L’INSTITUTFOURIER DE ANNALES

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A L

E S

E D L ’IN IT ST T U

F O U R

ANNALES

DE

L’INSTITUT FOURIER

Arzu BOYSAL & Michèle VERGNE

Paradan’s wall crossing formula for partition functions and Khovanski-Pukhlikov differential operator

Tome 59, n^o5 (2009), p. 1715-1752.

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PARADAN’S WALL CROSSING FORMULA FOR PARTITION FUNCTIONS AND

KHOVANSKI-PUKHLIKOV DIFFERENTIAL OPERATOR

by Arzu BOYSAL & Michèle VERGNE

Abstract. — LetP(s)be a family of rational polytopes parametrized by in- equations. It is known that the volume ofP(s)is a locally polynomial function of the parameters. Similarly, the number of integral points in P(s) is a locally quasi-polynomial function of the parameters. Paul-Émile Paradan proved a jump formula for this function, when crossing a wall. In this article, we give an algebraic proof of this formula. Furthermore, we give a residue formula for the jump, which enables us to compute it.

Résumé. — SoitP(s)une famille de polytopes rationnels paramétrés par des inéquations. On sait que le volume de P(s) est une fonction localement polynomiale des paramètres. Similairement, le nombre de points entiers dans P(s)est une fonction localement quasi-polynomiale des paramètres. Paul-Émile Paradan a donné une formule de saut pour cette fonction, lorsqu’on traverse un mur. Dans cet article, nous donnons une démonstration algébrique de ces formules de saut.

Nous exprimons aussi le saut, à l’aide d’une formule de résidus, ce qui permet de le calculer.

1. Introduction

The function computing the number of ways one can decompose a vector as a linear combination with nonnegative integral coefficients of a fixed finite set of integral vectors is called apartition function. This problem can be expressed in terms of polytopes as follows. LetAbe arbyNintegral matrix with column vectorsφ₁, . . . , φN, and assume that the elementsφk

generate the lattice Z^r. Let a ∈ Z^r be a r-dimensional integral column vector and let P(Φ, a) := {y ∈ R^N_>0 | Ay = a} be the convex polytope

Keywords:Polytopes, toric varieties.

Math. classification:52B20, 14M25.

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associated toΦ = [φ₁, φ₂, . . . , φ_N]anda. The functiona→ |P(Φ, a)∩Z^N| will be called the partition function k(Φ)(a). It is intuitively clear that k(Φ)(a) is related to the volume function vol(Φ)(a) = volume(P(Φ, a)).

The latter varies polynomially as a function of a, provided the polytope P(Φ, a)does not change ‘shape’, that is, whenavaries in a chambercfor A. In short, there is a decomposition ofR^r in closure of chambers ci and polynomial functionsv(Φ,ci)such that the functionvol(Φ)(a)coincide with the polynomial functionv(Φ,c_i)(a)on each conec_i. Similarly, there exists quasi-polynomial functions k(Φ,ci) onZ^r such that the function k(Φ)(a) coincide with the quasi-polynomial functionk(Φ,c_i)(a)onc_i∩Z^r.

When c1 and c2 are adjacent chambers, P.-E. Paradan [8] gave a re- markable formula for the quasi-polynomial functionk(Φ,c₁)−k(Φ,c₂)as a convolution of distributions. His proof relies on indices of transversally elliptic operators. There is an analogous formula forv(Φ,c1)−v(Φ,c2).

In this note, we give an elementary algebraic proof of Paradan’s convolution formula for the jumps. We also express k(Φ,c1)−k(Φ,c2) and v(Φ,c₁)−v(Φ,c₂)by one-dimensional residue formulae. More generally, in Theorem 4.2, we give a formula for the convolution of a polynomial den- sity on a hyperplane with several Heaviside functions (and of the discrete analogue in Theorem 5.1 and Theorem 6.7). These formulae can be used to compute such objects easily.

Let us describe our residue formulae.

Let c1 and c2 be two adjacent chambers lying on two sides of a hyper- planeW (determined by a primitive vectorE). Define Φ₀ = Φ∩W. The intersection ofc1andc2is contained in the closure of a chamberc12ofΦ0.

Theorem 1.1.

• Letv12=v(Φ0,c12)be the polynomial function onW associated to the chamberc12 ofΦ0. Let V₁₂ be any polynomial function onR^r extendingv12. Then, ifhE,c1i>0, we have fora∈R^r

v(Φ,c1)(a)−v(Φ,c2)(a)

= Res_z=0

V₁₂(∂_x)· e^ha,x+zEi Q

φ∈ΦrΦ₀hφ, x+zEi

x=0

.

• SupposeΦis unimodular. Let k₁₂ =k(Φ₀,c₁₂) be the polynomial function onW associated to the chamberc12ofΦ0. LetK12be any polynomial function onR^r extendingk₁₂. Then, if hE,c₁i>0, we have fora∈R^r

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k(Φ,c1)(a)−k(Φ,c2)(a)

= Resz=0

K12(∂x)· e^ha,x+zEi Q

φ∈ΦrΦ₀(1−e^−hφ,x+zEi)

x=0

. In fact we will give a general version of the second part of this theorem in Section 6, whereΦis not necessarily unimodular.

Our proof of the residue formulae for the jumps relies on an easy induction argument.

It is immediate to see that both formulae for the jumps k(Φ,c₁)− k(Φ,c2) and v(Φ,c1)−v(Φ,c2) are related by the application of a gen- eralized Khovanski-Pukhlikov differential operator [4], [7], [3].

We also demonstrate in various examples how to use these formulae to compute the functionsv(Φ,c)andk(Φ,c).

Another algebraic proof of Paradan’s jump formulae, as a consequence of difference equations, was obtained recently in [6].

