Scaling limit of random planar maps Lecture 1. Olivier Bernardi, - - PowerPoint PPT Presentation

Scaling limit of random planar maps Lecture 1.

Olivier Bernardi, CNRS, Université Paris-Sud Workshop on randomness and enumeration Temuco, November 2008

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Planar maps

A planar map is a connected planar graph embedded in the sphere and considered up to deformation.

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Goals

We consider maps as discrete metric spaces: (V, d). Question: What metric space is the limit of random maps ?

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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 ?

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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 1: Maps, Bijection with well-labelled trees, continuous trees and maps. Lecture 2: Gromov-Hausdorff topology on metric spaces, convergence of random trees toward the Continuum Random Tree, convergence of random maps toward the Brownian map.

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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 (91). Brownian map: Marckert & Mokkadem (06). Convergence of random maps: Le Gall (07). . . .Bouttier, Di Francesco, Guitter, Miermont, Paulin, Weill. . .

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Maps

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Rooted maps

A planar map is a connected planar graph embedded in the sphere and considered up to deformation.

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A map is rooted by distinguishing a corner.

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Quadrangulations

A quadrangulation is a planar map such that every faces has degree 4. Proposition: A quadrangulation with n faces has 2n edges and n + 2 vertices. proof: Incidence relation faces/edges + Euler relation.

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Quadrangulations

A quadrangulation is a planar map such that every faces has degree 4. Remark: Quadrangulations are bipartite (since faces generate all cycles).

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Quadrangulations and general maps

Proposition: Maps with n edges are in bijection with quadrangulations with n faces. ⇐ ⇒

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Quadrangulations and general maps

Proposition: Maps with n edges are in bijection with quadrangulations with n faces.

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Quadrangulations and general maps

Proposition: Maps with n edges are in bijection with quadrangulations with n faces.

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Quadrangulations and general maps

Proposition: Maps with n edges are in bijection with quadrangulations with n faces.

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Quadrangulations and general maps

Proposition: Maps with n edges are in bijection with quadrangulations with n faces.

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Counting maps

Theorem [Tutte 63]: The number of rooted quadrangulations with n faces is qn = 2 · 3n (n + 1)(n + 2) 2n n

• .

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Counting maps

Theorem [Tutte 63]: The number of rooted quadrangulations with n faces is qn = 2 · 3n (n + 1)(n + 2) 2n n

• .

Remarks:

• The asymptotic behavior qn ∼ c n−5/2 ρn is typical.
• The generating function G(z) =

M∈M qnzn is algebraic:

1 − 16z + (18z − 1)G(z) − 27z2G(z)2 = 0.

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Counting maps

Theorem [Tutte 63]: The number of rooted quadrangulations with n faces is qn = 2 · 3n (n + 1)(n + 2) 2n 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].

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A bijection of Schaeffer

Quadrangulations ⇔ Well-labelled trees

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Well-labelled trees

A rooted plane tree is a rooted planar map with a single face. There are Cn =

1 n+1

2n

n

• trees with n edges.

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Well-labelled trees

A well-labelled tree is a tree with positive labels such that

• the root vertex is labelled 1.
• the difference of labels between adjacent vertices is −1, 0 or 1.

2 2 1 3 2 4 1 2 1 2 1

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From quadrangulations to trees

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From quadrangulations to trees

Step 1. Compute the graph distance from the vertex incident to the root-edge.

3 2 1 2 1 2 2 3 1 3

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From quadrangulations to trees

Step 2. Create an edge of the tree for each face of the quadrangulation. i+1 i i i i+1 i+1 i+1 i+2

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From quadrangulations to trees

Step 2. Create an edge of the tree for each face of the quadrangulation.

3 2 1 2 1 2 2 3 1 3

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From quadrangulations to trees

Step 2. Create an edge of the tree for each face of the quadrangulation.

3 2 1 2 1 2 2 3 1 3

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Bijection

Theorem [Schaeffer 98]: This construction is a bijection between quadrangulations with n faces and well-labelled trees with n edges. ⇐ ⇒

2 2 1 3 2 4 1 2 1 2 1

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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 + 2 vertices

⇒ The image has n edges and n + 1 vertices. i i−1

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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 + 2 vertices

⇒ The image has n edges and n + 1 vertices.

• The image has no cycle, hence it is a tree.

i i−1 i−1

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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 + 2 vertices

⇒ The image has n edges and n + 1 vertices.

• The image has no cycle, hence it is a tree.
• It is well-labelled.

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From trees to quadrangulations

2 2 1 3 2 4 1 2 1 2 1

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From trees to quadrangulations

2 2 1 3 2 4 1 2 1 2 1

Step 1. Add an isolated vertex labelled 0.

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From trees to quadrangulations

2 2 1 3 2 4 1 2 1 2 1

Step 2. Join every corner labelled i to the next corner labelled i−1 around the tree.

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From trees to quadrangulations

2 2 1 3 2 4 1 2 1 2 1

One obtains a quadrangulation with labels indicating the distance from the vertex adjacent to the root-edge.

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Counting ?

It does not seem easy to count well-labelled trees. qn = 2 · 3n (n + 1)(n + 2) 2n n

• .

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Counting ?

It does not seem easy to count well-labelled trees. But it is easy to see that there are tn =

3n (n+1)

2n

n

• labelled trees:
• The minimum of labels is 1.
• The difference of labels between adjacent vertices is −1,0,1.

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Extended bijection

Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that . . .

2 3 3 4 2 1 2 1 3 2 1

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Extended bijection

Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that . . .

2 3 3 4 2 2 1 2 1 2 1

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Extended bijection

Theorem: Labelled trees are in bijection with rooted quadrangulations with a marked vertex, such that . . .

