AN INTRODUCTION TO ORIENTATIONS ON MAPS 2nd lecture May, 16th 2017

Transcription

1 AN INTRODUCTION TO ORIENTATIONS ON MAPS 2nd lecture May, 16th 2017 Marie Albenque (CNRS, LIX, École Polytechnique) Mini-school on Random Maps and the Gaussian Free Field ProbabLyon

2 Plan Yesterday : Construction of orientations, existence, uniqueness 1 - Some definitions : maps, orientations. 2 - Existence of orientations 3 - Flip and flop : the lattice of orientations Today : Applications : graph drawings, couplings, bijections 1 - Couplings and spanning trees. 2 - Schnyder woods and graph drawings. 3 - Orientations and blossoming trees Thursday : Why should you care? 1 - Higher genus 2 - Scaling limits for simple triangulations. 3 - Scaling limits for maps with an orientation?

3 Spanning trees and orientations [Propp 93] A rooted plane map M

4 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T

5 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T Goal : For a fixed M, what can we say about the structure of spanning trees? Can we endow them with an orientation structure?

6 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M Goal : For a fixed M, what can we say about the structure of spanning trees? Can we endow them with an orientation structure? Convention :

7 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M Goal : For a fixed M, what can we say about the structure of spanning trees? Can we endow them with an orientation structure? Convention :

8 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M Goal : For a fixed M, what can we say about the structure of spanning trees? Can we endow them with an orientation structure? Convention :

9 Spanning trees and orientations [Propp 93] A rooted plane map M M = completion of M = superimposition of M and M

10 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M M = completion of M = superimposition of M and M Convention :

11 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M M = completion of M = superimposition of M and M For a spanning tree T of M, we define the α T -orientation of M by : { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Convention :

12 Spanning trees and orientations [Propp 93] A rooted plane map M A spanning tree T T = spanning tree of M M = completion of M = superimposition of M and M For a spanning tree T of M, we define the α T -orientation of M by : { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Convention : And reciprocally?

13 Spanning trees and orientations Proposition : [Propp 93, Felsner 04] The spanning trees of M are in bijection with the α T -orientations of M. { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3

14 Spanning trees and orientations Proposition : [Propp 93, Felsner 04] The spanning trees of M are in bijection with the α T -orientations of M. { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 = deg( ) 1

15 Spanning trees and orientations Proposition : [Propp 93, Felsner 04] The spanning trees of M are in bijection with the α T -orientations of M. { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 At each vertex, 2 possible configurations :

16 Spanning trees and orientations Proposition : [Propp 93, Felsner 04] The spanning trees of M are in bijection with the α T -orientations of M. { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 At each vertex, 2 possible configurations :

17 Spanning trees and orientations Proposition : [Propp 93, Felsner 04] The spanning trees of M are in bijection with the α T -orientations of M. { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 At each vertex, 2 possible configurations : {blue edges} = V (M) 1 To prove that it is a tree, enough to prove that there is no cycles (exercise!).

18 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid?

19 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex.

20 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex.?

21 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex.?

22 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex.?

23 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex.?

24 Lattice structure on the set of spanning trees { α T ( ) = α T ( ) = 0 α T ( ) = α T ( ) = 1 α T ( ) = 3 Which cycles are essential? Which edges are rigid? To make our life easier, assume that M is bridgeless. Property : The rigid edges are the edges incident to the root vertex. Property : The essential cycles are the facial cycles of M, without root vertices.

25 Lattice structure on the set of spanning trees Property : The essential cycles are the facial cycles of M, without root vertices.

