Residue Formulae for Verlinde Sums, and for Number of Integral Points in Convex Rational Polytopes

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 Mich`ele Vergne Notes by Sylvie Paycha

1 Introduction

These two lectures are dedicated to two topics and their contents are independent. In the first lecture on Bernoulli series and Verlinde sums, I present some algebraic formulae concerning a (fascinating) subject of which I am rather 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-negative integer, and the fact that these functions are polynomials is not obvious at all from the definition of w(k). There is a common geometric concept underlying 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 can associate to M a quantized vector space Q(M,ω). The Riemann-Roch theorem asserts in particular that the dimension of Q(M,kω) is a polynomial in k with leading term kdim M/2 vol(M), where vol(M) is the symplectic volume of M. The underlying manifold M in the first topic is the moduli space of flat connections on a Riemann surface with holonomy t around one hole. Verlinde sums yield the dimension of Q(M,kω) while Bernoulli series compute the volume of such manifolds. Underlying manifolds in the second topic are toric manifolds. An integral polytope P determines a toric manifold M of dimension 2 dim P together with a symplectic form ω on M. The volume of the polytope P is the volume of M and the number of integral points in kP is the dimension of Q(M,kω). It is a polynomial in k with leading term kdim P vol(P ). In both cases, inspired by the Riemann-Roch theorem, it is possible to develop a purely algebraic Riemann-Roch calculus, and to prove directly a beautiful relation between Bernoulli series and Verlinde sums, as well as between volumes of polytopes and number of integral points belonging to the polytope. The main idea of this purely algebraic relation in the first case is an idea of A. Szenes while in the second case it goes back to G. Khovanskii and A.V. Pukhlikov.

2 We have a sort of Riemann-Roch relation between two functions, one depending on a continuous parameter t, the volume, and the other on a discrete parameter (the integer k), the dimension of a vector space. What about showing directly the relation between these two quantities by ”computing” them explicitly? Thus my aim in both lectures is to give a hint of ”explicit residue formulae” for all these quantities (Bernoulli series, Verlinde sums, volume of polytopes, number of integral points in polytopes), formulae which allows both concrete calculations, and quick proofs of the Riemann-Roch relations. Many thanks to Sylvie Paycha, for writing and expanding these notes, to Velleda Baldoni-Silva for illustrating them and to Charles Cochet, Shrawan Kumar 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 Bernoulli series to Verlinde sums. I hope to show in particular that residue formulae for series or sums are very eﬃcient tools to relate these sums and calculate them. The one dimensional case is presented here in detail and is already very 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 in algebraic geometry (Durham 1993), 241-253, London Math. Soc. Lecture Note 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’s formula (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 residue formulae) of articles of Jeﬀrey-Kirwan and Bismut-Labourie on the Verlinde formulae. References for the geometry underlying this calculation are given at 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 the ﬁrst k + 1 numbers, and to compare them to the corresponding integral k m km+1 0 x dx = m+1 . R For m = 0, we have:

00 +10 +20 + ··· + k0 = k +1. For m = 1, we have:

k2 + k 0+1+2+3+ ··· + k = . 2 For m = 2, we have: k3 k2 k 02 +12 +22 +32 + ··· + k2 = + + . 3 2 6 5 For m = 3, we have:

k4 k3 k2 03 +13 +23 +33 + ··· + k3 = + + . 4 2 4

··· This polynomial behavior is a general feature of such sums as the following theorem shows:

Theorem 1 (Jakob BERNOULLI -(1654-1705)) k m th For any positive integer m, the sum Sm(k)= a=0 a of the m -powers of the ﬁrst k +1 integers is given by a polynomialP formula in k.

