WEBVTT
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YUFEI ZHAO: OK.
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So let's get started.
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So we spent quite a bit
of time with graph theory
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in the first part
of this course,
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and today I want to
move beyond that.
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So we're going to talk
about more central topics
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and additive
combinatorics, starting
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with the Fourier analytic
proof of Roth's theorem.
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We discussed Roth's theorem,
and we gave a proof,
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during the course, using
similarities, graph regularity
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lemma, as well as the
triangle removal lemma.
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Today, I want to show
you a different approach
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to proving Roth's theorem that
goes through Fourier analysis.
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So this is a very
important proof,
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and it's one of the main tools
in additive combinatorics.
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Let me remind you what
Roth's theorem says.
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So Roth proved, in
1953, that if we
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write our sub-3 of interval n
to be the maximum size of a 3AP3
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subset of 1 through n, then
Roth showed that our 3 of n
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is little O of n.
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So in other words, if you
have a positive density
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subset of the
integers, then it must
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contain a three-term
arithmetic progression then.
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So what I said is equivalent
to the statement here.
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So previously, we gave a
proof using regularity.
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Actually the regularity approach
of Szemerédi was only found
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the '70s so Roth's original
proof was through Fourier
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analysis.
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And we'll see that tomorrow.
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Today, we'll see a toy
version of this proof.
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But it's not really
a toy version.
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It has the same ideas, but
in a slightly easier setting
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that has fewer technicalities.
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But before showing
you that, let me just
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discuss a bit of history
around Roth's theorem.
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We will show, next
time, the next lecture,
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we'll show the bound.
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Also by regularity, we get some
bound, which is little O then,
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but because of the use
of regularity lemmas,
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it's pretty poor dependence.
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We got something like
n over log star n.
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Next lecture, and basically
Roth's original proof,
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gives you a bound which
is n over log-log n.
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So it's a much more
reasonable bound.
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The current best
upper bound known
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has the form, essentially,
n over log n raised to 1
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plus little 1,
roughly n over log n.
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We do not know, or even
have great guesses on,
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what the answer should be.
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So the best lower bound,
and this is a construction
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that we saw earlier in
the course due to Behrend
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is of the form n over
E to the c root log n.
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It seems it may be very
difficult to improve
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this upper bound without
some genuine new ideas.
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On the other hand,
there is some evidence
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that the lower bound
might be closer
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to the truth in that there are
variants of the Roth problem
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for which we know that the lower
bound is basically the truth.
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What I want to do today
is look at a variant
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of this problem in what's
called a finite field model.
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And that basically
just means we're
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going to be looking at Roth's
theorem, not in the integers,
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but in some finite field
vector space, specifically F3
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to the n.
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So we're going to define
our sub-3 of this F3
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to the n to be the
maximum size of the 3AP3
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subset of this finite
field vector space.
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So the finite field
model is really useful.
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We're going to see this again
later in the course as well.
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Many of the ideas
and techniques that
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work for the real problem, so to
speak, many of those techniques
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also work in the
finite field model,
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but they are technically
simpler to execute.
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So this is often--
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you view it as a sandbox, a
playing ground, for testing out
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many of the ideas.
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And once you have
those ideas, then you
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can see if you can bring
them to the integer setting.
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And this is a very
successful program,
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and we'll see one aspect of
what happens when we do this.
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For this specific problem of
Roth's theorem, in F3 to the n,
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there are some nice
interpretations
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of what this problem means.
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So here's a pretty
easy fact that for--
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so n F3 to the n for
three elements, x, y, z,
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the following interpretation,
so what it means to be a 3AP,
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are equivalent.
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So x. y, z form a 3AP.
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So 3AP means the y is x
plus D. Z is x plus 2D.
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Equivalently, they satisfy this
equation, x minus 2y plus z
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equals zero.
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OK.
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In F3, minus 2 is plus 1.
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So it's the same as this
even nicer looking equation,
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x plus y plus z equals 2.
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It also turns out to be the
same as saying that x, y, z
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would lie on a line.
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But they are aligned.
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So the line has
three points in F3.
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And if you look at the
coordinate for every i,
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the i-th coordinate of x, y, z
are all distinct or all equal.
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So easy to check
all these things
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are equivalent to each other.
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And the last one is
a nice interpretation
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in terms of a game that
many of you know, Set.
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So in the game Set, you
have a bunch of cards.
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There have some number of
properties, n properties,
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like color, the number
of symbols, the shape.
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And you want to form
a set being three
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cards such that every
property, they're
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all same or all different.
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So that's exactly
this model over here.
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So what can we say
about this problem?
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What's the size of
the maximum subset
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of F3 to the n-th without 3-AP.
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If you look at the proof
that we did earlier
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in this course, the one
using triangle removal lemma,
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you see the proof
works verbatim.
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Previously, we
worked over Z mod n.
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Now, you work over a
different group, same proof.
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So triangle removal
lemma, it tells you
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that this r3 is always little
o of the size of the space.
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But we would like to do better.
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So this gives you
something like log star.
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It's not very good dependence.
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So we would like to do better.
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So what we will show
today, so this theorem
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is attributed to Meshulam.
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So in this case, the order of
history is somewhat reversed.
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So we'll see the
finite field toy model,
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but it historically
actually came afterwards.
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But you'll see that the Fourier
analytic proof that we'll
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see today, it's basically the
same proof in the two settings.
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So F is r3, we will prove a
bound, well, just like that.
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So much better than what you
get from the regularity method.
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In terms of--
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OK, so let me tell you a
bit more about the history
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of this problem in
terms of what we
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know in terms of upper
bounds and lower bounds.
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So let me say more about F3.
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So what's the best that
you might hope for?
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So the best lower
bound is due to E del.
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And it's some construction,
some very specific construction,
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which gives you a
bound that's something
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like 2.21 to the n-th.
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And the upper bound
is 3 to the n-th--
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on the-- 3 minus
little 1 to the n-th.
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So for a long time, it was
open whether the answer
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should be basically
roughly like 3
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to the n-th or some constant
less than 3 to the n-th.
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And improvements on the
upper bound were very slow,
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and/or some very difficult
works that nudge that down
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below just a little bit.
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And then a couple years
ago, a few years ago, there
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was this incredible
breakthrough,
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where in this paper that was
just a couple pages long,
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they managed to significantly
improve the upper bound
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to basically 2.76 to the n-th.
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So this was an incredible
breakthrough that
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happened just a few years ago.
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And we'll talk about this
proof in a couple of lectures.
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It turns out this proof,
which uses what's now
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called the polynomial method--
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so not Fourier analytic,
but a different method--
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unfortunately does
not seem to generalize
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to the original Roth's theorem.
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In fact, you shouldn't
expect it to generalize
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in a straightforward
way, because up there you
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we know that you do not have a
power saving, whereas here you
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have a power saving.
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So the exponent goes down.
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OK, so this is roughly the
history of this theorem.
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Any questions?
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AUDIENCE: Do we have
to have [INAUDIBLE]??
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YUFEI ZHAO: Yeah.
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So I'll-- So I can tell you this
is known as a Croot-Lev-Pach.
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So I'll say more about
it in a couple lectures.
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But this for F3 is due to
Ellenberg and Gijswijt.
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So I'll tell you more about
it in a couple lectures.
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What I want to focus on today
is the Fourier analytic nature
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of the proof that gives
you this bound up there,
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3 to the n over n.
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And it may seem like a
completely different topic
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compared to what
we've been doing
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so far in the course, which
is more about graph theory.
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But I want you to
think about what
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are the relationships
between what we'll see today
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and what we've seen so far.
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And there are lots
of connections.
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So even though the
proof may superficially
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look quite different,
many of these ideas
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about quasirandomness versus
structure will come up.
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And I want to present the
proof in a way that highlights
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the similarities between
what we did previously
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and this Fourier analytic proof.
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So let's talk
about the strategy.
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In the proof of the Szemerédi
graph regularity lemma,
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we had the strategy that we
called the energy increment
00:13:17.920 --> 00:13:19.590
strategy.
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So you start-- you want
to find a good partition.
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You start doing partitioning.
00:13:23.220 --> 00:13:26.580
And you keep track of this
thing called the energy--
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must go up at every step,
cannot go up forever,
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so has a bounded
number of steps.
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This strategy for Roth's theorem
is also an important strategy.
00:13:38.140 --> 00:13:40.270
It's a variant of
energy increment,
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but now density increment.
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So we start with a set--
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A subset of F3 to
the n-th, and we
00:14:00.560 --> 00:14:02.420
would like to
understand something
00:14:02.420 --> 00:14:07.160
about its structure
versus pseudorandomness
00:14:07.160 --> 00:14:10.040
in a way that is similar
to when we discussed
00:14:10.040 --> 00:14:13.670
the similar issue for graphs.
00:14:13.670 --> 00:14:16.280
In particular, there
will be this dichotomy
00:14:16.280 --> 00:14:24.460
that if A is in some
sense pseudorandom--
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so earlier, we saw what it means
for a graph to be pseudorandom.
00:14:28.020 --> 00:14:30.750
So now, what does it
mean for a subset of F3
00:14:30.750 --> 00:14:33.030
to the n-th to be pseudorandom.
00:14:33.030 --> 00:14:34.440
So we'll address that today.
