12:40.540
of the Balog-Szemeredi-Gowers
theorem, which
00:12:40.540 --> 00:12:43.660
is the main topic today.
00:12:43.660 --> 00:12:51.890
So the Balog-Szemeredi-Gowers
theorem says that if you have
00:12:51.890 --> 00:12:52.760
a set--
00:12:52.760 --> 00:12:56.060
so we're working always in
some arbitrary Abelian group.
00:12:56.060 --> 00:13:00.650
If you have a set
with large energy,
00:13:00.650 --> 00:13:07.280
then there exists some
subset A prime of A such
00:13:07.280 --> 00:13:13.460
that A prime is a fairly
large proportion of A.
00:13:13.460 --> 00:13:18.260
And here, by large I mean up to
polynomial changes in the error
00:13:18.260 --> 00:13:18.920
parameters.
00:13:22.010 --> 00:13:28.310
So this A prime is such that
A prime has small doubling.
00:13:34.340 --> 00:13:36.760
If you have large
additive energy,
00:13:36.760 --> 00:13:40.960
then I can pick out a large
piece with small doubling
00:13:40.960 --> 00:13:44.200
constant, and I only
loose a polynomial
00:13:44.200 --> 00:13:45.775
in the error factors.
00:13:48.200 --> 00:13:50.075
So that's the
Balog-Szemeredi-Gowers theorem,
00:13:50.075 --> 00:13:56.420
and it describes
this example up here.
00:13:56.420 --> 00:13:58.058
Any questions about
the statement?
00:14:01.200 --> 00:14:04.970
So what I will actually show you
is a slight variant, actually
00:14:04.970 --> 00:14:08.782
a more general statement, where,
instead of having one set,
00:14:08.782 --> 00:14:09.990
we're going to have two sets.
00:14:12.720 --> 00:14:16.880
So here's Balog-Szemeredi-Gowers
theorem version
00:14:16.880 --> 00:14:25.460
2, where now we have two sets.
00:14:25.460 --> 00:14:27.050
Again, A and B are--
00:14:27.050 --> 00:14:28.640
I'm not going to write any--
00:14:28.640 --> 00:14:29.690
I'm not going to write
it in this lecture,
00:14:29.690 --> 00:14:32.210
but A and B are always subsets
of some arbitrary Abelian
00:14:32.210 --> 00:14:32.710
group.
00:14:32.710 --> 00:14:34.790
So A and B both have
size of, at most,
00:14:34.790 --> 00:14:40.325
n, and the energy
between A and B is large.
00:14:44.670 --> 00:14:53.580
Then there exists a subset A
prime of A, B prime of B such
00:14:53.580 --> 00:14:59.820
that both A prime
and B prime are
00:14:59.820 --> 00:15:05.250
large fractions of
their parent set,
00:15:05.250 --> 00:15:14.070
and such that A prime
plus B prime is not
00:15:14.070 --> 00:15:15.790
too much bigger than n.
00:15:21.170 --> 00:15:24.150
It's not so obvious
why the second version
00:15:24.150 --> 00:15:26.020
implies the first version.
00:15:26.020 --> 00:15:29.030
So you can say, well, take
A and B to be the same.
00:15:29.030 --> 00:15:31.580
But then the
conclusion gives you
00:15:31.580 --> 00:15:36.200
possibly two different
subsets, A prime and B prime.
00:15:36.200 --> 00:15:39.980
But the first version,
I only want one subset
00:15:39.980 --> 00:15:43.320
that has small doubling.
00:15:43.320 --> 00:15:45.270
So, fortunately,
the second version
00:15:45.270 --> 00:15:47.782
does imply the first version.
00:15:47.782 --> 00:15:48.490
So let's see why.
00:15:54.020 --> 00:15:58.610
The second version implies the
first version because, if we--
00:16:03.090 --> 00:16:06.350
so there's a tool
that we introduced
00:16:06.350 --> 00:16:08.630
early on when we discussed
Freiman's theorem,
00:16:08.630 --> 00:16:14.390
and this is the Ruzsa
triangle inequality.
00:16:14.390 --> 00:16:16.410
So the spirit of Ruzsa
triangle inequality
00:16:16.410 --> 00:16:19.680
is it allows you to
relate, to sort of go
00:16:19.680 --> 00:16:23.010
back and forth between different
sumsets in different sets.
00:16:23.010 --> 00:16:31.250
So by Ruzsa triangle inequality,
if we apply the second version
00:16:31.250 --> 00:16:34.535
with A equals to B, then--
00:16:37.750 --> 00:16:40.290
and we pick out this
A prime and B prime,
00:16:40.290 --> 00:16:43.050
then we see that A
prime plus A prime
00:16:43.050 --> 00:16:54.370
is, at most, A prime plus B
prime squared over B prime.
00:16:54.370 --> 00:16:56.290
Well, actually, this uses the--
00:16:56.290 --> 00:16:58.420
vice versa it uses a
slightly stronger version
00:16:58.420 --> 00:17:01.910
that we had to use
Plunnecke-Ruzsa key lemma
00:17:01.910 --> 00:17:02.650
to prove.
00:17:02.650 --> 00:17:04.089
But you can come up--
00:17:04.089 --> 00:17:06.730
I mean, if you don't care
about the precise loss
00:17:06.730 --> 00:17:09.040
in the polynomial
factors, you can also
00:17:09.040 --> 00:17:10.780
use the basic Ruzsa
triangle inequality
00:17:10.780 --> 00:17:13.270
to deduce a similar statement.
00:17:13.270 --> 00:17:14.920
This is easier to deduce.
00:17:14.920 --> 00:17:16.300
So you have that.
00:17:16.300 --> 00:17:19.990
And now, the second
version tells you
00:17:19.990 --> 00:17:24.150
that the numerator
is, at most, poly kn,
00:17:24.150 --> 00:17:30.340
and the denominator
is, at most-- at least,
00:17:30.340 --> 00:17:33.170
n divided by poly k.
00:17:33.170 --> 00:17:40.440
Remember, over here, to get this
hypothesis, we automatically
00:17:40.440 --> 00:17:45.830
have that the size
of A and B are not
00:17:45.830 --> 00:17:47.580
two much smaller than n.
00:17:47.580 --> 00:17:50.700
Or else this cannot be true.
00:17:50.700 --> 00:17:58.200
So putting all these estimates
together, we get that.
00:17:58.200 --> 00:18:02.705
So these two versions, they
are equivalent to each other.
00:18:02.705 --> 00:18:04.080
Second version
implies the first.
00:18:04.080 --> 00:18:06.240
The second one is stronger.
00:18:06.240 --> 00:18:08.292
The first one is
slightly more useful.
00:18:08.292 --> 00:18:09.750
They're not
necessarily equivalent,
00:18:09.750 --> 00:18:13.190
but the second one is stronger.
00:18:13.190 --> 00:18:16.370
Any questions?
00:18:16.370 --> 00:18:17.847
All right.
00:18:17.847 --> 00:18:19.680
So this is a
Balog-Szemeredi-Gowers theorem.
00:18:19.680 --> 00:18:21.570
So the content of
today's lecture
00:18:21.570 --> 00:18:24.930
is to show you how to
prove this theorem.
00:18:24.930 --> 00:18:26.940
A remark about the
naming of this theorem.
00:18:26.940 --> 00:18:29.100
So you might notice that
these three letters do not
00:18:29.100 --> 00:18:31.738
coming in alphabetical order.
00:18:31.738 --> 00:18:33.780
And the reason is that
this theorem was initially
00:18:33.780 --> 00:18:37.380
approved by Balog and
Szemeredi, but using
00:18:37.380 --> 00:18:40.890
a more involved
method that didn't
00:18:40.890 --> 00:18:43.470
give polynomial high bounds.
00:18:43.470 --> 00:18:47.190
And Gowers, in his proof
of Szemeredi's theorem,
00:18:47.190 --> 00:18:49.530
his new proof of Szemeredi's
theorem with good bounds,
00:18:49.530 --> 00:18:51.030
he required--
00:18:51.030 --> 00:18:52.470
well, he looked
into this theorem
00:18:52.470 --> 00:18:54.930
and gave a new
proof that resulted
00:18:54.930 --> 00:18:56.990
in this polynomial type bounds.
