25062
>> Yuval Peres: Okay. Good afternoon.
tell us about the Margulis expanders.
We're happy to have James Lee
>> James Lee: I'm going to tell you what I'm going to say. So -[laughter] -- no, that makes sense. As I was teaching a course on
spectral graph theory in the spring, and I showed all the students that
are random regular graph is an expander, and I want to do an explicit
construction. And so you hear a lot of things about explicit
construction of expander graphs and zigzag product, but I wanted to do
something that was really sort of, I was hoping I could do something
elementary enough you could do it in one lecture and be completely
understood.
I actually that actually Rhino Dahmler [phonetic] wrote this blog entry
a few years ago that upset Avi Vigerson [phonetic] because Ryan said
everybody says that the zigzag expanders are the first one who could
not be understood, but actually the Gabber-Galil analysis is just a few
pages of analysis -- and Avi got upset because he said nobody ever said
they were the first understandable examples and so on.
So then I looked at the Gabber-Galil analysis and unfortunately it's
elementary, but there's three or four pages of calculations which don't
seem amenable to a lecture. And then there's a follow-up paper of Jim
and Maruka [phonetic] where the proof is more elementary because there
are no complex numbers.
But it's a little bit more complicated. And then even in this sort of
the expander survey of Horey Lineal and Vigerson [phonetic]; they say
here's the analysis of the Margulis graphs of the few pages, but it's
still subtle and mysterious.
So it just turns out that if you take various pieces from various
places and put them together in the right way, then the proof is
almost -- it's almost disillusioningly simple, almost trivial, which is
kind of, which is kind of -- yeah. You always -- when you think about
expanders, you think there's always something deep lying behind there.
Like some kind of exponential sum or a sum product theorem or some
representation theory.
So the point today is that hopefully everything -- almost
say it's like it's very easy to follow. So this would be
minutes at a very leisurely pace. So let me just say who
what's going on here. So in 1973, Margulis presented the
of an explicit family of expanders.
everything I
like 25, 30
are the -first example
And then -- but the analysis used, quote/unquote, deep fax from
representation theory. And then Gabber-Galil in, I guess their general
paper was in '81, their Fox paper was in '79, gave an elementary proof
using just sort of essentially a bunch of inequalities with complex
numbers.
And then Jim Bow and Maruka [phonetic] in '85, but in -- actually, this
was in Fox '79. I think this was in Fox '83, gave an even more
elementary proof, as I said, with no complex numbers but slightly more
complicated.
So today I want to present just what seems to be a proof that has
almost no ideas in it. Okay. So let me remind you what's going on.
We have a graph and a subset. Let's define the expansion constant of
the subset with respect to the graph, and, by the way, of course ask me
questions at any time. And also I don't claim anything here as new.
And I heard a noise when I said that. I think they stopped filming.
Okay. So the expansion constant of the subset S is just a number of
S -- edges that cross from S with the compliment from the size of S and
let's define also overloading notation a bit H of G to be the expansion
constant of G.
So look over all subset size at most half of the expansion of that
subset. Okay. So that's the expansion. That's for G. Let's also
define the corresponding note for functions. So if F is a function in
front of vertices, let's say to the complex numbers for the sake of
what I'm doing here, then the Rayleigh quotient rests -- it's not
Rayleigh, it's Rayleigh, according to the world, that's the correct
pronunciation.
But all right. Okay. Here's the Rayleigh quotient and finally the
spec and eigenvalue of the Laplacian of this graph is just the minimum
of this Rayleigh quotient over all functions that sum to zero. Okay.
So this is the setup. And then let's define what an expander family
is.
So an expander family is a sequence of deregular graphs such that the
second eigenvalues are uniformly bounded or equivalently the expansion
constants of the graph are uniformly modeled. The goal today is to
present an elementary family of graphs with an elementary analysis
showing that they're expanders.
Okay. And I need one thing for this to be elementary. So I assume
everybody sort of believes the following connection between eigenvalues
and expansions. But let me say it in a way that I need it. And again
I'm assuming that you know this sort of that this seems like so natural
to you that you wouldn't question it.
Because that makes the proof -- in order for the things I said about
triviality to be true, that has to hold. So what is this? So we have
a graph G, a subset of the vertices U. Okay. This graph can be -doesn't have to be finite. This graph could be an infinite graph, what
I'm saying here, graph G, U and we have a function from the vertices.