00:14:34.440 --> 00:14:37.010
If A pseudorandom--
00:14:37.010 --> 00:14:43.470
OK, so the short answer is
that it is Fourier uniform--
00:14:43.470 --> 00:14:48.120
in other words, all
Fourier coefficients small.
00:14:51.220 --> 00:14:53.470
That's what pseudorandom
will refer to.
00:14:53.470 --> 00:14:56.720
So then there is
a counting lemma.
00:15:01.720 --> 00:15:03.760
And the counting lemma
will in particular
00:15:03.760 --> 00:15:07.160
imply that A has lots of 3-APs.
00:15:13.700 --> 00:15:16.880
So then you find your 3-AP.
00:15:16.880 --> 00:15:19.410
If this is not the case, then--
00:15:19.410 --> 00:15:21.890
so what's the opposite
of Fourier uniform--
00:15:21.890 --> 00:15:25.890
is that A now has some
large Fourier coefficient.
00:15:36.190 --> 00:15:39.250
And what we'll do is
to use this Fourier
00:15:39.250 --> 00:15:47.080
coefficient to extract some
codimension 1 affine subspace--
00:15:51.960 --> 00:15:57.420
it's also called a hyperplane--
00:15:57.420 --> 00:16:06.470
where the density of A
goes up significantly,
00:16:06.470 --> 00:16:09.126
if you restrict to
that sub hyperplane.
00:16:14.090 --> 00:16:15.500
And you can repeat this process.
00:16:15.500 --> 00:16:18.920
Now, restrict to this
hyperplane and ask yourself
00:16:18.920 --> 00:16:20.660
the same question.
00:16:20.660 --> 00:16:25.220
Is A, when restricted to this
hyperplane, pseudorandom?
00:16:25.220 --> 00:16:27.830
In which case, we find APs.
00:16:27.830 --> 00:16:31.230
Or is A restricted
to this hyperplane,
00:16:31.230 --> 00:16:33.420
does it have a large
Fourier coefficient?
00:16:33.420 --> 00:16:34.920
In which case, we
restrict further.
00:16:38.400 --> 00:16:42.970
And each time you iterate, you
obtain a density increment.
00:16:42.970 --> 00:16:50.130
And the density increment
cannot go on forever,
00:16:50.130 --> 00:16:53.400
because your total
density is at most 1.
00:16:57.890 --> 00:16:59.860
So the number of
steps must be bounded.
00:17:02.590 --> 00:17:04.290
So that's the strategy.
00:17:04.290 --> 00:17:06.750
So this should remind you
somewhat of the energy
00:17:06.750 --> 00:17:10.210
increment strategy from
Szemerédi's regularity lemma,
00:17:10.210 --> 00:17:12.210
although there are some
fundamental differences.
00:17:12.210 --> 00:17:13.790
We're not doing partitionings.
00:17:16.603 --> 00:17:18.020
Any questions about
this strategy?
00:17:21.470 --> 00:17:23.810
OK.
00:17:23.810 --> 00:17:26.099
I want to tell you
about Fourier analysis.
00:17:34.740 --> 00:17:37.460
So probably, all of you
have seen some version
00:17:37.460 --> 00:17:40.410
of Fourier analysis, maybe
in your calculus class
00:17:40.410 --> 00:17:43.140
with Fourier series and whatnot.
00:17:43.140 --> 00:17:45.690
So you play with formulas,
and solve some differential
00:17:45.690 --> 00:17:46.930
equations.
00:17:46.930 --> 00:17:49.830
So I want to give you
more than just a bunch
00:17:49.830 --> 00:17:53.110
of ways about handling
Fourier coefficients,
00:17:53.110 --> 00:17:55.600
a way to think about
Fourier analysis.
00:17:55.600 --> 00:17:58.920
So think of this as a crash
course about Fourier analysis
00:17:58.920 --> 00:18:01.200
from the perspective
of combinatorics.
00:18:04.002 --> 00:18:05.460
And Fourier analysis,
I think, it's
00:18:05.460 --> 00:18:08.160
much easier if you
work in a finite group,
00:18:08.160 --> 00:18:11.880
in a finite abelian group,
which is what we're doing here.
00:18:11.880 --> 00:18:13.690
Many of the
technicalities go away.
00:18:13.690 --> 00:18:17.960
So we'll be looking specifically
at Fourier analysis in F3
00:18:17.960 --> 00:18:21.840
to the n-th, although
the 3 can be any prime.
00:18:21.840 --> 00:18:24.270
So it's really the same.
00:18:24.270 --> 00:18:29.700
So the main actors
in Fourier analysis
00:18:29.700 --> 00:18:31.293
are the Fourier characters.
00:18:35.370 --> 00:18:39.840
The Fourier characters
are denoted gamma sub r.
00:18:39.840 --> 00:18:42.420
And they're characters
on the group,
00:18:42.420 --> 00:18:45.030
meaning that they
are maps which--
00:18:45.030 --> 00:18:47.850
so they turn out, happen
to be homomorphisms for the
00:18:47.850 --> 00:18:50.520
multiplicative group under--
00:18:50.520 --> 00:18:53.760
so C under multiplication.
00:18:53.760 --> 00:19:02.850
And they're indexed by r,
which also elements of F3
00:19:02.850 --> 00:19:03.510
to the n-th.
00:19:03.510 --> 00:19:05.670
So I'm going to be
fairly concrete here.
00:19:05.670 --> 00:19:07.520
There are ways to do
this more abstractly.
00:19:07.520 --> 00:19:08.880
But I'll be fairly concrete.
00:19:08.880 --> 00:19:12.670
So it's defined by
gamma sub r evaluated
00:19:12.670 --> 00:19:19.470
on x equals to omega raised
to r dot product x, where
00:19:19.470 --> 00:19:24.450
here omega is a
third root of unity
00:19:24.450 --> 00:19:28.060
and the dot is a dot product.
00:19:33.050 --> 00:19:33.550
So--
00:19:42.660 --> 00:19:45.090
So that's the definition
of the Fourier transform--
00:19:45.090 --> 00:19:47.400
sorry, that's the definition
of the Fourier characters.
00:19:47.400 --> 00:19:49.140
And once you have the
Fourier characters,
00:19:49.140 --> 00:19:56.830
you can have this Fourier
transform, just defined
00:19:56.830 --> 00:19:58.010
as follows.
00:19:58.010 --> 00:20:00.086
If you start with a function--
00:20:03.814 --> 00:20:07.540
let's say, a complex-valued
function on your space--
00:20:07.540 --> 00:20:13.550
then I define the
Fourier transform
00:20:13.550 --> 00:20:20.080
to be another function,
like that, defined
00:20:20.080 --> 00:20:22.180
by the following formula.
00:20:34.240 --> 00:20:37.178
So that's the formula for
the Fourier transform.
00:20:40.770 --> 00:20:44.220
It is basically
the inner product
00:20:44.220 --> 00:20:47.520
between F and the
Fourier character.
00:20:47.520 --> 00:20:50.220
So let me make the
comment here, I
00:20:50.220 --> 00:20:52.440
think this is actually a
pretty important comment,
00:20:52.440 --> 00:20:54.003
about the normalization.
00:20:58.350 --> 00:21:03.390
Now, when you first learn
Fourier transforms, usually
00:21:03.390 --> 00:21:07.350
in the reals, there are all
these questions about what
00:21:07.350 --> 00:21:08.730
number to put in the exponent.
00:21:08.730 --> 00:21:09.660
Is it 2 pi?
00:21:09.660 --> 00:21:10.690
Is it root 2 pi?
00:21:10.690 --> 00:21:12.000
Is it some other thing?
00:21:12.000 --> 00:21:17.770
And somehow, one answer
is better than the others.
00:21:17.770 --> 00:21:19.873
And the same thing is
true here in groups.
00:21:19.873 --> 00:21:21.790
So we'll stick with the
following convention--
00:21:21.790 --> 00:21:23.957
and I want all of you to
stick with this convention,
00:21:23.957 --> 00:21:26.320
otherwise, we'll confuse
ourselves to no end--
00:21:26.320 --> 00:21:27.680
is that for a finite group--
00:21:34.940 --> 00:21:36.930
actually, let me,
start of a board.
00:21:36.930 --> 00:21:43.788
So the convention is
that, in a finite group,
00:21:43.788 --> 00:21:46.310
the Fourier transform
is defined--
00:21:46.310 --> 00:21:49.650
and more generally, anything
you do in the physical space,
00:21:49.650 --> 00:21:56.320
we always use the
averaging measure.
00:21:59.590 --> 00:22:02.660
Don't sum, always average
in the physical space.
00:22:02.660 --> 00:22:10.040
And in the frequency
space, always use sums,
00:22:10.040 --> 00:22:12.860
use the counting measure.
00:22:12.860 --> 00:22:14.458
Keep this in mind.
00:22:14.458 --> 00:22:16.250
Any of these questions
about normalization,
00:22:16.250 --> 00:22:17.667
if you stick with
this convention,
00:22:17.667 --> 00:22:19.545
things will become much easier.
00:22:19.545 --> 00:22:21.670
So there won't be any of
these questions about when
00:22:21.670 --> 00:22:23.390
you take the inverse
Fourier transform,
00:22:23.390 --> 00:22:25.750
do I put an extra
factor in front or not,
00:22:25.750 --> 00:22:27.691
if you stick with this
correct convention.
