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YUFEI ZHAO: Last time, we
considered the relationship

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between pseudo-random graphs
and their eigenvalues.

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And the main message is that
the smaller your second largest

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eigenvalue is, the more
pseudo-random a graph is.

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In particular, we were
looking at this class

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of graphs that are d-regular--
they are somewhat easier

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to think about.

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And there is a
limit to how small

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the second largest
eigenvalue can be.

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And that was given by
the Alon-Boppana bound.

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You should think of d
here as a fixed number--

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so here, d is a fixed constant.

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Then, as the number of
vertices becomes large,

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the second largest eigenvalue
of a d-regular graph cannot be

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less than this
quantity over here.

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So this is the
limit to how small

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this second largest
eigenvalue can be.

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And last time, we gave
a proof of this bound

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by constructing an
appropriate function that

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witnesses this lambda 2.

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We also gave a
second proof which

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proves a slightly
weaker result, which

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is that the second largest
eigenvalue in absolute value

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is at least this quantity.

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So in spirit, it amounts to
roughly the same result--

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although technically,
it's a little bit weaker.

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And that one we proved
by counting walks.

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And also at the
end of last time,

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I remarked that
this number here--

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the fundamental
significance of this number

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is that it is the
spectral radius

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of the infinite d-regular tree.

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So that's why this
number is here.

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Of course, we proved
some lower bound.

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But you can always
ask the question,

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is this the best
possible lower bound?

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Maybe it's possible to prove
a somewhat higher bound.

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And that turns out
not to be the case.

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So that's the first
thing that we'll

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see today is some discussions--

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I won't show you any proofs--
but some discussions on why

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this number is best possible.

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And this is a very interesting
area of graph theory--

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goes under the name
of Ramanujan graphs.

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So I'll explain the history
in a second, why they're

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called Ramanujan graphs.

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Ramanujan did not
study these graphs,

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but they are called
them for good reasons.

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So by definition, a Ramanujan
graph is a d-regular graph,

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such that, if you look at its
eigenvalue of the adjacency

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matrix, as above, the second
largest eigenvalue in absolute

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value is, at most,
that bound up there--

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2 root d minus 1.

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So it's the best possible
constant you could put here

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so that there still exists
infinitely many d-regular

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Ramanujan graphs for fixed d--

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and the size of the
graph going to infinity.

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And the last time, we also
introduced some terminology.

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Let me just repeat that here.

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So this is, in other
words, an nd lambda graph,

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with lambda at most
2 root d minus 1.

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Now, it is not hard to obtain
a single example of a Ramanujan

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graph.

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So I just want some
graph such that--

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or the top eigenvalue is d.

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I want the other
ones to be small.

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So for example, if you get
this click, it's d-regular.

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Here, the top eigenvalue is d.

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And if it's not
too hard to compute

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that all the other eigenvalues
are equal to exactly minus 1.

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So this is an easy computation.

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But the point is that I
want to construct graphs--

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I want to understand
whether they

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are graphs where d is fixed.

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So this is somehow
not a good example.

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What we really want is fixed
d and n going to infinity.

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Large number of vertices.

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And the main open conjecture in
this area is that for every d,

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there exists infinitely many
Ramanujan d-regular graphs.

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So let me tell you
some partial results

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and also explain the
history of why they're

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called Ramanujan graphs.

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So the first paper
where this name

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appeared and coined this name--

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and I'll explain the
reason in a second--

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is this important result of
Lubotzky, Phillips, and Sarnak.

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From the late '80s.

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So their paper was
titled Ramanujan graphs.

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So they proved that
this conjecture is true.

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So the conjecture is true for
all d, such that d minus 1

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is a prime number.

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I should also remark
that the same result

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was proved independently by
Margulis at the same time.

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Their construction of this graph
is a specific Cayley graph.

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So they gave an explicit
construction of a Cayley graph,

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with the group being the
projective special linear

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group--

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PSL 2 q.

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So, some group-- and this
group actually comes up a lot.

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It's a group with lots of nice
pseudo-randomness properties.

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And to verify that,
the corresponding graph

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has the desired
eigenvalue properties,

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they had to invoke
some deep results

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from number theory
that were related

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to Ramanujan conjectures.

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So that's why they called
these graphs Ramanujan graphs.

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And that name stuck.

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So these papers, they
proved that these graphs

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exist for some special values
of d-- namely, when d minus 1

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is a prime.

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There was a later
generalization in the '90s,

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by Morgenstern, generalizing
such constructions

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showing that you can also take
d minus 1 to be a prime power.

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And really, that's
pretty much it.

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For all the other
values of d, it

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is open whether there exists
infinitely many d-regular

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Ramanujan graphs.

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In particular, for d equal
to 7, it is still open.

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Do there exist infinitely many
semi-regular Ramanujan graphs?

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No.

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What about a random graph?

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If I take a random
graph, what is

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the size of its second
largest eigenvalue?

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And there is a difficult
theorem of Friedman--

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I say difficult,
because the paper itself

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is more than 100 pages long--

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that if you take a fixed
d, then a random end

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vertex d-regular graph.

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So what does this mean?

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So the easiest way to explain
the random d-regular graph

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is that you look at the set of
all possible d-regular graphs

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on a fixed number of
vertices and you pick one

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uniformly at random.

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So random-- such graph
is almost Ramanujan,

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in the following sense--

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that the second largest
eigenvalue in absolute value

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is, at most, 2 root d minus
1 plus some small arrow

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little 1, where the
little 1 goes to 0

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as n goes to infinity.

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So in other words, this
constant cannot be improved,

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but this result doesn't tell
you that any of these graphs are

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Ramanujan.

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Experimental evidence
suggests that if you

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take, for fixed value of d--

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let's say d equals to
7 or d equals to 3--

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if you take a random
d-regular graph,

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then a specific percentage of
those graphs are Ramanujan.

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So the second largest
eigenvalue has

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some empirical
distribution, at least

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from computer experiments,
where some specific fraction--

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I don't remember
exactly, but let's

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say 40% of three
regular graphs--

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is expected in the
limit to be Ramanujan.

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So that appears to
be quite difficult.

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We have no idea how to even
approach such conjectures.

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There were some exciting
recent breakthroughs

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in the past few years concerning
a variant, a somewhat weakening

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of this problem--

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a bipartite analogue
of Ramanujan graphs.

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Now, in a bipartite graph--

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so all bipartite graphs
have the property

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that its eigenvalues,
its spectrum,

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is symmetric around 0.

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Its smallest
eigenvalue is minus d.

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So if you plot all
the eigenvalues,

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it's symmetric around 0.

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This is not a hard fact to see--

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I encourage you
to think about it.

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And that's because, if
you have an eigenvector--

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so it lifts
somewhere on the left

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and somewhere on the right--

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I can form another
eigenvector, which

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is obtained by flipping
the signs on one part.

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If the first eigenvector
has eigenvalue lambda,

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then the second one has
eigenvalue minus lambda.

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So the eigenvalues come
in symmetric pairs.

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So by definition, a
bipartite Ramanujan graph

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is one where it's
a bipartite graph

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and I only require that the
second largest eigenvalue

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is less than 2 root d minus 1.

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Everything's symmetric
around the origin.

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So this is by definition.

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If you start with
a Ramanujan graph,

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I can use it to create a
bipartite Ramanujan graph,

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because if I look
at this 2 lift--

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so where there's
this construction--

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this means if I start with
some graph, G-- so for example,

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if G is this graph
here, what I want to do

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is take two copies
of this graph,

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think about having them
on two sheets of paper,

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one on top of the other.

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And I draw all the
edges criss-crossed.

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So that's G cross K 2.

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This is G.

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You should convince
yourself that if G

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has eigenvalues lambda then
G cross K 2 has eigenvalues.

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The original spectrum,
as well, it's

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symmetric-- so it's negation.

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So if G is Ramanujan,
then G cross

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K 2 is a bipartite
Ramanujan graph.

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So it's a weaker concept--
if you have Ramanujan graphs,

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then you have bipartite
Ramanujan graphs-- but not

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in reverse.

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But, still a problem
of do there exist

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d-regular bipartite Ramanujan
graphs is still interesting.

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It's a somewhat weaker problem,
but it's still interesting.

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And there was a
major breakthrough

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00:13:20,020 --> 00:13:32,070
a few years ago by Marcus
Spielman and Srivastava,

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showing that for
all fixed d, there

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00:13:34,860 --> 00:13:42,370
exist infinitely many d-regular
bipartite Ramanujan graphs.

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00:13:47,860 --> 00:13:50,950
And unlike the earlier
work of Lubotzky, Phillips,

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00:13:50,950 --> 00:13:52,030
and Sarnak--

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00:13:52,030 --> 00:13:55,480
which, the earlier work was
an explicit construction

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of a Cayley graph--

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00:13:57,400 --> 00:14:01,650
this construction here is a
probabilistic construction.

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00:14:01,650 --> 00:14:03,970
It uses some very
nice tools that they

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00:14:03,970 --> 00:14:06,770
called interlacing families.

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00:14:06,770 --> 00:14:09,460
So it showed,
probabilistically, using

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00:14:09,460 --> 00:14:12,190
a very clever
randomized construction,

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00:14:12,190 --> 00:14:14,060
that these graphs exist.

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00:14:14,060 --> 00:14:17,770
So it's not just take a usual
d-regular random bipartite

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00:14:17,770 --> 00:14:21,442
graph, but there's some clever
constructions of randomness.

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And this is more
or less the state

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of knowledge regarding the
existence of Ramanujan graphs.

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00:14:30,680 --> 00:14:34,250
Again, the big open problem
is that there exists

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00:14:34,250 --> 00:14:36,530
d-regular Ramanujan graphs.

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00:14:36,530 --> 00:14:39,080
For every d, there
are infinitely many

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00:14:39,080 --> 00:14:42,810
such Ramanujan graphs.

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00:14:42,810 --> 00:14:43,786
Yeah.

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AUDIENCE: So in the
conception of G cross K 2,

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lambda 1 is equal to d?

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Or is equal to
original [INAUDIBLE]..

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00:14:56,103 --> 00:14:57,520
YUFEI ZHAO: Right,
so the question

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00:14:57,520 --> 00:15:01,390
is, if you start with
a d-regular graph

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00:15:01,390 --> 00:15:09,180
and take this construction,
the spectrum has 1 d

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00:15:09,180 --> 00:15:12,180
and it also has a minus d.

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00:15:12,180 --> 00:15:14,550
If your graph is
bipartite, its spectrum

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is symmetric around the origin.

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00:15:16,390 --> 00:15:19,180
So you always have
d and minus d.

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00:15:19,180 --> 00:15:21,510
So a bipartite graph
can never be Ramanujan.

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00:15:21,510 --> 00:15:24,400
But the definition of a
bipartite Ramanujan graph

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00:15:24,400 --> 00:15:28,950
is just that I only require that
the remaining eigenvalues sit

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00:15:28,950 --> 00:15:30,520
in that interval.

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00:15:30,520 --> 00:15:33,790
I'm OK with having minus d here.

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00:15:33,790 --> 00:15:36,940
So that's by definition of
a bipartite Ramanujan graph.