So I need the complex numbers here. Let's say the complex numbers, of
course, it doesn't matter for this for what I'm about to say.
Such that the support of F -- so the place, all the places where F are
non-zero is contained inside U. In fact, it's not clear that I need
this to set U. So let's just say it this way. I have a graph and a
function.
And the function is bounded in L2. So FU squared is less than
infinity. Then there exists a finite subset of the vertices and in
fact this subset is a subset of the support of F.
So this subset has nodes on which contains only nodes on which F takes
a non-zero value. So there exists a subset, such the expansion of the
subset is at most square root 2 times the Rayleigh quotient of F. So
given any function on the graph, which is bounded in L2 there exists a
sub support of F whose expansion is at most -- bounded in terms of the
Rayleigh quotient.
Okay. So now we're ready to start the proof of the theorem. So I'll
state a theorem in a second. First, let me just introduce -- let me
sort of state the main technical lemma, which is very simple. So
suppose S and T are a plane to itself defined by -- so let's say X is
XY maps XY to X plus YY and T of XY is the same thing in the other
coordinate. XY plus X. And also need the inverse of these functions.
Let me just write them for sake of not having to think about what they
are. Okay. Here are the inverse of these functions. So I have ST, S
inverse and T inverse.
And let me define a graph based on these functions. So the graph will
be -- will have vertex set, which is the integer lattice, and the edges
of the graph will be the following: Will connect a vertex XY to -let's do SXYTXY S inverse XY and T inverse XY. Okay. So this graph
has degree at most four. The origin, for instance, has you can think
of it as having four self loops, the origin has no adjacent things.
And now here's the main ->>: Inverse [inaudible].
>> James Lee: Yes, absolutely. Thank you. I planted that so I could
check if you were paying attention. Okay. So now here's the main
technical theorem. Let's call it a lemma. Okay.
So I'm going to prove sort of -- that there's a sequence of expander
graphs. So we need something to expand. So this graph is going to be
under our expanding object. And here's the main proof. The main lemma
for any subset of the integers that doesn't contain zero. So remember
zero doesn't have any neighbors. So let's admit zero. The number of
edges from A to its complement is at least the size of A. Okay?
So in some sense this is infinite graph but sometimes it's infinite
graph is expanding. Okay. This is the claim. And this is the
really -- this is the expander part of the whole proof. Okay. So
let's prove this. The proof is really simple.
So here's the proof. So let's -- here's the plane. This is the plane.
Let's break it up into four quadrants, Q1, Q2, Q3 and Q4, and I want to
partition the plane except for the origin. So I'll include the Y -the X axis in Q1. I don't know if this notation makes any sense. So
just to be clear, Q1 is the set of pairs XY such that X is bigger than
0 and Y is bigger than or equal to 0. That's Q1. It's the quadrant
with the positive X axis.
And similarly for Q2. Q2 is just
and Q 3 is 180 degree rotation of
plane, I mean the integer lattice
sub I to be the intersection of A
the set into quadrants.
in fact the 90 degree rotation of Q1
Q1. So these four sets partition the
minus the origin. And I'll define A
with the Ith quadrant. I'll break
Okay. So have to remember not to write in this no man's land over
here. Here's the claim that I'm going to prove that the number of
edges from A1 to the complement intersected with the first quadrant.
These are edges that go from A1 outside of A1 all inside the first
quadrant. This is at least the size of A1. If I prove that I've
proved a lemma, because this applies separately to A1, A2, A3, and A4
and it's really without loss. This graph is completely -- this graph
is invariant under rotations by 90 degrees.
So I have the -- if I flip or exchange the coordinates the graph is
invariant. So I really think the proof of the first quadrant,
everything is symmetric. And here's the proof for the first quadrant.
Okay. It's simple enough to do here. The first thing is that if I
apply S or T to A1 then I stay inside the first quadrant. That's
straightforward. I have nonnegative coordinates. I add to them I keep
my nonnegative coordinates. That's good. The third claim is actually
they're disjoint. And the reason they're disjoint is because if I draw
the line Y equals X, then S maps everything above this line and T maps
everything below this line, and if you think about the boundary. So
points here will get mapped by which one changes the Y coordinate?