00:22:30.520 --> 00:22:33.580
So with that convention
in mind, what the Fourier
00:22:33.580 --> 00:22:40.510
transform really is is inner
product between F and a Fourier
00:22:40.510 --> 00:22:41.478
character.
00:22:49.810 --> 00:22:51.670
There are some
important properties
00:22:51.670 --> 00:22:52.760
of the Fourier transform.
00:22:52.760 --> 00:22:54.760
So let me go through a
few of the key properties
00:22:54.760 --> 00:22:55.540
that we'll need.
00:23:04.940 --> 00:23:07.970
The first one is pretty easy.
00:23:07.970 --> 00:23:14.250
What is the meaning of the
0-th Fourier coefficient?
00:23:14.250 --> 00:23:24.020
You plug it in, and you see that
it is just the average of F.
00:23:24.020 --> 00:23:27.800
So 0-th coefficient
is the average of F.
00:23:27.800 --> 00:23:32.270
The second fact goes
under one of two names,
00:23:32.270 --> 00:23:34.870
and they're often
used interchangeably--
00:23:34.870 --> 00:23:36.110
Plancherel or Parseval.
00:23:41.340 --> 00:23:48.560
And it says that if you
look at the inner product
00:23:48.560 --> 00:23:57.080
in the physical space,
then this product
00:23:57.080 --> 00:24:02.216
is preserved, if you take
the Fourier transform.
00:24:05.617 --> 00:24:07.700
But now, of course, you're
in the frequency space,
00:24:07.700 --> 00:24:10.070
so you should sum instead
of doing the inner product.
00:24:15.180 --> 00:24:20.490
So this identity can be proved
in a fairly straightforward way
00:24:20.490 --> 00:24:24.360
by plugging in what the
definition is for the Fourier
00:24:24.360 --> 00:24:25.418
transform.
00:24:25.418 --> 00:24:26.960
This is a straightforward
computation
00:24:26.960 --> 00:24:27.850
I'm not going to
do on the board,
00:24:27.850 --> 00:24:29.970
but I highly encourage you
to actually do at home,
00:24:29.970 --> 00:24:33.685
just to do it once to make sure
you understand how it goes.
00:24:33.685 --> 00:24:35.310
But there is also a
more conceptual way
00:24:35.310 --> 00:24:38.410
to understand this identity.
00:24:38.410 --> 00:24:41.490
And that's because--
now, this is also
00:24:41.490 --> 00:24:44.670
important to understand what
the Fourier transform is.
00:24:44.670 --> 00:24:46.980
It's not just some magical
formula somebody wrote down,
00:24:46.980 --> 00:24:50.080
like this is a very
natural operation.
00:24:50.080 --> 00:24:59.280
It's because the characters,
the set of characters,
00:24:59.280 --> 00:25:02.620
is an orthonormal basis.
00:25:02.620 --> 00:25:10.815
So the Fourier characters
form an orthonormal basis.
00:25:15.750 --> 00:25:18.420
As a result, what
the Fourier transform
00:25:18.420 --> 00:25:22.310
is is a unitary change of basis.
00:25:35.510 --> 00:25:36.320
You can check.
00:25:36.320 --> 00:25:37.737
It's very
straightforward to check
00:25:37.737 --> 00:25:39.800
that the Fourier
characters, indeed, form
00:25:39.800 --> 00:25:45.210
a orthonormal basis, because--
00:25:45.210 --> 00:25:51.650
well, you can evaluate the inner
product between two Fourier
00:25:51.650 --> 00:25:53.480
characters.
00:25:53.480 --> 00:25:55.730
So remember, in
the physical space,
00:25:55.730 --> 00:25:57.786
we're always doing averaging.
00:26:01.050 --> 00:26:06.240
And so now, I'll
just write down first
00:26:06.240 --> 00:26:08.310
what I mean by
the inner product.
00:26:11.100 --> 00:26:12.920
So that's the inner product.
00:26:12.920 --> 00:26:21.977
And by the definition of
the Fourier character,
00:26:21.977 --> 00:26:22.560
you have that.
00:26:25.410 --> 00:26:29.010
So think about what
this expectation is--
00:26:29.010 --> 00:26:37.470
unless r equals to s, in which
case, this expectation is 1.
00:26:37.470 --> 00:26:39.240
Unless that is the
case, you always
00:26:39.240 --> 00:26:43.360
have some coordinate
of x in the exponent.
00:26:43.360 --> 00:26:46.230
So as you average over
all possibilities,
00:26:46.230 --> 00:26:47.270
they average out to 0.
00:26:58.020 --> 00:27:01.050
So this calculation shows you
that the Fourier characters
00:27:01.050 --> 00:27:02.880
form a orthonormal basis.
00:27:02.880 --> 00:27:05.910
And a basic fact you
know from linear algebra
00:27:05.910 --> 00:27:07.470
is that if you do
a change of basis,
00:27:07.470 --> 00:27:10.510
if you do a unitary
change of basis,
00:27:10.510 --> 00:27:13.640
then inner product is preserved.
00:27:13.640 --> 00:27:15.430
It's like a rotation.
00:27:15.430 --> 00:27:18.150
It's the same-- so you're not
changing the inner product.
00:27:18.150 --> 00:27:19.660
So the inner
product is preserved
00:27:19.660 --> 00:27:21.640
under this change of basis.
00:27:21.640 --> 00:27:24.280
And that's why
Plancherel is true.
00:27:27.780 --> 00:27:30.870
Another important
thing is what's
00:27:30.870 --> 00:27:34.684
known as the Fourier
inversion formula.
00:27:34.684 --> 00:27:39.720
The Fourier transform tells
you how to go from a function
00:27:39.720 --> 00:27:42.730
to the Fourier transform.
00:27:42.730 --> 00:27:46.020
Well, now, if you are given
the Fourier transform,
00:27:46.020 --> 00:27:48.090
how do you go back?
00:27:48.090 --> 00:27:50.040
There's a formula
which tells you
00:27:50.040 --> 00:28:01.350
that you can go back by the
following formula there.
00:28:01.350 --> 00:28:04.370
So that's the Fourier
inversion formula.
00:28:04.370 --> 00:28:07.110
It allows you to
do this inversion.
00:28:07.110 --> 00:28:09.540
And again, it's one of
these formulas where
00:28:09.540 --> 00:28:12.990
I encourage you to try it
out yourself by plugging
00:28:12.990 --> 00:28:16.770
in the formula and expanding.
00:28:16.770 --> 00:28:19.660
And it's pretty easy to check.
00:28:19.660 --> 00:28:22.090
It's much easier in the finite
field setting, by the way.
00:28:22.090 --> 00:28:25.630
So if you use the usual Fourier
transform on the real line,
00:28:25.630 --> 00:28:27.070
there are some
technicalities even
00:28:27.070 --> 00:28:28.690
to prove the Fourier inversion.
00:28:28.690 --> 00:28:31.070
But in finite groups,
it's almost trivial.
00:28:31.070 --> 00:28:33.170
You expand, and then you'll see.
00:28:33.170 --> 00:28:35.025
So it's very easy to prove.
00:28:35.025 --> 00:28:37.150
But you can also see this
Fourier inversion formula
00:28:37.150 --> 00:28:39.040
more conceptually,
because you're
00:28:39.040 --> 00:28:41.550
in a unitary change of basis.
00:28:41.550 --> 00:28:45.300
So to go back, well, think about
what it means in linear algebra
00:28:45.300 --> 00:28:48.990
to revert a unitary
transformation.
00:28:48.990 --> 00:28:52.530
You simply multiply
the coefficients
00:28:52.530 --> 00:28:56.490
with the coordinates.
00:28:56.490 --> 00:29:00.335
Orthogonal, orthonormal
change of basis.
00:29:00.335 --> 00:29:06.430
Finally, Fourier
transform behaves while
00:29:06.430 --> 00:29:07.900
under convolution.
00:29:07.900 --> 00:29:15.430
So by convolution, we
define the convolution
00:29:15.430 --> 00:29:19.320
of two functions, f and g,
using the following formula.
00:29:27.510 --> 00:29:38.090
And so then the claim is that
the Fourier transform behaves
00:29:38.090 --> 00:29:39.800
very well under convolution.
00:29:39.800 --> 00:29:44.250
It's basically multiplicative
under convolution.
00:29:44.250 --> 00:29:45.860
So what this means
is, if I put in--
00:29:50.490 --> 00:29:54.670
so it's pointwise
true everywhere.
00:29:54.670 --> 00:29:59.590
Again, very easy proof, because
I just evaluate the left hand
00:29:59.590 --> 00:30:04.850
side, see what--
00:30:04.850 --> 00:30:07.730
plug in the formula for
the Fourier transform.
00:30:07.730 --> 00:30:16.170
And I find it's that.
00:30:16.170 --> 00:30:19.268
And now, I plug in the
formula for convolution.
00:30:40.450 --> 00:30:45.140
So now, you can do a
change of variables.
00:30:45.140 --> 00:30:48.865
And then you-- it's
not hard to see.
00:30:48.865 --> 00:30:50.740
You eventually end up
at the right hand side.
