Computing the moments of the GOEbijectively

Probability Seminar at MIT, February 2011

Olivier Bernardi (MIT)

λ1 λ3 λ4 λ5 λ6λ7λ2

1 28

4

Computing the moments of the GOEbijectively

Probability Seminar at MIT, February 2011

Olivier Bernardi (MIT)

λ1 λ3 λ4λ5 λ6λ7λ2

1 28

4

2+3=5; n

A combinatorial problem

Surfaces from a polygon

We consider the different ways of gluing the sides of a 2n-gon in pairs.

Surfaces from a polygon

We consider the different ways of gluing the sides of a 2n-gon in pairs.

The gluing of two sides can either be orientable (giving a cylinder) ornon-orientable (giving a Mobius strip).

Orientable gluing Non-orientable gluing

The surface obtained is orientable if and only if each gluing isorientable.

Surfaces from a polygon

There are (2n− 1)!! = (2n− 1)(2n− 3) · · · 3 ways of obtaining anorientable surface.There are 2n(2n− 1)!! ways of obtaining a general surface.

We consider the different ways of gluing the sides of a 2n-gon in pairs.

Surfaces from a polygon

There are (2n− 1)!! = (2n− 1)(2n− 3) · · · 3 ways of obtaining anorientable surface.There are 2n(2n− 1)!! ways of obtaining a general surface.

We consider the different ways of gluing the sides of a 2n-gon in pairs.

Question: How many ways are there to obtain each surface (consid-ered up to homeomorphism) ?

Surfaces from a polygon

There are (2n− 1)!! = (2n− 1)(2n− 3) · · · 3 ways of obtaining anorientable surface.There are 2n(2n− 1)!! ways of obtaining a general surface.

We consider the different ways of gluing the sides of a 2n-gon in pairs.

Question: How many ways are there to obtain each surface (consid-ered up to homeomorphism) ?

Example: The number of ways of getting the sphere is the Catalannumber Cat(n) = 1

n+1

(2nn

).

Surfaces from a polygon

We consider the different ways of gluing the sides of a 2n-gon in pairs.

By the Euler relation, the type of the surface is t = n+ 1−#vertices.

Question: How many ways are there to obtain a surface of type t ?

Type: 00 1 2 3 4

The Gaussian Orthogonal Ensemble

The GOE

S =

si,j = sj,i

Let Sp be the set of real symmetric matrices of dimension p× p.

We define a random variable S in Sp by choosing the entries si,j fori ≤ j to be independent centered Gaussian variables with variance2 if i = j and variance 1 if i < j (and then setting si,j = sj,i for i > j).

The GOE

Hence, the distribution γ of S over Sp has density κ exp(−tr(S2)/4)with respect to the Lebesgue measure dS :=

∏i≤j dsi,j .

S =

si,j = sj,i

Let Sp be the set of real symmetric matrices of dimension p× p.

We define a random variable S in Sp by choosing the entries si,j fori ≤ j to be independent centered Gaussian variables with variance2 if i = j and variance 1 if i < j (and then setting si,j = sj,i for i > j).

The GOE is the probability space (Sp, γ).

Eigenvalues of the GOE

S =si,j = sj,i

Question: What is the distribution of λ := λU , with U uniform in [p]?

Let λ1 ≤ λ2 ≤ · · · ≤ λp be the eigenvalues of S.

λ1 λ3 λ4 λ5 λ6λ7λ2

Eigenvalues of the GOE

The nth moment of λ is

⟨1

p

p∑i=1

λni

⟩=

1

p〈tr(Sn)〉.

S =si,j = sj,i

Question: What is the distribution of λ := λU , with U uniform in [p]?

Let λ1 ≤ λ2 ≤ · · · ≤ λp be the eigenvalues of S.

λ1 λ3 λ4 λ5 λ6λ7λ2

Remark. Odd moments are 0 by symmetry.

