Residue formulae for Verlinde sums, and for

number of integral points in convex rational

polytopes

Congress of the European Women in Mathematics. Malta August 2001.Lectures by Michele Vergne

Notes by Sylvie Paycha

1

Introduction

These two lectures are dedicated to two topics and their contents are inde-pendent. In the first lecture on Bernoulli series and Verlinde sums, I presentsome algebraic formulae concerning a (fascinating) subject of which I amrather ignorant. I believe I am more knowledgeable on the second topic:volume and number of integral points of rational polytopes. In both topics,we shall encounter polynomial functions w(k) of k, where k is a non-negativeinteger, and the fact that these functions are polynomials is not obvious atall from the definition of w(k). There is a common geometric concept un-derlying this fact for both topics: compact symplectic manifolds. If (M,ω)is a compact symplectic manifold and if the form ω is integral, then we canassociate to M a quantized vector space Q(M,ω). The Riemann-Roch theo-rem asserts in particular that the dimension of Q(M,kω) is a polynomial ink with leading term kdim M/2 vol(M), where vol(M) is the symplectic volumeof M .

The underlying manifold M in the first topic is the moduli space of flatconnections on a Riemann surface with holonomy t around one hole. Verlindesums yield the dimension of Q(M,kω) while Bernoulli series compute thevolume of such manifolds.

Underlying manifolds in the second topic are toric manifolds. An integralpolytope P determines a toric manifold M of dimension 2 dimP together witha symplectic form ω on M . The volume of the polytope P is the volume ofM and the number of integral points in kP is the dimension of Q(M,kω). Itis a polynomial in k with leading term kdim P vol(P ).

In both cases, inspired by the Riemann-Roch theorem, it is possible todevelop a purely algebraic Riemann-Roch calculus, and to prove directlya beautiful relation between Bernoulli series and Verlinde sums, as well asbetween volumes of polytopes and number of integral points belonging to thepolytope. The main idea of this purely algebraic relation in the first case isan idea of A. Szenes while in the second case it goes back to G. Khovanskiiand A.V. Pukhlikov.

2

We have a sort of Riemann-Roch relation between two functions, onedepending on a continuous parameter t, the volume, and the other on a dis-crete parameter (the integer k), the dimension of a vector space. What aboutshowing directly the relation between these two quantities by ”computing”them explicitly? Thus my aim in both lectures is to give a hint of ”explicitresidue formulae” for all these quantities (Bernoulli series, Verlinde sums, vol-ume of polytopes, number of integral points in polytopes), formulae whichallows both concrete calculations, and quick proofs of the Riemann-Rochrelations.

Many thanks to Sylvie Paycha, for writing and expanding these notes, toVelleda Baldoni-Silva for illustrating them and to Charles Cochet, ShrawanKumar and Andras Szenes for improvements or corrections.

3

Lecture 1

Residue formulae for Bernoulli polynomials and

Verlinde sums

4

1 Introduction

In this lecture, I will present some simple sum formulae, which relate Bernoulliseries to Verlinde sums. I hope to show in particular that residue formulaefor series or sums are very efficient tools to relate these sums and calculatethem. The one dimensional case is presented here in detail and is alreadyvery amusing (and amazing).

These lectures are mainly based on the following articles by A. Szenes(all available on ArXiv):

–[Sz 1]: The combinatorics of the Verlinde formulae. Vector bundles inalgebraic geometry (Durham 1993), 241-253, London Math. Soc. LectureNote Ser., 208, Cambridge Univ. Press, Cambridge, 1995.

–[Sz 2]: Iterated residues and multiple Bernoulli polynomials. Internat.Math. Res. Notices 1998, 18, pp 937–956.

–[Sz 3]: A residue formula for rational trigonometric sums and Verlinde’sformula (math CO/0109038)

I do not explain here the underlying geometry. These lectures may (also)serve as an introduction to the algebraic aspects (multi-dimensional residueformulae) of articles of Jeffrey-Kirwan and Bismut-Labourie on the Verlindeformulae. References for the geometry underlying this calculation are givenat the end of this lecture.

2 Bernoulli’s theorem

We have all tried to work out formulae for sums of the m-th powers of thefirst k + 1 numbers, and to compare them to the corresponding integral∫ k

0xmdx = km+1

m+1.

For m = 0, we have:

00 + 10 + 20 + · · · + k0 = k + 1.

For m = 1, we have:

0 + 1 + 2 + 3 + · · · + k =k2 + k

2.

For m = 2, we have:

02 + 12 + 22 + 32 + · · · + k2 =k3

3+

k2

2+

k

6.

5

For m = 3, we have:

03 + 13 + 23 + 33 + · · · + k3 =k4

4+

k3

2+

k2

4.

· · ·

This polynomial behavior is a general feature of such sums as the followingtheorem shows:

Theorem 1 (Jakob BERNOULLI -(1654-1705))For any positive integer m, the sum Sm(k) =

∑ka=0 am of the mth-powers

of the first k + 1 integers is given by a polynomial formula in k.

We will prove this theorem and relate the sum Sm(k) to the Bernoullipolynomials.

Given a real number t, the Bernoulli polynomials B(m, t) are defined bythe following relation:

zetz

ez − 1=

∞∑

m=0

B(m, t)zm

m!.

This means, we expand z etz

ez−1in a Taylor series at z = 0. The coefficient

of zm

m!in this Taylor series depends polynomially on t, and is defined to be

the Bernoulli polynomial B(m, t). The first ones are

B(0, t) = 1,

B(1, t) = t −1

2,

B(2, t) = t2 − t +1

6,

B(3, t) = t3 −3

2t2 +

1

2t,

B(4, t) = t4 − 2t3 + t2 −1

30,

· · ·

6

It follows immediately from the definition of the Bernoulli polynomialsthat

d

dtB(p, t) = pB(p − 1, t),

∫ 1

0

B(p, t)dt = 0 if p > 1.

(Notice also that (−1)pB2p = (−1)pB(2p, 0) is positive.)The Bernoulli number Bm is defined to be B(m, 0). Bernoulli numbers

satisfy the following relation:

z

ez − 1=

∞∑

m=0

Bmzm

m!.

Since1

1 − e−z=

1

ez − 1+ 1,

we have∑∞

m=0 Bmzm

m!+ z =

∑∞m=0 Bm

(−z)m

m!so that the Bernoulli numbers

are equal to 0 for m odd, except when m = 1 in which case we have B1 = −12.

Now remark that

1

(m + 1)!(B(m + 1, t + 1) − B(m + 1, t))

is the coefficient of zm+1 in the Taylor series of

z(e(t+1)z

ez − 1−

etz

ez − 1) = z

etz(ez − 1)

ez − 1= zetz.

Thus we obtain:

B(m + 1, t + 1) − B(m + 1, t) = (m + 1)tm.

We write this equality for t = 0, 1, . . . , k:

B(m + 1, 1) − B(m + 1, 0) = (m + 1)0m

B(m + 1, 2) − B(m + 1, 1) = (m + 1)1m

· · ·

B(m + 1, k) − B(m + 1, k − 1) = (m + 1)(k − 1)m,

B(m + 1, k + 1) − B(m + 1, k) = (m + 1)km.

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Adding up these equations, we obtain

B(m + 1, k + 1) − B(m + 1, 0) = (m + 1)Sm(k).

As

ze(t+1)z

ez − 1= (−z)

e((−t)(−z))

e(−z) − 1,

it follows that B(m, t + 1) = (−1)mB(m,−t).Thus we obtain Bernoulli’s Theorem:

Proposition 2 The function k 7→ Sm(k) is given by the polynomial formulain k:

Sm(k) =1

m + 1((−1)m+1B(m + 1,−k) − Bm+1).

3 Bernoulli series and residues

In order to describe the Bernoulli polynomials in terms of series, it is use-ful to consider rational functions of the type φ(z) = E(z)

zp where E(z) is apolynomial. The sum over all non-zero integers n ∈ Z, n 6= 0:

B(φ)(t) =∑

n6=0

φ(2iπn)e2iπnt

converges absolutely at each real number t if p is a sufficiently large integersince

∑

n6=01

nα converges for α sufficiently large. It defines a periodic functionof t. It always converges as a generalized function of t, which is smooth whent /∈ Z. For 0 < t < 1 and p sufficiently large, the residue formula in the planeyields:

B(φ)(t) = residuex=0(φ(x)ext

1 − ex).

(For −1 < t < 0, the formula is B(φ)(t) = −residuex=0(φ(x) ext

1−e−x ).)In particular, t 7→ B(φ)(t) is given by a polynomial formula whenever

0 < t < 1.

From the definition of B(p, t), it follows that

B(p, t) = −p! residuex=0(x−p ext

1 − ex)

so that setting φ(z) = z−p, we obtain, for 0 < t < 1, the following formulafor the Bernoulli polynomial B(p, t):

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Proposition 3 Let 0 < t < 1. Then, we have:

B(p, t)

p!= −residuex=0(x

−p ext

1 − ex) = −

∑

n6=0

e2iπnt

(2iπn)p.

It is important to notice that the expression on the right hand side isperiodic with respect to t, while the left hand side is polynomial. We willcall the right hand side the Bernoulli series. The Bernoulli polynomial andthe Bernoulli series coincide ONLY on the interval 0 < t < 1. If p ≥ 2, thesum in the right hand side is absolutely convergent and the formula is validfor all 0 ≤ t ≤ 1.

In particular, for p = 2g and t = 0, we get:

B2g = −(2g)!∑

n6=0

1

(2iπn)2g

= 2(−1)g+1(2g)!(2π)−2gζ(2g)

where ζ(k) =∑∞

n=1 n−k is the zeta function at point k.For g = 1, as B2 = 1

6, we have

∞∑

n=1

1

n2=

π2

6.

4 Trigonometric sums

We now consider a closely related sum, which will be a special case of aVerlinde sum. Let p be an integer and let

F (z) =1

(1 − z)p.

Given a positive integer t, let us consider the expression:

W (p, t)(k) =∑

ωk=1, ω 6=1

ωtF (ω) =∑

1≤n≤(k−1)

e2iπnt/k

(1 − e2iπn/k)p.

It is very similar to the formula for 0 < t < k:

9

kp B(p, t/k)

p!= −

∑

n6=0

e2iπnt/k

(2iπn/k)p.

In fact, when k tends to ∞, it is not difficult to see that k−pW (p, t)(k)tends to 0 if p is odd, while if p is even, k−pW (p, t)(k) tends to −Bp

p!.

It is not at all obvious from its definition that the function k 7→ W (p, t)(k)is polynomial in k. We will prove it now.

Theorem 4 Assume that t and p are integers such that 0 ≤ t ≤ p, andk ≥ 1. Then the function k 7→ W (p, t)(k) is given by a polynomial formulain k. If p is even, or if p = 1, this polynomial is of degree p with highestdegree term −kp Bp

p!. If p is odd and p 6= 1, this polynomial is of degree less

or equal to p − 1.For all p ≥ 1, we have the residue formula:

W (p, t)(k) = −k residuex=0

(

etx

(1 − ex)p

1

(1 − e−kx)dx

)

.

Proof. Consider the 1-form k zt

(1−z)pzk

(1−zk)dzz. From the conditions 0 ≤ t ≤ p,

and k ≥ 1, it follows that this form has no residues at 0 and ∞. Poles of thefactor 1

(1−zk)arise at z = e2iπn/k (0 ≤ n ≤ (k − 1)). They are simple when

n 6= 0, and their residues add up to the sum −W (p, t)(k). As the sum ofresidues of this 1-form is equal to 0, we obtain from the residue theorem

W (p, t)(k) = k residuez=1

(

zt

(1 − z)p

zk

(1 − zk)

dz

z

)

.

A change of variable z = ex in the residue yields:

W (p, t)(k) = −k residuex=0

(

etx

(1 − ex)p

1

(1 − e−kx)dx

)

.

This last expression depends on k via the Laurent expression of k1−e−kx . Recall

from the definition of the Bernoulli numbers that

kx

ekx − 1=

∞∑

r=0

Br(kx)r

r!.

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Hence

W (p, t)(k) = −∞

∑

r=0

BrRr(p, t)kr

r!

where

Rr(p, t) = (−1)rresiduex=0

(

etx

(1 − ex)pxr−1dx

)

so that Rr(p, t) vanishes for large r. In fact, as 1(1−ex)p has a pole at 0 of order

p, we see that Rr(p, t) vanishes for r > p, so that W (p, t)(k) is a polynomialof degree less or equal than p. More precisely, it is of degree p for even p andof degree less or equal to p − 1 for odd p, p 6= 1. For p even or p = 1, thehighest degree term is −Bp

kp

p!. If p is odd, p 6= 1, the term of degree p in k is

equal to 0 as Bp = 0, while the term of degree (p−1) is −(p/2− t)Bp−1kp−1

(p−1)!.