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00:15:39,900 --> 00:15:41,131
Any more questions?

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00:15:44,220 --> 00:15:49,880
All right, so combining
a Alon-Boppana bound

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00:15:49,880 --> 00:15:52,970
and both the existence
of Ramanujan graphs

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00:15:52,970 --> 00:15:54,710
and also Friedman's
difficult result

250
00:15:54,710 --> 00:15:57,530
that random graph
is almost Ramanujan,

251
00:15:57,530 --> 00:16:01,640
we see that this
2 root d minus 1--

252
00:16:01,640 --> 00:16:03,050
that number there is optimal.

253
00:16:06,520 --> 00:16:09,300
So that's the extent in
which a d-regular graph

254
00:16:09,300 --> 00:16:10,580
can be pseudo-random.

255
00:16:14,580 --> 00:16:16,330
Now, the rest of
this lecture, I want

256
00:16:16,330 --> 00:16:18,670
to move onto a somewhat
different topic,

257
00:16:18,670 --> 00:16:22,210
but still concerning sparse
pseudo-random graphs.

258
00:16:22,210 --> 00:16:24,730
Basically, I want to tell you
what I did for my PhD thesis.

259
00:16:28,140 --> 00:16:33,820
So, so far we've been talking
about pseudo-random graphs,

260
00:16:33,820 --> 00:16:36,640
but let's combine it with the
topic in the previous chapter--

261
00:16:36,640 --> 00:16:39,940
namely, similarities
regularity lemma.

262
00:16:39,940 --> 00:16:44,590
And we can ask, can we
apply the regularity method

263
00:16:44,590 --> 00:16:46,570
to sparse graphs?

264
00:16:49,260 --> 00:16:51,360
So when we talk about
similarities regularity,

265
00:16:51,360 --> 00:16:54,180
I kept emphasizing that it's
really about dense graphs,

266
00:16:54,180 --> 00:16:56,400
because there are these
error terms which are little

267
00:16:56,400 --> 00:16:57,240
and squared.

268
00:16:57,240 --> 00:16:58,740
And if your graph
is already sparse,

269
00:16:58,740 --> 00:17:01,190
that error term
eats up everything.

270
00:17:01,190 --> 00:17:05,260
So for sparse graphs, you
need to be extra careful.

271
00:17:05,260 --> 00:17:08,530
So I want to explore the
idea of a sparse regularity.

272
00:17:08,530 --> 00:17:11,380
And here, sparse
just means not dense.

273
00:17:11,380 --> 00:17:20,210
So sparse means x
density, little 1.

274
00:17:20,210 --> 00:17:24,000
So, the opposite of dense.

275
00:17:24,000 --> 00:17:26,597
We saw the triangle
removal [INAUDIBLE]..

276
00:17:35,680 --> 00:17:37,290
So, let me remind
you the statement.

277
00:17:37,290 --> 00:17:46,320
It says that for every epsilon,
there exists some delta, such

278
00:17:46,320 --> 00:18:01,720
that if G has a small
number of triangles,

279
00:18:01,720 --> 00:18:16,880
then G can be made
triangle-free by removing

280
00:18:16,880 --> 00:18:17,950
a small number of edges.

281
00:18:31,250 --> 00:18:35,030
I would like to state a sparse
version of this theorem that

282
00:18:35,030 --> 00:18:39,650
works for graphs where I'm
looking at sub constant x

283
00:18:39,650 --> 00:18:42,222
densities.

284
00:18:42,222 --> 00:18:43,930
So roughly, this is
how it's going to go.

285
00:18:46,900 --> 00:18:50,610
I'm going to put in
these extra p factors.

286
00:18:50,610 --> 00:18:52,360
And you should think
of p as some quantity

287
00:18:52,360 --> 00:18:54,960
that goes to 0 with n.

288
00:19:00,240 --> 00:19:04,170
So, think of p as
the general scale.

289
00:19:04,170 --> 00:19:07,380
So that's the x density
scale we're thinking of.

290
00:19:07,380 --> 00:19:10,260
I would like to
say that if G has

291
00:19:10,260 --> 00:19:13,800
less than that many
triangles, then

292
00:19:13,800 --> 00:19:17,985
G can be made free by deleting
a small number of edges.

293
00:19:17,985 --> 00:19:19,740
But what does small mean here?

294
00:19:19,740 --> 00:19:22,620
Small should be relative
to the scale of x

295
00:19:22,620 --> 00:19:24,550
densities you're looking at.

296
00:19:24,550 --> 00:19:30,060
So in this case, we should add
an extra factor of p over here.

297
00:19:30,060 --> 00:19:32,370
So that's the kind of
statement I would like,

298
00:19:32,370 --> 00:19:35,940
but of course, this is
too good to be true,

299
00:19:35,940 --> 00:19:37,980
because we haven't
really modified anything.

300
00:19:37,980 --> 00:19:41,370
If you read the statement,
it's just completely false.

301
00:19:41,370 --> 00:19:45,780
So I would like adding some
conditions, some hypotheses,

302
00:19:45,780 --> 00:19:49,680
that would make such
a statement true.

303
00:19:49,680 --> 00:19:52,630
And hypothesis is going to
be roughly along those lines.

304
00:19:52,630 --> 00:19:55,440
So I'm going to call this a
meta-theorem, because I won't

305
00:19:55,440 --> 00:19:58,320
state the hypothesis precisely.

306
00:19:58,320 --> 00:20:00,510
But roughly, it will
be along the lines

307
00:20:00,510 --> 00:20:08,440
that, if gamma is
some, say, sufficiently

308
00:20:08,440 --> 00:20:29,350
pseudo-random graph and
vertices and x density, p, and G

309
00:20:29,350 --> 00:20:38,030
is a subgraph of
gamma, then I want

310
00:20:38,030 --> 00:20:42,830
to say that G has this triangle
removal property, relatively

311
00:20:42,830 --> 00:20:45,700
inside gamma.

312
00:20:45,700 --> 00:20:47,630
And this is true.

313
00:20:47,630 --> 00:20:49,330
Well, it is true
if you're putting

314
00:20:49,330 --> 00:20:52,870
the appropriate, sufficiently
pseudo-random condition.

315
00:20:52,870 --> 00:20:54,035
So I'm leaving here--

316
00:20:54,035 --> 00:20:55,910
I'll tell you more later
what this should be.

317
00:20:58,560 --> 00:21:00,750
So this is a kind of
statement that I would like.

318
00:21:00,750 --> 00:21:05,160
So a sparse extension of
the triangle removal lemma

319
00:21:05,160 --> 00:21:08,580
says that, if you have a
sufficiently pseudo-random

320
00:21:08,580 --> 00:21:12,090
host, or you think of this
gamma as a host graph,

321
00:21:12,090 --> 00:21:17,760
then inside that host, relative
to the density of this host,

322
00:21:17,760 --> 00:21:21,510
everything should
behave nicely, as you

323
00:21:21,510 --> 00:21:25,040
would expect in the dense case.

324
00:21:25,040 --> 00:21:29,420
The dense case is also a
special case of the sparse case,

325
00:21:29,420 --> 00:21:34,130
because if we took gamma
to be the complete graph--

326
00:21:34,130 --> 00:21:37,560
which is also pseudo-random,
it's everything--

327
00:21:37,560 --> 00:21:41,730
it's uniform-- then
this is also true.

328
00:21:41,730 --> 00:21:44,060
And that's triangle removal
among the dense case,

329
00:21:44,060 --> 00:21:48,520
but we want this
sparse extension.

330
00:21:48,520 --> 00:21:49,235
Question.

331
00:21:49,235 --> 00:21:53,690
AUDIENCE: Where does the
c come into [INAUDIBLE]??

332
00:21:53,690 --> 00:21:57,570
YUFEI ZHAO: Where
does p come in to--

333
00:21:57,570 --> 00:22:01,900
so p here is the edge
density of gamma.

334
00:22:01,900 --> 00:22:04,470
AUDIENCE: [INAUDIBLE]

335
00:22:04,470 --> 00:22:05,750
YUFEI ZHAO: Yeah.

336
00:22:05,750 --> 00:22:08,360
So again, I'm not really
stating this so precisely,

337
00:22:08,360 --> 00:22:12,490
but you should think
of p as something

338
00:22:12,490 --> 00:22:16,240
that could decay with n--

339
00:22:16,240 --> 00:22:20,440
not too quickly, but decay
at n to the minus some

340
00:22:20,440 --> 00:22:23,380
small constant.

341
00:22:23,380 --> 00:22:24,190
Yeah.

342
00:22:24,190 --> 00:22:27,107
AUDIENCE: Delta here
doesn't depend on gamma?

343
00:22:27,107 --> 00:22:27,940
YUFEI ZHAO: Correct.

344
00:22:27,940 --> 00:22:30,820
So the question is, what
does delta depend on?

345
00:22:30,820 --> 00:22:36,250
So here, delta depends
on only epsilon.

346
00:22:36,250 --> 00:22:38,830
And in fact, what
we would like--

347
00:22:38,830 --> 00:22:41,180
and this will indeed
be basically true--

348
00:22:41,180 --> 00:22:44,140
is that delta is more
or less the same delta

349
00:22:44,140 --> 00:22:46,383
from the original
triangle removal lemma.

350
00:22:49,624 --> 00:22:51,476
Yeah.

351
00:22:51,476 --> 00:22:53,660
AUDIENCE: If G is
any graph, what's

352
00:22:53,660 --> 00:22:57,772
stopping you from
making it a subgraph

353
00:22:57,772 --> 00:23:01,750
of some large [INAUDIBLE]?

354
00:23:01,750 --> 00:23:04,690
YUFEI ZHAO: So the question
is, if G is some arbitrary

355
00:23:04,690 --> 00:23:06,890
graph, what's to
stop you from making

356
00:23:06,890 --> 00:23:10,030
it a subgraph of a large
pseudo-random graph?

357
00:23:10,030 --> 00:23:11,710
And there's a great question.

358
00:23:11,710 --> 00:23:13,390
If I give you a
graph, G, can you

359
00:23:13,390 --> 00:23:16,190
test whether G satisfies
the hypothesis?

360
00:23:16,190 --> 00:23:18,190
Because the conclusion
doesn't depend on gamma.

361
00:23:18,190 --> 00:23:21,040
The conclusion is only
on G, but the hypothesis

362
00:23:21,040 --> 00:23:23,200
requires us to gamma.

363
00:23:23,200 --> 00:23:25,673
And so my two answers
to that is, one,

364
00:23:25,673 --> 00:23:27,340
you cannot always
embed it in the gamma.

365
00:23:33,037 --> 00:23:34,829
I guess the easier
answer is the conclusion

366
00:23:34,829 --> 00:23:38,273
is false with other hypothesis.

367
00:23:38,273 --> 00:23:40,190
So you cannot always
embed it in such a gamma.

368
00:23:40,190 --> 00:23:43,162
But it is somewhat
difficult to test.

369
00:23:43,162 --> 00:23:45,120
I don't know a good way
to test whether a given

370
00:23:45,120 --> 00:23:46,850
G lies in such a gamma.