Computing the 2nth moment using the Wick formula

We want the expectation of tr(S2n) =∑

i1,i2,...,i2n∈[p]

si1,i2si2,i3 · · · si2n,i1 .

Since the si,j are Gaussian, the Wick formula gives

〈si1,i2si2,i3 · · · si2n,i1〉 =∑

π matching on [2n]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩.

Computing the 2nth moment using the Wick formula

We want the expectation of tr(S2n) =∑

i1,i2,...,i2n∈[p]

si1,i2si2,i3 · · · si2n,i1 .

Since the si,j are Gaussian, the Wick formula gives

〈si1,i2si2,i3 · · · si2n,i1〉 =∑

π matching on [2n]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩.

⟨tr(S2n)

⟩=

∑π matching on [2n]

∑i1...i2n∈[p]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩.

Contribution of matching π?

Computing the 2nth moment using the Wick formula

We want the expectation of tr(S2n) =∑

i1,i2,...,i2n∈[p]

si1,i2si2,i3 · · · si2n,i1 .

Since the si,j are Gaussian, the Wick formula gives

〈si1,i2si2,i3 · · · si2n,i1〉 =∑

π matching on [2n]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩.

⟨tr(S2n)

⟩=

∑π matching on [2n]

∑i1...i2n∈[p]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩.

Hint:⟨sik,ik+1

sil,il+1

⟩= 0 except if (ik, ik+1) = (il, il+1) or (il+1, il).i1

i2

i3

i4i5

i2n

i1i2

i3

i4i5

i2n

N O

Contribution of matching π?

Computing the 2nth moment using the Wick formula

i1i2

i3

i4i5

i2n

i1i2

i3

i4i5

i2n

N O

⟨tr(S2n)

⟩p=

∑π matching on [2n]

∑i1...i2n∈[p]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩=

∑π matching on [2n]

ε:π→{O,N}

∑i1...i2n∈[p]

∏{k,l}∈ε−1(N)

1ik=il,ik+1=il+1

×∏

{k,l}∈ε−1(O)

1ik=il+1,ik+1=il

=∑

gluing of 2n-gon

p#vertices.

Computing the 2nth moment using the Wick formula

Summary: The 2nth moment of λ in GOE(p) is1

p

∑gluings of

2n-gon

p#vertices.

⟨tr(S2n)

⟩p=

∑π matching on [2n]

∑i1...i2n∈[p]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩=

∑π matching on [2n]

ε:π→{O,N}

∑i1...i2n∈[p]

∏{k,l}∈ε−1(N)

1ik=il,ik+1=il+1

×∏

{k,l}∈ε−1(O)

1ik=il+1,ik+1=il

=∑

gluing of 2n-gon

p#vertices.

Computing the 2nth moment using the Wick formula

Summary: The 2nth moment of λ in GOE(p) is1

p

∑gluings of

2n-gon

p#vertices.

⟨tr(S2n)

⟩p=

∑π matching on [2n]

∑i1...i2n∈[p]

∏{k,l}∈π

⟨sik,ik+1

sil,il+1

⟩=

∑π matching on [2n]

ε:π→{O,N}

∑i1...i2n∈[p]

∏{k,l}∈ε−1(N)

1ik=il,ik+1=il+1

×∏

{k,l}∈ε−1(O)

1ik=il+1,ik+1=il

=∑

gluing of 2n-gon

p#vertices.

Remark. When p→∞ the 2nth moment of λ√p which is∑

gluings of 2n-gon

p#vertices−n−1 tends to Cat(n). ⇒ Semi-circle law.

Back to the polygon

Surfaces from a polygon: state of the art

Question: How many ways of gluing 2n-gon into a surface of type t ?

Type: 0 1 2 3 4

Surfaces from a polygon: state of the art

Question: How many ways of gluing 2n-gon into a surface of type t ?

v = #vertices

n = #edges

11 15 5 241 52 22 5509 690 374 93 14

ηv(n)

Surfaces from a polygon: state of the art

[Harer, Zagier 86] Results for orientable surfaces:Formula for

∑v εv(n)pv and recurrence formula for εv(n).