5 A relation between Bernoulli series and trigono-

metric sums

Define:

V (q, k) =k−1∑

n=1

1

4q(sin(πn/k))2q.

As we will see later on, V (q, k) is a special case of a Verlinde sum.We have V (q, k) = (−1)qW (2q, q)(k). Indeed

(−1)qW (2q, q)(k) = (−1)q∑

1≤n≤(k−1)

e2iπnq/k

(1 − e2iπn/k)2q

=k−1∑

n=1

1

(1 − e2iπn/k)q(1 − e−2iπn/k)q

=k−1∑

n=1

1

4q(sin(πn/k))2q.

Thus, from Theorem 4, we obtain that k 7→ V (q, k) is a polynomial in k,of degree 2q. Notice that V (q, k) is a sum of positive real numbers so thatV (q, k) is positive.

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The polynomial k 7→ V (q, k) is of degree 2q and its highest degree termis

(−1)q+1 B2q

(2q)!k2q.

(In our conventions for Bernoulli numbers, (−1)q+1B2q is positive.)There is a more precise relation between the polynomial function t 7→

B(2q, t) and the polynomial function k 7→ V (q, k). Consider the Taylorseries at the origin of the function of x

A(q, x) : =(x/2)2q

(sinh(x/2))2q

= 1 −q

12x2 + (

q

1440+

q2

288)x4 + · · · .

Substitute x = ∂∂t

in A(q, x), and consider A(q, ∂) as a series of differential

operators in powers of ∂ := ∂∂t

. The action of A(q, ∂) on a polynomial functionof t is well defined.

Theorem 5

V (q, k) = (−1)(q+1) k2q

2q!(A(q, ∂/k) · B(2q, t))|t=0.

From this expression, we see again that the highest degree term of thepolynomial function k 7→ V (q, k) is (−1)q+1B2q

k2q

2q!.

(Remark. In the context of the Verlinde formula, this theorem is closelyrelated to the Riemann-Roch theorem on the manifold Mg := M(SU(2), g)of flat connections on vector bundles of rank 2 on a Riemann surface ofgenus g. This manifold is provided with a line bundle L. The above expres-sion arises when calculating the integral

∫

Mgch(Lk−2)A(Mg) (where ch is the

Chern character and A(Mg) is the A genus). This evaluates the dimension ofthe space of so-called generalized theta functions: the dimension of the spaceof holomorphic sections of the holomorphic line bundle Lk−2 over Mg. Weshall come back to this analogy later.)Proof. How to prove this theorem: Consider the residue formula for theBernoulli polynomial

B(2q, t) = −(2q!) residuex=0

(

x−2q ext

(1 − ex)

)

.

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We can apply the series A(q, ∂/k) to this expression. Under the residue,any analytic function is automatically replaced by its Taylor series. Thus weobtain

(−1)q+1 k2q

2q!A(q, ∂/k)B(2q, t) = (−1)qk2qresiduex=0

(

A(q, x/k)x−2q ext

(1 − ex)

)

= residuex=0

(

1

(1 − ex/k)q(1 − e−x/k)q

ext

(1 − ex)

)

.

Thus at t = 0, we obtain

(−1)q+1 k2q

2q!

(

A(q, ∂/k)B(2q, t))

|t=0

= residuex=0

(

1

(1 − ex/k)q(1 − e−x/k)q

1

(1 − ex)

)

.

The change of variables x 7→ −kx leads to

(−1)q+1 k2q

2q!

(

A(q, ∂/k)B(2q, t))

|t=0

= −k residuex=0

(

1

(1 − ex)q(1 − e−x)q

1

(1 − e−kx)

)

= −(−1)q k residuex=0

(

eqx

(1 − ex)2q

1

(1 − e−kx)

)

.

We recognize here the residue expression given in Theorem 4 for (−1)qW (2q, q)(k) =V (q, k). Thus we obtain

(−1)q+1k2q

2q!

(

A(q, ∂/k)B(2q, t))

|t=0 = V (q, k)

=k−1∑

n=1

1

(1 − e2iπn/k)q(1 − e−2iπn/k)q.

This proves Theorem 5.

It is amusing to give a false proof of this theorem, by interverting differ-entiation and summations:

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Indeed applying formally A(q, ∂/k) to the sum −∑

n6=0e2iπnt

(2iπn)2q expressing

the Bernoulli polynomial 12q!

B(2q, t), we would obtain:

(−1)(q+1) k2q

2q!

(

A(q, ∂/k)B(2q, t))

|t=0 = (−1)qk2q∑

n6=0

A(q, 2iπn/k)1

(2iπn)2q

=∑

n6=0

1

(1 − e2iπn/k)q(1 − e−2iπn/k)q.

This last expression is highly divergent, for at least two reasons: first, forn 6= 0 multiple of k, the term to add to the sum is equal to ∞, second, all theterms in the arithmetic progression n+kj give the same summand. However,it gives a hint of why this operator A(q, ∂/k) occurs in the comparison. The”renormalized” sum consists in restricting the sum to 0 < n < k, a set ofrepresentatives of the non-zero elements of Z/kZ.

The function V (q, k) has some remarkable integral property (which fol-lows from the representation theory of SL(2, C)). Indeed, the function

Ver(q, k) := (2(k + 2))qV (q, k + 2) = 2−q(k + 2)q

k∑

n=0

1

sin((n + 1)π/(k + 2))2q

takes positive integral values on integers k. Notice that for k = 0, we haveVer(q, 0) = 1.

We have

Ver(1, k) =1

6(k + 1)(k + 2)(k + 3),

Ver(2, k) =1

180(k + 2)2(k + 3)(k + 1)(k2 + 4k + 15),

Ver(3, k) =1

7560(k + 2)3(k + 1)(k + 3)(2k4 + 16k3 + 71k2 + 156k + 315),

. . .

You can indeed verify on a few numbers k the amazing fact that thesefunctions take integral values on integers.

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6 Preliminaries on semi-simple Lie algebras

The Verlinde sums we saw corresponded to the Lie group SL(2, C) withLie algebra sl(2) and compact form SU(2). Before introducing more generalVerlinde sums, we need to recall some basic facts on the representation theoryof semi-simple Lie algebras.

A Lie algebra g over a field C is called semi-simple if its Cartan-Killingform 〈·, ·〉 : g × g → C defined by

〈X,Y 〉 = tr(ad(X)ad(Y ))

is non-degenerate. Any semi-simple Lie algebra is the direct product of simpleLie algebras (a Lie algebra g is called simple if it has no proper ideals). Anexample of simple Lie algebra is the algebra sl(n) of all n× n matrices withtrace equal to 0. Then, up to normalization, the Killing form is 〈X,Y 〉 =tr(XY ) where tr is the ordinary trace.

In particular sl(2) := {

(

x1 x2

x3 −x1

)

| x1, x2, x3 ∈ C} is a simple Lie alge-

bra.In order to describe the finite dimensional representations of g, we need

to introduce a few preliminary definitions.A Cartan subalgebra of a semi-simple Lie algebra g is a subalgebra h of

g such that:i) h is abelian and every element X ∈ h is such that the transformation

ad(X) is diagonalizable,ii) h is its own normalizer in g i.e. {X ∈ g| [X, h] ⊂ h} = h.Such an algebra h is a maximal abelian subalgebra of g and is unique up

to conjugacy. The dimension r of h is called the rank of g. In the case ofsl(n), the algebra h is the set of diagonal matrices (with zero trace). Therank of sl(n) is n − 1.

For sl(2), we write an element t of h as

t =

(

t1 00 −t1

)

.

The restriction of 〈·, ·〉 to h × h is non-degenerate. We can then identifyh and its dual h∗.

A root is an element α ∈ h∗, α 6= 0, such that the corresponding rootspace

gα := {X ∈ g| [H,X] = α(H)X for all H ∈ h}

15

is non-zero.Let ∆ := {α ∈ h∗| α is a root}. A positive system of roots is a set

∆+ ⊂ ∆ such that

∆+ ∩ (−∆+) = ∅,

∆+ ∪ (−∆+) = ∆,

and such that for any α, β ∈ ∆+, α + β ∈ ∆ ⇒ α + β ∈ ∆+.A simple system of roots is a set S ⊂ ∆+, S = {α1, . . . , αr}, such that

given α ∈ ∆+, there are uniquely defined non-negative integers mi (i =1, . . . , r) such that α = m1α1 + · · · + mrαr.

Let S = {α1, . . . , αr} be a simple system of roots; we set Hi to bethe unique element of h such that λ(Hi) = 2

〈αi,αi〉〈λ, αi〉, for any λ ∈ h∗.

This element is well defined, since 〈·, ·〉h×h is non-degenerate. Thus we haveαi(Hi) = 2.

We denote by hR the real vector space of h spanned by the elements Hi

of the complex Cartan subalgebra h. An element λ ∈ h∗ is called an integralweight if λ(Hi) is an integer for all 1 ≤ i ≤ r. It is called dominant if λ(Hi)is real and non-negative for all 1 ≤ i ≤ r.

We denote by P ⊂ h∗R

the lattice of integral weights. We denote by Qits dual lattice in hR: the lattice Q is exactly the set of elements t ∈ hR,where all integral weights take integral values. A weight λ is called regular,if 〈λ, α〉 6= 0 for all α ∈ ∆. We denote by Preg ⊂ h∗

Rthe set of regular integral

weights.The set

Γ := {m1α1 + · · · + mrαr| mi non − negative integers}

induces a partial ordering on the weights: λ ≤ λ′ whenever λ′ − λ ∈ Γ.Given a finite-dimensional representation U of g in a complex vector space

V , a weight λ of U is an element λ ∈ h∗ such that

Vλ := {v ∈ V | U(H)v = λ(H)v for all H ∈ h}

is not reduced to zero. We have V =∑

λ∈h∗ Vλ. All weights of a finite-dimensional representation are integral weights.

If V is an irreducible finite-dimensional representation of g, then V has aunique highest weight λ (all other weights λ′ of the representation V satisfy

16

λ′ < λ). This weight λ is an integral and dominant weight. Reciprocally,consider the dominant cone

D := {λ ∈ h∗R| λ(Hi) non − negative integer for 1 ≤ i ≤ r}

of all dominant integral weights. Given λ ∈ D, there is exactly one equiva-lence class of finite dimensional irreducible representations of g admitting λas its highest weight. Let Uλ be a representative of this class. In other words,the representations Uλ, λ ∈ D, exhaust all the irreducible representations offinite dimension of g up to equivalence.

7 Witten series

One interesting generalization of the Bernoulli series are the Witten seriesB(p, g)(t). Here t is an element of hR, and

B(p, g)(t) =∑

λ∈Preg

e(2iπλ,t)

(∏

α∈∆+ 2iπ〈α, λ〉)p.

This series converges for p sufficiently large. For any p, it is well definedas a generalized function of t. The function of t ∈ hR defined by this seriesis periodic with respect to the lattice Q. On hR/Q (represented by a domainin hR), the expression B(p, g)(t) above is polynomial in sectors delimited byhyperplanes. On each sector, these series can in fact be expressed in termsof the Bernoulli polynomials in one variable.

In the case of sl(2) one recovers:

B(p, t1) = −B(p, sl(2))(t) = −∑

n6=0

e2iπt1n

(2iπn)p.

Here

t ∈ hR =

(

t1 00 −t1

)

with t1 ∈]0, 1[.Similarly to the case of sl(2), a residue formula due to A. Szenes ([Sz1])

can be given for these infinite sums, a formula which allows to calculate themeffectively and to prove their polynomial behavior in sectors.

17

Consider the example of sl(3): we get

B(q, sl(3))(t1, t2) =∑

n1 6=0,n2 6=0,n1+n2 6=0

e2iπ(t1n1+(t1+t2)n2)

(2iπn1)q(2iπn2)q(2iπ(n1 + n2))q.

Here the point (t1, t2) ∈ R2 represents the diagonal matrice in hR

t1 0 00 t2 00 0 −(t1 + t2)

.

