Scaling limit of random planar maps Lecture 1.

Transcription

1 Scaling limit of random planar maps Lecture. Olivier Bernardi, CNRS, Université Paris-Sud Workshop on randomness and enumeration Temuco, Olivier Bernardi p./3

2 Planar maps A planar map is a connected planar graph embedded in the sphere and considered up to deformation. = Olivier Bernardi p./3

3 Goals We consider maps as discrete metric spaces: (V, d). Question: What metric space is the limit of random maps? Olivier Bernardi p.3/3

4 Goals Question: What random metric space is the limit in distribution, for the Gromov Hausdorff topology, of uniformly random, rescaled maps of size n, when n goes to infinity? Olivier Bernardi p.3/3

5 Goals Question: What random metric space is the limit in distribution, for the Gromov Hausdorff topology, of uniformly random, rescaled maps of size n, when n goes to infinity? Lecture : Maps, Bijection with well-labelled trees, continuous trees and maps. Lecture : Gromov-Hausdorff topology on metric spaces, convergence of random trees toward the Continuum Random Tree, convergence of random maps toward the Brownian map. Olivier Bernardi p.3/3

6 Goals Question: What random metric space is the limit in distribution, for the Gromov Hausdorff topology, of uniformly random, rescaled maps of size n, when n goes to infinity? References: Bijection: Schaeffer Ph.D. Thesis (98). Distribution of distances: Chassaing & Schaeffer (04). Continuum Random Tree: Aldous (9). Brownian map: Marckert & Mokkadem (06). Convergence of random maps: Le Gall (07)....Bouttier, Di Francesco, Guitter, Miermont, Paulin, Weill... Olivier Bernardi p.3/3

7 Maps Olivier Bernardi p.4/3

8 Rooted maps A planar map is a connected planar graph embedded in the sphere and considered up to deformation. = A map is rooted by distinguishing a corner. Olivier Bernardi p.5/3

9 Quadrangulations A quadrangulation is a planar map such that every faces has degree 4. Proposition: A quadrangulation with n faces has n edges and n + vertices. proof: Incidence relation faces/edges + Euler relation. Olivier Bernardi p.6/3

10 Quadrangulations A quadrangulation is a planar map such that every faces has degree 4. Remark: Quadrangulations are bipartite (since faces generate all cycles). Olivier Bernardi p.6/3

11 Quadrangulations and general maps Proposition: Maps with n edges are in bijection with quadrangulations with n faces. Olivier Bernardi p.7/3

12 Quadrangulations and general maps Proposition: Maps with n edges are in bijection with quadrangulations with n faces. Olivier Bernardi p.7/3

13 Quadrangulations and general maps Proposition: Maps with n edges are in bijection with quadrangulations with n faces. Olivier Bernardi p.7/3

14 Quadrangulations and general maps Proposition: Maps with n edges are in bijection with quadrangulations with n faces. Olivier Bernardi p.7/3

15 Quadrangulations and general maps Proposition: Maps with n edges are in bijection with quadrangulations with n faces. Olivier Bernardi p.7/3

16 Counting maps Theorem [Tutte 63]: The number of rooted ( ) 3 n n quadrangulations with n faces is q n =. (n + )(n + ) n Olivier Bernardi p.8/3

17 Counting maps Theorem [Tutte 63]: The number of rooted ( ) 3 n n quadrangulations with n faces is q n =. (n + )(n + ) n Remarks: The asymptotic behavior q n cn 5/ ρ n is typical. The generating function G(z) = M M q nz n is algebraic: 6z + (8z )G(z) 7z G(z) = 0. Olivier Bernardi p.8/3

18 Counting maps Theorem [Tutte 63]: The number of rooted ( ) 3 n n quadrangulations with n faces is q n =. (n + )(n + ) n Main methods for counting maps: Generating function approach [Tutte 63]: Encoding a recurrence relation via generating functions. Matrix integrals [Brézin-Itzykson-Parisi-Zuber 78]: Interpreting maps as the Feynman diagrams. Computation of characters [Goulden-Jackson]: Interpreting maps as products of permutations. Bijections with decorated trees [Schaeffer 98]. Olivier Bernardi p.8/3

19 A bijection of Schaeffer Quadrangulations Well-labelled trees Olivier Bernardi p.9/3

20 Well-labelled trees A rooted plane tree is a rooted planar map with a single face. There are C n = n+( n n ) trees with n edges. Olivier Bernardi p.0/3

