>> Eyal Lubetzky: Okay. We're very happy to have Noga here. He was a frequent
visitor in the past, had a break of a few years, but we hope he'll resume the frequent visits
from now on. And he'll be telling us about the eliminating cycles in the torus.
>> Noga Alon: Okay. Thanks. Right. So I promise to eliminate cycles in the torus.
And I'm going to do it on the white board. So we'll have -- maybe I'll write the title:
Eliminating Cycles in the Torus.
And I want to first define the torus -- actually, the tori I will be speaking about, because
there will be three of them. Incidentally, please do not hesitate to stop me whenever
something doesn't make sense.
Okay. So we'll have a definition, say two of the tori will be discrete, it will be graphs,
and one of them will be the usual continuous torus. So these are definitions. They are
notations. The CMD1. That's a notation for one graph. So that's a graph.
The vertices will be a ZM to the D. So sequences of length D of elements of the group,
modulo of integers, modulo M. And I have to define when two are adjacent, so X1 up to
XD is adjacent to Y1 up to YD.
These are two typical vertices if and only, so let's write it like this, so the sum of XI
minus YI mod M, and we take the absolute value and we write here 1, and what this
means is that the two vertices, or the two vectors, sequences, are equal in all coordinates,
and in one coordinate they are consecutive modulo M.
So for the two -- for D equals 2, this is a -- the picture is this. And it's a torus, so it's
usual we identify top and bottom and left and right. So this is this graph.
>>: [inaudible]
[laughter]
>> Noga Alon: Didn't get it?
>>: [inaudible] both ends of [inaudible].
>> Noga Alon: Uh-huh. Yeah, yeah. Okay. Uh-huh. I agree.
Okay. So this is one. Then we'll have a similar graph of a torus, so this will be CMD
infinity. So that's also a graph. And the vertices are the same, CMD. And now two are
adjacent if and only if -- and we take an infinity norm, so that would be the maximum
over I between 1 and D of the same thing, XI minus YI mod M, like this, equals 1. And
maybe I'll make it smaller.
So to be adjacent in each coordinate they have to be either equal or adjacent [inaudible]
torus. And this is the picture. So it's another graph.
And then we have the continuous torus. So this will be a TD, this will be a -- so, if you
want, it's RD modulo CD, but it's just a continuous torus. And the picture for dimension
2 is this.
So these are three tori. And we want to ask something about them. And afterwards I'll
tell you why we are doing it. So these are problems that have been considered before.
So I want to define what is a nontrivial cycle in each of these tori. So the cycle is
nontrivial if it is not homotopic to a point, but I'll just write informally if it wraps around;
namely, we cannot contract it to a point in the torus. So, for example, here is -- I just see
the cycle. So I'll do it in the continuous torus the same.
So, for example, just to practice it, this is a cycle, right, because it starts here, goes here.
This is identical to this because we identify topic [inaudible] and then it goes here and
this is identical to this. And this you'll see that it wraps around actually in both
dimensions here. And in this cycle we can look at some other cycle. So this one is a -- is
trivial. This is not nontrivial because it doesn't wrap around.
And, in fact, the only thing we will want to use about cycles that wrap around is that their
projection on at least one coordinate is full. This is not an if and only if, but it goes in
one direction, if a cycle wraps around, then this implies that the projection on some
coordinate, maybe on more than one coordinate, is full.
So this is something that we use. And then the final definition is what is called a spine,
so a spine will be -- but, again, we'll have three of them, so we'll define it vertex spine.
Vertex spine.
So this will apply to the discrete cases, to any two of the graphs. So this will be a set of
vertices that intersect, so intersecting each nontrivial cycle. And, similarly, we will have
an edge spine. So, as you can guess, this will be a set of edges intersecting each
nontrivial cycle.