Question: How many ways of gluing 2n-gon into a surface of type t ?

v = #vertices

n = #edges

11 15 5 241 52 22 5509 690 374 93 14

v = #vertices

n = #edges

10 11 0 20 10 0 521 0 70 0 14

εv(n)

ηv(n)

Surfaces from a polygon: state of the art

[Harer, Zagier 86] Results for orientable surfaces:Formula for

∑v εv(n)pv and recurrence formula for εv(n).

Question: How many ways of gluing 2n-gon into a surface of type t ?

v = #vertices

n = #edges

11 15 5 241 52 22 5509 690 374 93 14

ηv(n)

Related work: Jackson, Adrianov, Zagier, Mehta, Haagerup-Thorbjornsen, Kerov,Ledoux, Lass, Goulden-Nica, Schaeffer-Vassilieva, Morales-Vassilieva, Chapuy. . .

∑v

εv(n)pv =

p∑q=1

(pq

)2q−1

( n

q − 1

)(2n− 1)!!

(n+ 1)εv(n) = (4n− 2)εv−1(n− 1) + (n− 1)(2n− 1)(2n− 3)εv(n− 2).

Surfaces from a polygon: state of the art

[Harer, Zagier 86] Results for orientable surfaces:Formula for

∑v εv(n)pv and recurrence formula for εv(n).

Question: How many ways of gluing 2n-gon into a surface of type t ?

v = #vertices

n = #edges

11 15 5 241 52 22 5509 690 374 93 14

[Goulden, Jackson 97] Formula for∑v ηv(n)pv.

[Ledoux 09] Recurrence formula for ηv(n).

ηv(n)

Surfaces from a polygon: state of the art

[Harer, Zagier 86] Results for orientable surfaces:Formula for

∑v εv(n)pv and recurrence formula for εv(n).

Question: How many ways of gluing 2n-gon into a surface of type t ?

v = #vertices

n = #edges

11 15 5 241 52 22 5509 690 374 93 14

[Goulden, Jackson 97] Formula for∑v ηv(n)pv.

[Ledoux 09] Recurrence formula for ηv(n).

ηv(n)

[B., Chapuy 10] Asymptotic ηt(n) ∼n→∞ ctn3(t−1)/24n,

where ct =

2t−2

√6t−1

(t−1)!!if t odd,

3·2t−2

√π√6t(t−1)!!

∑t/2−1i=1

(2ii

)16−i if t even.

Results

Theorem [B.]: The number of ways of gluings a 2n-gon with verticescolored using every color in [q] is

n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Results

That is, Pq,r is the coefficient of xqyr in the series P (x, y) defined by:27P 4 − (36x+ 36y − 1)P 3

+(24x2y + 24xy2 − 16x3 − 16y3 + 8x2 + 8y2 + 46xy − x− y)P 2

+xy(16x2 + 16y2 − 64xy − 8x− 8y + 1)P−x2y2(16x2 + 16y2 − 32xy − 8x− 8y + 1) = 0.

where Pq,r is the number of planar maps with q vertices and r faces.

Results

Theorem [B.]: The number of ways of gluings a 2n-gon with verticescolored using every color in [q] is

n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Results

where Pq,r is the number of planar maps with q vertices and r faces.

Theorem [Harer-Zagier 86]: The number of orientable ways ofgluings a 2n-gon with vertices colored using every color in [q] is

2q−1(

n

q − 1

)(2n− 1)!!

Results

Theorem [B.]: The number of ways of gluings a 2n-gon with verticescolored using every color in [q] is

n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Results

⇒ The 2nth moment of λ in the GOE(p) = 1p

∑gluings of

2n-gonp#vertices

=1

p#gluings of 2n-gon with vertices colored in [p],

=1

p

p∑q=1

(p

q

)n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

where Pq,r is the number of planar maps with q vertices and r faces.