00:18:56.990 --> 00:18:59.700
And it is that idea that
we're going to see today.
00:19:09.567 --> 00:19:11.650
So this course is called
graph theory and additive
00:19:11.650 --> 00:19:12.850
combinatorics.
00:19:12.850 --> 00:19:15.490
And the last two
topics of this course--
00:19:15.490 --> 00:19:17.680
today being
Balog-Szemeredi-Gowers,
00:19:17.680 --> 00:19:20.740
and tomorrow we're going to
see sum-product problem--
00:19:20.740 --> 00:19:23.500
are both great
examples of problems
00:19:23.500 --> 00:19:26.740
in additive combinatorics
where tools from graph theory
00:19:26.740 --> 00:19:29.960
play an important role
in their solutions.
00:19:29.960 --> 00:19:33.910
So it's a nice combination
of the subject where we see
00:19:33.910 --> 00:19:36.350
both topics at the same time.
00:19:36.350 --> 00:19:39.190
So I want to show you the proof
of Balog-Szemeredi-Gowers,
00:19:39.190 --> 00:19:41.890
and the proof goes
via a graph analog.
00:19:41.890 --> 00:19:44.860
So I'm going to state for
you a graphical version
00:19:44.860 --> 00:19:48.960
of the Balog-Szemeredi-Gowers
theorem.
00:19:48.960 --> 00:19:50.520
And it goes like this.
00:19:50.520 --> 00:20:02.860
If G is a bipartite graph
between vertex sets A and B--
00:20:02.860 --> 00:20:06.140
and here A and B are still
subsets of the Abelian group--
00:20:17.740 --> 00:20:24.910
we define this restricted
sumset, A plus sub G of B,
00:20:24.910 --> 00:20:33.820
to be the set of
sums where I'm only
00:20:33.820 --> 00:20:40.256
taking sums across edges in g.
00:20:44.540 --> 00:20:47.500
So, in particular, if G is
the complete bipartite graph,
00:20:47.500 --> 00:20:50.100
then this is the usual sumset.
00:20:50.100 --> 00:20:54.390
But now I may allow
G to be a subset
00:20:54.390 --> 00:20:56.220
of the complete bipartite graph.
00:20:56.220 --> 00:20:59.430
So only taking some
but not all of the--
00:20:59.430 --> 00:21:01.920
only taking-- yes, some of
this sums but not all of them.
00:21:04.600 --> 00:21:12.940
The graphical version of
Balog-Szemeredi-Gowers
00:21:12.940 --> 00:21:23.290
says that if you have A and B
be subsets of an Abelian group,
00:21:23.290 --> 00:21:27.610
both having size,
at most, n, and G
00:21:27.610 --> 00:21:35.770
is a bipartite graph
between A and B,
00:21:35.770 --> 00:21:42.460
such that G has lots of
edges, has at least n squared
00:21:42.460 --> 00:21:43.570
over k edges.
00:21:47.290 --> 00:21:56.090
If the restricted sumset
between A and B is small--
00:21:56.090 --> 00:22:02.470
So here we're not looking at all
the sums but a large fraction
00:22:02.470 --> 00:22:04.540
of the possible pairwise sums.
00:22:04.540 --> 00:22:07.040
If that sumset has
small size, this
00:22:07.040 --> 00:22:10.210
is kind of like a restricted
doubling constant.
00:22:10.210 --> 00:22:16.090
Then there exists
A prime, subset
00:22:16.090 --> 00:22:26.530
of A, B prime, subset of
B, with A prime and B prime
00:22:26.530 --> 00:22:32.830
both fairly large fractions
of their parent set,
00:22:32.830 --> 00:22:36.970
and such that the unrestricted
sumset between A prime and B
00:22:36.970 --> 00:22:40.270
prime is not too large.
00:22:48.020 --> 00:22:50.390
So let me say it again.
00:22:50.390 --> 00:22:52.760
So we have a fairly dense--
00:22:52.760 --> 00:22:55.180
so a constant
fraction edge density,
00:22:55.180 --> 00:22:59.480
a fairly dense bipartite
graph between A and B. A and B
00:22:59.480 --> 00:23:02.660
are subsets of
the Abelian group.
00:23:02.660 --> 00:23:08.700
Then-- and such that the
restricted sumset is small.
00:23:08.700 --> 00:23:14.840
Then I can restrict A and B to
subsets, fairly large subsets,
00:23:14.840 --> 00:23:19.070
so that the complete sumset
between the subsets A prime
00:23:19.070 --> 00:23:21.170
and B prime is small.
00:23:26.180 --> 00:23:29.720
Let me show you why the
graphical version of BSG
00:23:29.720 --> 00:23:33.270
implies the version of
BSG I stated up there.
00:23:50.630 --> 00:23:54.511
But, so why do we care about
this graphical version?
00:23:54.511 --> 00:23:59.530
Well, suppose we-- so we
have all of these hypotheses.
00:23:59.530 --> 00:24:08.030
Let's write-- so we have all
of those hypotheses up there.
00:24:08.030 --> 00:24:11.938
So let's write r
to be r sub A comma
00:24:11.938 --> 00:24:16.930
B, so I don't have to carry
the subscripts all around.
00:24:16.930 --> 00:24:17.920
What do you think--
00:24:17.920 --> 00:24:20.760
so I start with
A and B up there,
00:24:20.760 --> 00:24:24.660
and I need to
construct that graph G.
00:24:24.660 --> 00:24:26.340
So what should we
choose as our graph?
00:24:30.940 --> 00:24:34.460
Let's consider the popular sums.
00:24:40.370 --> 00:24:44.900
So the popular sums are
going to be elements
00:24:44.900 --> 00:24:50.390
in the complete
sumset such that it
00:24:50.390 --> 00:24:54.530
is represented as a sum
in many different ways.
00:25:02.760 --> 00:25:07.840
And we're going to take
edges that correspond
00:25:07.840 --> 00:25:12.760
to these popular sums.
00:25:12.760 --> 00:25:26.670
So let's consider
bipartite graph G such
00:25:26.670 --> 00:25:39.770
that A comma B is an edge if and
only A plus B is a popular sum.
00:25:46.900 --> 00:25:50.390
So let's verify some
of the hypotheses.
00:25:50.390 --> 00:25:53.110
So we're going to
assume graph BSG,
00:25:53.110 --> 00:25:57.340
and let's verify the
hypothesis in graph BSG.
00:25:57.340 --> 00:25:59.500
On one hand, because
each element of S
00:25:59.500 --> 00:26:05.170
is a popular sum, if we
consider its multiplicity,
00:26:05.170 --> 00:26:13.750
we find that the size of S
multiplied by n over 2k, lower
00:26:13.750 --> 00:26:19.245
bound be size of A
times the size of B.
00:26:19.245 --> 00:26:26.750
So if you think about all the
different pairs in A and B,
00:26:26.750 --> 00:26:31.780
each sum here, each popular
sum, contributes this many times
00:26:31.780 --> 00:26:36.330
to this A cross B.
00:26:36.330 --> 00:26:41.370
So, as a result, because
size of A and size of B
00:26:41.370 --> 00:26:44.880
are both, at most, n, we
find that the size of S
00:26:44.880 --> 00:26:46.180
is, at most, 2kn.
00:26:49.382 --> 00:26:51.780
And if you think
about what G is,
00:26:51.780 --> 00:27:00.840
then this implies also that the
restricted sumset of A and B
00:27:00.840 --> 00:27:02.310
across this graph G--
00:27:02.310 --> 00:27:04.080
which only requires
the popular sums.
00:27:04.080 --> 00:27:10.718
So the restricted sumset is
precisely the popular sums.
00:27:10.718 --> 00:27:13.660
So restricted sumset
is not too large.
00:27:18.930 --> 00:27:19.900
OK, good.
00:27:19.900 --> 00:27:24.020
So we got one of the conditions,
that the restricted sumset
00:27:24.020 --> 00:27:25.910
is not too large.
00:27:25.910 --> 00:27:30.150
And now we want to show that
this graph has lots of edges.
00:27:30.150 --> 00:27:31.360
It has lots of edges.