2. Partition functions 2.1. Definitions and notations

Let U be a r-dimensional real vector space and V be its dual vector space. We assume that V is equipped with a lattice Γ. We will usually denote byx the variable inU and bya the variable in V. We will see an elementP of S(V)both as a polynomial function onU and a differential operator onV via the relationP(∂_a)e^ha,xi=P(x)e^ha,xi.

Let Φ = [φ1, φ2, . . . , φN]be a sequence of non-zero, not necessarily dis- tinct, linear forms onU lying in an open half space. Assume that all the φk ∈ Φ belong to the lattice Γ. We denote by hΦi the linear span of Φ.

ThenΦgenerates a lattice in hΦi. We denote this lattice byZΦ⊂Γ.

We consider R^N with basis (ω₁, . . . , ω_N) and let A be the linear map from R^N to the vector space hΦi defined by A(ωk) = φk, 1 6 k 6 N. The vectorsφ_k are the column vectors of the matrixA, and the mapA is surjective ontohΦi. Fora∈ hΦi, we consider the convex polytope

P(Φ, a) :=n

t= (t1, t2, . . . , tN)∈R^N_>0|At=ao . In other words,

P(Φ, a) =n

t= (t₁, t₂, . . . , t_N)∈R^N_>0|X

i

t_iφ_i=ao .

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Any polytope can be realized as a polytope P(Φ, a).

LetC(Φ)⊂ hΦibe the cone generated by{φ1, . . . , φN}. The coneC(Φ) is a pointed polyhedral cone. The dual coneC(Φ)^∗ ofC(Φ) is defined by C(Φ)^∗ ={x∈U | hφ, xi>0 for allφ∈Φ} and its interior is non-empty.

The polytope P(Φ, a) is empty if a is not in C(Φ). If a ∈ hΦi is in the relative interior of the coneC(Φ), then the polytopeP(Φ, a)has dimension d:=N−dim(hΦi).

We choosedxonhΦi^∗and denote bydathe dual measure onhΦi. Letdt be the Lebesgue measure onR^N. The vector spaceKer(A) =A⁻¹(0)is of dimensiond=N−dim(hΦi)and it is equipped with the quotient Lebesgue measuredt/dasatisfying(dt/da)∧da=dt. Fora∈ hΦi, A⁻¹(a)is an affine space parallel to Ker(A), thus also equipped with the Lebesgue measure dt/da. Volumes of subsets ofA⁻¹(a) are computed with this measure. In particular we can define for anya∈ hΦi, the numbervol(Φ)(a, dx)as being the volume of the convex setP(Φ, a)in the affine space A⁻¹(a)equipped with the measuredt/da. Ifdx is rescaled by c >0, then vol(Φ)(a, cdx) = cvol(Φ)(a, dx). By definition, if the dimension of P(Φ, a) is less than d, vol(Φ)(a, dx)is equal to0.

Definition 2.1. — LethΦibe the subspace ofV generated byΦ.

• Ifa∈ hΦi, definevol(Φ, dx)(a) = volume(P(Φ, a), dt/da).

• Ifa∈ hΦi, definek(Φ)(a) =|P(Φ, a)∩Z^N|.

We extend the definition of the functions vol(Φ, dx)(a)and k(Φ)(a)as functions onV by defining vol(Φ, dx)(a) = 0 if a /∈ hΦi, k(Φ)(a) = 0 if a /∈ hΦi.

Clearly,vol(Φ, dx)(a) = 0ifais not inC(Φ)and k(Φ, a) = 0ifais not inZΦ∩C(Φ).

In the rest of this article, we will formulate many of our statements when Φgenerates V, as we can always reduce to this case replacing eventually V byhΦi.

If Φ = [φ1, φ2, . . . , φr] consists of linearly independent vectors, then the set P(Φ, a) is just one point when a ∈ C(Φ) and is empty when a is not in the closed cone C(Φ). Thus the function vol(Φ)(a, dx) is just the characteristic function of the closed cone C(Φ) multiplied by |det(Φ)|⁻¹ where the determinant is computed with respect to the Lebesque measure da. Similarly, the functionk(Φ)(a)is the characteristic function ofC(Φ)∩ Pr

i=1Zφi.

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Lemma 2.2. — Assume Φ = Φ⁰∪ {φ} whereΦ⁰ generateshΦi. Then vol(Φ, dx)(a) =

Z

t>0

vol(Φ⁰, dx)(a−tφ)dt for anya∈V.

Proof. — Indeed, decomposeΦ = [φ,Φ⁰]. Then P(Φ, a) =

[t, t⁰] ; t>0, t⁰ ∈P(Φ⁰, a−tφ) .

The proof follows by Fubini.

By induction, we obtain the following corollary.

Corollary 2.3. — The functionvol(Φ, dx)(a)is continuous onC(Φ).

For an elementγin V, define the translation operatorτ(γ)on functions k(a)onV by the formula: ifa∈V, then

(τ(γ)k)(a) =k(a−γ).

The difference operatorD(γ) = 1−τ(γ) acts on functionsk(a) onV by the formula:

(D(γ)k)(a) =k(a)−k(a−γ).

The following lemma is obvious from the definition.

Lemma 2.4. — Letφ∈Φanda∈Γ. Then k(Φ)(a) =

∞

X

n=0

k(Φr{φ})(a−nφ).

The following relation follows immediately.

Lemma 2.5. — Letφ∈Φanda∈Γ. Then (D(φ)k(Φ))(a) =k(Φr{φ})(a).

In particular,(D(φ)k(Φ))(a)is equal to 0 ifais not in the subspace ofV generated byΦr{φ}.

Lemma 2.6. — Assume Φ generates V. Let W be a hyperplane in V such that W ∩C(Φ) is a facet of C(Φ). Let Φ0 be the sequence Φ∩W which spansW. Ifa∈W, thenk(Φ)(a) =k(Φ₀)(a).