Will get mapped by T up to the diagonal, and of course they'll also get
mapped to themselves by S.
And points -- but points on the diagonal get mapped off the diagonal.
So it's really the case that their images are disjoint. The image of S
and T in this quadrant are disjoint because again S maps everything up
here including this line to here and T maps everything here to here.
Okay. So that's the end of the proof. Because that implies -- let me
do it here. By the way, tell me if it's -- I won't write over here
anymore. Because that implies that.
>>: [inaudible].
>> James Lee: Okay. That implies that this equals this but S and T
are bijections. So this is just twice A1. So when I apply S and T,
the set gets bigger by a factor of two. So I have at least A1 edges
coming out of A1. So that's the end of the proof of the claim and the
end of the proof of the lemma.
So it's really just that -- it's really that you take a set and sort
of -- okay. Half the set goes up here and sort of another copy of the
set goes down here. That happens in every one of the four quadrants.
You get twice as many vertices. So you get an expander.
Okay. So I claim that's it. I mean the proof is not finished. We
haven't even defined a family of graphs yet. But this is the main
technical part of the proof. And in fact this is the part that
corresponds to the mysterious subtle argument in all the other proofs
that involve a few pages of 4A analysis, just this thing.
So now let's see what I want to do.
>>: [inaudible].
>> James Lee: No, no, so of course we haven't -- but you'll see
everything else is now just sort of -- I mean is like parlor tricks.
But there's nothing deep that happens now.
>>: The other question is there's [inaudible].
>> James Lee:
What's that?
>>: At this point.
>> James Lee: All the proofs have some kind of a fundamental flow. If
you try to use discrete Fourier transform, then everything has to be
messy. If you use continuous Fourier transform and you don't use the
discrete Cheeger inequality, then this proof looks much harder than it
actually is. I think that's the essence of it. Although, could argue
that it's -- yeah, that's essentially why it's -- yeah. So okay. All
right. So let's --
>>: You use unique Cheeger even if you just -- I mean now you proved
the combinatorial expansion. If you stick to the combinatorial
expansion ->> James Lee: So maybe you'll say at some point there's something
subtle that happens. But at some point we'll have to pass to -- we'll
have to pass to functions not sets, just because basically because if
you take the Fourier transform of a set you won't get a set in the
Fourier basis. Okay. Now I said Fourier like eight times, so it seems
like something complicated is going to happen but it's not true.
Okay. So let me -- so is this okay? Okay. So let me define -- okay.
We have the infinite graph here. So let me define a finite family of
graphs. It's the most obvious thing you can think of, which is take
everything mod N. And that's the infinite family of graphs. Okay. So
the vertex set is just going to be discrete Torus. And the edge set,
now you have to be a little careful. The edge set is going to be these
edges plus the neighbor edges. So the edge set is XY is adjacent to X
plus minus 1Y XY plus minus 1. X plus minus YY. So these are the S
edges and X -- okay. That's the graph. And now let's even state the
main theorem here. Theorem lambda two. Okay. There exists a C such
as all N lambda NG is bigger than C that's the main theorem.
So now let's prove it. Okay. I'll keep discrete Cheeger inequality
where I need it. Okay. So now you -- so people might take issue with
this part of the proof but I claim this actually makes the proof less
interesting. So let me -- so I'm going to want to use the continuous
sort of the two Torus, and I'm going to think about the Hilbert space
of functions here. So this is the space of all functions from the
Torus to the complex numbers such that the two norm is bounded.
Okay. So, again, the two Torus is just -- right. The unit squared
with the sides identified. Okay. That's the two Torus. And now let
me define -- this is just a number. There's not going to be any
operator here. But lambda two in the analogous way to be the minimum
over functions here of -- okay. So I should take F minus F compose
with T, F minus F composed with -- sorry, S and T. And subject to the
constraint when you integrate F over the Torus equals 0. So this is
the second eigenvalue of the Torus with this strange SNT. But this is
just a number.
>>: [inaudible].
>> James Lee:
Which maze of things?
>>: Torus and the ->> James Lee: Oh, yeah. So this is typesetting issue that I never -that I just encountered now. I just -- my blackboard T looks really
bad. Looks like a pi. I wasn't sure how to -- I see. So let's -- I
guess we've already used -- let's replace T by something else.
