>> David Wilson: Welcome back. We'll have a sequence of three talks in this
session. The first will be by Ken Stephenson on the rigidity of circle packing.
Ken.
>> Ken Stephenson: It's certainly a privilege to speak here, and I want to thank
the organizers. It's going to be wonderful to hear the stories. Each of us has our
modes of interaction with Oded. I was not a direct co-author, but talked with him
many times, and I'm looking forward to seeing those various gems that we're
used to seeing come out of his work and conversations with him.
And one of those is what I'll be talking about today I'm talking about circle
packings. And I think that circle packings, if you look at the record, have a solid
claim on Oded's affections early in his career.
And we talk about beauty in mathematics in various ways. This is maybe a
superficial one, but I can't get over the lovely pictures you see in circle packing.
A Doyle spiral. This is one of Oded's square grid packings, a couple of different
versions of the error function, early stages of developing an error function
discretely.
This is a conformal welding using circle packing. Another connection with Oded's
work is the squared rectangle on the left and the circle packing version on the
right. Same combinatorics and more or less the same rectangle.
And just in general, the number of ways in which circle packings appear in the
study of conformal structures. And that's sort of the underlying theme here. And
with Oded's work on circle packing was the connection with conformal geometry.
And so I'm going to give just a little background because I know that not
everyone knows about circle packing or knows the basis of it or potentially what it
contains. And so just the basic definition. Circle packing is a configuration of
circles with a specified pattern of tangencies. And as with everything that Oded
worked on, there are variations. We might be working in a Finsler metric, we
might be working with overlapping circle packings. I'm going to stick with
tangency pattern.
And so the idea is that someone comes up and gives you a combinatoric
situation purely combinatoric, like a triangulation, and then there's and then
there's this theorem, and we'll quota couple of theorems on the existence of
circle packings having those combinatorics.
So I'll start with the sphere, which we're going to work on mostly today. So here
I've used a random point process to put points on a randomly on a sphere, and
I've taken the Delaunay triangulation of those points. So this is what I would call
a random triangulation of the sphere.
Now, every triangulation of the sphere has associated a circle packing. The color
coding is just about the degree, so there's nothing special about that. It's a little
prettier sometimes. But the point is that with every vertex in the triangulation,
there's associated a circle. And if two vertices are connected by an edge, those
circles are required to be tangent to one another.
And so what happens is an arbitrary combinatoric situation is given a geometry
by being required to fit together with these circles. And so the circle set change
the geometry but respect the combinatorics of the original pattern.
And so we're going to talk about the basic existence and uniqueness results just
to get started. So the first result is the Koebe-Andreev-Thurston theorem. As Bill
Thurston told me one time, and you'll have to check with him whether he still
says this, he proved this in the morning, found a proof in the afternoon that -from Andreev's work but then it was several years later that it was found to have
been proven by Koebe back in 19 -- the 1930s.
So that says given any triangulation of a topological sphere, there exists a
univalent. So univalent simply means that the circles don't overlap. There's a
univalent circle packing of the Riemann sphere having the combinatorics of that
triangulation. And what we're going to talk about is the rigidity, which is sort of
uniqueness version of things.
So moreover, this packing is unique up to Mobius transformations and
inversions, and we're going to cut out inversions, we're going to think of oriented
objects. So given the triangulation, there's a rigidity here and they get a circle
packing associated with it, and it's unique up to Mobius transformation.
That's been extended considerably. So the more general theorem is given a
triangulation of any topological surface, there exists a conformal structure on that
surface making it a Riemann surface, and there's a univalent circle packing in the
intrinsic metric, the constant curvature metric for that Riemann surface that fails.
So that's a little technical thing in the disk you don't want to be shrunk down. So
you want to be sort of a maximal packing in that Riemann surface.
So given any triangulation, the surface might be compact, open, the triangulation
might be finite, certainly in a compact case or infinite with boundary, without
boundary, et cetera. Given any triangulation of an oriented surface, you can put
a structure on it and a circle packing compatible with that structure in the surface.
And again, there's a uniqueness statement. The conformal structure is uniquely
determined by the combinatorics. And the circle packing is unique up to
conformal automorphisms of that surface.
So these are the uniqueness type of results or rigidity I want to talk about. And
the upshot for me is that circle packs come along and they take a purely
combinatorial situation and endow it with a geometry.
And the important thing to under is that this is not just any geometry. It has the
local rigidity, which is the sort of localized version of the rigidity globally we're
going to talk about. And if we allowed ourselves to move off the sphere and put
some boundary in or allow some branch points, we have a global flexibility. And
in my mind it's this interaction of the flexibility globally and the local rigidity which
puts it sort of in the category of analytic function theory, which is where I came
from. So we all like to recast things in our own favorite way, but to me this is
where conformal geometry comes in.
And so this is a particularly familiar geometry that we get. And I call it discrete
conformal geometry. And that's Oded's work started in proving the uniqueness
and existence of these packings and studying the relations between the discrete
geometry and classical conformal geometry.
The next speaker, Zheng-Xu He, will be speaking more about this.
And so what I wanted to do was to talk about how this came up and actually do a
proof. So the setting was this. Bill Thurston suggested that if you take a region
omega and you fill it with a circle packing and then using Kurbis theorem you
repack that in the unit disk that from this, the mapping between these two regions
-- so you put this circle packing in here. Now, this has the same combinatorics,
so this is a circle packing with the same neighbor relationships, and we have now
a map between the unit disk and our original region omega.
And Thurston suggested that this was a discrete conformal map, or I think he
called it finite Riemann map. And so the theory developed around studying the
conjecture which was that under refinement, that is if you fill the region with
smaller and smaller circles and increase the complexity and do these maps and
appropriately normalize, they will converge in the limit to the classical conformal
map of the disk on to omega.
Now, this was proven shortly after the conjecture was was made by Rodin and
Sullivan, and it's been vastly extended to other situations, other combinatorics in
particular, and a lot of that work in fact is Oded's work on more general
combinatorics, for example. But sort of a catch phrase is that under refinement
objects in the discrete world of circle packing invariably it seems converge to
their classical counterparts. So if you build something discretely that looks like a
classical object and you refine it and keep it looking like that classical object, it
will actually converge to it. So it's somehow capturing the same geometry.
But what I want to look at in particular, there are a huge number of papers, I'll list
some of the end, of aspects of this that Oded worked on. But I want to talk about
a particular result. And so it's the KAT theorem here, the
Koebe-Andreev-Thurston theorem, but it generalizes beyond that case. But I'll
prove that.
