>> Yuval Peres: Alright, good afternoon everyone. We’re delighted to have Rick Kenyon here. I’ve
heard a short version of this talk before and already, based on that, you’re in for a treat. Rick, please.
>> Rick Kenyon: Thanks Yuval. This is joint work with my colleague, Aaron Abrams. And please feel free
to interrupt if you have questions or comments or ideas. So I want to start with just a little story, a sort
of motivational story. If you have a string of lights—right—all identical light bulbs, you hook them up in
series and you put a battery across from one side to the other then they’ll all glow at the same rate, at
the same brightness. But if you have a more complicated grid like suppose, for example, if you have an
electric blanket or something—right—and you have a bunch of resisters in some sort of network, some
graph, here it’s a planar graph, it doesn’t have to be planar but—you know—and you can hook up your
battery at two locations but then not all… even if the bulbs are identical the amount of current that the
amount of voltage dropped that one bulb is gonna see is a function of all the other resistances and—you
know—they won’t all be the same brightness. Alright, so one of the questions one can ask is, “Can you
design a set of bulbs or set of conductances on the edges that are resistances, put in some—you know—
stronger bulbs here and weaker bulbs there or something so that all the bulbs will glow at the same
brightness, and by glowing at the same brightness I mean the energy dissipated across each resister is
the same, okay? So that is a solution in the linear case but—you know—one may wonder if it’s possible
and how many solutions are there and so on. So this is… so to put this in a mathematical setting here,
hopefully this is familiar to everybody—you know—stop me if you… if it’s not familiar, but I’m gonna
start with a graph G—right—a set of vertices and edges and on each edge I’m gonna have a positive real
number called the conductance. And I’m gonna… there’s a subset of the vertices B called the boundary
vertices and for most of the talk it’s just gonna consist of two points, one at the top and one at the
bottom, the boundary vertices. And on the boundary I have some function, let’s call it u, some fixed
function, fixed once and for all. And now we’re looking for a function f on the rest of the graph, f on the
vertices, which is harmonic, and harmonic means that the value at every point is the weighted average
of the neighboring values, so weighted by the conductances on the edges. So if all the conductances are
one then you know it’s just gonna be the actual average of the neighboring values. And I want, of
course, the function on the boundary to be equal to u. So that’s a classical Dirichlet problem, here’s the,
by the way, the underlying equation for harmonicity is this weighted Laplacian, so the Laplacian of the
function f that we’re looking for is the sum of the f of x minus f of y. So if you look at all neighbors, for
each x you look at the neighbors y of x and you multiply f of x minus f of y times the conductance there
and you want that to be equal zero at every internal vertex x.
Alright, and standard fact is that the, you know, there’s a unique harmonic function with those boundary
values and it’s given by… it’s the function which minimizes the so-called Dirichlet energy. The Dirichlet
energy is, by definition, the sum over all the edges of the graph, the conductance times the difference of
the function on the two vertices, right? So for every edge you look at the difference of the function, you
square it, multiply by the conductance and you sum that up over all the edges, you get the Dirichlet
energy. And we’re gonna call this quantity the edge energy, the thing you’re summing up.
And so one more ingredient we need to discuss is the so called orientation of the edges. Whenever you
have a harmonic function on a graph there’s an induced current flow, in fact, if you think about the—
you know—the current in the light bulb example—right—the current is flowing from higher voltage to
lower voltage, so on every edge there’s an orientation which tells you which direction the current is
flowing. You know, if I put voltage one here and zero there and I let the thing relax to the equilibrium
state, then the current will flow, this will be a source of current, this will be a sink of current and then at
the intermediate vertices, well there won’t be any sources in sinks in the graph because the current has
to be… there’s no loss of current in the interior, right? So each interior vertex has to have some
incoming and some outgoing arrows. Okay, well there may be some edges which have no current but at
least if you choose sort of generic conductances and that won’t happen. I mean, having zero current is
sort of a nongeneric situation.
Alright, so that’s an orientation of the edges, it’s an acyclic orientation, meaning that—you know—it
goes downhill, down potential, and we’re gonna be interested in the set of acyclic orientations which
could potentially arise from these boundary values. So remember we have a fixed set of boundary
vertices and a fixed function u on the boundary and given that data there’s a certain set of acyclic
orientations which can happen and those… it’s not hard to show that those are obtained from, well
okay, I should have expanded this a teeny bit, but let’s call capital sigma the set of orientations, acyclic
orientations sigma, what I mean is by just a choice of orientation for each edge, such that, well, which
arise from functions on the vertices which have no interior extrema and are equal to u on the boundary.