Results

Theorem [B.]: The number of ways of gluings a 2n-gon with verticescolored using every color in [q] is

n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Results

⇒ The 2nth moment of λ in the GOE(p) = 1p

∑gluings of

2n-gonp#vertices

=1

p#gluings of 2n-gon with vertices colored in [p],

=1

p

p∑q=1

(p

q

)n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Theorem [Goulden, Jackson 97] This number is

n!n∑

k=0

22n−kn∑

r=0

(n− 12

n− r

)(k + r − 1

k

)( p−12

r

)+ (2n− 1)!!

p−1∑q=1

2q−1(p− 1

q

)( n

q − 1

).

where Pq,r is the number of planar maps with q vertices and r faces.

Results

Corollary [recovering Ledoux 09]: The number ηv(n) of ways ofgluings a 2n-gon into a surfaces of type n+ 1− v satisfies:

(n+ 1) ηv(n)=(4n− 1) (2 ηv−1(n−1)− ηv(n−1))+(2n− 3)

((10n2 − 9n) ηv(n−2) + 8 ηv−1(n− 2)− 8 ηv−2(n−2)

)+5(2n− 3)(2n− 4)(2n− 5) (ηv(n−3)− 2 ηv−1(n−3))−2(2n− 3)(2n− 4)(2n− 5)(2n− 6)(2n− 7) ηv(n−4).

Results

Corollary [recovering Ledoux 09]: The number ηv(n) of ways ofgluings a 2n-gon into a surfaces of type n+ 1− v satisfies:

(n+ 1) ηv(n)=(4n− 1) (2 ηv−1(n−1)− ηv(n−1))+(2n− 3)

((10n2 − 9n) ηv(n−2) + 8 ηv−1(n− 2)− 8 ηv−2(n−2)

)+5(2n− 3)(2n− 4)(2n− 5) (ηv(n−3)− 2 ηv−1(n−3))−2(2n− 3)(2n− 4)(2n− 5)(2n− 6)(2n− 7) ηv(n−4).

[Harer, Zagier 86]: The number εv(n) of orientable ways of gluingsa 2n-gon into a surfaces of type n+ 1− v satisfies:

(n+ 1)εv(n) = (4n− 2)εv−1(n− 1) + (n− 1)(2n− 1)(2n− 3)εv(n− 2).

Crash course on maps

Definition: A map is a graph embedded in a surface obtained bygluing together polygons (i.e. the faces are simply connected).

Crash course on maps

Definition: A map is a graph embedded in a surface obtained bygluing together polygons (i.e. the faces are simply connected).Maps are considered up to homeomorphism.

6==

Crash course on maps

Definition: A map is a graph embedded in a surface obtained bygluing together polygons (i.e. the faces are simply connected).Maps are considered up to homeomorphism.

6==

A planar map is a map on the sphere.A unicellular map is a map with 1 face.A rooted map is a map with a marked edge-direction+side.

Example: the rooted unicellular planar mapsare the ”rooted plane trees”.

Crash course on maps

Definition: A map is a graph embedded in a surface obtained bygluing together polygons (i.e. the faces are simply connected).Maps are considered up to homeomorphism.

6==

Embedding Thorem: Rooted maps on orientable surfaces are inbijection with graph+rooted rotation system, that is,- a total order on the half-edges around a “root-vertex”,- a cyclic order on the half-edges around the other vertices.

Bijections for q-colored unicellular maps

A tree-rooted map is a map on an orientable surface with a markedspanning tree.A planar-rooted map is a map on an orientable surface with amarked planar connected spanning submap.

Remark. The number of tree-rooted maps with q vertices and n edges

is Cat(q − 1)

(2n

2q

)(2n− 2q + 1)!! =

2q−1

q!

(n

q − 1

)(2n− 1)!!

