Maps, Branched Covers and the KP Hierarchyipgoulde/cmsfrednbju10.pdf · Maps, Branched Covers and...

Maps, Branched Covers and the KP Hierarchy

Ian Goulden (with David Jackson, Sean Carrell)

Department of Combinatorics and OptimizationUniversity of Waterloo

June 5, 2010

Outline

I symmetric functions, Bernstein operators;

KP hierarchy,integrable systems; Plucker relations, algebraic geometry

I transitive factorizations in the symmetric group,combinatorics; branched covers, Hurwitz numbers, geometry

I rooted hypermaps, rooted maps, rooted triangulations;asymptotics of maps

Outline

I symmetric functions, Bernstein operators; KP hierarchy,integrable systems;

Plucker relations, algebraic geometry



Outline

I symmetric functions, Bernstein operators; KP hierarchy,integrable systems; Plucker relations, algebraic geometry



Outline


I transitive factorizations in the symmetric group,combinatorics;

branched covers, Hurwitz numbers, geometry


Outline




Outline



I rooted hypermaps, rooted maps, rooted triangulations;

asymptotics of maps

Outline




Consider the diagram of a partition, for example, for the partitionλ = (5, 4, 2) below:

EES

ES

EES

S

EES

The code of this partition is the two-way infinite sequence

. . .SSEESEESESEE . . . .

The code of the partition λ(i) is obtained by switching the ith Efrom the left in the code of λ to S , for i ≥ 1.The code of the partition λ(−i) is obtained by switching the ith Sfrom the right in the code of λ to E , for i ≥ 1.


EES

ES

EES

S

EES





EES

ES

EES

S

EES



The code of the partition λ(i) is obtained by switching the ith Efrom the left in the code of λ to S , for i ≥ 1.

The code of the partition λ(−i) is obtained by switching the ith Sfrom the right in the code of λ to E , for i ≥ 1.


EES

ES

EES

S

EES




Define the Bernstein Operators B(t), Bn on symmetric functions by

B(t) =∑n∈Z

Bntn = exp

∑k≥1

tk

kpk

exp

−∑i≥1

t−i ∂

∂pi

,

where pk is the kth power sum symmetric function in a countableset of indeterminates (regarded as algebraicly independentindeterminates).

NOTE: If we define weight(pm11 pm2

2 · · · ) = m1 + m2 + . . . = M,then B(t)pm1

1 pm22 · · · is a Laurent series in t, with minimum degree

−M.

Define the Bernstein Operators B(t), Bn on symmetric functions by

B(t) =∑n∈Z

Bntn = exp

∑k≥1

tk

kpk

exp

−∑i≥1

t−i ∂

∂pi

,

where pk is the kth power sum symmetric function in a countableset of indeterminates (regarded as algebraicly independentindeterminates).NOTE: If we define weight(pm1

1 pm22 · · · ) = m1 + m2 + . . . = M,

then B(t)pm11 pm2

2 · · · is a Laurent series in t, with minimum degree−M.

For symmetric functions f , g , h, with Hall inner product 〈·, ·〉,define f ⊥ by

〈f ⊥g , h〉 = 〈g , fh〉.

The we have

p⊥i = i∂

∂pi, i ≥ 1,

∑j≥0

hj tj = exp

∑k≥1

tk

kpk

,∑m≥0

emtm = exp

−∑i≥1

(−t)i

ipi

.

Together, these give

B(t) =∑

j ,m≥0

(−1)mt j−mhje⊥m .


〈f ⊥g , h〉 = 〈g , fh〉.

The we have

p⊥i = i∂

∂pi, i ≥ 1,

∑j≥0

hj tj = exp

∑k≥1

tk

kpk

,∑m≥0

emtm = exp

−∑i≥1

(−t)i

ipi

.


B(t) =∑

j ,m≥0



〈f ⊥g , h〉 = 〈g , fh〉.

The we have

p⊥i = i∂

∂pi, i ≥ 1,

∑j≥0

hj tj = exp

∑k≥1

tk

kpk

,∑m≥0

emtm = exp

−∑i≥1

(−t)i

ipi

.