21 Well-labelled trees A well-labelled tree is a tree with positive labels such that the root vertex is labelled. the difference of labels between adjacent vertices is, 0 or. 4 3 Olivier Bernardi p.0/3

22 From quadrangulations to trees Olivier Bernardi p./3

23 From quadrangulations to trees Step. Compute the graph distance from the vertex incident to the root-edge Olivier Bernardi p./3

24 From quadrangulations to trees Step. Create an edge of the tree for each face of the quadrangulation. i+ i i+ i+ i i+ i i+ Olivier Bernardi p./3

25 From quadrangulations to trees Step. Create an edge of the tree for each face of the quadrangulation Olivier Bernardi p./3

26 From quadrangulations to trees Step. Create an edge of the tree for each face of the quadrangulation Olivier Bernardi p./3

27 Bijection Theorem [Schaeffer 98]: This construction is a bijection between quadrangulations with n faces and well-labelled trees with n edges. 4 3 Olivier Bernardi p./3

28 Bijection Theorem [Schaeffer 98]: This construction is a bijection between quadrangulations with n faces and well-labelled trees with n edges. Proof that one obtains a well-labelled tree: The quadrangulation has n faces and n + vertices The image has n edges and n + vertices. i i Olivier Bernardi p./3

29 Bijection Theorem [Schaeffer 98]: This construction is a bijection between quadrangulations with n faces and well-labelled trees with n edges. Proof that one obtains a well-labelled tree: The quadrangulation has n faces and n + vertices The image has n edges and n + vertices. The image has no cycle, hence it is a tree. i i i Olivier Bernardi p./3

30 Bijection Theorem [Schaeffer 98]: This construction is a bijection between quadrangulations with n faces and well-labelled trees with n edges. Proof that one obtains a well-labelled tree: The quadrangulation has n faces and n + vertices The image has n edges and n + vertices. The image has no cycle, hence it is a tree. It is well-labelled. Olivier Bernardi p./3

31 From trees to quadrangulations 4 3 Olivier Bernardi p.3/3

32 From trees to quadrangulations Step. Add an isolated vertex labelled 0. Olivier Bernardi p.3/3

33 From trees to quadrangulations Step. Join every corner labelled i to the next corner labelled i around the tree. Olivier Bernardi p.3/3

34 From trees to quadrangulations One obtains a quadrangulation with labels indicating the distance from the vertex adjacent to the root-edge. Olivier Bernardi p.3/3

35 Counting? It does not seem easy to count well-labelled ( ) trees. 3 n n q n =. (n + )(n + ) n Olivier Bernardi p.4/3

36 Counting? It does not seem easy to count well-labelled trees. But it is easy to see that there are t n = 3n (n+)( n n ) labelled trees: The minimum of labels is. The difference of labels between adjacent vertices is,0,. Olivier Bernardi p.4/3

37 Extended bijection Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that Olivier Bernardi p.5/3

38 Extended bijection Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that Olivier Bernardi p.5/3

39 Extended bijection Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that Olivier Bernardi p.5/3

40 Extended bijection Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that the root-edge is oriented away from the root-vertex. Olivier Bernardi p.5/3

41 Extended bijection Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that the root-edge is oriented away from the root-vertex. There are n + q n such rooted+marked quadrangulations. q n = ( ) n + t 3 n n n =. (n + )(n + ) n The bijection gives a linear random generation algorithm for rooted quadrangulation. Olivier Bernardi p.5/3

42 Continuous trees and maps Olivier Bernardi p.6/3

43 Dyck words A Dyck word of size n is a sequence D = d 0 d...d n of non-negative integers satisfying d 0 = d n = 0 and d i d i = ± for all i =...n n Olivier Bernardi p.7/3

44 Bijection: Trees Dyck words Bijection between trees of size n and Dyck word of size n: turn around the tree and record the heights of each corner. Remark: If the Dyck word D = d 0 d...d n encodes the tree T, then two indices i < j correspond to the same vertex of T if and only if d i = d j = inf i k j d k. Olivier Bernardi p.8/3

45 From sequences to functions It is convenient to consider Dyck words as continuous functions from [0, ] to R + : take the piecewise linear function f such that f( i n ) = d i. Olivier Bernardi p.9/3

46 From sequences to functions It is convenient to consider Dyck words as continuous functions from [0, ] to R + : take the piecewise linear function f such that f( i n ) = d i. Remarks: The function f is obtained from the tree T by turning around the tree at speed n. Two reals 0 s < t correspond to the same point of the tree if and only if f(s) = f(t) = inf s x t f(x). Olivier Bernardi p.9/3