Proof. — AsW∩C(Φ)is a facet ofC(Φ), ifa∈W∩C(Φ), any solution ofa=PN

i=1y_iφ_i withy_i>0 will havey_i= 0forφ_i∈/W. The following lemma is also obtained immediately from Fubini’s theorem applied to the integralR

R^N_>0e^−hPN

i=1t_iφ_i,xidt₁dt₂· · ·dt_N decomposed along the fibers of the mapA:R^N_>0→C(Φ), or to the analogous discrete sum.

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Lemma 2.7. — Forxin the interior ofC(Φ)^∗, Z

C(Φ)

vol(Φ, dx)(a)e^−ha,xida= 1 Q

φ∈Φhφ, xi, X

a∈C(Φ)∩Γ

k(Φ)(a)e^−ha,xi= 1 Q

φ∈Φ1−e^−hφ,xi.

2.2. Chambers and the qualitative behavior of partition functions

In this section, we assume thatΦgeneratesV. For any subsetν ofΦ, we denote byC(ν)the closed cone generated byν. We denote byC(Φ)_singthe union of the conesC(ν)whereνis any subset ofΦof cardinality strictly less thanr = dim(V). By definition, the set C(Φ)_reg ofΦ-regular elements is the complement ofC(Φ)sing. A connected component ofC(Φ)regis called a chamber. We remark that, according to our definition, the exterior ofC(Φ) is itself a chamber denoted bycext. The chambers contained inC(Φ)will be called interior chambers. Ifcis a chamber, andσis a basis ofV contained inΦ, then eitherc⊂C(σ), or c∩C(σ) =∅, as the boundary of C(σ)does not intersectc.

Let Φ⁰ ⊂Φ be such thatΦ⁰ generatesV. Ifc is a chamber forΦ, there exists a unique chamberc⁰ forΦ⁰ such thatc⊂c⁰.

A wall ofΦis a (real) hyperplane generated byr−1linearly independent elements ofΦ. It is clear that the boundary of a chambercis contained in an union of walls.

We now define the notion of a quasi-polynomial function on the latticeΓ.

LetΓ^∗be the dual lattice ofΓ. An elementx∈U gives rise to the exponen- tial functione_x(a) =e^2iπhx,aionΓ. Remark that the functione_x(a)depends only of the class ofx(still denoted byx) in the torus T(Γ) :=U/Γ^∗.

LetM be a positive integer. A quasi-polynomial function with periodM onΓis a functionK onΓ of the formK(a) =P

x∈Fex(a)Px(a) whereF is a finite set of points ofU such that M F ⊂Γ^∗ and Px are polynomial functions onV. Then the restriction of the function K to cosets h+MΓ ofΓ/MΓcoincide with the restriction toh+MΓof a polynomial function onV. If the degree of the polynomialP_x(a)is less or equal tokfor allx∈F, we say thatKis a quasi-polynomial function of degree kand periodM.

If Γ =Zandγ∈C^∗ is aM^th root of unity, the functionn7→n^kγⁿ is a quasi-polynomial function onZof periodM and degreek.

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If C is an affine closed cone in V with non empty interior, a quasi- polynomial function onΓvanishing onΓ∩Cis identically equal to0onΓ.

If γ∈Γ, the difference operator D(γ)k(a) =k(a)−k(a−γ)leaves the space of quasi-polynomial functions onΓstable.

The following theorem is well known (see [4], [3], [9], [5]). See a simple proof in [6].

Proposition 2.8. — Letcbe an interior chamber ofC(Φ).

• There exists a unique homogeneous polynomial functionv(Φ, dx,c) of degreedonV such that, fora∈c,

vol(Φ, dx)(a) =v(Φ, dx,c)(a).

• There exists a unique quasi-polynomial functionk(Φ,c)onΓ such that, fora∈c∩Γ,

k(Φ)(a) =k(Φ,c)(a).

Remark 2.9. — The sequenceΦis called unimodular if, for any subsetσ ofΦforming a basis ofV, the subsetσis a basis ofZΦ. In other words, we have|det(σ)|= 1, where the determinant is computed using the volumeda giving volume1 to a fundamental domain forZΦ. In this particular case, the functionk(Φ,c)is polynomial onZΦ.

In the next lemma, we list differential equations satisfied by the polynomial functionv(Φ, dx,c).

Lemma 2.10. — Let φ ∈ Φ. If Φr{φ} does not generate V, then

∂(φ)v(Φ, dx,c) = 0.

If Φr{φ} generates V, let c⁰ be the chamber of Φr{φ} containing c, then∂(φ)v(Φ, dx,c) =v(Φr{φ}, dx,c⁰).

Proof. — IfΦ₀= Φr{φ}is contained in a wallW, thenV =W ⊕Rφ, and it is immediate to see that an interior chambercforΦis of the form c=c0+R>0φ, wherec0is a chamber forΦ0. Ifa=w+tφwithw∈W and t >0, then vol(Φ, dx,c)(w+tφ) = vol(Φ₀, dx₀,c0)(w), with dx₀dφ=dx.

This proves the first statement.

To prove the second statement, ifa∈c, we use the following relation (as given in Lemma 2.2)

vol(Φ, dx)(a)−vol(Φ, dx)(a−φ) = Z

t=0

vol(Φ₀, dx)(a−tφ)dt.

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Corollary 2.11. — Let Φ0 ⊂ Φ such that Φ0 does not generate V.

Then

Y

φ∈ΦrΦ₀

∂(φ)

v(Φ, dx,c) = 0.

In the next lemma, we list difference equations satisfied by the quasi- polynomial functionk(Φ,c).

Lemma 2.12. — Let φ ∈ Φ. If Φr{φ} does not generate V, then D(φ)k(Φ,c) = 0.

If Φr{φ} generates V, let c⁰ be the chamber of Φr{φ} containing c, thenD(φ)k(Φ,c) =k(Φr{φ},c⁰)

Proof. — By Lemma 2.5, the functionk(Φ)satisfiesD(φ)k(Φ) =k(Φr {φ}). Considering this relation on an affine subconeS ofcsuch thatS−φ does not touch the boundary ofc, we obtain the relations of the lemma.