00:30:54.760 --> 00:30:56.260
So these are some
of the properties.
00:30:56.260 --> 00:30:58.658
So there are
important properties
00:30:58.658 --> 00:30:59.700
of the Fourier transform.
00:30:59.700 --> 00:31:01.743
So this is something
that, whenever
00:31:01.743 --> 00:31:03.160
you learn about
Fourier transform,
00:31:03.160 --> 00:31:06.720
you always see these
few properties.
00:31:06.720 --> 00:31:08.230
And so we'll use them.
00:31:08.230 --> 00:31:10.690
But we'll also need
another property
00:31:10.690 --> 00:31:14.440
that is specific to the
analysis of 3-term arithmetic
00:31:14.440 --> 00:31:15.410
progressions.
00:31:17.990 --> 00:31:22.165
So what does Fourier transform
have to do with 3-APs?
00:31:22.165 --> 00:31:24.040
So we want to use it to
prove Roth's theorem.
00:31:24.040 --> 00:31:27.400
So we better have
some tool that allows
00:31:27.400 --> 00:31:29.330
us to analyze the number 3-APs.
00:31:29.330 --> 00:31:45.250
And here is a key identity
relating Fourier with 3-APs.
00:31:45.250 --> 00:31:56.410
And it's that, if you
have three functions,
00:31:56.410 --> 00:32:02.920
then the following quantity,
which relates the number
00:32:02.920 --> 00:32:06.840
of 3-APs--
00:32:14.880 --> 00:32:18.720
So this function basically
counts the number 3-APs,
00:32:18.720 --> 00:32:21.500
if your f, g, and h are
indicator functions of a set.
00:32:24.240 --> 00:32:27.960
I want to express this formula
in terms of the Fourier
00:32:27.960 --> 00:32:29.835
transforms of these functions.
00:32:32.550 --> 00:32:35.880
The formula turns out
to be fairly simple,
00:32:35.880 --> 00:32:45.970
that it is simply that.
00:32:45.970 --> 00:32:50.190
So it's a single sum
over the r's of f hat
00:32:50.190 --> 00:32:55.240
of r, g hat of minus
2r, and h hat of r.
00:32:55.240 --> 00:32:57.880
You might wonder why I put a
minus 2 here, because minus 2
00:32:57.880 --> 00:33:00.520
is r, and it looks
certainly much nicer
00:33:00.520 --> 00:33:02.212
with just r in there.
00:33:02.212 --> 00:33:05.080
And that is true.
00:33:05.080 --> 00:33:08.170
This formula as written is
true for over any group.
00:33:13.940 --> 00:33:16.340
And our proof will show it.
00:33:16.340 --> 00:33:18.560
So it's not really about
F3 at all, but any group.
00:33:25.300 --> 00:33:29.880
So let me prove this for you
in a couple of different ways.
00:33:29.880 --> 00:33:39.520
So the first proof is
basically a straightforward no
00:33:39.520 --> 00:33:46.240
thinking involved proof, as
in we apply these formula
00:33:46.240 --> 00:33:48.250
for using either
Fourier inversion
00:33:48.250 --> 00:33:53.820
or the inverse Fourier
transform, and plug it in,
00:33:53.820 --> 00:33:56.545
and expand, and check.
00:33:56.545 --> 00:33:58.170
So it's worth doing
at least this once.
00:33:58.170 --> 00:34:01.980
So let's do this
together at least once.
00:34:01.980 --> 00:34:04.420
But this something that is
a fairly straightforward
00:34:04.420 --> 00:34:05.580
computation.
00:34:05.580 --> 00:34:10.900
The left hand side
can be expanded
00:34:10.900 --> 00:34:12.340
using Fourier inversion.
00:34:14.920 --> 00:34:21.389
so r1 f hat r1 omega
to the minus r1
00:34:21.389 --> 00:34:32.280
dot x, and sum over r2 g
hat r2 omega to the minus r2
00:34:32.280 --> 00:34:38.199
dot x plus y, and
then finally sum
00:34:38.199 --> 00:34:47.550
over r3 h hat of r3 omega to
the minus r3 dot x plus 2y.
00:34:50.989 --> 00:34:54.770
So I'm using Fourier
inversion, replace f, g, and h
00:34:54.770 --> 00:34:58.800
by their Fourier transforms.
00:34:58.800 --> 00:35:02.280
Oh, sorry, there should be--
00:35:02.280 --> 00:35:03.800
yeah, so no minus.
00:35:06.960 --> 00:35:10.620
So now, we exchange
sums and expectations,
00:35:10.620 --> 00:35:18.265
do a switch in the order of
summation, so r1, r2, r3 and f
00:35:18.265 --> 00:35:27.050
hat of r1 g hat
of r2 h hat of r3.
00:35:27.050 --> 00:35:33.860
And you have this expectation
over x and y and omega
00:35:33.860 --> 00:35:37.970
of x dot r1 plus r2 plus r3.
00:35:40.610 --> 00:35:45.340
In fact, I can even write
the x and y separately,
00:35:45.340 --> 00:35:57.180
y omega to the y
dot r2 plus 2r3.
00:35:57.180 --> 00:35:59.540
So just rearranging.
00:35:59.540 --> 00:36:01.570
And now, you see
that as you take
00:36:01.570 --> 00:36:09.110
expectation over x, this
expectation is equal to 1,
00:36:09.110 --> 00:36:19.130
if r1 plus r2 plus r3 is
equal to 0, and 0 otherwise.
00:36:19.130 --> 00:36:23.930
And likewise, the third
expectation is either 1 or 0,
00:36:23.930 --> 00:36:28.610
depending on the
sums of r2 and r3.
00:36:33.650 --> 00:36:36.760
So the only terms that
remain, after you take out
00:36:36.760 --> 00:36:43.920
these 0's, are cases where
both these two equations
00:36:43.920 --> 00:36:44.730
are satisfied.
00:36:47.350 --> 00:36:55.150
And then you see that the only
remaining terms are basically
00:36:55.150 --> 00:37:02.410
the ones given in the sum
on the right hand side.
00:37:05.410 --> 00:37:06.000
OK?
00:37:06.000 --> 00:37:07.930
So that's the proof.
00:37:07.930 --> 00:37:10.363
Pretty straightforward, you
plug in Fourier inversion.
00:37:13.540 --> 00:37:17.073
I want to show you a
different proof that
00:37:17.073 --> 00:37:19.240
hopefully will be more
familiar and more conceptual.
00:37:19.240 --> 00:37:21.657
Now, it doesn't involve carrying
through this calculation,
00:37:21.657 --> 00:37:26.380
even though this is not
at all hard calculation.
00:37:26.380 --> 00:37:31.610
But first, let me rewrite
the formula up there.
00:37:31.610 --> 00:37:37.000
So in F3, it will be
convenient, and so
00:37:37.000 --> 00:37:39.190
the formula is actually
slightly easier
00:37:39.190 --> 00:37:42.700
to interpreting F3,
in F3, the identity
00:37:42.700 --> 00:37:47.110
says that, if you
look at the quantity--
00:37:52.460 --> 00:37:53.020
I need.
00:37:58.732 --> 00:38:04.293
So let me give you a second
proof that works just in F3,
00:38:04.293 --> 00:38:06.210
but you can modify it
to work in other groups.
00:38:06.210 --> 00:38:08.950
But in F3, it's
particularly nice.
00:38:08.950 --> 00:38:14.240
The left hand side, you
see, the left hand side,
00:38:14.240 --> 00:38:21.980
I can rewrite it as
the following form,
00:38:21.980 --> 00:38:24.840
where I sum over--
00:38:24.840 --> 00:38:29.430
well, I take expectation over
all triples x, y, z that sum
00:38:29.430 --> 00:38:29.930
to 0.
00:38:32.600 --> 00:38:36.470
Because a 3-AP is the
same as three elements,
00:38:36.470 --> 00:38:38.720
three points in the
vector space summing to 0.
00:38:41.830 --> 00:38:46.000
But now, you see
that this quantity
00:38:46.000 --> 00:38:54.310
is the same as the
convolution evaluated at 0--
00:38:57.640 --> 00:39:00.610
so if you extend the definition
of convolution to more than one
00:39:00.610 --> 00:39:03.630
func-- more than two functions.
00:39:03.630 --> 00:39:05.540
But now, we apply
Fourier inversion.
00:39:09.650 --> 00:39:11.540
And we find that--
00:39:21.780 --> 00:39:22.660
OK.
00:39:22.660 --> 00:39:26.940
So by Fourier inversion,
you have that.
00:39:26.940 --> 00:39:30.700
But now, by the identity that
relates the Fourier transform
00:39:30.700 --> 00:39:41.560
and inversion, you have that.
00:39:41.560 --> 00:39:45.410
And that's the proof, because
minus 2 r is the same as r.
00:39:48.470 --> 00:39:52.100
So it's shorter, because we're
using some properties here
00:39:52.100 --> 00:39:54.310
about convolution and--
00:39:54.310 --> 00:39:56.229
yeah, so about the convolution.
00:40:03.720 --> 00:40:05.580
That formula up
there is, of course,
00:40:05.580 --> 00:40:07.530
related to counting 3-APs.