So the claim here is that if I have a circle packing of the sphere like the one
pictured, call it P, that any other circle packing of this sphere with the same
combinatoric, same number of circles, same neighbor relationships as this one,
will in fact be a Mobius image of it. So that's uniqueness.
And the crucial tool in this is a very deep result, which is why I'm presenting this,
because some people may not realize this, that two circles can intersect in at
most two points. So we'll exploit that deep property.
So we can actually pretty much prove this whole thing here. So the setup part
one, at least. So we have our packing, P. I'm going to identify mutually tangent
triple there. And I'm going to take my competitor, my other packing. I can't draw
it because of course it's just P. But we have a presumed competitor packing.
And I identify combinatorially the same triple. And with a Mobius transformation I
can get that triple of circles to be the exact same triple of circles as in P.
So I've normalized things. And that's where the Mobius transformation ultimately
is going to come in. I have to show the rest of the circles on the right are the
same as the circles here. And I'm going to presume that the very next one that
fails to be the case. So here is one sort of extending that triple a neighboring
circle and in the competing packing that neighboring circle is a different size.
This doesn't happen. They have to be the same. So that's the first part of my
setup.
Now I just put the infinity inside that interstice and I project both these packings
down to the plane, and I get juxtaposed configurations. Now, I've drawn the
circles for one of them, but the other circles are in there as well for the other
packing. So everything is now inside this interstice with the two circles that
differed, the red and blue there, as shown.
One more small adjustment. I'm going to take the packing with the blue circle
and I'm going to scale it away from A just a little. And so we end up in the
situation on the right there. So I have these two packings with the same
combinatorics in that situation now. And it's that collection of circles that I'm
going to study a little more. So let's focus in on a small patch of this packing.
And I'm going to identify what I call the elements for lack of a better name. The
elements of the packing are -- well, the circles. So every circle is an element of
this packing and the interstice. So every interstice is an element.
Throw those together, I have E, the collection of elements of this packing. Now,
there's another packing around, which in this region looks like this. You can see
sort of which circles are identified in the combinatorics, but they're shifted from
one another. And so we have a collection of elements for this other packing, E
prime.
And in particular we're going to look at the triangular elements I'll call them.
These are the interstice elements. And the interstice elements in the two
competing packings here are shown, corresponding elements. And the dots
represent the tangency points of the circles. And those are identified from the
blue to the red, you know, which of those dots correspond to the other in the
other packing.
So what we are going to do now is compare these elements. And somehow, I
don't know exactly how Oded did this when he first saw this. He had a sort of
hands on approach as I think is common. He did not look over the literature and
try to find something that did his work for him. He just dug in. He knew it was
true, and he figured out direct method of showing it.
So he came up with the notion of compatibility. It's not the same as here, but it's
-- that was the genesis. We're going to use a fixed point index. So I'm going to
assume that I have two spell curves. The red and the blue, gamma and sigma.
And I'm going to define a homeomorphism between this, these two curves. I'm
going to set up this little terminology. So this is a fixed point free
homeomorphism orientation preserving between these two curves. And I'm
going to define the index.
And I think it would be appropriate maybe to just explain the -- what this index is.
So I actually have blue and red here. This is a very simple case. And so of
course the interactions could be more complicated. But I have these two curves,
and I'm going to take a function which maps each point on this curve to some
point on the other curve.
In the picture I start with these points of intersection. And let me describe this
index in a very simple way. Suppose that you are -- your friend is walking around
gamma and you are walking around sigma and you have an elastic string
between you. As you walk around, the only requirement is that you not both be
at one of these intersection points at the same time. That's the fixed-point free.
The other requirement is that your friend walks in the positive direction, you move
in the positive direction. And the winding number is that at the end of the day
whether that string winds around you when you get back to your initial points. So
your friend walks around one curve, you walk around the other, and what's the
purpose of the function? Well, the function says that I on my curve get to choose
how to walk around that curve and try to control that winding.
So given the curve gamma, I might be able to specify a function which tells me
how to walk around the curve sigma so that the winding has a certain property.
And that's what we're going to use is that winding number. So compatibility
results. If you have two circles as your curves, because they can only intersect
twice, it's not difficult to show that every fixed point-free homeomorphism has an
index that's greater than or equal to zero. Okay.
So there's no way that you can fail, no matter how you choose to walk around
your path compared to your friend on the other path. The index is always greater
than or equal to zero.
It's a little more complicated on these interstice elements. But the basic idea is
that in addition to the conditions on the map being a fixed-point free
homeomorphism, we're also going to require that when your friend is at A, you're
at A prime, when your friend is at B, you're at B prime and when your friend is at
C, you're at C prime. So that compatibility. And then the theorem is that given
these two topological triangles, their exists a fixed-point free homeomorphism
with index greater than or equal to zero.
And that's actually fun to figure out because it's a very straightforward simple
arguments to get this done. You can teach this in an honors undergraduate
course. It's a lot of fun.
So here's the proof. For every interstice element, that's the triangular ones, we
choose a map between the element in P, and the corresponding element in P
prime. And by the previous slide on the right-hand side, we're able to choose
that by force to have index greater than or equal to zero.
Now, on the circle elements, it turns out that your map is already defined
because every circle the boundary is covered by interstices, by edges of
interstices. And for each of those interstices, you've already defined a map to the
corresponding edge of the interstice and the other packing, and so this already
defines a map on the circles. And so you have maps on all the elements now.
And by the way, this in our overall configuration, there are some interstices
around the boundary, and so this defines a map on those, the outer boundary
there, because those edges are not shared by any other interstices, so they're
sort of left over. And there's a induced homeomorphism fixed-point free on the
outer boundaries of the configurations. So now we throw together this sum and
take about account of cancelations. Because when you're doing a winding
number, it's a change in argument. And if you're on an edge of an interstice
that's shared with a circle, the change in arguments when you're thinking of it as
a triangular element is canceled by the change in element because of orientation
when you think of it is as part of the neighboring circle element. So everything
inside gets canceled. This is a familiar argument to most of us. But the result is
that this index induced on the boundary is greater than or equal to zero.
But there's the boundary. All the edges of interstices and the circles and the
elements inside have been taken care of. And we have this homeomorphism on
the boundary and if you just isotope that, it's this configuration. And it's very
quick to see that the index is negative one.
