random stepped surfaces Richard Kenyon (Yale) <latexit - - PowerPoint PPT Presentation

random stepped surfaces

the hidden geometry of

Richard Kenyon (Yale)

Here σ(s, t) = √ 1 + s2 + t2 is the “surface tension”.

min

h

ZZ

U

q 1 + h2

x + h2 y dx dy

Minimize area:

Re ✓Z f(z)(1 − g(z)2) dz, i Z f(z)(1 + g(z)2) dz, Z f(z)g(z)dz ◆

Weierstrass-Enneper parameterization of minimal surfaces

...

Let f, g be (arbitrary) analytic functions, then

parameterizes a minimal surface in R3.

h : R2 → R

“Lozenge” tilings

lozenge tilings are stepped surfaces

What happens for large system size?

Dimer cover of a graph: perfect matching of the vertices.

The Dimer Model

Dimers on honeycomb lattice

Thm[Kasteleyn 1965] For G a part of the honeycomb graph, let K be the bipartite adjacency matrix, Kwb = ( 1 b ∼ w else. Then | det K| is the number of dimer covers.

Example: K is 12 × 12; det K = 20.

Proof:

det K = X

σ∈Sn

(−1)σkw1bσ(1)kw2bσ(2) . . . kwnbσn

Large random tilings seem to have a “shape”

A typical surface lies close to a certain nonrandom surface

Thm[Cohn,K,Propp (2000)] The function h : R → R describing the limit shape is the unique minimizer of the surface tension integral

Lozenge tiling limit shape

min

h

ZZ

R

σ(hx, hy) dx dy

0.0 0.5 1.0 0.0 0.5 1.0 0.3 0.2 0.1 0.0

The surface tension σ(s, t)

.

= volH3(ideal tetrahedron with dihedral angles (πs, πt, π(1 − s − t)))

σ(s, t)

s =density of green t =density of blue for each slope (s, t) there is an associated growth rate (entropy) −σ(s, t):

(Number of tilings) = e−Area·σ(s,t)(1+o(1))

1 − s − t =density of red

Alternatively, give tiles weights a, b, c depending on orientation.

Then choose a tiling with probability proportional to its weight aNabNbcNc.

{

How is σ calculated?

is the Legendre dual of F

First compute the “free energy” F (the normalized log-of-determinant).

F(a, b, c) = 1 4π2 ZZ

[0,2π]2 log(a + beiθ + ceiφ) dθ dφ

eigenvalues of K

σ(s, t) = −F(a, b, c) + (1 − s − t) log(a) + s log(b) + t log(c)

a

{

F(a, b, c) = 1 4π2 ZZ

[0,2π]2 log(a + beiθ + ceiφ) dθ dφ

eigenvalues of K

There is a special point (θ, φ) where a + beiθ + ceiφ = 0.

1

z = − b aeiθ

a

beiθ

ceiφ

−beiθ

z

πs πt

a

b

c

How are weights related to densities?

π(1 − s − t)

If a ≥ b + c, system is “frozen” in all red.

In particular,

if b ≥ a + c, system is frozen in all blue;

if c ≥ a + b, system is frozen in all green.

s =density of green t =density of blue 1 − s − t =density of red

How to solve the variational problem? The Euler-Lagrange equation for a gradient model is

div (r σ(r h)) =

x,y

x,y

s,t

...

Thm[K-Okounkov] z = z(x, y) is a minimizer iff y = z z − 1x + f(z) for some analytic function f.

Idea (Amp` ere): rewrite in terms of the conformal coordinate z (or w).

...a nonlinear PDE.

One can solve the minimization problem for “polygonal” boundary conditions...

in which case the arctic boundary is an algebraic curve.

Arctic circle theorem [Jockusch-Propp-Shor, Cohn-Larsen-Propp]

x(t − a1)(t − a2)(t − a3) (t − c1)(t − c2)(t − c3) + y (t − b1)(t − b2)(t − b3) (t − c1)(t − c2)(t − c3) = 1 a1 < b1 < c1 < a2 < · · · < c3 The arctic boundary is the envelope of a pencil of lines containing all boundary edges in order. t ∈ R ∪ {∞}

For t ∈ H, get z = z(x, y) in the interior.

Lozenges and monotone nonintersecting lattice paths

traffic flow?

time

space

space

time

During each time unit, speed equals 0 or 1.

r = 1

r = .1

if we penalize changing speed by a factor r per corner...

pressure waves

(large penalty)

(no penalty)

0.2 0.4 0.6 0.8 1.0 0.2 0.4 0.6 0.8 1.0

1 − s − t + (1 − r−2)st = 0

plot of −σ

s

t

p r e s s u r e w a v e s

• b

• w

• c

• D

• t

• 4

For r < 1, the surface tension has a different form:

r = 1 r = .1 r = 10 Simulations

r = .25

No limit shape: limit is random.

thank you for your attention!

thank you for your attention!