371
00:23:46,850 --> 00:23:48,600
I will motivate this
theorem in a second--

372
00:23:48,600 --> 00:23:52,160
why we care about
results of this form.

373
00:23:52,160 --> 00:23:52,735
Yes.

374
00:23:52,735 --> 00:23:57,020
AUDIENCE: Don't all sufficiently
large pseudo-random graphs--

375
00:23:57,020 --> 00:23:59,953
say, with respect to the
number of vertices of G--

376
00:23:59,953 --> 00:24:03,780
contain copies of every G?

377
00:24:03,780 --> 00:24:05,610
YUFEI ZHAO: So the
question is, if you

378
00:24:05,610 --> 00:24:09,410
start with a sufficiently
large pseudo-random gamma,

379
00:24:09,410 --> 00:24:10,950
does it contain every copy of G?

380
00:24:10,950 --> 00:24:14,190
And the answer is no, because G
has the same number of vertices

381
00:24:14,190 --> 00:24:16,530
as gamma.

382
00:24:16,530 --> 00:24:18,330
A sufficiently
pseudo-random-- again,

383
00:24:18,330 --> 00:24:20,622
I haven't told you what
sufficiently pseudo-random even

384
00:24:20,622 --> 00:24:21,280
means yet.

385
00:24:21,280 --> 00:24:25,890
But you should think of it as
controlling small patterns.

386
00:24:25,890 --> 00:24:28,200
But here, G is a
much larger graph.

387
00:24:28,200 --> 00:24:30,700
It's the same size,
it's just maybe,

388
00:24:30,700 --> 00:24:33,240
let's say, half of the edges.

389
00:24:33,240 --> 00:24:36,990
So what you should think about
is starting with gamma being,

390
00:24:36,990 --> 00:24:38,400
let's say, a random graph.

391
00:24:38,400 --> 00:24:41,310
And I delete
adversarially, let's say,

392
00:24:41,310 --> 00:24:43,720
half of the edges of gamma.

393
00:24:43,720 --> 00:24:47,220
And you get G.

394
00:24:47,220 --> 00:24:48,280
So let me go on.

395
00:24:48,280 --> 00:24:51,480
And please ask more questions.

396
00:24:51,480 --> 00:24:53,460
I won't really prove
anything today,

397
00:24:53,460 --> 00:24:55,290
but it's really
meant to give you

398
00:24:55,290 --> 00:25:00,090
an idea of what this
line of work is about.

399
00:25:00,090 --> 00:25:05,000
And I also want to motivate
it by explaining why we care

400
00:25:05,000 --> 00:25:06,590
about these kind of theorems.

401
00:25:06,590 --> 00:25:08,995
So first observation
is that it is not

402
00:25:08,995 --> 00:25:11,120
true with all the hypothesis--
hopefully all of you

403
00:25:11,120 --> 00:25:13,850
see this as obviously
too good to be true.

404
00:25:13,850 --> 00:25:18,530
But will also see some
specific examples.

405
00:25:18,530 --> 00:25:19,840
Here's a specific example.

406
00:25:19,840 --> 00:25:27,400
So this is not true
without this gamma.

407
00:25:27,400 --> 00:25:30,650
So for example, you
can have this graph, G.

408
00:25:30,650 --> 00:25:32,710
And we already saw
this construction that

409
00:25:32,710 --> 00:25:37,540
came from Behren's construction,
where you have n vertices

410
00:25:37,540 --> 00:25:49,780
and n to the 2 minus little 1
edges, where every edge belongs

411
00:25:49,780 --> 00:25:51,130
to exactly one triangle.

412
00:25:57,400 --> 00:26:00,060
If you plug in this
graph into this theorem,

413
00:26:00,060 --> 00:26:01,800
with all the yellow stuff--

414
00:26:01,800 --> 00:26:03,400
if you add in this p--

415
00:26:03,400 --> 00:26:04,660
you see it's false.

416
00:26:04,660 --> 00:26:08,125
You just cannot remove--

417
00:26:08,125 --> 00:26:08,625
anyway.

418
00:26:11,860 --> 00:26:17,010
In what context can we expect
such a sparse triangle removal

419
00:26:17,010 --> 00:26:18,500
lemma to be true?

420
00:26:18,500 --> 00:26:20,470
One setting for
which it is true--

421
00:26:20,470 --> 00:26:23,500
and this was a result that was
proved about 10 years ago--

422
00:26:23,500 --> 00:26:26,730
is that if your gamma
is a truly random graph.

423
00:26:26,730 --> 00:26:35,170
So this is true
for a random gamma

424
00:26:35,170 --> 00:26:42,140
if p is sufficiently large
and roughly it's at least--

425
00:26:42,140 --> 00:26:45,830
so there's some constant
such that if p is at least c

426
00:26:45,830 --> 00:26:47,413
over [INAUDIBLE],,
then it is true.

427
00:26:47,413 --> 00:26:49,205
So this is the result
of Conlon and Gowers.

428
00:26:56,686 --> 00:26:57,186
Yeah.

429
00:26:57,186 --> 00:27:00,540
AUDIENCE: Is this random
in the Erdos-Rényi sense?

430
00:27:00,540 --> 00:27:03,712
YUFEI ZHAO: So this is random
in the Erdos-Rényi sense.

431
00:27:03,712 --> 00:27:04,920
So, Erdos-Rényi random graph.

432
00:27:07,610 --> 00:27:09,710
But this is not the
main motivating reason

433
00:27:09,710 --> 00:27:12,170
why I would like to talk
about this technique.

434
00:27:12,170 --> 00:27:17,210
The main motivating example
is the Green-Tao theorem.

435
00:27:25,180 --> 00:27:29,090
So I remind you that
the Green-Tao theorem

436
00:27:29,090 --> 00:27:32,840
says that the primes contain
arbitrarily long arithmetic

437
00:27:32,840 --> 00:27:33,848
progressions.

438
00:27:52,280 --> 00:27:55,020
So the Green-Tao theorem
is, in some sense,

439
00:27:55,020 --> 00:27:57,300
an extension of
Szemeredi's theorem,

440
00:27:57,300 --> 00:27:59,072
but a sparse extension.

441
00:27:59,072 --> 00:28:00,780
Szemeredi's theorem
tells you that if you

442
00:28:00,780 --> 00:28:04,090
have a positive density
subset of the integers, then

443
00:28:04,090 --> 00:28:07,670
it contains long
arithmetic progressions.

444
00:28:07,670 --> 00:28:09,200
But here, the primes--

445
00:28:09,200 --> 00:28:13,690
we know from prime
number theorem

446
00:28:13,690 --> 00:28:19,750
that the density of
the primes up to n

447
00:28:19,750 --> 00:28:24,580
decays, like 1 over log n.

448
00:28:24,580 --> 00:28:27,490
So it's a sparse
set, but we would

449
00:28:27,490 --> 00:28:30,923
like to know that it has
all of these patterns.

450
00:28:30,923 --> 00:28:33,340
It turns out the primes are,
in some sense, pseudo-random.

451
00:28:33,340 --> 00:28:35,850
But that's a difficult
result to prove.

452
00:28:35,850 --> 00:28:38,830
And that was proved after
Green-Tao proved their initial

453
00:28:38,830 --> 00:28:40,032
theorem--

454
00:28:40,032 --> 00:28:44,590
so, by later works of Green
and Tao and also Ziegler.

455
00:28:44,590 --> 00:28:46,360
But the original strategy--

456
00:28:46,360 --> 00:28:49,350
and also the later strategy for
the stronger result, as well.

457
00:28:49,350 --> 00:28:51,760
But the strategy for the
Green-Tao theorem is this--

458
00:28:51,760 --> 00:28:56,020
you start with the primes
and you embed the primes

459
00:28:56,020 --> 00:28:57,370
in a somewhat larger set.

460
00:29:02,360 --> 00:29:05,210
You start with the
primes and you embed it

461
00:29:05,210 --> 00:29:09,260
in a somewhat larger set,
which we'll call, informally,

462
00:29:09,260 --> 00:29:12,580
pseudoprimes.

463
00:29:12,580 --> 00:29:17,014
And these m, roughly
speaking, numbers

464
00:29:17,014 --> 00:29:21,195
with no small prime divisors.

465
00:29:34,000 --> 00:29:37,360
Because these numbers
are somewhat smoother

466
00:29:37,360 --> 00:29:39,160
compared to the
primes, they're easier

467
00:29:39,160 --> 00:29:42,830
to analyze by analytic
number theory methods,

468
00:29:42,830 --> 00:29:44,950
especially coming
from sieve theory.

469
00:29:44,950 --> 00:29:49,390
And it is easier, although
still highly nontrivial,

470
00:29:49,390 --> 00:29:52,630
to show that these pseudoprimes
are, in some sense,

471
00:29:52,630 --> 00:29:53,860
pseudo-random.

472
00:30:02,020 --> 00:30:04,490
And that's the kind
of pseudo-random host

473
00:30:04,490 --> 00:30:08,510
that corresponds with
the gamma over there.

474
00:30:08,510 --> 00:30:11,560
So the Green-Tao strategy
is to start with the primes,

475
00:30:11,560 --> 00:30:14,980
build a slightly larger
set so that the prime sit

476
00:30:14,980 --> 00:30:18,460
inside the pseudoprimes in
a relatively dense manner.

477
00:30:26,480 --> 00:30:30,930
So it has high relative density.

478
00:30:30,930 --> 00:30:34,610
And then, if you had this
kind of strategy for a sparse

479
00:30:34,610 --> 00:30:36,140
triangle removal lemma--

480
00:30:36,140 --> 00:30:40,010
but imagine you also had it
for various other extensions

481
00:30:40,010 --> 00:30:43,160
of sparse hypergraph removal
lemma, which allows you

482
00:30:43,160 --> 00:30:45,050
to prove Szemeredi's theorem.

483
00:30:45,050 --> 00:30:48,000
And now you can use
it in that setting.

484
00:30:48,000 --> 00:30:52,220
Then you can prove Szemeredi's
theorem in the primes.

485
00:30:55,060 --> 00:30:57,520
That's the theorem and
that's the approach.

486
00:30:57,520 --> 00:30:59,740
And that's one of the
reasons, at least for me,

487
00:30:59,740 --> 00:31:04,090
why something like a sparse
triangle removal lemma

488
00:31:04,090 --> 00:31:07,810
plays a central role in
these kind of problems.

489
00:31:12,500 --> 00:31:15,790
So I want to say
more about how you

490
00:31:15,790 --> 00:31:18,040
might go about proving
this type of result

491
00:31:18,040 --> 00:31:23,030
and also what pseudo-random
graph means over here.

492
00:31:23,030 --> 00:31:28,150
So, remember the strategy for
proving the triangle removal

493
00:31:28,150 --> 00:31:28,840
lemma.

494
00:31:28,840 --> 00:31:32,580
And of course, all of you guys
are working on this problem set

495
00:31:32,580 --> 00:31:35,470
and so the method of
regularity hopefully

496
00:31:35,470 --> 00:31:39,140
should be very familiar to
you by the end of this week.

497
00:31:39,140 --> 00:31:43,510
But let me remind you that
there are three main steps, one

498
00:31:43,510 --> 00:31:49,650
being to partition your graph
using the regularity lemma.