So look at functions which are equal to the u on the boundary, have no interior extrema, no maxes or
mins inside, and then whose such that the signature of the derivative on an edge, I mean, the direction
of which the current would flow under f has to be equal to sigma.
Okay, so that’s a finite set of acyclic orientations, acyclic with only sinks and sources on the boundary. Is
that definition clear, what I mean by those kind of orientations? If you only have one source and one
sink like in this example, then it’s the set of all acyclic orientations with only that source and only that
sink.
Okay, so we are going back to our original problem, we want to choose conductances so that we get
whatever desired vector of energies we want, right? So let’s call this map capital Psi, the map from—
right—a vector of conductances which are positive real numbers to a vector of energies, right? So the
energies are always gonna be non-negative—you know—for some special choices of conductances you
may get some current zero and then the energy will be zero. So you have to include the left… the zero
possibility over there. And just by way of example—right—here’s the simplest nontrivial graph you
might consider, right? It’s got two boundary vertices, v zero and v one, and—you know—we’re just
gonna put… u is gonna be one here and zero there, so you think of the current as… think of this circuit as
being hooked up to a one volt battery across those two vertices, and let’s call a through e are the five
conductances, right? And then you can write down an equation for the energies as a function of those
five variables, a through e, the five conductances. And—you know—it’s a little bit complicated, but
there it is. Right, that’s the energy on edge one, energy on edge two, and so on, energy up to energy
five.
And now, how do you invert this equation, right? Now if I give you a set of energies, can you solve this
for a one through… I mean a through e? That’s the question, it’s just an algebraic… algebra question
now, right? Invert this map and—you know—I did, I just—you know—for illustration I did it in this
particular case. It’s not actually so easy, right? I wouldn’t want to do this by hand, but the computer
could handle this example, and I just… if you look at this equation here, so there’s… I took five energies,
little e one through e five, and I solve for a. So this is an equation, well it’s a polynomial and square root
of a—quadratic polynomial and square root of a—which determines a as a function of the five energies.
And well, it turns out…
>>: There’s some homogeneity right? So if we scale everything up we should have the…?
>> Rick Kenyon: That’s right. If you scale all the conductances by a constant then…
>>: So you get the same energy.
>> Rick Kenyon: You get the same energy scale also by the constant squared. So well, by the same
constant connect. So yeah, so there’s one fewer parameter than is apparent, than look… but none the
less. This is kind of interesting if you stare at this equation a little bit, well, for one thing, it’s not
completely obvious but there are two roots. It’s a quadratic equation—you know—if there’s one root
then there’s gonna be another root. Both roots are positive, right? ‘Cause you can tell that these are—
you know—the constant and the quadratic coefficient are positive numbers, right? So, in fact, there are
two solutions to a. In fact, if you invert this for any set of energies there are exactly two solutions, which
also happens to be the number of acyclic orientations of this graph. You know, if I look at an acyclic
orientation which has a source here and a sink there, then there’s two possibilities, right? The center
guy and be direct… I mean, these… a and b have to be down, d and e have to be down, c can go either
way. So it’s not really surprising that there are two solutions here, but there are. And—you know—
that’s the general statement, is that if you have any acyclic orientation which are compatible with the
boundary and whatever your favorite choice of energies are which are positive—I forgot to say positive
here, strictly positive— then there’s the unique choice of conductances which realizes that data, okay?
There’s a unique choice of conductances for which the associated harmonic function realizes this data.
So in particular, if you want all the energies to be one, for example, then the number of different choices
of conductances, the number of solutions to that equation is exactly the number of acyclic orientations,
which—you know—some people are interested in—you know—counting acyclic orientations of graphs,
this is sort of one way to do it is to find the number of solutions to this equation. Alright, and so…
>>: David Alvis had an idea for a book, “Complicated Solutions to Easy Questions.” [laughter]
>> Rick Kenyon: This is actually an NP-hard question, so—you know—you shouldn’t dismiss it too
quickly. [laughter] But—you know—it’s a good… I understand, alright. Okay so, well so, if you like
number theory algebraic geometry you may be interested in some topological features of this rational
map, the rational map of the type that I—you know—gave an example on the previous slide. You can
think of it as just as a map from complex numbers to complex numbers—right—you don’t have to
restrict to positive real numbers, and then it has a—you know—you can show that actually the
topological degree is exactly the number of solutions, the number of acyclic orientations. So this
theorem is a little stronger than theorem one because it says that there are no other solutions besides…
other than the real ones, the ones that are given by theorem one.