Script T is dangerous.
right.
How does one do script.
That's a J.
All
>>: It's like a ->> James Lee:
It's --
>>: Use the other direction.
>> James Lee: The other direction, like this? That looks like -- I'm
going to make up a symbol. Math Cal T, is that all right? I'm not
sure I can ->>: [inaudible].
>> James Lee:
All right.
>>: Put a tau.
>> James Lee: I'll use the blackboard. It looks like -- it's not
going to need to use a notation more than a few seconds. Looks like a
pi. Everybody has to get over this fact. I apologize. All right. If
you have bad eyesight like me, then just squint and it looks better.
Okay. Right. So here's the -- this is just a number. This is not
second eigenvalue or anything. I'm just using it for analogy. And
here's the second main lemma, which is that the second and final main
lemma, that there exists an epsilon such that for all N lambda 2 of GN
is at least lambda 2 of the Torus. With these S and T operators.
Okay. Okay.
So this should be reassuring, because it implies that we're not going
to be using any number theory in our analysis of this object. You
could look at this X plus Y, Y plus X, oh, this is somehow using the
pseudorandom nature of the primes with respect to, I don't know, but if
we replace it by Torus, there's really no, the modulus N has no effect.
It's nice it tells you why the graphs are all expanders because they're
essentially just converging to the Torus. Okay? So let me prove this.
The proof is very simple.
And again I claim that sort of this is -- you should be happy to see
this, because this means that nothing -- you know, once you're here
there's nothing tricky going on. There's no number theory at some
point, right? Okay. So what's the proof? So take any function from
the vertex of this set of graph GN. Say the real line -- here it's
fine to take it to the real line. Okay. Because I guess I defined it
in terms of complex numbers. It doesn't -- it doesn't matter. Okay.
Such that the sum is equal to zero. And now I just want to define an
extension of this to the Torus, and I'm going to define it in the
simplest possible, I'll define it in a very simple way at least. So
here's the Torus. I can put ZN on the Torus, I can put the grid lines
corresponding to sort of mod N. And then so I want to define some
extension of F to the whole Torus. So I think about my graph as
sitting inside the Torus. I want to define this extension. So suppose
that -- okay. So suppose that I have some square with corners U1, U 2,
U 3, U 4, and I have some Z in the middle here and I want to define the
extension to this point Z. You can do this in any sufficiently nice
way you want. Just interpolate. I guess the cleanest way to do it is
something like this. So I'll explain what this says in a second. So
I'm just going to write it as an average of the four points. How am I
writing the average? Well, here it looks like the L infinity to norm
looks nicely. So basically for every point I just look at the L
infinity, sort of the maximum of the X and Y distances to the four
corners and I use those for the weights on the F. And now it has a
nice property that if I look at a point here, the extension only
depends on U 4 and U 3 because there's no weight corresponding to G1
and G2. So in fact it's consistent along all the -- this is
consistently defines the extension F. This is not so important. This
is just some set of -- define F, right, define F ->>: But I don't see why is F -- Q1 all ->>: So here the infinity known to 1 is U1.
goes away.
So the coefficient of U1
>>: Take any smooth function that vanishes when you -- okay. This is
not -- okay. So now we just have to -- now we just want to prove that
the sort of the Rayleigh quotient for F tilde is at most the Rayleigh
quotient for F. So we need to verify -- so let's first verify that F
integrates to zero. That's straightforward just by the symmetry when
you integrate you pick up every corner with the same weight and the sum
of the S to zero. So that's zero. That's easy. In fact, the full
number is S and T. The second property I claim is that integral F
squared over Torus is at least some constant times N squared times the
sum of FU squared. I claim this. And this is again very simple. Take
any square -- say take the maximum corner and take a little box around
it. You know, a box of side length one-eighth, then in here the
integral of the square F tilde will be proportional to the F of U, F of
U1 squared.
And the volume of the little box is a constant time over N squared. So
when you integrate F tilde in the box you pick up the maximum corner at
least. And so sum over all the squares, you get this kind of thing.
