Counting colored maps: algebraicity results ArXiv: 0909.1695 - - PowerPoint PPT Presentation

Counting colored maps: algebraicity results

ArXiv: 0909.1695 Olivier Bernardi, MIT Joint work with Mireille Bousquet-Mélou IHP 2009

IHP 2009 Olivier Bernardi – p.1/25

Outline

• 1. Potts polynomial.
• 2. Functional equation for Potts model (easy part).
• 3. Solving equations (hard part).
• 4. Results and open questions.

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Potts polynomial

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Potts model

A q-coloring of G = (V, E) is a function c : V → {1, 2, . . . , q}. m(c) = 2 An edge is monochromatic if its endpoints have the same color.

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Potts model

A q-coloring of G = (V, E) is a function c : V → {1, 2, . . . , q}. m(c) = 2 The Potts polynomial (partition function of the Potts model) is PG(q, u) =

• c:V →[q]

um(c), where m(c) is the number of monochromatic edges.

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Potts model

A q-coloring of G = (V, E) is a function c : V → {1, 2, . . . , q}. m(c) = 2 The Potts polynomial (partition function of the Potts model) is PG(q, u) =

• c:V →[q]

um(c), where m(c) is the number of monochromatic edges. Remark: The chromatic polynomial PG(q, 0) counts proper colorings.

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Potts polynomial

Fact: The Potts polynomial PG(q, u) =

c:V →[q] um(c), is a

polynomial in q, u satisfying : PG(q, u) = PG\e(q, u) + (u − 1) PG/e(q, u). G e

Contraction

G\e G/e

Deletion

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Potts polynomial

Fact: [Fortuin and Kastelein 72] The Potts polynomial and Tutte polynomial are equivalent.

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Potts polynomial

Fact: [Fortuin and Kastelein 72] The Potts polynomial and Tutte polynomial are equivalent.

• c:V →[q]

um(c) =

• c:V →[q]
• (i,j)∈E

(1 + δ(ci, cj)(u − 1)) =

• c:V →[q]
• S⊆E
• (i,j)∈S

δ(ci, cj)(u − 1) =

• S⊆E
• c:V →[q]
• (i,j)∈S

δ(ci, cj)(u − 1) =

• S⊆E

qk(S)(u − 1)|S| where k(S) is the number of connected components.

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Potts polynomial

Fact: [Fortuin and Kastelein 72] The Potts polynomial and Tutte polynomial are equivalent. Remarks:

• The Potts model of a planar graph G and of its dual graph

G∗ are related (by PG∗(q, u) = (u−1)e(G)

qv(G)−1 PG(q, 1 + q/(u − 1))).

• the Potts polynomial can be specialized to count various

structures: spanning trees, forests, connected subgraphs, acyclic orientations, score vectors, bipolar orientations, sandpile configurations...

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Maps

A planar map is an embedding of a connected planar graph in the sphere, considered up to continuous deformation.

=

=

(I indicate the rooting by pointing a corner)

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Potts model on Maps

The partition function of the (annealed) Potts model on maps is G(q, u, z) =

• M map

PM(q, u)z|M|. Phase transitions can be characterized by analyzing the singularities of G(q, u, z).

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Potts model on Maps

The partition function of the (annealed) Potts model on maps is G(q, u, z) =

• M map

PM(q, u)z|M|. Remark: The series G(q, u, z) contains (as specializations)

• the GF of maps G(1, 1, z),
• the GF of properly q-colored maps G(q, 0, z),
• the GF of tree-rooted maps (spanning trees),
• the GF of Baxter numbers (bipolar orientations),...

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Potts model on Maps

The partition function of the (annealed) Potts model on maps is G(q, u, z) =

• M map

PM(q, u)z|M|. Question: For which values of q, u is G(q, u, z) algebraic ? (meaning P(G(q, u, z), z) = 0 for a polynomial P = 0)

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Potts model on Maps

The partition function of the (annealed) Potts model on maps is G(q, u, z) =

• M map

PM(q, u)z|M|. Question: For which values of q, u is G(q, u, z) algebraic ? Known:

• GF of maps is algebraic [Tutte].
• GF of tree-rooted maps or Baxter numbers are not algebraic.
• GF of properly colored triangulation T(q, 0, z) is algebraic

for q = 2 + 2 cos(2π/m) [Tutte / Richmond, Odlyzko 83].

• Results in [Bonnet, Eynard 99] suggests that T(q, u, z) is

algebraic for q = 2 + 2 cos(kπ/m).

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Potts model on Maps

The partition function of the (annealed) Potts model on maps is G(q, u, z) =

• M map

PM(q, u)z|M|. Question: For which values of q, u is G(q, u, z) algebraic ? Thm [B., MBM]: The GF G(q, u, z) of the Potts model on planar maps is algebraic for q = 0, 4 of the form q = 2 + 2 cos(kπ/m). The same is true for the GF concerning triangulations. Examples: q = 1, 2, 3, 2 + √ 2, 2 + √ 3...