>>: In a way, but it doesn’t imply theorem one.
>> Rick Kenyon: No, it doesn’t imply… no, okay, so I shouldn’t say stronger, I should say it gives you
more information about what’s going on. But in particular, if you plug in, if you have rational boundary
values u and suppose that you’re interested in finding rational energies, for example, all the energies
one which is sort of the natural solution, a natural condition that you’re looking for, then the—you
know—you’re solving this algebraic system of polynomial equations for every Galois automorphism of
the—you know—absolute Galois group, the group of totally… the group of algebraic numbers will give
you a new solution. Because all the… because the rational function has rational coefficients, the Galois
group permutes the solutions, in fact, the Galois group… the relevant Galois group here is just the group
of the totally real algebraic numbers, because you never see… everything is always real. The solutions
are always real; you never have to deal with complex numbers. Anyway, that’s just something that
might interest number theorists. Okay, any questions about the statement?
So in order to illustrate or draw pictures of these solutions there’s a nice construction due to, well, this
four authored paper, Brooks, Smith, Stone, and Tutte, and they called it a Smith diagram, apparently it
was discovered by one of those four people in the late thirties and…
>>: Which one?
>> Rick Kenyon: Smith. [laughter] Sorry. Okay, so suppose I have a planar graph now and it’s got a
harmonic function on it—right—and so from that you can draw a tiling of a rectangle, well okay, this is a
case when the boundary just consists… again, just consists of two vertices, the v zero and v one. Then
you can draw a tiling of the rectangle which geometrically realizes that harmonic function, and it goes as
follows: you, well, for each edge in the graph you’re gonna have a rectangle over here, and I’ve labeled
the rectangles according to the edges over there, and for each vertex over here there’s a horizontal
segment in the—sort of maximal horizontal segment—in the union of the rectangle boundaries. And so
each rectangle over here has a upper boundary and a lower boundary and that corresponds to the two
vertices which that edge connects. And the—you know—the lower boundary is the v zero and the
upper boundary is v one, and the… so far it’s just that’s the topological description of the tiling, but then
once you have a harmonic function on the graph then you can assign a height to all the rectangle
boundaries which is the corresponding voltage of the interior vertices. The value of the harmonic
function tells you the height, the y coordinate here, and so the voltage is the y coordinate and the width
of the rectangle is the amount of current going along that edge. So the—you know—you have these
boundary values one and zero and then you solve for the harmonic function, each edge has a certain
current, so it has a certain… that corresponds to… that gives you the the width of the rectangle. And—
you know—for each vertex, the rectangles above it correspond to the current coming in and the
rectangles below correspond to current coming out, and the sum of the current coming in equals the
sum of the current coming out. Does that make sense? So the sum of the widths above each edge is
equal to the sum of the widths below the edge. That’s the…
>>: Kirchhoff’s formula.
>> Rick Kenyon: Kirchhoff, no-loss-of-current rule. Okay, and so if you know the width and the height
you can get the aspect ratio of the rectangle that corresponds to the conductance, the width divided by
the height is the conductance. That’s why we’ve been calling the conductance. And the energy is the
height times the width, that’s the area of the rectangle. So the—you know—if you ask for a solution
where the combinatorics fits the graph and all the rectangles have the same area, this is… you’ve solved
for the harmonic function which has the correct… all the energies equal. Okay, is that clear? Is that…
how many people know this construction already? Half you. Okay, and of course you can read off the
orientation from the tiling because you know—you know—for each vertex—you know—which vertices
are above it and which are below it.
Okay, so now we can just play the game. Here’s a particular graph, sort of my running example, it’s
got… happens to have twelve basic orientations, and for each orientation the theorem says there is a—
because it is a planar graph—there is a tiling of a rectangle with all the rectangles of area one. And
that’s one of them. So it’s kind of a cute way to relate the energy… fixed energy problem to the area
one rectangulation problem. And what you see in this example is that, well okay, so it’s significantly
more complicated to solve this system in this case and—you know—if you look at the width of the first
rectangle there, the amount of current going along that first edge, it’s the root of some nasty, quite
large polynomial which—you know—Mathematica spit out for me and—you know—I didn’t spend much
time thinking about this, but each of these coefficients here has some combinatorial meaning—some
mysterious combinatorial meaning; it’s the… each of these coefficients is a positive polynomial in the
energies. Here all the energies are one, so if it comes out to be an integer, but if I’d written it out as
with twelve variable energies, e one through e twelve, then you were to see this is a… that this
coefficient is a positive polynomial in all those things with—you know…
>>: Integer coefficients?