B(t) =∑

j ,m≥0


Let sλ be the Schur symmetric function indexed by partition λ.The Pieri Rules are

hjsλ =∑

µ

sµ, emsλ =∑

ν

sν ,

summed over µ such that µ− λ is a horizontal j-strip, and over νsuch that ν − λ is a vertical m-strip.

For example, the boxes with ‘x’ form a horizontal 3-strip; the boxeswith ‘y’ form a vertical 3-strip:

y

y

y

x

xx

Let sλ be the Schur symmetric function indexed by partition λ.The Pieri Rules are

hjsλ =∑

µ

sµ, emsλ =∑

ν

sν ,

summed over µ such that µ− λ is a horizontal j-strip, and over νsuch that ν − λ is a vertical m-strip.For example, the boxes with ‘x’ form a horizontal 3-strip; the boxeswith ‘y’ form a vertical 3-strip:

y

y

y

x

xx

Theorem (G., Carrell)

B(t)sλ =∑i≥1

(−1)|λ|−|λ(i)|+i−1t |λ

(i)|−|λ|sλ(i) .

PROOF: From the Pieri Rules, B(t) =∑

j ,m≥0(−1)mt j−mhje⊥m

acts on sλ by first removing a vertical m-strip, and then adding ahorizontal j-strip. Proceeds with a sign-reversing involution tocancel the contributions of all shapes that are not of the form λ(i)

for some i ≥ 1.


B(t)sλ =∑i≥1

(−1)|λ|−|λ(i)|+i−1t |λ

(i)|−|λ|sλ(i) .

PROOF: From the Pieri Rules, B(t) =∑

j ,m≥0(−1)mt j−mhje⊥m

acts on sλ by first removing a vertical m-strip, and then adding ahorizontal j-strip. Proceeds with a sign-reversing involution tocancel the contributions of all shapes that are not of the form λ(i)

for some i ≥ 1.


B(t)sλ =∑i≥1

(−1)|λ|−|λ(i)|+i−1t |λ

(i)|−|λ|sλ(i) .

Corollary

For scalars aλ, λ ∈ P (the set of all partitions),

B(t)∑λ∈P

aλsλ =∑β∈P

sβ∑j≥1

(−1)j−1t |β|−|β(−j)|aβ(−j) .


B(t)sλ =∑i≥1

(−1)|λ|−|λ(i)|+i−1t |λ

(i)|−|λ|sλ(i) .

Corollary

For scalars aλ, λ ∈ P (the set of all partitions),

B(t)∑λ∈P

aλsλ =∑β∈P

sβ∑j≥1

(−1)j−1t |β|−|β(−j)|aβ(−j) .

KP Hierarchy

Consider two independent sets of indeterminates p = (p1, p2, . . .)and p = (p1, p2, . . .). The a formal power series τ is a τ -functionfor the KP hierarchy is and only if it satisfies

[t−1] exp

∑k≥1

tk

k(pk − pk)

exp

−∑i≥1

t−i

(∂

∂pi− ∂

∂pi

) τ(p)τ(p).

= 0.

Here, [t−1] is the coefficient of t−1.

KP Hierarchy


[t−1] exp

∑k≥1

tk

k(pk − pk)

exp

−∑i≥1

t−i

(∂

∂pi− ∂

∂pi

) τ(p)τ(p).

= 0.

Here, [t−1] is the coefficient of t−1.

KP Hierarchy


[t−1] exp

∑k≥1

tk

k(pk − pk)

exp

−∑i≥1

t−i

(∂

∂pi− ∂

∂pi

) τ(p)τ(p)

= 0.

In terms of the Bernstein Operators, this equation is

[t−1] (B(t)τ(p)) ·(B⊥(t−1)τ(p)

)= 0.

KP Hierarchy


[t−1] exp

∑k≥1

tk

k(pk − pk)

exp

−∑i≥1

t−i

(∂

∂pi− ∂

∂pi

) τ(p)τ(p)

= 0.

In terms of the Bernstein Operators, this equation is

[t−1] (B(t)τ(p)) ·(B⊥(t−1)τ(p)

)= 0.

Now, for τ(p) =∑

λ∈P aλsλ, so τ(p) =∑

λ∈P aλsλ, we obtain∑α,β∈P

sαsβ∑m,k

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

where the inner sum is over m, k ≥ 1 such that|α(m)|+ |β(−k)| = |α|+ |β|+ 1. Thus this inner sum must equal 0for each pair α, β ∈ P, and this is called the Plucker Relation.

