Computing the moments of the GOE - MIT Mathematicsmath.mit.edu/~bernardi/slides/slides-GOE.pdf ·...
Transcript of Computing the moments of the GOE - MIT Mathematicsmath.mit.edu/~bernardi/slides/slides-GOE.pdf ·...
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
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 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.
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.
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
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 ). �