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Functional equations for colored maps

(a.k.a. loop equations)

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Generatingfunctionology

Class A (+size function) → Generating function A(z) =

• A∈A

z|A| =

• n≥0

anzn. Recursive description of A → Equation for A(z)

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Generatingfunctionology

Class A (+size function) → Generating function A(z) =

• A∈A

z|A| =

• n≥0

anzn. Recursive description of A → Equation for A(z) Combinatorial description → generating function Disjoint union C = A ⊎ B → C(z) = A(z) + B(z) Cartesian product C = A × B → C(z) = A(z) × B(z) Sequence C = Seq(A) → C(z) = 1 1 − A(z) . . . . . .

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Example: plane trees

Generating function of rooted plane trees: A(z) =

• n

an zn. =

⊎

→ A(z) = 1 + zA(z)2.

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Example: plane trees

Generating function of rooted plane trees: A(z) =

• n

an zn. =

⊎

→ A(z) = 1 + zA(z)2. The GF of plane trees is algebraic ! More generally, classes of trees defined by (finite) degree constraints are algebraic.

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Recursive description for maps [Tutte 63]

G(z) =

• M∈M

ze(M). + + =

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Recursive description for maps [Tutte 63]

G(z) =

• M∈M

ze(M). + + = G(z) = 1 +

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Recursive description for maps [Tutte 63]

G(z) =

• M∈M

ze(M). + + = G(z) = 1 + zG(z)2 +

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Recursive description for maps [Tutte 63]

G(z) =

• M∈M

ze(M). + + = G(z) = 1 + zG(z)2 + ? We are forced to take the degree of the root-face d f into account.

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Recursive description for maps [Tutte 63]

G(x, z) =

• M∈M

xd

f(M)ze(M).

+ + = G(y, z) = 1 + y2zG(y, z)2 + yz yG(y, z) − G(1, z) y − 1

• .
• A small map M corresponds to from d

f(M) + 1 big maps xkzn−1 ❀ xzn + . . . + xk+1zn = xznxk+1 − 1 x − 1

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Recursive description for maps [Tutte 63]

G(x, z) =

• M∈M

xd

f(M)ze(M).

+ + = Remarks:

• To describe maps by root-deletion we were forced to

record the root-face degree.

• To describe maps by root-contraction we would be forced

to record the root-vertex degree.

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Equation for Potts model on maps [Tutte 71]

G(x, y) ≡

• M∈M

xd

f(M)yd v(M)ze(M)PM(q, u)

q . + + = +

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Equation for Potts model on maps [Tutte 71]

G(x, y) ≡

• M∈M

xd

f(M)yd v(M)ze(M)PM(q, u)

q . + + = + G(x, y) = 1 + (q−1+u)x2yzG(x, y)G(x, 1) + uxy2zG(x, y)G(1, y) +

• xyz
• xG(x,y)−G(1,y)

x−1

• − xyzG(x, y)G(1, y)
• +(u−1)
• xyz
• xG(x,y)−G(x,1)

y−1

• − xyzG(x, y)G(x, 1)
• .

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Other equations

Properly colored triangulations [Tutte 73]:

T(x, y) = (q − 1)y + xyzT(x, y)T(x, 1) + yz T(x, y) − T(0, y) x − xy2z T(x, y) − T(x, 1) y − 1 .

Potts model on cubic maps [Eynard, Bonnet 99]:

T(x, y) − T0(y) x − T(x, y) − T0(x) y + (u − 1)z(xT0(x) − yT0(y))T(x, y) = (u − 1)z „T(x, y) − T0(x) − yT1(x) y2 − T(x, y) − T0(y) − xT1(xy) x2 « .

Alternatively, Potts model on triangulations [B., MBM]:

T(x, y) = 1 + x2z(q + u − 1)T(x, y)T(x, 0) + uxz (T2(y) + 2 yT1(y)) T(x, y) +yz (T(x, y) − 1 − xT1(y)T(x, y)) + z ` T(x, y) − 1 − xT1(y) − x2T(x, y)T2(y) ´ x + x2z2 (u − 1) yuT(x, y)T(x, 0) 1 − yuz + xz (u − 1) (T(x, y) − T(x, 0)) (1 − yuz) y .

There exists equations for Potts model on p-angulations for any p [B., MBM].

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Solving functional equations

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Functional vs algebraic equations

G(y, z) = 1 + y2zG(y, z)2 + yz yG(y, z) − G(1, z) y − 1

• .