499
00:31:49,650 --> 00:31:51,640
The second one, to clean.

500
00:31:51,640 --> 00:31:53,230
And the third one, the count.

501
00:31:56,350 --> 00:32:05,360
And I want to explain where the
sparse regularity method fails.

502
00:32:05,360 --> 00:32:07,777
So you can try to do
everything the same and then--

503
00:32:07,777 --> 00:32:09,860
so what happens if you try
to do all these things?

504
00:32:13,790 --> 00:32:19,620
So first, let's talk about
sparse regularity lemma.

505
00:32:30,710 --> 00:32:32,680
So let me remind
you-- previously,

506
00:32:32,680 --> 00:32:43,230
we said that a pair of
vertices is epsilon regular

507
00:32:43,230 --> 00:32:52,150
if, for every subset
U of A and W of B--

508
00:32:52,150 --> 00:32:53,020
neither too small.

509
00:32:56,800 --> 00:33:08,950
So if neither are
too small, one has

510
00:33:08,950 --> 00:33:13,600
that the number of
edges between U and W

511
00:33:13,600 --> 00:33:19,000
differs from what
you would expect.

512
00:33:19,000 --> 00:33:25,870
So the x density between U and
W is close to what you expect,

513
00:33:25,870 --> 00:33:30,450
which is the ordinal edge
density between A and B.

514
00:33:30,450 --> 00:33:32,930
So they differ by no
more than epsilon.

515
00:33:36,800 --> 00:33:39,470
So this should be a
familiar definition.

516
00:33:39,470 --> 00:33:41,690
What we would like
is to modify it

517
00:33:41,690 --> 00:33:43,720
to work for the sparse setting.

518
00:33:43,720 --> 00:33:46,730
And for that, I'm going to
add in an extra p factor.

519
00:33:46,730 --> 00:33:51,350
So I'm going to say
epsilon, p regular.

520
00:33:51,350 --> 00:33:54,260
Well, this condition here--
now, oh, the densities

521
00:33:54,260 --> 00:33:56,900
are on the scale of p.

522
00:33:56,900 --> 00:33:58,980
Which goes to 0 as
n goes to infinity.

523
00:33:58,980 --> 00:34:01,040
So in what does the
property compare them?

524
00:34:01,040 --> 00:34:03,950
I should add an
extra factor of p

525
00:34:03,950 --> 00:34:06,690
to put everything
on the right scale.

526
00:34:06,690 --> 00:34:07,880
Otherwise, this is too weak.

527
00:34:16,840 --> 00:34:19,460
And given this
definition here, we

528
00:34:19,460 --> 00:34:29,580
can say that a
partition of vertices

529
00:34:29,580 --> 00:34:41,610
is epsilon regular if all part
but at most epsilon fraction

530
00:34:41,610 --> 00:34:47,489
pairs is epsilon regular--

531
00:34:47,489 --> 00:34:52,219
so an equitable partition.

532
00:34:52,219 --> 00:34:54,510
And I would modify it
to the sparse setting

533
00:34:54,510 --> 00:34:57,990
by changing the appropriate
notion of regular

534
00:34:57,990 --> 00:35:03,600
to the sparse version, where
I'm looking at scales of p.

535
00:35:03,600 --> 00:35:06,367
I still require at
most epsilon fraction--

536
00:35:06,367 --> 00:35:07,200
that stays the same.

537
00:35:07,200 --> 00:35:08,908
That's not affected
by the density scale.

538
00:35:11,540 --> 00:35:23,070
Previously, we had the
irregularity lemma,

539
00:35:23,070 --> 00:35:27,340
which said that
for every epsilon,

540
00:35:27,340 --> 00:35:40,390
there exists some M such
that every graph has

541
00:35:40,390 --> 00:35:49,330
an epsilon regular partition
into at most M parts.

542
00:35:54,030 --> 00:36:01,750
And the sparse version
would say that if your graph

543
00:36:01,750 --> 00:36:09,325
has x density at most p--

544
00:36:09,325 --> 00:36:11,780
and here, all of these
constants are negotiable.

545
00:36:11,780 --> 00:36:14,570
So when I say p, I really
could mean 100 times p.

546
00:36:14,570 --> 00:36:16,400
You just change these constants.

547
00:36:16,400 --> 00:36:22,040
So if it's most p, then it has
an epsilon, p regular partition

548
00:36:22,040 --> 00:36:24,320
into at most m parts.

549
00:36:24,320 --> 00:36:26,600
Here, m depends only on epsilon.

550
00:36:29,730 --> 00:36:34,300
So previously, I wrote down the
sparse triangle removal lemma.

551
00:36:34,300 --> 00:36:37,740
And I wrote down the
statement and it was false--

552
00:36:37,740 --> 00:36:41,290
with all the additional
hypotheses, it was false.

553
00:36:41,290 --> 00:36:44,190
It turns out that this
is actually true--

554
00:36:44,190 --> 00:36:47,010
the version of the
sparse regularity lemma,

555
00:36:47,010 --> 00:36:49,950
which sounds almost too
good to be true, initially.

556
00:36:49,950 --> 00:36:56,130
We are adding in a
whole lot of sparsity

557
00:36:56,130 --> 00:37:00,450
and sparsity seems to be
more difficult to deal with.

558
00:37:00,450 --> 00:37:03,750
And the reason why I
think sparsity is harder

559
00:37:03,750 --> 00:37:06,810
to deal with is
that, in some sense,

560
00:37:06,810 --> 00:37:09,030
there are a lot
more sparse graphs

561
00:37:09,030 --> 00:37:11,500
than there are dense graphs.

562
00:37:11,500 --> 00:37:16,050
So let me pause for a
second and explain that.

563
00:37:16,050 --> 00:37:18,900
It is not true
that, in some sense,

564
00:37:18,900 --> 00:37:21,330
there are more sparse graphs.

565
00:37:21,330 --> 00:37:23,280
Because if you just count--

566
00:37:23,280 --> 00:37:25,870
once you have sparser things,
there are fewer of them.

567
00:37:25,870 --> 00:37:29,250
But I mean in terms of
the actual complexity

568
00:37:29,250 --> 00:37:31,200
of the structures
that can come up.

569
00:37:31,200 --> 00:37:33,600
When you have sparser
objects, there's

570
00:37:33,600 --> 00:37:35,940
a lot more that can happen.

571
00:37:35,940 --> 00:37:39,590
In dense objects,
Szemeredi's regularity lemma

572
00:37:39,590 --> 00:37:42,830
tells us, in some sense,
that the amount of complexity

573
00:37:42,830 --> 00:37:45,260
in the graph is bounded.

574
00:37:45,260 --> 00:37:49,270
But that's not the
case for sparse graphs.

575
00:37:49,270 --> 00:37:50,770
In any case, we
still have some kind

576
00:37:50,770 --> 00:37:53,400
of sparse regularity lemma.

577
00:37:53,400 --> 00:37:55,710
And this version
here, as written,

578
00:37:55,710 --> 00:37:58,770
is literally true if you have
the appropriate definitions--

579
00:37:58,770 --> 00:38:02,240
and more or less, we have
those definitions up there.

580
00:38:02,240 --> 00:38:06,580
But I want to say
that it's misleading.

581
00:38:06,580 --> 00:38:09,683
This is true, but misleading.

582
00:38:15,250 --> 00:38:17,320
And the reason why
it is misleading

583
00:38:17,320 --> 00:38:19,420
is that, in a sparse
graph, you can

584
00:38:19,420 --> 00:38:23,570
have lots of
intricate structures

585
00:38:23,570 --> 00:38:26,950
that are hidden in
your irregular parts.

586
00:38:26,950 --> 00:38:46,480
It could be that most edges are
inside irregular pairs, which

587
00:38:46,480 --> 00:38:48,970
would make the irregularity
lemma a somewhat

588
00:38:48,970 --> 00:38:53,210
useless statement, because
when you do the cleaning step,

589
00:38:53,210 --> 00:38:55,010
you delete all of your edges.

590
00:38:55,010 --> 00:38:56,010
And you don't want that.

591
00:38:59,800 --> 00:39:01,690
But in any case, it is true--

592
00:39:01,690 --> 00:39:03,480
and I'll comment on
the proof in a second.

593
00:39:03,480 --> 00:39:06,390
But the way I want you to think
about the sparse regularity

594
00:39:06,390 --> 00:39:11,620
lemma is that it
should work when--

595
00:39:11,620 --> 00:39:14,680
so before jumping to
that, a specific example

596
00:39:14,680 --> 00:39:17,330
where this happens
is, for example,

597
00:39:17,330 --> 00:39:25,690
if your graph G is a click on a
sublinear fraction of vertices.

598
00:39:25,690 --> 00:39:28,100
Somehow, you might
care about that click.

599
00:39:28,100 --> 00:39:30,640
So that's a pretty important
object in the graph.

600
00:39:30,640 --> 00:39:33,250
But when you do the sparse
regularity partition,

601
00:39:33,250 --> 00:39:35,380
it could be that the
entire click is hidden

602
00:39:35,380 --> 00:39:39,450
inside an irregular part.

603
00:39:39,450 --> 00:39:40,720
And you just don't see it--

604
00:39:40,720 --> 00:39:41,950
that information gets lost.

605
00:39:45,230 --> 00:39:48,260
The proper way to think about
the sparse regularity lemma

606
00:39:48,260 --> 00:39:51,800
is to think about
graphs, G, that satisfy

607
00:39:51,800 --> 00:39:54,720
some additional hypotheses.

608
00:39:54,720 --> 00:40:07,810
So in practice, g is assumed to
satisfy some upper regularity

609
00:40:07,810 --> 00:40:08,778
condition.

610
00:40:17,570 --> 00:40:19,880
And an example of
such an hypothesis

611
00:40:19,880 --> 00:40:25,610
is something called
no dense spots,

612
00:40:25,610 --> 00:40:28,580
meaning that it doesn't have
a really dense component,

613
00:40:28,580 --> 00:40:33,220
like in the case of a click on
a very small number of vertices.

614
00:40:33,220 --> 00:40:36,240
So no dense spots--

615
00:40:36,240 --> 00:40:41,140
one definition could be
that there exists some eta--

616
00:40:41,140 --> 00:40:43,990
and here, just as in
quasi-random graphs,

617
00:40:43,990 --> 00:40:46,030
I'm thinking of
sequences going to 0.

618
00:40:46,030 --> 00:40:47,890
So there exists
eta sequence going

619
00:40:47,890 --> 00:40:58,250
to 0 and a constant,
c, such that for all

620
00:40:58,250 --> 00:41:01,760
set x in the graph--

621
00:41:01,760 --> 00:41:11,470
let's say X and Y. If X and Y
have size at least eta fraction

622
00:41:11,470 --> 00:41:18,380
of V, then the density
between X and Y

623
00:41:18,380 --> 00:41:22,760
is bounded by at most
a constant factor,

624
00:41:22,760 --> 00:41:27,180
compared to the overall density,
p, that we're looking at.

625
00:41:27,180 --> 00:41:30,350
So in other words, no
small piece of the graph

626
00:41:30,350 --> 00:41:31,910
has too many edges.