3. Two polynomial functions 3.1. Residue formula

LetL be the space of Laurent series in one variablez:

L:=

f(z) = X

k>k₀

f_kz^k

.

Forf ∈ L, we denote byResz=0f(z) the coefficient f₋₁ ofz⁻¹. If g is a germ of meromorphic function atz= 0, thenggives rise to an element ofL by considering the Laurent series atz= 0and we still denote byResz=0g its residue atz= 0. Ifg= _dz^df, thenResz=0g= 0.

With the notation of Section 2.1, let E be a vector in U. It defines a hyperplaneW ={a∈V | ha, Ei= 0} inV.

Definition 3.1. — LetP be a polynomial function onV and letΨbe a sequence of vectors not belonging toW. We define, fora∈V,

• Pol(P,Ψ, E)(a) = Resz=0

P(∂x)· e^ha,x+zEi Q

ψ∈Ψhψ, x+zEi

x=0

.

• Par(P,Ψ, E)(a) = Resz=0

ψ∈Ψ(1−e^−hψ,x+zEi)

x=0

. It is easy to see thatPol(P,Ψ, E)(a)as well asPar(P,Ψ, E)(a)are polynomial functions ofa∈V.

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Lemma 3.2. — The functions Pol(P,Ψ, E) and Par(P,Ψ, E) depend only on the restrictionpofP toW.

Proof. — Ifp= 0, thenP =EQwhereQis a polynomial function onV. Then

P(∂x)F(x+zE) = d

dQ(∂x)F(E+x+zE)=0

= d

dQ(∂x)F(x+ (z+)E)=0

= d

dz(Q(∂x)F(x+zE))

so that the residue Resz=0 vanishes on the function z 7→ P(∂x)F(x+

zE)x=0.

We can then give the following definitions.

Definition 3.3. — Letpbe a polynomial function onW. We define Pol(p,Ψ, E) := Pol(P,Ψ, E)

whereP is any polynomial on V extendingp.

We define

Par(p,Ψ, E) := Par(P,Ψ, E) whereP is any polynomial on V extendingp.

In the following, given polynomialsp, q, . . .onW, we denote byP, Q, . . . polynomials onV extendingp, q, . . .

3.2. Some properties

Let us list some properties satisfied by the function Pol(p,Ψ, E).

We first remark that if we replace ψ in Ψ by cψ with c 6= 0, then Pol(p,Ψ, E)becomes ¹_cPol(p,Ψ, E).

We now discuss how Pol(p,Ψ, E) transforms under the action of differ- entiation.

Proposition 3.4. — Letψ∈Ψ. Then

∂(ψ) Pol(p,Ψ, E) = Pol p,Ψr{ψ}, E .

Letw∈W. Then

∂(w) Pol(p,Ψ, E) = Pol ∂(w)p,Ψ, E .

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Proof. — The first formula follows immediately from the definition.

For the second part of the proposition, we will use the following lemma, which is implied by the relationP(∂_x)hx, wi − hx, wiP(∂_x) = (∂(w)P)(∂_x).

Lemma 3.5. — For any functionJ(x)ofx∈U,

P(∂x)hx, wiJ(x)

x=0

=

(∂(w)P)(∂x)J(x)

x=0

. Now, ashw, Ei= 0, forJ(x, z) = Q ¹

ψ∈Ψhψ,x+zEi, we have

∂(w) Res_z=0

P(∂_x)e^ha,x+zEiJ(x, z)

x=0

= Res_z=0

P(∂_x)hw, xie^ha,x+zEiJ(x, z)

x=0

= Res_z=0

(∂(w)P)(∂_x)e^ha,x+zEiJ(x, z)

x=0

.

So we obtain the formula of the proposition.

Lemma 3.6.

• IfΨ ={ψ}, then forw∈W andt∈R,

Pol(p,{ψ}, E)(w+tψ) = Par(p,{ψ}, E)(w+tψ) = p(w) hψ, Ei.

• If |Ψ| > 1, then the restriction of Pol(p,Ψ, E) to W vanishes of order|Ψ| −1.

Proof. — Let U₀ = {x | hψ, xi = 0}. We write U = U₀ ⊕RE. The spaceS(U0)is isomorphic to the space of polynomial functions on W. We may chooseP in S(U0). We write x=x0+x1E, with x0 ∈U0. In these coordinateshψ, x+zEi= (x₁+z)hψ, Eiis independent ofx₀. So we can setx1= 0in the formula

Resz=0

P(∂x₀)· e^ha,x+zEi (x₁+z)hψ, Ei

x=0

and the residue is computed for a function that have a simple pole atz= 0.

The formula follows. The other points are also easy to prove.

Let us list some difference equations satisfied by the function Par(p,Ψ, E).

Proposition 3.7. — Letψ∈Ψ. Then

D(ψ) Par(p,Ψ, E) = Par(p,Ψr{ψ}, E).

Letw∈W. Then

τ(w) Par(p,Ψ, E) = Par(τ(w)p,Ψ, E).

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Proof. — The first formula follows immediately from the definition.

The translation operatorτ(w)satisfies the relation P(∂x)e^−hw,xi=e^−hw,xi(τ(w)P)(∂x).

Thus, the second formula follows from the same argument as in the proof

of the second item in Proposition 3.4.

4. Wall crossing formula for the volume

In this section, we give two formulae for the jump of the volume function across a wall. The first one uses convolutions of Heaviside distributions and is in the spirit of Paradan’s formula ([8], Theorem 5.2) for the jump of the partition function. The second one is a one dimensional residue formula.

4.1. Inversion formula

We will need some formulae for Laplace transforms in dimension 1. For z >0 andk>0an integer, we have

(4.1) 1

z^k+1 = Z ∞

0

t^k k!e^−tzdt.