00:27:36.210 --> 00:27:39.120
And here's where we would need
to use the hypothesis that,
00:27:39.120 --> 00:27:44.166
between A and B, originally
there is large additive energy.
00:27:44.166 --> 00:27:49.980
And the point here is that
these unpopular sums cannot
00:27:49.980 --> 00:27:55.140
contribute very much to the
additive energy in total,
00:27:55.140 --> 00:27:58.240
because each one of
them is unpopular.
00:27:58.240 --> 00:28:01.960
So the dominant contributions
to the additive energy
00:28:01.960 --> 00:28:05.280
are going to come
from the popular sums,
00:28:05.280 --> 00:28:08.910
and we're going to use that to
show that G has lots of edges.
00:28:12.660 --> 00:28:16.980
So let's lower bound the number
of edges of G by first showing
00:28:16.980 --> 00:28:18.820
that--
00:28:18.820 --> 00:28:35.030
so we'll show that the unpopular
sums contribute very little
00:28:35.030 --> 00:28:42.210
to the additive energy
between A and B. Indeed,
00:28:42.210 --> 00:28:49.860
the sums of the squares
of the r's, if for x
00:28:49.860 --> 00:28:58.130
not in popular sums,
is upper bounded by--
00:28:58.130 --> 00:29:01.170
well, claim that
it is upper bounded
00:29:01.170 --> 00:29:09.910
by the following quantity,
that n over 2k times n squared.
00:29:14.520 --> 00:29:19.470
Because I can take
out one factor r,
00:29:19.470 --> 00:29:24.180
upper bound by this
number, just by definition,
00:29:24.180 --> 00:29:27.820
and the sums of the
r's is n squared.
00:29:32.540 --> 00:29:39.150
So you have this additive
energy between A and B.
00:29:39.150 --> 00:29:41.190
I know that it is
large by hypothesis.
00:29:45.940 --> 00:29:48.310
Whereas, I also know
that I can write it
00:29:48.310 --> 00:29:52.570
as a sum of the squares
of the r's, which
00:29:52.570 --> 00:30:00.550
I can break into the
popular contributions
00:30:00.550 --> 00:30:02.530
and the unpopular contributions.
00:30:05.533 --> 00:30:06.950
And, hopefully,
this should all be
00:30:06.950 --> 00:30:09.470
somewhat reminiscent of
basically all these proofs
00:30:09.470 --> 00:30:11.200
that we did so far
in this course,
00:30:11.200 --> 00:30:14.750
where we separate a sum
into the dominant terms
00:30:14.750 --> 00:30:16.510
and the minor terms.
00:30:16.510 --> 00:30:20.010
This came up in Fourier
analysis in particular.
00:30:20.010 --> 00:30:24.320
So we do this
splitting, and we upper
00:30:24.320 --> 00:30:28.820
bound the unpopular
contributions by the estimate
00:30:28.820 --> 00:30:29.890
from just now.
00:30:36.810 --> 00:30:40.800
So, as a result, bringing
this small error term,
00:30:40.800 --> 00:30:44.610
it doesn't cancel
much of the energy.
00:30:44.610 --> 00:30:52.350
So we still have a lower bound
on the sum of the squares
00:30:52.350 --> 00:30:56.590
of the r's in the popular sums.
00:31:00.010 --> 00:31:04.240
But I can also give a fairly
trivial upper bound to a single
00:31:04.240 --> 00:31:08.050
r, namely it cannot
be bigger than n.
00:31:16.220 --> 00:31:23.860
And so the number
of edges of G--
00:31:23.860 --> 00:31:27.480
so what's the number
of edges of G?
00:31:27.480 --> 00:31:28.260
Look at that.
00:31:28.260 --> 00:31:33.470
Each x here contributes
rx many edges.
00:31:33.470 --> 00:31:36.750
So the number of edges of G is
simply the sums of these rx's.
00:31:41.310 --> 00:31:42.690
Which is quite large.
00:31:49.740 --> 00:31:56.070
So the hypothesis of
graph BSG are satisfied.
00:31:56.070 --> 00:31:59.850
And so we can use the
conclusion of graph BSG, which
00:31:59.850 --> 00:32:02.730
is the conclusion that
we're looking for in BSG.
00:32:11.520 --> 00:32:12.532
Any questions?
00:32:17.095 --> 00:32:17.595
Good.
00:32:17.595 --> 00:32:19.860
So the remaining
task is to prove
00:32:19.860 --> 00:32:23.160
the graphical version of BSG.
00:32:23.160 --> 00:32:26.040
So let's take a
quick break, and when
00:32:26.040 --> 00:32:30.030
we come back we'll
focus on this theorem,
00:32:30.030 --> 00:32:35.140
and it has some nice
graph theoretic arguments.
00:32:35.140 --> 00:32:37.430
OK, let's continue.
00:32:37.430 --> 00:32:42.230
We've reduced the proof of the
Balog-Szemeredi-Gowers theorem
00:32:42.230 --> 00:32:44.540
to the following
graphical result.
00:32:44.540 --> 00:32:46.170
Well, it's not just
graphical, right?
00:32:46.170 --> 00:32:49.370
Still-- we're still inside
some an Abelian group,
00:32:49.370 --> 00:32:52.570
still looking at some set
in some Abelian group,
00:32:52.570 --> 00:32:57.140
but, certainly, now it has
a graph attached to it.
00:32:57.140 --> 00:33:01.410
Let me show this theorem
through several steps.
00:33:01.410 --> 00:33:04.700
First, something called
a path of length 2 lemma.
00:33:15.502 --> 00:33:17.860
So the path of length
2 lemma, the statement
00:33:17.860 --> 00:33:21.340
is that you start
with a graph G which
00:33:21.340 --> 00:33:27.130
is a bipartite graph
between vertex sets A and B.
00:33:27.130 --> 00:33:29.050
And now A and B no longer need--
00:33:29.050 --> 00:33:30.100
they're just sets.
00:33:30.100 --> 00:33:31.150
They're just vertex sets.
00:33:31.150 --> 00:33:34.550
We're not going to have sums.
00:33:34.550 --> 00:33:38.175
And the number of edges is
at least a constant fraction
00:33:38.175 --> 00:33:39.175
of the maximum possible.
00:33:45.570 --> 00:33:48.620
Then the conclusion
is that there
00:33:48.620 --> 00:33:55.385
exists some U, a subset of A,
such that U is fairly large.
00:33:59.650 --> 00:34:10.199
And between most pairs
of elements of U--
00:34:10.199 --> 00:34:24.880
so between 1 minus epsilon
fraction of pairs of U--
00:34:24.880 --> 00:34:30.840
there are lots of
common neighbors.
00:34:30.840 --> 00:34:36.650
So at least epsilon
delta squared
00:34:36.650 --> 00:34:46.230
B over 2 common neighbors.
00:34:46.230 --> 00:34:58.550
So you start with this bipartite
graph A and B. Lots of edges.
00:34:58.550 --> 00:35:01.520
And we would like
to show that there
00:35:01.520 --> 00:35:08.840
exists a pretty large subset U
such that between most pairs--
00:35:08.840 --> 00:35:11.150
all but an epsilon fraction--
00:35:11.150 --> 00:35:12.980
of ordered pairs--
they could be the same,
00:35:12.980 --> 00:35:15.350
but it doesn't really matter--
00:35:15.350 --> 00:35:22.600
the number of paths of length
2 between these two vertices
00:35:22.600 --> 00:35:24.920
is quite large.
00:35:24.920 --> 00:35:28.430
So they have lots
of common neighbors.
00:35:28.430 --> 00:35:30.440
Where have we seen
something like this before?
00:35:30.440 --> 00:35:30.890
There's a question?
00:35:30.890 --> 00:35:32.694
AUDIENCE: Is there a
[INAUDIBLE] epsilon?
00:35:32.694 --> 00:35:33.600
YUFEI ZHAO: Ah, yes.
00:35:33.600 --> 00:35:37.156
So for every epsilon
and every delta.
00:35:37.156 --> 00:35:43.634
So let epsilon,
delta be parameters.
00:35:48.080 --> 00:35:51.860
Where have we seen
something like this before?