>> Rick Kenyon: With integer coefficients, yeah. And so there is some interesting combinatorial
meaning which I wish I… but I don’t know what it is. I mean, you kind of saw that in the previous
example here, right? Each of these coefficients is… has all the coefficients, all the coefficients are
integers and they’re all… they’re the same sign—the signs alternate, but… Okay, and as I said, in this
case there are twelve orientations, right? So there are twelve rectangulations, all sort of combinatorially
equivalent to each other, but—you know—if you go back to this polynomial, this polynomial has twelve
real roots and the… one of those roots is the width of this tile, and the other eleven roots are the widths
of the tile labeled number… tile number one in the other twelve tilings. Because the—you know—the
Galois group permutes the roots of that polynomial, it permutes the tilings as well and it will move one
of these tilings to some other tiling. I mean, it permutes these twelve objects, always taking—you
know—if you look at all the widths of tile one, there’s twelve widths and they together form the roots of
that polynomial. It’s kind of a cute, very intriguing number theory connection. I don’t know if that… this
is a very useful observation or not, but the… and you know…
Well, how many acyclic orientations does a graph have? I mean what is the size of this thing, this
number of compatible acyclic orientations of graph? And well, so this is something that people have
studied and it’s in this… if we can add an edge between v zero and v one if there’s not one already, it
doesn’t really affect anything about the problem. Then this number, the number of acyclic orientations
compatible with these two boundary vertices is called the chromatic invariant of the graph, it’s some—
you know—reasonably well studied graph invariant. It’s the—you know—derivative at one of the
chromatic polynomial, if you happen to know what the chromatic polynomial is, or if you know what the
Tutte polynomial is, it’s the derivative at zero, the x derivative at zero. But in particular it’s an NP-hard
thing to compute. But you know the… yeah, in fact—you know—when I just took this graph somewhat
randomly, I plugged it into the Mathematica, it spit out this degree twelve equation and I knew at that
point that there were twelve solutions. So somehow it counted the number of solutions, it computed
the number of solutions to this NP-hard problem by some mysterious means, right? But let me… let’s
move on.
So there was this theorem that the main theorem says that—you know—for every acyclic orientation
and set of energies there is a solution. So… and the proof really just fits on one slide here. We’re
looking for a harmonic function, right? Here’s the Laplacian, but—you know—we’re not giving the
conductances, the conductances are unknowns, but we can rewrite this ce times dh as e over dh,
because the energy, the energy remember, is the conductance times the dh squared. And so instead of
looking at our standard harmonic equation here—right—we can rewrite the equation like this because
the energies are what we’re given. So now we have… now our unknowns become the values of the
harmonic function. So we have to solve this new equation, zero equals… well, it’s a system of
equations—right—one for each vertex x, the sum over the neighbors of x, h of x minus h of y, inversed,
times the energy. So we call this the enharmonic equation which is short for energy harmonic because
we’re given the energies rather than the conductances.
And well, how do you solve this equation? Now it’s in nonlinear, right? We have to—you know—
unfortunately it’s no longer linear, but it still has some nice properties and one of which is that the
solutions are critical points of a certain functional, just like in the case of the harmonic equation. Alright,
we had the Dirichlet energy was our nice functional and the harmonic function was the minimal of the
Dirichlet energy. Here the solutions are critical points of the following functional, which is the product
over the edges of the—you know h of x minus h of y to the power of the energy E, E sub e.
Well, why is that true? Well—you know—just differentiate this. Take the log rhythmic derivative of
M—you know—and you get that for every… with respect to h of x, and you get this, this equation. Well,
except for the… you have to worry about the absolute value signs which, well, what’s going on at those
absolute value signs? Well, that’s when the—you know—the thing is not differentiable, but it’s only not
differentiable when h of x equals h of y, right? And that’s a place where there’s some edge with energy
zero. So as long as you avoid the places where energy’s zero this is not… you’re not gonna have
problems. And, in fact, let’s… we can restrict ourselves to a certain set of functions h, let’s restrict
ourselves to the set of functions h, which have the correct orientation on each edge, right? Remember
we have some boundary conditions and if we fix the orientation ahead of time, we’re looking for a
solution which has this particular orientation. So we only need to consider functions h such that the
value here and the value here, this one is bigger than that one. So that essentially determines for each
absolute value sign which one is bigger, h of x or h of y. But that’s… and it’s also a polytope; it’s a
polytope of the set of functions with a fixed sigma… I mean a fixed difference along each edge. And on
that polytope this function is, in fact, strictly concave because it’s… well, if you take a log rhythm it’s a
sum of log of h of x minus y… log is a concave function, it’s the sum of concave functions, so it’s concave.