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.

For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2),

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2),

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2),

and if β = (1), then we have β(−1) = (),β(−2) = (2), β(−3) = (2, 1),

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2), and if β = (1), then we have β(−1) = (),β(−2) = (2), β(−3) = (2, 1),

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2), and if β = (1), then we have β(−1) = (),β(−2) = (2), β(−3) = (2, 1),

so α = (2), β = (1) give the Pluckerrelation

−a(1)a(2,1) + a(1,1)a(2) + a(2,2)a() = 0.

Plucker Relations

∑m,k≥1

(−1)|α|−|α(m)|+m+kaα(m)aβ(−k) = 0,

such that |α(m)|+ |β(−k)| = |α|+ |β|+ 1.For example, if α = (2), then we have α(1) = (1), α(2) = (1, 1),α(3) = (2, 2), and if β = (1), then we have β(−1) = (),β(−2) = (2), β(−3) = (2, 1), so α = (2), β = (1) give the Pluckerrelation

−a(1)a(2,1) + a(1,1)a(2) + a(2,2)a() = 0.

This gives an algebraic combinatorics proof of the classical resultfor the KP hierarchy:

∑λ∈P aλsλ is a τ -function for the KP

hierarchy if and only if {aλ}λ∈P satisfies the Plucker relations,

ifand only if log

(∑λ∈P aλsλ

)satisfies the (quadratic) pde’s in the

KP hierarchy:

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0,

F3,2 − F4,1 + 16F2,1,1,1 + F1,1F2,1 = 0,

F4,2 − F5,1 + 14F3,1,1,1 − 1

120F1,1,1,1,1,1 + F1,1F3,1 + 12F 2

2,1

−18F 2

1,1,1 − 112F1,1F1,1,1,1 = 0.

Here F2,1 denotes ∂2

∂p1∂p2F , and the usual physics variables are

xi = pi/i .



hierarchy if and only if {aλ}λ∈P satisfies the Plucker relations, ifand only if log

(∑λ∈P aλsλ


KP hierarchy:

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0,

F3,2 − F4,1 + 16F2,1,1,1 + F1,1F2,1 = 0,

F4,2 − F5,1 + 14F3,1,1,1 − 1

120F1,1,1,1,1,1 + F1,1F3,1 + 12F 2

2,1

−18F 2

1,1,1 − 112F1,1F1,1,1,1 = 0.



xi = pi/i .




(∑λ∈P aλsλ


KP hierarchy:

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0,

F3,2 − F4,1 + 16F2,1,1,1 + F1,1F2,1 = 0,

F4,2 − F5,1 + 14F3,1,1,1 − 1

120F1,1,1,1,1,1 + F1,1F3,1 + 12F 2

2,1

−18F 2

1,1,1 − 112F1,1F1,1,1,1 = 0.



xi = pi/i .




(∑λ∈P aλsλ


KP hierarchy:

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0,

F3,2 − F4,1 + 16F2,1,1,1 + F1,1F2,1 = 0,

F4,2 − F5,1 + 14F3,1,1,1 − 1

120F1,1,1,1,1,1 + F1,1F3,1 + 12F 2

2,1

−18F 2

1,1,1 − 112F1,1F1,1,1,1 = 0.



xi = pi/i .

Content

In the diagram of a partition, the content of box w is c(w) = j − i ,where w is in row i and column j .

For example, in the diagram ofλ = (4, 3, 1) below, each box contains its content:

1

1 2 3

-2

-1 0

0

We shall consider content products of the form∏

w∈λ yc(w). Forexample, when λ = (4, 3, 1), this product is equal to

y−2y−1y20 y2

1 y2y3.

Content

In the diagram of a partition, the content of box w is c(w) = j − i ,where w is in row i and column j . For example, in the diagram ofλ = (4, 3, 1) below, each box contains its content:

1

1 2 3

-2

-1 0

0



y−2y−1y20 y2

1 y2y3.

Content


1

1 2 3

-2

-1 0

0


w∈λ yc(w).

Forexample, when λ = (4, 3, 1), this product is equal to

y−2y−1y20 y2

1 y2y3.