We will prove this theorem and relate the sum Sm(k) to the Bernoulli polynomials. Given a real number t, the Bernoulli polynomials B(m, t) are deﬁned by the following relation:

etz ∞ zm z = B(m, t) . ez − 1 m! mX=0 etz This means, we expand z ez−1 in a Taylor series at z = 0. The coefficient zm of 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, 1 B(1,t) = t − , 2 1 B(2,t) = t2 − t + , 6 3 1 B(3,t) = t3 − t2 + t, 2 2 1 B(4,t) = t4 − 2t3 + t2 − , 30 ···

6 It follows immediately from the definition of the Bernoulli polynomials that d B(p, t) = pB(p − 1,t), dt 1 B(p, t)dt = 0 if p> 1. Z0 p p (Notice also that (−1) B2p =(−1) B(2p, 0) is positive.) The Bernoulli number Bm is defined to be B(m, 0). Bernoulli numbers satisfy the following relation: z ∞ zm = B . ez − 1 m m! mX=0 Since 1 1 = +1, 1 − e−z ez − 1 ∞ zm ∞ (−z)m we have m=0 Bm m! + z = m=0 Bm m! so that the Bernoulli numbers 1 are equalP to 0 for m odd, exceptP when m = 1 in which case we have B1 = − 2 . Now remark that 1 (B(m +1,t + 1) − B(m +1,t)) (m + 1)! is the coefficient of zm+1 in the Taylor series of e(t+1)z etz etz(ez − 1) z( − )= z = zetz. ez − 1 ez − 1 ez − 1 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.

7 Adding up these equations, we obtain

B(m +1, k + 1) − B(m +1, 0)=(m + 1)Sm(k). As e(t+1)z e((−t)(−z)) z =(−z) , ez − 1 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 formula in k: 1 S (k)= ((−1)m+1B(m +1, −k) − B ). m m +1 m+1 3 Bernoulli series and residues

In order to describe the Bernoulli polynomials in terms of series, it is use- E(z) ful to consider rational functions of the type φ(z) = zp where E(z) is a polynomial. The sum over all non-zero integers n ∈ Z, n =6 0: B(φ)(t)= φ(2iπn)e2iπnt Xn6=0 converges absolutely at each real number t if p is a sufficiently large integer 1 since n6=0 nα converges for α sufficiently large. It defines a periodic function of t. ItP always converges as a generalized function of t, which is smooth when t∈ / Z. For 0

From the deﬁnition of B(p, t), it follows that ext B(p, t)= −p! residue (x−p ) x=0 1 − ex so that setting φ(z) = z−p, we obtain, for 0

8 Proposition 3 Let 0

B(p, t) ext e2iπnt = −residue (x−p )= − . p! x=0 1 − ex (2iπn)p Xn6=0

It is important to notice that the expression on the right hand side is periodic with respect to t, while the left hand side is polynomial. We will call the right hand side the Bernoulli series. The Bernoulli polynomial and the Bernoulli series coincide ONLY on the interval 0

∞ −k where ζ(k)= n=1 n is the zeta function at point k. 1 For g = 1,P as B2 = 6 , we have

∞ 1 π2 = . n2 6 Xn=1 4 Trigonometric sums

We now consider a closely related sum, which will be a special case of a Verlinde sum. Let p be an integer and let 1 F (z)= . (1 − z)p

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

e2iπnt/k W (p, t)(k)= ωtF (ω)= . (1 − e2iπn/k)p ωk=1X,ω6=1 1≤nX≤(k−1)

It is very similar to the formula for 0

9 B(p, t/k) e2iπnt/k kp = − . p! (2iπn/k)p Xn6=0 In fact, when k tends to ∞, it is not diﬃcult to see that k−pW (p, t)(k) −p Bp tends to 0 if p is odd, while if p is even, k W (p, t)(k) tends to − 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, and k ≥ 1. Then the function k 7→ W (p, t)(k) is given by a polynomial formula in k. If p is even, or if p = 1, this polynomial is of degree p with highest p Bp degree term −k 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:

etx 1 W (p, t)(k)= −k residue dx . x=0 (1 − ex)p (1 − e−kx)

zt zk dz Proof. Consider the 1-form k (1−z)p (1−zk) z . From the conditions 0 ≤ t ≤ p, and k ≥ 1, it follows that this form has no residues at 0 and ∞. Poles of the 1 2iπn/k factor (1−zk) arise at z = e (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 of residues of this 1-form is equal to 0, we obtain from the residue theorem

zt zk dz W (p, t)(k)= k residue . z=1 (1 − z)p (1 − zk) z

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

etx 1 W (p, t)(k)= −k residue dx . x=0 (1 − ex)p (1 − e−kx)