Consider the Laplace transform L(p)(z) =

Z

R⁺

e^−tzp(t)dt.

Assume thatp(t) =P

ikc_ikp_i,k(t)is a linear combination of the functions pi,k(t) =e^−txⁱ^t_k!^k. We assume thatxi>0. Then, the integral definingL(p) is convergent. We have

(4.2) L(p)(z) =X

i,k

cik

(z+xi)^k+1. The following inversion formula is immediate to verify.

Lemma 4.1. — LetR >0. Assume that|x_i|< Rfor alli. Then we have

(4.3) p(t) = 1

2iπ Z

|z|=R

L(p)(z)e^tzdz.

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Reciprocally, if p is a continuous function on R such that L(p)(z) is convergent and given by Formula (4.2), thenpis given by Equation (4.3).

Ifp(t) =P

kckt^k

k! is a polynomial (that is all the elementsxiare equal to 0), thenL(p)(z)is the Laurent polynomialP

kckz^−k−1, and the inversion formula above reads

(4.4) p(t) = Resz=0L(p)(z)e^tz.

4.2. Convolution of measures

LetE∈Ube a non zero linear form onV andW ⊂V the corresponding hyperplane. LetV⁺={a∈V | ha, Ei>0}andV⁻={a∈V | ha, Ei<0}

denote the corresponding open half spaces. Let∆⁺ = [α1, α2, . . . , αQ]be a sequence of vectors contained inV⁺. Consider the span h∆⁺iof ∆⁺. We choose a Lebesgue measuredaonh∆⁺iwith dual measuredxonh∆⁺i^∗. We define the continuous functionv(∆⁺, dx)(a) on the coneC(∆⁺) ⊂ h∆⁺i such that, forx∈C(∆⁺)^∗, we have

(4.5) 1

Q

α∈∆⁺hα, xi= Z

C(∆⁺)

v(∆⁺, dx)(a)e^−ha,xida.

By Lemma 2.7, v(∆⁺, dx)(a) = vol(∆⁺, dx)(a).

We choose the Lebesgue measure dw =da/dt onW ∩ h∆⁺i wheret = ha, Ei. The measuredwdetermines a measure on all affine spacesW∩(a+ h∆⁺i).

Theorem 4.2. — Letpbe a polynomial function onW. We define (p∗v(∆⁺, dx))(a) =

Z

W∩(a+h∆⁺i)

p(w)v(∆⁺, dx)(a−w)dw.

Then, fora∈V⁺, we have

(p∗v(∆⁺, dx))(a) = Pol(p,∆⁺, E)(a)fora∈V⁺.

Remark that p∗v(∆⁺, dx) depends only on the choice of E. Indeed, p∗v(∆⁺, dx)is the convolution of two functions, one of which depends on the measuredx, while the convolution depends on the measuredw. We see that finally this depends only of the choice ofE.

We also remark that, for fixeda∈V⁺, the integral definingp∗v(∆⁺, dx) is in fact over the compact setW ∩(a−C(∆⁺))where v(∆⁺, dx)(a−w) is not equal to zero .

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Proof. — We decomposeW =W₀⊕W₁, whereW₀=W∩ h∆⁺i. Then, we can writea∈V asa=tF+w0+w1, withhF, Ei= 1. Ifp(w0+w1) = p₀(w₀)p₁(w₁), we see that (p∗v(∆⁺, dx))(tF +w₀+w₁) = p₁(w₁)(p₀∗ v(∆⁺, dx))(tF+w0). Hence, it is sufficient to prove the proposition in the case where∆⁺ generatesV. Then

(p∗v(∆⁺, dx))(a) = Z

W

p(w)v(∆⁺, dx)(a−w)dw.

The polynomial nature of (p∗v(∆⁺, dx))(a) is clear intuitively. In any case, we will prove the explicit formula of the proposition, which gives a polynomial formula for(p∗v(∆⁺, dx))(a).

We need to compute, fora∈V⁺,I(a) :=R

Wp(w)v(∆⁺, dx)(a−w)dw.

This integral is over a compact subset ofW. LetP ∈S(U)be a polynomial function onV extendingp. We may write

I(a) =

P(∂_x)· Z

W

v(∆⁺, dx)(a−w)e^hw,xidw

x=0

.

Define

gx(a) = Z

W

v(∆⁺, dx)(a−w)e^−ha−w,xidw.

Thengx(a)depends analytically on the variablex∈U, and we have

(4.6) I(a) =

P(∂x)·e^ha,xigx(a)

|x=0. The function a 7→ gx(a) = R

Wv(∆⁺, dx)(a−w)e^−ha−w,xidw is a continuous function of a moduloW, that is, it is a continuous function of the variable t = ha, Ei >0 when a ∈ V⁺. We then write gx(t) = gx(tF) = R

Wv(∆⁺, dx)(tF −w)e^{−htF−w,xi}dw.

To identify the functiong_x(t), we compute its Laplace transform in one variable. Letz >0. Ifxis in C(∆⁺)^∗, the integral defining L(gx) is convergent and we have

L(g_x)(z) = Z

t>0

e^−tzg_x(tF)dt= Z

V⁺

e^−ha,zEiv(∆⁺, dx)(a)e^−ha,xida

= 1

Q

α∈∆⁺hα, x+zEi

by Formula (4.5). Here hα, x+zEi=d_αz+hα, xi withd_α =hα, Ei>0 andhα, xi>0.

Thus, by partial fraction decomposition,L(g_x)(z)is a function of the type given by Formula (4.2). By the inversion formula for the Laplace transform

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in one variable, we obtain that (forxsmall enough) gx(a) = 1

2iπ Z

|z|=1

e^ha,zEi Q

α∈∆⁺hα, x+zEidz.

Thus Formula (4.6) becomes I(a) =P(∂x)·

e^ha,xi 1

2iπ Z

|z|=1

e^ha,zEi Q

α∈∆⁺hα, x+zEidz

x=0

= 1 2iπ

Z

|z|=1

α∈∆⁺hα, x+zEi

x=0

dz.