627
00:41:41,440 --> 00:41:50,980
And with that notion
of the no dense spots,

628
00:41:50,980 --> 00:41:55,090
we can now prove the
sparse regularity lemma

629
00:41:55,090 --> 00:41:58,090
under that additional
hypothesis.

630
00:41:58,090 --> 00:42:00,700
And basically, the
proof is the same

631
00:42:00,700 --> 00:42:04,030
as the usual
semi-regularity lemma proof

632
00:42:04,030 --> 00:42:06,420
that we saw a few weeks ago.

633
00:42:09,360 --> 00:42:16,746
So if you have proof
of sparse regularity

634
00:42:16,746 --> 00:42:19,540
with no dense spots--

635
00:42:23,530 --> 00:42:24,210
hypothesis.

636
00:42:27,150 --> 00:42:32,115
OK, so I claim this as the
same proof as Szemeredi's

637
00:42:32,115 --> 00:42:34,230
irregularity lemma.

638
00:42:34,230 --> 00:42:41,015
And the reason is that in the
energy increment argument,

639
00:42:41,015 --> 00:42:42,140
you do everything the same.

640
00:42:42,140 --> 00:42:43,807
You do partitioning
if it's not regular.

641
00:42:43,807 --> 00:42:47,410
You refine and you keep going.

642
00:42:47,410 --> 00:42:50,410
In the energy
increment argument,

643
00:42:50,410 --> 00:42:52,690
one key property we
used was that the energy

644
00:42:52,690 --> 00:42:55,000
was bounded between 0 and 1.

645
00:42:55,000 --> 00:42:59,820
And every time, you went
up by epsilon to the fifth.

646
00:42:59,820 --> 00:43:04,110
And now the energy
increment argument--

647
00:43:04,110 --> 00:43:10,890
that each step,
the energy goes up

648
00:43:10,890 --> 00:43:16,140
by something which
is like epsilon,

649
00:43:16,140 --> 00:43:21,910
let's say, to the
fifth and p squared.

650
00:43:21,910 --> 00:43:25,860
The energy is some kind
of mean square density,

651
00:43:25,860 --> 00:43:28,070
so this p squared
should play a role.

652
00:43:34,030 --> 00:43:37,630
So if you only knew that,
then the number of iterations

653
00:43:37,630 --> 00:43:39,850
might depend on p--

654
00:43:39,850 --> 00:43:41,730
it might depend on n.

655
00:43:41,730 --> 00:43:45,360
So, not a constant-- and
that would be an issue.

656
00:43:45,360 --> 00:43:50,150
However, if you have
no dense spots--

657
00:43:50,150 --> 00:43:57,190
so, because no dense spots--

658
00:43:57,190 --> 00:44:08,080
the final energy, I claim,
is, at most, something

659
00:44:08,080 --> 00:44:11,350
like C squared p squared.

660
00:44:11,350 --> 00:44:15,430
Maybe some small error,
because of all the epsilons

661
00:44:15,430 --> 00:44:19,680
flowing around, but
that's the final energy.

662
00:44:19,680 --> 00:44:23,850
So you still have a
bounded number of steps.

663
00:44:28,730 --> 00:44:33,650
So the bound only
depends on epsilon.

664
00:44:33,650 --> 00:44:35,970
So the entire proof
runs through just fine.

665
00:44:40,500 --> 00:44:43,520
OK, so having the
right hypothesis helps.

666
00:44:43,520 --> 00:44:45,830
But then I said, the
more general version

667
00:44:45,830 --> 00:44:49,350
without the hypothesis
is still true.

668
00:44:49,350 --> 00:44:51,220
So how come that is the case?

669
00:44:51,220 --> 00:44:53,470
Because if you do this proof,
you run into the issue--

670
00:44:53,470 --> 00:44:57,680
you cannot control the
number of iterations.

671
00:44:57,680 --> 00:45:02,150
So here's a trick introduced
by Alex Scott, who came up

672
00:45:02,150 --> 00:45:03,400
with that version there.

673
00:45:06,740 --> 00:45:13,280
So this is a nice trick, which
is that, instead of using

674
00:45:13,280 --> 00:45:14,170
x squared--

675
00:45:14,170 --> 00:45:18,640
the function as energy--

676
00:45:18,640 --> 00:45:22,780
let's consider a somewhat
different function.

677
00:45:22,780 --> 00:45:29,220
So the function I want to use
is fe of x which is initially

678
00:45:29,220 --> 00:45:32,220
quadratic-- so,
initially x squared--

679
00:45:32,220 --> 00:45:35,850
but up to a specific point.

680
00:45:35,850 --> 00:45:36,480
Let's say 2.

681
00:45:40,120 --> 00:45:42,520
And then after this
point, I make it linear.

682
00:45:51,560 --> 00:45:54,030
So that's the function
I'm going to take.

683
00:45:54,030 --> 00:45:56,940
Now, this function has a
couple of nice properties.

684
00:45:56,940 --> 00:46:03,110
One is that you also have this
boosting, this energy increment

685
00:46:03,110 --> 00:46:08,780
step, because for all
random variables, x--

686
00:46:12,450 --> 00:46:14,780
so x is a non-negative
random variable.

687
00:46:14,780 --> 00:46:16,970
So think of this as edge
densities between parts

688
00:46:16,970 --> 00:46:18,770
on the refinement.

689
00:46:18,770 --> 00:46:26,100
If the mean of x is,
at most, 1, then,

690
00:46:26,100 --> 00:46:35,250
if you look at this
energy, it increases

691
00:46:35,250 --> 00:46:41,475
if x has a large variance.

692
00:46:44,810 --> 00:46:48,710
Previously, when we used fe
as square, this was true.

693
00:46:48,710 --> 00:46:50,390
So this is true
with equal to 1--

694
00:46:50,390 --> 00:46:52,610
in fact, that's the
definition of variance.

695
00:46:52,610 --> 00:46:55,700
But this inequality is also
true for this function, fe--

696
00:46:55,700 --> 00:46:59,210
so that when you do the
irregularity breaking,

697
00:46:59,210 --> 00:47:01,610
if you have irregular
parts, then you

698
00:47:01,610 --> 00:47:05,060
have some variance in
the edge densities.

699
00:47:05,060 --> 00:47:08,390
So you would get
an energy boost.

700
00:47:08,390 --> 00:47:11,900
But the other thing
is that we are

701
00:47:11,900 --> 00:47:16,880
no longer worried about
the final energy being

702
00:47:16,880 --> 00:47:20,060
much higher than the individual
potential contributions.

703
00:47:20,060 --> 00:47:28,360
Because, if you end up having
lots of high density pieces,

704
00:47:28,360 --> 00:47:31,150
they would contribute a lot.

705
00:47:31,150 --> 00:47:44,180
So, in other words,
the expectation

706
00:47:44,180 --> 00:47:49,630
for the second thing is
that the expectation of fe

707
00:47:49,630 --> 00:48:03,410
is upper-bounded by, let's say,
4 times the expectation of x.

708
00:48:03,410 --> 00:48:07,040
And so this inequality there
would cap the number of steps

709
00:48:07,040 --> 00:48:08,570
you would have to do.

710
00:48:08,570 --> 00:48:11,660
You would never actually end
up having too many iterations.

711
00:48:16,640 --> 00:48:19,730
So this is a discussion of
the sparse regularity lemma.

712
00:48:19,730 --> 00:48:21,980
And the main message here
is that the regularity lemma

713
00:48:21,980 --> 00:48:23,990
itself is not so difficult--

714
00:48:23,990 --> 00:48:27,090
that's largely the same as
Szemeredi's regularity lemma.

715
00:48:27,090 --> 00:48:29,270
And so that's actually not
the most difficult part

716
00:48:29,270 --> 00:48:31,580
of sparse triangle
removal lemma.

717
00:48:31,580 --> 00:48:35,840
The difficulty lies in the other
step in the regularity method--

718
00:48:35,840 --> 00:48:38,220
namely, the counting step.

719
00:48:38,220 --> 00:48:40,820
And we already alluded
to this in the past.

720
00:48:40,820 --> 00:48:43,010
The point is that there
is no counting lemma

721
00:48:43,010 --> 00:48:46,540
for sparse regular graphs.

722
00:48:46,540 --> 00:48:49,150
And we already saw
an example where,

723
00:48:49,150 --> 00:48:52,750
if you start with a
random graph which

724
00:48:52,750 --> 00:48:55,030
has a small number
of triangles and I

725
00:48:55,030 --> 00:48:57,760
delete a small number
of edges corresponding

726
00:48:57,760 --> 00:48:59,590
to those triangles--

727
00:48:59,590 --> 00:49:04,300
one, I do not affect
its quasi-randomness.

728
00:49:04,300 --> 00:49:06,040
But two, there's no
triangles anymore,

729
00:49:06,040 --> 00:49:08,660
so there's no triangle
counting lemma.

730
00:49:08,660 --> 00:49:11,290
And that's a serious obstacle,
because you need this counting

731
00:49:11,290 --> 00:49:12,560
step.

732
00:49:12,560 --> 00:49:15,100
So what I would
like to explain next

733
00:49:15,100 --> 00:49:20,190
is how you can salvage that
and use this hypothesis here

734
00:49:20,190 --> 00:49:24,040
written in yellow to obtain a
counting lemma so that you can

735
00:49:24,040 --> 00:49:26,290
complete this
regularity method that

736
00:49:26,290 --> 00:49:29,200
would allow you to prove the
sparse triangle removal lemma.

737
00:49:29,200 --> 00:49:31,210
And a similar kind of
technique can allow you

738
00:49:31,210 --> 00:49:33,860
to do the Green-Tao theorem.

739
00:49:33,860 --> 00:49:36,970
So let's take a quick break.

740
00:49:36,970 --> 00:49:38,140
OK, any questions so far.

741
00:49:41,695 --> 00:49:43,320
So let's talk about
the counting lemma.

742
00:49:51,940 --> 00:49:55,240
So, the first case of the
counting lemma we considered

743
00:49:55,240 --> 00:49:56,710
was the triangle counting lemma.

744
00:50:03,170 --> 00:50:05,190
So remember what it says.

745
00:50:05,190 --> 00:50:08,870
If you have 3 vertex sets--

746
00:50:08,870 --> 00:50:18,090
V1, V2, V3-- such that,
between each pair,

747
00:50:18,090 --> 00:50:19,770
it is epsilon regular.

748
00:50:24,880 --> 00:50:32,100
And edge density--
that's for simplicity's

749
00:50:32,100 --> 00:50:35,090
sake-- they all have
the same edge density.

750
00:50:35,090 --> 00:50:36,690
Actually, they can be different.

751
00:50:36,690 --> 00:50:40,530
So d sub ij-- so possibly
different edge densities.

752
00:50:40,530 --> 00:50:42,120
But I have the set-up.

753
00:50:42,120 --> 00:50:44,610
And then the triangle
counting lemma

754
00:50:44,610 --> 00:50:51,220
tells us that the
number of triangles

755
00:50:51,220 --> 00:50:59,280
with one vertex in
each part is basically

756
00:50:59,280 --> 00:51:02,730
what you would expect
in the random case--

757
00:51:02,730 --> 00:51:05,110
namely, multiplying
these three edge

758
00:51:05,110 --> 00:51:10,530
densities together,
plus a small error,

759
00:51:10,530 --> 00:51:15,880
and then multiplying the
vertex sets' sizes together.