Changing n1 to n + m, and n2 to −m this is also equal to

(−1)q∑

n6=0,m6=0,n+m6=0

e2iπ(t1n−t2m)

(2iπn)q(2iπm)q(2iπ(n + m))q,

The residue formula depends on the position of (t1, t2) in sectors (as in the

one dimensional case, the residue formula for the similar sum∑

n6=0e2iπnt

(2iπn)p

was only valid for 0 < t < 1). It reads

(−1)qB(q, su(3))(t1, t2) =

−residuex=0

(

residuey=0e{t1}x−{t2}y

xqyq(x + y)q

1

(1 − ex)(1 − e−y)

)

+residuex=0

(

residuey=0e{t1+t2}x+{t2}y

xqyq(x + y)q

1

(1 − ex)(1 − ey)

)

.

Here {t} denotes t − [t] where [t] is the integral part of t.For example, we have B(2, su(3))(0, 0) = − 1

30240.

Very naively, looking at the sum of residues at the points (x = 2iπn, y =2iπm), the first iterated residue: −residuex=0(residuey=0·) should alreadylead to

∑

n6=0,m6=0,n+m6=0

e2iπ(t1n−t2m)

(2iπn)q(2iπm)q(2iπ(n + m))q

and the second residue should lead to the sum

∑

n6=0,m6=0,n+m6=0

e2iπ(t1+t2)n+(t2)(n+m)

(2iπn)q(2iπm)q(2iπ(n + m))q

18

which is equal, as seen from the first formula for B(q, sl(3))(t1, t2), afterchanging n1 and n2.

But, in fact the two iterated residues are not equal, and Szenes formulashows that the correct answer is the sum of both iterated residues. Further-more we must be careful with the order in which we take residues.

( Remark: Let G be the compact simply connected group whose Lie algebrais a compact form of g. Up to some normalization, for p = 2g−1, the Wittenseries computes the symplectic volume of the manifold M(G, g, t) of modulispace of flat connections on a G-bundle on a Riemann surface of genus gwith one hole (the variable t ∈ hR parametrizes the holonomy of the flatconnection around the hole). We describe this manifold in Section 9, andgive some references for its geometric meaning.)

8 Verlinde sums

One interesting generalization of the sum

V (q, k) =k−1∑

n=1

1

4q(sin(πn/k))2q

considered in Section 5 is the Verlinde sum, that we are going to describe.Let g be a simple Lie algebra of rank r. Let S be a simple system of

roots, ∆+ the corresponding positive system of roots and D the associatedcone of dominant integral weights. Let ρ = 1

2

∑

α∈∆+ α.Let θ be the highest root. Let us also introduce the set (called an alcove):

Ak :=

{

λ ∈ D| 2〈λ, θ〉

〈θ, θ〉≤ k

}

(the definition of Ak does not depend of the choice of the scalar product.)

Let h be the dual Coxeter number h = 2 〈ρ,θ〉〈θ,θ〉

+ 1. When g = sl(n), thenh = n.

Consider the scalar product < ·, · > such that < θ, θ >= 2.Let q be an integer. The Verlinde sum is

19

Ver(q, g)(k) = cqG(k + h)rq

∑

u∈Ak

1∏

α∈∆+(1 − e2iπ<α,u+ρ>

k+h )q(1 − e−2iπ<α,u+ρ>

k+h )q

= cqG(k + h)rq

∑

u∈Ak

1∏

α∈∆+(2 sin(π<α,u+ρ>k+h

))2q.

Here cG is an explicit constant depending on G. It is such that Ver(1, g)(0) =1.

When g = sl(2), Ver(q, g)(k) is the polynomial Ver(q, k) we considered inSection 4. Similarly to this case, there is a residue formula (due to Szenes)for the Verlinde sum, which allows to show its polynomial behavior in k andto compare it to the Witten series.

For g = sl(3), the above sum is

Ver(q, sl(3))(k) = 3q(k + 3)2q

×∑

n1≥0,n2≥0,n1+n2≤k

(

8 sin(π(n1 + 1)

k + 3) sin(

π(n2 + 1)

k + 3) sin(

π(n1 + n2 + 2)

k + 3)

)−2q

.

It is known (and amazing) that the function k 7→Ver(q, g)(k) takes in-tegral values on integers: it is the dimension of a vector space V arisingas the space of generalized theta functions which is the space of holomor-phic sections of the holomorphic line bundle L over the manifold M(G, g) =M(G, g, 0)) of a Riemann surface of genus g = q +1 with central charge k. Itwas proved independently by Beauville-Laszlo, Faltings, Kumar-Narasimhan-Ramanathan that the space V is isomorphic to the space L of conformalblocks extensively studied by Tsuchiya-Ueno-Yamada. From their factoriza-tion theorem, one obtains the Verlinde formula giving the dimension of thespace V as Ver(q, g)(k).

There is a relation between the Witten series and the Verlinde sum: theVerlinde sum is obtained from the Witten series by applying a series of differ-ential operators A(q, ∂/(k+h)) to the Witten series. The form of the wantedoperator A(q, ∂/(k + h)) can be guessed from the same intuitive argumentof “differentiating” under the sum sign. The correct argument follows imme-diately from the explicit residue formula. It allows to compute the integral∫

M(G,g)ch(Lk)A(M(G, g)) (ch is the Chern character and A(M(G, g)) is the

A genus), at least when k is sufficiently large.

20

It is known that there exists an integer d > 0 such that the functionk 7→ Ver(q, g)(dk) is polynomial in k. What is not known is the value of thesmallest possible d. It is known that d = 1 for g = sl(n).

The following conjecture on d is natural in view of Kumar-Narasimhanwork on the Picard group of M(g,G, 0) (Math. Ann. 308 (1997) 155-173):

d = 1 for Cr,

d = 2 for Br(r ≥ 3), Dr(r ≥ 4), G2,

d = 6 for F4, E6,

d = 12 for E7,

d = 60 for E8,

where Cr, Br, Dr are the series of classical simple Lie algebras and G2, F4, E6, E7, E8

are the exceptional Lie algebras. (From the above cited work of Kumar-Narasimhan, it follows that d is smaller or equal to these values).

21

9 Geometry beyond. Very few references

The underlying geometric object to the theory of Witten series and Verlindesums is the manifold M(G, g, t). Here G is a compact (simply connected)Lie group with Lie algebra g, g is a positive integer and t is an elementof the Cartan subalgebra hR of the complex semi-simple Lie algebra gC. Ifu1, u2 ∈ G, we denote [u1, u2] = u1u2u

−11 u−1

2 .The manifold M(G, g, t) is defined to be:

M(G, g, t) = {(u1, v1, . . . , ug, vg)| ui, vi ∈ G such that

g∏

i=1

[ui, vi] = e2iπt}/T.

Here T denotes the maximal compact torus of G with Lie algebra ihR. Thenotation /T means that we identify the (2g)-tuples (u1, v1, . . . , ug, vg) and(u′

1, v′1, . . . , u

′g, v

′g) if there exists t ∈ T such that u′

i = tuit−1 and v′

i = tvit−1.

• The main reference on the geometric properties of the manifold M(G, g, t)for G = Un is:

M. Atiyah and R. Bott: The Yang-Mills equation on a Riemann surface.Phil. Trans. R. Soc. A 1982 308.

• The computation of the volume of M(G, g, t) when G = SU(2) is dueto M. Thaddeus, who related volumes and Bernoulli numbers.

M. Thaddeus: Conformal field theory and the cohomology of the mod-uli space of stable bundles. J. Differential Geom. 1992, pp 131-149.

In the general case, the volume of M(G, g, t) is determined in the formof Witten series in:

E. Witten: On quantum gauge theories in two dimensions. Comm.Math. Phys. 1991, 141, pp 153-209.

• A simple description of the manifold M(G, g, t) together with the com-putation of its symplectic form is in

–A. Alekseev, A. Malkin, E. Meinrenken: Lie group-valued momentmaps. J. Differential Geom 1998, 48, pp 445-495

A quick computation of Witten formulae for its symplectic volume isin:

–A. Alekseev, E. Meinrenken and C. Woodward: Duistermaat-Heckmandistributions for group valued moment maps. ArXiv math DG/9903087.

22

• The Verlinde formula was conjectured by:

E. Verlinde: Fusion rules and modular transformations in 2D-conformalfield theory. Nuclear Physics B 1988 300, pp 360-376.

• It was proved using fusion rules, by

– A. Beauville and Y. Laszlo (for SL(n)): Conformal blocks and gener-alized theta functions. Comm. Math. Phys. 1994 164, pp 385–419.

– G. Faltings: A proof of the Verlinde formula. J. of Alg. Geometry.1994, 3, pp 347–374.

– S. Kumar, M.S. Narasimhan and A. Ramanathan: Infinite Grass-mannian and moduli spaces of G-bundles. Math. Annal. 1994, 300,pp 41–75.

–Tsuchiya-Ueno-Yamada: Conformal field theory on universal familyof stable curves with gauge symmetry. Adv. Studies in Pure Math.1989, 19, pp 459–566.

• Results on computation of intersection numbers (and thus the Riemann-Roch formula) are obtained via multi-dimensional residues in:

–Lisa Jeffrey and Frances Kirwan (for SL(n)): Intersection theory onmoduli spaces of moduli spaces of holomorphic vector bundles of ar-bitrary rank on a Riemann surface. Ann. of Math. 1998, 148, pp109–196.

• A proof of the Verlinde formula for general compact simply connectedgroups and any number of holes (but with restrictions on k) is obtainedvia the Riemann-Roch theorem and multi-dimensional residue calculusin:

–Jean-Michel Bismut and Francois Labourie. Symplectic geometry andthe Verlinde formulae. Surveys in differential geometry: differentialgeometry inspired by string theory pp 97-311, Surv. Differ. Geom., 5,Int. Press, Boston, MA , 1999.

• Another approach, which handles the general case, can be found in

– A. Alekseev, E. Meinrenken and C. Woodward: Formulae of Verlindetype for non-simply-connected groups. (ArXiv SG/0005047).

23

Lecture 2

Residue formulae for volumes and number of

integral points in convex rational polytopes

1

Introduction

As second topic, I shall present here some recent results on the number ofpoints with integral coordinates in convex rational polytopes. My own inter-est in this topic comes from my efforts to understand the relations betweensymplectic manifolds and group representations, the existence of such rela-tions being the Credo of quantum mechanics. Quantum mechanics enablesus to associate discrete quantities to some geometric objects. For example,the volume of a compact symplectic manifold (M,ω), with an integral closednon-degenerate 2-form ω, has a discrete analogue which is the dimension ofthe vector space given by the quantum model Q(M,ω) for M . It is importantto understand the relation between both quantities. The dimension q(k) ofthe vector space Q(M,kω) is a polynomial in k, and the volume of the man-ifold M is the limit of k− dim M/2q(k) when k tends to ∞. The full expressionfor the dimension of Q(M,kω) is the content of the Riemann-Roch theorem,which expresses this integer q(k) as an integral over M . A similar comparisonproblem is the following: if P ⊂ R

n is a convex polytope with integral (orrational) vertices, can we compare the number |P ∩ Z

n| of points in P withintegral coordinates and the volume of P ? It is clear that the volume of Pis obtained as the limit when k tends to ∞ of k−n|kP ∩ Z

n|. Can we givemore precise relations ?

There is a dictionary between polytopes and some classes of compactsymplectic manifolds (toric manifolds). This dictionary inspired the formu-lation of several results, starting with the fascinating formula of Khovanskii-Pukhlikov (see Theorem ??). The point of view of these lectures will be al-gebraic and based on generating functions combined with an algebraic recipedue to Jeffrey-Kirwan ([?]) for the inversion of Laplace transforms. We shallonly give some references on the correspondence between polytopes and sym-plectic geometry in the last section.

I hope to show, in this lecture, that calculating the volume of a convexrational polytope or calculating the number of points with integral coor-dinates inside this polytope are similar problems (both difficult). Convexpolytopes arise in many fields of mathematics: symplectic geometry, repre-sentation theory, algebraic geometry. Computing volumes is important. Itis also important to study integral points in rational polytopes: they are theintegral solutions of systems of linear inequations with integral coefficients.For example, solving the equation 5x + 10y + 20z = 105 with x, y, z positiveintegral numbers is an equation that we see everyday, when buying an item

2

of value 105 cents, with coins of 5 cents, 10 cents and 20 cents (the numberof solutions is 36). Similarly regulating flows in networks is reduced to linearinequalities. Notice that already the problem of finding if there exists anintegral point in a rational polytope is non-trivial (a polytope is called ratio-nal if its vertices have rational coordinates). H. Lenstra ([?]) showed in 1983that there is, for a fixed dimension n, a polynomial time algorithm checkingif a rational polytope P in R

n contains an integral point: |P ∩ Zn| 6= ∅, and

A. Barvinok ([?]) showed in 1994 that there is, for a fixed dimension n, apolynomial time algorithm giving the number |P ∩ Z

n| of integral points inP .