And because it’s, well, strictly concave and on this polytope, and on the boundary of that polytope it
goes to minus infinity because—you know—the log of zero is minus infinity when you get to… when you
tend toward the boundary. So I guess the relevant picture here is that you look at the space of all
possible functions h, it’s divided up into these polytopes like that. And on each polytope the function is
strictly concave and therefore has a unique… it goes to minus infinity on the boundaries and therefore it
has a unique maximum. One inside each… one for each polytope. Okay, so that gives you for every
orientation, that is, for every polytope, a unique maximum, maximizer. Okay and I just want to highlight
this equation. This is gonna be the most important equation here, the enharmonic equation.
Okay. Well okay, so theorem two is not also very short. If you recall, theorem two says essentially that
there are no complex solutions, all the solutions are real. How do we see that? Well, here’s the
equation we’re trying to satisfy. The energies are positive real numbers. We’re just trying to show that
any solution to this equation has h real. And there’s a well-known theorem, the Gauss-Lucas Theorem
says that if you have any polynomial p the roots of the derivative of p are contained in the convex hull of
the roots of p itself… anyway, some classical theorem, the proof is like one line; it’s on the next slide. So
once you accept this theorem then you can see that if h satisfies this, then h, h of x is a root of p prime
of z, where p of z is this polynomial here. You take z minus the h of y, where h of y runs over the
neighbors of x. Think of that as a function of z it’s a… well, it’s almost a polynomial except these may
be—you know—real numbers, but looks kind of like a polynomial, generalized polynomial. And—right—
if you just differentiate p logarithmically you get that equation for h of x. So that implies that by the
Gauss-Lucas theorem that h of x has to be in the convex hull of the neighboring values the h of ys. So on
my graph—you know—for every point, the value of h at any point is inside the convex hull of the
boundary val… its neighboring values. But all the boundary values are real numbers and therefore
everything has to be real, right? If anybody had a positive imaginary part for example, then you could
find a neighbor which had at least as large imaginary part, and so on. Essentially, the maximum
principle.
Okay. Is there a clock here? Somebody want to see the proof of Gauss-Lucas theorem? [laughter]
Somebody showed this to me. Was it not… was it somebody in this room? Was it you, Ayel?
>>: No.
>> Rick Kenyon: Okay. Yeah, well you can find it on Wikipedia. It’s just one line. Here’s my enharmonic
equation. I just took the complex conjugate—right—but it’s the same, because the energies are real, it’s
also true, right? Conjugate of zero to zero. But then I multiplied the numerator and denominator by h
of x minus h of y—right—so the denominator now becomes real, this is real and h of x minus h of y,
those may be complex, but then you put the h of x on the other side, then it has a real coefficient here
and so what you see is that h of x is a convex combination of the h of ys, right? This sum is equal to that
sum without those guys. So the h of x is a convex combination of the h of ys and therefore it’s in the
convex hull. Okay. That’s pretty cute and very easy.
Alright, let’s see. Maybe I’ll skip this guy. Well okay, we can… I already gave you an example, but the
nice thing about this variational principle, this maximization scheme, is that it’s very easy to construct
explicit examples. Now, so what I did here is I took a grid, a forty by forty square grid, right? I turned it
forty-five degrees and then—you know—I don’t know why, but I took one source here and one sink
there. So you think of this as a very dense forty by forty grid, and I oriented all the arrows—you know—
south and west, what used to be south and west. And then, given that orientation you can ask, “Well,
how do I find conductances so that all the energies are one?” Or in other words, can I find a tiling of a
rectangle which has the combinatorics of the square grid such that all the rectangles that area one?
Well, there it is. It’s a unique solution once you fix the orientation, right? Now, it’s just a few seconds
on your computer to maximize this thing. But, what can you say about… I mean, there are lots of
questions if you like geometry. Now—you know—is this sort of converging to some interesting
mapping? So this is really a… you can think of this as some sort of mapping from the diamond shape to
the square which—right—which, you know, the top vertex sort of goes to the top boundary but—you
know—it’s sort of continuous inside, each little unit of area here goes to some unit of area in the square.