Just go around here. A has to go to A prime, B has to go to B prime, C has to go
to C prime, and D has to go to D prime. So it's a negative winding. And we're
done. Competing packing fails, and we have uniqueness. So to me personally,
this is the prettiest proof I know pretty much in mathematics. It's all very
elementary and it shows you an amazing result. It applies further than just
triangulations of the sphere. It shows, for example, that circle packings that fill
the plane are unique and that's where the proof of Rodin and Sullivan depended
on, to get this whole topic going.
So that's the proof. And I want to say a little more though about where this leads.
So let's put a bookend to the conjecture that started this topic for a lot of us,
Thurston's conjecture about these conformal maps because Oded and Zheng-Xu
have proven by further study of the distortion of these circle packings the local
rigidity that the Koebe-Andreev-Thurston theorem on circle packings of the
sphere actually implies the Riemann Mapping Theorem for plane regions.
So there's no higher accolade in -- for a complex analyst then to say you have a
new proof of the Riemann Mapping Theorem, okay.
And a final word about other topics. There are too many. I didn't want to rush
through them. But here's another example of something that that Oded saw
which I really still can't quite believe, although I've studied the proof, and it's true.
So here is one of my favorite packings. It's heptagonal packing. Every circle has
seven neighbors. There are infinitely many generations of circles here, all in the
unit disk, all actually the same hyperbolic radius.
So I'll take this just as an example packing. Infinite hyperbolic packing. And I
can sort of repack that in the square. So that's the same combinatorics. But of
course I'm lying to you. There are not infinitely many circles either place. And I
actually don't know if that packing, if I can construct that packing in reality. But
I've gotten close enough for government work.
But they have a theorem, Schramm and He again, have a theorem that given any
Jordan domain their exists a univalent circle packing with these combinatorics
which fills that region. So not just a square, but any simply, take your most
horrible Jordan curve, you can fill the inside with a packing of these
combinatorics where every circle has seven neighbors. I just can't imagine that.
And in fact, I can't construct them. It's not a constructive proof. But it would be
appropriate maybe. By the way, there's another uniqueness result here. It would
be appropriate in the circumstances, however, to be able to pack a simply
connected region. Like this [inaudible]. So there's one that I suggest I'd like to
see a packing of that with every circle having seven neighbors.
Okay. Here's some of the papers to indicate Oded's early work on this topic. It's
just gorgeous stuff. As I said earlier, you just enjoy reading these. And often I
think you -- if you're lucky, you see the idea, and the extra stuff that has to be
imposed later doesn't bother you. Those ideas are really in there. So thanks to
you and thanks to Oded.
[applause].
>> David Wilson: Are there any questions?
>>: Well, I don't have a question, however, I have a comment. [inaudible]
already set you know, Oded wouldn't just settle for one proof but he produced
several usually. And his is the most important theorem about circumventing and
supplying, he actually produced like five proofs. And the one you use that's
probably the most important because as you said it applies to also to infinite
packings, and they're beautiful. But one proof I actually found that I wasn't aware
of, I found only very recently when I prepared [inaudible] article, and it's
[inaudible] and I found it actually on Wikipedia. [laughter].
So we all know that that was a very [inaudible] I mean it really was a very -- he
loved to contribute to the Wikipedia and he contributed also a lot to the circle
packing on Wikipedia. And the proof that is on there is so short and so beautiful
and so elegant that you wouldn't believe it. I thought it must be false. [laughter].
No, really, it's very short. Maybe [inaudible]. But then I wasn't quite sure. It
definitely had Oded's handwriting all over it. And also there was no credit, no
nothing [laughter] and then I went to the -- you know, when you see the history
file who contribute of course there were a lot of Oded entries, and I wasn't so
sure you know I just put in my article I believe that this proof is due to Oded, and
then I got some comments and with [inaudible] we actually chased down, we
looked at every paper on the entry [inaudible] it's an incredibly short elegant like
five line or three line proof of the rigidity in the finite setting. Anyway, sorry.
>> Ken Stephenson: Well, can I keep this as my favorite anyway? [laughter].
>> David Wilson: [inaudible].
>> Ken Stephenson: If I could say one more thing. My last correspondence with
Oded was about a year ago, and I'm working on some random triangulation stuff,
and I have -- I suggested something about convergence, sort of random
convergence, a conformal structure, and the reply was certainly seemed like it
could be true, but it looked very hard to prove. And so needless to say, I moved
on to other things. [laughter]. Because if his inside didn't really convince him
and he said it was hard to prove, I think I'll take his word on that.
>> David Wilson: Any other questions or comments? Let's thank the speaker
again.
[applause] we'll take a quick break of about three minutes.
The second talk in this theories of three on circle packing will be given by
Zheng-Xu He. He'll speak about disk packings and conformal maps.
>> Zheng-Xu He: Okay. Thank you. So it was really a hard choice for me.
There are so many things to choose about -- to choose to talk about when I was
invited to talk here. Anyway, finally I came down to talk about just one proof.
Let's see. So basically this is built on what Ken has already described, so the
relationship between circle packings. So I want -- we want to call it disk packing
for simplicity of definitions. Desk packing and the conformal maps. So basically
how disk packings can approximate conformal maps.
So here was the original theorem. Actually Thurston's conjecture that was
proved by Rodin-Sullivan. So basically we have two disk packings, one in any
domain and the other one in a unit disk. So basically the theorem says what the
theorem says is that the one-to-one correspondence between disk one in one
packing to disks in another packing approximates a conformal mapping.
So basically we have a few -- anyway, this is the full statement. So we have two
packings, one in an arbitrary domain, the other one in the unit disk. And we have
some conditions. Those are basically some trivial conditions that -- see. So on
the boundary disks basically we want the disk packing to fill the whole domains.
And also the third condition is a normalization condition so that the limit packing
will not push to what the boundary.
So under those conditions subsequence of F sub N which is basically the
correspondence map at that -- so subsequence of F sub-N should converge
locally uniformly to a Riemann map.
So there are a few problems. So the first one is to show the limit map is a
conformal map. And this was achieved by -- first achieved by Rodin and
Sullivan, and they proved a Rigidity Theorem. This is by itself very interesting
theorem, Rigidity Theorem of infinite hexagonal packing.
So basically if you have another packing -- if you have any infinite disk packing in
the plane that is combinatorially equivalent to the infinite hexagonal disk packing,
then all disks of the packing must have identical radius. So this is a very
beautiful rigidity result.