k This last expression depends on k via the Laurent expression of 1−e−kx . Recall from the deﬁnition of the Bernoulli numbers that

kx ∞ (kx)r = B . ekx − 1 r r! Xr=0

10 Hence ∞ kr W (p, t)(k)= − B R (p, t) r r r! Xr=0 where etx R (p, t)=(−1)rresidue xr−1dx r x=0 (1 − ex)p 1 so that Rr(p, t) vanishes for large r. In fact, as (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 polynomial of degree less or equal than p. More precisely, it is of degree p for even p and of degree less or equal to p − 1 for odd p, p =6 1. For p even or p = 1, the kp highest degree term is −Bp p! . If p is odd, p =6 1, the term of degree p in k is kp−1 equal to 0 as Bp = 0, while the term of degree (p−1) is −(p/2−t)Bp−1 (p−1)! .

5 A relation between Bernoulli series and trigonometric sums

Deﬁne:

k−1 1 V (q, k)= . 4q(sin(πn/k))2q Xn=1 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

e2iπnq/k (−1)qW (2q, q)(k) = (−1)q (1 − e2iπn/k)2q 1≤nX≤(k−1) k−1 1 = (1 − e2iπn/k)q(1 − e−2iπn/k)q Xn=1 k−1 1 = . 4q(sin(πn/k))2q Xn=1 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 that V (q, k) is positive.

11 The polynomial k 7→ V (q, k) is of degree 2q and its highest degree term is B (−1)q+1 2q k2q. (2q)! q+1 (In our conventions for Bernoulli numbers, (−1) B2q 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 Taylor series at the origin of the function of x (x/2)2q Aˆ(q, x): = (sinh(x/2))2q q q q2 = 1 − x2 +( + )x4 + ··· . 12 1440 288 ∂ ˆ ˆ Substitute x = ∂t in A(q, x), and consider A(q,∂) as a series of diﬀerential ∂ ˆ operators in powers of ∂ := ∂t . The action of A(q,∂) on a polynomial function of t is well deﬁned.

Theorem 5 k2q V (q, k)=(−1)(q+1) (Aˆ(q,∂/k) · B(2q,t))| . 2q! t=0

From this expression, we see again that the highest degree term of the q+1 k2q polynomial function k 7→ V (q, k) is (−1) B2q 2q! . (Remark. In the context of the Verlinde formula, this theorem is closely related 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 of genus g. This manifold is provided with a line bundle L. The above expression arises when calculating the integral ch(Lk−2)Aˆ(M ) (where ch is the Mg g

Chern character and Aˆ(Mg) is the Aˆ genus).R This evaluates the dimension of the space of so-called generalized theta functions: the dimension of the space k−2 of holomorphic sections of the holomorphic line bundle L over Mg. We shall come back to this analogy later.) Proof. How to prove this theorem: Consider the residue formula for the Bernoulli polynomial ext B(2q,t)= −(2q!) residue x−2q . x=0 (1 − ex)

12 We can apply the series Aˆ(q,∂/k) to this expression. Under the residue, any analytic function is automatically replaced by its Taylor series. Thus we obtain k2q ext (−1)q+1 Aˆ(q,∂/k)B(2q,t) = (−1)qk2qresidue Aˆ(q, x/k)x−2q 2q! x=0 (1 − ex) 1 ext = residue . x=0 (1 − ex/k)q(1 − e−x/k)q (1 − ex)

Thus at t = 0, we obtain

2q q+1 k (−1) Aˆ(q,∂/k)B(2q,t) |t=0 2q! 1 1 = residue . x=0 (1 − ex/k)q(1 − e−x/k)q (1 − ex) The change of variables x 7→ −kx leads to

2q q+1 k (−1) Aˆ(q,∂/k)B(2q,t) |t=0 2q!

1 1 = −k residue x=0 (1 − ex)q(1 − e−x)q (1 − e−kx) eqx 1 = −(−1)q k residue . x=0 (1 − ex)2q (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

2q q+1 k (−1) Aˆ(q,∂/k)B(2q,t) |t=0 = V (q, k) 2q! k−1 1 = . (1 − e2iπn/k)q(1 − e−2iπn/k)q Xn=1 This proves Theorem 5.