00:40:07.530 --> 00:40:22.080
Because if f, g, and h are
all indicators of some set,
00:40:22.080 --> 00:40:25.710
then the left hand
side is the same
00:40:25.710 --> 00:40:35.535
as basically the number of
triples of elements in A
00:40:35.535 --> 00:40:38.940
whose sum is equal to 0.
00:40:42.210 --> 00:40:46.680
And the right hand
side is the sum
00:40:46.680 --> 00:40:52.184
of the third power of
the Fourier coefficients.
00:40:56.360 --> 00:40:59.980
And this formula should
look somewhat familiar,
00:40:59.980 --> 00:41:03.940
because we also used
this kind of formula
00:41:03.940 --> 00:41:07.610
back when we discussed
spectral graph theory.
00:41:07.610 --> 00:41:11.350
And remember, the third
moment of the eigenvalues
00:41:11.350 --> 00:41:16.330
is the trace of the third
power, which counts closed
00:41:16.330 --> 00:41:18.880
walks in Cayley graph.
00:41:18.880 --> 00:41:21.100
So this is actually
the same formula.
00:41:21.100 --> 00:41:26.210
So in the case if
A is symmetric,
00:41:26.210 --> 00:41:34.460
let's say, then this is
the same as the formula
00:41:34.460 --> 00:41:47.530
that counts closed
walks of length three
00:41:47.530 --> 00:41:48.660
in the Cayley graph.
00:41:53.945 --> 00:41:55.820
The point of this comment
is just to tell you
00:41:55.820 --> 00:41:57.440
that Fourier
transform is somehow
00:41:57.440 --> 00:42:02.160
it's not this brand new concept
that we've never seen before.
00:42:02.160 --> 00:42:04.400
It is intimately tied
to many of the things
00:42:04.400 --> 00:42:08.866
that we have seen earlier in
this course but in disguise.
00:42:08.866 --> 00:42:11.390
So it is related to the
spectral graph theory
00:42:11.390 --> 00:42:13.640
that we discussed at length
earlier in this course.
00:42:21.290 --> 00:42:23.080
Now that we have the
Fourier transform,
00:42:23.080 --> 00:42:28.420
I want to develop some
machinery to prove
00:42:28.420 --> 00:42:32.930
Roth's theorem following
the strategy up there.
00:42:32.930 --> 00:42:34.100
So let's take a quick break.
00:42:34.100 --> 00:42:36.392
And then when we come back,
we'll prove Roth's theorem.
00:42:40.570 --> 00:42:41.500
Any questions so far?
00:42:45.204 --> 00:42:50.596
AUDIENCE: So for
this one, you said--
00:42:50.596 --> 00:42:53.150
is it like we only
used the fact that it's
00:42:53.150 --> 00:42:57.255
F3 to the n-th at the end of the
proof, like in that last step?
00:42:57.255 --> 00:42:57.880
YUFEI ZHAO: OK.
00:42:57.880 --> 00:43:01.040
So question is, where do we
use that in F3 to the n-th?
00:43:01.040 --> 00:43:04.970
This formula here holds in
every finite abelian group,
00:43:04.970 --> 00:43:07.190
if you use the correct
definition of Fourier
00:43:07.190 --> 00:43:10.240
transform with the
averaging normalization.
00:43:10.240 --> 00:43:15.020
So in the other formula where
you replace minus 2 by 1,
00:43:15.020 --> 00:43:17.890
that requires F3.
00:43:17.890 --> 00:43:18.850
But you can--
00:43:18.850 --> 00:43:22.280
I mean, you can follow
the proof and come up
00:43:22.280 --> 00:43:27.080
with a similar formula
for every equation.
00:43:27.080 --> 00:43:28.580
So there's a general
principle here,
00:43:28.580 --> 00:43:30.600
which I'll discuss
more at length
00:43:30.600 --> 00:43:34.120
in a bit, that for
patterns that are
00:43:34.120 --> 00:43:36.760
governed by a single equation--
00:43:36.760 --> 00:43:41.260
in this case, 3-APs, x
minus 2y plus z equal to 0--
00:43:41.260 --> 00:43:43.900
patterns that can be
governed by a single equation
00:43:43.900 --> 00:43:46.196
can be controlled by
a Fourier transform.
00:43:49.770 --> 00:43:53.210
So let's begin our proof,
the Fourier analytic proof
00:43:53.210 --> 00:43:55.636
of Roth's theorem
in F3 to the n-th.
00:43:55.636 --> 00:43:57.695
AUDIENCE: So at
the end, you said
00:43:57.695 --> 00:43:59.445
it was going to be
connected with counting
00:43:59.445 --> 00:44:00.362
the [INAUDIBLE] graph.
00:44:00.362 --> 00:44:03.045
Does this mean that
the Fourier transform
00:44:03.045 --> 00:44:06.982
of the indicator of A, those
are exactly the eigenvalues
00:44:06.982 --> 00:44:08.304
of the Cayley graph?
00:44:08.304 --> 00:44:09.387
Or is it like [INAUDIBLE]?
00:44:09.387 --> 00:44:12.050
YUFEI ZHAO: OK, so you're asking
about the final step, where
00:44:12.050 --> 00:44:13.505
we're talking about--
00:44:13.505 --> 00:44:15.380
so I mentioned that
there was this connection
00:44:15.380 --> 00:44:18.170
between counting walks in graphs
and spectral graph theory.
00:44:18.170 --> 00:44:27.230
So you can check that, if you
have a subset A of an abelian
00:44:27.230 --> 00:44:31.840
group, then the
Fourier transforms of A
00:44:31.840 --> 00:44:39.180
are exactly the eigenvalues
of the Cayley graph.
00:44:44.090 --> 00:44:46.350
AUDIENCE: So then
I guess, have we
00:44:46.350 --> 00:44:48.810
done anything so far
that could have been done
00:44:48.810 --> 00:44:50.950
in a spectral way yet?
00:44:50.950 --> 00:44:54.190
Well, I guess, where is
the Fourier analysis better
00:44:54.190 --> 00:44:55.555
than the spectral [INAUDIBLE]?
00:44:55.555 --> 00:44:56.280
YUFEI ZHAO: OK.
00:44:56.280 --> 00:44:57.950
Question, where is
the Fourier analysis
00:44:57.950 --> 00:44:59.130
better than the spectral posts?
00:44:59.130 --> 00:45:00.170
Well, let's see the proof first.
00:45:00.170 --> 00:45:01.520
And then you'll see, yeah.
00:45:01.520 --> 00:45:03.470
So there's no graphs anymore.
00:45:03.470 --> 00:45:05.535
So we're going to work
inside F3 to the n-th.
00:45:08.270 --> 00:45:10.528
But just like the proof
of regularity in counting,
00:45:10.528 --> 00:45:12.070
we're going to have
a counting lemma.
00:45:12.070 --> 00:45:14.240
So all of these are analytic.
00:45:14.240 --> 00:45:19.243
And at this point, they should
be very familiar to you.
00:45:19.243 --> 00:45:20.660
They may come in
a different form.
00:45:20.660 --> 00:45:22.790
They may be dressed
in different clothing.
00:45:22.790 --> 00:45:24.738
But it's still a counting lemma.
00:45:24.738 --> 00:45:25.280
So let's see.
00:45:34.360 --> 00:45:36.190
The counting lemma,
in this case,
00:45:36.190 --> 00:45:44.170
says that if you are in the
setting of A F3 to the n-th--
00:45:44.170 --> 00:45:47.470
and I'm going to
throughout write
00:45:47.470 --> 00:45:56.460
the density of A as
alpha, then let me write,
00:45:56.460 --> 00:46:06.420
let me define this lambda 3
of A to be the function which
00:46:06.420 --> 00:46:17.470
basically counts 3-APs in
A but with the averaging
00:46:17.470 --> 00:46:18.762
normalization.
00:46:21.660 --> 00:46:23.590
So this is-- we
saw this earlier.
00:46:26.550 --> 00:46:33.550
So the counting lemma says that
this normalized number of 3-APs
00:46:33.550 --> 00:46:34.900
in A--
00:46:34.900 --> 00:46:37.970
so including trivial 3-APs;
that's why this is a nice
00:46:37.970 --> 00:46:39.650
analytic expression--
00:46:39.650 --> 00:46:46.490
differs from what you might
guess based on density alone.
00:46:46.490 --> 00:46:50.960
This difference should be small
if all the nonzero Fourier
00:46:50.960 --> 00:46:53.920
coefficients of A are small.
00:47:04.190 --> 00:47:07.330
So in this strategy, I said
that if-- so the counting
00:47:07.330 --> 00:47:10.330
lemma tells you, if
A is Fourier uniform,
00:47:10.330 --> 00:47:11.550
then it is pseudorandom.
00:47:11.550 --> 00:47:13.630
And this is where it comes in.
00:47:13.630 --> 00:47:16.940
If A has all small
Fourier coefficients,
00:47:16.940 --> 00:47:19.890
then you have a counting
lemma, which tells you
00:47:19.890 --> 00:47:21.810
that the counts
of A should not be
00:47:21.810 --> 00:47:25.980
so different from the
guess based on density.
00:47:34.290 --> 00:47:36.880
So the proof is very short.
00:47:36.880 --> 00:47:41.310
It's based on the identity
that we saw earlier.