26 Lattice structure on the set of spanning trees Property : The essential cycles are the facial cycles of M, without root vertices. v

27 Lattice structure on the set of spanning trees Property : The essential cycles are the facial cycles of M, without root vertices. v

28 Lattice structure on the set of spanning trees Property : The essential cycles are the facial cycles of M, without root vertices. v

29 Lattice structure on the set of spanning trees Property : The essential cycles are the facial cycles of M, without root vertices. v FLIP v

30 Example of a lattice M Applications : Coupling from the past for distributive lattices. easy to sample a spanning tree (Wilson s algorithm), gives a way to sample some perfect matchings (a.k.a. dimer models) [Kenyon, Propp, Wilson]

31 Schnyder woods Simple triangulation endowed with a 3-orientation.

32 Schnyder woods Simple triangulation endowed with a 3-orientation.

33 Schnyder woods Simple triangulation endowed with a 3-orientation. Consider the middle -paths

34 Schnyder woods Simple triangulation endowed with a 3-orientation. Consider the middle -paths

35 Schnyder woods Simple triangulation endowed with a 3-orientation. Consider the middle -paths

36 Schnyder woods Simple triangulation endowed with a 3-orientation. Consider the middle -paths

37 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths

38 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths

39 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths

40 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths

41 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths

42 Schnyder woods Property : The middle-paths are self-avoiding + the 3 middle paths starting at a given vertex do not intersect one another. Simple triangulation endowed with a 3-orientation. Consider the middle -paths Around each inner vertex : Property : The subset of edges of a given color is a spanning tree of the inner vertices (+ one outer vertex) of the triangulation. This coloring is a Schnyder wood. [Schnyder 89]

43 Application to straight-line drawing. a 2 a 1 a 3 Goal : Find a representation of the plane map in which all the edges are straight lines and where the vertices lie on a grid.

44 Application to straight-line drawing. a 2 a 1 a 3 Goal : Find a representation of the plane map in which all the edges are straight lines and where the vertices lie on a grid.

45 Application to straight-line drawing. a 2 a 2 v a 1 a 3 a 1 a 3 Goal : Find a representation of the plane map in which all the edges are straight lines and where the vertices lie on a grid. Algo : Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), ( F (M), 0).

46 Application to straight-line drawing. a 2 a 2 v a 1 a 3 a 1 a 3 Goal : Find a representation of the plane map in which all the edges are straight lines and where the vertices lie on a grid. Algo : Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ).

47 Application to straight-line drawing. a 2 a 2 v a 1 R 2 (v) F (R 2 (v)) = 6 a 3 a 1 a 3 Algo : Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ).

48 Application to straight-line drawing. R 3 (v) F (R 3 (v)) = 8 a 2 a 2 v 6 v a 1 R 2 (v) F (R 2 (v)) = 6 a 3 a 1 8 a 3 Algo : Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ).

49 Application to straight-line drawing. a 2 a 2 v 6 v a 1 a 3 a 1 8 a 3 Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), Algo : ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ). Why is it working?

50 Application to straight-line drawing. R 3 (v 1 ) R 3 (v) a 2 a 2 v 1 v 6 v a 1 R 2 (v 1 ) R 2 (v) a 3 a 1 v 1 8 a 3 Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), Algo : ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ). Why is it working?

51 Application to straight-line drawing. R 3 (v 1 ) R 3 (v) a 2 a 2 v 1 v 6 v a 1 R 2 (v 1 ) R 2 (v) a 3 a 1 v 1 8 a 3 Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), Algo : ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ). Why is it working?

52 Application to straight-line drawing. a 2 a 2 v 2 v 2 v 6 v v 1 v 3 3 v 1 v 3 a 1 a 3 a a 3 Put the vertices a 1, a 2 et a 3 at positions (0, 0), (0, F (M) ), Algo : ( F (M), 0). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ). Why is it working?

53 Application to straight-line drawing. a 2 a 2 v a 1 a 3 a 1 a 3 Theorem ([Schnyder 89]) : This algorithm Put theproduces vertices a 1 straightline, a 2 et a 3 at drawing positions of the (0, 0), triangulation (0, F (M) ), Algo where: ( F all the (M), vertices 0). belong to a grid of size F (M) F (M). Put each inner vertex v at position ( F (R 3 (v)), F (R 2 (v)) ).