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

13 ˆ e2iπnt Indeed applying formally A(q,∂/k) to the sum − n6=0 (2iπn)2q expressing 1 the Bernoulli polynomial 2q! B(2q,t), we would obtain:P

k2q 1 (−1)(q+1) Aˆ(q,∂/k)B(2q,t) | = (−1)qk2q Aˆ(q, 2iπn/k) 2q! t=0 (2iπn)2q Xn6=0 1 = . (1 − e2iπn/k)q(1 − e−2iπn/k)q Xn6=0

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

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

k 1 Ver(q, k) := (2(k + 2))qV (q, k +2)=2−q(k + 2)q sin((n + 1)π/(k + 2))2q Xn=0 takes positive integral values on integers k. Notice that for k = 0, we have Ver(q, 0) = 1. We have 1 Ver(1, k) = (k + 1)(k + 2)(k + 3), 6 1 Ver(2, k) = (k + 2)2(k + 3)(k + 1)(k2 +4k + 15), 180 1 Ver(3, k) = (k + 2)3(k + 1)(k + 3)(2k4 + 16k3 + 71k2 + 156k + 315), 7560 ...

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

14 6 Preliminaries on semi-simple Lie algebras

The Verlinde sums we saw corresponded to the Lie group SL(2, C) with Lie algebra sl(2) and compact form SU(2). Before introducing more general Verlinde sums, we need to recall some basic facts on the representation theory of semi-simple Lie algebras. A Lie algebra g over a field C is called semi-simple if its Cartan-Killing form h·, ·i : g × g → C defined by hX, Y i = tr(ad(X)ad(Y )) is non-degenerate. Any semi-simple Lie algebra is the direct product of simple Lie algebras (a Lie algebra g is called simple if it has no proper ideals). An example of simple Lie algebra is the algebra sl(n) of all n × n matrices with trace equal to 0. Then, up to normalization, the Killing form is hX, Y i = tr(XY ) where tr is the ordinary trace. x1 x2 In particular sl(2) := { | x1, x2, x3 ∈ C} is a simple Lie alge- x3 −x1 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 of sl(n), the algebra h is the set of diagonal matrices (with zero trace). The rank of sl(n) is n − 1. For sl(2), we write an element t of h as t 0 t = 1 . 0 −t1 The restriction of h·, ·i to h × h is non-degenerate. We can then identify h and its dual h∗. A root is an element α ∈ h∗, α =6 0, such that the corresponding root space 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 be the unique element of h such that λ(H ) = 2 hλ,α i, for any λ ∈ h∗. i hαi,αii i This element is well deﬁned, since h·, ·ih×h is non-degenerate. Thus we have αi(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 integral weight 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 ⊂ hR the lattice of integral weights. We denote by Q its 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 hλ,αi= 6 0 for all α ∈ ∆. We denote by Preg ⊂ hR the 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.P If V is an irreducible ﬁnite-dimensional representation of g, then V has a unique 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 ∈ hR| λ(Hi) non − negative integer for 1 ≤ i ≤ r} of all dominant integral weights. Given λ ∈ D, there is exactly one equivalence class of ﬁnite dimensional irreducible representations of g admitting λ as its highest weight. Let Uλ be a representative of this class. In other words, the representations Uλ, λ ∈ D, exhaust all the irreducible representations of ﬁnite dimension of g up to equivalence.

7 Witten series

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

e(2iπλ,t) g B(p, )(t)= p . ( + 2iπhα, λi) λ∈XPreg α∈∆ Q This series converges for p sufficiently large. For any p, it is well defined as a generalized function of t. The function of t ∈ hR defined by this series is periodic with respect to the lattice Q. On hR/Q (represented by a domain in hR), the expression B(p, g)(t) above is polynomial in sectors delimited by hyperplanes. On each sector, these series can in fact be expressed in terms of the Bernoulli polynomials in one variable. In the case of sl(2) one recovers:

e2iπt1n B(p, t )= −B(p,sl(2))(t)= − . 1 (2iπn)p Xn6=0 Here t1 0 t ∈ hR = 0 −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 them eﬀectively and to prove their polynomial behavior in sectors.