00:35:51.860 --> 00:35:54.620
So in a bipartite graph
with lots of edges,
00:35:54.620 --> 00:35:59.180
I want to find a large
subset of one of the parts
00:35:59.180 --> 00:36:02.270
so that every pair of
elements, or almost
00:36:02.270 --> 00:36:05.842
every pair of elements, have
lots of common neighbors.
00:36:11.254 --> 00:36:12.238
Yes.
00:36:12.238 --> 00:36:13.570
AUDIENCE: [INAUDIBLE].
00:36:13.570 --> 00:36:15.070
YUFEI ZHAO: Dependent
random choice.
00:36:15.070 --> 00:36:17.150
So in the very first
chapter of this course,
00:36:17.150 --> 00:36:19.340
when we did extremal
graph theory
00:36:19.340 --> 00:36:21.530
forbidding bipartite
subgraphs, there
00:36:21.530 --> 00:36:26.900
was a technique for proving the
extremal number, upper bounds,
00:36:26.900 --> 00:36:29.540
for bipartite graphs
of bounded degree.
00:36:29.540 --> 00:36:32.810
And there we used something
called dependent random choice
00:36:32.810 --> 00:36:35.860
that had a conclusion that
was very similar flavor.
00:36:35.860 --> 00:36:39.848
So there, we had every pair--
so a fairly large, but not as
00:36:39.848 --> 00:36:41.390
large as this-- a
fairly large subset
00:36:41.390 --> 00:36:45.640
where every pair of elements
had lots of common neighbors.
00:36:45.640 --> 00:36:48.230
For every couple, every
k couple of vertices,
00:36:48.230 --> 00:36:50.330
have lots of common neighbors.
00:36:50.330 --> 00:36:51.470
So it's very similar.
00:36:51.470 --> 00:36:53.960
In fact, it's the
same type of technique
00:36:53.960 --> 00:36:56.480
that we'll use to prove
this lemma over here.
00:37:00.390 --> 00:37:05.030
So who remembers how
dependent random choice goes?
00:37:05.030 --> 00:37:09.120
So the idea is that we
are going to choose U
00:37:09.120 --> 00:37:11.200
not uniformly at random.
00:37:11.200 --> 00:37:12.872
So that's not going to work.
00:37:12.872 --> 00:37:15.860
Going to choose it in
a dependent random way.
00:37:15.860 --> 00:37:19.630
So I want elements of U to
have lots of common neighbors,
00:37:19.630 --> 00:37:20.720
typically.
00:37:20.720 --> 00:37:24.950
So one way to guarantee
this is to choose U to be
00:37:24.950 --> 00:37:28.550
a neighborhood from the right.
00:37:28.550 --> 00:37:33.640
So pick a random
element in B and choose
00:37:33.640 --> 00:37:37.300
U to be its neighborhood.
00:37:37.300 --> 00:37:39.230
So let's do that.
00:37:39.230 --> 00:37:41.210
So we're going to use
dependent random choice.
00:37:47.505 --> 00:37:49.380
See, everything in the
course comes together.
00:37:56.000 --> 00:38:04.480
So let's pick v an element
of B uniformly at random.
00:38:09.580 --> 00:38:15.440
And let U be the neighborhood
v. So, first of all,
00:38:15.440 --> 00:38:19.100
by linearity of
expectations, the size of U
00:38:19.100 --> 00:38:27.600
is at least delta of A. So
because the average degree
00:38:27.600 --> 00:38:32.472
from the right from B is
at least delta A just based
00:38:32.472 --> 00:38:33.430
on the number of edges.
00:38:36.550 --> 00:38:43.560
If you have two vertices
a and a prime in A
00:38:43.560 --> 00:39:04.220
with a small number of common
neighbors, then the size of--
00:39:04.220 --> 00:39:07.020
so sorry.
00:39:07.020 --> 00:39:10.520
Let me-- I skipped ahead a bit.
00:39:10.520 --> 00:39:15.170
So if a and a prime have a small
number of common neighbors,
00:39:15.170 --> 00:39:22.730
then the probability that
a and a prime both lie in U
00:39:22.730 --> 00:39:25.580
should be quite small.
00:39:25.580 --> 00:39:28.670
Because if they both had--
00:39:28.670 --> 00:39:33.040
if a and a prime have a small
number of common neighbors,
00:39:33.040 --> 00:39:36.863
in order for a and a prime
to be included in this U,
00:39:36.863 --> 00:39:37.780
you must have chosen--
00:39:41.550 --> 00:39:44.920
so suppose this were
their common neighbor.
00:39:44.920 --> 00:39:49.550
Then in order that a and
a prime be contained in U,
00:39:49.550 --> 00:39:54.470
it must have chosen this v to be
inside the common neighborhood
00:39:54.470 --> 00:39:55.300
of a and a prime.
00:39:57.840 --> 00:39:59.940
Which is unlikely
if a and a prime
00:39:59.940 --> 00:40:02.940
had a small number
of common neighbors.
00:40:02.940 --> 00:40:08.600
So this probability is, at most,
epsilon delta squared over 2.
00:40:12.460 --> 00:40:15.520
Just think about how
U is constructed.
00:40:15.520 --> 00:40:26.740
So if we let x be the number
of a and a primes in U cross U
00:40:26.740 --> 00:40:34.800
with, at most, epsilon
delta squared over 2 times B
00:40:34.800 --> 00:40:42.900
common neighbors, then, by
linearity of expectations,
00:40:42.900 --> 00:40:46.300
the expectation of x is--
00:40:46.300 --> 00:40:54.420
well, by summing up all of these
probabilities of a and a prime,
00:40:54.420 --> 00:40:56.890
both being in U--
00:40:56.890 --> 00:41:02.010
so this is, at most,
epsilon delta squared
00:41:02.010 --> 00:41:04.740
over 2 times size of A squared.
00:41:08.250 --> 00:41:12.030
So, typically, at
least in expectation,
00:41:12.030 --> 00:41:16.260
you do not expect very
many pairs of elements in U
00:41:16.260 --> 00:41:20.510
with few common neighbors.
00:41:20.510 --> 00:41:22.940
But we can also turn
such an estimate
00:41:22.940 --> 00:41:24.830
into a specific instance.
00:41:28.015 --> 00:41:33.840
And the way to do this is to
consider the quantity size of U
00:41:33.840 --> 00:41:39.280
squared minus x over epsilon.
00:41:39.280 --> 00:41:43.070
Well, first of all, we can
lower bound this quantity,
00:41:43.070 --> 00:41:47.630
because the size of
second moment of U
00:41:47.630 --> 00:41:53.030
is at least the first
moment of U squared.
00:41:53.030 --> 00:42:02.450
And we also know that the
size of x in expectation
00:42:02.450 --> 00:42:04.830
is not very large.
00:42:04.830 --> 00:42:07.700
So the whole expression
can be lower bounded
00:42:07.700 --> 00:42:15.910
by delta squared over 2
times the size of A squared.
00:42:25.630 --> 00:42:26.935
So this is epsilon, sorry.
00:42:30.120 --> 00:42:34.880
Therefore, there is
some concrete instance
00:42:34.880 --> 00:42:39.110
of this randomness resulting
in some specific U such
00:42:39.110 --> 00:42:41.180
that this inequality holds.
00:42:41.180 --> 00:42:54.330
So there exists some U such
that this inequality holds.
00:42:54.330 --> 00:43:03.310
And, in particular, we find
that the size of U is at least--
00:43:03.310 --> 00:43:05.110
just forget about
this minus term--
00:43:05.110 --> 00:43:08.950
is at least that right-hand
side, square root.
00:43:08.950 --> 00:43:11.500
So, in particular, the
size of U is at least
00:43:11.500 --> 00:43:14.415
delta over 2 times
the size of A.
00:43:14.415 --> 00:43:18.307
And, just looking at the
left-hand side, which
00:43:18.307 --> 00:43:20.890
must be a non-negative quantity
because the right-hand side is
00:43:20.890 --> 00:43:26.800
non-negative, we find that
x is, at most, an epsilon
00:43:26.800 --> 00:43:29.786
fraction of U squared.
00:43:34.480 --> 00:43:38.730
So putting these
together, we arrive
00:43:38.730 --> 00:43:42.280
at the path of length 2 lemma.