54 Representation by homothetic triangles. A B 1 A C F 1 3 B D E 2 G C 3 F D 2 G E [Motivated by some ideas of Felsner]

55 Representation by homothetic triangles. A B 1 A C F 1 3 B D E 2 G C 3 F D 2 G E [Motivated by some ideas of Felsner]

56 Representation by homothetic triangles. B 1 2 A D E F 3 G C 4 I H 6 K M 5 L J [Motivated by some ideas of Felsner]

57 Representation by homothetic triangles. B 1 2 A 1 B 2 C A D 4 E F I 3 G C D 4 E F 3 I 5 J G H 6 K M 5 L J H 6 M L [Motivated by some ideas of Felsner]

58 Maps enumeration 4-regular maps Simple triangulations (neither loops nor multiple edges )

59 Maps enumeration 4-regular maps Number of rooted 4-regular planar maps with n vertices : ( ) R n = 2 3 n 2n n + 2 n + 1 n [Tutte, 62], [Schaeffer 97] Simple triangulations (neither loops nor multiple edges )

60 Maps enumeration 4-regular maps Number of rooted 4-regular planar maps with n vertices : ( ) R n = 2 3 n 2n n + 2 n + 1 n [Tutte, 62], [Schaeffer 97] Simple triangulations (neither loops nor multiple edges ) Number of simple triangulations with n + 2 vertices : n = 2 (4n 3)! n!(3n 1)! [Tutte, 62], [Poulalhon-Schaeffer 05]

61 Maps enumeration 4-regular maps Number of rooted 4-regular planar maps with n vertices : ( ) R n = 2 3 n 2n n + 2 n + 1 n [Tutte, 62], [Schaeffer 97] Simple triangulations (neither loops nor multiple edges ) Number of simple triangulations with n + 2 vertices : n = Many methods to count maps : 2 (4n 3)! n!(3n 1)! [Tutte, 62], [Poulalhon-Schaeffer 05] Recursive decomposition [Tutte 60s +...], matrix integrals [t Hooft ] Bijective proofs [Cori-Vauquelin-Schaeffer, Bouttier-diFrancesco-Guitter, Bernardi, Fusy, Poulalhon,... ] = bijections btw maps and labeled trees or btw maps and blossoming trees.

62 Maps enumeration 4-regular maps Number of rooted 4-regular planar maps with n vertices : ( ) R n = 2 3 n 2n n + 2 n + 1 n [Tutte, 62], [Schaeffer 97] Simple triangulations (neither loops nor multiple edges ) Number of simple triangulations with n + 2 vertices : n = Many methods to count maps : 2 (4n 3)! n!(3n 1)! [Tutte, 62], [Poulalhon-Schaeffer 05] Recursive decomposition [Tutte 60s +...], matrix integrals [t Hooft ] Bijective proofs [Cori-Vauquelin-Schaeffer, Bouttier-diFrancesco-Guitter, Bernardi, Fusy, Poulalhon,... ] = bijections btw maps and labeled trees or btw maps and blossoming trees.

63 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

64 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

65 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

66 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

67 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

68 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

69 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

70 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems

71 What is a blossoming tree? A blossoming tree is a plane tree whose vertices can carry opening stems or closing stems (or both) such that : # opening stems = # closing stems A plane map can canonically associated to any blossoming tree by making all closures clockwise.

72 What is a blossoming tree? A plane map can be canonically associated to any blossoming tree by making all closures clockwise. If the edges of the tree are oriented + closure edges oriented naturally orientation of the map without ccw cycles.

73 What is a blossoming tree? A plane map can be canonically associated to any blossoming tree by making all closures clockwise. If the edges of the tree are oriented towards its root + closure edges oriented naturally accessible orientation of the map without ccw cycles.

74 Can we always map a map to a blossoming tree? Theorem [Bernardi 07], [A., Poulalhon 14+] : Let M be a rooted plane map endowed with an accessible and minimal (= without ccw cycles) orientation. Then, there exists a unique rooted blossoming tree whose closure is M endowed with the same orientation.

