Chapter III Solution of loop equations In this chapter, we solve the loop equations (Tutte’s equations), we compute rather explicitly the generating functions counting maps of given genus and boundaries. We are first going to solve them for planar maps with 1 boundary (the disc, i.e. planar rooted maps), then two boundaries (the cylinder), and then arbitrary genus and arbitrary number of boundaries. The disc case (planar rooted maps) was already done by Tutte []. Generating functions for higher topologies have been computed more recently []. In this chapter, we shall show that, amazingly, the first two cases (disc and cylinder) are in fact more irregular than the general case. This is a general feature in geometry of surfaces: unstable surfaces with Euler characteristics χ≥0 are more irregular than stable surfaces (χ < 0). There is a deep algebraic geometry reason to that, because stable Riemann surfaces have a finite group of automorphims, and the volume of their moduli space is welldefined, whereas unstable ones need to be ”renormalized”, see chapter V. Physicists would say that sums over unstable surfaces χ ≥ 0 involve zero modes, whereas stable surfaces χ < 0 have no zero modes.

1

Disc amplitude

Discs are planar (g = 0) maps with one boundary (k = 1), or also ”rooted maps” (see section I.2.6). In that case, the loop equation eq.(II-5-3) reduces to Tutte’s equation eq.(I-3-1) of chapter I. In this section, we solve Tutte’s equation for discs: l−1 ! j=0

(0) (0) Tj Tl−1−j

=

(0) Tl+1

−

d ! j=3

(0)

tj Tl+j−1 41

,

(0)

T0

=t

(III-1-1)

(0)

where Tl is the generating function of planar maps (g = 0) with 1 boundary (k = 1) of perimeter l: Tl

1.1

(0)

= t δl,0 +

∞ !

!

tv

v=0

n (Σ)

t3 3

n (Σ)

. . . td d

.

(0)

Σ∈M1 (v), l1 (Σ)=l

Solving Tutte’s equation

It is more convenient to rewrite Tutte’s equation eq.(III-1-1) in terms of the resolvent (0) W1 (see def.2.2): ∞ t ! 1 (0) (0) W1 (x) = + Tl l+1 x l=1 x (0)

and Tutte’s equation can be written as eq.(II-5-4) for W1 : " #2 (0) (0) (0) W1 (x) = V $ (x)W1 (x) − P1 (x)

(III-1-2)

where we recall that: $

V (x) = x −

d !

j−1

tj x

,

(0) P1 (x)

j=3

=t−

j−2 d ! ! j=3 l=0

(0)

tj Tj−l−2 xl

(0)

P1 (x) is a polynomial in x of degree d − 2, we have: (0) P1 (x)

" # (0) $ = Pol V (x) W1 (x)

where Pol means the polynomial part of the Laurent series at ∞, indeed, the left hand side of eq.III-1-2 goes to zero at ∞. Equation eq.III-1-2 is called the spectral curve, we will develop the notion of spectral curves in chapter VII. Solving the second degree equation eq.III-1-2, yields: (0) W1 (x)

1 = 2

$

& % (0) $ 2 V (x) − V (x) − 4P1 (x) . $

(0)

In other words, if we knew how to determine the polynomial P1 , i.e. the coefficients (0) (0) (0) (0) T1 , . . . , Td−2 , then we would have determined W1 , i.e. Tl for every l. We do it below. 42

1.2

A useful lemma (0)

The following lemma allows to determine the polynomial P1 . It is very useful, and it is known under various names in the combinatorics or physics litterature. In combinatorics it is more or less equivalent to Brown’s lemma [], and in physics it is called the 1-cut assumption (although it is not an assumption, it is derived here). (0)

Lemma 1.1 The polynomial V $ (x)2 − 4P1 (x) has only one pair of simple zeroes, all the other zeroes are even. More precisely, there exist α, γ 2 and M(x) which are formal power series in t, and M(x) is a polynomial of x such that: (0)

V $ (x)2 − 4P1 (x) = (M(x))2 (x − a)(x − b) with a = α + 2γ, b = α − 2γ, and such that α = O(t)

,

γ 2 = t + O(t2)

,

M(x) =

V $ (x) + O(t) x

The meaning of this lemma will become clearer in chapter IV, where we discuss cases which do not satisfy this Lemma. As we shall see below, this lemma determines (0) the polynomial P1 uniquely. proof: (0) (0) We have T0 = t, and if l ≥ 1, recall that Tl counts maps Tl