The functional equation (with catalytic variable x)

• determines G(x, z) and G(1, z) uniquely,
• does not directly give access to asymptotic.

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Functional vs algebraic equations

G(y, z) = 1 + y2zG(y, z)2 + yz yG(y, z) − G(1, z) y − 1

• .

The functional equation (with catalytic variable x)

• determines G(x, z) and G(1, z) uniquely,
• does not directly give access to asymptotic.

By contrast, asymptotic informations can be deduced almost automatically from the algebraic equation 1 − 16z + (18z − 1)G − 27z2G2 = 0. satisfied by G ≡ G(1, z).

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Equations with 1 catalytic variable

G(x, z) = 1 + x2zG(x, z)2 + xz xG(x, z) − G(1, z) x − 1

• .

Linear case: Kernel method [Knuth 68,. . .] Quadratic case (1 unknown function): Quadratic method [Tutte,Brown 65] General case: P(F(x, z), F1(z), .., Fk(z), x, z) = 0 [MBM & Jehanne 06]

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Equations with 1 catalytic variable

Thm [MBM & Jehanne 06]: Suppose that the series F(x, z), F1(z), .., Fk(z) are related by

Pol(F(x, z), ∆1(F), . . . , ∆k(F), x, z) = 0,

where ∆j = (x − a)−j

• F(x, z) −

j−1

• i=0

(x − a)iF (i)(a) i!

• .

Then, the series F(x, z), F1(z), .., Fk(z) are algebraic. (+general strategy for obtaining the equation.)

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Equations with 1 catalytic variable

Thm [MBM & Jehanne 06]: Suppose that the series F(x, z), F1(z), .., Fk(z) are related by

Pol(F(x, z), ∆1(F), . . . , ∆k(F), x, z) = 0,

where ∆j = (x − a)−j

• F(x, z) −

j−1

• i=0

(x − a)iF (i)(a) i!

• .

Then, the series F(x, z), F1(z), .., Fk(z) are algebraic. (+general strategy for obtaining the equation.) ⇒ Any class of maps defined by degree constraints is algebraic.

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Equations with 1 catalytic variable

One starts with P(F(x, z), F1(z), .., Fk(z), x, z) = 0. Method [MBM-Jehanne 06]:

• 1. Search (find) k series X1(z), . . . , Xk(z) such that

P ′

F(F(Xi(z), z), F1(z), .., Fk(z), Xi(z), z) = 0.

• 2. These series then also satisfy:

P ′

x(F(Xi(z), z), F1(z), .., Fk(z), Xi(z), z) = 0.

• 3. This is a system of 3k polynomial equations for 3k

unknowns F(Xi(z), z), Xi(z), Fi(z), i = 1 . . . k. The system can be solved by resultants or Groebner basis techniques.

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Equations with 2 catalytic variables

T(x, y) = q(q−1)yz + xy q T(x, y)T(x, 1) + yz T(x, y) − T(0, y) x − xy2z T(x, y) − T(x, 1) y − 1 .

Linear case: Obstinate kernel methods [MBM & Petkovsek 03] Polynomial case: [Tutte] (unique example) The proof is long !

[Tutte 73] Chromatic sums for rooted planar triangulations : the cases λ = 1 and λ = 2. [Tutte 73] Chromatic sums for rooted planar triangulations, II : the case λ = τ + 1. [Tutte 73] Chromatic sums for rooted planar triangulations, III : the case λ = 3. [Tutte 73] Chromatic sums for rooted planar triangulations, IV : the case λ = ∞. [Tutte 74] Chromatic sums for rooted planar triangulations, V : special equations. [Tutte 78] On a pair of functional equations of combinatorial interest. [Tutte 82] Chromatic solutions. [Tutte 82] Chromatic solutions II. [Tutte 84] Map-colourings and differential equations.

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Equations with 2 catalytic variables

T(x, y) = q(q−1)yz + xy q T(x, y)T(x, 1) + yz T(x, y) − T(0, y) x − xy2z T(x, y) − T(x, 1) y − 1 .

Linear case: Obstinate kernel methods [MBM & Petkovsek 03] Polynomial case: [Tutte] (unique example)

Synthesis article : [Tutte: Chromatic sums revisited 95]

From Physics literature: Potts model and O(n) model on triangulations

[Eynard, Zinn-Justin 92, Eynard, Kristjansen 95, Bonnet, Eynard 99]

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Solving the Potts model on maps (sketch)

Equation for G(x, y) ≡ G(x, y; q, u, t) has the form K(x, y)G(x, y) = R(x, y), where K(x, y) and R(x, y) involve q, u, z, x, y, G(x, 1), G(1, y).

• 1. We find two series Y1, Y2 in q, u, x, z such that

K(x, Y1) = K(x, Y2) = 0.

