Expansion complexes for finite subdivision rules

Transcription

1 Expansion complexes for finite subdivision rules W. Floyd (joint work with J. Cannon and W. Parry) Department of Mathematics Virginia Tech Spring Topology and Dynamics Conference : March, 2013

2 Definition of a finite subdivision rule R C-F-P, Finite subdivision rules, Conform. Geom. Dyn. 5 (2001), (electronic). finite subdivision complex S R S R is the union of its closed 2-cells. Each 2-cell is modeled on an n-gon (called a tile type) with n 3. subdivision R(S R ) of S R subdivision map σ R : R(S R ) S R σ R is cellular and takes each open cell homeomorphically onto an open cell. R-complex: a 2-complex X which is the closure of its 2-cells, together with a structure map h : X S R which takes each open cell homeomorphically onto an open cell One can use a finite subdivision rule to recursively subdivide R-complexes. R(X) is the subdivision of X.

3 Example 1. The pentagonal subdivision rule The subdivision of the tile type and the first three subdivisions. (The subdivisions are drawn using Stephenson s CirclePack).

4 Example 2. The dodecahedral subdivision rule The subdivisions of the three tile types. Note that there are two edge types.

5 The second subdivision of the quadrilateral

6 The third subdivision of the quadrilateral

7 Expansion complexes An expansion R-complex is an R-complex X with structure map f : X S R such that X is homeomorphic to R 2 and there is an orientation-preserving homeomorphism ϕ: X X with σ R f = f ϕ. Expansion complexes arise as direct limits of sequences of subdivisions.

8 The pentagonal expansion complex P.L. Bowers and K. Stephenson, A regular pentagonal tiling of the plane, Conform. Geom. Dyn. 1 (1997), (electronic.)

9 An expansion complex for the dodecahedral fsr

10 This subdivision rule on the sphere at infinity The dodecahedral subdivision rule comes from the recursion at infinity for a Kleinian group. The figure was drawn using Weeks s program SnapPea.

11 Cannon s Conjecture Cannon s Conjecture: If G is a Gromov-hyperbolic discrete group whose space at infinity is S 2, then G acts properly discontinuously, cocompactly, and isometrically on H 3. While a primary motivation for this was Thurston s Hyperbolization Conjecture, even after Perelman s proof of the Geometrization Conjecture this conjecture is still open. How do you proceed from combinatorial/topological information to analytic information? Given a sequence of subdivisions of a tiling, how do you understand/control the shapes of tiles? When can you realize the subdivisions so that the subtiles stay almost round?

12 Weight functions, combinatorial moduli shingling (locally-finite covering by compact, connected sets) T on a surface S, ring (or quadrilateral) R S weight function ρ on T : ρ: T R 0 ρ-length of a curve, ρ-height H ρ of R, ρ-area A ρ of R, ρ-circumference C ρ of R moduli M ρ = H 2 ρ /A ρ and m ρ = A ρ /C 2 ρ moduli M(R) = sup ρ H 2 ρ /A ρ and m(r) = inf ρ A ρ /C 2 ρ The sup and inf exist, and are unique up to scaling. (This follows from compactness and convexity.)

13 Optimal weight functions - an example

14 Combinatorial Riemann Mapping Theorem Now consider a sequence of shinglings of S. Axiom 1. Nondegeneration, comparability of asymptotic combinatorial moduli Axiom 2. Existence of local rings with large moduli conformal sequence of shinglings: Axioms 1 and 2, plus mesh locally approaching 0. Theorem (C): If {S i } is a conformal sequence of shinglings on a topological surface S and R is a ring in S, then R has a metric which makes it a right-circular annulus such that analytic moduli and asymptotic combinatorial moduli on R are uniformly comparable. J. W. Cannon, The combinatorial Riemann mapping theorem, Acta Math. 173 (1994),

