Computing the moments of the GOE
bijectively
Olivier Bernardi (MIT)
4
8
λ1
λ2 λ3
λ4 λ5
1
2
λ6 λ7
Probability Seminar at MIT, February 2011
Computing the moments of the GOE
bijectively
Olivier Bernardi (MIT)
n
2+3=5;
4
8
λ1
λ2λ3 λ4 λ5
1
2
λ6λ7
Probability Seminar at MIT, February 2011
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) or
non-orientable (giving a Möbius strip).
Orientable gluing
Non-orientable gluing
The surface obtained is orientable if and only if each gluing is
orientable.
Surfaces from a polygon
We consider the different ways of gluing the sides of a 2n-gon in pairs.
There are (2n − 1)!! = (2n − 1)(2n − 3) · · · 3 ways of obtaining an
orientable surface.
There are 2n (2n − 1)!! ways of obtaining a general surface.
Surfaces from a polygon
We consider the different ways of gluing the sides of a 2n-gon in pairs.
There are (2n − 1)!! = (2n − 1)(2n − 3) · · · 3 ways of obtaining an
orientable surface.
There are 2n (2n − 1)!! ways of obtaining a general surface.
Question: How many ways are there to obtain each surface (considered up to homeomorphism) ?
Surfaces from a polygon
We consider the different ways of gluing the sides of a 2n-gon in pairs.
There are (2n − 1)!! = (2n − 1)(2n − 3) · · · 3 ways of obtaining an
orientable surface.
There are 2n (2n − 1)!! ways of obtaining a general surface.
Question: How many ways are there to obtain each surface (considered up to homeomorphism) ?
Example: The number of
ways of getting the sphere is the Catalan
2n
1
number Cat(n) = n+1
n .
Surfaces from a polygon
We consider the different ways of gluing the sides of a 2n-gon in pairs.
Question: How many ways are there to obtain a surface of type t ?
Type: 0
1
2
3
4
By the Euler relation, the type of the surface is t = n + 1 − #vertices.
The Gaussian Orthogonal Ensemble
The GOE
Let Sp be the set of real symmetric matrices of dimension p × p.
si,j = sj,i
S=
We define a random variable S in Sp by choosing the entries si,j for
i ≤ j to be independent centered Gaussian variables with variance
2 if i = j and variance 1 if i < j (and then setting si,j = sj,i for i > j).
The GOE
Let Sp be the set of real symmetric matrices of dimension p × p.
si,j = sj,i
S=
We define a random variable S in Sp by choosing the entries si,j for
i ≤ j to be independent centered Gaussian variables with variance
2 if i = j and variance 1 if i < j (and then setting si,j = sj,i for i > j).
2
Hence, the distribution γ of S over Sp has density
κ
exp(−tr(S
)/4)
Q
with respect to the Lebesgue measure dS := i≤j dsi,j .
The GOE is the probability space (Sp , γ).
Eigenvalues of the GOE
Let λ1 ≤ λ2 ≤ · · · ≤ λp be the eigenvalues of S.
si,j = sj,i
S=
λ1
λ2 λ3
λ4 λ5
λ6 λ7
Question: What is the distribution of λ := λU , with U uniform in [p]?
Eigenvalues of the GOE
Let λ1 ≤ λ2 ≤ · · · ≤ λp be the eigenvalues of S.
si,j = sj,i
S=
λ1
λ2 λ3
λ4 λ5
λ6 λ7
Question: What is the distribution of λ := λU , with U uniform in [p]?
* p
+
1X n
1
The nth moment of λ is
λi = htr(S n )i.
p i=1
p
Remark. Odd moments are 0 by symmetry.
Computing the 2nth moment using the Wick formula
X
We want the expectation of tr(S 2n ) =
si1 ,i2 si2 ,i3 · · · si2n ,i1 .
i1 ,i2 ,...,i2n ∈[p]
Since the si,j are Gaussian, the Wick formula gives
X
Y hsi1 ,i2 si2 ,i3 · · · si2n ,i1 i =
sik ,ik+1 sil ,il+1 .