(0)

= t δl,0 +

∞ ! v=1

tv

!

n (Σ)

t3 3

n (Σ)

. . . td d

(0)

Σ∈M1 (v), l1 (Σ)=l

where v is the number of vertices of the maps. Since our maps are discs, i.e. g = 0 and k = 1 boundary, the Euler characteristics constraint eq.(I-2-1) implies (remember that l ≥ 1): d l 1! v =1+ + (j − 2) nj ≥ 2 (III-1-3) 2 2 j=3 (0)

(0)

i.e. Tl is a power series which starts as O(t2 ) for l ≥ 1. Therefore P1 (x) vanishes at t = 0, it is a power series in t which starts at order 1 in t: (0) P1 (x)

= t(1 −

d !

tj xj−2 ) + O(t2 , xd−3 ) = t

j=3

V $ (x) + O(t2 , xd−3 ) x (0)

This implies that, to leading order in t, the zeroes of V $ (x)2 − 4P1 (x) are of the form: % (0) √ P1 (Xi ) + o( t) , V $ (Xi ) = 0 , i = 1, . . . , deg V $ ∼ Xi ± 2 V $$ (Xi ) 43

√ (0) and they are formal power series in t. In other words, the zeroes of V $ (x)2 − 4P1 (x) come by pairs [ai , bi ] centered around the zeroes Xi of V $ (x), and their distance to Xi √ 2 is of order at most O( t). Notice that a√ i + bi and (ai − bi ) are formal power series of t (whereas ai and bi are power series of t). In particular, notice that one of the zeroes of V $ (x), is X1 = 0, and we have (0) V $$ (0) = 1, and P1 (0) = tV $$ (0) + O(t2 ), thus: √ √ √ √ a1 ∼ 2 t + o( t) , b1 ∼ −2 t + o( t) (0)

And for the other zeroes of V $ , we have ∀ i = 2, . . . , deg V $ , P1 (Xi ) = 0 + O(t2 ) thus: ai ∼ Xi + O(t)

,

bi ∼ Xi + O(t)

(0)

Then, notice that for given v, M1 (v) is a finite set (see theorem 2.1 in chapter I), and thus, there is a maximum perimeter l ≤ 2v − 1 (see eq.(III-1-3)). In other words, (0) W1 (x) is, order by order in t, a polynomial in 1/x (of degree at most 2v), and we have: ( ' 1 1 if C encircles 0 (0) W1 (x)dx = . (III-1-4) 0 otherwise 2iπ C

This equality holds for any contour C in the complex plane, order by order in t. Assume that ai (= bi , and thus there exists mi ≥ 0 and Ci (= 0 such that ai − bi = Ci tmi +1/2 (1 + O(t)).

Choose a contour C (independent of t), which surrounds the zero Xi of V $ (x), then, (0) order by order in t, the contour C surrounds the pair% [ai , bi ] of zeroes of V $ (x)2 −4P1 (x). ) (0) One easily computes that the contour integral C V $ (x)2 − 4P1 (x)dx behaves like !!

i) Ci2 t2mi +1 (1 + O(t)) at small t, thus it does not vanish1 . iπ V (X 4 This shows that the assumption ai (= bi was wrong and therefore this proves that if (0) i (= 1 we must have ai = bi to all orders in t. Therefore V $ (x)2 − 4P1 (x) has only one pair [a1 , b1 ] of simple zeroes, all the others come by pairs:

(0)

V $ (x)2 − 4P1 (x) = M(x)2 (x − a1 )(x − b1 ) We have proved the Lemma. ! Remark 1.1 In chapter IV, we are going to consider a situation where this lemma does not hold, i.e. we will have more odd zeroes centered around the other zeroes of V $ . That situation is called multi-cut solution of loop equations. In chapter IV, we are going to see what is the combinatorics meaning of solutions of loop equations for which this lemma is not valid. 1

For simplicity, we assume that V !! (Xi ) (= 0. The lemma remains true when V !! (Xi ) = 0 but for the proof, one needs to go further in the Taylor expansion...