15 The Cannon-Swenson Theorem Theorem (C-Swenson): In the setting of Cannon s conjecture, it suffices to prove that the sequence {D(n)} n N of disks at infinity is conformal. Furthermore, the D(n) s satisfy a linear recursion. J. W. Cannon, E. L. Swenson, Recognizing constant curvature groups in dimension 3, Trans. Amer. Math. Soc. 350 (1998),

16 Applications of expansion complexes The pentagonal subdivision rule is conformal; the dihedral symmetry makes showing this much easier. For potential applications, you would like to be able to make use of rotational symmetry. The intuition is that for Cannon s Conjecture expansion complexes correspond to tangent spaces at infinity, and at fixed points of loxodromic elements you will see rotational (but not dihedral) symmetry. Theorem: If a one-tile rotationally invariant finite subdivision rule has bounded valence and mesh approaching zero, then it has an invariant (partial) conformal structure. Theorem: If a finite subdivision rule R has bounded valence, mesh approaching zero, and an invariant (partial) conformal structure, then it is conformal.

17 Example 3, a one-tile rotationally invariant fsr

18 A key observation for rotational invariance

19 The expansion complex One can put a piecewise conformal structure on the expansion complex X with regular pentagons, and then use power maps to extend over the vertices. (This is inspired by the Bowers-Stephenson construction.) The expansion map agrees with a conformal map on the vertices. One can conjugate to get a new fsr for which this conformal map is the expansion map. The subdivision map is conformal with respect to the induced conformal structure on the subdivsion complex. C-F-P, I, Conform. Geom. Dyn (2006) (electronic) C-F-P, II, Conform. Geom. Dyn (2006) (electronic)

20 Growth series for expansion complexes Suppose X is an expansion complex and S is a seed for X. You can define a norm on the tiles of X by assigning norm 0 to the tiles of S, assigning norm 1 to the tiles of X \ int(s) that intersect S in at least an edge, and continuing recursively. For n 0, let s n be the number of tiles of norm n and let b n be the number of tiles of norm at most n. Maria Ramirez-Solano asked last fall whether, for the s n pentagonal expansion complex, lim = 0? n b n More generally, what kind of growth can an expansion complex have? And how nice is the growth series s n x n? n=0

21 Growth series for expansion complexes Suppose R is a finite subdivision rule with bounded valence and mesh approaching zero. Let X be an expansion complex for R, let S be a seed of X, and let p be the center (fixed point of the expansion map). There exist a > 0 and b > 1 such that for any n N, dist(p, R n (S)) ab n. There exist c > 0 and d > 1 such that for any n N, #(B(ab n )) cd n. #(R n (S)) cd n. lim (cd n ) 1/(abn) = 1, so the growth of the series is n subexponential.

22 Example 4

23 Example 4 The growth series s n x n = n=0 4(1 + x + 3x 2 + 3x 3 + 9x 4 + 9x 5 + 9x 6 + 9x x ). s n n ln(3)/ ln(2), with equality if n is a power of 2.

24 The type problem How do you tell if an expansion complex is parabolic or hyperbolic? (Here the type can be defined in terms of circle packings or in terms of conformal structures on the complex.) Bowers and Stephenson showed that the pentagonal expansion complex is parabolic by putting a conformal structure on it for which the expansion map is conformal. A conformal fsr cannot have a hyperbolic expansion complex, but in general the type problem is delicate, as shown by the following example.

25 Example 5

26 A hyperbolic expansion complex

27 Part of a parabolic expansion complex

28 Riemann surface laminations This fsr has bounded valence and mesh approaching zero, and is irreducible. Any compact subcomplex of one of those two expansion complexes is isomorphic to a subcomplex of the other. This example can be viewed in the context of Riemann surface laminations. In a survey paper, Ghys gave an example of Kenyon s of a Riemann surface lamination that has a hyperbolic leaf and parabolic leaves, and Blanc generalized this greatly in his thesis. E. Ghys, Laminations par surfaces de riemann, Panoramas & Synthèses 8 (2000), E. Blanc, Propriétés génériques de laminations, Ph.D. thesis, Université Lyon-1, 2001.