00:43:42.280 --> 00:43:43.830
So let me go through it again.
00:43:43.830 --> 00:43:46.100
So this is the dependent
random choice method,
00:43:46.100 --> 00:43:50.480
where we're going to--
we want to find this U,
00:43:50.480 --> 00:43:52.430
where most pairs
of vertices in U
00:43:52.430 --> 00:43:55.920
have lots of common neighbors.
00:43:55.920 --> 00:43:58.640
So we start from the right side.
00:43:58.640 --> 00:44:02.480
We start from B, pick a
uniform random vertex, which
00:44:02.480 --> 00:44:08.598
you call v, and let U be
the neighborhood of v.
00:44:08.598 --> 00:44:11.140
And I claim that this
U, typically, should
00:44:11.140 --> 00:44:13.170
have the desired property.
00:44:13.170 --> 00:44:18.160
And the reason is that, if
you have a pair of vertices
00:44:18.160 --> 00:44:24.030
on the left that do not
have many common neighbors,
00:44:24.030 --> 00:44:27.360
then I claim it is highly
unlikely that these two
00:44:27.360 --> 00:44:33.360
vertices both appear in U.
Because for them to both appear
00:44:33.360 --> 00:44:38.310
in U, your v have been selected
inside the common neighborhood
00:44:38.310 --> 00:44:42.390
of a and a prime, which is
unlikely if a and a prime
00:44:42.390 --> 00:44:46.550
have few common neighbors.
00:44:46.550 --> 00:44:50.050
So, as a result,
the expected number
00:44:50.050 --> 00:44:58.338
of pairs in U with small number
of common neighbors is small.
00:44:58.338 --> 00:45:00.130
And, already, that's
a very good indication
00:45:00.130 --> 00:45:01.020
that we're on the right track.
00:45:01.020 --> 00:45:02.670
And, to finish
things off, we look
00:45:02.670 --> 00:45:07.830
at this expression, which we
can lower bound by convexity.
00:45:07.830 --> 00:45:10.890
And we know the size of U
in expectation is large.
00:45:10.890 --> 00:45:13.470
And, also, the size of
x, that we just saw,
00:45:13.470 --> 00:45:17.260
is small in expectation.
00:45:17.260 --> 00:45:19.890
So you have this
inequality over here.
00:45:19.890 --> 00:45:21.690
And because there's
an expectation,
00:45:21.690 --> 00:45:25.560
it implies that there's some
specific instance such that,
00:45:25.560 --> 00:45:28.800
without the expectation,
the inequality holds.
00:45:28.800 --> 00:45:30.570
So take that specific instance.
00:45:30.570 --> 00:45:34.740
We obtain some U such that
this inequality is true,
00:45:34.740 --> 00:45:37.800
which simultaneously
implies that U is large
00:45:37.800 --> 00:45:40.910
and x, the number of
bad pairs, is small.
00:45:43.796 --> 00:45:47.850
So that was dependent
random choice.
00:45:47.850 --> 00:45:48.953
Any questions?
00:45:51.731 --> 00:45:54.046
All right.
00:45:54.046 --> 00:45:56.250
So that was the path
of length 2 lemma.
00:45:56.250 --> 00:45:57.900
So it tells us I
can take a large set
00:45:57.900 --> 00:46:02.590
with lots of paths of length 2
between most pairs of vertices.
00:46:02.590 --> 00:46:07.888
Let's upgrade this lemma to
a path of length 3 lemma.
00:46:18.850 --> 00:46:20.640
So, in the path
of length 3 lemma,
00:46:20.640 --> 00:46:27.378
we start with a bipartite graph,
as before, between A and B.
00:46:27.378 --> 00:46:33.970
So G is a bipartite
between A and B.
00:46:33.970 --> 00:46:39.230
And, as before, we have a
lot of edges between A and B.
00:46:39.230 --> 00:46:42.690
It's the delta fraction
of all possible edges.
00:46:42.690 --> 00:46:50.840
Then the conclusion is that
there exists A prime in A and B
00:46:50.840 --> 00:46:57.270
prime subset of B such
that A prime and B
00:46:57.270 --> 00:47:01.560
prime are both large
fractions of their parent set.
00:47:08.070 --> 00:47:15.820
And now, the-- and,
furthermore, every pair
00:47:15.820 --> 00:47:24.390
between A prime and
B prime is joined
00:47:24.390 --> 00:47:28.895
by many paths of length 3.
00:47:36.820 --> 00:47:39.300
So a path of length 3
means there's 3 edges.
00:47:42.610 --> 00:47:50.130
And, here, this eta is basically
the original error term
00:47:50.130 --> 00:47:51.690
up to a polynomial change.
00:48:00.270 --> 00:48:05.020
So starting with this bipartite
graph that's fairly dense,
00:48:05.020 --> 00:48:08.500
the lemma tells us
that we can find
00:48:08.500 --> 00:48:13.870
some large A prime
and large B prime so
00:48:13.870 --> 00:48:17.440
that between every vertex in
A prime and every vertex in B
00:48:17.440 --> 00:48:21.960
prime, there are lots of paths
of length 3 between them.
00:48:28.530 --> 00:48:29.263
Every time.
00:48:33.500 --> 00:48:37.215
So we should think about
all of these constants as--
00:48:37.215 --> 00:48:39.830
plus you only make polynomial
changes in the constants,
00:48:39.830 --> 00:48:42.290
we're happy.
00:48:42.290 --> 00:48:46.560
Here, eta is a polynomial
change in the delta.
00:48:46.560 --> 00:48:49.130
There's a convention which I
like which is not universal,
00:48:49.130 --> 00:48:51.860
but it's often solved,
unlike this convention.
00:48:51.860 --> 00:48:53.930
It's the difference
between the little c
00:48:53.930 --> 00:48:56.630
and the big C is that
a little c is better
00:48:56.630 --> 00:48:59.440
if you make it smaller,
and a big C is better--
00:48:59.440 --> 00:49:02.420
I mean, it's better in
the sense that if this
00:49:02.420 --> 00:49:04.970
is true for little
c and big C, and you
00:49:04.970 --> 00:49:10.050
make little c smaller and big C
bigger, then it is still true.
00:49:10.050 --> 00:49:12.570
So big C is a sufficiently
large constant,
00:49:12.570 --> 00:49:15.436
and little c is a
sufficiently small constant.
00:49:15.436 --> 00:49:16.422
Just a--
00:49:30.740 --> 00:49:36.650
So let's see the path of
length 3 lemma, see it's proof.
00:49:36.650 --> 00:49:39.460
We're going to use the
path of length 2 lemma,
00:49:39.460 --> 00:49:42.070
but we need a bit of
preparation first.
00:49:42.070 --> 00:49:46.930
So the proof has some nice
ideas, but it's also--
00:49:46.930 --> 00:49:50.740
some parts of it are slightly
tedious, so bear with me.
00:49:50.740 --> 00:49:58.690
So we're going to construct
a chain of subsets A--
00:49:58.690 --> 00:50:02.717
inside A. So A1, A2, A3.
00:50:02.717 --> 00:50:04.800
And this is just because
there's a few cleaning up
00:50:04.800 --> 00:50:08.100
steps that need to be done.
00:50:08.100 --> 00:50:19.880
Let's call two
vertices in A friendly
00:50:19.880 --> 00:50:23.980
if they have lots
of common neighbors.
00:50:23.980 --> 00:50:25.430
And, precisely,
we're going to say
00:50:25.430 --> 00:50:28.750
they're friendly if they
have more than delta
00:50:28.750 --> 00:50:34.770
squared over 80 times the
size of B common neighbors.
00:50:41.770 --> 00:50:46.590
Let me construct this sequence
of subsets as follows.
00:50:46.590 --> 00:50:53.870
First, let A1 be all
the vertices in A
00:50:53.870 --> 00:50:58.200
with degree not too small.
00:50:58.200 --> 00:51:02.500
So this is in preparation.
00:51:02.500 --> 00:51:05.950
So it will make our life
quite a bit easier later on.
00:51:05.950 --> 00:51:09.100
Let's just trim all the
really small degree vertices
00:51:09.100 --> 00:51:11.850
so that we don't have
to think about them.
