linear spaces of tilings Richard Kenyon (Brown University) Thursday, May 12, 16
Rectangle tilings come in linear families (polytopes) y 2 2 y 1 x 2 x 3 x 1 y 1 < y 2 { P x 1 < x 2 , x 3 Thursday, May 12, 16
GUE minors Given such a polytope, one can make a random tiling by choosing a Lebesgue random point Thursday, May 12, 16
Smith diagram of a planar network [BSST 1939] (with a harmonic function) v 1 3 1 1. 2 2. 4. 6 4 3. 5. 9 5 8 7 7. 6. 8. 10 10. 9. 12 11 11. 12. v 0 vertex = horizontal line voltage = y -coordinate edge = rectangle current = width (width/height) conductance = aspect ratio energy = area Thursday, May 12, 16
As we change conductances, the polytope can change: the polytope is defined by direction of current flow in the network These directions form a bipolar orientation of the network. 2 [K,Abrams] Thm: There is one fixed-area rectangulation for each bipolar orientation. 1 ⇣ ⌘ √ 19 + 73 36 1 √ 36(19 − 73) Thursday, May 12, 16
[0 , ∞ ) E (0 , ∞ ) E | J log | = 1 energies/areas conductances m to 1 ” o e m K o h = “ J { x, y } union of polytopes Thursday, May 12, 16
[K,Abrams] Thm: There is one fixed-area rectangulation for each bipolar orientation. The corresp. harmonic functions are the solutions of the enharmonic eqn : 1 X f ( v ) − f ( u ) = 0 u ∼ v f ( v 0 ) = 0 f ( v 1 ) = 1 . Thursday, May 12, 16
1 3 3 1 3 2 9 5 6 5 2 2 1 4 6 3 1 2 7 8 4 4 7 9 6 7 1011 10 9 8 10 11 5 12 6 4 8 7 10 9 5 8 11 12 12 11 12 3 1 2 2 2 3 1 2 2 4 6 1 6 3 4 4 6 3 1 1 3 9 6 4 5 7 8 9 5 8 7 8 7 4 6 8 7 9 5 7 8 10 10 10 9 5 10 10 9 5 12 11 11 12 12 11 11 12 12 11 3 1 1 3 3 1 1 2 9 4 2 2 3 4 6 6 4 2 6 9 5 6 8 4 5 9 8 5 7 5 8 7 10 9 12 8 7 10 10 10 7 11 12 12 11 12 11 11 Thursday, May 12, 16
A random bipolar orientation of a random graph: e γ h dx 2 + e − γ h dy 2 ? Thursday, May 12, 16
T-graphs with fixed slopes come in linear families (polytopes) Thursday, May 12, 16
Polygons (or closed polygonal curves) with fixed edge slopes Thurston: Given a convex n -gon, the space of closed polygonal curves = R n − 2 . with the same edge slopes is ∼ On this space the signed area is a quadratic form of signature (1 , n − 3). x 2 Proof by picture: ⇤ x 1 x 3 A = C 3 x 2 3 − C 1 x 2 1 − C 2 x 2 2 Thursday, May 12, 16
For fixed area, there are two components to the space, called orientations: triangle quadrilateral pentagon Thursday, May 12, 16
Fixing area= 1, each component is isometric to H n − 3 . The space of area-1 convex polygons is a convex polytope R = R ( P ) in H n − 3 “Butterfly moves” are hyperbolic isometries (reflections in the sides of R ). Shape of R depends on slopes of sides of P : parallel sides of P implies side of R “at infinity”. Thursday, May 12, 16
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. Thursday, May 12, 16
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. Thursday, May 12, 16
(if nonempty) Thm: For each choice of orientation, the set of possible areas is homeomorphic to a closed ball of dimension F . Proof: The map Ψ : { intercepts } → { areas } is a local homeomorphism because D Ψ is a Kasteleyn matrix for the underlying bipartite graph. (which has dimer covers!) Injectivity of Ψ follows from convexity: given two tilings with same areas and same orientations, their average has greater area for each tile. ⇤ Thursday, May 12, 16
Thursday, May 12, 16
Conclusion: for rectangulations, polytopes ↔ bipolar orientations of network for generic slopes, polytopes ↔ orientations (of white vertices) Q. what about intermediate cases? Thursday, May 12, 16
Many nontrivial facts can be proved using networks... √ 2 − 1 1 1 Q1. Can P be tiled with squares? (no) 1 1 √ 2 − 1 Q2. Can Q be tiled with rectangles of rational area? 1 − 2 1 / 3 (no) 2 1 2 2 1 / 3 1 2 Thursday, May 12, 16
thank you for your attention! Thursday, May 12, 16
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