I shall explain here explicit and very similar formulae for both the volumeand the number of integral points in rational convex polytopes. The formulafor the volume is deduced in a straightforward way from the inversion for-mula of Jeffrey-Kirwan ([?]) for the Laplace transform. The formula for thenumber of points given in ([?]) follows from a multidimensional residue the-orem, deduced from the one dimensional residue theorem with the help of aresult due to A. Szenes ([?]) on separation of variables. From these formulae,the relations obtained by Khovanskii-Pukhlikov ([?]) and more generally byBrion-Vergne ([?]), Cappell-Shaneson ([?]), Guillemin ([?]) between volumesand the number of integral points will become clear. The residue formulaefor volume and number of points have a theoretical interest: for example, theperiodic-polynomial dependence of the formula in terms of the inequalitiesdefining the polytope is clear from the given formulae. They can also be usedfor the computation of these quantities, at least for network polytopes. Thisis work in progress with Velleda Baldoni and Jesus de Loera.

3

a

b

c

0

Figure 1: Tetrahedron

1 Definition of the Ehrhart polynomial

By definition, a convex polytope in Rn is the convex hull of a finite number

of points in Rn, or equivalently a compact set of R

n defined by linear inequa-tions. A third way of representing a convex polytopes is as a set of positivesolutions to linear equations: i.e. the first positive quadrant intersected witha linear space.

Example. The Tetrahedron. The convex hull in R3 of the points

0 = (0, 0, 0), A = (a, 0, 0), B = (0, b, 0), C = (0, 0, c) is also described by theinequations

x ≥ 0, y ≥ 0, z ≥ 0 andx

a+

y

b+

z

c≤ 1

or is also clearly isomorphic to the convex polytope in R4 described by

x ≥ 0, y ≥ 0, z ≥ 0, h ≥ 0 andx

a+

y

b+

z

c+ h = 1.

A convex polytope is called integral if its vertices have integral coordi-nates.

A convex polytope is called rational if its vertices have rational coordi-nates.

Our topic of discussion here is the calculation of the volume of convexpolytopes, together with the calculation of the number of points with integralcoordinates in an integral convex polytope P .

4

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

|(k∆ ∩ Z2)| = (k+1)(k+2)

2

k

k

0 0

k = 8

Volume(k∆) = k2/2

Let us start with the example of the standard simplex ∆ in Rn, with

vertices 0, e1, e2, . . . , en. The polytope k∆ is defined by the inequations

k∆ = {x1 ≥ 0, . . . , xn ≥ 0| x1 + x2 + · · · + xn ≤ k}.

It is not difficult to show that the volume of k∆ is

vol(k∆) =kn

n!.

The number of integral points in k∆ is given (as we shall see shortly) by

pn(k) =(k + 1)(k + 2) · · · (k + n)

n!.

We refer to Section ?? for the relation between points in k∆ and a basisof vector spaces V (k) attached to P .

Thus we see that the discrete analogue to the monomial xn

n!giving the

volume of k∆ for x = k is the polynomial pn(x) = (x+1)(x+2)···(x+n)n!

whichgives the number of integral points in k∆ for x = k. This polynomial pn(x)has same leading term xn

n!, but it has the advantage that it takes integral

values on all integers, and any polynomial which takes integral values on allintegers is a linear combination with integral coefficients of such polynomialspn(x).

5

More generally, to any integral convex polytope P is associated a poly-nomial function: the Ehrhart polynomial ([?],[?],[?]).

Theorem 1 (The Ehrhart theorem). Let P ⊂ Rn be an integral convex

polytope with non-empty interior. Then the function on N defined by

iP (k) = cardinal(kP ∩ Zn), k = 0, 1, 2, . . .

is given by a polynomial expression in k called the Ehrhart polynomial. The

first two leading terms of this polynomial are vol(P )kn+ 12volZ(δP )kn−1+· · · .

The constant term of the Ehrhart polynomial iP (k) is equal to 1.

Here volZ(δP ) is the sum of the volumes of the faces of the boundary δPof the polytope P . The volume of a face F of the boundary δP is computedusing a normalized measure built from the Lebesgue measure on the affinespace AF spanned by the face. The normalization is done in such a way thatthe fundamental domain of the lattice Z

n ∩ AF has volume 1.The Ehrhart function iP (k) is our second example of a function of k, the

polynomial feature of which does not follow evidently from its definition, thefirst example (given in Lecture 1) being the Verlinde sums k 7→ Ver(q, k). Af-ter some further comments, we shall give here the proof of Ehrhart’s theorem,following Ehrhart.

The following theorem directly generalizes Bernoulli Theorem on sums ofthe m-th powers of the first k + 1 numbers (see Lecture 1, Section 1).

Theorem 2 Let P ⊂ Rn be an integral convex polytope with non-empty

interior. Let f be a polynomial function on Rn homogeneous of degree m.

Then

k 7→∑

ξ∈kP∩Zn

f(ξ)

is a polynomial function of k. The leading term of this polynomial is kn+m∫

Pf(x)dx.

We also note here the important reciprocity law for the Ehrhart polyno-mial.

Theorem 3 (Reciprocity law). Let P ⊂ Rn be an integral convex polytope

with non-empty interior. Let P 0 be the interior of P . Define

iP 0(k) = cardinal(kP 0 ∩ Zn)

to be the number of integral elements in the interior of kP . Then

iP 0(k) = (−1)niP (−k).

6

k

k

k

|(0, k) ∩ Z| = k − 1

|[0, k] ∩ Z| = k + 1

lenght [0, k] = k

0

0

0

Figure 2: Reciprocity law for the interval [0,k]

The obvious example of the reciprocity law is when n = 1 and P = [0, 1]is the unit interval. Then iP (k), the number of integral points in [0, k], is(k +1) and the number iP 0(k) is the number of integers strictly greater than0 and strictly less than k. We have

iP 0 = (k − 1) = −iP (−k).

Ehrhart’s theorem stating that the number of integral points in the dilatedpolytope kP is a polynomial in k, with leading term kn vol(P ) is far frombeing obvious.

Let us give an example involving dilatations in just one direction, wherewe shall see that even the asymptotics between the volume and the numberof integral points is only true under dilations in all directions.

Example: The hanging pyramid.

Let m ∈ R+. The hanging pyramid Pm is the convex hull in R

3 of thevertices s0 = (0, 0, 0), s1 = (1, 0, 0), s2 = (0, 1, 0), and s3 = (1, 1,m).

Its volume depends on m:

vol(Pm) = m/6.

Assume that m is an integer, then Pm is an integral convex polytope, butthere are no integral points in Pm other than its 4 vertices. Whatever the

7

Vol(Pm) = m/6 |Pm ∩ Z3| = 4

Figure 3: Hanging pyramid

value of m might be:i(Pm) = 4

where we have set i(P ) := iP (1) taking k = 1. Indeed, as the full pyramidprojects on the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, any integral point in Pm

is above one of the 4 points (0, 0), (0, 1), (1, 0), (1, 1), thus is one of the 4vertices. The asymptotics in m of the volume and the number of integralpoints are therefore very different in this example.

Before going to the proof of Ehrhart’s theorem, let us have a look at theformulae relating the number of points in an integral convex polytope andits volume in dimension 1 and 2:

In dimension 1: Let us consider the polytope given by the intervalP = [0, 1]. Then

iP (k) = k + 1

8

which relates to the length by

iP (k) = k length(P ) + 1.

In dimension 2: Pick’s theorem tells us that the number of points in aconvex polytope with integral vertices dilated by k is given by:

iP (k) = k2 vol(P )+k

2(number of integral points on the boundary of P ) +1.

2 Number of integral points in Simplices

To prove Ehrhart’s theorem on the polynomial behavior of the function iP (k),it is sufficient to prove it for simplices, by decomposing an integral convexpolytope in unions and differences of integral simplices. Thus we consider(n+1) points in Z

n and the convex polytope given by the convex hull of thesepoints. Ehrhart gave a formula for the number of integral points containedin this set.

Let us start with the standard simplex ∆ in Rn, with vertices 0, e1, e2, . . . , en.

The polytope k∆ is defined by the inequations

x1 ≥ 0, . . . , xn ≥ 0, x1 + x2 + · · · + xn ≤ k.

We need to find the number of integer solutions of the inequations x1 ≥0, . . . , xn ≥ 0, x1 + x2 + · · · + xn ≤ k. In other words, we have to find thenumber p(n, k) of solutions in non-negative integers (x0, x1, . . . , xn) of theequation x0 + x1 + x2 + · · · + xn = k.

We first prove:

Theorem 4 For any integer k ≥ −n, the number p(n, k) of solutions in

non-negative integers (x0, x1, . . . , xn) of the equation

x0 + x1 + x2 + · · · + xn = k

is

p(n, k) =(k + 1)(k + 2) · · · (k + n)

n!.

9

Proof. The result is obvious for k = −1,−2, . . . ,−n, both sides of theformula being equal to 0. Let us set k ≥ 0 and consider the generatingfunction:

Zk(t) =∞∑

k=0

p(n, k)e−kt, t > 0.

We shall recover p(n, k) as the coefficient of e−kt in Zk(t).We have

Zk(t) =∑

(x0,...,xn)∈(Z+)n+1

e−t(x0+x1+···+xn)

=n∏

i=0

∞∑

xi=0

e−txi

=1

(1 − e−t)n+1.

Differentiating∑∞

k=0 zk = 11−z

, we get 1(1−z)n+1 =

∑∞k=0

(

k+nn

)

zk and hence

1

(1 − e−t)n+1=

∞∑

k=0

(

k + n

n

)

e−kt

with(

k+nn

)

= (k+1)(k+2)···(k+n)n!

.

Corollary 5

i∆(k) = cardinal(k∆ ∩ Zn) =

(k + 1)(k + 2) · · · (k + n)

n!.

Notice that i∆(k) is indeed a polynomial in k with constant term 1 and

leading terms 1n!

kn + 12

(n+1)(n−1)!

kn−1 + · · · as announced in Ehrhart’s theorem

since vol(∆) = 1n!

and the boundary of ∆ is the union of (n + 1) standardsimplices, each of volume 1

(n−1)!.

Let us now generalize this to the case of a general simplex P in Rn. We

give a formula due to Ehrhart for the number of points in kP .

10

Theorem 6 Let P be a simplex in Rn with integral vertices α0, α1, . . . , αn.

The number of integer points in kP is

iP (k) =n∑

r=0

cardinal(�(P, r))

(

k − r + n

n

)

where for an integer r between 0 and n

�(P, r) =

{

u ∈ [0, 1[n+1|n∑

i=0

ui = r,

n∑

i=0

uiαi ∈ Zn

}

which is a finite set. Thus iP (k) is a polynomial in k.

Proof. Let α0, α1, . . . , αn be the vertices of P which we assume to haveinteger coordinates. An element of kP is an element of the convex hull ofthe points kα0, kα1, . . . , kαn. It can be uniquely written as

w = x0α0 + x1α1 + · · · + xnαn

with x0 + x1 + · · · + xn = k. By taking the integral part ki of xi, the pointw can be written in an unique way as:

w = (u0α0 + u1α1 + · · · + unαn) + (k0α0 + k1α1 + · · · + knαn)

with ki non-negative integers, 0 ≤ ui < 1, and∑n

i=0 ki +∑n

i=0 ui = k. Asthe elements αi are in Z

n, the point w is in Zn if and only if

u0α0 + u1α1 + · · · + unαn ∈ Zn.

Notice that the number∑n

i=0 ui = k−∑n

i=0 ki is a non-negative integer. Thisinteger r is less or equal to n, as the value (n + 1) for r would compel all ui

to be equal to 1, while we have ui < 1.Conversely, let 0 ≤ r ≤ n be an integer. Now, if the non-negative integers

ki satisfyk0 + k1 + · · · + kn = k − r

and u ∈ �(P, r), the point

(u0α0 + u1α1 + · · · + unαn) + (k0α0 + k1α1 + · · · + knαn)

11

lies in kP . We enumerate this way all the points in kP . Since r ≤ n andk ≥ 0, the integer k − r is greater or equal to −n and it follows from whatwe have proven before that the number of solutions of the equation

k0 + k1 + · · · + kn = k − r

is(

k−r+nn

)

, for any k ≥ 0.We therefore obtain Ehrhart’s formula:

iP (k) =n∑

r=0

cardinal(�(P, r))

(

k − r + n

n

)

.