00:51:11.850 --> 00:51:15.870
So you trim all the
small degree vertices.
00:51:15.870 --> 00:51:20.420
And think about how
many edges you trim.
00:51:20.420 --> 00:51:25.700
You cannot trim so many edges,
because each time you trim such
00:51:25.700 --> 00:51:30.100
a vertex, you only get rid
of a small number of edges.
00:51:30.100 --> 00:51:34.300
So, in the end, at least half
of the original set of edges
00:51:34.300 --> 00:51:36.690
must remain.
00:51:36.690 --> 00:51:43.320
And, as a result, the
size of A1 is at least
00:51:43.320 --> 00:51:50.590
a delta over 2 fraction of
the original vertex set.
00:51:50.590 --> 00:51:53.050
Otherwise, you could
not have contained half
00:51:53.050 --> 00:51:57.460
of the original set of edges.
00:51:57.460 --> 00:51:59.380
So this is the
first trimming step.
00:52:02.180 --> 00:52:07.390
So we got rid of some edges, but
we got rid of fewer than half
00:52:07.390 --> 00:52:10.940
of the original edges.
00:52:10.940 --> 00:52:15.720
And because now you have
a minimum degree on A1,
00:52:15.720 --> 00:52:18.670
the number of edges
between A1 and B
00:52:18.670 --> 00:52:22.660
is quite large,
still quite large.
00:52:22.660 --> 00:52:27.040
So think about passing
down to A1 now.
00:52:27.040 --> 00:52:31.530
In the second step, we are going
to apply the path of length 2
00:52:31.530 --> 00:52:34.480
lemma to this A1.
00:52:34.480 --> 00:52:41.880
So A2 is going to be
constructed from--
00:52:41.880 --> 00:52:50.170
so using the path
of length 2 lemma,
00:52:50.170 --> 00:52:56.620
specifically with parameter
epsilon being delta over 10.
00:52:56.620 --> 00:52:59.440
Although, remember, now
the density of the graph
00:52:59.440 --> 00:53:01.760
went from delta to delta over 2.
00:53:01.760 --> 00:53:04.245
Again, if you don't care
about the specific numbers,
00:53:04.245 --> 00:53:05.620
they're all
polynomials in delta.
00:53:05.620 --> 00:53:06.703
So don't worry about them.
00:53:06.703 --> 00:53:08.590
Everything's poly delta.
00:53:08.590 --> 00:53:11.860
So we're going to apply
the path of length 2 lemma
00:53:11.860 --> 00:53:16.240
to find this subset A2.
00:53:16.240 --> 00:53:25.660
And it has the property
that A2 is quite large,
00:53:25.660 --> 00:53:45.320
and all but a small fraction
of pairs in A2 are friendly.
00:53:54.580 --> 00:53:59.540
So we passed down to, first,
trimming small degree vertices,
00:53:59.540 --> 00:54:02.120
and then passed
down further to A2,
00:54:02.120 --> 00:54:06.020
where all but a small
fraction of elements in A2,
00:54:06.020 --> 00:54:08.563
or all but a small
fraction of the pairs
00:54:08.563 --> 00:54:10.230
are friendly to each
other, meaning they
00:54:10.230 --> 00:54:11.770
have lots of common neighbors.
00:54:15.020 --> 00:54:16.870
And now let's look
at the other side.
00:54:16.870 --> 00:54:21.610
Let's look at B. So
we're in this situation
00:54:21.610 --> 00:54:24.520
now where you have--
00:54:27.760 --> 00:54:31.390
so we're now in a situation
where you've passed down
00:54:31.390 --> 00:54:42.630
to A2 and in B, where, because
of what we did initially,
00:54:42.630 --> 00:54:47.410
every vertex in here
have large degree.
00:54:47.410 --> 00:54:53.830
So there's this minimum
degree condition
00:54:53.830 --> 00:54:57.190
from every vertex on the left.
00:54:57.190 --> 00:55:00.250
So the average degree
is still very high.
00:55:02.960 --> 00:55:07.020
As a result, the
average degree from B
00:55:07.020 --> 00:55:09.280
is going to be quite high.
00:55:09.280 --> 00:55:13.720
So let's focus on the B side
and pick out vertices in B
00:55:13.720 --> 00:55:16.570
that have high degree.
00:55:16.570 --> 00:55:23.390
So let's B1 denote
vertices in B such
00:55:23.390 --> 00:55:30.530
that the degree from B
to A2 is at least half
00:55:30.530 --> 00:55:33.390
of what you expect
based on average degree.
00:55:37.850 --> 00:55:41.390
And, as before, the same
logic as the A1 step.
00:55:41.390 --> 00:55:52.760
We see that B1 has large size,
is a large fraction of B.
00:55:52.760 --> 00:55:57.410
And now we pass
down to this B1 set.
00:56:04.214 --> 00:56:17.970
Now, finally, let's consider
A3 to be vertices in A2
00:56:17.970 --> 00:56:21.490
where a is friendly.
00:56:21.490 --> 00:56:28.650
So vertices a in A2 such that
a is friendly to at least 1
00:56:28.650 --> 00:56:31.210
over delta over--
00:56:31.210 --> 00:56:38.740
so 1 minus delta over
5 fraction of A2.
00:56:42.590 --> 00:56:50.430
So we saw that, in A2, most
pairs of vertices are friendly.
00:56:50.430 --> 00:57:00.000
So most, meaning all but
a delta over 10 fraction.
00:57:00.000 --> 00:57:05.770
So if we consider
vertices which are
00:57:05.770 --> 00:57:10.740
unfriendly to many
other vertices in A2,
00:57:10.740 --> 00:57:13.560
there aren't so many of them.
00:57:13.560 --> 00:57:16.440
If there were many of them,
you couldn't have had that.
00:57:16.440 --> 00:57:18.850
So that's why I
constructed this set
00:57:18.850 --> 00:57:23.540
A3 consisting of
elements in A2 that
00:57:23.540 --> 00:57:27.110
are friendly to many elements.
00:57:27.110 --> 00:57:32.990
And the size of A3 is at
least half of that of A2.
00:57:40.974 --> 00:57:50.510
So we have this A3 inside.
00:57:50.510 --> 00:57:51.696
All right.
00:57:51.696 --> 00:57:57.235
And now I claim that we can
take A3 and B as our final sets,
00:57:57.235 --> 00:58:01.510
and that between every vertex
in A3 and every vertex in B1,
00:58:01.510 --> 00:58:04.990
I claim there must be
lots of paths of length 3.
00:58:04.990 --> 00:58:07.420
But, first, let's
check their sizes.
00:58:07.420 --> 00:58:09.820
I mean, the sizes all should
be OK, because we never
00:58:09.820 --> 00:58:11.960
lost too much at each step.
00:58:11.960 --> 00:58:13.740
If you only care about
polynomial factors,
00:58:13.740 --> 00:58:15.990
well, you already see that
we never lost anything more
00:58:15.990 --> 00:58:17.590
than a polynomial factor.
00:58:17.590 --> 00:58:20.510
But just to be precise, the
size of A3 is at least--
00:58:20.510 --> 00:58:24.320
so if you count up the
factor lost at each step,
00:58:24.320 --> 00:58:29.230
so it's 1/2 delta
over 4 delta over 2.
00:58:29.230 --> 00:58:34.480
So it's at least delta
squared over 16 fraction
00:58:34.480 --> 00:58:36.870
of the original set A.
00:58:36.870 --> 00:58:44.320
And now, if we
consider a comma b
00:58:44.320 --> 00:58:49.650
to be an arbitrary
pair in A3 cross B1,
00:58:49.650 --> 00:58:53.140
I claim that there
must be many paths.
00:58:53.140 --> 00:58:59.090
Because by using-- so what
properties do we know?
00:58:59.090 --> 00:59:05.920
We know that b is adjacent
to a large fraction.
00:59:05.920 --> 00:59:10.300
So here large means at least
delta over 4-- so bounded
00:59:10.300 --> 00:59:16.080
below-- a large fraction of A2.
00:59:16.080 --> 00:59:16.580
Yes.
00:59:16.580 --> 00:59:17.302
So I apologize.