So if we delete these set of edges, say, no nontrivial cycles remain, and same for the
vertices here. And then we have a continuous spine, so maybe I'll just call it a spine. So
that will be a surface. I mean, we are in dimension D, it will be a D minus one
dimensional surface. So surface. And, again, intersecting each nontrivial cycle.
>>: [inaudible] nontrivial?
>> Noga Alon: Yeah. So it's good question. Let's look at it and try to -- so the question
is if a -- let's look at the horizontal line. Actually, a horizontal line is a cycle in itself,
right? It's a nontrivial cycle. But it's not a spine, right, because it does not intersect any
other horizontal line, for example. So it doesn't intersect this one.
But these ->>: [inaudible]
>> Noga Alon: The question was?
>>: A nontrivial cycle or ->> Noga Alon: Yeah. Sure. This is a nontrivial cycle.
>>: Maybe the definition should be noncontractible.
>> Noga Alon: Yeah, it can be noncontractible cycle if you want. So it's, yeah, any
cycle that you cannot contract to a point continuously on the -- so definitely a horizontal
line is there.
And these two things together are a spine, right, in this, because -- okay. We'll see.
Now, what is the problems that we want? So problem. I want to determine or estimate a
mean possible size. Well, of course, size may be vertices, may be edges, or may be D
minus one dimensional volume depending on -- so that's why of spine. We want to know
what it is. So how small can it be, can such a spine be.
Now, why would we want to look at such a thing? Okay. So let's say something about
[inaudible]. So I mentioned three reasons. Each of them has some nice story behind it,
but probably I'll not tell you much about it because it will take too much time. So unless
you will really push me to do so. So I'll just kind of -- I'm saying more or less the
buzzwords, but, again, on each of them I can tell you more.
So region L1, as I said, some versions of this problem have been considered before. The
first one maybe is in a paper by Saks, Samorodnitsky and Zosin. And their motivation
came from integrality gaps in a directed multicommodity problem.
And basically they show -- specifically they look at vertex spines in this second graph.
I'll come to this later. They show that every vertex spine here is big. You have to delete
many vertices to destroy your cycles.
But, fractionally, whatever this means, I didn't even define it, but fractionally you don't
have to delete much, you don't have to put too many weights to get a fractional spine if
you define what a fractional spine means correctly, and therefore there is some large
integrality gap. And it was somehow interesting for them. I promise this will not make
more sense, so I hope that -- [laughter] but okay.
Now, here is the second one. It's a -- there's a story, is a -- it's even longer and is
interesting I think. So that's what is called a parallel repetition of the odd cycle game.
And, again, I don't want to talk too much about it. So maybe some of you heard
something about it and maybe not.
But it's somehow a nice -- and this -- the connection between this and one of these
problems was pointed out by Feige and Kindler and Ryan O'Donnell. So these are all
papers. They're all from a few years ago, so maybe '05 or something.
So let's leave it like that. The last one, I'll say something about it. So this is a -- what
they call foam -- maybe -- I didn't say here, but maybe let's -- I'll just write it. We have
here five -- we really have five different problems, right, because we have two graphs,
so -- [laughter] so it's the first theorem I can prove -- well, I would have proved if I would
have time.
So we have two graphs. Then for each of the graphs we have vertex spine and edge
spine. And then we have the continuous one. So there are really five different problems.
And all of them, besides one, are interesting.
Okay. So foam, let's look at just to explain what this means. So this has to do with soap
bubbles. Let's look at the -- so base spine or smallest continuous spine for D equals 2,
that's only one that is known. So for D equals -- well, for D smaller than 2, it's not
interesting. Even for D equal 3, it's not known what the smallest is.
But for D equals 2, this was done by Cho [phonetic] in '89. And it turns out that -- so
here it's just length, right, D minus 1 is 1. So you take here is the midpoints of these.
And this is a middle here. And then you take the point here that -- so that every angle
here is 120 degrees, and you do it symmetrically here.