This expression is polynomial in k.

The sets �(P, r) are not very easy to determine. Furthermore, the de-composition of P in integral simplices is also not immediate. So the formulaabove has mainly a theoretical interest. We shall give later on more specificresult on the Ehrhart polynomial.

Let P be a rational convex polytope: the vertices of P have rationalcoordinates, instead of integral coordinates. Ehrhart analyzed more generallythe behavior in k of the function k 7→ iP (k) = cardinal(kP ∩Z

n) and provedthat this function is given by a“periodic-polynomial”, i.e. is a polynomialfunction in k with coefficients periodic functions of k. In other words, thereexists an integer d, such that the function k 7→ iP (dk) (k ≥ 0) is polynomial,as well as all functions iP (dk + j) where j is an integer such that 0 ≤ j <d. For example, if P := [0, 1

2], then, with d = 2, iP (2k) = k + 1, while

iP (2k + 1) = k + 1. These two formulae can be assembled together to givethe periodic-polynomial function iP (k) = k

2+ 1 − 1

4(1 − (−1)k).

For a given rational polytope P , it is not easy to determine the smallestvalue p of the integer d with the property that functions k 7→ iP (dk + j) arepolynomials in k. Of course, when P has integral vertices, we just provedthat p = 1. Similarly, if q is an integer such that, for all vertices s of P ,qs ∈ Z

n, then p divides q. It may happen that p is strictly smaller than q.

Example. Stanley’s pyramid. Consider the following example givenby Stanley ([?]): let P be the convex polytope in R

3 with vertices O =(0, 0, 0), A = (1, 0, 0),B = (0, 1, 0), D = (1, 1, 0) and C = (1

2, 0, 1

2). Then P

is not integral, however

iP (k) =(k + 1)(k + 2)(k + 3)

3!

12

A B

C

D

1/2

0

Figure 4: Stanley’s pyramid

is a polynomial function.

3 Some examples of Ehrhart polynomials

I first give some easy examples, where the brutal formula above can be cal-culated.

Example 1. Let P be the standard simplex with α0 = 0, α1 = e1,. . . ,αn = en.

Then �(P, 0) = {0}, the other sets �(P, r) are empty. We have theformula that we used:

ip(k) =

(

k + n

n

)

.

Example 2. Let P be the triangle in R2 with vertices α0 = 0, α1 = e1,

α2 = 2 e2.Then �(P, 0) = {0}, �(P, 1) has 1 element and the others are empty. We

13

have the formula:

iP (k) =

(

k + 2

2

)

+

(

k + 1

2

)

= (k + 1)2.

(The volume of P is 1.)

Example 3. Let P be the simplex in R3 with vertices α0 = 0, α1 = e1,

α2 = 2 e2, α3 = 3 e3.Then �(P, 0) = {0}, �(P, 1) has 4 elements and �(P, 2) has 1 element

so that we have the formula:

iP (k) =

(

k + 3

3

)

+ 4

(

k + 2

3

)

+

(

k + 1

3

)

= (k + 1)3.

(The volume of P is 1.)

Example 4. Let P the simplex in R4 with vertices α0 = 0, α1 = e1,

α2 = 2 e2, α3 = 3 e3, α4 = 4 e4.Then �(P, 0) = {0} has 1 element, �(P, 1) has 12 elements, �(P, 2) has

11 elements. We obtain

iP (k) = (k + 1)(k + 2)(k2 +4

3k +

1

2).

(The volume of P is 1.)

Example 5: The hanging pyramid Pm. Let Pm be the simplex withvertices α0 = (0, 0, 0), α1 = (1, 0, 0), α2 = (0, 1, 0), α3 = (1, 1,m).

The set �(P, 0) has 1 element, �(P, 1) has 0 elements and �(P, 2) has(m − 1) elements. We obtain

iPm(k) =

1

6(k + 1)(k + 2)(k + 3) + (m − 1)

1

6(k − 1)(k)(k + 1)

=m

6k3 + k2 + (12 − m)

k

6+ 1.

Notice on this last formula that we indeed have iPm(1) = m

6+1+ 12−m

6+1 =

4. But we see here that the coefficient of k in the Ehrhart polynomial is anegative number, when m is large enough. In particular, the coefficients

of Ehrhart polynomials are not necessarily positive.

14

Vol(kPm) = mk3

6|k(Pm) ∩ Z3| =

mk3

6+ k2 + (12 − m)k

6+ 1

Figure 5: Hanging pyramid

15

Finally, we include here a formula due to Mordell ([?]) for the number ofpoints in the polytope P (a, b, c) defined by the vertices 0, A = a e1, B = b e2,C = c e3 where a, b, c are relatively prime integers. The volume of P (a, b, c)is abc/6.

The number of integral points iP (a,b,c)(1) with integral coordinates inP (a, b, c) is not a rational function of a, b, c. Here is the Ehrhart polynomialfor the number of points with integral coordinates contained in kP (a, b, c):

iP (a,b,c)(k) =1

6abck3 + (

1

4(bc + ca + ab + 1))k2

+(1

4(a + b + c + 3) +

1

12(bc

a+

ca

b+

ab

c+

1

abc))k

−(s(bc, a) + s(ca, b) + s(ab, c))k + 1.

Here s(p, q) is the Dedekind sum, defined by

s(p, q) =

q∑

i=1

((i

q))((

pi

q))

with ((x)) = 0 if x is an integer and ((x)) = x − [x] − 12

otherwise.

4 The magic square polytope Magic(n) and

the Chan-Robbins-Yuen polytope

Let us now consider a special polytope given by the convex set Magic(n)formed by the doubly stochastic (n × n)-matrices. These are matrices (xij)with non-negative entries and such that, on each line and each column, thecoefficients add up to 1. Thus Magic(n) is defined as the intersection ofthe positive quadrant in R

n2

, cut by the 2n linear equations∑

i xij = 1for all 1 ≤ j ≤ n and

∑

j xij = 1 for all 1 ≤ i ≤ n. Clearly the sum ofcoefficients in all lines is equal to the sum of coefficients on all columns andits value is n. Thus this polytope is of dimension n2 − (2n − 1) = (n − 1)2.It is not difficult ([?]) to see that Magic(n) is also the convex hull of the n!permutation matrices so that its vertices are the n! permutation matrices.In particular Magic(n) is an integral convex polytope.

16

Example. The set Magic(5) is the set of matrices

x11 x12 x13 x14 x15

x21 x22 x23 x24 x25

x31 x32 x33 x34 x35

x41 x42 x43 x44 x45

x51 x52 x53 x54 x55

with xij ≥ 0 andx11 + x12 + x13 + x14 + x15 = 1,

x21 + x22 + x23 + x24 + x25 = 1,

. . .

x11 + x21 + x31 + x41 + x51 = 1,

x12 + x22 + x32 + x42 + x52 = 1,

. . .

As follows from the general theory of the Ehrhart polynomial, the numberof integral elements in kMagic(n) is a polynomial of the form

mnk(n−1)2 + · · · + 1,

where mn is the volume of Magic(n). Finding iMagic(n)(k) boils down tocounting magic squares: square (n× n)-matrices filled up with non-negativeintegers and such that the lines and the columns add up to k.

Here is an example of an element of 8 Magic(3).

4 3 12 4 22 1 5

.

We haveiMagic(1)(k) = 1,

iMagic(2)(k) = k + 1,

iMagic(3)(k) =

(

k + 4

4

)

+

(

k + 3

4

)

+

(

k + 2

4

)

,

Notice that:iMagic(n)(0) = 1,

17

iMagic(n)(1) = n!

(number of permutation matrices.)The Ehrhart polynomial iMagic(n)(k) is known only when n ≤ 9. A com-

putation due to Chan-Robbins (co/9806076) gives a formula for iMagic(n)(k)when n ≤ 8. The recent calculation by Beck and Pixton (co/0202267) of theEhrhart polynomial for n = 9 requires 325 days of computer time on a 1GHzPC running under Linux. The leading term of the Ehrhart polynomial (thevolume) requires 15 seconds for n = 7, 54 minutes for n = 8, 317 hours forn = 9.

We now consider the Chan-Robbins-Yuen polytope CRYn, namely thesubset of the set of (n × n) doubly stochastic matrices consisting of thosewith just one non-zero line above the diagonal permitted. For example, CRY5

is the set of matrices

x11 x12 0 0 0x21 x22 x23 0 0x31 x32 x33 x34 0x41 x42 x43 x44 x45

x51 x52 x53 x54 x55

with xij ≥ 0 and∑

i

xij = 1, for all 1 ≤ j ≤ 5,

and∑

j

xij = 1, for all 1 ≤ i ≤ 5.

The dimension of this convex polytope is n(n− 1)/2. The correspondingEhrhart polynomial is not known but the leading term is known (Chan-Robbins-Yuen ([?]) and Zeilberger ([?])). It is given by

(n−2∏

i=1

(2i)!

(i + 1)!i!)

kn(n−1)/2

(n(n − 1)/2)!.

We shall see why this calculation is possible in Section ??.

18

5 Brion’s formulae

The volume of a convex polytope and the Ehrhart polynomial are both dif-ficult to compute. There are very few cases for which they are explicitlyknown. One can however hope to find some relations between them.

A relation together with a way to compute the volume and the num-ber of integral convex polytopes follows from Brion’s formulae ([?]), see also([?],[?],[?]). This formula has been used by Barvinok ([?]) to give an al-gorithm to compute in polynomial time (when the dimension is fixed) thenumber of integral points in a rational convex polytope.

I shall give Brion’s formulae here since they are very beautiful. However,we shall indicate later a method of generating function and separation ofvariables, which leads to a direct comparison between volumes and numberof points via residue formulae. Furthermore, the residue formulae seem toyield a more efficient computational tool than Brion’s formulae, at least fortransportation polytopes.

Michel Brion provided a formula for integral of exponential functionsover convex polytopes or for sums of exponentials over integral points. Theygeneralize the following formulae:

• For any a, b ∈ R∫ b

a

exydx = −eay

y+

eby

y

which follows by integration.

• For a, b integers

b∑

u=a,u∈Z

euy =eay

1 − ey+

eby

1 − e−y.

This last result follows from writing integers between a and b as differencesof integers strictly greater than b and integers greater or equal than a andsumming arithmetic progressions.

The first integral formula generalizes from the interval polytope [a, b] toa general convex polytope P ⊂ R

n. Let P be a convex polytope in Rn, and

let V(P ) be the finite set of its vertices. We assume (this is the generic case)that at each vertex s ∈ V(P ) start n edges as

1, as2, . . . , a

sn of the polytope P :

vectors asi are elements of R

n such that near s the convex polytope P is the

19

Figure 6: Non-generic pyramid versus generic pyramids

set of points of the forms s+∑n

i=1 tiasi , with ti ≥ 0 (ti small). In other words,

the tangent cone at s to P is the affine cone s +∑n

i=1 R+as

i .We identify the exterior product Λn

Rn with R, and |as

1∧as2∧· · ·∧as

n| is, bydefinition, the absolute value of the determinant of the (n×n)-matrix with as

i

as column vectors. Elements asi are defined here only up to proportionality.

Then, we have:

(−1)n

∫

P

e〈x,y〉dx =∑

s∈V(P )

|as1 ∧ · · · ∧ as

n|e〈s,y〉

〈as1, y〉 · · · 〈a

sn, y〉

.

Notice that this formula does not depend of the choice of the length ofthe edges as

i passing through the vertex s.With n = 1, setting P = [a, b], then there are 2 vertices a and b. We can

choose aa1 = b − a, ab

1 = a − b, and the above formula yields back

∫ b

a

exy = −

(

|(b − a)|eay

(b − a)y+

|(a − b)|eay

(a − b)y

)

= −eay

y+

eby

y.

In particular, Brion’s formula gives a formula for vol(P ) knowing thevertices s and the edges through s.

vol(P ) = (−1)n∑

s∈V(P )

|as1 ∧ · · · ∧ as

n|〈s, y〉n

〈as1, y〉 · · · 〈a

sn, y〉

20

for any generic y.Brion established similar formulae for sums of exponentials. To simplify

the statement of Brion’s formula, we assume that the convex polytope P isan integral convex polytope. Furthermore, we assume that at each vertex s,we can choose integral vectors as

i ∈ Zn such that |as

1 ∧ · · ·∧asn| = 1, and such

that, as before, the tangent cone at s to P is the affine cone s +∑n

i=1 R+as

i .We call such a convex polytope a Delzant polytope.