All right. So then that last that you need to verify is that F -- so
feel free to argue with me at any point about this is elementary, but
hopefully all this is really like -- okay. It's very -- all right. So
the last point is to prove that this is at most the constant times
that.
This is again very easy. So what's the reason for this? Well, take
any square, any other square, and suppose I have Z here and up here is
S of Z. So now basically I want to say that whenever the Torus has to
pay over here, this sum over here will also have to pay.
So how do you do that? Well, the first thing is just notice in the
graph -- so there are eight points here corresponding to the eight
corners. Any one of these points can reach any other point in the
graph in at most five steps. So you might have to take S somewhere and
then take an edge to get back here and then come up here. But most
five steps any point here can reach any point here. This is the only
place, by the way, where we're going to use -- where we use the fact
that we put these edges in there. It's just to make the graph look
like a Torus. So what it means is that so now I want to say that say
this distance is say F tilde has a difference of delta between here and
here. I want to say there exists at least one corner -- there exists
at least two corners whose distance is at least delta over four. And
the reason for that is just that well suppose all of these corners
within delta over four and all of these are within delta over four, the
value of F tilde is the convex combination of the corners. So that
means that the value of F tilde here and here is also within delta over
four. So we've -- so there must be two corners that differ a lot.
And since all the corners are connected by path of at length at most 5,
some edge in this sum has to pay for this difference. Okay. And
everything I wrote here is correct except for the fact that I want,
when you integrate this only has volume one over N squared. So you
have to pay -- you have to get more N squared. So I claim that's
the -- again, basically I claim that this lemma was trivial. Like you
just extend it to F and just observe that it holds. And now, okay,
good. So now the proof -- now we're just left to show that this thing
is bigger than zero. So first let me do it and then I'll explain it.
The proof really only has about one minute left in it. Modulo the
explanation. So what I'm going to use now for the final step is the
Fourier transform, the classical Fourier transform from the Torus to
this. So by L2 Z this is the set of all state conflicts value
functions on the square such that -- okay. Okay. So I'm going to use
this Fourier transform.
And the point is just that this Fourier transform is linear map and
it's an isometry. So now here's what I'll do. So it's a linear map
and it's an isometry.
So what it means is that let's just apply it in here. So how would
applying it look. So let's write F hat for the Fourier transform. The
Fourier transform is linear isometry. I can apply it -- I can take
hats in all these places and none of the distances change.
And it's linear. So I can distribute the hats. And then I don't take
a hat here. But this condition is equivalent to -- I'll see it in a
second -- that the Fourier transform vanishes at zero. Same thing as
saying neglization [phonetic], coefficient of aero is zero. So now you
have this and now there's a claim that I'll justify in a second. You
need the fact that when you apply the Fourier transform to a shift,
it's a shift in the Fourier basis as well. So what does it mean?
The claim is that this is the same thing as -- I guess it turns out
it's T inverse. And I'll argue this in a second but let me finish the
proof first.
Okay. My claim is this. We'll prove this in a second. Now with this
in hand we can just rewrite this. So I'll rewrite it. Lambda two
equals the minimum over F hat of -- okay. Sum over Z and the lattice F
of Z minus F of T inverse Z squared plus F of Z minus -- there should
be hats all over here. S inverse Z squared, divided by some F hat
squared. Subject to the constraint that F hat of zero equals zero.
Okay. All right. So I'm just rewriting -- so I took Fourier
transforms and then I used my claim to write that the Fourier transform
of F hat S is just actually S hat composed with T inverse.
Now, the point is that this is -- I guess we could just wrap it -- this
is exactly the minimum Rayleigh quotient of F hat with respect to our
original infinite graph G on the integers. Our graph that had edges
defined by S and T. Subject to the constraint that is equal to zero.
This is just the Rayleigh quotient of our original graph G. . But now
we know that this is at least half by the lemma combined with the
discrete Cheeger inequality. The discrete Cheeger inequality says that
for the Rayleigh quotient -- what does it say? The Rayleigh quotient
of any F that vanishes at zero is at least the expansion of sets that
don't -- at least half times the square of the expansion of sets that
don't contain zero. This is half times one squared where one is the
expansion constant of sets that don't contain zero. Okay. So I'll
justify the claim in a second.
But that's the proof. You prove this fact about the integer lattice.