00:47:41.310 --> 00:47:47.040
The 3-AP count of A by
the identity earlier
00:47:47.040 --> 00:47:51.890
is simply the third power
of the Fourier transforms.
00:47:55.180 --> 00:47:58.585
And all of these calculations
should be reminiscent,
00:47:58.585 --> 00:48:00.460
because we've done these
kind of calculations
00:48:00.460 --> 00:48:03.460
in some form or another
earlier in this course.
00:48:03.460 --> 00:48:06.370
So we're going to
separate out the main term
00:48:06.370 --> 00:48:07.690
and subsequent terms.
00:48:07.690 --> 00:48:13.750
So the main term is the one
corresponding to r equals to 0.
00:48:13.750 --> 00:48:15.970
So that's the density.
00:48:15.970 --> 00:48:18.640
And all the other
terms I'm going
00:48:18.640 --> 00:48:24.310
to lump together into this sum.
00:48:24.310 --> 00:48:33.610
So we now know
that the difference
00:48:33.610 --> 00:48:36.850
we're trying to bound
is upper bounded
00:48:36.850 --> 00:48:47.850
by the third moment of the
absolute values of the Fourier
00:48:47.850 --> 00:48:50.730
transform.
00:48:50.730 --> 00:48:53.040
And I want to upper
bound this quantity here,
00:48:53.040 --> 00:48:58.320
assuming that all of these
Fourier coefficients are small.
00:48:58.320 --> 00:49:00.540
We've also done this kind
of calculations before.
00:49:00.540 --> 00:49:01.957
So where have we
seen this before?
00:49:06.310 --> 00:49:08.240
We saw this calculation
earlier in the class
00:49:08.240 --> 00:49:09.700
with the 3 replaced by a 4.
00:49:15.640 --> 00:49:18.160
So in counting four cycles
in a proof equivalent
00:49:18.160 --> 00:49:21.940
of quasirandomness, we said
that, if all the eigenvalues
00:49:21.940 --> 00:49:24.010
other than the top
one are small, then
00:49:24.010 --> 00:49:26.980
you can count four cycles.
00:49:26.980 --> 00:49:28.320
It's the same proof.
00:49:28.320 --> 00:49:30.370
And remember, in
that proof, there
00:49:30.370 --> 00:49:34.510
was a important trick,
where you do not uniformly
00:49:34.510 --> 00:49:38.470
bound each term by the
max, because then you lose.
00:49:38.470 --> 00:49:40.900
You lose by an extra factor
of n that you don't want.
00:49:40.900 --> 00:49:45.100
So you only take out one factor.
00:49:45.100 --> 00:49:47.260
So you take out one factor.
00:49:53.840 --> 00:49:56.900
And you keep the rest in there.
00:49:56.900 --> 00:49:59.980
In fact, I can be
more generous and even
00:49:59.980 --> 00:50:02.860
throw the r equal
to 0 term back in.
00:50:06.460 --> 00:50:10.950
And now, by Plancherel--
00:50:10.950 --> 00:50:15.290
so by Plancherel/Parseval,
this here
00:50:15.290 --> 00:50:22.740
is equal to the expectation
of the indicator
00:50:22.740 --> 00:50:23.910
function of A squared.
00:50:23.910 --> 00:50:25.775
So you take this--
00:50:25.775 --> 00:50:30.170
you go back to physical space,
and that's simply the density.
00:50:30.170 --> 00:50:33.710
So then that proves the theorem.
00:50:39.175 --> 00:50:41.550
So the moral of the counting
lemma is the same as the one
00:50:41.550 --> 00:50:43.550
that we've seen before
when we discussed graphs.
00:50:43.550 --> 00:50:46.440
If you're in pseudorandom,
then you have good counting.
00:50:46.440 --> 00:50:49.890
And here, pseudorandom
means having small Fourier
00:50:49.890 --> 00:50:55.040
coefficients, uniformly
small Fourier coefficients.
00:50:55.040 --> 00:50:57.890
So now, let's begin the
proof of Roth's theorem.
00:50:57.890 --> 00:51:01.858
The Roth's theorem proof
will have three steps.
00:51:01.858 --> 00:51:08.455
The first step, we will
observe that if you're
00:51:08.455 --> 00:51:14.700
in the 3-AP-free set, then
there exists a large Fourier
00:51:14.700 --> 00:51:15.731
coefficient.
00:51:26.050 --> 00:51:28.990
Throughout, I'm going to
use uppercase N to denote
00:51:28.990 --> 00:51:32.620
the size of the ambient group.
00:51:32.620 --> 00:51:36.560
And specifically, we
will prove the following.
00:51:36.560 --> 00:51:41.320
And also throughout, A is
a subset of F3 to the n-th
00:51:41.320 --> 00:51:45.075
and with density alpha.
00:51:45.075 --> 00:51:47.420
So I'll keep this convention
throughout this proof.
00:51:50.380 --> 00:52:05.230
We will show that if A is
3-AP-free and N is at least 2
00:52:05.230 --> 00:52:09.030
to the alpha to the minus 2--
00:52:09.030 --> 00:52:11.680
so N is at least
somewhat large--
00:52:11.680 --> 00:52:16.750
then there exists
a nonzero r such
00:52:16.750 --> 00:52:23.230
that the r-th Fourier
coefficient is at least alpha
00:52:23.230 --> 00:52:24.520
squared over 2.
00:52:27.450 --> 00:52:30.900
If you're 3-AP-free,
free then provided
00:52:30.900 --> 00:52:34.290
that you're working in a
large enough ambient space,
00:52:34.290 --> 00:52:37.763
you always have some
large Fourier coefficient.
00:52:40.601 --> 00:52:43.720
So the proof is
essentially-- well,
00:52:43.720 --> 00:52:50.213
this claim is essentially a
corollary of counting lemma,
00:52:50.213 --> 00:52:51.130
by the counting lemma.
00:52:54.820 --> 00:52:57.860
And using the fact
that in a 3-AP-free
00:52:57.860 --> 00:53:02.611
set, what is the
quantity lambda 3 of A?
00:53:02.611 --> 00:53:08.170
Up there, you only have
the trivial 3-APs present.
00:53:08.170 --> 00:53:11.590
So this quantity
lambda must then
00:53:11.590 --> 00:53:17.770
be the size of A divided by N
squared or alpha over N, which
00:53:17.770 --> 00:53:22.100
are precisely counting
the trivial 3-APs.
00:53:26.040 --> 00:53:28.220
So by the counting
lemma, then we
00:53:28.220 --> 00:53:32.540
have that the upper
bound on the right hand
00:53:32.540 --> 00:53:43.120
side, which we now write
alpha max over nonzero r's, is
00:53:43.120 --> 00:53:49.840
at least alpha cubed minus
the lambda 3 term, which
00:53:49.840 --> 00:53:55.370
should be alpha
over N. So provided
00:53:55.370 --> 00:53:57.380
that N is large enough--
00:53:57.380 --> 00:53:58.890
big N is large enough--
00:53:58.890 --> 00:54:02.450
then the trivial 3-APs should
not contribute very much.
00:54:02.450 --> 00:54:06.770
So I can lower bound the
right hand side by, let's say,
00:54:06.770 --> 00:54:07.570
alpha 3--
00:54:07.570 --> 00:54:10.020
alpha cubed over 2.
00:54:10.020 --> 00:54:13.945
So then you deduce
the conclusion.
00:54:13.945 --> 00:54:16.070
I want you to think about
how this proof is related
00:54:16.070 --> 00:54:19.270
to Szemerédi's graph
regularity lemma.
00:54:19.270 --> 00:54:21.020
The analogy will break
down at some point.
00:54:21.020 --> 00:54:23.780
But we've seen this
step before as well.
00:54:23.780 --> 00:54:30.050
From lack of 3-APs, you
extract some useful information
00:54:30.050 --> 00:54:32.980
and from this will
extract some structure.
00:54:32.980 --> 00:54:34.430
And the structure here--
00:54:34.430 --> 00:54:36.590
and this is where the
proof now diverges
00:54:36.590 --> 00:54:40.190
from that of regularity--
00:54:40.190 --> 00:54:46.540
having a large
Fourier coefficient
00:54:46.540 --> 00:54:56.946
will now imply a density
increment on a hyperplane.
00:55:04.730 --> 00:55:06.950
Specifically, if you have--
00:55:11.390 --> 00:55:15.070
so keeping the same
convention as before,
00:55:15.070 --> 00:55:21.100
if the Fourier
coefficient of A at r
00:55:21.100 --> 00:55:29.310
is at least delta
for some nonzero r,
00:55:29.310 --> 00:55:42.160
then A has density at least
alpha plus delta over 2
00:55:42.160 --> 00:55:44.460
when restricted to a hyperplane.
00:55:55.280 --> 00:55:57.640
So if you have a large
Fourier coefficient,
00:55:57.640 --> 00:56:00.730
then I can pass down to a
smaller part of the space
00:56:00.730 --> 00:56:03.518
where the density of A
goes up significantly.
00:56:10.210 --> 00:56:11.980
To see why this is
true, let's go back
00:56:11.980 --> 00:56:15.910
to the definition of the
Fourier coefficient, the Fourier
00:56:15.910 --> 00:56:16.510
transform.