And this is a -- it turns out that this is the base spine. And if you repeat it, so you can
repeat it periodically, let's take another torus or square another one, another one, and they
just draw this same spine in each of them.
So maybe it looks like this, here it looks like this, and here it looks like this. So this is
what you get in the middle. I mean, in this picture. And if I would do it periodically, we
would get it everywhere. You get this, say, bounded body. And it is bounded exactly
because this thing is a spine.
So if it wouldn't have been a spine, if it would be a cycle that is not blocked, then
somehow this thing would not be bounded. We would be able to go to infinity without
crossing any boundary.
But this thing will be a tile in the surface area of the length of the perimeter in this case,
of the tile is exactly twice the length of the spine. Because somehow every part of the
spine is counted once here and once here.
So and this is something that tiles because of the periodicity. It tiles the plane with
respect to the usual lattice. And if the spine is small, it is something, the tile is a plane,
which has a small surface area, or here a small perimeter, but in higher dimension it will
be a small surface area.
And why foam. Because there the soap bubbles, it is believed that what's a -- or the
physical model is that what they tried to optimize is the surface area subject to all the
constraints that they have to satisfy in the specific scenario.
So here if we think that we blow a soap bubble in the middle of every lattice point and if
they all have to be the same, so it would be some periodic tiling that we field the plane,
but because they tried to minimize the surface area, they want to find the tile that tiles as
a plane with respect to this lattice and has the smallest possible surface area or perimeter
in this case in higher dimension. It will be D minus one dimensional thing.
And we want to understand how it is. And in particular you see in big dimension. So
one way to tile is just to take a cube, to tile by cubes and boxes.
And in the surface area will be a D, 2D. Right? Will not be interested in constant factor,
so it will behave like D.
Now, on the other hand, if we tile, then the volume of each tile is 1, so by the usual
isoperimetric inequality, if we would even forget that this is a tiling, if we have in
D-dimensional located in space anybody whose volume is one, then the surface area is at
least square root D times some constant. So you can compute what is surface area of a
sphere.
And, therefore, the smallest possible surface area of something that tiles the space with
respect to the usual lattice is something between D and square root D, and we want to
know how it behaves. So can it really look like a ball in terms of surface area and still
tile the space with respect to the usual lattice.
Seems it cannot be really like a ball, but we want to understand it.
Okay. So this is about the motivation. And let me tell you something about what is
known previously and what I want to show you today.
So, as we said, we have these five problems. Now, one of them is trivial, so this one is
easy. Smallest edge spine. Edge spine in a CMD1 is of a size exactly 1 over M fraction
of all edges.
Namely, we have this torus here, so what we could do is we just cut -- we cut these edges
and we cut these edges. And what's good about this is that from every horizontal and
every vertical nontrivial cycle it -- we cut exactly one of the edges.
Well, we have to cut at least one edge from each of them, and we cut exactly one. And
you can see -- well, you can't see, but basically you can see that this is a spine, that
every -- okay.
And it's obvious why it is the smallest possible, right, because the whole torus here is just
the edge is joint union of those nontrivial cycles that look pretty trivial anyway. And
maybe it's the horizontal. The vertical is the ones in each -- so in each of them we have
to delete at least one edge. And if we manage to delete in each of them exactly one edge,
then, of course, that's the smallest. Okay. That's simple.
Now, here is one that also is known precisely, but that's already a result. So that was -so this will be the vertex spine in CMD infinity.
So there also the base spine is kind of the natural one that you can think about. You just
cut the boundaries. So that's a result by Bollobas [inaudible] also Kindler and Leader and
also Ryan O'Donnell, also a few years ago.
So they proved that a smallest vertex spine in this CMD infinity is of size -- here we
know exactly what it is, so it is of size exactly M to the D minus M minus 1 to the D;
namely, all vertices with a zero coordinate.