Content


1

1 2 3

-2

-1 0

0



y−2y−1y20 y2

1 y2y3.

TheoremFor indeterminates yi , i ∈ Z, and qi , i ≥ 1, algebraicly independentof each other and of pi , i ≥ 1,

{sλ(q1, q2, . . .)∏w∈λ

yc(w)}λ∈P

satisfies the Plucker relations,

so

log

(∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

yc(w)

)

satisfies the KP hierarchy (with differentiation in p1, p2, . . .).

TheoremFor indeterminates yi , i ∈ Z, and qi , i ≥ 1, algebraicly independentof each other and of pi , i ≥ 1,

{sλ(q1, q2, . . .)∏w∈λ

yc(w)}λ∈P

satisfies the Plucker relations, so

log

(∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

yc(w)

)

satisfies the KP hierarchy (with differentiation in p1, p2, . . .).

A Combinatorial ProblemFor partitions α, β of n ≥ 1, and nonnegative integers a1, a2, . . .

(where∑

i≥1 ai is finite), define N(a1,a2,...)α,β to be the number of

tuples (σ, γ, π1, π2, . . .) of permutations on {1, . . . , n} such that

I σ is in the conjugacy class Cα, γ ∈ Cβ, n −#cycles(πi ) = ai ,i ≥ 1,

I σγπ1π2 · · · = identity,

I 〈σ, γ, π1, π2, . . .〉 acts transitively on {1, . . . , n}.

Theorem

∑|α|=|β|=n≥1

∑a1,a2,...≥0

N(a1,a2,...)α,β

n!pαqβua1

1 ua22 · · ·

= log

∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

∏i≥1

(1 + uic(w))

,

which satisfies the KP hierarchy.


(where∑






Theorem

∑|α|=|β|=n≥1

∑a1,a2,...≥0

N(a1,a2,...)α,β

n!pαqβua1

1 ua22 · · ·

= log

∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

∏i≥1

(1 + uic(w))

,



(where∑






Theorem

∑|α|=|β|=n≥1

∑a1,a2,...≥0

N(a1,a2,...)α,β

n!pαqβua1

1 ua22 · · ·

= log

∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

∏i≥1

(1 + uic(w))

,



(where∑






Theorem

∑|α|=|β|=n≥1

∑a1,a2,...≥0

N(a1,a2,...)α,β

n!pαqβua1

1 ua22 · · ·

= log

∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

∏i≥1

(1 + uic(w))

,



(where∑






Theorem

∑|α|=|β|=n≥1

∑a1,a2,...≥0

N(a1,a2,...)α,β

n!pαqβua1

1 ua22 · · ·

= log

∑λ∈P

sλ(p1, p2, . . .)sλ(q1, q2, . . .)∏w∈λ

∏i≥1

(1 + uic(w))

,


Special cases of the combinatorial problem

I Branched covers of the sphere with branch points0,∞,X1,X2, . . ., at which we have branching σ, γ, π1, π2, . . .,respectively.

(The product equal to the identity permutationis a monodromy condition, and the transitivity conditionmeans that the covers are connected.) The genus g of thecover is given by a1 + a2 + . . . = l(α) + l(β) + 2g − 2, fromthe Riemann-Hurwitz formula.

I Double Hurwitz numbers. Here ai = 1 fori = 1, . . . ,= l(α) + l(β) + 2g − 2, and ai = 0 otherwise (Notethat ai = 1 means that πi is a transposition, so there is simplebranching at Xi .)

I Single Hurwitz numbers. Here, in addition, l(β) = n, so γ isthe identity permutation, so branching is trivial at ∞.

I Bousquet-Melou/Schaeffer numbers. Here ai = 0 for i > m(where m is a fixed parameter), and l(β) = n, g = 0.

I Rooted hypermaps (2-face coloured) with n edges.


I Branched covers of the sphere with branch points0,∞,X1,X2, . . ., at which we have branching σ, γ, π1, π2, . . .,respectively. (The product equal to the identity permutationis a monodromy condition, and the transitivity conditionmeans that the covers are connected.)

The genus g of thecover is given by a1 + a2 + . . . = l(α) + l(β) + 2g − 2, fromthe Riemann-Hurwitz formula.