• 2. We combine them with R(x, Y1) = R(x, Y2) = 0 to obtain

I(Y1) = I(Y2)

and J(Y1) = J(Y2)

where the invariants I(y), J(y) contain q, u, z, y, G(1, y). Works only for q = 2 + 2 cos(2kπ/m).

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Solving the Potts model on maps (sketch)

...where the invariants I(y), J(y) contain q, u, z, y, G(1, y).

• 3. A theorem shows that

J(y) =

m

• i=1

aiI(y)i where series ai’s depend on u, z (but not on y).

• 4. Asymptotic expansion at y = 1 gives ai in terms of ∂iG(1,y)

∂yi

. Moreover, conditions [MBM,Jehanne 06] are satisfied ⇒ Algebraicity.

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Explicit solutions q = 2

Thm: The GF of the 2-states Potts model on maps satisfies

G(1, 1; 2, u, z) = 1 + 3uS − 3uS2 − u2S3 (1 − 2S + 2u2S3 − u2S4)2 ×

• u3S6+2u2(1−u)S5+u(1−6u)S4−u(1−5u)S3+(1+2u)S2−(3+u)S+1
• .

where S = z + O(z2) is the series satisfying

S = z

• 1 + 3uS − 3uS2 − u2S32

1 − 2S + 2u2S3 − u2S4 .

Similar results for triangulations, recovering results from [Boulatov, Kazakov 87, MBM, Schaeffer 03]

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Explicit solutions q = 3

Maple is too weak to solve the system for general u. Thm: The GF of properly 3-colored maps is G(1, 1; 3, 0, z) = (1 + 2S) (1 − 2S2 − 4S3 − 4S4) (1 − 2S3)2 . where S = z + O(z2) is the series satisfying z = S(1 − 2S3) (1 + 2S)3 . Similar result for triangulations is not interesting (Eulerian triangulations) but...

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Conjecture for 3-colored cubic maps

+4294967296 ` 280335535308800 z2 − 25398219177984 z + 446991689475 ´ C9 −1024 ` 379991218559385600000 z4 − 188284129271105978368 z3 + 74426563120993402880 z2 −3460024309515976704 z + 60644726921050599) C8 −1024 ` 855256650185747464192 z5 + 198557240861845880832 z4 + 7030700057733103616 z3 −2005025500677518336 z2 + 65719379546147724 z − 1261082394855783 ´ C7 −64 ` 13794761675403801133056 z6 + 1749420037224685109248 z5 − 278771160986127695872 z4 +3443220359730862080 z3 + 294527021649617744 z2 − 12400864344288084 z + 58608117981429 −16 ` 32829338688610212249600 z7 − 541704013946292273152 z6 − 549137038895633924096 z5 +41876669882140680192 z4 − 936289577498747840 z3 +12987916499676352 z2 + 208517314053540 z − 54447680943015 ´ C5 −32 ` 124515522497539473408 z9 + 6242274275823592669184 z8 − 898808183791057633280 z7 −5275329284641325056 z6 + 6539785066149118976 z5 − 361493662811609868 z4 +9979948894517522 z3 − 432679480767965 z2 + 6248694091833 z + 378858660750 ´ C4 −8 ` 747093134985236840448 z10 + 5932367633073989222400 z9 − 1529736206124490686464 z8 +132585839072566050816 z7 − 3048630269218258944 z6 − 135087570198766176 z5 +5706147748413032 z4 − 229584590608200 z3 + 23755821897083 z2 − 152875558308 z − 277386 + ` −3361919107433565782016 z11 − 6012198464670331305984 z10 + 2332964327872863928320 z −341248528343609901056 z8 + 24933054438553903104 z7 − 994662704339242816 z6 +33270083406272816 z5 − 1608971168541300 z4 + 7467003627448 z3 +5037279798640 z2 − 194388001728 z + 808501760 ´ C2 +z ` −840479776858391445504 z11 − 157618519659107057664 z10 + 157170928122096254976 z9 −34691457904249143296 z8 + 3785139252232855552 z7 − 224694559056638912 z6 +6999136302319904 z5 − 197576502742812 z4 + 19551640345287 z3 −1347626230088 z2 + 40099744688 z − 404250880 ´ C −4 z4 ` 19698744770118549504 z9 − 8025289374453202944 z8 + 1366977099830657024 z7 −120213529404735488 z6 + 5234026490678784 z5 − 86995002866345 z4

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What’s next ?

Tutte went one step further. Thm[Tutte 84]: The GF of q-colored triangulations satisfies 2q2(1−q)z+(qz+10H−6zH′)H′′+q(4−q)(20H−18zH′+9z2H′′) = 0 where H ≡ z2T(q, 0, √z). ⇒ Are there bijections for colored maps ?

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Merci de votre attention.

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