Then pass the finite graphs by taking quotients move from your family
of finite graphs to the single, just to the Torus, and then just apply
Fourier transforms and the problem you are left with after you apply
Fourier transforms is exactly the problem on this lattice.
Okay. So I mean I think the proof is simple enough that you can
actually remember it. Like you can recall it later. So I guess let me
just argue about -- let me just remind people. So here I'll do it
right here. What's going on. Okay. So what does the Fourier
transform look like? Okay. So, first of all, what's the Fourier
basis? So for integers M and N you look at functions 2 pi IMXNY,
that's the Fourier basis. And then the point is that you can write
every F as a sum over M and N. The actual fact that this holds was not
proved until actually I wish I remember the history of this. But my
feeling is that for functions of this form that are very, very simple,
it's not too hard to show that.
Okay. So the point is you have this -- you can write any function in
the Fourier basis where this Fourier coefficient is just the inner
product of F and ->>: The emergence of [inaudible].
>> James Lee:
Right.
>>: The original or everything like that.
>> James Lee:
Right.
>>: Don't need.
>>: No, just need the sum of it.
>>: Right.
>>: Just need the fact this is an isometry.
>> James Lee: We just really need the fact that this is -- it's really
simple, right? It's not just -- it's the fact it's Hilbert space and
you have orthogonality of the characters.
>>: The completeness. The basis of the function which is orthogonal to
all the -- not just the fact that ->>: We don't even need that.
We just need the [inaudible] this part.
>> James Lee: And listen none of it matters even if you had some error
you can take the error to be epsilon and take the epsilon to be 0. You
actually can use, do everything with bounded error.
So okay.
So do you disagree?
>>: The fact that the error goes to 0 is completeness.
So --
>> James Lee: Proving it for this function there goes to 0 is binary
trivial. This function is just like, like you can make it piecewise,
very nice.
>>: But the point left when you have the minimum over F hat. So
suppose -- so the isometry fact is that the fact that the isometry is
on two --
>> James Lee: No, you need it to be injected. Everything here can
be -- so I'm proving that this minimum is large, which is stronger than
proving that the image of this thing under the Fourier transform is
large. Right? But in any case, so now you can see that the first
Fourier coefficient is just, it's just the integral of F because it's
constant function. And then what's the other fact you need to see.
Just that if you take -- if you compose -- on-under the Fourier basis
for F and what you get is this with S is this, because now X gets
replaced by X plus Y and so Y picks up an M. So you get that. Okay.
And similarly M plus NN.
And this implies that -- well, okay. So what does it imply? So look
at the Fourier transform of F composed with S. This is sum over NM of
this. But now we have to write MN plus M and then just change the
variables and we get this.
So this is why -- okay. So that proves this claim that when you take
the Fourier transform of F composed with S, then what you get is hat F
composed with T inverse. And similarly for S. So it's really just -these shifts just act as shifts in the Fourier basis. Okay. That's
the proof. Any questions? And in particular I'm curious to know why
people think that was simple.
We didn't do anything hard.
>>: So this thing about the lambda two of the Torus must be some
exercising the [inaudible] geometry or something?
>> James Lee:
not --
No, because this S and T, these are not like -- this is
>>: Not standard.
>> James Lee: No. These are some strange operators. I'm not sure
what you would want to consider those. They don't reflect the geometry
of the -- in some sense that's where the expansion, the
pseudorandomness comes from how do you start with a two-dimensional
object and get an expander you have to have some operators that don't
behave or are not geometrically well behaved. These things jump across
the space.
>>: Right.
Just is it too [inaudible].
>> James Lee: Okay. But okay. So that's -- it seems to be -- okay.
At least if you sort of -- if this is second nature to you, then all
the steps of the proof really just flow very nicely. Okay. Maybe you
guys have a difference of opinion.
>>: I see it well that you sort of apply -- you don't get your graph
really as a quotient of Z squared. But you take your graph. You
approximate Torus within the Fourier transform gets you to Z squared.