44

1.3

1-cut solution, Zhukovsky’s variable

Therefore, the solution of loop equation III-1-2, is: # * 1" $ (0) W1 (x) = V (x) − M(x) (x − a)(x − b) 2

with

$

V (x) = x −

d !

j−1

tj x

,

j=3

√ √ a = 2 t + o( t)

,

(III-1-5)

V $ (x) + O(t) M(x) = x √ √ b = −2 t + o( t)

Instead of x, it is more convenient to use another more appropriate variable of expansion z, with the help of Zhukovsky’s transformation: x(z) = which has the property that *

(0)

a+b a−b 1 + (z + ) 2 4 z

* (x − a)(x − b) is a rational function of z:

(x(z) − a)(x(z) − b) =

1 a−b (z − ) 4 z

and thus W1 (x) is a rational function of z. Zhukovsky’s transformation maps the x–plane cut along the segment [b, a] to the exterior of the unit disc in the z–plane, and the points a, b to 1, −1. It maps ∞ to ∞, and the other sheet of the x–plane is mapped to the interior of the unit disc:

x b

z

a

−1

1

Zhukovsky was a discoverer of the aerodynamics of wings, and he invented that transformation in order to transform conformally an infinitely thin wing profile, into a circular wing profile, for which equations of aerodynamics are much easier to solve. (0)

In Zhukovsky’s variable z, W1 polynomial in 1/z of the form:

is a rational function of z, and in fact it is a

(0)

W1 (x(z)) =

d−1 !

αk z −k

k=1

Indeed eq.(III-1-5) gives clearly a Laurent polynomial in z and 1/z, and there can be (0) no positive power of z because by definition W1 (x) contains no positive power of x at 45

large x, and we take the convention that large x corresponds to large z. The coefficients αk ’s, as well as a and b can be determined as follows, expand V $ (x) into powers of z: d−1 !

V $ (x(z)) =

uk (z k + z −k )

k=0

and expand y = − 12 M(x) y(z) = − (0)

Since W1

*

(x − a)(x − b):

d−1 * 1 1 ! M(x(z)) (x(z) − a)(x(z) − b) = u˜k (z k − z −k ) 2 2 k=1

(III-1-6)

= 12 V $ + y must have no positive powers, we find: uk = u˜k

and thus: (0) W1 (x(z))

=

d−1 !

uk z −k

k=1

In addition we must have

u0 = 0 (0)

and, since W1 (x) =

t x

+ O(1/x2), we must have: u1 =

4t a−b

Let us summarize those results in the following recipe: Theorem 1.1 Disc amplitude. For any α and γ, let x(z) = α + γ(z + 1/z), then expand: $

V (x(z)) =

d−1 !

uk (z k + z −k )

(III-1-7)

k=0

(the uk ’s are polynomial in α and γ). Then α and γ are uniquely determined by: u0 = 0

,

u1 =

t γ

,

α = O(t) , γ =

√

t(1 + O(t)). (0)

Then one has a = α + 2γ, b = α − 2γ, and the resolvent W1 , i.e. the generating function of planar rooted maps is: (0)

W1 (x(z)) =

d−1 ! k=1

46

uk z −k .

Examples of applications of this theorem are given in the next subsections, where we compute explicitly the case of quadrangulations and triangulations. More explicitly, one can expand eq.(III-1-7), and write the uk ’s as explicit polynomials of α and γ: uk = αδk,0 + γδk,1 −

d−1 (l+k)/2 ! ! l=2

tl+1

j=k

l! γ 2j−k αl+k−2j j!j − k!l + k − 2j!

In particular with k = 0 and k = 1, we see that α and γ are determined by two algebraic equations: 0 = u0 = α − d

d−1 ! l !

tl+j+1

l=1 j=0

(l + j)! γ 2j αl−j j!j!(l − j)!

(III-1-8)

l

!! t (l + j − 1)! = u1 = γ − γ 2j−1 αl−j tl+j γ j!(j − 1)!(l − j)! l=2 j=1

(III-1-9)

Those two algebraic equations yield a finite number of solutions for α and γ, and it is easy to see that all of them but one diverge at small t, and there is a unique solution such that α ∼ O(t) and γ 2 ∼ t + O(t2 ) at small t. It is easy to solve those equations order by order in t. To the first few orders we have: α = 2t3 γ 2 + (t3 α2 + 6t4 αγ 2 + 6t5 γ 4 ) + . . . γ 2 = t + (2t3 γ 2 α + 3t4 γ 4 ) + (3t4 α2 γ 2 + 12t5 αγ 4 + 20t6 γ 6 ) + . . . i.e. γ 2 = t + t2 (4t23 + 3t4 ) + O(t3 ) . . . α = t(2t3 ) + t2 (12t33 + 12t3 t4 ) + O(t3) . . . Bipartite maps Bipartite maps are those containing only unmarked faces of even perimeters. In other words we choose all t2k+1 = 0. In that case V $ (x) is an odd function of x: ! t2k x2k−1 . V $ (x) = x − k≥2