π matching on [2n] {k,l}∈π
Computing the 2nth moment using the Wick formula
X
We want the expectation of tr(S 2n ) =
si1 ,i2 si2 ,i3 · · · si2n ,i1 .
i1 ,i2 ,...,i2n ∈[p]
Since the si,j are Gaussian, the Wick formula gives
X
Y hsi1 ,i2 si2 ,i3 · · · si2n ,i1 i =
sik ,ik+1 sil ,il+1 .
π matching on [2n] {k,l}∈π
tr(S
2n
) =
X
X
Y sik ,ik+1 sil ,il+1 .
π matching on [2n] i1 ...i2n ∈[p] {k,l}∈π
Contribution of matching π?
Computing the 2nth moment using the Wick formula
X
We want the expectation of tr(S 2n ) =
si1 ,i2 si2 ,i3 · · · si2n ,i1 .
i1 ,i2 ,...,i2n ∈[p]
Since the si,j are Gaussian, the Wick formula gives
X
Y hsi1 ,i2 si2 ,i3 · · · si2n ,i1 i =
sik ,ik+1 sil ,il+1 .
π matching on [2n] {k,l}∈π
tr(S
2n
) =
X
X
Y sik ,ik+1 sil ,il+1 .
π matching on [2n] i1 ...i2n ∈[p] {k,l}∈π
Contribution of matching π?
Hint: sik ,ik+1 sil ,il+1 = 0 except if (ik , ik+1 ) = (il , il+1 ) or (il+1 , il ).
i2n
i1
i2
i2n
i1
i3
i5
i4
i2
i3
N
i5
i4
O
Computing the 2nth moment using the Wick formula
X
X
Y 2n
sik ,ik+1 sil ,il+1
tr(S ) p =
π matching on [2n] i1 ...i2n ∈[p] {k,l}∈π
=
π
X
X
matching on
[2n]
Y
1ik =il ,ik+1 =il+1
i1 ...i2n ∈[p] {k,l}∈−1 (N )
:π→{O,N }
×
Y
1ik =il+1 ,ik+1 =il
{k,l}∈−1 (O)
=
X
p#vertices .
gluing of 2n-gon
i2n
i1
i2
i2n
i1
i3
i5
i4
i2
i3
N
i5
i4
O
Computing the 2nth moment using the Wick formula
X
X
Y 2n
sik ,ik+1 sil ,il+1
tr(S ) p =
π matching on [2n] i1 ...i2n ∈[p] {k,l}∈π
=
π
X
matching on
X
[2n]
Y
1ik =il ,ik+1 =il+1
i1 ...i2n ∈[p] {k,l}∈−1 (N )
:π→{O,N }
×
Y
1ik =il+1 ,ik+1 =il
{k,l}∈−1 (O)
=
X
p#vertices .
gluing of 2n-gon
1
Summary: The 2nth moment of λ in GOE(p) is
p
X
gluings of
2n-gon
p#vertices .
Computing the 2nth moment using the Wick formula
X
X
Y 2n
sik ,ik+1 sil ,il+1
tr(S ) p =
π matching on [2n] i1 ...i2n ∈[p] {k,l}∈π
X
=
π
matching on
X
[2n]
Y
1ik =il ,ik+1 =il+1
i1 ...i2n ∈[p] {k,l}∈−1 (N )
:π→{O,N }
×
Y
1ik =il+1 ,ik+1 =il
{k,l}∈−1 (O)
X
=
p#vertices .
gluing of 2n-gon
1
Summary: The 2nth moment of λ in GOE(p) is
p
Remark.
X
X
gluings of
2n-gon
When p → ∞ the 2nth moment of
p#vertices−n−1 tends to Cat(n).
gluings of 2n-gon
p#vertices .
λ
√
p
which is
⇒ 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 ?
1
1
5
41
n = #edges 509
v = #vertices
1
5
2
52 22 5
690 374 93
ηv (n)
14
Surfaces from a polygon: state of the art
Question: How many ways of gluing 2n-gon into a surface of type t ?