760
00:51:20,050 --> 00:51:22,450
So what we would
like is a statement

761
00:51:22,450 --> 00:51:28,610
that says that if you
have epsilon p regular

762
00:51:28,610 --> 00:51:33,590
and x densities now
at scale p, then

763
00:51:33,590 --> 00:51:36,610
we would want the
same thing to be true.

764
00:51:36,610 --> 00:51:39,140
Here, I should add
an extra p cubed,

765
00:51:39,140 --> 00:51:42,040
because that's the densities
we're working with.

766
00:51:42,040 --> 00:51:44,910
And I want some error here--

767
00:51:44,910 --> 00:51:48,200
OK, I can even let you
take some other epsilon.

768
00:51:48,200 --> 00:51:52,550
But small changes are OK.

769
00:51:52,550 --> 00:51:54,610
So that's the kind of
statement we want--

770
00:51:54,610 --> 00:51:56,250
and this is false.

771
00:51:56,250 --> 00:51:58,490
So this is completely false.

772
00:51:58,490 --> 00:52:00,740
And the example
that I said earlier

773
00:52:00,740 --> 00:52:07,130
was one of these examples
where you have a random graph.

774
00:52:07,130 --> 00:52:11,810
So this initial
version is false,

775
00:52:11,810 --> 00:52:18,870
because if you take a G and
p, with p somewhat less than 1

776
00:52:18,870 --> 00:52:24,080
over root n, and then remove
an edge from each triangle--

777
00:52:24,080 --> 00:52:26,010
or just remove all
the triangles--

778
00:52:26,010 --> 00:52:30,920
then you have a graph which
is still fairly pseudo-random,

779
00:52:30,920 --> 00:52:33,600
but it has no triangles.

780
00:52:33,600 --> 00:52:36,170
So you cannot have
a counting lemma.

781
00:52:36,170 --> 00:52:38,300
So there's another example
which, in some sense,

782
00:52:38,300 --> 00:52:40,590
is even better than
this random example.

783
00:52:40,590 --> 00:52:42,690
And it's a somewhat
mysterious example

784
00:52:42,690 --> 00:52:51,530
due to a law that gives
you a pseudo-random gamma.

785
00:52:51,530 --> 00:52:55,070
So it's, in some sense, an
optimally pseudo-random gamma,

786
00:52:55,070 --> 00:53:07,100
such that it is d-regular with
d on the order of n to the 3/2s.

787
00:53:07,100 --> 00:53:11,870
And it's an nd lambda
graph, where lambda

788
00:53:11,870 --> 00:53:15,340
is on the order of root d.

789
00:53:15,340 --> 00:53:16,900
Because here, d
is not a constant.

790
00:53:16,900 --> 00:53:19,060
But even in this case,
roughly speaking,

791
00:53:19,060 --> 00:53:23,030
this is as pseudo-random
as you can expect.

792
00:53:23,030 --> 00:53:26,600
So the second eigenvalue
is roughly square root

793
00:53:26,600 --> 00:53:28,610
of the degree.

794
00:53:28,610 --> 00:53:30,830
And yet, this graph
is triangle free.

795
00:53:37,420 --> 00:53:40,510
So you have some graph which,
for all the other kinds

796
00:53:40,510 --> 00:53:42,910
of pseudo-randomness
is very nice.

797
00:53:42,910 --> 00:53:45,650
So it has all the nice
pseudo-randomness properties,

798
00:53:45,650 --> 00:53:46,900
yet it is still triangle free.

799
00:53:46,900 --> 00:53:47,400
It's sparse.

800
00:53:50,190 --> 00:53:52,330
So the triangle
counting lemma is not

801
00:53:52,330 --> 00:53:54,040
true without
additional hypotheses.

802
00:53:56,660 --> 00:54:00,230
So I would like to add in some
hypotheses to make it true.

803
00:54:00,230 --> 00:54:03,010
And I would like a theorem.

804
00:54:03,010 --> 00:54:08,710
So again, I'm going to put
as a meta-theorem, which

805
00:54:08,710 --> 00:54:17,700
says that if you assume that G
is a subgraph of a sufficiently

806
00:54:17,700 --> 00:54:36,310
pseudo-random gamma and
gamma has edge density p,

807
00:54:36,310 --> 00:54:37,750
then the conclusion is true.

808
00:54:40,383 --> 00:54:41,550
And this is indeed the case.

809
00:54:47,640 --> 00:54:50,870
And I would like to tell
you what is the sufficiently

810
00:54:50,870 --> 00:54:52,310
pseudo-random--

811
00:54:52,310 --> 00:54:53,810
what does that hypothesis mean?

812
00:54:53,810 --> 00:54:56,600
So that at least you have
some complete theorem to take.

813
00:55:00,408 --> 00:55:02,200
There are several
versions of this theorem,

814
00:55:02,200 --> 00:55:05,110
so let me give you one
which I really like,

815
00:55:05,110 --> 00:55:08,910
because it has a fairly
clean hypothesis.

816
00:55:08,910 --> 00:55:14,350
And the version is that the
pseudo-randomness condition--

817
00:55:14,350 --> 00:55:16,840
so here it is.

818
00:55:16,840 --> 00:55:23,770
So, a sufficient
pseudo-randomness hypothesis

819
00:55:23,770 --> 00:55:35,550
on gamma, which is that gamma
has the correct number--

820
00:55:35,550 --> 00:55:39,730
"correct" in quotes, because
this is somewhat normative.

821
00:55:39,730 --> 00:55:42,040
So what I'm really
saying is it has,

822
00:55:42,040 --> 00:55:45,040
compared to a random case,
what you would expect.

823
00:55:45,040 --> 00:55:55,390
Densities of all
subgraphs of K--

824
00:55:55,390 --> 00:55:56,575
2, 2, 2.

825
00:56:15,160 --> 00:56:19,750
Having correct density of
H means having H density.

826
00:56:22,720 --> 00:56:30,130
1 plus little 1 times p, raised
to the number of edges of H,

827
00:56:30,130 --> 00:56:32,290
which is what you would
expect in a random case.

828
00:56:36,230 --> 00:56:38,540
So you should think
of there, again,

829
00:56:38,540 --> 00:56:41,265
not being just one graph,
but a sequence of graphs.

830
00:56:41,265 --> 00:56:42,890
You can also equivalently
write it down

831
00:56:42,890 --> 00:56:45,380
in terms of deltas and epsilons
having error parameters.

832
00:56:45,380 --> 00:56:47,900
But I like to think of it
having a sequence of graphs,

833
00:56:47,900 --> 00:56:51,260
just as in what we did
for quasi-random graphs.

834
00:56:51,260 --> 00:56:56,540
If your gamma has this
pseudo-randomness condition,

835
00:56:56,540 --> 00:57:00,688
which is we're in
this sparse setting.

836
00:57:00,688 --> 00:57:02,230
So if you try to
compare this to what

837
00:57:02,230 --> 00:57:05,260
we did for quasi-random
graphs, you might get confused.

838
00:57:05,260 --> 00:57:07,360
Because there, having
the correct C4 count

839
00:57:07,360 --> 00:57:09,520
already implies everything.

840
00:57:09,520 --> 00:57:11,710
This condition, it
actually does already

841
00:57:11,710 --> 00:57:15,464
include having the
correct C4 count.

842
00:57:15,464 --> 00:57:19,140
So K 2, 2, 2 is this
graph over here.

843
00:57:21,730 --> 00:57:27,480
And I'm saying that if it
has the correct density of H,

844
00:57:27,480 --> 00:57:38,450
whenever H is a
subgraph, of K 2, 2, 2--

845
00:57:38,450 --> 00:57:41,880
then it has a correct density.

846
00:57:41,880 --> 00:57:45,552
So in particular, it already
has a C4 count, but I want more.

847
00:57:45,552 --> 00:57:47,350
And it turns out this
is genuinely more,

848
00:57:47,350 --> 00:57:50,110
because in a sparse setting,
having the correct C4

849
00:57:50,110 --> 00:57:55,431
count is not equivalent to other
notions of pseudo-randomness.

850
00:57:58,380 --> 00:58:00,520
So this is a hypothesis.

851
00:58:00,520 --> 00:58:03,256
So if I start with a
sequence of gammas,

852
00:58:03,256 --> 00:58:07,190
I have the correct
counts of K 2, 2, 2s

853
00:58:07,190 --> 00:58:09,860
as well as subgraphs
of K 2, 2, 2s.

854
00:58:09,860 --> 00:58:14,210
Then I claim that that
pseudo-random host is good

855
00:58:14,210 --> 00:58:18,260
enough to have a
counting lemma--

856
00:58:18,260 --> 00:58:19,691
at least for triangles.

857
00:58:22,990 --> 00:58:24,067
Any questions?

858
00:58:30,150 --> 00:58:33,190
Now, you might want to ask for
some intuitions about where

859
00:58:33,190 --> 00:58:35,230
this condition comes from.

860
00:58:35,230 --> 00:58:38,460
The proof itself
takes a few pages.

861
00:58:38,460 --> 00:58:39,850
I won't try to do it here.

862
00:58:39,850 --> 00:58:44,080
I might try to give you some
intuition how the proof might

863
00:58:44,080 --> 00:58:46,090
go and also what
are the difficulties

864
00:58:46,090 --> 00:58:50,520
you might run into when you
try to execute this proof.

865
00:58:50,520 --> 00:58:54,000
But, at least how
I think of it is

866
00:58:54,000 --> 00:59:00,930
that this K 2, 2, 2 condition
plays a role similar to how

867
00:59:00,930 --> 00:59:03,870
previously, in dense
quasi-random graphs,

868
00:59:03,870 --> 00:59:06,540
we had this somewhat
magical looking

869
00:59:06,540 --> 00:59:09,860
C4 condition,
which can be viewed

870
00:59:09,860 --> 00:59:16,280
as a doubled version of an edge.

871
00:59:16,280 --> 00:59:19,430
So actually, the technical
name is called a blow-up.

872
00:59:19,430 --> 00:59:22,180
It's a blow-up of an edge.

873
00:59:22,180 --> 00:59:30,280
Whereas the K 2, 2, 2 condition
is a 2 blow-up of a triangle.

874
00:59:35,090 --> 00:59:41,210
And this 2 blow-up hypothesis is
some kind of a graph theoretic

875
00:59:41,210 --> 00:59:43,220
analogue of controlling
second moment.

876
00:59:53,720 --> 00:59:56,390
Just as knowing the variance
of a random variable--

877
00:59:56,390 --> 00:59:57,920
knowing its second moment--

878
00:59:57,920 --> 01:00:00,200
helps you to control
the concentration

879
01:00:00,200 --> 01:00:03,770
of that random variable, showing
that it's fairly concentrated.

880
01:00:03,770 --> 01:00:07,130
And it turns out that having
this graphical second moment

881
01:00:07,130 --> 01:00:11,600
in this sense also allows you
to control its properties so

882
01:00:11,600 --> 01:00:14,610
that you can have nice tools,
like the counting lemma.