47 Trees as quotients of [0, ] Let f be a continuous function from [0, ] to R + such that f(0)=f()=0. We denote s f t if f(s) = f(t) = inf s x t f(x) and define T f as the quotient of [0, ] by the relation f. Olivier Bernardi p.0/3

48 Trees as quotients of [0, ] Let f be a continuous function from [0, ] to R + such that f(0)=f()=0. We denote s f t if f(s) = f(t) = inf s x t f(x) and define T f as the quotient of [0, ] by the relation f. Ancestor relation u v is defined by min(u) < min(v) max(v) < max(u). Distance d f (u,v) = f(u) + f(v) f(ρ) where ρ is the greatest common ancestors. Olivier Bernardi p.0/3

49 Trees as quotients of [0, ] Let f be a continuous function from [0, ] to R + such that f(0)=f()=0. We denote s f t if f(s) = f(t) = inf s x t f(x) and define T f as the quotient of [0, ] by the relation f. We call (T f,d f ) a (rooted, plane) real tree. Prop: A rooted metric space (T,d) can be represented by a real tree if and only if there exists a geodesic path between any pair of elements in T and there exists no other simple path. Olivier Bernardi p.0/3

50 Trees as quotients of [0, ] Let f be a continuous function from [0, ] to R + such that f(0)=f()=0. We denote s f t if f(s) = f(t) = inf s x t f(x) and define T f as the quotient of [0, ] by the relation f. Next lecture: We will see that the rescaled uniformly random discrete tree converges to the Continuum Random Tree (CRT) encoded by the Brownian excursion. Olivier Bernardi p.0/3

51 Real quadrangulations? By the bijection of Schaeffer, quandrangulations are in bijection with well labelled trees. 4 3 Vertices of the tree are identified with non-marked vertices of the quadrangulation. Olivier Bernardi p./3

52 Real quadrangulations? For two vertices u,v of the tree T, we define d 0 Q(u,v) = l(u) + l(v) + min(l(w) : w u T v). u v There are always two shortest paths u T v. Olivier Bernardi p./3

53 Real quadrangulations? For two vertices u,v of the tree T, we define d 0 Q(u,v) = l(u) + l(v) + min(l(w) : w u T v). Prop: The distance d Q between two vertices u,v is at most d 0 Q (u,v). u l(u) min(l(w)) l(v) v Olivier Bernardi p./3

54 Real quadrangulations? For two vertices u,v of the tree T, we define d 0 Q(u,v) = l(u) + l(v) + min(l(w) : w u T v). Prop: The distance d Q between two vertices u,v is at most d 0 Q (u,v). Hence, d Q (u,v) d Q(u,v) = min d 0 u=u 0,u,...,u k Q(u i,u i+ ). =v i Olivier Bernardi p./3

55 Quadrangulations as quotient of trees Let T f be a real tree and let g be a continuous function from T f to R + such that g(ρ) = 0. For two points u,v of T f, we define D 0 (u,v) = g(u) + g(v) inf(g(w) : w u T v). D (u,v) = inf D u=u 0,u,...,u k Q(u 0 i,u i+ ). =v i Olivier Bernardi p./3

56 Quadrangulations as quotient of trees Let T f be a real tree and let g be a continuous function from T f to R + such that g(ρ) = 0. For two points u,v of T f, we define D 0 (u,v) = g(u) + g(v) inf(g(w) : w u T v). D (u,v) = inf D u=u 0,u,...,u k Q(u 0 i,u i+ ). =v i We denote u g v if D (u,v) = 0, we define T f,g as the quotient of T f by the relation g, and call real quadrangulation the metric space (T f,g,d ). Olivier Bernardi p./3

57 Quadrangulations as quotient of trees Let T f be a real tree and let g be a continuous function from T f to R + such that g(ρ) = 0. For two points u,v of T f, we define D 0 (u,v) = g(u) + g(v) inf(g(w) : w u T v). D (u,v) = inf D u=u 0,u,...,u k Q(u 0 i,u i+ ). =v i Next lecture: We will see that the rescaled uniformly random quadrangulation converges (at least along subsequences) to a random metric space (T e,g,d), where T e is the Continuum Random Tree, g is a Gaussian process on T e conditioned to be positive, D is a distance on T e,g which is less than D. Olivier Bernardi p./3

58 End of Lecture... Olivier Bernardi p.3/3