So the argument goes so if you have any -- if you have a disk packing as in the
picture here, so you can take a limit here. And the limit tagging as angles the
infinite must be a regular hexagonal packing. So all circles in the limit should
have the same radii. So as a consequence, the limit mapping must be
conformal. So this is basically a consequence of the rigidity theorem of infinite
packing.
Now, so the argument was only good for C0 convergence and that use is actually
quasi conformal maps, convergence of quasi conformal maps. And also you
need Bounded valence as to be able to take limit.
So actually I want to do a small digression here. So this rigidity theorem for
hexagonal, infinite hexagonal packing was actually generalized for all locally
finite, infinite disk packing in the plane with triangular interstices. So was done
when Oded was a PhD student. And then later this was generalized for disk
patterns when the disks not only a tangent, they may intersect in the angle. So
this is true not just for disk packings, but also for disk patterns with overlapping.
Okay.
And there are some further methods by other mathematicians as well, Ken, who
spoke here earlier and others. Anyway, they worked under strict restriction that
the radii of all packings are bounded between, basically from below and from
above. So -- and also only C0 convergence was proved. So this was in
late '80s.
And I first became interested in the packing when I was a PhD student under
Professor Freedman, and I think in my thesis I proved the C1 convergence and
this can be generalized to prove C2 convergence as well. So that's how I
became interested in circle packings.
Now, the problems for ours was after all of this work was to generalize to
arbitrary patterns. So this is actually a picture made by Oded. So what happens
with arbitrary patterns? There's no restriction at all on the combinatorics.
There's no bound on the valence, so a circle can be tangent to many -- as many
other circles as you like. So some circles can be really small.
So and what about the convergence of derivatives in such case? So in was a
open problem in the early '90s. So first of all, how do we define derivatives in this
case? Because in the regular pattern you can do like numerical analysis. So
what happens if you have a general pattern?
Well, we can use the Mobius transformations, so the idea is for any two triplets of
mutually tangent disks, there is a unique fractional linear transformation that
sends the first triplet to the second. So basically you take the tangency points to
the tangency points, and those three tangency points determine the fractional
linear transformations. Okay.
So -- and in particular the map, the fractional linear transformation is -- can be
defined on the interstice. So triangular interstice. Basically that's the shaded
area. Those are triangular. This is union of triangular interstices.
So we can define a map from the support of P sub-N, that's the packing in the
first domain to the support of Q sub-N, the packing in the second domain. So
this is P sub-N, that's Q sub-N here. And so that the map is conformal on every
triangular interstice. So because it is conformal, so we can define the first order
in a second order derivatives for the map from the support of one packing to the
support of the other. But you cannot define on the disks, but you can define it -you can define it on the union of interstices. So that's the derivatives we are
talking about.
And the question for you would be what about the third order derivative? So we'll
talk about that a little toward the end. So this is the main theorem. So we have a
pair of domains here. So half circle, and that's -- I don't know how to call that.
Anyway, it's like a cone domain. So we have packing in the first domain, P sub-N
and another packing, Q sub-N that's combinatorially equivalent, the same
number of circles and the same tangency relationships.
So we have a map F sub-N from the support of the first packaging to the support
of the second packing so that maps as before the map triangle collar interstices
to triangular interstices, and it maps the disks on to the disks of Q sub-N. So F
sub-N is only conformal in the interstices. So it's not conformal in the disks,
okay. So it's just a continuous map.
Now, those are generic assumptions basically that that spherical diameters of the
disk are small and also the boundary disks are very close to the boundary of the
domain. And finally a normalization condition that the map will not converge to
the boundary basically. So there's a compact subset so a fixed point, a Q naught
will be mapped to that subset. So not everything will [inaudible] to the boundary.
So the condition is that a subsequence of F sub-N will converge locally uniformly
to a conformal homeomorphism, okay. So this is the C0 convergence. So this
does not depend on the combinatorial pattern at all. And furthermore, we have
convergence of the first order and the second order derivatives. So again, those
relationships are only restricted to the triangular interstices because -- because
those are not defined in the disks. They are only continuous functions, but they
are defined. They are actually fractional linear transformations in the set of
triangular interstices. Okay.
So a corollary would be that the absolute value of the derivative of the Riemann
map -- of the conformal map may be approximated by the ratio of radii of the
corresponding disks. It does not matter if the disk is small or big. So this has
really nice control on the size of the disks. Okay.
So the main points -- actually those are broken out into three points. So first of
all to though that the limit map of the subsequence is a conformal map, and
again the difficulty is that there's no limit disk packing because the combinatorics
is rather arbitrary. And the uniform convergence of the derivatives first order and
the second order. Derivatives.
Okay. So actually those -- the first two points can be proved by some idea in my
PhD thesis mainly I mean some idea there, and applying -- actually that's after I
learned the concept of transband -- boundary extreme length from Oded. I mean
he taught me about extreme length concept. Actually he got it from Jim Canon,
some paper. Or maybe he got it before that. But actually the -- he eventually
found that concept was also in Jim Canon's paper.
So he taught me that and I used that idea to take care of some technical difficulty
and then I was able to -- and then I wasn't able to prove number three, and then I
talked to Oded about it. Initially he didn't believe that I had a proof. He thought
my method wouldn't work. Then he said let me think about it, and then a few
days later he came up with his own proof. So -- and then we wrote a paper
together on this. So I -- certainly I was saved from writing my cumbersome proof.
He wrote the whole paper basically and put my name. It was really nice for him
to put my name on that.
Anyway, so the argument was really simple and can be understood by any
college student with some training in complex analysis. It's really very simple
idea.
>>: Where is it [inaudible] [laughter].
>> Zheng-Xu He: Well, I left the colleges already, so -- anyway. So this
argument starts with a circle fixed point lemma, which actually Ken talked about
earlier, and this circle fixed point lemma actually was used extensively in the
proof of Koebe conjecture for countably connected domains. This is another one
-- another one of Oded's joint papers.
And this is a very powerful tool. And actually I learned this fixed point lemma
from Oded. He was working on the rigidity of infinite disk packings, and they was
able to come up with a proof. But that proof was -- we were able to use it for so
called circle domains as well. So that's another story. So I don't have time to
talk about that here.
So the fixed point index lemma was actually presented earlier by Ken, and so I
would rephrase it this way here. So we have two circles, C1 and C2 in the plane
that are positively oriented, and we have F1, F2 orientation preserving
homeomorphisms. So basically those are two curves, F1, F2 are basically two
curves that have images that are circles. And assume that F1, F2 is never zero
on C, so basically there's no fixed point. This -- sorry, this no points for -- no zero
point for the difference. Then the winding number of the curve, F1 minus F2
around a zero is non-negative.