You see what we -- this is this number right because it's the total number of vertices is M
to the D and the ones which you do not have a zero coordinate is a minus 1 to the D, and,
again, so this says that you kind of take the -- maybe the most natural vertex spine or the
ones that are here, these and these.
So that's a big step we can do. And, as I said, that's a -- that's already a proof. So they
make some effort. Actually, there is a remark here. There is a very short proof, a nice
proof, but this is something we observed with [inaudible], a student in Tel Aviv, that take
this -- so I write -- you know, [inaudible] write one line.
There are these one-line proofs, but usually the trouble is that these one-line proofs, if
you want to show the one line, you need a few pages of preparation before the one line,
so I won't really show you.
But really it's somehow -- in some sense it's a one-line proof. And it's even -- which even
gives a -- which also gives all cases of equality and it uses something that maybe in
retrospect is not so surprising.
So using the Brewman-Kovsky inequality [phonetic], inequality which says that for
bodies in the D dimensional space, if we look at the volume of a -- the sum of two bodies,
in the cough ski sum, it's at least the volume of A to the 1 over D plus the volume of B,
so the 1 over D.
And somehow there is a natural way to represent these things as we really replace every
vertex of the graph by a small box that represents this vertex. And, okay, we say it is this
one line, but I'm not going to say it. But it's very short proof.
Okay. Now, so far, so these two results, one you say -- one is trivial, the other is -- as
you say it's very natural, so the smallest spine is what you think a smallest spine is, and
how would you cut, you cut in each dimension and that's it.
Now, the three others are all different. So they are all somewhat surprisingly small, and
they are surprising in two respects. So in one respect they are smaller than maybe what
we would expect, maybe they are not what we get from the natural one.
And also in all these three cases we don't see the spine. So there will be some proofs if
they exist and if they are small, but we -- but the proofs are not explicit, so we don't see
them. And it's -- in all these cases.
Okay. So let's mention those. So maybe the first one is, again, in the paper by these
guys, BKLO, so they say that a smallest vertex spine in a CMD1, so that was the one for
which the edge spine was easy, if we want vertex spine, is of size -- well, now we don't
know exactly, but, at most, some constant, D raised to the power log base two of three
halves over M, so this would be fraction, of all vertices.
Okay. So always we need to take at least a 1 over M fraction of the vertices or edges,
you know, these spines, because of looking at, let's say, horizontal nontrivial cycles.
But what they say and kind of the obvious thing to do, which is what happened in the
other two spines, is to take it in every dimension separately. And then we would get a
fraction D over M, but this gives something like a D to the .6 maybe over M. So it is
somewhat smaller than what we expect, and, as I say, we don't really see it from the
proof, but there is a spine that is smaller.
Now, the next one -- so, as I said, so we still have ->>: I don't understand exactly what the big O means, so this is as D [inaudible].
>> Noga Alon: Yes. So big O just means absolute constant independent of anything,
like 5 1/2. So not a function of M and not a function of D. Right. And actually ->>: That's interesting, because all of them have 1 over M, right?
>> Noga Alon: Yeah, yeah. So you mean -- right. Yeah.
>>: [inaudible]
>> Noga Alon: But actually, since you ask it, I think that specifically this will come later
on. Specifically in their bound actually the constant is 1. So could be a -- yeah. But later
I'm going to have all kinds of constants, and I do not care about absolute constant. So
whenever I write O, it's absolute constant.
Okay. Now, here is a -- so this is something done by Ryan Raz. So, again, he formulate
it for the parallel repetition problem, but it's equivalent to what I say now.
This somehow was surprising and this says that the smallest, what, edge spine in CMD
infinity is of size -- and, again, we want as a fraction at most. And here definitely there is
a constant, but it's again an absolute -- just an absolute constant.
And instead of -- so we said that kind of the trivial upper bound would be D over M if we
cut in all directions, but actually gets a square root D over M fraction of all edges. So,
again, it's small. And, again, we don't -- from the proof, we don't really see it.