Eq.(III-1-8) becomes: 

0 = u0 = α 1 −

This implies that

d/2 l−1 ! ! l=2 j=0

t2l



(2l − 1)! γ 2j α2l−2j−2 = α(1 − O(t)). j!j!(2l − 2j − 1)! α = 0. 47

The equation eq.(III-1-9) for γ is now a polynomial equation for γ 2 : 2

t=γ −

d/2 !

t2l

l=2

(2l − 1)! 2l γ . l!(l − 1)!

We have (with the convention t2 = −1): γ

2

= t−

∞ !

k

t

k=2

t2k−2 P ai

k−1 !

/ (l + k)! (2k − 1 − 2 ai )! / l! k! (k − 1 − ai )!2

!

l=0 a1 +...+al (0) = − Res xl W1 (x) dx ∂tl l l z→∞ and that the initial conditions (tl = 0) agree. • Let us compute ∂/∂tl of eq.III-4-4. First, notice that at fixed x(z) we have: ∂z ∂α ∂γ 1 d x(z) = 0 = x$ (z) + + (z + ), d tl ∂tl ∂tl ∂tl z and thus

∂z dz " ∂α ∂γ 1 # =− + (z + ) ∂tl dx ∂tl ∂tl z which implies after dividing by z: ∂ ln z d ln z " ∂α ∂γ 1 # =− + (z + ) ∂tl dx ∂tl ∂tl z

and applying d/dx:

d ∂ ∂ " dz # d " dz " ∂α ∂γ 1 ## ln z = =− + (z + ) . dx ∂tl ∂tl zdx dx zdx ∂tl ∂tl z / l 2 Since V (x) = x2 − dl=3 tl xl , we have:

dz # 1 # xl dz dz " ∂α ∂γ ∂ " Res V (x) = − Res + Res V $ (x) + (z + ) ∞ ∞ ∞ ∂tl z l z z ∂tl ∂tl z 75

where we integrated the second term by parts. Since we have V $ (x(z)) = z −i ), we have: 0 = 2 u0 = − Res V $ (x) ∞

therefore:

dz z

,

/d−1 i=0

ui (z i +

t dz = u1 = − Res V $ (x) dz = − Res V $ (x) 2 ∞ ∞ γ z

dz # xl dz 2t ∂γ ∂ " Res V (x) = − Res − ∞ ∞ ∂tl z l z γ ∂tl

i.e.

# dz xl dz ∂ " Res V (x) + t ln γ 2 = − Res = r0 ∞ ∞ ∂tl z l z where we have defined l ! 1 l x = r0 + rj (z j + z −j ) = r0 + r+ (z) + r− (z) l j=1

,

r± (z) =

l ! j=1

rj z ±j = r∓ (1/z).

(0)

For k = 1 and g = 0, eq.III-4-2, together with ω2 (z1 , z2 ) = 1/(z1 − z2 )2 , becomes: 1 (0) ∂W1 (x) 11 1 (0) 1 = − Res W2 (x, x2 ) xl2 dx2 1 ∂tl l x2 →∞ x $ & 1 1 1 − xl2 dx2 = − Res l x2 →∞ x$ (z)x$ (z2 )(z − z2 )2 (x − x2 )2 $ & ∞ 1 ! 1 z j−1 xj−1 = − j Res − xl2 dx2 l j=1 x2 →∞ x$ (z)x$ (z2 ) z2j+1 xj+1 2 j−1 ! z 1 Res j j+1 x(z2 )l dz2 = −xl−1 − $ z →∞ l x (z) 2 z2 j (0)

Since x$ (z)W1 (x) contains only negative powers of z, we have: 1 (0) 1 1 ∂W (x) 1 1 $ x (z) 1 = −(x(z)l−1 x$ (z))− = − ((x(z))l )$− = −r− (z)$ (III-4-5) 1 ∂tl l x