So this is basically the same lemma Ken was taking about earlier. And I like to
mention the proofs because it's really relevant to what we're going to do next. So
we have four cases, the bound disjoint interiors where one is inside the other.
And so the case two and case three the winding number is 1 because you can
shrink a point in the smaller circle to a point.
And case four is really more interesting, and it takes a little bit more work. So
when the two circles overlap into two points. Anyway, so there are two
observations. So C1 and C2 really do not need to be circles. They can be any
Jordan curves as far as the intersecting at most two points. And that's by I mean
a simple homotopy argument.
And the second is a -- this is not needed but so the lemma is still valid if F sub-K
-- so F1, F2 are locally non-decreasing functions, they don't have to be
homeomorphisms. They actually they can be of different topological degrees.
They don't even have to have the same degrees.
Okay. So -- and here's a basic topological lemma. I think that's implied in Ken's
talk. So this is basically I think in any complex analysis book, textbook. So we
have a bigger curve right here, okay. So this is a bigger curve. And this is like
close Jordan domain by -- bounded by the bigger curve J. And we have J1, J2, J
sub-N being Jordan curves inside the domain which bound mutually disjoint
interiors. So those are actually boundary of disks in our applications. So they
bound disjoint interiors.
And A be the closed subset obtained from B by removing the interiors of those
little curves, okay, basically those little disks. So in our application A is a union of
closed triangular interstices, okay. So just they are not disks.
And all curves are positively oriented. So now G is a complex value continuous
function that is never zero in A. So A is like the set that union of interstices, the
disks are removed. Then the winding number of the curve G around zero of the
bigger curve J around zero equals -- is equal to the sum of winding numbers of
the smaller curves around zero. That's because there's no zero in A, okay.
And this is like a homology in dimension 1, okay. So now what if G has some
isolated zeros in the interior of A? Then you have to add up -- I mean, the sum
right here, you have to add up the degree of zeros of G in those isolated zeros.
So that's the -- what I call adjusted topological lemma. So the winding number
around zero is equal to the sum of winding numbers of the curves plus the sum
of degrees of zeros in the interiors of A. And in our application G is going to be
holomorphic. So the degree of zero is always positive, at least a one, okay?
And also those curves have -- those curves are going to be circles, and the -and the G is going to be in difference of two like maps as in lemma one, so this is
-- so those -- those -- the winding numbers is always non negative in our
application. Because we take G to be the difference of two maps. So we're
going to see that later.
Now, so I'm going to do a quick proof of the main theorem. So basically we have
F sub-N which is a from support of P sub-N to support of Q sub-N and then maps
triangular interstices conformally on to the 1s of Q sub-N. Okay. So F sub-N is
before, it's a holomorphic function in the interior of the triangular interstices. So
those are continuous. So the first step of the proof is to show that F sub-N's are
uniformly bi-Lipschitz in the interior of the domain.
And this is by using so called Schwarz Pick lemma for mappings of disk
packings. I think the first -- it first appeared in a paper by Beardon and
Stephenson, and the idea is that if you -- if F contracts a bigger disk packing into
a smaller one then the derivative must be bounded by one in absolute value.
So basically if the disk packing is contracted from a bigger one to a smaller one.
So this shows that the derivative in absolute value is bounded between those
numbers when the domains are bounded domains, and the general case may be
reduced to that case. Okay. So that's basically step 1.
Now, step 2 is to show that the second derivative is also uniformly bounded in
the interior of D. So remember the second derivative is actually defined by the
fractional linear transformations that send interstices to interstices so basically
we have to show -- we have to show that -- we have to show that the fractional
linear transformations is uniformly bounded. The second derivative of the
fractional linear transformation is uniformly bounded.
So we take such a fractional linear transformation so what we do is we're going
to take some Jordan curve in closing Z naught and then we change the Jordan
curve a little bit, so we take a Jordan curve that contains Z naught, and then we
change it a little bit so that it is still contained -- it's contained in the union of
interstices, so -- and then the boundedness of the second derivative follows from
the following facts: First of all, the boundedness of the zero's derivative, if you
like. That's because F sub-N converges by step one, okay. And they be the
boundedness of the first order derivative, so that's because F sub-N is
bi-Lipschitz, so we have boundedness of the first order derivative. And then the
following lemma.
So remember that F sub-N is a fractional linear transformation, so it only has
three degrees of freedom. So those two conditions one and two already place
two conditions. So we just need one more condition to imply the boundedness of
the second derivative. And basically the third condition is this. It's a topological
condition right here. So the image of the Jordan curve by F sub-N intersects the
bounded Jordan curve of F sub-N, of J sub-N. Remember F sub-N because it's
bi-Lipschitz that converges. So this is bounded. So that means that actually
places a strong condition on the fractional linear transformation F sub-N.
Okay. So lemma 3 is the key. Now, how do we prove that? That's just by the
fixed point index idea. So basically we assume those are disjoint, so both are
simple loops in the Riemann sphere. So suppose that disjoint but they share a
common point. So this is a common point. Because they are identical in that
interstice. That means the images share a point. Actually they share many
points. But the boundaries are disjoint, so one must be contained in another. So
as a consequence, the June is actually the bigger sets. So basically one is a
subset of another.
That means there's a point which is outside of the union, and we can map that
point to the point of infinite by a fractional linear transformation. So we can
replace the packing, image packing by L, Q sub-N, so the image by the linear
fractional transformation, so basically we can assume those two sets are
bounded sets, they are contained in the plane. Because they miss the point of
the infinite.
So now those -- so this is a technical step. So basically they are in the plane.
Now, those two maps it's a complex value map, and we know the image of the
Jordan curve by one is in the interior or the other one is in the interior. So the
fixed point index of the difference must be one on that Jordan curve.
But on the other hand, those two maps agree in that interstice. So they
completely agree in that interstice. And the lemma 4 says that if that's the case,
then the winding number around zero is at least three. So we get a contradiction.
So the contradiction shows that the two sets must intersect each other.
So now the lemma 4 basically says that the winding number of the difference is
at least three. The assumption is that the two maps agree in that interstice, okay.
So that's the only assumption, the key assumption. Because they agree in the
interstice, so how are we going to talk about the difference? A difference is
actually zero in that interstice but the idea of getting the winding number is just
give this map a perturbation. So basically we're going to talk about the following
map. So we perturb F sub-N by a fractional linear transformation so that this is
fixed point of three in the -- on the boundary and also achieves three fixed points.