An integral convex polytope is rarely a Delzant polytope:

Example. Let T (r) be the triangle in R2 with vertices A = (0, 0), B =

(1, r), C = (1, 0). It is a Delzant polytope, if and only if |r| = 1.

Let us however state Brion’s formula in this case:

Theorem 7 Let P be a Delzant polytope in Rn. Then we have

∑

ξ∈P∩Zn

e〈ξ,y〉 =∑

s∈V(P )

e〈s,y〉

(1 − e〈as1,y〉) · · · (1 − e〈as

n,y〉).

Example. Brion’s formula to integrate an exponential over the standardsimplex ∆ dilated by k in R

2 is

∫

k∆

ey1x1+y2x2dx1dx2 =1

y1y2

+eky1

(−y1)(y2 − y1)+

eky2

(−y2)(y1 − y2),

while the formula to sum up an exponential on all the integral points (p1, p2)in k∆ is

∑

(p1,p2)∈k∆

ey1p1+y2p2 =1

(1 − ey1)(1 − ey2)+

eky1

(1 − e−y1)(1 − ey2−y1)+

eky2

(1 − e−y2)(1 − ey1−y2).

6 Partition polytopes

We now introduce more general families of convex polytopes than just thedilated polytopes kP .

21

k

k

0

Figure 7: Simplex and edges through vertices

Let P and Q be two convex polytopes in Rn. The Minkowski sum is

defined asP + Q = {x + y| x ∈ P, y ∈ Q}.

The following theorem, proved in ([?]), is a generalization of Ehrhart’stheorem.

Theorem 8 • Let P1, P2, . . . , Pr be convex polytopes in Rn. Then the

function

v(t1, .., tr) = volume(t1P1 + t2P2 + · · · + trPr)

is a polynomial function of (t1, t2, . . . , tr) ∈ (R+)r.

• Assume P1, P2, . . . , Pr are integral convex polytopes. Then the function

on (Z+)r given by

i(k1, k2, . . . , kr) = cardinal((k1P1 + · · · + krPr) ∩ Zn)

is a polynomial function of (k1, k2, . . . , kr).

We shall study more general families of convex polytopes.

22

LetΦ = [α1, α2, . . . , αN ]

be a sequence of N vectors in Rn (elements αi may not be all distinct). We

assume that all vectors αi lie strictly on the same side of a hyperplane, sothat Φ generates an acute cone C(Φ) = {

∑Ni=1 tiαi| ti ≥ 0} in R

n (an acutecone is a cone which does not contain any straight line).

Let {w1, w2, . . . , wN} be the canonical basis of RN . We define a linear

map from RN to R

n by:

AΦ(x1, x2, . . . , xN) =N∑

i=1

xiαi.

We may write the linear map A := AΦ as a (n×N)-matrix A with columnvectors the vectors αi.

For a ∈ C(Φ) define

PΦ(a) = A−1Φ (a) ∩ R

N+ ,

which is the intersection of an affine space with the standard quadrant.In other words PΦ(a) consists of all solutions of the equation Ax = a,

where x is a N -vector with non-negative coordinates. The set PΦ(a) is aconvex polytope called partition polytope.

Example. Transportation polytopes

Let R2n with canonical basis e1, . . . , en, f1, . . . , fn and let Φ be the set of

n2 vectorsΦ = [(ei + fj)| 1 ≤ i, j ≤ n].

ThenA(xij) =

∑

xij(ei + fj).

The convex polytope PΦ(a1e1 +a2e2 + · · ·+anen +b1f1 +b2f2 + · · ·+bnfn) canbe identified to the set of (n×n)-matrices (xij) with non-negative coefficientsand with given sums of coefficients in each row and given sums of coefficientsof each column. Indeed the vector equation

∑

xij(ei + fj) = (a1e1 + a2e2 +· · ·+anen +b1f1 +b2f2 + · · ·+bnfn) is equivalent to the series of 2n equations:

n∑

j=1

xij = ai,

23

n∑

i=1

xij = bj.

Of course, to get a solution, we need that∑n

i=1 ai =∑n

j=1 bj.When a1 = a2 = · · · = an = b1 = b2 = · · · = bn = 1, we recover the poly-

tope Magic(n). For general (ai, bj) (such that∑

ai =∑

bj), this polytope iscalled a transportation polytope, and is very important in studying flow innetworks.

If Φ is a sequence of vectors in Zn, then PΦ(a) is a rational convex poly-

tope.

Example: LetA = (6, 10, 15),

thenA(x1, x2, x3) = 6x1 + 10x2 + 15x3.

The polytope PΦ(k) is the convex hull of its 3 vertices

(k

6, 0, 0), (0,

k

10, 0), (0, 0,

k

15).

If A is surjective, then the polytopes PΦ(a) are of dimension N −n, for ain the interior of C(Φ). The polytopes PΦ(a) are contained in affine spacesparallel to E := Ker A. If we denote by f1, f2, ..., fN the restrictions of thelinear forms x1, x2, . . . , xN on R

N to the subspace E, then for u ∈ RN , the

polytope PΦ(Au) is isomorphic to the polytope in E = Ker A described bythe inequations

Q(u) := {y ∈ E| 〈fi, y〉 + ui ≥ 0}.

Indeed, if y ∈ Q(u), the point u + y lies in P (Au). Some of the inequalitiesabove might be irrelevant. If Au = Av, polytopes Q(u) and Q(v) are justtranslations of each other in the space KerA by the vector u − v.

Consider the polytope PΦ(A(u + hwk)) = PΦ(Au + hαk). Then the poly-tope Q(u + hwk) is isomorphic to the polytope PΦ(Au + hαk) and is definedby the same inequations as the polytope Q(u) except that one of the in-equalities 〈fi, y〉 + ui ≥ 0 has been replaced for i = k by the inequality〈fk, y〉 + (uk + h) ≥ 0. In particular all polytopes PΦ(a + h), when a isgeneric and h varies in a small neighborhood of 0, have parallel faces.

Example

24

We draw the pictures of the polytopes Q(Au) and their small deformationsfor the matrix:

A =

(

1 0 0 10 1 1 1

)

.

The space Ker A is isomorphic to R2 with basis b1 = (w3 − w2), b2 =

(w4 − w1 − w2). We write y ∈ Ker A as y = y1b1 + y2b2. To describe thefamily Q(u), we may vary u in a supplementary subspace to Ker A. Wechoose u = u1w1 + u2w2.

The 4 equations describing Q(u) are

−y2 + u1 ≥ 0,

−(y1 + y2) + u2 ≥ 0,

y1 ≥ 0,

y2 ≥ 0.

There are 6 cases leading to different polytopes, depending on the stratifica-tion of the cone C(Φ) in chambers (a topic that we shall discuss after thisexample).

• u1 = u2 = 0.

Then Q(0, 0) = {0}

• u1 = 0, u2 > 0

Then Q(0, u2) is the interval [0, u2]b1.

• u1 > 0, u2 = 0

Then Q(u1, 0) is the interval [0, u1]b2.

• u2 > u1.

Then Q(u1, u2) has 4 vertices

0, A = u1b2, B = u2b1, C = (u2 − u1)b1 + u1b2 and is a trapeze.

• u1 = u2 = u

Then Q(u1, u2) is a triangle with vertices 0, A = ub2, B = ub1.

25

u2 − u1

u1 − u2

A

A

u1

u2

u2u2

00

C

BB

u2 > u1 u1 > u2

Figure 8: Variation of the partition polytopes: Q(u)

• u2 < u1

Then Q(u1, u2) is a triangle with vertices 0, A = u2b2, B = u2b1.

Notice that in the last two cases, the equation −y2 + u1 ≥ 0 is irrelevantand does not produce a face of Q(u).

Conversely, any convex polytope described by inequations can be realizedcanonically as a member of a family of partition polytopes, which containsalso its small deformations obtained by moving faces parallel to themselves.

Let us, following Gelfand-Kapranov-Zelevinski ([?]), decompose the coneC(Φ) as a union of “chambers”. By definition, a chamber is a connectedcomponent of the open subset of C(Φ) obtained by removing the boundariesof all the cones C(σ) spanned by subsets σ of Φ forming a basis of V ∗. Inthe case of the matrix

A =

(

1 0 0 10 1 1 1

)

26

0

12(e1 − e3)

e2 − e3

12e2

12e2

13e1

e3

e1 − e2

Figure 9: Chambers for A3

then α1 = e1 = α3, α2 = e2 and α4 = e1 + e2. The cone C(Φ) is thefirst quadrant, but to define chambers, we have to remove the half-linesR

+e1, R+e2, R

+(e1 + e2). This leads to the two chambers {x1 > 0, x2 >0, x1 > x2} or {x1 > 0, x2 > 0, x1 < x2}.

When a varies inside a chamber, the combinatorial nature of the polytopePΦ(a) remains the same. But, as we have already seen in the last example,the combinatorial nature of the polytope PΦ(a) varies when a crosses the wallof a chamber. If {P1, P2, . . . , Pr} is a set of convex polytopes in R

n, then thefamily of polytopes obtained by taking their Minkowski sums

{t1P1 + · · · + trPr}

with ti ≥ 0 can be embedded as a subset of a family of partition polytopesPΦ(a) where a varies in the closure of a chamber C of C(Φ).

Figure 9 is the drawing of chambers for the matrix

A :=

1 0 0 1 0 10 1 0 1 1 10 0 1 0 1 1

For a general matrix A, it is difficult to describe chambers of the coneC(Φ). The program PUNTOS available on the homepage of Jesus de Loera

27

(www.math.ucdavis.edu/deloera) gives an algorithm to compute them, basedon Gelfand-Kapranov-Zelevinski theory.

7 Generating functions

The method we used to compute the volume as well as number of points ofa rational convex polytopes is based on generating functions.

We are in the setting of partition polytopes, with a surjective map A :R

N → Rn. We denote by (w1, w2, . . . , wN) the canonical basis of R

N , byαk = A(wk) and by Φ := [α1, α2, . . . , αN ]. We assume that all vectors αk liesstrictly on an open half-space delimited by an hyperplane. Then the dualcone to the cone C(Φ) generated by Φ is an open cone: it consists of allvectors z ∈ R

n such that 〈z, αk〉 > 0, for all 1 ≤ k ≤ N .For a ∈ C(Φ), let

i(a) = cardinal(PΦ(a) ∩ ZN)

andv(a) = vol(PΦ(a)).

To compute i(a) and v(a) we shall use generating functions.

Proposition 9 Let z in the dual cone to C(Φ), then

∫

C(Φ)

v(a)e−〈a,z〉da =1

∏

α∈Φ〈α, z〉,

∑

a∈C(Φ)∩Zn

i(a)e−〈a,z〉 =1

∏

α∈Φ(1 − e−〈α,z〉).

Proof. Let z in the dual cone to C(Φ) and let us compute

F (z) :=

∫

RN+

e−〈A(x),z〉dx.

We first write A(x) =∑N

i=1 xiαi. The integral reads

28

F (z) =

∫

RN+

e−〈P

xiαi,z〉dx1dx2 · · · dxN

=N∏

i=1

∫

R+

e−xi〈αi,z〉dxi

=N∏

i=1

1

〈αi, z〉.

On the other hand, we can use Fubini formula. We first integrate overthe x such that A(x) = a, then we integrate on a. Thus

F (z) =

∫

a∈C(Φ)

(

∫

{x∈RN+ | A(x)=a}

e−〈A(x),z〉dx

)

da.

The set {x ∈ RN+ | A(x) = a} is our partition polytope PΦ(a) and the

integral over this set of e−〈A(x),z〉 is e−〈a,z〉 vol(PΦ(a)). We thus obtain thefirst formula of the proposition.

The second formula arises in the same way by calculating in two differentways the sum

∑

x∈ZN+

e−〈A(x),z〉.

The problem of computing v(a) and i(a) now boils down to computingthe inverse on the right hand side of the two equations of Proposition ??.A similar problem was considered by Jeffrey and Kirwan ([?]), who, in thecontext of arrangements of hyperplanes, found an efficient calculus for theinversion of Laplace transforms.

8 Inversion of Laplace transforms and residue

formulae

We explain our method ([?]) in the very simple case where n = 1.Let n = 1, and let G be the space of functions of the form

f(z) :=P (z)

zM,

29

where P is a polynomial such that f(z) tends to 0 when z tends to ∞. Thenthere exists a polynomial function v(h) such that

f(z) =

∫

h≥0

v(h)e−hzdh

for z > 0.I claim that v is given by the residue formula

v(h) = residuex=0(f(x)ehx),

which is obvious to check. Indeed, f(z) is of the form∑M

p=1up

zp , and we may

directly check the formula for 1zp =

∫

h≥0hp−1

(p−1)!dh.