The number of planar-rooted maps with q vertices, r faces, and n

edges is Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

Bijections for q-colored unicellular maps

Thm [B.] There is a bijection between rooted unicellular map coloredusing every color in [q] with n edges on orientable surfaces and tree-rooted maps with q labelled vertices and n edges.

Moreover, number of edges between colors i and j → number of edgesbetween vertices i and j.

1 28

4

Corollary [Harer-Zagier] The number of [q]-colored rooted unicellular

maps with n edges on orientable surfaces is: 2q−1(

n

q − 1

)(2n− 1)!!

Previous combinatorial proofs: [Lass 01], [Goulden,Nica 05].Bipartite case: [Schaeffer,Vassilieva 08], [Morales,Vassilieva 09].

Bijections for q-colored unicellular maps

Thm [B.] There is a q!r!21−r-to-1 correspondence Φ between rootedunicellular maps on general surfaces colored using every color in [q]with n edges and planar-rooted maps with q vertices, r faces and nedges.

Moreover, number of edges incident to color i → degree of vertex i.

Corollary:The number of [q]-colored rooted unicellular maps with nedges on general surfaces is:

n−q+2∑r=1

q!r!

2r−1Pq,r

(2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!

where Pq,r is the number of planar maps with q vertices and r faces.

Sketch of the proof

Idea 1: Colored unicellular map ↔ bi-Eulerian tour

Def. Let G = (V,E) be a graph. A bi-Eulerian tour is a walk startingand ending at the same vertex and using every edge twice.

1

2

36

4

910

5

118

7

12

Idea 1: Colored unicellular map ↔ bi-Eulerian tour

Def. Let G = (V,E) be a graph. A bi-Eulerian tour is a walk startingand ending at the same vertex and using every edge twice.

Lemma [adapting Lass 01]. [q]-colored unicellular maps are in bijec-tion with pairs (G,E), where G is a graph with vertex set [q] and Eis a bi-Eulerian tour.

Moreover, the map is on an orientable surface if and only if no edge isused twice in the same direction.

12

3

4

567

8

9

10

1112

1

2

36

4

910

5

118

7

12

Idea 2: BEST Theorem

BEST Theorem. Let ~G = (V,A) be a directed graph such that everyvertex has as many ingoing and outgoing arcs.The Eulerian tours of ~G starting and ending at v are in bijection withpairs (T,O) made of a spanning tree T oriented toward v + orderingO around each vertex of the outgoing arc not in T .

v

Idea 2: BEST Theorem

BEST Theorem. Let ~G = (V,A) be a directed graph such that everyvertex has as many ingoing and outgoing arcs.The Eulerian tours of ~G starting and ending at v are in bijection withpairs (T,O) made of a spanning tree T oriented toward v + orderingO around each vertex of the outgoing arc not in T .

v

1

4

5102

9

712

113

68

v

12

3

12

3

4

1

2

Idea 2: BEST Theorem

BEST Theorem. Let ~G = (V,A) be a directed graph such that everyvertex has as many ingoing and outgoing arcs.The Eulerian tours of ~G starting and ending at v are in bijection withpairs (T,O) made of a spanning tree T oriented toward v + orderingO around each vertex of the outgoing arc not in T .

Corollary 1. Let G be a graph. The bi-Eulerian tours of G whichnever take an edge twice in the same direction are in bijection withpairs (T,R) made of a spanning tree + rooted rotation system.←→ tree-rooted maps having underlying graph G.

6

25

431 872

13

123

4

+ embedding Theorem

Idea 2: BEST Theorem

⇒ Bijection between rooted unicellular maps colored using everycolor in [q] with n edges on orientable surfaces and tree-rooted mapswith q labelled vertices and n edges.

1

3

2

6

5

6

7

8

4

2

5

431 87

2

13

12

34

+ embedding Theorem

Idea 2: BEST Theorem

A bi-oriented tree-rooted map is a rooted map on orientable surface+ spanning tree + partial orientation such that• indegree=outdegree for every vertex• the oriented edges in the spanning tree toward parent.

