On the Riemann surface type of Random Planar Maps

Transcription

1 On the Riemann surface type of Random Planar Maps Department of Mathematics University of Washington Seattle, WA gill or March 24, 2011

2 UIPT Infinite Necklace Riemann Surface Unfocused Question: Given a bag of infinitely many equilateral triangles of unit size, glue them together randomly. When you are done, what sort of surface do you have? Answer of course depends on what we mean by random and surface.

3 UIPT Infinite Necklace Riemann Surface Uniform Infinite Planar Triangulation (Angel, Schramm 04) τ n is the uniform probability measure on all triangulations of S 2 with n vertices together with a distinguished root triangle. metric: two triangulations are 1/(1 + k) apart if the k-neighborhoods of the root are equivalent graphs. (0-neighborhood=root vertex, (k + 1)-neighborhood = k neighborhood + all faces adjacent to vertices in k-neighborhood along with their edges and vertices) There exists a probability measure τ = UIPT which is the distributional limit of the measures τ n as n with respect to the metric above. The limit is almost surely a triangulation of the plane.

4 UIPT Infinite Necklace Riemann Surface A triangulation of the 2-sphere

5 UIPT Infinite Necklace Riemann Surface How far are these two disc triangulations apart?

6 UIPT Infinite Necklace Riemann Surface UIPT snippet (figure from O. Angel)

7 UIPT Infinite Necklace Riemann Surface Infinite Necklace (Sheffield) Law of this random surface is given by bi-infinite sequences of {B, R, b, r} chosen independently with equal probability. Upper half plane necklace for the sequence BRbRRbBBrrRBRR.

8 UIPT Infinite Necklace Riemann Surface We must make sure each point is in a conformal chart and that charts are compatible. We call the Riemann surface associated with triangulation T in this way R(T ).

9 UIPT Infinite Necklace Riemann Surface Koebe Uniformization Theorem: Every simply connected Riemann surface is conformally equivalent to either the unit disc, the complex plane, or the Riemann sphere. In our context, we will not be dealing with spherical surfaces. D = hyperbolic C = parabolic B. M. transient B. M. recurrent

10 UIPT Infinite Necklace Riemann Surface Small portions of a parabolic and hyperbolic Riemann surface

11 Interstice Packing Condition for Interstice Packing to be parabolic Some essential definitions: unbiased: a probability measure is unbiased if, conditioned on an unrooted triangulation, the root is uniformly distributed. disc triangulation: a planar graph whose faces are all triangles and the union of all faces, vertices, and edges is simply connected. one end: the complement with respect to any finite subgraph contains exactly one infinite component.

12 Interstice Packing Condition for Interstice Packing to be parabolic Limit of this sequence will be 2-ended

13 Interstice Packing Condition for Interstice Packing to be parabolic Theorem (G, Rohde) Suppose (T, o) is a subsequential distributional limit of a sequence (T n, o n ) of random finite unbiased disc triangulations (limit with respect to above metric). Suppose further that (T, o) has one end almost surely and that dgr(o n, T n ) in law. Then R(T ) is parabolic almost surely. Corollary 1 UIPT is parabolic a.s. (BM is recurrent) Corollary 2 The infinite necklace is parabolic a.s. (BM is recurrent) What about Simple Random Walk on Graphs?

14 Interstice Packing Condition for Interstice Packing to be parabolic Strategy 1 Use Koebe uniformization theorem to associate with each finite disc triangulation a packing of compact topological discs in C. 2 Find a necessary and sufficient condition for such a packing to be associated with a parabolic infinite disc triangulation. 3 Use Montel s theorem, Prokhorov s theorem, and a magical lemma of Benjamini and Schramm to show we meet 2. (this step involves introducing a finer topology and using complex analysis tools to show that we still have (subsequential) convergence)

15 Interstice Packing Condition for Interstice Packing to be parabolic For each face f F for a triangulation, call the small center triangle I f, its center c f. We call these interstice triangles.

16 Interstice Packing Condition for Interstice Packing to be parabolic Koebe Distortion Theorem Let K be a compact subdomain of a domain D, f : D C a conformal map. Then for x, y, z K f (x) f (y) f (x) f (z) C x y K,D x z where the constant C depends only on K and D. Every interstice triangle (almost) has this flat picutre!

17 Interstice Packing Condition for Interstice Packing to be parabolic Main conclusion from Koebe distortion: Geometry Controlled: Under a conformal map the set of center images has a limit point the set of interstice images has a (non-interior to an interstrice) limit point.

18 Interstice Packing Condition for Interstice Packing to be parabolic First two terms in a sequence of packings whose limit packing has limit points but where centers of discs are limit point free

19 Interstice Packing Condition for Interstice Packing to be parabolic Proposition: Let φ be a conformal map of R(T ), where T is a disc triangulation, into C. Then R is parabolic {φ(c f )} f F has one (finitely many) limit point(s) in C dgr(v, T ) = for some vertex v.

20 Interstice Packing Condition for Interstice Packing to be parabolic Magic Lemma (Benjamini, Schramm, EJP 01) = {φ(c f )} f F has at most one limit point in C a.s. So criteria in Proposition are satisfied and R(T ) is parabolic as it can have either 0 or -many limit points in C. This lemma uses unbiasedness in a crucial way!

21 For a finite set of points V C and a point v V the isolation radius of v is ρ v = min{ v w : w V \ {v}} and for δ > 0 and s 1 we say that v is (δ, s)-supported if inf V (D(v, ρ v /δ) \ D(p, ρ v δ))} s p C

22 Lemma (Benjamini, Schramm 01) For every δ (0, 1) there is a constant c = c(δ) such that for every finite set V C and every s 2 the proportion of (δ, s) - supported points is < c s.

23 Background Two perspectives, on the likely perspective, limit points vanish!