1
1
5
41
n = #edges 509
v = #vertices
1
5
2
52 22 5
690 374 93
ηv (n)
14
[Harer, Zagier
for orientable surfaces:
P 86] Results
Formula for v v (n)pv and recurrence formula for v (n).
1
0
1
0
n = #edges 21
v = #vertices
1
0
10
0
2
0
70
v (n)
5
0
14
Surfaces from a polygon: state of the art
Question: How many ways of gluing 2n-gon into a surface of type t ?
1
1
5
41
n = #edges 509
v = #vertices
1
5
2
52 22 5
690 374 93
ηv (n)
14
[Harer, Zagier
for orientable surfaces:
P 86] Results
Formula for v v (n)pv and recurrence formula for v (n).
X
v
v (n)pv =
p X
p
q=1
n (2n − 1)!!
2q−1
q−1
q
(n + 1)v (n) = (4n − 2)v−1 (n − 1) + (n − 1)(2n − 1)(2n − 3)v (n − 2).
Related work: Jackson, Adrianov, Zagier, Mehta, Haagerup-Thorbjornsen, Kerov,
Ledoux, Lass, Goulden-Nica, Schaeffer-Vassilieva, Morales-Vassilieva, Chapuy. . .
Surfaces from a polygon: state of the art
Question: How many ways of gluing 2n-gon into a surface of type t ?
1
1
5
41
n = #edges 509
v = #vertices
1
5
2
52 22 5
690 374 93
ηv (n)
14
[Harer, Zagier
for orientable surfaces:
P 86] Results
Formula for v v (n)pv and recurrence formula for v (n).
[Goulden, Jackson 97] Formula for v ηv (n)pv .
[Ledoux 09] Recurrence formula for ηv (n).
P
Surfaces from a polygon: state of the art
Question: How many ways of gluing 2n-gon into a surface of type t ?
1
1
5
41
n = #edges 509
v = #vertices
1
5
2
52 22 5
690 374 93
ηv (n)
14
[Harer, Zagier
for orientable surfaces:
P 86] Results
Formula for v v (n)pv and recurrence formula for v (n).
[Goulden, Jackson 97] Formula for v ηv (n)pv .
[Ledoux 09] Recurrence formula for ηv (n).
t
3(t−1)/2 n
[B., Chapuy
10]
Asymptotic
η
(n)
∼
c
n
4 ,
n→∞
t

t−2
2
 √ t−1
if t odd,
6
(t−1)!!
where ct =
Pt/2−1 2i −i
t−2
3·2
 √ √t
if t even.
i=1
i 16
P
π 6 (t−1)!!
Results
Theorem [B.]: The number of ways of gluings a 2n-gon with vertices
colored using every color in [q] is
n−q+2
X q!r!
2n
Pq,r
(2n − 2q − 2r + 1)!!
r−1
2q + 2r − 4
2
r=1
where Pq,r is the number of planar maps with q vertices and r faces.
That is, Pq,r is the coefficient of xq y r in the series P (x, y) defined by:
27P 4 − (36x + 36y − 1)P 3
+(24x2 y + 24xy 2 − 16x3 − 16y 3 + 8x2 + 8y 2 + 46xy − x − y)P 2
+xy(16x2 + 16y 2 − 64xy − 8x − 8y + 1)P
−x2 y 2 (16x2 + 16y 2 − 32xy − 8x − 8y + 1) = 0.
Results
Theorem [B.]: The number of ways of gluings a 2n-gon with vertices
colored using every color in [q] is
n−q+2
X q!r!
2n
Pq,r
(2n − 2q − 2r + 1)!!
r−1
2q + 2r − 4
2
r=1
where Pq,r is the number of planar maps with q vertices and r faces.
Theorem [Harer-Zagier 86]: The number of orientable ways of
gluings a 2n-gon with vertices colored using every color in [q] is
n
q−1
2
(2n − 1)!!
q−1
Results
Theorem [B.]: The number of ways of gluings a 2n-gon with vertices
colored using every color in [q] is
n−q+2
X q!r!