It’s an area preserving mapping, but—you know—I don’t actually know what mapping that is it
converges to. It’d be kind of interesting to find out. But—you know—you can play around with these
things because it’s very easy to do local deformations, right? Here I just imposed a row shape on that
previous picture and then I suppose you’re interested in some—you know—property of that picture
near some location, you can just re-solve for the tiling which increases the area of some rectangles near
the area… near the region of interest, right? That’s what I did here is I just blew up the areas of some of
those rectangles according to some—you know—and so this part of the picture blows up, the rest of the
picture gets a little bit smaller, but it’s an interesting sort of deformation which is—you know—not
conformal which is… that’s why it’s… I think it’s kind of interesting and novel, right? Okay, and yeah, you
can… here I just playing around with these things. This is a sequence. You know, if I just change the
area of one of the rectangles, you can see how the other rectangles sort of move in that… under that
deformation.
So it turns out that there is a nice scaling limit of these mappings, right? If I take a larger and larger,
finer and finer grid and I do the same mapping into a rectangle, I can see what happens as the mesh size
goes to zero and you do get some very interesting solution in the end. That is, the functions converge
and they converge to some analytic functions which satisfy the following: so twisted Cauchy-Riemann
equations. You know, here u and v are the x and y coordinates of the image, x and y are the x and y
coordinates of the domain, and the u and v have to satisfy this system of PDEs here which resemble the
Cauchy-Riemann equations for complex analysis but they’re nonlinear. Yes?
>>: Does it say that these permutations that you showed in the slide where there were three by four of
them, that if you make a small modification then it’s kind of local? These…
>> Rick Kenyon: I understand. Yes, if you make a small modification of the orientation then I believe—I
haven’t heard anything to this effect—but I’m pretty sure it’s true that the picture will just change
locally, like you say. That is, the effect of… so what I did with the big picture is I took the sort of a nice
periodic orientation where all the edges were oriented south and west. But what Ayel is suggesting is—
you know—what happens if I just change the orientation of a single edge? Will the picture change in a
drastic way or will it change in a local way? And I think it just changes in a very small location, sort of
exponential decay of influence there. But I don’t know how to analyze that.
So yeah, so the usual Cauchy-Riemann equations are over here. There’s some nice linear things. Here
I’ve got the… that’s the constant conductance, Laplacian or—you know—analytic functions. Here are
the Cauchy-Riemann equations in the constant energy case—you know—they do resemble it but—you
know—here the associated Laplacian is our friendly Laplacian. Here there is a Laplacian nonlinear… I
don’t know what to call that guy. If you have some good terminology for these things I’d be… I’d
appreciate it. We call it the enharmonic Laplacian, but I don’t know if that’s any good either. [laughs]
Yeah, so both the real and imaginary part here satisfy the same… just like in this case—right—the real
and imaginary parts, the u and v, both satisfy the same Laplacian equation. Same thing here, but that
doesn’t mean I know how to solve this equation. Okay and—you know—so just playing around here, I
took a random orientation of the square grid and I packed it in a rectangle with area one rectangles. Of
course, it’s not so easy to find a random orientation. This is… there are no—you know—good algorithms
for sampling a random orientation of a large graph, random acyclic orientation, but so I cheated a little
bit here, but it’s the best I could do. [laughs] I used some Glauber dynamics and waited a long time.
Yes?
>>: Is this rotated? Wouldn’t we expect to see the very top line and the very bottom line to be the full
width? Look, all the previous examples the very top one and the very bottom one are… it is a slit, the
full, the graph…
>> Rick Kenyon: I think what I did in this case is I—I apologize—I took different boundary conditions,
which is that I glued all the lower… the left and lower boundary to a single vertex and all the upper and
right boundary to a single vertex. Don’t look too closely. [laughter] But one thing you can do which…
>>: You just flipped and censored and moved the created cycle?
>> Rick Kenyon: Yes. That’s right. Right, so this is a well-known way to connect the space of
orientations, bipolar orientations, which is to choose an edge and reverse its orientation if it’s allowed.
One thing that is known that’s a little bit…
>>: So this picture creates a bipolar disorder? [laughter]
>> Rick Kenyon: Exactly. So if you don’t fix the graph, but if you pick a random graph with a random
bipolar orientation—I should’ve said bi… I should start saying bipolar because bipolar means there’s a
single source and a single sink—then these objects are in bijection with some well-known combinatorial
objects called Baxter permutations. People know how to count them, people know how to generate
them, and here’s an exact random sample of a triangulation with, I don’t know, forty thousand
rectangles of a square—all of equal area. So I don’t know if you can see anything interesting going on
here but I thought it was kind of pretty. [laughs] But yeah, so there are things you can see in this case,
but not… which we can’t see in this case is a grid.