And, finally, that would be the last one in the previous results. So that would be -- the
last one that was left was a continuous spine. And so that's really -- as we said, it's a
minimum surface area of the body that tiles the space with respect to the usual lattice.
So that was a recent paper by Kindler and the same O'Donnell and now Rao and
Wigderson. And that's also from maybe last year.
And they proved that the smallest continuous spine is a very at most, again, some
absolute constant square root D. So really it behaves like the surface up to a constant like
the surface area of the ball and not like the surface area of a cube. So it's kind of very
small.
And, okay, so this is what's known and what is new. So something I want to show you.
So here's something we proved with [inaudible] from Tel Aviv. So this one, actually, one
can -- it can be much smaller. So this will be one result. I'll tell you something about it.
So this will be the smallest. Again, a vertex spine in CMD1 is of size at most, so then
there is some absolute constant. And here instead of this D to the .6, it will be only log
D. It will be really small and, again, a fraction of all vertices.
But maybe what's more interesting -- so I'll tell you something about this. But basically
we get -- so each of these, say, three results it gives surprisingly small spines that we
don't see, is each of the proofs is very different from the other two. And what we have is
one approach that gives all of them. So basically ->>: [inaudible]
>> Noga Alon: So up to a logarithmic term this one is known to be [inaudible]. But
there is some logarithmic terms that are not clear. But this one is tile. Yeah. Because of
size of parametric. So this one is tiled up to a constant factor, that one is tiled up to a
logarithmic term, this one I guess is tiled up to a logarithmic term because 1 over M is a
trivial. But it's not clear actually.
>>: [inaudible]
>> Noga Alon: No. You mean if you can get in one of -- in this one something from an
isoperimetric, no, it's from some other. Yeah. Yeah. Because somehow ->>: [inaudible]
>> Noga Alon: Yes. Specifically for this one it's a 1.5 over M. So somehow in
dimension two it is known exactly. In dimension two it's 1.5 over M. And whatever you
have for some dimension, it is monotone because you can cut. Like in three dimension
you can look at the layers in each one.
Okay. But what I want to tell you is that this is a plus same approach, same approach.
Gives other two small spines as well. Right? So it will give an alternative proof of this
result of Rao's and an alternative proof of this. And they will be very simple, all of them.
Now, what is this approach? So generally steps of proof of each of them, two steps only.
So there will be step A and step B. So step A will be find or prove existence, say, prove
existence of a subgraph or body, so depends if we are talking -- if we are in the discrete
case or in this continuous space.
So we want to prove that there is a subgraph or some body with no -- in the torus, right,
with no -- okay. Right. In torus. With no nontrivial cycle. No nontrivial cycle, a small
boundary, whatever small and whatever boundary means, so vertex boundary or
[inaudible] boundary or continuous boundary.
And how do we prove it, this existence. So this will be using an appropriate
isoperimetric inequality. So I show you one of them. But in another case, so in the
continuous case it will be actually the simplest. In the continuous case it will just be
what's known as Cheeger's inequality, so it's a known inequality. You kind of substitute
it and you will get this existence.
And then the spine -- so this means that we are somewhere in the torus and we get a -some body inside it, and it has a small boundary. This is a boundary. And it doesn't
contain any nontrivial cycle. But it's still -- there is no spine here, right?
So this will be step B, which will be simple. So I'll show you step B, is that the spine is
obtained by pieces of random shifts of -- maybe pieces of boundaries of random shifts of
this subgraph or body.
Okay. So, again, this does not -- so this will become clear. But this is basically how the
proof goes.
And I want to show you a -- just have enough time to show you a -- one of them.
So, as I said, they're all essentially the same. So I want to describe one of the proofs.
And maybe the nicest will be to try to describe this one, because they're -- indeed it's a
new estimate. But, as I said, basically all of them are the same.