00:59:17.302 --> 00:59:19.260
When I say the word large,
depending on context
00:59:19.260 --> 00:59:22.640
it can mean bigger than delta,
or it could mean at least 1
00:59:22.640 --> 00:59:23.370
minus delta.
00:59:23.370 --> 00:59:25.380
So you look at
what I write down.
00:59:25.380 --> 00:59:31.070
So b is adjacent to at least
delta over 4 fraction of A2.
00:59:31.070 --> 00:59:39.300
At the same time, we know that
a is friendly to at least 1
00:59:39.300 --> 00:59:43.940
minus delta over
5 fraction of A2.
00:59:49.000 --> 00:59:54.070
So these two sets, they must
overlap by at least a delta
00:59:54.070 --> 00:59:55.060
over 20 fraction.
01:00:00.351 --> 01:00:05.260
So let's take a vertex b.
01:00:05.260 --> 01:00:13.100
So you-- so it's adjacent
to many vertices here.
01:00:13.100 --> 01:00:17.526
And if you look
at a vertex in A,
01:00:17.526 --> 01:00:21.240
it's friendly to
a large fraction.
01:00:21.240 --> 01:00:25.570
So, in particular, it's
friendly to all these elements
01:00:25.570 --> 01:00:26.070
over here.
01:00:28.760 --> 01:00:34.840
So, to finish off, what
does it mean for a--
01:00:34.840 --> 01:00:37.600
this is-- this vertex is a.
01:00:37.600 --> 01:00:38.810
This vertex is b.
01:00:38.810 --> 01:00:40.970
What does it mean for
a to be friendly to all
01:00:40.970 --> 01:00:42.830
of these shaded elements?
01:00:42.830 --> 01:00:47.510
It means that there are
lots of paths from a
01:00:47.510 --> 01:00:51.930
to each of these elements.
01:00:51.930 --> 01:00:55.974
And then you can finish off
the paths going back to b.
01:00:55.974 --> 01:00:56.750
Yes.
01:00:56.750 --> 01:01:00.092
AUDIENCE: The shaded stuff is
allowed to be outside of A3?
01:01:00.092 --> 01:01:01.800
YUFEI ZHAO: No. the
shaded-- the question
01:01:01.800 --> 01:01:04.040
is, is the shaded stuff
allowed to be outside of A3?
01:01:04.040 --> 01:01:04.540
No.
01:01:04.540 --> 01:01:06.870
The shaded things are inside A3.
01:01:06.870 --> 01:01:10.900
So we're looking at
intersections within A3.
01:01:14.550 --> 01:01:15.090
No, sorry.
01:01:17.055 --> 01:01:18.180
Actually, no, you're right.
01:01:18.180 --> 01:01:20.530
So the shaded things
can be outside A3.
01:01:20.530 --> 01:01:22.450
So shaded things
can be outside A3.
01:01:22.450 --> 01:01:23.130
I apologize.
01:01:23.130 --> 01:01:25.650
So everything now is in A2.
01:01:28.800 --> 01:01:35.100
So b is adjacent to a
large fraction of A2.
01:01:35.100 --> 01:01:43.800
And a here is friendly to some
part of the neighbors of b.
01:01:43.800 --> 01:01:48.920
So you can complete
paths like that.
01:01:52.750 --> 01:01:53.250
Yes.
01:01:53.250 --> 01:01:54.880
So only the starting
and ending points
01:01:54.880 --> 01:01:56.410
have to be in A
prime and B prime.
01:01:56.410 --> 01:01:58.970
Everything else, they can go
outside of the A prime and B
01:01:58.970 --> 01:02:00.666
prime.
01:02:00.666 --> 01:02:03.970
Yes, thank you.
01:02:03.970 --> 01:02:22.516
So the number of paths from
a to B to A2 back to b is--
01:02:22.516 --> 01:02:24.960
let's see if I can
stay within B1--
01:02:24.960 --> 01:02:26.010
so is at least--
01:02:34.110 --> 01:02:34.610
yes.
01:02:34.610 --> 01:02:36.530
So it's-- sorry.
01:02:36.530 --> 01:02:44.070
This is B. So it's at least
delta over 20 times A2 times
01:02:44.070 --> 01:02:49.680
delta over delta squared
over 80 times B. So
01:02:49.680 --> 01:02:52.470
if you don't care about
polynomial factors in delta,
01:02:52.470 --> 01:02:55.000
then you see that--
01:02:58.080 --> 01:03:00.140
the point is there's
a large fraction of--
01:03:02.418 --> 01:03:03.460
there are a lot of paths.
01:03:03.460 --> 01:03:07.000
So there are a lot of paths
between each little a and each
01:03:07.000 --> 01:03:09.670
little b by the
construction we've done.
01:03:15.190 --> 01:03:16.630
So let me just do a recap.
01:03:16.630 --> 01:03:19.990
So there were quite a few
details in this proof,
01:03:19.990 --> 01:03:22.300
and some of them have
to do with cleaning up.
01:03:22.300 --> 01:03:24.640
Because it's not so
nice to work with graphs
01:03:24.640 --> 01:03:26.850
that just have large
average degree.
01:03:26.850 --> 01:03:28.480
It's much nicer to
work with graphs
01:03:28.480 --> 01:03:29.920
with large minimum degree.
01:03:29.920 --> 01:03:33.580
So there are a couple of steps
here to take care of vertices
01:03:33.580 --> 01:03:34.990
with small degrees.
01:03:34.990 --> 01:03:39.290
So we started with, between
A and B, lots of edges.
01:03:39.290 --> 01:03:42.410
And we trim vertices
from A with small degree.
01:03:42.410 --> 01:03:45.250
So we get A1.
01:03:45.250 --> 01:03:48.970
And then we apply the path
of length 2 lemma to get A2.
01:03:48.970 --> 01:03:52.838
So inside A2, most
pairs of vertices
01:03:52.838 --> 01:03:54.630
have lots of common
neighbors, but not all.
01:03:57.510 --> 01:04:01.860
We then go back to
B to get B1, which
01:04:01.860 --> 01:04:04.940
has large minimum degree to A2.
01:04:07.650 --> 01:04:12.030
And then A3 looks
at vertices in A
01:04:12.030 --> 01:04:16.490
with many friendly
companions in A2.
01:04:20.200 --> 01:04:24.100
And A3 is large, and I claim
that between every vertex in A3
01:04:24.100 --> 01:04:28.120
and every vertex in B, you
have many paths of length 3.
01:04:28.120 --> 01:04:32.340
Because if you start
with a vertex in A3,
01:04:32.340 --> 01:04:35.430
it has many friendly companions.
01:04:35.430 --> 01:04:41.640
So many here means at least 1
minus delta over 5 fraction.
01:04:41.640 --> 01:04:49.470
Whereas every vertex in B1
has lots of neighbors in A2,
01:04:49.470 --> 01:04:53.430
where lots means at
least delta over 4.
01:04:53.430 --> 01:04:56.610
So there's
necessarily an overlap
01:04:56.610 --> 01:04:59.230
of at least delta over 20.
01:04:59.230 --> 01:05:01.750
And for that overlap,
we can create
01:05:01.750 --> 01:05:07.510
lots of paths going
through this overlap from A
01:05:07.510 --> 01:05:12.947
to B. Any questions?
01:05:16.220 --> 01:05:16.780
OK, great.
01:05:16.780 --> 01:05:21.940
So let's put everything together
to prove the graphical version
01:05:21.940 --> 01:05:23.704
of Balog-Szemeredi-Gowers.
01:05:31.040 --> 01:05:32.730
So we'll prove the
graphical version
01:05:32.730 --> 01:05:34.452
of Balog-Szemeredi-Gowers.
01:05:42.660 --> 01:05:46.920
So by-- so, first, note
that the hypothesis
01:05:46.920 --> 01:05:49.450
of Balog-Szemeredi-Gowers
already
01:05:49.450 --> 01:05:53.730
implies that the size
of A and the size of B
01:05:53.730 --> 01:05:55.650
are not too small.
01:05:58.910 --> 01:06:03.287
Because, otherwise, you couldn't
have had n squared over k edges
01:06:03.287 --> 01:06:03.870
to begin with.