883
01:00:20,500 --> 01:00:22,450
So let me explain some
of the difficulties.

884
01:00:22,450 --> 01:00:26,390
If you try to run the original
proof of the triangle removal

885
01:00:26,390 --> 01:00:31,010
lemma for the sparse
setting, what happens?

886
01:00:31,010 --> 01:00:36,690
So if you start with a vertex--

887
01:00:36,690 --> 01:00:39,940
so remember how the proof of
triangle removal lemma went.

888
01:00:39,940 --> 01:00:46,520
You start with this set-up
and you pick a typical vertex.

889
01:00:46,520 --> 01:00:51,390
This typical vertex has lots
of neighbors to the left

890
01:00:51,390 --> 01:00:53,980
and lots of neighbors
to the right.

891
01:00:53,980 --> 01:00:58,290
And here, a lot means
roughly the edge density

892
01:00:58,290 --> 01:01:00,570
times the number of vertices--

893
01:01:00,570 --> 01:01:04,630
and a lot of vertices over here.

894
01:01:04,630 --> 01:01:06,780
And then you say that,
because these are two fairly

895
01:01:06,780 --> 01:01:11,280
large vertex sets, there are
lots of edges between them

896
01:01:11,280 --> 01:01:16,980
by the hypotheses on
epsilon regularity,

897
01:01:16,980 --> 01:01:19,660
between the bottom two sets.

898
01:01:19,660 --> 01:01:22,120
But now, in the
sparse setting, we

899
01:01:22,120 --> 01:01:26,580
have an additional factor of p.

900
01:01:26,580 --> 01:01:31,180
So these two sets
are now quite small.

901
01:01:31,180 --> 01:01:33,040
They're much smaller
than what you

902
01:01:33,040 --> 01:01:36,280
can guarantee from the
definition of epsilon,

903
01:01:36,280 --> 01:01:38,170
p regular.

904
01:01:38,170 --> 01:01:42,340
So you cannot conclude from
them being epsilon regular that

905
01:01:42,340 --> 01:01:46,120
there are enough edges between
these two very small sets.

906
01:01:46,120 --> 01:01:49,390
So the strategy of proving
the triangle removal lemma

907
01:01:49,390 --> 01:01:51,410
breaks down in the
sparse setting.

908
01:02:00,950 --> 01:02:04,040
In general-- not
just for triangles,

909
01:02:04,040 --> 01:02:09,010
but for other H's as well--

910
01:02:09,010 --> 01:02:12,760
we also have this
counting lemma.

911
01:02:12,760 --> 01:02:14,210
So, the sparse counting lemma.

912
01:02:22,245 --> 01:02:24,990
And also the triangle case,
which I stated earlier.

913
01:02:24,990 --> 01:02:29,660
So this is drawing work due
to David Colin, Jacob Fox,

914
01:02:29,660 --> 01:02:31,820
and myself.

915
01:02:31,820 --> 01:02:33,840
Says that there
is a county lemma.

916
01:02:33,840 --> 01:02:35,370
So let me be very informal.

917
01:02:35,370 --> 01:02:52,230
So, that there exists a sparse
counting lemma for counting H,

918
01:02:52,230 --> 01:02:56,580
in this set-up as before.

919
01:02:56,580 --> 01:03:10,640
If gamma has a
pseudo-random property

920
01:03:10,640 --> 01:03:29,140
of containing the correct
density of all subgraphs

921
01:03:29,140 --> 01:03:40,260
of the 2 blow-up of H.

922
01:03:40,260 --> 01:03:44,370
Just as in the triangle,
the 2 blow-up is K 2, 2, 2.

923
01:03:44,370 --> 01:03:52,950
In general, the 2 blow-up
takes a graph, H, and then

924
01:03:52,950 --> 01:03:58,920
doubles every vertex
and puts in four edges

925
01:03:58,920 --> 01:04:02,940
between each pair of vertices.

926
01:04:02,940 --> 01:04:11,420
So that's the 2 blow-up of H.

927
01:04:11,420 --> 01:04:16,010
If your gamma has pseudo-random
properties concerning

928
01:04:16,010 --> 01:04:18,530
counting subgraphs
of this 2 blow-up,

929
01:04:18,530 --> 01:04:22,850
then you can obtain a
counting lemma for H itself.

930
01:04:26,710 --> 01:04:27,628
Any questions?

931
01:04:32,410 --> 01:04:35,090
OK, so let's take this counting
lemma for granted for now.

932
01:04:38,690 --> 01:04:45,760
How do we proceed to proving the
sparse triangle removal lemma?

933
01:04:45,760 --> 01:04:48,730
Well, I claim that actually
it's the same proof where

934
01:04:48,730 --> 01:04:52,840
you run the usual simulated
regularity proof of triangle

935
01:04:52,840 --> 01:04:53,980
removal lemma.

936
01:04:53,980 --> 01:04:55,840
But now, with all
of these extra tools

937
01:04:55,840 --> 01:04:58,180
and these extra
hypotheses, you then

938
01:04:58,180 --> 01:05:01,100
would obtain the
sparse triangle removal

939
01:05:01,100 --> 01:05:03,640
lemma, which I stated earlier.

940
01:05:03,640 --> 01:05:05,560
And the hypothesis
that I left out--

941
01:05:05,560 --> 01:05:08,540
the sufficiently pseudo-random
hypothesis on gamma--

942
01:05:08,540 --> 01:05:12,520
is precisely this
hypothesis over here,

943
01:05:12,520 --> 01:05:14,070
as required by the
counting lemma.

944
01:05:23,370 --> 01:05:25,560
And once you have
that, then you can

945
01:05:25,560 --> 01:05:31,980
proceed to prove a relative
version of Roth's theorem--

946
01:05:31,980 --> 01:05:35,070
and also, by extension,
two hyper-graphs--

947
01:05:35,070 --> 01:05:37,953
also a relative version
of Szemeredi's theorem.

948
01:05:41,200 --> 01:05:43,680
So, recall that the Roth's
theorem tells you that if you

949
01:05:43,680 --> 01:05:45,540
have a sufficiently large--

950
01:05:45,540 --> 01:05:49,920
so let me first write
down Roth's theorem.

951
01:05:49,920 --> 01:05:53,440
And then I'll add in the extra
relative things in yellow.

952
01:05:53,440 --> 01:05:59,340
So if I start with A,
the subset of z mod N,

953
01:05:59,340 --> 01:06:06,640
such that A has size--

954
01:06:06,640 --> 01:06:08,170
at least delta n.

955
01:06:14,500 --> 01:06:16,870
So then, Roth's
theorem tells us that A

956
01:06:16,870 --> 01:06:19,780
contains at least one three-term
arithmetic progression.

957
01:06:19,780 --> 01:06:23,020
But actually, you can
boost that theorem.

958
01:06:23,020 --> 01:06:25,720
And you've seen some
examples of this in homework.

959
01:06:25,720 --> 01:06:27,910
And also our proofs also
do this exact same thing.

960
01:06:27,910 --> 01:06:30,640
If you look at any of the
proofs that we've seen so far,

961
01:06:30,640 --> 01:06:35,110
it tells us that A not only
contains one single 3Ap,

962
01:06:35,110 --> 01:06:46,070
but it contains many
3Ap's, where C is

963
01:06:46,070 --> 01:06:48,693
some number that is positive.

964
01:06:53,700 --> 01:06:55,470
So you can obtain
this by the versions

965
01:06:55,470 --> 01:06:58,273
we've seen before, either
by looking at a proof--

966
01:06:58,273 --> 01:06:59,940
problem is in the the
proof gift stack--

967
01:06:59,940 --> 01:07:04,275
or by using the black box
version of Roth's theorem.

968
01:07:04,275 --> 01:07:06,150
And then there's a super
saturation argument,

969
01:07:06,150 --> 01:07:10,220
which is similar to things
you've done in the homework.

970
01:07:10,220 --> 01:07:12,390
What we would like is
a relative version.

971
01:07:16,130 --> 01:07:22,020
And a relative version will
say that if you have a set, S,

972
01:07:22,020 --> 01:07:24,500
which is sufficiently
pseudo-random.

973
01:07:32,000 --> 01:07:35,390
And S has density, p.

974
01:07:38,990 --> 01:07:40,600
Here, [INAUDIBLE].

975
01:07:43,860 --> 01:07:54,260
And now A is a subset of S.
And A has size at least delta,

976
01:07:54,260 --> 01:08:01,280
that of S. Then, A contains
still lots of 3Ap's, but I

977
01:08:01,280 --> 01:08:03,580
need to modify the
quantity, because I

978
01:08:03,580 --> 01:08:04,880
am looking at density, p.

979
01:08:09,580 --> 01:08:11,920
So this statement is
also true if you're

980
01:08:11,920 --> 01:08:15,470
putting the appropriate
hypothesis into sufficiently

981
01:08:15,470 --> 01:08:17,560
pseudo-random.

982
01:08:17,560 --> 01:08:21,149
And what should
those hypotheses be?

983
01:08:21,149 --> 01:08:23,670
So think about the proof
of Roth's theorem--

984
01:08:23,670 --> 01:08:25,770
the one that we've done--

985
01:08:25,770 --> 01:08:27,120
where you set up a graph.

986
01:08:30,620 --> 01:08:31,710
So, you set up this graph.

987
01:08:34,760 --> 01:08:38,779
So, one way to do
this is that you

988
01:08:38,779 --> 01:08:42,920
say that you put in edges
between the three parts--

989
01:08:42,920 --> 01:08:44,930
x, y, and z.

990
01:08:44,930 --> 01:08:49,830
So the vertex sets are
all given by z mod N.

991
01:08:49,830 --> 01:09:03,770
And you put in an edge between
x and y, if 2x plus y lies in S.

992
01:09:03,770 --> 01:09:07,819
Pulling the edge between
x and z-- if x minus z

993
01:09:07,819 --> 01:09:11,420
lies in S and a third
edge between y and z,

994
01:09:11,420 --> 01:09:19,420
if minus y minus 2z
lies in S. So this

995
01:09:19,420 --> 01:09:24,220
is a graph that we constructed
in the proof of Roth's theorem.

996
01:09:24,220 --> 01:09:29,520
And when you construct this
graph, either for S or for A--

997
01:09:29,520 --> 01:09:30,760
as we did before--

998
01:09:30,760 --> 01:09:34,710
then we see that the
triangles in this graph

999
01:09:34,710 --> 01:09:38,720
correspond precisely to
the 3Ap's in the set.

1000
01:09:41,550 --> 01:09:46,380
So, looking at the triangle
counting lemma and triangle

1001
01:09:46,380 --> 01:09:48,330
removal lemma-- the
sparse versions--

1002
01:09:48,330 --> 01:09:52,290
then you can read out what type
of pseudo-randomness conditions

1003
01:09:52,290 --> 01:09:55,000
you would like on S--

1004
01:09:55,000 --> 01:09:57,690
so, from this graph.