I Branched covers of the sphere with branch points0,∞,X1,X2, . . ., at which we have branching σ, γ, π1, π2, . . .,respectively. (The product equal to the identity permutationis a monodromy condition, and the transitivity conditionmeans that the covers are connected.) The genus g of thecover is given by a1 + a2 + . . . = l(α) + l(β) + 2g − 2, fromthe Riemann-Hurwitz formula.







I Double Hurwitz numbers. Here ai = 1 fori = 1, . . . ,= l(α) + l(β) + 2g − 2, and ai = 0 otherwise

(Notethat ai = 1 means that πi is a transposition, so there is simplebranching at Xi .)




























The green faces are hyperedges, the white faces are hyperfaces.

The green faces are hyperedges, the white faces are hyperfaces.

36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9), G = (1 8 6)(2 9 4)(3 5 7),W = (1 5 8)(2 6 3 9)(4 7), VGW = identity

〈V ,G ,W 〉 acts transitively on {1, . . . , 9} (the map is connected).

36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9),

G = (1 8 6)(2 9 4)(3 5 7),W = (1 5 8)(2 6 3 9)(4 7), VGW = identity


36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9), G = (1 8 6)(2 9 4)(3 5 7),

W = (1 5 8)(2 6 3 9)(4 7), VGW = identity


36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9), G = (1 8 6)(2 9 4)(3 5 7),W = (1 5 8)(2 6 3 9)(4 7),

VGW = identity


36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9), G = (1 8 6)(2 9 4)(3 5 7),W = (1 5 8)(2 6 3 9)(4 7), VGW = identity


36

2

9

4

7

8

5

1

V = (1 2 7)(3 4)(5 6)(8)(9), G = (1 8 6)(2 9 4)(3 5 7),W = (1 5 8)(2 6 3 9)(4 7), VGW = identity


Corollary

Let Hgα,β be the number of rooted hypermaps in a surface of genus

g, with vertex degrees specified by the parts of α, hyperedgedegrees specified by the parts of β, and

H =∑ Hg

α,β

dpαqβz l(α)+l(β)+2g−2.

Then H is a solution to the KP hierarchy.

(We determine thenumber of hyperfaces by Euler’s formula.)

Proof.We have u1 = z , u2 = u3 = · · · = 0, and (d−1)!

d! = 1d .

Corollary



H =∑ Hg

α,β


Then H is a solution to the KP hierarchy. (We determine thenumber of hyperfaces by Euler’s formula.)


d! = 1d .

Corollary



H =∑ Hg

α,β


Then H is a solution to the KP hierarchy. (We determine thenumber of hyperfaces by Euler’s formula.)


d! = 1d .

For example, H satisfies the pde

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.

Now, in H, substitute

I p4 = p5 = · · · = 0 – this removes vertices of degrees 4 ormore,

I q1 = q3 = q4 = · · · = 0 – this removes hyperedges of degreesother than 2.

The result is a generating series for rooted maps with vertices ofdegree at most 3. Note that loops and multiple edges are allowed.


F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.






F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.






F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.




The result is a generating series for rooted maps with vertices ofdegree at most 3.

Note that loops and multiple edges are allowed.


F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.





But, rooted maps with vertices of degree at most 3 can beuniquely constructed from rooted maps with all vertices of degree3 (cubic maps).

A rooted cubic map in the plane with 8 faces.

But, rooted maps with vertices of degree at most 3 can beuniquely constructed from rooted maps with all vertices of degree3 (cubic maps).

A rooted cubic map in the plane with 8 faces.

Rooted cubic trees, in which all vertices have degrees 1 or 3.

The generating series T (x), with respect to non-root vertices,satisfies quadratic equation:

T = x + xT 2.

Rooted cubic trees, in which all vertices have degrees 1 or 3.

The generating series T (x), with respect to non-root vertices,satisfies quadratic equation:

T = x + xT 2.

, , , , .. .

...,,,

, , , , .. .