So let's see one example. So I want some isoperimetric inequality. Now, if I talk about
this vertex spine in CMD1, then I need some vertex isometric inequalities, or all kinds of
known qualities. This will be something that is kind of simpler. So it's more or less the
simplest.
So here is an isoperimetric inequality for a graph. It is general. It talks about graphs in
general. It's kind of independent of what I talked about. So even if you lost me or if you
have not been listening, you don't have to, but you can listen to this again.
So this will be a lemma. It's a general thing about graphs. So all these things are similar
to the kind of inequalities that you deal with when you talk about expanders, about the
ratio between the so-called isoperimetric constant or Cheeger constant and the
eigenvalues. But this will be a simpler one.
So this will be a discrete vertex isoperimetric inequality with the Dirichlet boundary
condition, if you want to give it a long name.
So here it is. If G is a general graph, U, a subset of the nonempty, which is sometimes
called nonboundary, it's good to think about it as a boundary because in our example it
will be the boundary of the torus, maybe, and we define C to be the minimum or the
[inaudible]. It's finite. So the minimum over all the W, which are subsets of a V minus
U. So we take sets that do not touch the boundary.
And what we look is that the number of vertices -- I'll define it. So NW minus W, these
are all the neighbors of vertices in W, but which lie outside W. And we divide by W.
Okay.
Okay, where, as I said, N of W is a set of all neighbors, X and V. There is YW so that
XY is an edge. So if my set here is Ws and the neighborhood are all the ones that are
connected to W, so we want to know how small can the boundary of the set be if the set
does not touch the boundary U, is a fraction of the original set W and we want to bound it
by something.
So if this is C and if F is some function from V to the real numbers -- no negative real
numbers, let's say, and F owns the boundary on U, it is identically zero, and the
boundary, then this C is at most some expression.
So it will be the sum of FU over all U and V. And here it will be the sum, again, over all
U and V of the maximum over all V in NU union U. So we take all the neighbors of U
and U itself and we take FV minus FU, we maximize over all these spheres.
This is always nonnegative, this thing, because in this set I have U itself, so F of U minus
F of U is zero, we can get only something bigger. And the claim is that C is at most this
thing.
Now, why -- so this is some isoperimetric inequality, tells us something about it bounds
this C this in terms of some function. Why is it good for us. Because if we can find a
function for the case that we care about for which this thing is small, then it will prove
the existence of a body or a subgraph with small a boundary.
So this is what -- and this is not surprising, the expression here. As I said, the proof of
this is not difficult. I'll not show it, but it's really not difficult. But it's not surprising
because if the function F is a 01 function, a characteristic function of a set, then what is
written here is exactly the background write because you see so then you will get -- well,
here, again, if F is a characteristic function of some set W, here you get a volume or the
size of W because you get 1 for each vertex of W. And here you get 1 if and only if this
U is a vertex on which the value is zero but it's a neighbor of someone in which the value
is was 1.
So it's in the boundary. So then this is -- namely, if I would have known to choose the
base F, then I could have gotten equality here. But because it will be -- F will be a
characteristic function of the W, that gives us the minimal. But normally we don't know
to do that. And the ability to choose a more continuous F enables us to get some bound.
So, as I said, proving this is not difficult. Let's leave it and -- but the corollary would
be -- corollary in CM. So we are in this step A. So I want in CMD1 to get a subgraph
which does not contain nontrivial cycles and have a small vertex boundary.
The claim is that in CMD1 there is a W set of vertices. Set of vertices containing no
nontrivial cycle. And the N of W, the boundary, the set of vertices outside of W that
have neighbors in W divided by W is small, is at most maybe 2 or 3 times a log D over
M.
And how do we prove this. So we just want to do this previous thing in a graph -independent graph G which is this CMD1. It will be more convenient. Convenient. So
it's convenient to consider a, let's say, C2MD1 on the set of vertices.