17 Consider the example of sl(3): we get

e2iπ(t1n1+(t1+t2)n2) B(q,sl(3))(t1,t2)= q q q . (2iπn1) (2iπn2) (2iπ(n1 + n2)) n16=0,n26=0X,n1+n26=0

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

t1 0 0 0 t2 0  . 0 0 −(t + t )  1 2 

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

e2iπ(t1n−t2m) (−1)q , (2iπn)q(2iπm)q(2iπ(n + m))q n6=0,m6=0X,n+m6=0

The residue formula depends on the position of (t1,t2) in sectors (as in the e2iπnt one dimensional case, the residue formula for the similar sum n6=0 (2iπn)p was only valid for 0

q (−1) B(q,su(3))(t1,t2)=

e{t1}x−{t2}y 1 −residue residue x=0 y=0 xqyq(x + y)q (1 − ex)(1 − e−y) e{t1+t2}x+{t2}y 1 +residue residue . x=0 y=0 xqyq(x + y)q (1 − ex)(1 − ey) Here {t} denotes t − [t] where [t] is the integral part of t. 1 For example, we have B(2, su(3))(0, 0) = − 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 already lead to e2iπ(t1n−t2m) (2iπn)q(2iπm)q(2iπ(n + m))q n6=0,m6=0X,n+m6=0 and the second residue should lead to the sum

e2iπ(t1+t2)n+(t2)(n+m) (2iπn)q(2iπm)q(2iπ(n + m))q n6=0,m6=0X,n+m6=0

18 which is equal, as seen from the first formula for B(q,sl(3))(t1,t2), after changing n1 and n2. But, in fact the two iterated residues are not equal, and Szenes formula shows 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 algebra is a compact form of g. Up to some normalization, for p =2g−1, the Witten series computes the symplectic volume of the manifold M(G,g,t) of moduli space of flat connections on a G-bundle on a Riemann surface of genus g with one hole (the variable t ∈ hR parametrizes the holonomy of the flat connection around the hole). We describe this manifold in Section 9, and give some references for its geometric meaning.)

8 Verlinde sums

One interesting generalization of the sum

k−1 1 V (q, k)= 4q(sin(πn/k))2q Xn=1 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 associated 1 cone of dominant integral weights. Let ρ = 2 α∈∆+ α. Let θ be the highest root. Let us also introduceP the set (called an alcove): hλ, θi A := λ ∈ D| 2 ≤ k k hθ, θi

(the deﬁnition of Ak does not depend of the choice of the scalar product.) hρ,θi Let h be the dual Coxeter number h = 2 hθ,θi +1. When g = sl(n), then h = n. Consider the scalar product < ·, · > such that <θ,θ>= 2. Let q be an integer. The Verlinde sum is

19 q rq 1 Ver(q, g)(k) = cG(k + h) 2iπ<α,u+ρ> 2iπ<α,u+ρ> + q − + q + k h k h uX∈Ak α∈∆ (1 − e ) (1 − e ) Q 1 = cq (k + h)rq . G π<α,u+ρ> 2q + (2 sin( )) uX∈Ak α∈∆ k+h Q 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 in Section 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 and to compare it to the Witten series. For g = sl(3), the above sum is

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

π(n + 1) π(n + 1) π(n + n + 2) −2q × 8 sin( 1 ) sin( 2 ) sin( 1 2 ) . k +3 k +3 k +3 n1≥0,n2≥X0,n1+n2≤k It is known (and amazing) that the function k 7→Ver(q, g)(k) takes integral values on integers: it is the dimension of a vector space V arising as the space of generalized theta functions which is the space of holomorphic 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. It was proved independently by Beauville-Laszlo, Faltings, Kumar-Narasimhan- Ramanathan that the space V is isomorphic to the space L of conformal blocks extensively studied by Tsuchiya-Ueno-Yamada. From their factoriza- tion theorem, one obtains the Verlinde formula giving the dimension of the space V as Ver(q, g)(k). There is a relation between the Witten series and the Verlinde sum: the Verlinde sum is obtained from the Witten series by applying a series of differential operators Aˆ(q,∂/(k +h)) to the Witten series. The form of the wanted operator Aˆ(q,∂/(k + h)) can be guessed from the same intuitive argument of “differentiating” under the sum sign. The correct argument follows immediately from the explicit residue formula. It allows to compute the integral k ˆ ˆ M(G,g) ch(L )A(M(G, g)) (ch is the Chern character and A(M(G, g)) is the ARˆ genus), at least when k is sufficiently large.