Okay. So this is the -- so I'm going to skip this part.
Basically the idea is that the two maps agree in the completely grow in the
interstice. So you can always give a perturbation on one of the map F sub-N by
a fractional linear transformation. So you'll get three mixed points. Okay. So G,
this new G vanishes at three points, and that little G sub-N vanishes in the
interstice. So you can do a perturbation.
So after the perturbation, because the map has three fixed points, so the fixed
point index must be at least three. So that's basically the idea here. Okay.
So now -- so basically at this -- at this time -- so we have -- so we have shown
the boundedness of the second order derivative. Now, because of that, we can
take limits and also we can take limits of the first order derivative and use a -and use some kind of path integral idea and take limits. We can show that F is
differentialable and its derivative actually is the limit of F sub-N. So this is some
elementary step.
No, the final step, actually this is going to be the key and it's very simple, those
that actually the second order derivatives also converges. So this is -- well, this
uses the same lemma, lemma 4 here, the winding number idea. So the proof
goes as follows: So basically we have -- sorry. Okay. So I'll finish in two
minutes.
So basically we have T sub-N. That's a -- the triangular interstice that contains
some sequence of points. And F sub-N are the fractional linear transformation.
So we have to show that the second derivative converges to the limit. The
second derivative of the limit. So we're going to -- again, we're going to use the
fixed point index okay? So basically we're going to use the difference, so F
minus N.
So we take a little curve and we show that this [inaudible] curve has one number
around zero at minimum three. And this curve J, the reason is because -- the
reason is because F and M, they agree in some interstice inside, so the winding
number has to be at least three by lemma four. And F minus N, we take J to be
so small that this -- this has no other zero. So that implies that the second order
derivatives must be the same at P. Okay.
Okay. So finally I just want to comment a little bit on the convergence of higher
order derivatives. So first of all we don't know how to define this for arbitrary
combinatorial pattern. But you can still do it for the regular packing for the
hexagonal pattern because you can define using numerical analysis basically
idea to define the second order, third order derivatives. And this was actually
proved to be convergent.
But we don't know how to do this for higher orders.
Finally I just want to re-call my last telephone call to Oded, and that was almost
six years ago, and I was leaving for China. And he asked me why would I go to
China. He said was it due to 9/11 or something? I said no, I said maybe there's
a lot of misreports in the US about events in China. Actually China is changing a
lot. So and he told me that he wished to do some further work with me and
maybe have some more collaboration and he was looking forward to that.
Unfortunately that didn't realize. And I was really -- I felt really at a loss when I
heard about Oded's passing away last year. Because we lost one of my best
collaborators, actually best two. The other one was my PhD advisor. Anyway, I
feel really sorry, and I was really grateful for the invitation of the organization -organizing committee to give me a chance to talk here and share my experience.
Okay. Appreciate it. Sorry for the delay.
[applause].
>> David Wilson: Are there questions? Michael?
>>: I was trying to recall a little bit from those early days when [inaudible] NSF
post-doc at UC San Diego and John [inaudible] was there finishing up his thesis
that day. Peter Doyle was there, and these three guys would talk about
interstices [inaudible]. I knew there was a lot of interesting stuff there. I always
stayed out of those collaborations because I didn't want to have to try to say that
word during a lecture [laughter] ever in my life. I think a better word could be
found to [laughter] for so linguistically challenged people.
I was always impressed by Oded's beautiful point -- his reliance on elementary
topological arguments. He didn't want to know [inaudible] or anything like that.
[inaudible] where he wanted to live and he realized that there was a lot of
mathematics that would come out of you know understanding the degree of the
circle and the circle that we didn't have to get [inaudible] a few hundred years.
So, you know, I saw this amazing simplicity in Oded at the time.
But you know, I was always amazingly impressed by John [inaudible] analytic
work in that same period. In fact, I was so interested by that, I remember I told
Peter Doyle once that I was thinking that I was going to study analysis seriously,
and he said, look, you know John's phone number, that's all the analysis you'll
ever need. [laughter]. So I stuck with topology. [laughter].
>> David Wilson: Are there other questions or comments? If not, let's thank the
speaker again.
[applause].
>> David Wilson: We'll take a quick break here of three minutes and then start
up again. So 18 after. Pretty much okay. I'd like to introduce our third speaker,
Igor Pak, who will speak on caged eggs and the rigidity of convex polyhedra.
Igor.
>> Igor Pak: Thank you. So hi. So I'm actually deeply honored to be asked to
speak here today. It's really great honor. So thank your organizer for having me.
I'm going to talk about something -- I'm going to follow the same theme of circle
packings, but I'm going to talk a little bit more actually mostly I'm going to talk
about combinatorial aspects of one particular -- one particular paper which I find
extremely beautiful. So this is -- so this is a result. And like probably many of
you I knew the result for many, many years, but I never actually until a few years
ago, I never actually opened it like a normal person and read the proof. And
there's several reasons for that. But basically I -- you know, to be honest, I didn't
think there was a combinatorics in it. I thought there was something else. And
turns out that this particular paper is a wonderful combinatorial result, and I'm
going to talk a little about that, about what's inside the paper.
So this is meant to show that the here is a sphere and there is a convex
polytope, which I'm going to forget about polytope and think about just the cage
which holds this sphere. So I'm going to say that the polytope is mid scribed
around the sphere if actually all these edges are tangent to the sphere. Okay?
Are you with me?
>>: So in the picture it's not true.
>> Igor Pak: No.
>>: Okay.
>> Igor Pak: This is just a caged sphere. And that's how it all started, you know.
People actually looked at the caged [inaudible] particular caged spheres. But
this is -- this is very strong property which is hard to draw. [laughter].
Any way. So what Oded proved is this remarkable statement saying that for
every convex polyhedron there exists another one equivalent which mid scribes
the sphere. For every -- yes. So for every strictly convex point K, there is a
convex polyhedron which describes that convex body. So this is much more
general result than for the sphere. And there is a -- there are lots of stories
behind this result, so I'm going to start with a history and we'll see how this goes.
So here is sort of a basic motivation. Many, many years ago Steinitz observed
that if you asked for inscribed, inscribed or circumscribed polyhedra, that doesn't
work. So a real very fine standard theorem says that if you have a three
connected plane or graph, there is this [inaudible] polyhedron was that graph.