The dependance of v(h) in h is via the Taylor series of ehx at x = 0. Thefunction f(x) has a pole at x = 0 of order M , thus we need to take the Taylordevelopment of ehx only up to order M . In particular, it is clear that v(h) isa polynomial in h of degree less or equal to M − 1.

Let us turn to the calculation of i(a).Let M be the space of functions of the form

F (z) :=P (z)

(1 − z)M,

where P (z) is a polynomial in z. We assume that F (z) tends to 0 when ztends to ∞. Expand the function F (z) as a Taylor series at the origin. Thenthere exists a polynomial function i(k) such that

F (z) =∞∑

k=0

i(k)zk

for |z| < 1.I claim that i is given by the residue formula

i(k) = −residuez=1(F (z)z−k dz

z).

Indeed, integrating on a small circle near z = 0 the Taylor expansion of F ,we obtain

i(k) = residuez=0(F (z)z−k dz

z).

30

From our assumption, it follows that the 1-form F (z)z−k dzz

has no residue at∞. Thus we obtain, from the residue theorem on P1(C), that for any k ≥ 0,

i(k) = −residuez=1(F (z)z−k dz

z).

Writing z = e−x, we also obtain

i(k) = residuex=0(F (e−x)ekx),

which is strikingly similar to the formula

v(h) = residuex=0(f(x)ehx).

The function F (e−x) has a pole of order M at x = 0. So it is clear thati(k) is a polynomial in k of degree less or equal to M − 1.

A similar multidimensional residue formula is used in the final formulaeof Theorems ?? and ??. The same idea of moving a residue from a contournear z = 0 to a contour near z = 1 is involved in the proof of this formula.The cohomology of the complement of an union of hyperplanes is the crucialtool we need (or better an algebraic version of this cohomology).

In a very simple example, consider functions F (z1, z2) of the forms F (z1, z2) =1

(1−z1)M (1−z2)N (1−z1z2)Q . Here M,N,Q are positive integers. Expand

F (z1, z2) =∑

k1≥0,k2≥0

v(k1, k2)zk1

1 zk2

2

in Taylor series. Then there exists two polynomial functions v1(a1, a2) andv2(a1, a2) such that v1(a, a) = v2(a, a) and such that

v(k1, k2) = v1(k1, k2)

if k1 ≥ k2, whilev(k1, k2) = v2(k1, k2)

if k1 ≤ k2.Indeed, we have, from the Cauchy theorem,

v(k1, k2) = (1

2iπ)2

∫

|z1|=ǫ1,|z2|=ǫ2

1

(1 − z1)M(1 − z2)N(1 − z1z2)Qz−k1

1 z−k2

2

dz1

z1

dz2

z2

31

whatever small non-zero real numbers ǫ1, ǫ2 we choose. It is tempting, as inthe one-dimensional case, to use the other pole z1 = z2 = 1 to compute thisintegral. When choosing exponential coordinates e−x1 , e−x2 near this pole,we are led to consider the 2-form

1

(1 − e−x1)M(1 − e−x2)N(1 − e−(x1+x2))Qek1x1ek2x2dx1dx2.

In a neighborhood of (0, 0), the function

J(x1, x2) =1

(1 − e−x1)M(1 − e−x2)N(1 − e−(x1+x2))Q

has now poles on the hyperplanes {x1 = 0}, {x2 = 0} and {x1 + x2 = 0}.Thus its restriction to a cycle C(ǫ1, ǫ2) = {(x1, x2)| |x1| = ǫ1, |x2| = ǫ2} isholomorphic provided ǫ1 6= ǫ2. Now the cycles C(ǫ1, ǫ2) for ǫ1 > ǫ2 or ǫ2 > ǫ1

cannot be deformed to each other without coming across a pole of J(x1, x2).It is not difficult to show that we have

v1(a1, a2) = (1

2iπ)2

∫

C(ǫ1,ǫ2)

J(x1, x2)ea1x1+a2x2dx1dx2

with ǫ2 > ǫ1 > 0, while

v2(a1, a2) = (1

2iπ)2

∫

C(ǫ1,ǫ2)

J(x1, x2)ea1x1+a2x2dx1dx2,

with ǫ1 > ǫ2 > 0.In other words, the analogue of the residue formula for

i(k) = residuex=0(F (e−x)ekx),

for the Taylor series of F in one variable is replaced by the two formulae(both obviously polynomial in k1, k2.)

v1(k1, k2) = residuex2=0residuex1=0(F (e−x1 , e−x2)ek1x1+k2x2),

v2(k1, k2) = residuex1=0residuex2=0(F (e−x1 , e−x2)ek1x1+k2x2).

32

9 Arrangements of hyperplanes and the total

residue

Let V ∗ be a real vector space of dimension n with a set ∆ of non-zeros vectors.We assume ∆ symmetric, that is ∆ = −∆. To each α ∈ ∆ we associate ahyperplane {α = 0} in V .

Let R∆ be the ring of rational functions on VC with poles on the hyper-planes {α = 0}. If F ∈ R∆, we have, for z ∈ VC

F (z) =P (z)

∏

α∈∆〈α, z〉nα

for some polynomial P and non-negative integers nα.In dimension 1, with linear form α(z) = z, our space R∆ is just the

space of Laurent polynomials C[z, z−1]. The power zk is the derivative ofthe function 1

k+1zk+1 except when k = −1. Thus there is just a particular

function 1z

which has no primitive. We can write:

C[z, z−1] = C1

z⊕ ∂zC[z, z−1].

Let us come back to a general hyperplane arrangement and let us give adescription of R∆ which generalizes this decomposition of C[z, z−1].

Let σ be a subset of ∆, such that elements of σ form a basis of V ∗. Wewill say that σ is a basic subset of ∆. Setting

σ = {αi1 , αi2 , . . . , αin},

we define the ”simple fraction”

fσ(z) =1

〈αi1 , z〉〈αi2 , z〉 · · · 〈αin , z〉

and the space S∆ generated by simple fractions:

S∆ =∑

σ

Cfσ,

where σ runs over all basic subsets of ∆.

33

Choose a basis {e1, e2, . . . , en} of V . The partial derivative ∂i acts on R∆.We denote

∂R∆ =n∑

i=1

∂iR∆.

The following theorem is proved in ([?]).

Theorem 10 We have:

R∆ = S∆ ⊕ ∂R∆.

The space S∆ will be very important in our formulae. It is spanned bythe functions fσ, however the following examples show that the elements fσ

are not generally linearly independent.

Example. We consider V of dimension 2 with basis e1, e2. Consider theset

∆+ = {z1, z2, (z1 + z2)}

of linear forms on V := {z = z1e1 + z2e

2 }. Let ∆ = ∆+ ∪ ∆−. Thereare 3 basis of V ∗ formed with elements of ∆+, namely σ1 = {z1, z2}, σ2 ={z1, (z1 + z2)}, and σ3 = {z2, (z1 + z2)}. There is a linear relation betweenthe 3 corresponding simple fractions:

1

z1z2

=1

z1(z1 + z2)+

1

z2(z1 + z2).

Example. We consider the vector space Rn and the set of n(n+1) linear

forms:An = {±(zi − zj)| 1 ≤ i < j ≤ n} ∪ {±zi| 1 ≤ i ≤ n}.

The following proposition can be proved by induction on n.

Proposition 11 A basis {fw} of the space SAnis obtained as follows: set

fw(z) =1

(zw(1) − zw(2))(zw(2) − zw(3)) · · · (zw(n−1) − zw(n))zw(n)

where w is a permutation on {1, . . . , n}.

34

Then SAnhas a basis indexed by elements w of the permutation group

Σn and hence dim SAn= n!.

Let us come back to the general case. As

R∆ = S∆ ⊕ ∂R∆,

there is a projection:Tres : R∆ 7→ S∆

called the total residue. Notice that the total residue of F vanishes wheneverF is a sum of derivatives. For instance, the residue of the following function

z31

(z1 − z2)(z1 − z3)(z2 − z3)z1z2z3

−1

(z1 − z2)(z2 − z3)z3

+1

(z1 − z3)(z3 − z2)z2

vanishes. Indeed, we can verify that this function is equal to:

−∂2(z1 − 2z3)

z3(z1 − z3)(z2 − z3)+ ∂3

(z1 − 2z2)

z2(z1 − z2)(z3 − z2).

The total residue vanishes on homogeneous functions of R∆ which areof degree m, whenever m 6= −n. Thus we can extend the total residue tofunctions F (z) = P (z)

Q

α∈∆〈α,z〉nα, where P is a holomorphic function defined near

0.The most important tool in computations is the following linear form on

the space S∆.For each chamber C there exists a linear functional denoted by

φ 7→ 〈〈C, φ〉〉

on S∆, defined by〈〈C, | det σ|fσ〉〉 = 1

if C ⊂ C(σ)〈〈C, | det σ|fσ〉〉 = 0

otherwise.Via projections of R∆ on S∆, we identify linear forms on S∆ to linear

forms on R∆ vanishing on derivatives.Determining the chambers C and the linear form 〈〈C, φ〉〉 is difficult in

general. The linear form 〈〈C, φ〉〉 can be realized as an integration on a cyclein the space VC \ ∪α∈∆{α = 0} depending of the chamber C.

35

It can be easy to describe the linear form 〈〈C, φ〉〉 for some very specialcases such as

∆ := An := {±(ei − ej)| 1 ≤ i < j ≤ n} ∪ {±ei| 1 ≤ i ≤ n}

andΦ := A+

n := {(ei − ej)| 1 ≤ i < j ≤ n} ∪ {ei| 1 ≤ i ≤ n}.

The space R∆ consists of rational functions of (z1, z2, . . . , zn) with poles onthe hyperplanes {zi = zj} or {zi = 0}.

The cone C(Φ) is described as follows:

C(Φ) = {a1e1+a2e2+· · ·+anen| a1 ≥ 0, a1+a2 ≥ 0, . . . , a1+a2+· · ·+an ≥ 0}.

The number of chambers of the cone C(Φ) is not known (see de Loera-Sturmfels ([?]) for the description of chambers for small n). But in anydimension, one of the chambers of the cone C(Φ) above is

Cnice = {a1e1 + a2e

2 + · · · + anen| a1 > 0, a2 > 0, . . . , an > 0}.

Furthermore, in this case the linear form attached to this chamber canbe written in terms of iterated residues:

〈〈Cnice, φ〉〉 = residuez1=0 · · · residuezn=0(φ(z1, z2, . . . , zn)).

The linear forms attached to any chamber C are not too difficult to com-pute, as we know an explicit basis of the space SAn

.

Example. Let us consider the matrix:

A :=

1 0 0 1 0 10 1 0 −1 1 00 0 1 0 −1 −1

.

The system Φ spanned by this matrix is the system A+3 . There are 7

chambers.Our space R∆ consists of functions f(z1, z2, z3) with poles on {zi = 0} or

on {zi = zj}. Thus when taking a contour

C(ǫ1, ǫ2, ǫ3) := {|z1| = ǫ1, |z2| = ǫ2, |z3| = ǫ3},

provided that all ǫi are different, the function f(z1, z2, z3) is well defined onC(ǫ1, ǫ2, ǫ3). However, the function f(z1, z2, z3) with poles on {zi = zj}, wesee that the relative order on ǫ1, ǫ2, ǫ3 will lead to different cycles.

Let us give two examples for two different chambers.

36

0

12(e1 − e3)

e2 − e3

12e2

12e2

13e1

e3

e1 − e2

Figure 10: Chambers for A3

• Let C1 be the the chamber spanned by e1, e2, e3. (This is what we calledthe nice chamber). Then

〈〈C1, φ〉〉 = (1

2iπ)3

∫

C(ǫ1,ǫ2,ǫ3)

φ(z1, z2, z3)dz1dz2dz3

where ǫ1 > ǫ2 > ǫ3 > 0.

• Let C2 be the the chamber spanned by e1 − e2, e1, e3.

Then

〈〈C2, φ〉〉 = (1

2iπ)3

∫

C(ǫ1,ǫ2,ǫ3)

φ(z1, z2, z3)dz1dz2dz3

−(1

2iπ)3

∫

C(ǫ′1,ǫ′2,ǫ′3)

φ(z1, z2, z3)dz1dz2dz3

where ǫ1 > ǫ2 > ǫ3 > 0 and ǫ′2 > ǫ′1 > ǫ′3 > 0.