That implies:

∂ xl (0) (0) Res V (x) W1 (x) dx + 2 Res W (x) dx ∞ ∂tl ∞ l 1 (0) ∂W1 (x) xl (0) = Res V (x) dx + Res W (x) dx ∞ ∞ ∂tl l 1 (0) = Res V $ (x) r− (z) x$ (z)dz + Res (r0 + r+ (z)) W1 (x) dx = =

∞

∞

∞

∞

(0)

Res V $ (x) r− (z) x$ (z)dz + Res r+ (z) W1 (x) dx − tr0 (0)

(0)

Res (V $ (x) − W1 (x)) r− (z) x$ (z)dz + Res r+ (z) W1 (x) dx − tr0 ∞

∞

76

(0)

(0)

= Res W1 (x(1/z)) r− (z) x$ (z)dz + Res r+ (z) W1 (x) dx − tr0 ∞

∞

(0)

(0)

= Res W1 (x(z)) r+ (z) x$ (z) dz + Res r+ (z) W1 (x) dx − tr0 0

∞

(0)

(0)

= Res W1 (x(z)) r+ (z) dx + Res r+ (z) W1 (x) dx − tr0 0 ∞ = −tr0 (0)

indeed it gives −tr0 , since W1 (z)r+ (z)x$ (z) has no other poles than 0 and ∞. That implies: ∂ dz xl (0) (0) ( Res V (x)W1 dx + t Res V + t2 ln γ 2 ) = −2 Res W1 dx ∞ ∞ ∂tl ∞ z l which is the result we sought. • This shows that F0 is given by eq.(III-4-4), up to an integration constant independent of tk ’s. The constant can be computed with tk = 0 for k ≥ 3, i.e. a quadratic potential V (x) = −t2 2 x2 , which we have already studied in section.1.7, and for which 2 we should have F0 = − t2 ln (−t2 ). In that case we have (see the example section.1.7): u0 = 0 ,

u1 =

t γ

,

γ=

*

− t/t2

,

1 x(z) = γ(z + ) , z

y(z) =

t 1 (z − ) 2γ z

The expression eq.III-4-4 gives as expected:

!

# 1 " γ2 2 3t2 t2 u1 − 2tγu1 + + t2 ln (γ 2 /t) = − ln (−t2 ) 2 2 2 2

There are also nice expressions for the derivatives of F0 with respect to t: Theorem 4.5 Derivatives of F0 with respect to t: $ 2& dz γ Res V (x(z)) + t + t ln z→∞ t $ 2& z γ = t ln + t − 2v0 t $ 2& γ = ln t dz 1 " 1 1 # = Res 3 $ = + . z→±1 z x (z) y $ (z) 2γ y $(1) y $(−1)

∂F0 = ∂t

∂ 2 F0 ∂t2 1 ∂ 3 F0 + t ∂t3

proof: Specialize eq.(III-1-12) of lemma 1.2 to z = 1 and z = −1, at which x$ (z) vanishes, this implies: ∂γ ∂x(±1) ±1 ∂α ±2 = = $ , ∂t ∂t ∂t y (±1) 77

from which we get ∂γ 1 = ∂t 4

$

1 1 + $ $ y (1) y (−1)

&

,

∂α 1 = ∂t 2

$

1 1 − $ $ y (1) y (−1)

&

.

This also implies that x˙ = ∂x/∂t at fixed z is x˙ = α˙ + γ(z ˙ + 1/z) (we denote ˙ = ∂/∂t $ and = ∂/∂z), and applying the chain rule, we find that at fixed x we have ∂z 1 1 γ˙ =− $ (α˙ + γ(z ˙ + 1/z)) = − $ (α˙ + (x(z) − α)), ∂t x (z) x (z) γ and ∂ ln z 1 =− $ (α˙ + γ(z ˙ + 1/z)). ∂t z x (z) Then, use the first expression of F0 , namely: 2 F0 = Res V z→∞