Just to be easier to write down things, I want to put it so that the center is at zero. So it
will be minus MN to the D on this set of vertices. M at minus M are identical, so I wrote
it like this, so this will be the set of vertices. We define the boundary, boundary U will be
all vertices which are in the boundary with a coordinate plus or minus M.
And I have this torus and I want to -- so it's a graph, right, it's this graph that we
considered, and I want to define a -- so it's a graph and I want to apply to it this theorem.
So I define the boundary as all these thing here.
And now if I'll get here -- so W is a set of vertices that is contained and V is a set of
vertices minus the boundary without the green part.
If we are without the green part, then definitely W will not contain a nontrivial cycle
because the projection on each coordinate will be not full because we are missing the
boundary here.
So this will be -- okay. So we just need to define some F here that will be zero on the
green part, and so that this ratio will be small, and then this statement will tell us there is
a W with a small boundary, and, again, W does not contain nontrivial cycles because it
doesn't touch the boundaries of this torus.
>>: [inaudible] W could be a smaller box?
>> Noga Alon: It could, but then I'll be in trouble because these things do not scale. So
then the ratio will be very big somehow. Normally if you want the ratio to be small, you
want to take big things, relatively big things, right? Because the volume grows faster
than surface area or something. Okay. So right. So how -- so I just tell you what is
this -- so I define a -- anyway, I'm not going through the precise computation, but it's not
difficult.
So define F of this X1, X2 up to XD. This is a typical vertex of the torus. So this will be
the product of V of XI, I goes from 1 to D. It's a product over the coordinate in the
function fee [phonetic]. Yeah. I want to say something on state B. And I want to finish
on time.
So it turns out that a good fee is fee of X, will be zero, well, it has to be zero if X is plus
or minus M, that's the boundary. It otherwise can be 1 over D times 1 plus 2 log D over
M raised to the power M minus X. In any other case X is less than M.
And then we somehow do the computation here. And it's not difficult to see that the ratio
will be only log D over M. Basically you see -- well, you have to worry about the zeros,
but if you forget the zeros, so somehow for every vertex you hear there is some FU here
and then there is a term here.
If the term here is only 2 log 2 over M or something times the term here, then we are fine
because we sum over all of them. And this is basically what happens because if we are
enable -- enable means we change one coordinate and one coordinate exactly corresponds
to multiplying it by 1 plus 2 log D over M, so the difference between them would be log
D over M. And you have to look at it for a few minutes and convince yourself that's the
case. But it is correct.
Okay. So we now plug in star and write star here and get result. Right? Because an
upper bound for C is exactly the existence of W with a small boundary here.
>>: [inaudible] somehow not constructed? Or can you give us a clue --
>> Noga Alon: No, yeah, actually this part is not the big part, the nonconstructive part is
really part B and the trouble was part the -- also is that we have only three, four minutes
for it, but, yeah.
>>: But can you sort of give us some clue what the resulting W looks like?
>> Noga Alon: Yeah. So actually from the proof of these the resulting W is always level
set. It's a level set of this function. So but -- yeah. And, in fact, so you know, you more
or less know how it looks, W. Okay. But the trouble is now with the random shifts, so
this is the part that I want to try to say something about it.
So B, we have this W, and now we want to get the spine. And let me try to show you
quickly what is the spine. So take an infinite sequence, X1, X2, well, you'll stop at some
point. But this is in ZMD random. So each of them is random, independent element of
ZMD, just point in the torus.
Now, we have the W from before. Define a -- put WI will be XI plus W. Everything is
in the torus. So I'll write here [inaudible] in each coordinate.
So we take the random shift of this. This graph of the torus is a vertex transitive. If W
didn't contain a nontrivial cycle, none of these will contain nontrivial cycle.
Now we define this SI will be the neighbor -- the boundary of WI minus everything that
we saw before, the union over all J less than I of WJ union NWJ.
And then the spine will be just the union of all these sides. I mean, I wrote it here as an
infinite sequence, but, of course, after finitely many XIs with probability 1, this covers all
the torus and we can stop there.