01:06:08.060 --> 01:06:16.610
So by the path of
length 3 lemma,
01:06:16.610 --> 01:06:20.375
there exists A prime
in A and B prime
01:06:20.375 --> 01:06:24.980
in B with the
following properties.
01:06:24.980 --> 01:06:29.240
That A prime has a
large fraction of--
01:06:32.306 --> 01:06:36.110
so A prime and B prime
are both large in size.
01:06:40.460 --> 01:06:46.665
And for all vertices
a in A prime
01:06:46.665 --> 01:06:54.090
and vertices b in
B prime, there are
01:06:54.090 --> 01:06:59.040
lots of paths of length
3 between these vertices.
01:06:59.040 --> 01:07:05.010
So there are at least k
to the minus little o1--
01:07:05.010 --> 01:07:10.950
to the minus big
O1 times n squared
01:07:10.950 --> 01:07:18.050
pairs of intermediate
vertices a1,
01:07:18.050 --> 01:07:37.760
b1 in A cross B, such that
a b1 a1 b is a path in G.
01:07:37.760 --> 01:07:41.690
So let me draw the
situation for you.
01:07:48.920 --> 01:08:00.210
So we have A and B.
And so inside A and B,
01:08:00.210 --> 01:08:08.100
we have this fairly
large A prime
01:08:08.100 --> 01:08:14.026
and B prime, such that
for every little a
01:08:14.026 --> 01:08:23.470
and little b, there are
many paths like that
01:08:23.470 --> 01:08:26.890
going to b1 and a2.
01:08:30.439 --> 01:08:44.270
Let me set-- so let me set
x to be a plus b1, that sum,
01:08:44.270 --> 01:08:52.240
y to be a1 plus b1,
and z to be a1 plus b.
01:09:04.460 --> 01:09:21.380
So now notice that we can write
this a plus b in at least k
01:09:21.380 --> 01:09:27.350
to the minus big O1
times n squared ways
01:09:27.350 --> 01:09:39.410
as x minus y plus z by
following this path, where x, y,
01:09:39.410 --> 01:09:45.500
and z all lie in the
restricted sumset,
01:09:45.500 --> 01:09:49.350
because that's how the
restricted sumset is defined.
01:09:49.350 --> 01:09:54.229
So if you have an edge,
then the sum of the elements
01:09:54.229 --> 01:09:57.400
across on the two
ends, by definition,
01:09:57.400 --> 01:10:00.970
lies in the restricted sumset.
01:10:00.970 --> 01:10:02.900
So the path of length
3 lemma tells us
01:10:02.900 --> 01:10:06.320
that every pair a
and b, their sum
01:10:06.320 --> 01:10:12.290
can be written in many different
ways as this combination.
01:10:12.290 --> 01:10:21.360
As a result, we see that
A prime plus B prime--
01:10:21.360 --> 01:10:26.477
so this sum, if we consider sum
along with its multiplicity--
01:10:31.450 --> 01:10:36.760
so now we're really looking at
all the different sums as well
01:10:36.760 --> 01:10:43.890
as ways of writing the
sum as this combination--
01:10:43.890 --> 01:10:58.060
we see that it is bounded above
by the restricted sumset raised
01:10:58.060 --> 01:10:58.920
to the third power.
01:11:11.295 --> 01:11:13.740
Because each of these
choices, x, y, and z,
01:11:13.740 --> 01:11:16.330
they come from the
restricted sumset.
01:11:16.330 --> 01:11:19.150
But the hypothesis of
Balog-Szemeredi-Gowers,
01:11:19.150 --> 01:11:21.880
the graphical version, is
that the restricted sumset
01:11:21.880 --> 01:11:24.630
is small in size.
01:11:24.630 --> 01:11:32.430
So we can now upper bound
the restricted sumset
01:11:32.430 --> 01:11:36.580
by, basically, the--
01:11:36.580 --> 01:11:41.840
within a constant, within a
factor of the maximum possible.
01:11:41.840 --> 01:11:47.640
And now we are done,
because we have deduced
01:11:47.640 --> 01:11:52.770
that the complete sumset
between A prime and B prime
01:11:52.770 --> 01:12:02.320
is, at most, a constant
factor with change
01:12:02.320 --> 01:12:04.310
in constant by a polynomial.
01:12:04.310 --> 01:12:06.795
So a constant factor more
than the maximum possible.
01:12:11.000 --> 01:12:14.490
So it's, at mostly, k to
the big O1 poly k times n.
01:12:17.070 --> 01:12:20.310
So that proves the
graphical version
01:12:20.310 --> 01:12:22.890
of Balog-Szemeredi-Gowers.
01:12:22.890 --> 01:12:26.070
And because we showed earlier
that the graphical version
01:12:26.070 --> 01:12:28.560
of Balog-Szemeredi-Gowers
implies Balog-Szemeredi-Gowers,
01:12:28.560 --> 01:12:32.170
this shows the
Balog-Szemeredi-Gowers theorem.
01:12:32.170 --> 01:12:35.880
So let me recap some of
the ideas we saw today.
01:12:35.880 --> 01:12:38.260
And so the whole point
of Balog-Szemeredi-Gowers
01:12:38.260 --> 01:12:42.070
and all of these related lemmas
and theorems and variations
01:12:42.070 --> 01:12:47.050
is that you start with
something that has
01:12:47.050 --> 01:12:49.600
a lot of additive structure.
01:12:49.600 --> 01:12:54.440
Well, after we passed down to
graphs just a lot of edges.
01:12:54.440 --> 01:12:57.700
So you start with
a situation where
01:12:57.700 --> 01:13:02.325
you have kind of 1% goodness.
01:13:02.325 --> 01:13:03.700
And you want to
show that you can
01:13:03.700 --> 01:13:07.960
restrict to fairly
large subsets,
01:13:07.960 --> 01:13:10.610
so that you have perfection.
01:13:10.610 --> 01:13:14.510
So you have complete goodness
between these two sets.
01:13:14.510 --> 01:13:17.240
And this is what's going on
in both the graphical version
01:13:17.240 --> 01:13:18.920
and the additive version.
01:13:18.920 --> 01:13:21.560
So back to the graph
path of length 3 lemma.
01:13:21.560 --> 01:13:25.820
So we were able to boost the
path of length 2 lemma, which
01:13:25.820 --> 01:13:31.400
tells us something about
99% of the pairs having lots
01:13:31.400 --> 01:13:37.160
of common neighbors, to
100% of the pairs having
01:13:37.160 --> 01:13:41.010
lots of path of length 3.
01:13:41.010 --> 01:13:43.380
And in the additive
setting, we saw that
01:13:43.380 --> 01:13:47.880
by starting with a situation
where the hypothesis is
01:13:47.880 --> 01:13:51.210
somewhat patchy, so like
a 1% type hypothesis,
01:13:51.210 --> 01:13:54.750
we can pass down to
fairly large sets, where
01:13:54.750 --> 01:13:58.590
the complete sumset, starting
with just the restricted sumset
01:13:58.590 --> 01:14:01.110
being small, can pass
down to large sets
01:14:01.110 --> 01:14:03.540
where the complete
sumset is small.
01:14:03.540 --> 01:14:05.730
And this is an important
principle, that, often,
01:14:05.730 --> 01:14:11.210
when we have some typicality
by an appropriate argument--
01:14:11.210 --> 01:14:14.300
and, here, it's not at
all a trivial argument.
01:14:14.300 --> 01:14:15.890
So there's some
cleverness involved,
01:14:15.890 --> 01:14:18.120
that by doing some
kind of argument,
01:14:18.120 --> 01:14:21.950
we may be able to pass down
to some fairly large set
01:14:21.950 --> 01:14:25.130
where it's not typically
good, but everything's
01:14:25.130 --> 01:14:27.020
perfectly good.
01:14:27.020 --> 01:14:31.820
That's the spirit here of the
Balog-Szemeredi-Gowers theorem.
01:14:31.820 --> 01:14:35.010
So, next time, for the last
lecture of this course,
01:14:35.010 --> 01:14:38.600
I will tell you about
the sum-product problem,
01:14:38.600 --> 01:14:40.802
where the--
01:14:40.802 --> 01:14:44.500
there are also some graph-- very
nice graph theoretic inputs.