(0) (x)W1 (x)dx

3t2 + t2 ln + t Res V (x) d ln z + z→∞ 2

$

γ2 t

&

and take a derivative of each term with respect to t at fixed x, that gives & $ γ2 ∂F0 dz ∂ ln z γ˙ = 2 Res V (x) + t Res V (x) d + 2t + 2t ln + 2t2 . 2 z→∞ z→∞ ∂t z ∂t t γ Let us integrate the second term by parts, and get 2

dz dz γ˙ ∂F0 = 2 Res V (x) + t Res V $ (x) (α˙ + (x(z) − α)) z→∞ z→∞ ∂t z z γ 2 γ γ˙ +2t + 2t ln + 2t2 t γ γ2 dz = 2 Res V (x) + 2t + 2t ln z→∞ z t $ & t dz t γ˙ dz $ $ + (γ α˙ − αγ) ˙ Res V (x) + Res x V (x) + 2t z→∞ z→∞ γ z γ z

We have Res V $ (x)

z→∞

Res xV $ (x)

z→∞

∂ dz (0) = Res V $ (x) W1 (x) dx = 0, x→∞ z ∂t

dz ∂ ∂ (0) = Res xV $ (x) W1 (x) dx = − t2 = −2t. z ∂t x→∞ ∂t

This implies the result for ∂F0 /∂t. Then, one easily finds the last equality that ∂ 2 F0 /∂t2 = ln γ 2 /t. ! 78

4.4

genus 1 maps (1)

Genus 1 closed maps are elements of M0 . Their generating function is given by the following theorem. Theorem 4.6 The generating function of genus 1 maps is F1 = −

3 1 2 2 $ ln γ y (1) y $(−1)/t2 24

This result was derived many times, in particular in [?, ?, ?]. We leave the proof for the reader, it consists in checking that the derivatives with respect to the tl ’s are given (1) by eq.(III-4-1), with ω1 given by eq.(III-3-4). proof: Let us denote ∂/∂tk = ˙, and ∂/∂z =$ . From eq.(III-1-13) of lemma 1.2, we have x˙ y $ − y˙ x$ = Hk$

,

Hk (z) =

Taking the first derivative, we have

3 1 2 (x(z)k )+ − (x(z)k )− . 2k

x˙ $ y $ + x˙ y $$ − y˙ $ x$ − y˙ x$$ = Hk$$ and the second derivative x˙ $$ y $ + 2x˙ $ y $$ + x˙ y $$$ − y˙ $$ x$ − 2y˙ $ x$$ − y˙ x$$$ = Hk$$$ . At z = 1, we have x$ (z) = 0 and x˙ $ (1) = 0, therefore we get x(1) ˙ y $$(1) − y(1) ˙ x$$ (1) = Hk$$ (1) i.e.

y $$ (1) Hk$$ (1) − $$ , y(1) ˙ = x(1) ˙ x$$ (1) x (1)

and x˙ $$ (1) y $(1) + x(1) ˙ y $$$(1) − 2y˙ $(1) x$$ (1) − y(1) ˙ x$$$ (1) = Hk$$$ (1) i.e. Hk$$$ (1) 1 y $$$ (1) x$$$ (1) y˙ $(1) $$ = − $$ + x˙ (1) $$ + x(1) ˙ − y(1) ˙ y $(1) 2x (1) y $(1) 2x (1) 2x$$ (1) y $(1) 2x$$ (1) y $(1) y˙ $ (1) Hk$$$ (1) Hk$$ (1) x$$$ (1) $$ 1 = − + +x˙ (1) $$ +x(1) ˙ $ $$ $ $$ 2 $ y (1) 2x (1) y (1) 2x (1) y (1) 2x (1)

$

y $$$ (1) x$$$ (1)y $$(1) − 2x$$ (1) y $(1) 2x$$ (1)2 y $ (1)

Beside, we have H $ (1) x(1) ˙ = α˙ + 2γ˙ = $k y (1)

,

x˙ $$ (1) γ˙ 1 = = $$ x (1) γ 4γ 79

$

Hk$ (1) Hk$ (−1) − $ y $(1) y (−1)

&

.