Okay? So this is saying that you cannot make an additional requirement before
vertices being or flying on the unit sphere. That doesn't work. Okay? And
having everything midscribe, so having all edges midscribed is sort of another
way of making strong this classical standard theorem. And it's some of the
miracles that it works at all. If you just look at these pictures, there is no reason
why this should work.
So here is another less well known motivation geometry but actually quite an old
question that people ask. Suppose you look at all possible cages which hold the
unit sphere? What would be the total length of -- the total sum of edge lines that
you need? And so Fejes, Toth, Besicovitch, Eggleston and other people looked
at various packets of that. So if you want to -- so in particular, if you want to say
that unit sphere sits completely inside the polytope that's known and that has a
beautiful answer. That's exactly cube. Whatever it is for a cube.
But in this case, they conjecture that it's actually going to be a triangle prism and
doesn't work. There are better polyhedra, in fact the answer to this question is
still open. Maybe no one thought about it for 20 years, but still that's your
chance. The answer is still open.
>>: So the [inaudible].
>> Igor Pak: Yes. So for the open one. And ->>: So the open one was an open cage?
>> Igor Pak: Yes. So and the real motivation kind of comes from the circle
packing and the -- it sort of -- perhaps not immediate, this is only picture I was
able to draw, so what you do is you take a midscribe polyhedron and sort of cut
your sphere along the facets. What you get is you get a whole bunch of circles
which lie -- which lie on the sphere. And those circles form a circle packing. And
then we get exactly into those pictures that you've seen in two previous talks.
Lots and lots of circle packing. It kind of [inaudible] that way.
So in fact, in a particular case when your convex body sphere, that's exactly a
circle packing theorem. Now here is what doesn't work. It's sort of a strange
thing to put in a slide, but you'll see it in a second. So one sort of normal way to
prove the main result, let me show you the main result again. Would be to take
one of the standard proofs for the circle packing theorem, take the oldest one,
the one that people first came out with, the most natural one, and generalize it to
general convex body. And it's a natural way to do that, and I think it still might
work. So I'm not sure why it didn't work actually. In a sense that you just -- you
know, Oded tried, I guess, and it didn't work for him. So natural way would be to
say well, let's take -- let's say your polyhedron is simplicial, so all phases are
triangles.
Now, consider all possible position of vertices and for every edge, for every, you
know, let's look at the distance between that edge and the convex body. Maybe
square of the distance. If you want to be -- and then you do the usual things the
way people -- let's say you generalizing [inaudible] or some other variational
proof of the circle packing theorem. What you do is you prove some results
about this particular function, and you're done. So you prove that it's convex and
it attains the minimum exactly on what you want.
So somehow my way of making history is I believe Oded thought about it and
didn't work but he did figure out that he needed to generalize that to work with
non-simple polyhedra, okay? So he defines the following figures. He said well,
let's look at all possible -- you know, let's have not just vertices positions in the
space but both vertices and facets. So we have bunch of points and bunch of
planes. And the right functional should take into account both the edges being
tangent to your convex body and the vertices lying on the planes. Okay? That
would be the right idea. And you would need to take some kind of sum of
squares of the distances so it's unclear what exactly would work.
So since it doesn't -- since this approach doesn't work or seems not to work, I
believe what Schramm did is he came out with -- you know, I just heard about
existence of five, six hours of proofs, so I'm not an expert on all hours of proofs.
But He says in the paper well, I looked at some other proofs and here is
something. So he came out with the proof that was motivated, that was
motivated by another proof he did in previous paper. But I looked at the other
paper and the motivation's sort of farfetched. You cannot see how one proof
follows from another. It's not a derivative. So what he says is basically the
following. The most important part of the theorem is prove the rigidity result. I'm
going to -- and I'm going to illustrate the rigidity for this polyhedra, so let's say it's
going to say it's going to be a caging a sphere, and I'm going to fix this triangle
face. Okay?
Now what Schramm says is that there's only one way to position all the
remaining vertices so that this polyhedron cages sphere. So that all edges are
tangent. If you look at it from a combinatorial point of view, you're amazed.
There is no reason for that. Why would these vertices in the corner somehow
depend on this triangle in the beginning? That sounds strange.
On the other hand, if you take the picture that we just did with circles, you
remember this has the same set of vertices corresponding to the three circles
that Ken had in his talk. It's exactly the same picture, so there was a rigidity of a
circle packing that has to be rigidity here.
So now what he does is basically the following. He proves -- he proves the
rigidity result but he doesn't want the global result, he wants a local result, so
what happens when you change things infinitesimally? And then he thinks of
that, and then he wants to use interfunction theorem, so he wants to say that this
infinitesimal result is what exactly you know, saying the Jacobian of some
method nonzero. Okay? And then you can -- you can start moving things
around and eventually and actually he changes the convex body itself. He
changes the K. And eventually he gets what he wants.
So let me -- so what I'm going to do is I'm not going to cover the whole proof.
The proof has sort of technical parts, and which you wouldn't like, and analytic
parts that I personally don't care for. And I'm not qualified to speak about. So I'm
going to give you combinatorial part. And this is all you need from topology really
for that proof. This is saying roughly the following, that if you have a map let's
say both B and A manifolds of the same dimension, and if your map is injective,
continuous, and is proper, then it's homeomorphism. It's very important that they
have the same dimension, otherwise nothing works. Okay?
And so what I'm going to do is I'm going to destroy this topological principle on a
completely different theorem and then I'm going to go back and talk about how,
you know, the rigidity part of -- the rigidity part of Oded's proof. So here is a very
simple question. This you can ask in maybe high school, I don't know. So you
want to describe or you have fixed rays, fixed rays starting from the origin. And
you look at these polygons and you want to describe a sequence of angles of
angles alpha I of all those polygons. What are they? Do they form a nice set?
Okay?
And well, you know, you pretty quickly discover there is one linear relation and
that this one and there are a whole bunch of linear inequalities which they have
to satisfy. So what I mean is you cannot have all those angles to be
concentrating in one direction and there is basically zero here. That's impossible.
And this is where these inequalities come from. That's the only restriction. Now,
here is a key observation. And this is going to be sort of a rigidity result for the
angles on the rays. And the result is the following: If you have two polygons with
the same sequence of angles, then one has to be an extension of the other.
Okay? That sounds easy enough to have a short proof.
And this is the whole proof. You make one polygon to be tiny and so it fits inside
the other one and starts slowly expanding it. You expand it until it hits the bigger
one. Then there's going to be one of the angles smaller than the other. Done.