The function in the integrand has a Laurent series at z = 0 with polynomial coefficients inaof the form P

kg_k(a)z^k. Thus we obtain I(a) = Resz=0

α∈∆⁺hα, x+zEi

x=0

.

This shows thatI(a)coincide with the polynomial functionPol(p,∆⁺, E)

onV⁺.

We also remark that, ifpis homogeneous,Pol(p,∆⁺, E)is homogeneous inaof degree|∆⁺| −1 + deg(p).

4.3. The jump for the volume function

Letvol(Φ, dx)be the locally polynomial function on the coneC(Φ)generated byΦ. LetW be a wall, determined by a vectorE ∈U. LetV⁺ and V⁻denote the corresponding open half spaces. DefineΦ0= Φ∩W; this is a sequence of vectors inW spanning W.

Let c₁ ⊂ V⁺ and c₂ ⊂ V⁻ be two chambers on two sides of W and adjacent. Here, we mean thatc1∩c2 has non empty relative interior inW. Thusc1∩c2 is contained in the closure of a chamberc12ofΦ0. We choose the measuredwonW such thatda=dwdtwitht=ha, Ei. We write

Φ = [Φ0,Φ⁺,Φ⁻] whereΦ⁺= Φ∩V⁺ andΦ⁻= Φ∩V⁻.

Let

R+(Φ) = [φ|φ∈Φ⁺]∪[−φ|φ∈Φ⁻].

By construction, the sequenceR+(Φ)is contained inV⁺.

Theorem 4.3. — Let v₁₂ =v(Φ₀, dw,c₁₂) be the polynomial function onW associated to the chamber c12ofΦ0. Then, ifhE,c1i>0,

(4.7) v(Φ, dx,c1)−v(Φ, dx,c2) = Pol(v12,ΦrΦ0, E).

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Remark 4.4. — We have

Pol(v12,ΦrΦ0, E) = (−1)^|Φ⁻^|Pol v12, R+(Φ), E .

Thus, by results of the preceding section, the difference of the volume functionsv(Φ, dx,c1)−v(Φ, dx,c2)coincides onV⁺, up to sign, with the convolution of the polynomial measure v12(w)dw associated to the chamber c₁₂ and with the Heaviside distributions associated to the vectors ψ∈R+(Φ). This is in the line of Paradan’s description of the jump formula for partition functions ([8], Theorem 5.2).

Proof. — Denote byLeq(Φ)the left hand side and byReq(Φ) the right hand side of Equation (4.7) above.

We will first verify the claim in the theorem when there is only one vector φof Φ that does not lie in W. We can suppose that Φ⁺ ={φ} and that hE, φi = 1. Then the chamber c₁ is equal to c₁₂×R>0φ, while c₂ is the exterior chamber. In this case,v(Φ, dx,c1)(w+tφ) =v(Φ0,(dw)^∗,c12)(w) = v12(w), whilev(Φ, dx,c2) = 0. The equation (4.7) follows from the first item of Lemma 3.6.

If not, letφbe a vector inΦthat does not lie inW. We may assume that φ∈V⁺. Then the sequenceΦ⁰= Φr{φ}will still spanV. The intersection ofΦ⁰ withW isΦ0. If c⁰₁and c⁰₂ are the chambers of Φ⁰ containingc1 and c₂ respectively, they are adjacent with respect toW. AsΦ⁰∩W = Φ₀, the polynomialv₁₂⁰ attached toc12andΦ⁰∩W is equal tov12. By Lemma 2.10, we have

∂(φ)(v(Φ,c1)−v(Φ,c2)) =v(Φ⁰,c⁰₁)−v(Φ⁰,c⁰₂).

By Proposition 3.4,

∂(φ) Pol(v12,ΦrΦ0, E) = Pol(v12,Φ⁰rΦ0, E).

By induction, we obtain∂(φ)(Leq(Φ)−Req(Φ)) = 0.

This equation holds for anyφ. So we conclude thatLeq(Φ)−Req(Φ) is a constant. However, both are homogeneous polynomials of degree d. So ifd >0, we obtain that Leq(Φ) = Req(Φ). If d= 0, this means that the systemΦconsists of linearly independent vectors, and both sides are equal by direct calculation. This establishes the theorem.

Consider a vector spaceV with basis{ei, i: 1, . . . , r}; we denote its dual basis by{eⁱ}. The set

Φ(Br) ={ei, 16i6r}∪{ei+ej, 16i < j6r}∪{ei−ej, 16i < j6r}

is the set of positive roots for the system of typeB_r and generatesV. We will denote a vector a ∈ V by a = Pr

i=1aiei; it lies in C(Φ(Br)) if and

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Figure 4.1. Chambers ofB₂

only ifa1+· · ·+ai>0for alli: 1, . . . , r. This will be our notation for this root system in subsequent examples.

Example 4.5. — We consider the root system of typeB2(see Figure 4.1) with Φ = {e1, e₂, e₁ −e₂, e₁ +e₂}. We will calculate v(Φ,c) for all the chambers using our formula in Theorem 4.3 iteratively starting from the exterior chamber.

(i) Jump from the exterior chamber toc1(W =Re2): In this caseE=e¹, Φ0={e2},Φ⁺ ={e1+e₂, e₁, e₁−e₂}andΦ⁻=∅.

v(Φ,c1)(a)−v(Φ,cext)(a) = Pol(1,ΦrΦ0, E)(a)

= Resz=0

e^ha,x+ze¹ⁱ Q

φ∈Φ⁺∪Φ⁻hφ, x+ze¹i

x=0

= Resz=0

e^a¹^(x¹^+z)+a²^x²

(x₁+z+x₂)(x₁+z)(x₁+z−x₂)

x=0

v(Φ,c1)(a) = Resz=0

e^a¹^z z³

= 1 2a²₁.