&

Therefore 1 y˙ $(1) γ˙ Hk$$$ (1) Hk$$ (1) = − − + Hk$ (1) $ $ $ 24 y (1) 48γ 96γ y (1) 32γ y (1)

$

y $$$ (1) y $$(1) + 96γ y $(1)2 32γ y $(1)2

and, similarly at z = −1 we obtain 1 y˙ $(−1) γ˙ Hk$$$ (−1) Hk$$ (−1) = + − + Hk$ (−1) $ $ $ 24 y (−1) 48γ 96γ y (−1) 32γ y (−1)

$ −

&

y $$$ (−1) y $$(−1) + 96γ y $(−1)2 32γ y $(−1)2

From eq.(III-3-4), we have : ; y !!! (1) y !! (1) + 1 + ! ! 1 1 1 y (1) 3y (1) (1) ω1 (z) = + − 16γy $(1) (z − 1)4 (z − 1)3 2(z − 1)2 : !! (−1) !!! (−1) ; 1 − yy! (−1) + y3y! (−1) 1 1 1 − − − 16γy $ (−1) (z + 1)4 (z + 1)3 2(z + 1)2 and thus (1)

Res Hk (z) ω1 (z) dz : y !! (1) y !!! (1) ; 1 + ! (1) + 3y ! (1) 1 Hk$$$ (1) Hk$$ (1) y = + − Hk$ (1) 16γy $(1) 6 2 2 : !! (−1) 1 − yy! (−1) + 1 Hk$$$ (−1) Hk$$ (−1) $ − − − H (−1) k 16γy $ (−1) 6 2 2 z→±1

y !!! (−1) 3y ! (−1)

;

i.e. Res

z→±1

(1) Hk (z) ω1 (z) dz

1 = − 24 1 = − 24 1 = − 24

y˙ $ (1) 1 y˙ $ (−1) γ˙ 1 − + − y $ (1) 24 y $ (−1) 24γ 32 γ y˙ $ (1) 1 y˙ $ (−1) γ˙ γ˙ − + − $ $ y (1) 24 y (−1) 24γ 8 γ y˙ $ (1) 1 y˙ $ (−1) 2 γ˙ − − $ $ y (1) 24 y (−1) 24γ

$

Hk$ (1) Hk$ (−1) − $ y $ (1) y (−1)

&

Eventually, we arrive at 1 ∂ (1) ln γ 2 y $(1) y $(−1) = − Res Hk (z) ω1 (z) dz. z→±1 24 ∂tk (1)

The only poles of Hk (z) ω1 (z) are at z = 1, −1, 0, ∞, and thus, moving the integration contour we have: 1 ∂ (1) (1) ln γ 2 y $ (1) y $(−1) = Res Hk (z) ω1 (z) dz + Res Hk (z) ω1 (z) dz. ∞ 0 24 ∂tk 80

&

(1)

(1)

Near z = 0, we use the symmetry Hk (1/z) = −Hk (z) and ω1 (1/z) = z 2 ω1 (z), and thus (1)

(1)

(1)

Res Hk (z) ω1 (z) dz = Res Hk (1/z) ω1 (1/z) d(1/z) = Res Hk (z) ω1 (z) dz. 0

∞

∞

(1)

Then, near z = ∞, 2k Hk (z) = (x(z)k )+ − (x(z)k )− , and it is clear that (x(z)k )− ω1 (z) has no pole at z = ∞, thus: (1)

2 Res Hk (z) ω1 (z) dz = ∞

1 (1) Res x(z)k ω1 (z) dz. k ∞

This implies that 1 ∂ 1 ∂F1 (1) ln γ 2 y $ (1) y $(−1) = Res x(z)k ω1 (z) dz = − 24 ∂tk k ∞ ∂tk 1 where we used eq.(III-4-1). This proves that F1 + 24 ln γ 2 y $(1) y $(−1) is independent of tk , and can be computed when all tk = 0 ∀ k, i.e. for the Gaussian matrix model, 1 ln t2 . and we find that it is worth 24 !

4.5

Some general structure properties (g)

From the residue formula, it is easy to see that the stable ωn (z1 , . . . , zn ) are rational functions of the zi ’s, with poles only at z = ±1, of degree di + 2, such that di ≥ 0 and / th i di ≤ 6g − 6 + 2n. Moreover, they are rational fractions of γ and the (6g + 2n − 5) first derivatives of y(z) at z = ±1. More precisely:

=

ωn(g) (z1 , . . . , zn ) ! Pol(y $$(±1)/y $ (±1), . . . , y (6g+2n−5) (±1)/y $(±1)) di ,

P

di ≤6g+2n−4

y $(1)2g+n−2

y $(−1)2g+n−2

γ 4g+2n−4

where Pol is a polynomial, of total degree at most (6g − 6 + 2n − y (k) (±1)/y $(±1) is considered to be of degree k − 1. In particular if n = 0: Fg =

/