00:56:19.120 --> 00:56:25.040
So recall that Fourier transform
is given by the following
00:56:25.040 --> 00:56:32.030
formula, where I'm looking at
this expectation over points
00:56:32.030 --> 00:56:39.610
in F3 to the n-th together
with indicator of A multiplied
00:56:39.610 --> 00:56:43.840
by this Fourier character.
00:56:43.840 --> 00:56:49.840
And you see that this
function here, it
00:56:49.840 --> 00:57:01.090
is constant on cosets of
the hyperplane defined
00:57:01.090 --> 00:57:06.930
by the orthogonal
complement of r.
00:57:10.400 --> 00:57:14.550
So the value of this dot
product is this constant
00:57:14.550 --> 00:57:17.780
on the three hyperplanes.
00:57:17.780 --> 00:57:22.130
So I can rewrite this
expectation simply
00:57:22.130 --> 00:57:28.010
as 1/3 of alpha 0
plus alpha 1 omega
00:57:28.010 --> 00:57:37.220
plus alpha 2 omega squared,
where alpha 0, alpha 1, alpha 2
00:57:37.220 --> 00:57:48.451
are the densities of A on
the three cosets of r perp.
00:57:52.139 --> 00:57:56.697
So group this expectation
into these three hyperplanes.
00:57:59.320 --> 00:58:05.300
So now, you see that
if this guy is large,
00:58:05.300 --> 00:58:12.570
then you should expect that
alpha 0, alpha 1, and alpha 2
00:58:12.570 --> 00:58:15.550
are not all too
close to each other.
00:58:15.550 --> 00:58:19.260
So if they were all equal to
each other, you should get 0.
00:58:19.260 --> 00:58:22.380
But you should not expect them
to be too close to each other.
00:58:22.380 --> 00:58:27.150
In particular, we would want to
say that one of them goes up--
00:58:27.150 --> 00:58:29.520
is much bigger than alpha.
00:58:29.520 --> 00:58:32.730
So one of these, they must
be much bigger than alpha.
00:58:32.730 --> 00:58:34.230
That's an elementary inequality.
00:58:34.230 --> 00:58:36.522
This is something that I'm
sure I give you five minutes
00:58:36.522 --> 00:58:37.320
you can figure out.
00:58:37.320 --> 00:58:41.170
But let me show you a
small trick to show this.
00:58:41.170 --> 00:58:46.200
And the reason for this trick
is because in next lecture, when
00:58:46.200 --> 00:58:48.900
we look at Roth's theorem
over the integers,
00:58:48.900 --> 00:58:50.220
we'll need this extra trick.
00:58:55.290 --> 00:58:56.850
And the trick here is this.
00:58:56.850 --> 00:59:01.560
We now know that because
of the hypothesis,
00:59:01.560 --> 00:59:08.430
3 delta is lower bound to
the absolute value of alpha 0
00:59:08.430 --> 00:59:12.240
plus alpha 1 omega plus
alpha 2 omega squared.
00:59:16.010 --> 00:59:21.680
OK, so note here that the
average of the three alphas
00:59:21.680 --> 00:59:28.840
is equal to the original alpha
by definition of density.
00:59:28.840 --> 00:59:35.430
So this inner sum I can
read write like that.
00:59:44.620 --> 00:59:46.900
So the sum of the three
groups of unity add up to 0.
00:59:49.410 --> 01:00:02.260
And now, I apply
triangle inequality
01:00:02.260 --> 01:00:06.220
to extract the terms.
01:00:06.220 --> 01:00:07.820
So now, you should
already deduce
01:00:07.820 --> 01:00:12.080
that one of the alphas i's
has to be significantly
01:00:12.080 --> 01:00:14.157
larger than alpha.
01:00:14.157 --> 01:00:15.740
And has to be
significantly different,
01:00:15.740 --> 01:00:16.730
but there are only three times.
01:00:16.730 --> 01:00:18.620
One of them has to be
significantly larger.
01:00:18.620 --> 01:00:21.590
But let me do this
one extra trick, which
01:00:21.590 --> 01:00:24.740
we'll need next time,
which is that let
01:00:24.740 --> 01:00:34.850
me add an extra term like
that, which sums out to 0.
01:00:34.850 --> 01:00:39.894
But now, you see that each
summand is always nonnegative.
01:00:45.700 --> 01:00:49.120
So one of the--
01:00:52.050 --> 01:00:58.830
so there exists some j such
that delta lower abounds
01:00:58.830 --> 01:01:01.290
the j-th summand.
01:01:01.290 --> 01:01:06.960
And if you look at what
that means, then alpha lower
01:01:06.960 --> 01:01:11.370
bounds the j-th summand.
01:01:11.370 --> 01:01:15.588
So in particular, this sum
here should be nonnegative.
01:01:21.480 --> 01:01:22.480
So you have that.
01:01:22.480 --> 01:01:22.980
Good.
01:01:26.390 --> 01:01:30.413
So we obtained a density
increment of this hyperplane.
01:01:34.760 --> 01:01:42.130
And finally, I want to iterate
this density increment.
01:01:42.130 --> 01:01:44.749
So I want to iterate
this density increment--
01:02:01.400 --> 01:02:07.520
so summary so far
is that we have--
01:02:07.520 --> 01:02:20.520
so if A is 3-AP-free with
density alpha and N at least 2
01:02:20.520 --> 01:02:26.300
to the alpha to
the minus 2, then A
01:02:26.300 --> 01:02:35.510
has density at least alpha
plus alpha squared over 4
01:02:35.510 --> 01:02:37.175
on some hyperplane.
01:02:40.430 --> 01:02:44.740
So you combine step one and step
two, we obtain this conclusion.
01:02:48.340 --> 01:02:54.970
Well, I can now
repeat this operation.
01:02:54.970 --> 01:03:09.890
So I can repeat by restricting
A to this hyperplane.
01:03:09.890 --> 01:03:13.590
If A is originally 3-AP-free,
I restrict it to a hyperplane,
01:03:13.590 --> 01:03:15.460
it's still 3-AP-free.
01:03:15.460 --> 01:03:17.790
So I can keep going.
01:03:17.790 --> 01:03:20.400
I can keep going provided
that my space is still
01:03:20.400 --> 01:03:27.500
large enough, because I still
need this lower bound on F.
01:03:27.500 --> 01:03:29.640
So don't forget this one here.
01:03:29.640 --> 01:03:40.340
So I can keep going
as long as N is--
01:03:44.960 --> 01:03:47.110
so I'm using N sub
j to denote what
01:03:47.110 --> 01:03:50.240
happens after the j-th step.
01:03:50.240 --> 01:03:53.840
I can keep going as long
as this is still satisfied.
01:03:53.840 --> 01:03:56.540
But of course, you cannot
keep on going forever,
01:03:56.540 --> 01:04:00.030
because the density is bounded.
01:04:00.030 --> 01:04:01.970
So density cannot exceed 1.
01:04:01.970 --> 01:04:06.420
So these two will give you a
bound on the total dimension.
01:04:06.420 --> 01:04:08.750
So let's work this out.
01:04:08.750 --> 01:04:18.140
So let alpha i
denote the density
01:04:18.140 --> 01:04:25.770
after step i in this iteration.
01:04:25.770 --> 01:04:35.910
And we see from over here that
you start with density alpha,
01:04:35.910 --> 01:04:44.940
and each step you go up by an
increment, which is basically
01:04:44.940 --> 01:04:46.260
what--
01:04:46.260 --> 01:04:48.720
so you go up by some increment.
01:04:48.720 --> 01:04:51.570
And you want to know,
if you start with alpha,
01:04:51.570 --> 01:04:57.690
how many steps at most can you
take before you blow out 1.
01:05:00.440 --> 01:05:02.530
So can you give me some bound?
01:05:06.640 --> 01:05:08.160
So what's the maximum--
01:05:08.160 --> 01:05:09.590
at most, how many steps?
01:05:17.160 --> 01:05:24.184
So we know that the
density cannot exceed 1.
01:05:24.184 --> 01:05:27.523
AUDIENCE: 4 by alpha squared.
01:05:27.523 --> 01:05:31.920
YUFEI ZHAO: So you see
that you have at most 4
01:05:31.920 --> 01:05:38.850
over alpha squared steps,
because density is at most 1.
01:05:38.850 --> 01:05:44.710
And if you plug this in,
you get something which
01:05:44.710 --> 01:05:46.810
is not quite what I stated.
01:05:46.810 --> 01:05:50.920
So it turns out that
if you plug this in,
01:05:50.920 --> 01:05:53.620
you find that alpha is--
01:05:53.620 --> 01:05:57.730
you find that the
size of A is at most 3
01:05:57.730 --> 01:05:59.660
to the n-th over square root n.
01:06:03.210 --> 01:06:04.945
So let me do a
little bit better,
01:06:04.945 --> 01:06:09.730
so then simply seeing
that this term here is
01:06:09.730 --> 01:06:13.260
at least alpha squared over 4.
01:06:13.260 --> 01:06:15.970
And the point is that
when you increment,
01:06:15.970 --> 01:06:19.040
you increment faster and faster.
01:06:19.040 --> 01:06:21.680
So I can use that
to give a better
01:06:21.680 --> 01:06:24.420
bound on the number of steps.
01:06:24.420 --> 01:06:26.840
And here's the way to see it.