2n
Pq,r
(2n − 2q − 2r + 1)!!
r−1
2q + 2r − 4
2
r=1
where Pq,r is the number of planar maps with q vertices and r faces.
P
1
⇒ The 2nth moment of λ in the GOE(p) = p gluings of p#vertices
2n-gon
1
= #gluings of 2n-gon with vertices colored in [p],
p
p n−q+2
X
X
2n
1
p
q!r!
Pq,r
(2n − 2q − 2r + 1)!!
=
r−1
2q + 2r − 4
p q=1 q r=1 2
Results
Theorem [B.]: The number of ways of gluings a 2n-gon with vertices
colored using every color in [q] is
n−q+2
X q!r!
2n
Pq,r
(2n − 2q − 2r + 1)!!
r−1
2q + 2r − 4
2
r=1
where Pq,r is the number of planar maps with q vertices and r faces.
P
1
⇒ The 2nth moment of λ in the GOE(p) = p gluings of p#vertices
2n-gon
1
= #gluings of 2n-gon with vertices colored in [p],
p
p n−q+2
X
X
2n
1
p
q!r!
Pq,r
(2n − 2q − 2r + 1)!!
=
r−1
2q + 2r − 4
p q=1 q r=1 2
Theorem [Goulden, Jackson 97] This number is
p−1
n 1 p−1 p − 1 n X
X
n
−
k
+
r
−
1
2
2
.
22n−k
n!
+ (2n − 1)!!
2q−1
q−1
n−r
k
r
q
r=0
q=1
k=0
n
X
Results
Corollary [recovering Ledoux 09]: The number ηv (n) of ways of
gluings 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))
2
+(2n − 3) (10n − 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 of
gluings 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))
2
+(2n − 3) (10n − 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 v (n) of orientable ways of gluings
a 2n-gon into a surfaces of type n + 1 − v satisfies:
(n + 1)v (n) = (4n − 2)v−1 (n − 1) + (n − 1)(2n − 1)(2n − 3)v (n − 2).
Crash course on maps
Definition: A map is a graph embedded in a surface obtained by
gluing together polygons (i.e. the faces are simply connected).
Crash course on maps
Definition: A map is a graph embedded in a surface obtained by
gluing 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 by
gluing 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 maps
are the ”rooted plane trees”.
Crash course on maps
Definition: A map is a graph embedded in a surface obtained by
gluing together polygons (i.e. the faces are simply connected).
Maps are considered up to homeomorphism.
=
6=
Embedding Thorem: Rooted maps on orientable surfaces are in
bijection 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 marked
spanning tree.
A planar-rooted map is a map on an orientable surface with a
marked planar connected spanning submap.
Remark. Thenumber
with
q vertices
and n edges
of tree-rooted maps
2n
2q−1
n
is Cat(q − 1)
(2n − 2q + 1)!! =
(2n − 1)!!
2q
q! q − 1
The number of
planar-rooted
maps with q vertices, r faces, and n
2n
edges is Pq,r
(2n − 2q − 2r + 1)!!
2q + 2r − 4
Bijections for q-colored unicellular maps
Thm [B.] There is a bijection between rooted unicellular map colored
using every color in [q] with n edges on orientable surfaces and treerooted maps with q labelled vertices and n edges.
Moreover, number of edges between colors i and j → number of edges
between vertices i and j.
4
8
1
2
Corollary [Harer-Zagier] The number of [q]-colored
rooted
unicellular
n
q−1
maps with n edges on orientable surfaces is: 2
(2n − 1)!!
q−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 rooted
unicellular maps on general surfaces colored using every color in [q]
with n edges and planar-rooted maps with q vertices, r faces and n
edges.
Moreover, number of edges incident to color i → degree of vertex i.
Corollary:The number of [q]-colored rooted unicellular maps with n
edges on general surfaces is:
n−q+2
X q!r!