And—you know—just one last thing which I’m working on and hopefully will lead somewhere soon: can
we count the bipolar orientations of the grid, right? Remember the bipolar orientations are acyclic
orientations with no sources and sinks, or rather with just one source and one sink. And so here’s sort
of a scheme, plan of attack—right—using this technol…new technology, and so what I did is I took a
cylinder—right—and I glued the left… all the vertices on the left to a single vertex which is gonna be my
sink, and all the vertices on the right, that’s gonna be my source. And—right—so I’m interested in how
many—you know—orientations are there where there’s… with that sink and that source, that source
and that sink. And the theorem says all I have to do is count the number of solutions to that
enharmonic equation, right? Every solution gives you a unique orientation and vice versa. So let’s call X
i, j the value of the enharmonic function at an interior vertex i, j, and then we get… we can get a
recurrence relation—right—because the value here just depends on the four neighbors, right? The
value to the right is… I can think of that as a function of these four guys, it’s a rational function of these
four guys, so you get a recurrence X i plus 1, j is equal to X i, j plus this rational expression, right? So you
can start with some initial data here on the first row and then all the other points just depend on those…
on the values on the first row, and at the end you want them all to be equal to one X—you know—n plus
1, j. So using this recurrence we can write X i, j as a rational function of the X’s on the first row… on the
first column there. And there are techniques for counting the number of solutes… so then you get
some… at the end of the day you get some… I mean when you get to the end of the cylinder, you get
some system of equations; you have to set them all equal to one. And—you know—algebraic
geometers know how to do this—right—when I… if you want to compute the number of solutions to
some big polynomial system it’s the—you know— the mixed volume of their Newton polygon, Newton
polytopes, and the essential thing to do is to compute the growth rate of X i, j as a rational function of
the initial data. And I spent some time doing this calculation and… but unfortunately it came out too
big, so I probably made a mistake, but it seems to grow as 4 to the i which would indicate that there are
4 to the v, where v is the number of vertices orientations, but that’s clearly too big because—you
know—if you just orient each edge both either way then that’s already four per vertex, so that’s way too
many. So I probably just missed something obvious, but it seems, I mean, I feel like that should be
something I can complete sometime soon.
Okay, well unless people have questions I have just one last open, interesting open generalization—not
generalization but related problem which is about triangles, right? Since we dealt with rectangulations
with area one rectangles you can ask, “Is there’s a analogous result for triangulations?” In fact, the
reason I mention this is ‘cause this was actually our motivating problem, right? Aaron asked me this
question about triangles, which is sort of an old problem and I suggested that we look at rectangulations
‘cause they’re often easier. But so—right—for example, if I take this triangulization of rectangle can…
there are seven triangles; can I assign the areas arbitrarily? And the answer is no. And well, I’m not
gonna tell you why the answer is no, but here’s a related question, sort of well-known among people
who know it well, the theorem of Monsky which says that there… you can’t dissect a square into an odd
number of equal-area triangles.
>> [indiscernible]
>> Rick Kenyon: This is a fun question to think about if you don’t believe in and the reason it’s so
mysterious is that the proof uses the axiom of choice.
>>: Uses what?
>> Rick Kenyon: Axiom of choice.
>>: What? [laughter]
>>: Axiom of choice?
>> Rick Kenyon: The proof uses the axiom of choice.
>>: Axiom of choice.
>> Rick Kenyon: Axiom of choice, yeah. But—right—so, for example, here’s a triangulation with nine
triangles—right—and it does pretty well. [laughs] Right? So the area of each triangle is not quite one
ninth, but it’s pretty close.
>>: How close can it get to the function of the number of triangles?
>> Rick Kenyon: So there’s a paper… this was taken from a paper and I apologize for forgetting the
reference, but I have the reference on my computer. I will pull it up afterwards. So yeah, that’s an
interesting question. How close can you get as function of the number of triangles?
>>: If we can’t get arbitrarily close it won’t be some compactness issue or something that’s...