So this is -- so this is a kind of rigidity results that now you can take. Let me go
back to the theorem. So now if you think about it, you have a map phi from
coordinates of vertices along the rays into the angles. And that map is -- it
continues its injective from what we just proved. And a proffer that takes a little
bit of thinking. And it's actually -- both sets are connected. And then you're
done. Then from just an almost trivial rigidity results so to speak. I would say
uniqueness result.
You get the results that, you know, given the set of angles which decides those
inequalities, their exists a polygon with vertices on the rays which has those
angles. That's the idea. That's a very basic idea. And they are the same ideas
that Oded uses in his proof. So just to illustrate that this is -- this, you know,
angles is not a toy, this is the really theorem, and this is a polyhedra, so this is
three dimensional version of the same result for on the plane. Okay? Where
instead of angles you get curvatures of vertices, et cetera.
And his, Alexandrov's idea was that he should discretize. Suppose you have a
convex body, you take a curvature, project it to the unit sphere, and the question
is can you reconstruct the convex body from this Gaussian curvature? So this is
a discrete version of that result. Okay? And you can actually prove the
continuous version using the discrete version.
All right. Now I want to present a much easier result than what Schramm had.
But I'm going to show you this combinatorial tools that Schramm uses and
hopefully we'll get a flavor for what's exactly in that paper, where is the
combinatorics.
So I'm going to -- so this is Dehn's Rigidity Theorem, which people often confuse
with Cauchy Rigidity Theorem. And in fact, there are sort of -- neither of them
applies each other because the way Cauchy stated it's a uniqueness result. And
this is infinitesimal rigidity result. Okay? So both results are saying that if you
have two -- two polytopes -- if you have one convex polytope then this polytope is
rigid, so you cannot -- you cannot move vertices around so that all the edge
lengths are preserved for simplicial polytope. Now ->>: So why is [inaudible].
>> Igor Pak: Why is it weaker?
>>: [inaudible].
>> Igor Pak: No, no, it's not weaker. Cauchy doesn't -- Cauchy is a uniqueness
result which may be something that comes from the second order rigidity. Not
everything that's -- not everything that's unique is infinitesimal rigid.
>>: Well, this [inaudible] confirmation that doesn't come from rotation [inaudible].
>> Igor Pak: So, correct. So let me say it again. So both Cauchy and Dehn's
Rigidity Theorem show you that you can into the continuously deform, but neither
of them applies each other. Okay? Cauchy is a global -- is a global result. So
anyhow, let me define what's infinitesimal motion. I'm going to put the velocity
vector in every vertex and I'm going to say that infinitesimal motion is the
difference between velocity vectors are [inaudible] to the edge. Okay. And
infinitesimal rigidity -- every infinitesimal motion is actually rigid motion.
All right. It's important to understand. Clear what I just said? Yeah. You're not
following?
>>: I still don't under why it's not weaker than the [inaudible].
>> Igor Pak: Okay.
>>: It's unique and there could be [inaudible]. So that's [inaudible] Cauchy is not
weaker but ->> Igor Pak: I mean every true statement applies another, so -- but it's not
directly [inaudible], no. Anyhow, let me show you what exactly did Schramm do.
So what I'm going to do is I'm going to -- I'm going to weaken, weaken this
property, weaken this property of -- for infinitesimal motion. I'm going to allow a
lighter set of possible velocities. Okay? And I'm going to -- so all I'm going to do
is require them to have basically the same sign when projected on to the edge
vector. That's a much weaker property. So if they're orthogonal, they are both
orthogonal, but if one has a negative projection then so is another. Okay?
And I do that so I -- I'm going to orient all the edges depending on this sign.
Okay? So all that's important going to be the sign. Signs along the edges. And
then basically what you need is a completely combinatorial result saying that if
you have any kind of vectors -- if you have any kind of velocity vectors appointing
from each vertex so that -- so that one particular triangle face is fixed, then -- and
it's satisfies all these three properties, then the whole -- then the whole
orientation, there is no orientation -- edges are not reentered. We have -- we
have basically zeros everywhere. The whole -- the whole -- all velocity vectors
become zero. That's what you prove. Okay? And that's a combinatorial result.
So this is -- this is almost directly from Schramm's paper. What he does is he
looks at the orientations and he defines just almost like in the proof of Cauchy but
some difference also. He defines this full inversions and half inversions. So in
particular, so if you have -- if you have a edge oriented to -- into the vertex and
another oriented out of the vertex, that's going to be two halves. That's one full
inversion. But you can have lots of unoriented edges. So this is unoriented
edge. And you get -- and you get half inversion on each side. All right?
So that's -- that's what he does. And then -- and then a little combinatorial
observation is lemma one, which you can obtain by just basically looking at all
possibilities if you like, that this -- this set of Schramm's inversions is actually very
good in a sense that if there is at least one live vertex in a triangle, then the
whole triangle has at least one inversion. And live vertex, think of live vertex as
the one that's actually moved by those velocity vectors.
So that's -- believe it or not, that's half of the proof. So here is another half of the
proof. So previously we bounded a number of inversions from below. Now we're
going to bound from above by making the following duration, so looking at
triangles, I'm going to look at the vertices. And turns out you start looking at the
vertices there are at most two inversions around every live vertex. And again
you can just consider all the possibilities in each vertex. So I should probably
point out that you cannot have edge going in, out, in, out, simply because there is
this edge. All this edge is [inaudible] so let's say they're pointing towards the
black board. Then projection of velocity vector has to be sort of pointing in one
direction. You cannot have them alternating. And that's all you need to do for
the lemma 2, to just observe that. And the rest is all formal and double counting
argument. That's how the proof goes. And so I know you want to go to lunch.
Let's see. So let me just say that this is -- so I'll finish with an anecdote from that
paper.
So I actually looked at all these details and at some point I got really confused
because Oded spends like two paragraph signing some angles, cal calculating
some formula, and later on I realize that that's all his formula, so in addition to
proving many other things, including Dehn's rigidity theorem he proves all these
formal -- you know, using a non-standard new proves of Euler's formula. Okay.
So that is it. Thank you.
[applause].
>> David Wilson: Are there any questions or comments?
>>: These proofs are [inaudible].
>> Igor Pak: Not at all. Completely different. Dehn calculated the determinant is
a product of -- it's a beautiful, Dehn had a beautiful proof in 1950, and I read it
[inaudible]. It's also in my book. Or in German. Choose your language.
>> David Wilson: Other questions? If not, let's thank the speaker.
[applause]