linear spaces of tilings Richard Kenyon (Brown University) - - PowerPoint PPT Presentation

linear spaces of tilings

Richard Kenyon (Brown University)

Thursday, May 12, 16

Rectangle tilings come in linear families (polytopes) x1 x2 x3 y1 y2 y1 < y2 x1 < x2, x3

{

P

2

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Given such a polytope, one can make a random tiling by choosing a Lebesgue random point

GUE minors

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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

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

Smith diagram of a planar network

(with a harmonic function) [BSST 1939] vertex = horizontal line (width/height)

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1 36 ⇣ 19 + √ 73 ⌘

1 36(19 − √ 73)

As we change conductances, the polytope can change: the polytope is defined by direction of current flow in the network Thm: There is one fixed-area rectangulation for each bipolar orientation. These directions form a bipolar orientation of the network.

2

[K,Abrams]

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[0, ∞)E (0, ∞)E {x, y} conductances energies/areas m to 1 “ h

• m

e

• ”

union of polytopes |Jlog| = 1 J = K

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Thm: There is one fixed-area rectangulation for each bipolar orientation. [K,Abrams] X

u∼v

1 f(v) − f(u) = 0 f(v0) = 0 f(v1) = 1. The corresp. harmonic functions are the solutions of the enharmonic eqn:

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1 2 3 4 5 6 7 8 9 1011 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12 1 2 3 4 5 6 7 8 9 10 11 12

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

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• rientation of a

random graph: A random bipolar eγhdx2 + e−γhdy2 ?

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T-graphs with fixed slopes come in linear families (polytopes)

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Given a convex n-gon, the space of closed polygonal curves with the same edge slopes is ∼ = Rn−2. On this space the signed area is a quadratic form of signature (1, n − 3). Polygons (or closed polygonal curves) with fixed edge slopes Thurston: Proof by picture: ⇤

x1 x2 x3 A = C3x2

3 − C1x2 1 − C2x2 2

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For fixed area, there are two components to the space, called orientations: quadrilateral pentagon triangle

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“Butterfly moves” are hyperbolic isometries (reflections in the sides of R). The space of area-1 convex polygons is a convex polytope R = R(P) in Hn−3 Fixing area= 1, each component is isometric to Hn−3. Shape of R depends on slopes of sides of P: parallel sides of P implies side of R “at infinity”.

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Fix a tiling family (t-graph with fixed combinatorics and slopes) Thm: For generic slopes, there is exactly one (generalized) tiling for each choice of areas and tile orientations.

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• For example, if we fix the areas, in this case there are 16 generalized tilings

(8 up to 180 rotation). Reality conjecture: For rational slopes and areas, the Galois group permutes the solutions.

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because DΨ is a Kasteleyn matrix for the underlying bipartite graph. Injectivity of Ψ follows from convexity: given two tilings with same areas and same orientations, their average has greater area for each tile. ⇤ Proof: The map Ψ : {intercepts} → {areas} is a local homeomorphism Thm: For each choice of orientation, the set of possible areas is homeomorphic to a closed ball of dimension F. (if nonempty)

(which has dimer covers!)

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for generic slopes, polytopes ↔ orientations (of white vertices) for rectangulations, polytopes ↔ bipolar orientations of network

• Q. what about intermediate cases?

Conclusion:

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• Q1. Can P be tiled with squares?

√ 2 − 1 √ 2 − 1

1 1 1 1

• Q2. Can Q be tiled with rectangles of rational area?

21/3

1 − 21/3 2 1 2 1 2

(no) (no) Many nontrivial facts can be proved using networks...

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thank you for your attention!

Thursday, May 12, 16