+ embedding Theorem

Idea 2: BEST Theorem

A bi-oriented tree-rooted map is a rooted map on orientable surface+ spanning tree + partial orientation such that• indegree=outdegree for every vertex• the oriented edges in the spanning tree toward parent.

1

2

36

4

910

5

118

7

12

Corollary 2. Let G be a graph. There is a 1-to-2r correspondencebetween the bi-Eulerian tours of G and the bi-oriented tree-rootedmap on G having r oriented edges outside of the spanning tree.

+ embedding Theorem

Idea 3: cutting and pasting

Let B be a bi-oriented tree-rooted map. The planar-rooted mapP = Ψ(B) is obtained by cutting the oriented external edges in theirmiddle and regluing them according to the parenthesis system theyform around the tree (and then forgetting the tree + orientation).

Theorem: The mapping Ψ is r!-to-1 between bi-oriented tree-rootedmaps with r − 1 oriented edges outside of the spanning tree and planar-rooted map with r sub-faces.

(r−1)!-to-1 r-to-1

Idea 3: cutting and pasting

⇒ q!r!2r−1-to-1 correspondence between rooted unicellular mapscolored using every color in [q] with n edges on general surfaces andplanar-rooted maps with q vertices, r faces, and n edges.

1

2

36

4

910

5

118

7

12

q!-to-11-to-2r−1

r!-to-1

Recurrence relation

Toward the recurrence relation

We want to prove the recurrence of Ledoux:

(n+ 1) ηv(n)=(4n− 1) (2 ηv−1(n−1)− ηv(n−1))+(2n− 3)

((10n2 − 9n) ηv(n−2) + 8 ηv−1(n− 2)− 8 ηv−2(n−2)

)+5(2n− 3)(2n− 4)(2n− 5) (ηv(n−3)− 2 ηv−1(n−3))−2(2n− 3)(2n− 4)(2n− 5)(2n− 6)(2n− 7) ηv(n−4).

Toward the recurrence relation

We want to prove the recurrence of Ledoux:

(n+ 1) ηv(n)=(4n− 1) (2 ηv−1(n−1)− ηv(n−1))+(2n− 3)

((10n2 − 9n) ηv(n−2) + 8 ηv−1(n− 2)− 8 ηv−2(n−2)

)+5(2n− 3)(2n− 4)(2n− 5) (ηv(n−3)− 2 ηv−1(n−3))−2(2n− 3)(2n− 4)(2n− 5)(2n− 6)(2n− 7) ηv(n−4).

Let Vn(x) = 1(2n)!

∑n+1v=1 ηv(n)xn. The recurrence is equivalent to

−(2n)(2n− 1)(2n− 2)(n+ 1)Vn(x) + (4n− 1)(2n− 2)(2x− 1)Vn−1(x)+(10n2 − 9n+ 8x− 8x2)Vn−2(x) + 5(1− 2x)Vn−3(x)− 2Vn−4(x)= 0.

(*)

Toward the recurrence relation

We want to prove the recurrence of Ledoux:

(n+ 1) ηv(n)=(4n− 1) (2 ηv−1(n−1)− ηv(n−1))+(2n− 3)

((10n2 − 9n) ηv(n−2) + 8 ηv−1(n− 2)− 8 ηv−2(n−2)

)+5(2n− 3)(2n− 4)(2n− 5) (ηv(n−3)− 2 ηv−1(n−3))−2(2n− 3)(2n− 4)(2n− 5)(2n− 6)(2n− 7) ηv(n−4).

Let Vn(x) = 1(2n)!