01:06:26.840 --> 01:06:29.810
So let me-- we can do better.
01:06:33.600 --> 01:06:39.540
So starting at alpha, I then
now ask, how many steps do you
01:06:39.540 --> 01:06:42.594
need to take before it doubles?
01:06:42.594 --> 01:06:44.426
AUDIENCE: [INAUDIBLE]
01:06:44.426 --> 01:06:48.610
YUFEI ZHAO: It goes up
by alpha squared over 4.
01:06:48.610 --> 01:06:57.700
So it doubles after at
most 4 over alpha steps.
01:07:01.280 --> 01:07:08.860
And at which point
this alpha, new alpha,
01:07:08.860 --> 01:07:10.930
becomes at least the original--
01:07:10.930 --> 01:07:13.250
twice the original alpha?
01:07:13.250 --> 01:07:14.740
But now, you keep going.
01:07:14.740 --> 01:07:18.560
How many times does it
take to double again?
01:07:18.560 --> 01:07:21.030
2 over alpha, because the
alpha became twice as much.
01:07:26.760 --> 01:07:33.400
So it doubles again after
at most 2 over alpha steps.
01:07:33.400 --> 01:07:34.670
And then you keep going.
01:07:34.670 --> 01:07:38.390
The next iteration
is 1 over alpha.
01:07:38.390 --> 01:07:52.700
So in total-- so you see that
we must stop after at most 8
01:07:52.700 --> 01:07:56.000
over alpha steps.
01:07:58.570 --> 01:08:01.250
So the number of
times it doubles,
01:08:01.250 --> 01:08:04.830
actually it decreases by
at least half each time.
01:08:11.053 --> 01:08:11.970
So now, we know that--
01:08:14.860 --> 01:08:17.500
we see that the--
01:08:17.500 --> 01:08:19.609
so you keep on going.
01:08:19.609 --> 01:08:23.170
So you must stop after at
most 8 over alpha steps.
01:08:23.170 --> 01:08:26.859
What is the final density
when you have to stop,
01:08:26.859 --> 01:08:28.430
because when are
you forced to stop?
01:08:28.430 --> 01:08:32.399
You are forced to stop
if you run out of space.
01:08:32.399 --> 01:08:34.930
So you're forced to stop
when you run out of space.
01:08:34.930 --> 01:08:46.290
So if the process
terminates after m steps--
01:08:46.290 --> 01:09:04.305
so we're at density alpha m,
so then the final subspace
01:09:04.305 --> 01:09:16.100
has size less than 2 to the
alpha m raised to minus 2,
01:09:16.100 --> 01:09:19.130
which is--
01:09:19.130 --> 01:09:22.830
So now, I use this bound alpha.
01:09:25.490 --> 01:09:32.189
So the initial N is
upper bounded by what?
01:09:32.189 --> 01:09:34.250
So how many steps did you take?
01:09:34.250 --> 01:09:36.990
You took at most 8
over alpha steps.
01:09:36.990 --> 01:09:41.760
Each of those steps, you pass
down to codimension what?
01:09:41.760 --> 01:09:43.979
You lose a dimension
for each step.
01:09:43.979 --> 01:09:47.050
And the final subspace
has at least this--
01:09:47.050 --> 01:09:51.490
has at most that much space.
01:09:51.490 --> 01:10:01.300
So the final dimension is this,
basically log 1 over alpha.
01:10:01.300 --> 01:10:07.650
So put them together, we see
that the size of the space
01:10:07.650 --> 01:10:11.322
originally is at
most 1 over alpha.
01:10:11.322 --> 01:10:11.822
Yeah.
01:10:11.822 --> 01:10:14.466
AUDIENCE: Should this
be a lower case n?
01:10:14.466 --> 01:10:15.383
YUFEI ZHAO: Thank you.
01:10:15.383 --> 01:10:16.210
Yeah.
01:10:16.210 --> 01:10:19.499
This should be a lower
case n, so the dimension.
01:10:19.499 --> 01:10:19.999
Good.
01:10:22.780 --> 01:10:30.540
And OK, so then that's the
conclusion, that the density
01:10:30.540 --> 01:10:36.361
alpha is big O of 1 over n.
01:10:41.360 --> 01:10:44.990
That proves the main
theorem for today,
01:10:44.990 --> 01:10:48.940
so Roth's theorem
over F3 to the n-th.
01:10:48.940 --> 01:10:51.540
So we went through this
Fourier analytic proof.
01:10:51.540 --> 01:10:57.290
Next lecture, we will see
the same proof again but done
01:10:57.290 --> 01:11:00.800
in the integers for interval.
01:11:00.800 --> 01:11:05.460
And there, there are
some difficulties
01:11:05.460 --> 01:11:08.340
that we don't see over here.
01:11:08.340 --> 01:11:11.940
Because in the
finite field space,
01:11:11.940 --> 01:11:15.120
in the finite field
model, there's
01:11:15.120 --> 01:11:18.490
this very nice idea of
looking at subspaces,
01:11:18.490 --> 01:11:21.410
so looking at hyperplanes.
01:11:21.410 --> 01:11:25.820
Each Fourier coefficient gets
you down to one dimension less.
01:11:25.820 --> 01:11:29.730
But when you're working
in the integers,
01:11:29.730 --> 01:11:32.310
there are no
subspaces you can use.
01:11:34.860 --> 01:11:38.040
So we'll be looking
at ways to get around
01:11:38.040 --> 01:11:39.685
the lack of subspaces.
01:11:39.685 --> 01:11:41.310
And this is why I
said in the beginning
01:11:41.310 --> 01:11:43.790
that the finite
field model is often
01:11:43.790 --> 01:11:50.940
a very good playground for
additive combinatorics type
01:11:50.940 --> 01:11:53.940
techniques, especially
Fourier analytic techniques.
01:11:53.940 --> 01:11:55.830
Because in the additive--
01:11:55.830 --> 01:11:57.450
in all of these
techniques, they just
01:11:57.450 --> 01:11:59.050
come out to be much cleaner.
01:11:59.050 --> 01:12:01.560
If you're working in a
finite field setting,
01:12:01.560 --> 01:12:03.480
you have nice subspaces,
you have Fourier
01:12:03.480 --> 01:12:05.160
transform in a very clean way.
01:12:05.160 --> 01:12:08.130
The Fourier transform
always takes, in this case,
01:12:08.130 --> 01:12:10.220
one of three values.
01:12:10.220 --> 01:12:11.220
Everything's very clean.
01:12:11.220 --> 01:12:12.300
Everything's very simple.
01:12:12.300 --> 01:12:13.490
And you get to
see the idea here.
01:12:13.490 --> 01:12:15.615
You get to see the sense
of the increment argument.
01:12:18.210 --> 01:12:21.600
But once you
understand those ideas
01:12:21.600 --> 01:12:26.320
and you're willing to do
more work, then oftentimes,
01:12:26.320 --> 01:12:29.590
you can bring those
ideas to other settings,
01:12:29.590 --> 01:12:32.830
to other abelian
groups, to the integers,
01:12:32.830 --> 01:12:37.300
for instance, but with
more work in the--
01:12:37.300 --> 01:12:41.800
there are some extra ingredients
that you need to use.
01:12:41.800 --> 01:12:45.260
I mentioned that
there was a bound--
01:12:45.260 --> 01:12:46.330
OK, so initially--
01:12:49.812 --> 01:12:51.020
So next time, we'll see that.
01:12:51.020 --> 01:12:56.000
Next time, we'll see what
happens over the integers.
01:12:56.000 --> 01:12:58.490
Any questions?
01:12:58.490 --> 01:12:59.187
Yes.
01:12:59.187 --> 01:13:03.300
AUDIENCE: [INAUDIBLE]
01:13:03.300 --> 01:13:04.250
YUFEI ZHAO: OK, great.
01:13:04.250 --> 01:13:07.850
So question is why the process
must stop after at most 8
01:13:07.850 --> 01:13:08.840
over alpha steps?
01:13:12.430 --> 01:13:15.700
So you know that the density
doubles after this many steps,
01:13:15.700 --> 01:13:17.380
doubles again after
that many steps.
01:13:17.380 --> 01:13:20.710
So eventually, if it
keeps on doubling,
01:13:20.710 --> 01:13:24.360
it cannot keep on
doubling forever.
01:13:24.360 --> 01:13:26.570
So this process cannot
keep on doubling forever.
01:13:26.570 --> 01:13:28.820
So it must stop--
01:13:28.820 --> 01:13:39.560
so cannot double more than log
base 2 of 1 over alpha times.
01:13:42.980 --> 01:13:45.440
And that point,
you have to stop.
01:13:45.440 --> 01:13:47.270
So how many steps
have you taken?
01:13:47.270 --> 01:13:49.420
Well, you sum this
geometric series.
01:13:49.420 --> 01:13:55.940
So this-- and the
next thing is that you
01:13:55.940 --> 01:13:57.840
sum this geometric series.
01:13:57.840 --> 01:14:01.310
And that geometric series
sums to 8 over alpha.
01:14:10.230 --> 01:14:10.730
Great.
01:14:10.730 --> 01:14:12.280
So let's finish here.
01:14:12.280 --> 01:14:16.680
So next time, we'll see Roth's
proof of Roth's theorem.