>> James Lee: Yes there's two points. The lattice -- Z squared
expands. Again it's almost trivial, just because -- right? So in
general if you took the standard generators on the lattice, then, of
course, you wouldn't have an expander, because the boundary grows
linearly and the volume grows quadratically but the boundary grows
along with the size of the ball. So you are sort of fixing it by
making the boundary much thicker because you're taking these huge
jumps. So this is sort of second nature. Now the question I really
want to ask is when you take -- can any funny things happen when you
take torsions so somehow you get this miraculous thing where all these
jumps go to the same place. And once you get to here, I mean this sort
of says that nothing miraculous -- this says nothing miraculous is
happening. Because the N doesn't play any kind of role. So there's
two steps. The expansion for Z squared with these S and T operators
and then the fact that wraparound doesn't hurt the expansion.
Okay. So the fact that wraparound doesn't hurt the expansion, maybe
that's still a little mysterious.
>>: You have to give the function I think is you start off with just a
combinatorial expansion and that's what you end up with as well.
>> James Lee: So in fact you can -- there's another way to do the
proof. Actually, which is you could start with combinatorial
expansion. You could maintain combinatorial expansion and just go to
some kind of continuous expansion for sets in the Torus.
That also works. I think this is also simpler when you allow arbitrary
functions it's actually a little bit easier than playing around with a
set to make sure they all line up. But then when you take Fourier
transforms, eventually you get arbitrary functions again. So you can't
say the sets the whole time, it seems.
>>: So what does this have to do with Margulis's formulation?
>> James Lee:
That's a good question.
>>: [inaudible].
>> James Lee: Yeah. You know, this is one of the problems with this
paper being in Russia and also inaccessible because at some point I
looked at it -- no, no, there's an English translation. But then every
time -- the past five times I went back and looked at Margulis a paper
I don't get it so I no longer remember what -- I mean, you can see the
matrices that correspond to S and T. The question is, yeah. I want to
remember. So I'll tell you when I figure it out what Margulis's
construction was and what the representation theoretic machinery ->>: [inaudible].
>> James Lee: I guess it depends on whatever representation theoretic
result he was using, whether he needed primes. But here, yeah, there
are no primes, right?
>>: The deduction from [inaudible] property [inaudible] that was not ->> James Lee: Yes, that's -- so...I mean, in that sense you can say
it's actually -- I don't know the history of this very well. Is he the
one who observes that sort of [inaudible] and then you pass through
quotients gives you expanders?
>>: Gabber-Galil did this graph?
>> James Lee: Gabber-Galil did this graph, but for various reasons,
instead of plus minus one, they took plus minus two because when they
did the Fourier analysis it helped him have a square somewhere. So
they did -- they did essentially the same thing but now they
analyzed -- they analyzed -- they didn't use a -- they didn't pass the
set to analyze the Fourier transform. They just analyzed it in the -so the proof has a similar flavor where you sort of say that a function
has to grow a certain amount of the time with two-thirds of the time
blah, blah, blah. But it's nothing -- but it's a few pages of complex
analysis.
Not complex analysis, but you see things like bounding characters and
then there's a square and there's like oh we have to make this square
to have a trick in Fourier analysis. Anytime you have Fourier analysis
it's a little subtle when you just had a sequence of inequalities that
work out. So it's nice.
And in the discrete case you're completely screwed, because you
can't -- I mean, you can't get rid of the torsion. So again you can do
something which simulates sort of what happens in the continuous case,
but you can't get something this pretty if you take ->>: This discrete Cheeger and quality you get down to finite set. So
somehow -- because the lemma one works because you're only looking at
finite sets. So they are open to the outside and you get lots of
factors going out.
>> James Lee: But even here you'd be bounded in L2. You'd essentially
be finite. I mean, you can -- even if you weren't going to use
discrete Cheeger, just going to operate.
>>: Yes.
So you get this somehow.
>>: From Zook [phonetic] combinatorial.
>> James Lee:
Go on.
>>: Supposed to be elementary proof so this type of ->> James Lee: For the ZN? At some point a few months ago I looked
through everyone's lecture notes everywhere and tried to find the
simple proof. And in fact I should say that lemma one, there is a
paper of Yilm London [phonetic], which considers the same problem but
for continuous sets in the plane. And this proof it's very similar but
this proof is essentially an observation of Z veer about their paper.
But it's very but the point is complicated parts in all the other
proofs are really lemma one it's just lemma one plus the chigger
inequality. That's what.
>> Yuval Peres:
[applause]
Any other comments or questions?
Let's thank James.