So really in a picture what this looks, so we have this torus here, and then we have this
body. And we took the boundary, and then we took a random shift of it, and then we
took the boundary of it that lies outside. And then you take another random shift and
then -- and so on.
And you -- well, so some shifts can be modulo the torus, but I just -- and then the claim is
that this is a spine. That's easy to see. So no nontrivial cycle can be contained in one of
these guys, and because of this you can convince yourself. So I'll just write it easy,
because there is one nice point that I want to stress.
So easy is that S is the spine. And now here is the crucial fact. The fix, the fix X in the
torus, in ZMD.
Now let I be smallest the first time such that X belongs to WI union N of WI. So we
have some point here. It's here. You see so far it has not been covered. But now in some
shift it's going to covered.
And then the crucial fact that is kind of trivial but you have to stare at it for 30 seconds,
you don't have more, so then -- so X is equal to XI plus W, right, because these are the
points of WI.
And W in a W union NW is uniform random in W union N of W. Namely, by the time
this point will be covered by some shift of this, it can be covered by any point of this and
it's equally likely to be covered by any point.
And this is obvious, right, because for every point if you want it to be covered by this,
then the shift has to be exactly this. If you want it to be covered by this, the shift has to -so this is exactly one way in which it can be covered by every point.
And because of this, the probability that this point will lie in the boundary once it is
covered is exactly the ratio between the boundary and the -- so therefore the probability
that X belongs to SI, which will therefore be part of the spine S, is exactly -- exactly NW
minus W divided by W plus NW, I think, which is at most this log D over it.
So each point when it is covered it is -- it has only a small probability to lie in the spine.
And now just by linearity of expectation, expectation, expected size of spine is the total
number of vertices times this O log D over M. And that's what we wanted.
And so this is -- this is this case, as I said, the edge version, so you have to use some edge
isoperimetric inequality for graphs, and then you basically repeat the same thing. For the
continuous one, the isoperimetric inequality is Cheeger's inequality, you just write it
down.
There it's easier actually, and the same for the isoperimetric inequality because there it
comes from eigenvectors and eigenvalues of some Laplacian operator. And there you
know you don't have to be clever and invent in your function F, it's just the best
eigenvector. So you know what it is. For the continuous one, for example, it will be the
product of signs.
So just write it down. You do the computation. And you get -- and let me just write two
problems. Maybe I mentioned more. So two things that look interesting. One is that -is the log D [inaudible] in this CMD1 in the kind of proofs that I sketched. So maybe it's
a constant 1 over M. It's not clear.
>>: [inaudible]
>> Noga Alon: Is the log D what?
>>: [inaudible]
>> Noga Alon: So if we do it in this way with a function that is a product of all the
coordinates, then the log is tiled. But I don't know.
>>: [inaudible]
>> Noga Alon: Yeah, yeah. I don't know in general. So this is -- I mean, yeah,
otherwise if you don't -- if you don't [inaudible] on this function, it's the same question
really because if you delete the spine, then the thing that's left is a set whose boundary is
just a spine. So then it's the same question.
So this one -- the second which looks interesting is to see how any of these spines, how
they look like. And the really nonconstructive part is this random shift, that somehow we
don't have control.
>>: [inaudible]
>> Noga Alon: Yeah, yeah, not ->>: [inaudible]
>> Noga Alon: I can randomize if you think about M to the D as polynomial. If the size
of my graph is polynomial, okay.
But what I really want -- let's say it's a continuous one. So a continuous one is a pretty
strange body, right, so you have a tile, the tile is a space with respect to the usual lattice
and its surface area behaves like that of the ball, how does it look.
I have no idea how it looks. It kind of consists of pieces of things. I don't know. It's -yeah. So it will be nice to -- yeah.
Okay. So we are almost in time, and thanks for your patience.
[applause]