75 Can we always map a map to a blossoming tree? Theorem [Bernardi 07], [A., Poulalhon 14+] : Let M be a rooted plane map endowed with an accessible and minimal (= without ccw cycles) orientation. Then, there exists a unique rooted blossoming tree whose closure is M endowed with the same orientation.

76 Can we always map a map to a blossoming tree? Theorem [Bernardi 07], [A., Poulalhon 14+] : Let M be a rooted plane map endowed with an accessible and minimal (= without ccw cycles) orientation. Then, there exists a unique rooted blossoming tree whose closure is M endowed with the same orientation.

77 Can we always map a map to a blossoming tree? Theorem [Bernardi 07], [A., Poulalhon 14+] : Let M be a rooted plane map endowed with an accessible and minimal (= without ccw cycles) orientation. Then, there exists a unique rooted blossoming tree whose closure is M endowed with the same orientation. Proof by induction on the number of faces + identification of the closure edges...

78 Canonical orientations Each time a family of maps is characterized by the existence of some accessible orientations, we can apply this construction.

79 Canonical orientations Each time a family of maps is characterized by the existence of some accessible orientations, we can apply this construction. 4-regular maps 2 outgoing edges / vertex 2 ingoing edges / vertex

80 Canonical orientations Each time a family of maps is characterized by the existence of some accessible orientations, we can apply this construction. 4-regular maps 2 outgoing edges / vertex 2 ingoing edges / vertex A map is 4-regular iff it admits an orientation such that each vertex has outdegree 2 and indegree 2. Apply the general bijection to recover the result of [Schaeffer 97]

81 Canonical orientations Each time a family of maps is characterized by the existence of some accessible orientations, we can apply this construction. 4-regular maps Simple triangulations 2 outgoing edges / vertex 2 ingoing edges / vertex 3 outgoing edges / inner vertex 1 outgoing edge / outer vertex

82 Canonical orientations Each time a family of maps is characterized by the existence of some accessible orientations, we can apply this construction. 4-regular maps Simple triangulations 2 outgoing edges / vertex 2 ingoing edges / vertex 3 outgoing edges / inner vertex 1 outgoing edge / outer vertex A triangulation is simple iff it admits an orientation such that : each inner vertex has outdegree 3 each outer vertex has outdegree 1. General bijection gives the result of [Poulalhon, Schaeffer 05]

83 Bijective method Select a family of maps Maps with even degree = Eulerian maps

84 Bijective method Select a family of maps Find a characterization by orientations Maps with even degree = Eulerian maps Same in/outdegree

85 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

86 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

87 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

88 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

89 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

90 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Maps with even degree = Eulerian maps Same in/outdegree

91 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Apply the bijection. Maps with even degree = Eulerian maps Same in/outdegree

92 Bijective method Select a family of maps Find a characterization by orientations Consider the unique orientation without ccw cycles. Apply the bijection. Study the family of blossoming trees. Maps with even degree = Eulerian maps Same in/outdegree Trees with same indegree/outdegree.

93 References Reference on the theory of α-orientations [Felsner 04] (and also [Propp 93]). Application to straight-line drawing : [Schnyder 89] [Bonichon, Felsner, Mosbah 04] : refinement on Schnyder initial idea. [Fusy s PhD 07] Spanning trees and couplings : [Propp 93] [Kenyon, Propp, Wilson 00] Thank you! Bijections and orientations : [Bernardi 07] + [A.,Poulalhon 15] [Bernardi, Fusy 12] : unification of existing bijections relying on orientations. [Addario-Berry, A. +14] + [Bernardi, Collet, Fusy 14] : tracking of distances.

94 Exercise 1) Prove (using bijections with blossoming trees) that the number of rooted 4-regular maps with n vertices is : R n = 2 n + 2 ( 3 n n + 1 2n n ) [Tutte, 62], [Schaeffer 97]