1005
01:09:57,690 --> 01:10:00,330
So, we would like
a condition, which

1006
01:10:00,330 --> 01:10:02,400
says that this graph here--

1007
01:10:08,550 --> 01:10:11,570
which we'll call gamma sub S--

1008
01:10:11,570 --> 01:10:21,123
to have the earlier
pseudo-randomness hypotheses.

1009
01:10:26,060 --> 01:10:29,278
And you can spell this out.

1010
01:10:29,278 --> 01:10:30,390
And let's do that.

1011
01:10:30,390 --> 01:10:31,640
Let's actually spell this out.

1012
01:10:31,640 --> 01:10:33,940
So what does this mean?

1013
01:10:33,940 --> 01:10:38,610
What I mean is S, being
a subset of Z mod N--

1014
01:10:38,610 --> 01:10:45,860
we say that it satisfies
what's called a 3-linear forms

1015
01:10:45,860 --> 01:10:46,862
condition.

1016
01:10:55,226 --> 01:11:03,770
If, for uniformly
chosen random x0,

1017
01:11:03,770 --> 01:11:11,637
x1, y0, y1, z0, z1
elements of z mod nz.

1018
01:11:19,440 --> 01:11:21,540
Think about this K 2, 2, 2.

1019
01:11:21,540 --> 01:11:23,680
So draw a K 2, 2, 2 up there.

1020
01:11:23,680 --> 01:11:26,330
So what are the edges
corresponding to the K 2, 2, 2?

1021
01:11:26,330 --> 01:11:29,940
So they correspond to the
following expressions--

1022
01:11:29,940 --> 01:11:33,740
minus y0 minus 2z0--

1023
01:11:33,740 --> 01:11:37,170
minus y1 minus 2z0--

1024
01:11:37,170 --> 01:11:40,560
minus y0 minus 2z1--

1025
01:11:40,560 --> 01:11:43,070
minus y1 minus 2z1.

1026
01:11:43,070 --> 01:11:47,170
So those are the edges
corresponding to the bottom.

1027
01:11:47,170 --> 01:11:51,680
Draw C4 across the
bottom two vertex sets.

1028
01:11:51,680 --> 01:11:54,510
But then there are
two more columns.

1029
01:11:54,510 --> 01:11:56,370
And I'll just write
some examples,

1030
01:11:56,370 --> 01:11:57,690
but you can fill in the rest.

1031
01:12:06,670 --> 01:12:11,480
OK, so there are at
least 12 expressions.

1032
01:12:11,480 --> 01:12:24,460
And what we would like is that,
for random, the probability

1033
01:12:24,460 --> 01:12:30,340
that all of these numbers
are contained in S

1034
01:12:30,340 --> 01:12:44,434
is within 1 plus little 1
factor of the expectation,

1035
01:12:44,434 --> 01:12:48,220
if S were a random set.

1036
01:12:54,700 --> 01:12:58,130
In other words, in this
case, it's p raised to 12--

1037
01:12:58,130 --> 01:13:04,980
random set of density, p.

1038
01:13:04,980 --> 01:13:22,110
And furthermore, the same
holds if any subset of these 12

1039
01:13:22,110 --> 01:13:23,550
expressions are erased.

1040
01:13:40,270 --> 01:13:42,740
Now, I want you to use your
imagination and think about

1041
01:13:42,740 --> 01:13:48,090
what the theorem would look like
for not 3Ap's, but for 4Ap's--

1042
01:13:48,090 --> 01:13:49,960
and also for k-Ap's in general.

1043
01:13:49,960 --> 01:14:03,350
So there is a relative Szemeredi
theorem, which tells you

1044
01:14:03,350 --> 01:14:08,080
that if you start with S--

1045
01:14:08,080 --> 01:14:15,460
so here, we fix K. If
you start with this S,

1046
01:14:15,460 --> 01:14:19,177
that satisfies the
k-linear forms condition.

1047
01:14:24,050 --> 01:14:31,840
And A is a subset of S
that is fairly large.

1048
01:14:36,760 --> 01:14:43,350
Then A has k-Ap.

1049
01:14:43,350 --> 01:14:45,310
So I'm being
slightly sloppy here,

1050
01:14:45,310 --> 01:14:47,050
but that's the spirit
of the theorem--

1051
01:14:47,050 --> 01:14:48,880
that you have this
Szemeredi theorem

1052
01:14:48,880 --> 01:14:51,050
inside a sparse
pseudo-random set,

1053
01:14:51,050 --> 01:14:53,520
as long as the
pseudo-random set satisfies

1054
01:14:53,520 --> 01:14:55,290
this k-linear forms condition.

1055
01:14:55,290 --> 01:14:57,110
And that k-linear
forms condition

1056
01:14:57,110 --> 01:14:59,740
is an extension of this
3-linear forms condition, where

1057
01:14:59,740 --> 01:15:04,120
you write down the proof that
we saw for Szemeredi's theorem,

1058
01:15:04,120 --> 01:15:05,560
using hyper-graphs.

1059
01:15:05,560 --> 01:15:07,660
Write down the
corresponding linear forms--

1060
01:15:07,660 --> 01:15:10,736
you expand them out and then
you write down this statement.

1061
01:15:18,090 --> 01:15:23,180
So this is basically what
I did for my PhD thesis.

1062
01:15:23,180 --> 01:15:25,510
So we can ask, well, what
did Green and Tao do?

1063
01:15:25,510 --> 01:15:28,990
So they had the original
theorem back in 2006.

1064
01:15:28,990 --> 01:15:32,350
So their theorem, which also was
a relative Szemeredi theorem,

1065
01:15:32,350 --> 01:15:35,380
has some additional,
more technical hypotheses

1066
01:15:35,380 --> 01:15:39,190
known as correlation conditions,
which I won't get into.

1067
01:15:39,190 --> 01:15:43,990
But at the end of the day, they
constructed these pseudoprimes.

1068
01:15:43,990 --> 01:15:48,280
And then they verified that
those pseudoprimes satisfied

1069
01:15:48,280 --> 01:15:53,410
these required
pseudo-randomness hypotheses--

1070
01:15:53,410 --> 01:15:56,660
that those pseudoprimes
satisfied these linear forms

1071
01:15:56,660 --> 01:16:00,590
conditions, as well as
their now-extraneous

1072
01:16:00,590 --> 01:16:03,630
additional pseudo-randomness
hypotheses.

1073
01:16:03,630 --> 01:16:06,530
And then combining this
combinatorial theorem

1074
01:16:06,530 --> 01:16:08,210
with that number
adiabatic result.

1075
01:16:08,210 --> 01:16:14,760
You put them together, you
obtain the Green-Tao theorem,

1076
01:16:14,760 --> 01:16:18,860
which tells you not just that
the primes contain arbitrarily

1077
01:16:18,860 --> 01:16:22,910
long arithmetic progressions,
but any positive density

1078
01:16:22,910 --> 01:16:26,690
subset of the primes
also contains arbitrarily

1079
01:16:26,690 --> 01:16:30,245
long arithmetic progressions.

1080
01:16:30,245 --> 01:16:32,120
All of these theorems--
now, if you pass down

1081
01:16:32,120 --> 01:16:34,900
to a relatively dense subset,
it still remains true.

1082
01:16:38,180 --> 01:16:39,010
Any questions?

1083
01:16:42,370 --> 01:16:43,850
So this is the general method.

1084
01:16:43,850 --> 01:16:46,210
So the general method is you
have the sparse regularity

1085
01:16:46,210 --> 01:16:46,900
method.

1086
01:16:46,900 --> 01:16:49,540
And provided that you have
a good counting lemma,

1087
01:16:49,540 --> 01:16:52,840
you can transfer the entire
method to the sparse setting.

1088
01:16:52,840 --> 01:16:56,720
But getting the counting lemma
is often quite difficult.

1089
01:16:56,720 --> 01:16:59,950
And there are still
interesting open problems--

1090
01:16:59,950 --> 01:17:03,130
in particular, what kind of
pseudo-randomness hypotheses

1091
01:17:03,130 --> 01:17:06,000
do we really need?

1092
01:17:06,000 --> 01:17:08,430
Another thing is that
you don't actually

1093
01:17:08,430 --> 01:17:10,960
have to go through
regularity yourself.

1094
01:17:10,960 --> 01:17:12,930
So there is an additional
method-- which,

1095
01:17:12,930 --> 01:17:16,710
unfortunately I don't
have time to discuss--

1096
01:17:16,710 --> 01:17:23,040
called transference,
where the story I've told

1097
01:17:23,040 --> 01:17:26,700
you is that you
look at the proof

1098
01:17:26,700 --> 01:17:29,910
of Roth's theorem, the proof
of Szemeredi's theorem.

1099
01:17:29,910 --> 01:17:32,850
And you transfer the
methods of those proofs

1100
01:17:32,850 --> 01:17:34,740
to the sparse setting.

1101
01:17:34,740 --> 01:17:36,260
And you can do that.

1102
01:17:36,260 --> 01:17:39,200
But it turns out, you can
do something even better--

1103
01:17:39,200 --> 01:17:42,050
is that you can
transfer the results.

1104
01:17:42,050 --> 01:17:44,840
And this is what
happens in Green-Tao.

1105
01:17:44,840 --> 01:17:49,520
If you look at Szemeredi's
theorem as a black-box theorem

1106
01:17:49,520 --> 01:17:52,010
and you're happy
with its statement,

1107
01:17:52,010 --> 01:17:56,570
you can use these
methods to transfer

1108
01:17:56,570 --> 01:17:59,750
that result as a black
box without knowing

1109
01:17:59,750 --> 01:18:04,520
its proof to the sparse
pseudo-random setting.

1110
01:18:04,520 --> 01:18:07,520
And that sounds almost
too good to be true,

1111
01:18:07,520 --> 01:18:10,310
but it's worth
seeing how it goes.

1112
01:18:10,310 --> 01:18:14,520
And if you want to learn
more about this subject,

1113
01:18:14,520 --> 01:18:22,560
there's a survey by Colin Fox
and by myself called "Green-Tao

1114
01:18:22,560 --> 01:18:24,073
theorem, an exposition."

1115
01:18:34,430 --> 01:18:38,980
So, where you'll find a
self-contained complete proof

1116
01:18:38,980 --> 01:18:41,935
of the Green-Tao theorem,
except no modulo--

1117
01:18:41,935 --> 01:18:43,560
the proof of Szemeredi's
theorem, which

1118
01:18:43,560 --> 01:18:45,650
we've called as a black box.

1119
01:18:45,650 --> 01:18:48,070
But you'll see how the
transference method

1120
01:18:48,070 --> 01:18:48,843
works there.

1121
01:18:48,843 --> 01:18:50,260
And it involves
many of the things

1122
01:18:50,260 --> 01:18:52,580
that we've discussed
so far in this course,

1123
01:18:52,580 --> 01:18:55,960
including discussions of the
regularity method, the counting

1124
01:18:55,960 --> 01:18:56,470
lemma.

1125
01:18:56,470 --> 01:19:00,580
And it will contain a proof of
this sparse triangle counting

1126
01:19:00,580 --> 01:19:02,584
lemma.

1127
01:19:02,584 --> 01:19:03,510
OK, good.

1128
01:19:03,510 --> 01:19:05,390
We stop here.