...,,,

Let S = {(n, g) : n ≥ −1, 0 ≤ g ≤ 12(n + 1)} = {(−1, 0), (0, 0),

(1, 0), (1, 1), . . .},

and T (n, g) be the number of rootedtriangulations in a surface of genus g , with 2n faces, andc(n, g) = (3n + 2)T (n, g). Then

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),

for (n, g) ∈ S \ {(−1, 0)}, where the sum is over S × S, withi + j = n − 2 and h + k = g . The initial conditions arec(−1, 0) = 1

2 and c(n, g) = 0 for (n, g) /∈ S.

Let S = {(n, g) : n ≥ −1, 0 ≤ g ≤ 12(n + 1)} = {(−1, 0), (0, 0),

(1, 0), (1, 1), . . .}, and T (n, g) be the number of rootedtriangulations in a surface of genus g , with 2n faces,

andc(n, g) = (3n + 2)T (n, g). Then

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),


2 and c(n, g) = 0 for (n, g) /∈ S.

Let S = {(n, g) : n ≥ −1, 0 ≤ g ≤ 12(n + 1)} = {(−1, 0), (0, 0),

(1, 0), (1, 1), . . .}, and T (n, g) be the number of rootedtriangulations in a surface of genus g , with 2n faces, andc(n, g) = (3n + 2)T (n, g).

Then

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),


2 and c(n, g) = 0 for (n, g) /∈ S.

Let S = {(n, g) : n ≥ −1, 0 ≤ g ≤ 12(n + 1)} = {(−1, 0), (0, 0),

(1, 0), (1, 1), . . .}, and T (n, g) be the number of rootedtriangulations in a surface of genus g , with 2n faces, andc(n, g) = (3n + 2)T (n, g). Then

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),

for (n, g) ∈ S \ {(−1, 0)}, where the sum is over S × S, withi + j = n − 2 and h + k = g .

The initial conditions arec(−1, 0) = 1

2 and c(n, g) = 0 for (n, g) /∈ S.

Let S = {(n, g) : n ≥ −1, 0 ≤ g ≤ 12(n + 1)} = {(−1, 0), (0, 0),

(1, 0), (1, 1), . . .}, and T (n, g) be the number of rootedtriangulations in a surface of genus g , with 2n faces, andc(n, g) = (3n + 2)T (n, g). Then

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),


2 and c(n, g) = 0 for (n, g) /∈ S.

For example, T (1, 0) = 4,

obtained by the 1 + 3 rootings of themaps above.

For example, T (1, 0) = 4, obtained by the 1 + 3 rootings of themaps above.

For example, T (1, 1) = 1,

since there is only 1 rooting of the mapabove.

For example, T (1, 1) = 1, since there is only 1 rooting of the mapabove.

It is known that the asymptotic number of maps in many classes,rooted and unrooted (including rooted triangulations), is given by

α tg (βN)52 (g−1)γN ,

where N is the number of edges, α, β, γ are constants, and tg isdetermined implicitly.

Recently, our quadratic recurrence for triangulations has beenanalyzed by Bender, Gao, Richmond, and they have been able togive explicit asymptotics, and hence an explicit form for tg .

It is known that the asymptotic number of maps in many classes,rooted and unrooted (including rooted triangulations), is given by

α tg (βN)52 (g−1)γN ,

where N is the number of edges, α, β, γ are constants, and tg isdetermined implicitly.

Recently, our quadratic recurrence for triangulations has beenanalyzed by Bender, Gao, Richmond, and they have been able togive explicit asymptotics, and hence an explicit form for tg .

Problems

I Give a combinatorial proof that the generating series forrooted maps (with vertices of degrees at most 3 or not)satisfies the pde

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.

I Give a combinatorial proof that the number T (n, g) of rootedtriangulations in a surface of genus g with 2n faces satisfiesthe recurrence

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),

where c(n, g) = (3n + 2)T (n, g).

Problems

I Give a combinatorial proof that the generating series forrooted maps (with vertices of degrees at most 3 or not)satisfies the pde

F2,2 − F3,1 + 112F1,1,1,1 + 1

2F 21,1 = 0.

I Give a combinatorial proof that the number T (n, g) of rootedtriangulations in a surface of genus g with 2n faces satisfiesthe recurrence

c(n, g) =4(3n + 2)

n + 1

(n(3n − 2)c(n − 2, g − 1) +

∑c(i , h)c(j , k)

),

where c(n, g) = (3n + 2)T (n, g).

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