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(ii) Jump from c₁ toc₂ (W =R(e₁+e₂)): We have E=e¹−e², Φ0= {e1+e2},Φ⁺={e1, e1−e2} andΦ⁻={e2}.

v(Φ,c2)(a)−v(Φ,c1)(a) = Pol(1,ΦrΦ0, E)(a)

= Resz=0

e^a¹^(x¹^+z)+a²^(x²^−z) (x1+z)(x1−x2+ 2z)(x2−z)

x=0

=−Resz=0

e^(a¹^−a²^)z 2z³

=−1

4(a1−a2)². Using (i),v(Φ,c₂)(a) =¹₂a²₁−¹₄(a₁−a₂)²= ¹₄(a₁+a₂)²−¹₂a²₂.

(iii) Jump from c₂ to c₃ (W = Re₁): We have E = e², Φ0 = {e1}, Φ⁺={e2, e1+e2}andΦ⁻={e1−e2}.

v(Φ,c2)(a)−v(Φ,c3)(a) = Pol(1,ΦrΦ0, E)(a)

= Resz=0

e^a¹^x¹^+a²^(x²^+z)

(x2+z)(x1+x2+z)(x1−x2−z)

x=0

=−1 2a²₂.

Using (ii),v(Φ,c₃)(a) =¹₂a²₁−¹₄(a₁−a₂)²+¹₂a²₂=¹₄(a₁+a₂)².

5. Wall crossing formula for the partition function:

unimodular case

In this section, we compute the jumpk(Φ,c₁)−k(Φ,c₂)of the partition functionk(Φ)across a wall. In order to outline the main ideas in the proof, we will first consider the case whereΦis unimodular (see Remark 2.9 for the definition). We give two formulae. The first one is the convolution formula of Paradan ([8], Theorem 5.2). The second one is a one dimensional residue formula.

5.1. Discrete convolution

LetE∈U be a primitive element with respect toΓ^∗so thathE,Γi=Z. LetΨ be a sequence of vectors in Γ such that hψ, Ei 6= 0 for allψ ∈ Ψ.

ThusΨ = Ψ⁺∪Ψ⁻ withΨ⁺ = Ψ∩V⁺and Ψ⁻= Ψ∩V⁻. Define R+(Ψ) = [ψ|ψ∈Ψ⁺]∪[−ψ|ψ∈Ψ⁻].

Let

W :={a∈V | ha, Ei= 0},

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Γ0= Γ∩W.

We chooseF ∈Γsuch thathE, Fi= 1. We thus haveΓ = Γ0⊕ZF. LetΓ_>₀ be the set of elementsa∈Γsuch thatha, Ei>0. Let us define the functionK⁺(Ψ) onΓ_>0 such that we have, forx∈C(R+(Ψ))^∗,

(5.1) Y

ψ∈Ψ

1

1−e^−hψ,xi = X

a∈Γ_>0

K⁺(Ψ)(a)e^−ha,xi, that is we have written

1/(1−e^−ψ) =X

n>0

e^−nψ, ifψ∈Ψ⁺ and

1/(1−e^−ψ) =−e^ψ/(1−e^ψ) =−X

n>0

e^nψ ifψ∈Ψ⁻. Letκ−=P

ψ∈Ψ⁻ψso thathκ−, Ei=P

ψ∈Ψ⁻hψ, Eiis a strictly negative number if and onlyΨ⁻ is non empty. Then we have

K⁺(Ψ)(a) = (−1)^|Ψ⁻^|k(R₊(Ψ))(a−κ₋), wherek(R+(Ψ))is the partition function of the systemR+(Ψ).

The functionK⁺(Ψ)is supported on the pointed cone−κ−+C(R₊(Ψ)).

In particular the valueK⁺(Ψ)(0)is1ifΨ⁻is empty, or0ifΨ⁻is not empty.

Letqbe a polynomial function on Γ₀. Define fora∈Γ C(q,Ψ, E)(a) := X

w∈Γ0

q(w)K⁺(Ψ)(a−w).

The sum is over the finite setΓ0∩(a−C(R+(Ψ))).

Theorem 5.1. — AssumeΨ⁺is non empty. Assume that, for anyψ∈ Ψ, we havehψ, Ei=±1. Letq be a polynomial function onΓ0. Then, for a∈Γ_>0,

C(q,Ψ, E)(a) = Par(q,Ψ, E)(a).

Proof. — We need to compute, fora∈Γ_>0, S(a) := X

w∈Γ0

q(w)K⁺(Ψ)(a−w).

This sum is over a finite set.

LetQ∈S(U)be a polynomial function onV extendingq. We may write S(a) =

Q(∂x)· X

w∈Γ0

K⁺(Ψ)(a−w)e^hw,xi

x=0

.

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Define

Gx(a) = X

w∈Γ0

K⁺(Ψ)(a−w)e^−ha−w,xi.

ThenG_x(a)depends in an analytic way of the variablex∈U, and we have

(5.2) S(a) =

Q(∂x)·e^ha,xiGx(a)

|x=0. The functiona7→G_x(a) =P

w∈Γ0K⁺(Ψ)(a−w)e^−ha−w,xiis a function onΓ/Γ0=ZF. To identify the function Gx(nF), we compute its discrete Laplace transform in one variable. Letxbe inC(R₊(Ψ))^∗. Writeu=e^−z. We compute

Ldis(Gx)(u) =X

n>0

Gx(nF)uⁿ

=X

n>0

Gx(nF)e^−nz

=X

n,w

K⁺(Ψ)(nF−w)e^−hnF^−w,xie^−hnF,zEi

=X

a∈Γ

K⁺(Ψ)(a)e^−ha,xie^−ha,zEi

= 1

Q

= 1

Q

ψ∈Ψ(1−e^−hψ,xiu^hψ,Ei).