2n
Pq,r
(2n − 2q − 2r + 1)!!
r−1
2q
+
2r
−
4
2
r=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 starting
and ending at the same vertex and using every edge twice.
12
7
11
1
8
3
4
6
5
10
9
2
Idea 1: Colored unicellular map ↔ bi-Eulerian tour
Def. Let G = (V, E) be a graph. A bi-Eulerian tour is a walk starting
and ending at the same vertex and using every edge twice.
12
7
11
10
12 1
2
3
4
9
8
5
7
6
11
1
8
3
4
6
5
10
9
2
Lemma [adapting Lass 01]. [q]-colored unicellular maps are in bijection with pairs (G, E), where G is a graph with vertex set [q] and E
is a bi-Eulerian tour.
Moreover, the map is on an orientable surface if and only if no edge is
used twice in the same direction.
Idea 2: BEST Theorem
~ = (V, A) be a directed graph such that every
BEST Theorem. Let G
vertex has as many ingoing and outgoing arcs.
~ starting and ending at v are in bijection with
The Eulerian tours of G
pairs (T, O) made of a spanning tree T oriented toward v + ordering
O around each vertex of the outgoing arc not in T .
v
Idea 2: BEST Theorem
~ = (V, A) be a directed graph such that every
BEST Theorem. Let G
vertex has as many ingoing and outgoing arcs.
~ starting and ending at v are in bijection with
The Eulerian tours of G
pairs (T, O) made of a spanning tree T oriented toward v + ordering
O around each vertex of the outgoing arc not in T .
7
12
v
3
1
v3
11
1
6
4
2 4
8
5
10
2
2
2
9
1
3
1
Idea 2: BEST Theorem + embedding Theorem
~ = (V, A) be a directed graph such that every
BEST Theorem. Let G
vertex has as many ingoing and outgoing arcs.
~ starting and ending at v are in bijection with
The Eulerian tours of G
pairs (T, O) made of a spanning tree T oriented toward v + ordering
O around each vertex of the outgoing arc not in T .
Corollary 1. Let G be a graph. The bi-Eulerian tours of G which
never take an edge twice in the same direction are in bijection with
pairs (T, R) made of a spanning tree + rooted rotation system.
←→ tree-rooted maps having underlying graph G.
1
2 3
1 34 6 7 8
2
5
2
3
4
1
Idea 2: BEST Theorem + embedding Theorem
⇒ Bijection between rooted unicellular maps colored using every
color in [q] with n edges on orientable surfaces and tree-rooted maps
with q labelled vertices and n edges.
1
6
1 34 6 7 8
4
1
2
7
5
3
8
1
3
2
3
2
2
5
4
Idea 2: BEST Theorem + embedding 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.
Idea 2: BEST Theorem + embedding 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.
Corollary 2. Let G be a graph. There is a 1-to-2r correspondence
between the bi-Eulerian tours of G and the bi-oriented tree-rooted
map on G having r oriented edges outside of the spanning tree.
12
7
11
1
8
3
4
6
5
10
9
2
Idea 3: cutting and pasting
Let B be a bi-oriented tree-rooted map. The planar-rooted map
P = Ψ(B) is obtained by cutting the oriented external edges in their
middle and regluing them according to the parenthesis system they
form around the tree (and then forgetting the tree + orientation).
(r−1)!-to-1
r-to-1
Theorem: The mapping Ψ is r!-to-1 between bi-oriented tree-rooted
maps with r − 1 oriented edges outside of the spanning tree and planarrooted map with r sub-faces.
Idea 3: cutting and pasting
⇒ q!r!2r−1 -to-1 correspondence between rooted unicellular maps
colored using every color in [q] with n edges on general surfaces and
planar-rooted maps with q vertices, r faces, and n edges.
12
7
1-to-2r−1
q!-to-1
11
1
8
3
4
6
5
10
9
2
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))
2
+(2n − 3) (10n − 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))
2
+(2n − 3) (10n − 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)!
Pn+1
n
η
(n)x
.
v=1 v
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))
2
+(2n − 3) (10n − 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)!