∑n+1v=1 ηv(n)xn. The recurrence is equivalent to

−(2n)(2n− 1)(2n− 2)(n+ 1)Vn(x) + (4n− 1)(2n− 2)(2x− 1)Vn−1(x)+(10n2 − 9n+ 8x− 8x2)Vn−2(x) + 5(1− 2x)Vn−3(x)− 2Vn−4(x)= 0.

Vn(x) = 1(2n)

∑n+1q=1

(xq

)Un(q), where Un(q) is the number of unicellu-

lar maps with n edges colored using every color in [q].

⇒ my computer can translate (*) into a linear differential equation for

the series U(t, w) =∑n,q

Un(q)

(2n)!tnwq.

(*)

Toward the recurrence relation

We know Un(q)=

n−q+2∑r=1

q!r!

2r−1Pq,r

( 2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!.

⇒ U(t, w) =1

2exp

(t

2

)Q

(t

2, 2w

),

where Q(t, w)=∑

q,r>0

q! r!Pq,r

(2q + 2r − 4)!tq+r−2wq .

Toward the recurrence relation

We know Un(q)=

n−q+2∑r=1

q!r!

2r−1Pq,r

( 2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!.

⇒ U(t, w) =1

2exp

(t

2

)Q

(t

2, 2w

),

where Q(t, w)=∑

q,r>0

q! r!Pq,r

(2q + 2r − 4)!tq+r−2wq .

Equation (*) is equivalent to:(6tw + 4w2 − 36t− 7w − 6)Q(t, w)− (12t2 + 8tw + 7w − 25) ∂

∂tQ(t, w)

+w(w + 2)(8w + 6t− 7) ∂∂wQ(t, w) + (2t2 − 4tw + 37t+ 9) ∂2

∂t2Q(t, w)

−w(w + 2)(8t+ 7) ∂2

∂w∂tQ(t, w) + 2w2(w + 2)2 ∂2

∂w2Q(t, w)

+t(8t+ 11) ∂3

∂t3Q(t, w)− 4wt(w + 2) ∂3

∂w∂t2Q(t, w) + 2t2 ∂4

∂t4Q(t, w) = 0.

Toward the recurrence relation

We know Un(q)=

n−q+2∑r=1

q!r!

2r−1Pq,r

( 2n

2q + 2r − 4

)(2n− 2q − 2r + 1)!!.

⇒ U(t, w) =1

2exp

(t

2

)Q

(t

2, 2w

),

where Q(t, w)=∑

q,r>0

q! r!Pq,r

(2q + 2r − 4)!tq+r−2wq .

Equation (*) is equivalent to:

0 = (4y − 2x− 1)(72P (x, y)− 72x ∂

∂xP (x, y)− (72y − 2) ∂

∂yP (x, y)

)−72x2 ∂2

∂x2 P (x, y) +(8x2 − 6x− 24xy − 56y2 + 10y + 1

)∂2

∂y2 P (x, y)

+2x (4x− 80y − 1) ∂2

∂y∂xP (x, y)

+ (4x− 8y − 1)(y(4x+ 48y2 − 8y − 1

)∂3

∂y3 P (x, y) + 48x3 ∂3

∂x3 P (x, y))

+(4x− 8y − 1)(144x2y ∂3

∂y∂x2 P (x, y) + x(4x− 8y − 1 + 144y2

)∂3

∂y2∂xP (x, y)

),

where P (x, y) =∑

q,r>0

Pq,rxqyr.

(**)

Toward the recurrence relation

The differential equation (**) for the series P (x, y) of planar maps canbe checked using the algebraic equation:27P 4 − (36x+ 36y − 1)P 3

+(24x2y + 24xy2 − 16x3 − 16y3 + 8x2 + 8y2 + 46xy − x− y)P 2

+xy(16x2 + 16y2 − 64xy − 8x− 8y + 1)P−x2y2(16x2 + 16y2 − 32xy − 8x− 8y + 1) = 0.

(by expressing the partial derivatives of P as polynomials in P ). �

Thanks.

λ1 λ3 λ4 λ5 λ6λ7λ2