In the fourth equality, we have written any element a ∈ Γ_>0 as a = nF−w, with n >0 and w ∈ Γ0. The next equality is by the definition of the functionK⁺(Ψ)(a). Furthermore, we see that the sum is convergent when|u|<1.

As Ldis(Gx)(u) =P

n>0Gx(nF)uⁿ, Cauchy formula reads G_x(nF) = 1

2iπ Z

|u|=

u⁻ⁿL_dis(G_x)(u)du u

= 1 2iπ

Z

|u|=

u⁻ⁿ Q

ψ∈Ψ(1−e^−hψ,xiu^hψ,Ei) du

u. Thus we obtain for n=ha, Ei,

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S(a) =

Q(∂x)·e^ha,xi 1 2iπ

Z

|u|=

u⁻ⁿ Q

ψ∈Ψ(1−e^−hψ,xiu^hψ,Ei) du

u

x=0

= 1 2iπ

Z

|u|=

u⁻ⁿ

Q(∂_x)· e^ha,xi Q

ψ∈Ψ(1−e^−hψ,xiu^hψ,Ei)

x=0

du u . As at least one of thehψ, Eiis positive andn>0, it is easy to see that the function under the integrand has no pole atu=∞. Ashψ, Ei=±1, its poles are obtained foru= 0and u= 1. The integral on|u|=computes the residue atu= 0. We use the residue theorem so that−S(a) can also be computed as the residue foru= 1. We use the coordinateu=e^−znear u= 1, and we obtain

S(a) = Res_z=0

Q(∂_x)· e^ha,x+zEi Q

x=0

which establishes the formula in the theorem.

Finally, we compute the restriction toW ∩Γ ofPar(q,Ψ, E).

Lemma 5.2.

• If|Ψ⁻|=∅, then the restriction ofPar(q,Ψ, E)toW is equal toq.

• If|Ψ⁻|>0, then the restriction ofPar(q,Ψ, E)toW vanishes.

Proof. — The sum formula givesPar(q,Ψ, E)(w) =K⁺(Ψ)(0)q(w). Re- call that K⁺(Ψ)(0) vanishes as soon as |Ψ⁻| > 0; it is equal to 1 if

Ψ⁻=∅.

5.2. The jump for the partition function

LetΦbe a sequence of vectors spanning the latticeΓ. In this section we assume thatΦis unimodular and thatΓ =ZΦ.

Let k(Φ)(a) be the partition function. Then k(Φ)(a) coincides with a polynomial function on each chamber. We consider, as in Section 4.3, two adjacent chambersc1 andc2separated by a wallW. As before,Φ0 denotes W ∩Φ; it is also a unimodular system for the lattice Γ∩W. Let k₁₂ = k(Φ0,c12)be the polynomial function onW∩Γassociated to the chamber c₁₂ ofΦ₀. Consider the sequenceΨ = ΦrΦ₀. We chooseE∈U such that Ψ⁺ is non empty.

As the systemΦis assumed to be unimodular, the integersd_φ =hφ, Ei are equal to±1for anyφ∈Φnot inW.

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Theorem 5.3. — Letk₁₂ = k(Φ₀,c₁₂)be the polynomial function on W associated to the chamberc12. Then, ifhE,c1i>0, we have

(5.3) k(Φ,c1)−k(Φ,c2) = Par(k12,ΦrΦ0, E).

Remark 5.4. — By Theorem 5.1, the function Par(k12,ΦrΦ0, E) coincide, up to sign, on Γ_>₀ with the discrete convolution P

w∈W∩Γk₁₂(w) k(R+(Ψ))(a−κ₋−w)of the polynomial functionk12(w)onW by the partition function (shifted)k(R₊(Ψ)). Thus, our residue formula fork(Φ,c₁)−

k(Φ,c2)coincide with Paradan’s formula ([8], Theorem 5.2) for the jump of the partition function.

Proof. — Denote byLeq(Φ)the left hand side and byReq(Φ) the right hand side of Equation (5.3) above.

We check that D(φ)(Leq(Φ)−Req(Φ)) = 0for any φ∈Φ.

Let us first verify Equation (5.3) above when there is only one vector φ of Φ not in W. We can suppose that Ψ⁺ = {φ} and Ψ⁻ is empty. In this case the wallW is a facet of the coneC(Φ). The chamberc₁ is equal to c12×R>0φ, while c2 is the exterior chamber. It is easy to see that k(Φ,c₁)(w+tφ) =k₁₂(w), whereask(Φ,c₂) = 0. The equation follows from the first item of Lemma 3.6.

Suppose that this is not the case. Letφbe inΦ, and denoteΦ⁰ = Φr{φ}.

We study the difference equations satisfied byLeq(Φ)andReq(Φ). We have several cases to consider.

• φis not inW.

Then the sequenceΦ⁰ = Φr{φ} spansV and W is a wall for Φ⁰. The intersection ofΦ⁰withW isΦ0. Letc⁰₁andc⁰₂ be the chambers forΦ⁰containingc₁,c₂. Then, they are adjacent with respect toW. The chamberc12 remains the same. By Lemma 2.12, we have

D(φ)(k(Φ,c1)−k(Φ,c2)) =k(Φ⁰,c⁰₁)−k(Φ⁰,c⁰₂).

By Proposition 3.7,

D(φ) Par(k12,ΦrΦ0, E) = Par(k12,Φ⁰rΦ0, E).

By induction, we obtainD(φ)(Leq(Φ)−Req(Φ)) = 0.

• φis in W andΦ⁰₀= Φ0r{φ}spanW.

Then the sequenceΦ⁰ = Φr{φ} spansV, andW is a wall for the systemΦ⁰. Letc⁰₁ and c⁰₂ be the chambers forΦ⁰ containingc1,c2. Then, they are adjacent with respect to W. By Lemma 2.12, we have

D(φ)(k(Φ,c1)−k(Φ,c2)) =k(Φ⁰,c⁰₁)−k(Φ⁰,c⁰₂).