harmonic functions and the chromatic polynomial R. Kenyon (Brown) - - PowerPoint PPT Presentation

harmonic functions and the chromatic polynomial

• R. Kenyon (Brown)

based on joint work with A. Abrams, W. Lam

χ(n) = n(n − 1)(n − 2) χ satisfies a contraction-deletion rule: χG(n) = χG−e(n) − χG/e(n) but is #P-hard to compute in general.

(adjacent vertices have different colors)

The chromatic polynomial χG(n) of a graph G is the number of proper colorings with n colors.

A graph G = (V, E) c : E → R>0 the edge conductances B ⊂ V boundary vertices u : B → R boundary values Find f : V → R harmonic on V \ B and f|B = u.

The Dirichlet problem

1 2 3 4 5 6

0 = ∆f(x) = X

y∼x

ce(f(x) − f(y)) E(f) = X

e=xy

ce(f(x) − f(y))2

{

edge energy

f is the unique function with f|B = u minimizing the Dirichlet energy

1 2

3 4

5

6

7

8

9 10 11

12

A harmonic function induces a compatible orientation: an acyclic orientation with no internal sources or sinks, and no oriented paths from lower boundary values to higher boundary values. We let Σ be the set of compatible orientations How many are there? “current flows downhill”

x y 1

0.2 0.4 0.6 0.8 1.0 x 0.2 0.4 0.6 0.8 1.0 y

Then F = [

σ∈Σ

¯ Fσ where Fσ = {f ∈ F | sign(d f) = σ}. Let F ⊂ RV be the set of functions with boundary values u and no internal extrema.

The Fσ are convex polytopes.

Fixed energy problem:

Can we adjust edge conductances so that all bulbs burn with the same brightness?

a b c d e

a(bd + cd + ce + de)2 (ab + ac + ae + bc + bd + cd + ce + de)2 , b(ae + cd + ce + de)2 (ab + ac + ae + bc + bd + cd + ce + de)2 , c(bd − ae)2 (ab + ac + ae + bc + bd + cd + ce + de)2 , d(ab + ac + ae + bc)2 (ab + ac + ae + bc + bd + cd + ce + de)2 , e(ab + ac + bc + bd)2 (ab + ac + ae + bc + bd + cd + ce + de)2

( )

Ψ(a, b, c, d, e) = Example Ψ : (0, ∞)E → [0, ∞)E Let be the map from conductances to energies.

1

choice of conductances {ce} for which the associated harmonic function realizes this data. Ψ : (0, ∞)E → [0, ∞)E Let be the map from conductances to energies. Theorem 1: For any σ ∈ Σ and {Ee > 0} there is a unique = X

y∼x

Ee h(x) − h(y)

Thm 2: The harmonic functions h with energies {Ee} are the solutions to the system of equations Thm 3: The number of solutions N = |Σ| satisfies the contraction-deletion rule NG = NG−e + NG/e.

the enharmonic equation

“energy − harmonic”

exactly

∨

∀x ∈ V \ B,

Cor: Let Gk be obtained from G by adding k vertices as boundary, attached to all vertices of G. Then for any choice of k distinct boundary values, the number

• f compatible orientations is χG(1 − k).

G

π e

√ 2

|χ(1 − k)| = 1 |∆m| Z

∆m

Y

e=xy

(h(x) − h(y))2 Z dvol Lemma: The Jacobian determinant of Ψ has the form det JΨ = Y

e=xy

(h(x) − h(y))2. where the integral is over the m-simplex ∆m of conductances summing to 1, and Z = P

e=xy cxy(h(x) − h(y))2.

Corollary:

= X

y∼x

Ee h(x) − h(y) M(h) = Y

e

|h(x) − h(y)|Ee. ⇤ 0 = ∆h(x) = X

y∼x

ce(h(x) − h(y))

Proof of Theorems 1 and 2:

Note log M(h) is strictly concave on each polytope Fσ = {h | sign(dh) = σ}.

recall Exy = cxy(h(x) − h(y))2

.

(One needs also show that all solutions are real.)

Solutions of the enharmonic equation are critical points of the functional

Proof of Theorem 3: Recall that Ee = ce(h(x) − h(y))2 = IeVe where Ie = ce(h(x) − h(y)) and Ve = h(x) − h(y). When Ee → 0, either Ie → 0 or Ve → 0 (or both). In the first case, delete the edge; in the second contract the edge. Conversely, the operation of contracting or deleting is reversible by adding in an edge of small energy. ⇤

applications

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voltage = y-coordinate edge = rectangle current = width conductance = aspect ratio energy = area

Smith diagram of a planar network

(with a harmonic function)

v0 v1 3 2 1 12 11 6 4 9 8 7 5 10

This graph has 12 acyclic orientations with source at 1 and sink at 0. (|Σ| = 12.)

2315250000z12−107438625000z11+2230924692500z10−27361273241750z9+ 220350695004825z8−1225394593409700z7+4817113876088640z6−13468300499707200z5+ 26554002301384704z4−35985219877131264z3+31817913970765824z2−16489700865736704z+ 3791571715620864 = 0

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

v0 v1 3 2 1 12 11 6 4 9 8 7 5 10

width(1) is the root of a polynomial:

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1 2 3 4 5 6 7 8 9 10 11 12

500 1000 1500 2000 2500 3000 0.2 0.4 0.6 0.8 1.0

1

Z2, directed S&W

One of the ≈ 10599 area-1 rectangulations based on the 40 × 40 grid

(64/27)n2(1+o(1))

21/3

1 − 21/3 2 1 2 1 2

If a polygon P can be tiled with rectangles of rational area, horizontal lengths are in a totally real extension field of Q[v1, . . . , vk]. Cor.

Area 1 but not tileable.

Bipolar orientations on Z2 and square ice (with Miller, Sheffield, Wilson)

add a row of edges around the boundary oriented N and W

From each black vertex, the two outgoing green arrows separate the incoming and outgoing black arrows From each face center, outgoing green arrows point to the face max and min.

Bend outgoing edges right if from vertices, left if from faces.

The associated peano curve, colored according to distance traveled

(Conjectural) SLE12 scaling limit

Thank you