Pn+1
n
η
(n)x
.
v=1 v
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.
1
(2n)
Pn+1
x
q
Vn (x) =
Un (q), where Un (q) is the number of unicelluq=1
lar maps with n edges colored using every color in [q].
⇒ my computer can translate (*) into a linear differential equation for
X Un (q)
the series U (t, w) =
tn wq .
(2n)!
n,q
Toward the recurrence relation
n−q+2
X
q!r!
2n
We know Un (q) =
Pq,r
(2n − 2q − 2r + 1)!!.
r−1
2q + 2r − 4
2
r=1
t
t
1
Q
, 2w ,
⇒ U (t, w) = exp
2
2
2
X
q! r! Pq,r
where Q(t, w) =
tq+r−2 wq .
(2q + 2r − 4)!
q,r>0
Toward the recurrence relation
n−q+2
X
q!r!
2n
We know Un (q) =
Pq,r
(2n − 2q − 2r + 1)!!.
r−1
2q + 2r − 4
2
r=1
t
t
1
Q
, 2w ,
⇒ U (t, w) = exp
2
2
2
X
q! r! Pq,r
where Q(t, w) =
tq+r−2 wq .
(2q + 2r − 4)!
q,r>0
Equation (*) is equivalent to:
∂
(6tw + 4w2 − 36t − 7w − 6)Q(t, w) − (12t2 + 8tw + 7w − 25) ∂t
Q(t, w)
+w(w + 2)(8w + 6t −
−w(w + 2)(8t +
+t(8t +
∂
7) ∂w
Q(t, w)
∂2
7) ∂w∂t Q(t, w)
∂3
11) ∂t3 Q(t, w)
+
− 4wt(w +
+
(2t2
− 4tw + 37t +
∂2
9) ∂t2 Q(t, w)
∂2
2
+ 2) ∂w2 Q(t, w)
∂3
∂4
2
2) ∂w∂t2 Q(t, w) + 2t ∂t4 Q(t, w)
2w2 (w
= 0.
Toward the recurrence relation
n−q+2
X
q!r!
2n
We know Un (q) =
Pq,r
(2n − 2q − 2r + 1)!!.
r−1
2q + 2r − 4
2
r=1
t
t
1
Q
, 2w ,
⇒ U (t, w) = exp
2
2
2
X
q! r! Pq,r
where Q(t, w) =
tq+r−2 wq .
(2q + 2r − 4)!
q,r>0
Equation (*) is equivalent
to:
∂
∂
0 = (4y − 2x − 1) 72P (x, y) − 72x ∂x
P (x, y) − (72y − 2) ∂y
P (x, y)
∂2
∂2
2
2
2
−72x ∂x2 P (x, y) + 8x − 6x − 24xy − 56y + 10y + 1 ∂y2 P (x, y)
(**)+2x (4x − 80y −1)
∂2
P (x, y)
∂y∂x
4x + 48y 2 −
8y − 1
∂3
P (x, y)
∂y 3
+ (4x − 8y − 1) y
∂3
2
+ (4x − 8y − 1) 144x y ∂y∂x2 P (x, y) + x 4x − 8y
X
where P (x, y) =
Pq,r xq y r .
q,r>0
∂3
3
+ 48x ∂x3 P (x, y)
3
− 1 + 144y 2 ∂y∂2 ∂x P (x, y)
Toward the recurrence relation
The differential equation (**) for the series P (x, y) of planar maps can
be checked using the algebraic equation:
27P 4 − (36x + 36y − 1)P 3
+(24x2 y + 24xy 2 − 16x3 − 16y 3 + 8x2 + 8y 2 + 46xy − x − y)P 2
+xy(16x2 + 16y 2 − 64xy − 8x − 8y + 1)P
−x2 y 2 (16x2 + 16y 2 − 32xy − 8x − 8y + 1) = 0.
(by expressing the partial derivatives of P as polynomials in P ).
Thanks.
λ1
λ2λ3 λ4 λ5
λ6λ7