10 Jeffrey-Kirwan formula for the volume

Let Φ ⊂ V ∗ be a sequence of vectors belonging to ∆ and all lying on thesame side of some given hyperplane. We assume that Φ spans V ∗.

37

Let us give a formula for the volume of the polytope PΦ(a). A quick proofis given in ([?]).

Theorem 12 (Jeffrey-Kirwan formula)Let C be a chamber of C(Φ). Then for a ∈ C

vol(PΦ(a)) = 〈〈C, Trese〈a,z〉

〈α1, z〉 · · · 〈αN , z〉〉〉

=1

(N − n)!〈〈C, Tres

〈a, z〉N−n

〈α1, z〉 · · · 〈αN , z〉〉〉

which is a polynomial in a on the chamber C.

From this formula, following Aomoto’s proof of the calculation of Selberg-like integrals and the indication of Zeilberger ([?]), we show in ([?]) that itis possible to recover the conjectured formula for the volume of the Chan-Robbins-Yuen polytope.

Theorem 13 (Zeilberger).Let N := n(n − 1)/2. Then

vol(CRYn) =1

N !residuez1=0 · · · residuezn=0

zN1

z1z2 · · · zn

∏

i<j(zi − zj)

=1

N !

n−2∏

i=1

Ci

where the Ci = 2i!(i+1)!i!

are the Catalan numbers.

In the general case of a convex polytope realized as PΦ(a), the algorithmto compute its volume is the following. We choose ∆+ any subset of ∆containing Φ, and contained in a half-space. We compute (or better, weknow as in the case of the system An) a basis fσ of the space S∆, indexedby a finite set F of subsets of ∆+. Then, given a generic point a ∈ V ∗, wecompute if a belongs to the cone C(σ), only for those σ belonging to F .If C is the chamber containing a, this determines entirely the form 〈〈C, ·〉〉,as 〈〈C, | det σ|fσ〉〉 = 0 or 1 according to the fact that a ∈ C(σ) or not. Thenwe can realize the form 〈〈C, ·〉〉 as an iterated residue with respect to specialorders, entirely determined by our chamber C.

38

11 Residue formula for the number of inte-

gral points in rational polytopes

Let Φ be a sequence of Zn all lying on the same side of a given hyperplane

and spanning Rn. Then for a ∈ Z

n ∩ C(Φ), the polytope PΦ(a) is a rationalpolytope. Let me now state the relation between the partition functionand the number of points in the rational polytope PΦ(a) and indicate someproperties of the function i(a). Let us introduce a notation:

�(Φ) =∑

α∈Φ

[0, 1]α

Notice that the box �(Φ) grows larger as the set Φ gets larger. Further-more, as Φ spans V ∗, then for any chamber C of C(Φ), the open set C−�(Φ)contains the closure of the open set C.

The following qualitative theorem generalizes Theorem ??.

Theorem 14 Let Φ be a sequence of vectors in Zn all in the same side of an

hyperplane and spanning Rn. For z in the dual cone to C(Φ), the following

equality holds:

∑

a∈C(Φ)∩Zn

i(a)e−〈a,z〉 =1

∏

α∈Φ(1 − e−〈α,z〉)

where i(a) is the cardinal of the set of solutions in non-negative integers nk

of the equation

a = n1α1 + n2α2 + · · · + nNαN .

Then, for each chamber C of the cone C(Φ), there exists a periodic-

polynomial function iC on Rn such that i(a) = iC(a) for any a ∈ (C−�(Φ))∩

Zn.

The closure C of the chamber C is a subset of C − �(Φ). The periodic-polynomial behavior of the function i(a) on the set C is due to Sturmfels([?]).

In fact, we (i.e. Szenes and myself) proved in ([?]) an “explicit formula”for the function i(a). This formula is proven in a rather straightforward wayby a separation of variable argument due to A. Szenes ([?]) and the residuetheorem in one variable. We state it first in the unimodular case.

39

Let now ∆ be a set of vectors in Zn and Φ = [α1, . . . , αN ] a sequence

of elements of the set ∆. We assume that Φ spans Zn. We consider the

unimodular case where | det σ| = 1 for any subset σ of Φ consisting of nlinearly independent vectors. In this case the convex polytopes PΦ(a) haveintegral vertices, whenever a is in Z

n. We get almost the same formulae forthe volume or for the number iΦ(a) of integral points in PΦ(a).

Theorem 15 Let C be a chamber. For a ∈ (C − �(Φ)) ∩ Zn:

iΦ(a) = 〈〈C, Trese〈a,z〉

(1 − e−〈α1,z〉) · · · (1 − e−〈αN ,z〉)〉〉.

Both formulae for the volume v(a) (Theorem ??) and the number of pointsi(a) (Theorem ??) in terms of the linear form 〈〈C, ·〉〉 attached to C can beproven in a completely parallel way by reducing to the 1-dimensional case(treated in Section ??). The basic formula to reduce to the 1-dimensionalcase is an analogue of the two formulae below:

1

z1z2(z1 + z2)=

1

z21z2

−1

z21(z1 + z2)

,

1

(1 − z1)(1 − z2)(1 − z1z2)=

1

(1 − z1)2(1 − z2)+

1

(1 − z1)(1 − z−11 )(1 − z1z2)

.

Using the formula of Theorem ?? (together with a change of variable inresidues ([?])), with V. Baldoni, we have implemented a simple Maple pro-gram for calculations of number of integral points in transportation polytopesor more generally for networks. It computes number of points in (5 × 5)-transportation polytopes in approximately 4 hours. Previous algorithm werebased on Brion’s formula ([?]).

As an easy consequence of the two parallel formulae (Theorem ?? andTheorem ??), we recover the Riemann-Roch formula of Khovanskii-Pukhlikov([?]). Let us explain how.

Let α ∈ V ∗. We denote by ∂(α) the differential operator acting on func-tions on functions on V ∗ by

∂(α)P (h) =d

dǫP (h + ǫα)

∣

∣

ǫ=0.

40

Consider the function ToddN(z1, z2, . . . , zN) :=∏N

i=1zi

1−e−ziand its Taylor

series at the origin:

ToddN(z) := 1 +1

2

N∑

i=1

zi + · · · .

Let Φ = [α1, α2, . . . , αN ] our unimodular set of vectors. We now substitutezi = ∂(αi) in this series. We obtain a series of constant coefficients differentialoperators that we denote by

Todd(Φ, ∂) :=∏

α∈Φ

∂(α)

1 − e−∂(α).

We can apply this series of differential operators to a polynomial functionon V ∗.

Theorem 16 ( Khovanskii-Pukhlikov)Let Φ be an unimodular system of vectors in Z

n. Let C be a chamber of

C(Φ). Then, for a ∈ C ∩ Zn, we have the equality

i(a) = Todd(Φ, ∂)v(h)|h=a.

Proof. We argue in exactly the same way as in Lecture 1. To apply theseries of differential operators Todd(Φ, ∂) to the function v(h) given by theresidue formula

v(h) := 〈〈C, Trese〈h,z〉

〈α1, z〉 · · · 〈αN , z〉〉〉

we can commute the residue and the series (the residue operates automati-cally by truncation of series). Thus we obtain

Todd(Φ, ∂)v(h) = 〈〈C, TresN∏

i=1

(〈αi, z〉

1 − e−〈αi,z〉)

e〈h,z〉

〈α1, z〉 · · · 〈αN , z〉〉〉

= 〈〈C, Trese〈h,z〉

(1 − e−〈α1,z〉) · · · (1 − e−〈αN ,z〉)〉〉.

We recognize here the residue formula for i(a).

In the general case of a system Φ spanning Zn, but not necessarily uni-

modular, the formula for i(a) is periodic-polynomial over the sectors C−�(Φ).Here is the formula:

41

Theorem 17 Let C be a chamber of C(Φ). For a ∈ (C − �(Φ)) ∩ Zn:

iΦ(a) =∑

γ∈Rn/2πZn

〈〈C, Trese〈a,z+iγ〉

(1 − e−〈α1,z+iγ〉) · · · (1 − e−〈αN ,z+iγ〉)〉〉.

The infinite sum∑

γ∈Rn/2πZn means the sum over a finite set F of represen-tatives of γ, for which the function

z 7→e〈a,z+iγ〉

(1 − e−〈α1,z+iγ〉) · · · (1 − e−〈αN ,z+iγ〉)

has a non-zero total residue. (For γ generic, it is holomorphic at zero, thusits total residue is zero. It is easy to prove that indeed the set F describedabove is finite.)

This result implies the periodic-polynomial behavior of i(a) on the sectorC − �(Φ), which contains C. In particular, this formula along rays yieldsback Ehrhart’s theorem (Section 1).

As in the case of Khovanskii-Pukhlikhov, this residue formula implies theformula of Cappell-Shaneson ([?]) and Brion-Vergne ([?]) for the partitionfunction as derivatives of the volumes of nearby polytopes.

12 Polytopes and symplectic geometry: very

few references

Let P be a convex integral polytope in Rn. Let T be the torus S1 × S1 ×

· · ·×S1, i.e. the product of n groups of circular rotations {eiθ}. Under someconditions (the polytope has to be a Delzant polytope, see Section ??), thereexists a compact symplectic manifold MP of dimension 2n, an action of thegroup T on MP and a map from MP to P , such that the preimage of thepoint p ∈ P is an orbit of T . In other words, the polytope P is exactly theparameter space for the orbits of the action of commuting rotations on MP .The map f is the moment map. We can thus think of the manifold MP as asort of inflation of the polytope P . We refer to the lecture of Michele Audinin the proceedings of the 5th EWM meeting held in Luminy for more details.

The inflated symplectic manifold corresponding to the interval [−1, 1] isthe 2-sphere S ⊂ R

3 with radius 1. We project S on R via the height z.The image of S is the interval [−1, 1]. The rotation group T = S1 acts byrotation around the axis Oz.

42

MP

11

−1−1

f

P

Figure 11: Moment map

The inflated symplectic manifold corresponding to the standard simplex∆ ⊂ R

n is the projective space Pn(C). We realize Pn(C) as the space

{(z1, z2, . . . , zn+1)| |z1|2 + · · · + |zn|

2 + |zn+1|2 = 1}/{z 7→ eiθz}

with identification of all proportional points z and eiθz in the sphere S2n+1

in Cn+1, so that Pn(C) is of dimension 2n. Rotations are given by

(eiθ1 , eiθ2 , . . . , eiθn) · (z1, z2, . . . , zn, zn+1) = (eiθ1z1, eiθ2z2, . . . , e

iθnzn, zn+1).

The moment map F : Pn(C) → Rn is given by

F (z1, z2, . . . , zn, zn+1) = (|z1|2, |z2|

2, . . . , |zn|2).

It is clear that the image of points in Pn(C) are in ∆, indeed x1 = |z1|2 ≥

0, . . . , xn = |zn|2 ≥ 0, and x1 + x2 + · · · + xn = 1− |zn+1|

2 ≥ 0. Furthermoretwo points having the same image are conjugated by (eiθ1 , eiθ2 , . . . , eiθn).

The set of integral points in k∆n provides a basis for the vector space V (k)of homogeneous polynomials of degree k in (n + 1) variables (in other terms,V (k) is the space H0(Pn(C),O(k))), the point (p1, p2, · · · , pn) with pi ≥ 0

and p1 + p2 + · · · + pn ≤ k indexing the monomial zp1

1 zp2

2 · · · zpnn z

(k−Pn

i=1 pi)n+1 .

43

Here are a few references for relations between the set of integral pointsin polytopes and toric varieties.

• M. BRION. Points entiers dans les polytopes convexes. Expose 780.Seminaire Bourbaki, 1993-94. Asterisque 227 (1995).

• W. FULTON. Introduction to Toric Varieties. Annals of MathematicsStudies, 131. Princeton University Press, Princeton, NJ, 1993.

• V. GUILLEMIN. Moment maps and combinatorial invariants of Hamil-

tonian T -spaces. Birkhauser-Boston-Basel-Berlin. 1994. Progress inMathematics vol. 122.

The following review articles explain the general context of quantificationof symplectic manifolds and give further references.

• R. SJAMAAR. Symplectic reduction and Riemann-Roch formulas for

multiplicities. Bull. A. M. S. 1996, 38, pp 327–338.

• M. VERGNE. Convex polytopes and quantization of symplectic mani-

folds. Proc. Nat. Acad. Sci. USA 93 (1996) 25, 14238–14242.

• M. VERGNE. Quantification geometrique et reduction symplectique.

Seminaire Bourbaki, mars 2001.

References

[1] W. BALDONI-SILVA and M. VERGNE – Residues formulae for volumes

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