>> That’s right. You can get arbitrarily close. So there’s some gap, but we don’t… I don’t know what the
gap is. But sometimes there are solutions and sometimes there are combinatorially equivalent solutions
just like in the case of rectangles. So here’s two triangles which—you know—if you define the
combinatorics appropriately they are combinatorially equivalent, right? Each… I tried to label them so
you could kind of see what’s going on. Basically what you have to do is… so you see this edge, this edge
here which is adjacent to triangles four, six, and seven, there’s an edge over here which is adjacent to
triangles four, six, and seven too. But the—you know—you should think of this black edge as some sort
of degenerate triangle, right? It’s got three vertices on it and they sort of… this is also a triangle with
three vertices on it. So that… there’s some sense of how these are combinatorially equivalent and
sometimes you can get two solutions, sometimes you can get many solutions, but the number of solut—
unlike the case of rectangulations—the number of solutions is not always constant. Sometimes there
are complex solutions, sometimes there are no solutions. So this is… still, we still don’t really know
anything about this problem, but there is a nice analog of the harmonic function, uh…
>> [indiscernible]
>> Rick Kenyon: …correct the relation between the network and the tiling. And here, instead of talking
about a symmetric Markov chain we do an asymmetric Markov chain so you have a… you still have a
planar graph, each edge has a… in this case it’s a Mark… think of it as a Markov chain rather than a
resister network, but it’s… other than that it’s roughly… it’s fairly equivalent. Let’s have two outgoing
arrows from each vertex except for the top and the bottom and, well okay, I don’t know if I need to go…
>>: The green one and the red one.
>> Rick Kenyon: What?
>>: From the green one and the red one. Oh sorry, just the green one.
>> Rick Kenyon: What about the green one and the red one?
>>: Oh, no, sorry. Two out. Okay.
>> Rick Kenyon: Two out.
>>: The number of n is not specified.
>> Rick Kenyon: The number of n is not specified, but so the way you… the way that mapping works,
this mapping always works, is you take each triangle… let’s rotate the picture so that there are no
horizontal edges except maybe the bottom one in this picture. Okay? So just take and sort of generic
rotation so there are no horizontal edges, then each triangle has three vertices—you know—a top one,
a bottom one, and a middle one, and from the middle one you draw this dotted line. And so each... for
each triangle there there’s a vertex over here and you can think of the top part of the triangle as having
an outgoing arrow to the upper and an outgoing arrow to the lower one. So the green triangle, which
corresponds to the green vertex, connects to the purple one and to the bottom one, like here. And then
there’s a way to assign—you know—transition probabilities to the edges, let’s see, it should be down
here. Well, so every geometric concept over here corresponds to some probabilistic concept in this
network. The y coordinate over there is a harmonic function. Harmonic function is one which is—you
know—the value of the point is the average of… the weighted average of the neighbors. The… here—
you know—each edge has a slope and the slope corresponds to the winding number of a random walk.
You take a very long random walk where you add an edge from the top to the bottom and you ask for
the expected winding number around a face and that gives you the slope over there, and so on. The
areas are the energy. So there’s this kind of nice bijection just like in the previous case and one hope is
that we can use… some way use this bijection to say something interesting about the existence of or
nonexistence of fixed area rectang… triangulations in that case, but hasn’t been done yet. So I’m gonna
stop here. Thank you. [applause]
>>: Questions or comments?
>> Rick Kenyon: Yeah?
>>: So in the problem with dividing the square into equal-area triangles, is it…
>> Rick Kenyon: Yeah.
>>: … that if you assume the axiom of choice you actually can do it?
>> Rick Kenyon: No, no. The theorem is that there is no dissection.
>>: Right, but in order to prove that you need the axiom of choice. So if you assume the…
>>: The negation you’re obtaining…
>>: The negation of the axiom of choice.
>> Rick Kenyon: No one has… no one has that… well okay, to say that the proof uses the axiom of choice
is a little bit too strong because… well, is it too strong or not? No one has been able to give one.
[laughs] You can’t just be…
>>: So there are theorems where we know that you need the axiom of choice; this is not one of them,
right? You have no proof that you can’t prove it without the axiom of choice.
>> Rick Kenyon: That’s right. That’s right. It may be possible to prove it without the axiom of choice.
Thank you. [laugh]
>>: Every time you use integration—Lebesgue integration—you also use the axiom of choice.
>> Rick Kenyon: [laughter] No. My computer does not use the axiom of choice when it integrates a
function.
>>: Can you construct the Lebesgue measure without the axiom of choice?
>> Rick Kenyon: No, I never need to construct the Lebesgue...
>>: Not on all sets, but on triangles.
>>: Yeah, man.
>>: [indiscernible]
>>: Any other comments or questions? If not, let’s thank Rick again.
[applause]