2 3 3 4 2 2 1 2 1 2 1

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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.

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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 + 2

2 qn such rooted+marked quadrangulations. ⇒ qn = 2 n + 2tn = 2 · 3n (n + 1)(n + 2) 2n n

• .
• The bijection gives a linear random generation algorithm

for rooted quadrangulation.

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Continuous trees and maps

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Dyck words

A Dyck word of size n is a sequence D = d0d1 . . . d2n of non-negative integers satisfying d0 = d2n = 0 and di − di−1 = ±1 for all i = 1 . . . 2n.

1 2

. . . 2n

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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 = d0d1 . . . d2n encodes the tree T, then two indices i < j correspond to the same vertex of T if and only if di = dj = infi≤k≤j dk.

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From sequences to functions

It is convenient to consider Dyck words as continuous functions from [0, 1] to R+: take the piecewise linear function f such that f( i

2n) = di.

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From sequences to functions

It is convenient to consider Dyck words as continuous functions from [0, 1] to R+: take the piecewise linear function f such that f( i

2n) = di.

Remarks:

• The function f is obtained from the tree T by turning

around the tree at speed 2n.

• Two reals 0 ≤ s < t ≤ 1 correspond to the same point of

the tree if and only if f(s) = f(t) = infs≤x≤t f(x).

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Trees as quotients of [0, 1]

Let f be a continuous function from [0, 1] to R+ such that f(0)=f(1)=0. We denote s ∼f t if f(s) = f(t) = infs≤x≤t f(x) and define Tf as the quotient of [0, 1] by the relation ∼f.

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Trees as quotients of [0, 1]

Let f be a continuous function from [0, 1] to R+ such that f(0)=f(1)=0. We denote s ∼f t if f(s) = f(t) = infs≤x≤t f(x) and define Tf as the quotient of [0, 1] by the relation ∼f. ⇒ Ancestor relation u ≺ v is defined by min(u) < min(v) ≤ max(v) < max(u). ⇒ Distance df(u, v) = f(u) + f(v) − 2f(ρ) where ρ is the greatest common ancestors.

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Trees as quotients of [0, 1]

Let f be a continuous function from [0, 1] to R+ such that f(0)=f(1)=0. We denote s ∼f t if f(s) = f(t) = infs≤x≤t f(x) and define Tf as the quotient of [0, 1] by the relation ∼f. We call (Tf, df) 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

• ther simple path.

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Trees as quotients of [0, 1]

Let f be a continuous function from [0, 1] to R+ such that f(0)=f(1)=0. We denote s ∼f t if f(s) = f(t) = infs≤x≤t f(x) and define Tf as the quotient of [0, 1] 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.

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Real quadrangulations ?

By the bijection of Schaeffer, quandrangulations are in bijection with well labelled trees. ⇐ ⇒

2 2 1 3 2 4 1 2 1 2 1

Vertices of the tree are identified with non-marked vertices

• f the quadrangulation.

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Real quadrangulations ?

For two vertices u, v of the tree T, we define d0

Q(u, v) = ℓ(u) + ℓ(v) + 2 − 2 min(ℓ(w) : w ∈ u T v).

u v There are always two shortest paths u T v.

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Real quadrangulations ?

For two vertices u, v of the tree T, we define d0

Q(u, v) = ℓ(u) + ℓ(v) + 2 − 2 min(ℓ(w) : w ∈ u T v).

Prop: The distance dQ between two vertices u, v is at most d0

Q(u, v).

u v

ℓ(v) ℓ(u) min(ℓ(w))

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Real quadrangulations ?

For two vertices u, v of the tree T, we define d0

Q(u, v) = ℓ(u) + ℓ(v) + 2 − 2 min(ℓ(w) : w ∈ u T v).

Prop: The distance dQ between two vertices u, v is at most d0

Q(u, v).

Hence, dQ(u, v) ≤ d∗

Q(u, v) =

min

u=u0,u1,...,uk=v

• i

d0

Q(ui, ui+1).

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Quadrangulations as quotient of trees

Let Tf be a real tree and let g be a continuous function from Tf to R+ such that g(ρ) = 0. For two points u, v of Tf, we define

• D0(u, v) = g(u) + g(v) − 2 inf(g(w) : w ∈ u T v).
• D∗(u, v) =

inf

u=u0,u1,...,uk=v

• i

D0

Q(ui, ui+1).

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Quadrangulations as quotient of trees

Let Tf be a real tree and let g be a continuous function from Tf to R+ such that g(ρ) = 0. For two points u, v of Tf, we define

• D0(u, v) = g(u) + g(v) − 2 inf(g(w) : w ∈ u T v).
• D∗(u, v) =

inf

u=u0,u1,...,uk=v

• i

D0

Q(ui, ui+1).

We denote u ≈g v if D∗(u, v) = 0, we define Tf,g as the quotient of Tf by the relation ≈g, and call real quadrangulation the metric space (Tf,g, D∗).

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Quadrangulations as quotient of trees

Let Tf be a real tree and let g be a continuous function from Tf to R+ such that g(ρ) = 0. For two points u, v of Tf, we define

• D0(u, v) = g(u) + g(v) − 2 inf(g(w) : w ∈ u T v).
• D∗(u, v) =

inf

u=u0,u1,...,uk=v

• i

D0

Q(ui, ui+1).

Next lecture: We will see that the rescaled uniformly random quadrangulation converges (at least along subsequences) to a random metric space (Te,g, D), where

• Te is the Continuum Random Tree,
• g is a Gaussian process on Te conditioned to be positive,
• D is a distance on Te,g which is less than D∗.

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End of Lecture 1. . .

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