20 It is known that there exists an integer d > 0 such that the function k 7→ Ver(q, g)(dk) is polynomial in k. What is not known is the value of the smallest possible d. It is known that d = 1 for g = sl(n). The following conjecture on d is natural in view of Kumar-Narasimhan work on the Picard group of M(g, G, 0) (Math. Ann. 308 (1997) 155-173):

d = 1for Cr,

d = 2for Br(r ≥ 3),Dr(r ≥ 4), G2,

d = 6for F4,E6,

d = 12for E7,

d = 60for 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 Verlinde sums 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 element of the Cartan subalgebra hR of the complex semi-simple Lie algebra gC. If −1 −1 u1, u2 ∈ G, we denote [u1, u2]= u1u2u1 u2 . The manifold M(G,g,t) is defined to be: g 2iπt M(G,g,t)= {(u1, v1,...,ug, vg)| ui, vi ∈ G such that [ui, vi]= e }/T. Yi=1 Here T denotes the maximal compact torus of G with Lie algebra ihR. The notation /T means that we identify the (2g)-tuples (u1, v1,...,ug, vg) and ′ ′ ′ ′ ′ −1 ′ −1 (u1, v1,...,ug, vg) if there exists t ∈ T such that ui = tuit and vi = tvit . • 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 due to M. Thaddeus, who related volumes and Bernoulli numbers. M. Thaddeus: Conformal field theory and the cohomology of the moduli 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 form of 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 computation of its symplectic form is in –A. Alekseev, A. Malkin, E. Meinrenken: Lie group-valued moment maps. J. Differential Geom 1998, 48, pp 445-495 A quick computation of Witten formulae for its symplectic volume is in: –A. Alekseev, E. Meinrenken and C. Woodward: Duistermaat-Heckman distributions 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-conformal ﬁeld 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 generalized 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 family of 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 on moduli spaces of moduli spaces of holomorphic vector bundles of ar- bitrary rank on a Riemann surface. Ann. of Math. 1998, 148, pp 109–196.

• A proof of the Verlinde formula for general compact simply connected groups and any number of holes (but with restrictions on k) is obtained via the Riemann-Roch theorem and multi-dimensional residue calculus in: –Jean-Michel Bismut and Fran¸cois Labourie. Symplectic geometry and the Verlinde formulae. Surveys in differential geometry: differential geometry 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 Verlinde type 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 of points with integral coordinates in convex rational polytopes. My own interest in this topic comes from my efforts to understand the relations between symplectic manifolds and group representations, the existence of such relations being the Credo of quantum mechanics. Quantum mechanics enables us to associate discrete quantities to some geometric objects. For example, the volume of a compact symplectic manifold (M,ω), with an integral closed non-degenerate 2-form ω, has a discrete analogue which is the dimension of the vector space given by the quantum model Q(M,ω) for M. It is important to understand the relation between both quantities. The dimension q(k) of the vector space Q(M,kω) is a polynomial in k, and the volume of the manifold M is the limit of k− dim M/2q(k) when k tends to ∞. The full expression for 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 comparison problem is the following: if P ⊂ Rn is a convex polytope with integral (or rational) vertices, can we compare the number |P ∩ Zn| of points in P with integral coordinates and the volume of P ? It is clear that the volume of P is obtained as the limit when k tends to ∞ of k−n|kP ∩ Zn|. Can we give more precise relations ? There is a dictionary between polytopes and some classes of compact symplectic 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 algebraic and based on generating functions combined with an algebraic recipe due to Jeffrey-Kirwan ([?]) for the inversion of Laplace transforms. We shall only give some references on the correspondence between polytopes and symplectic geometry in the last section. I hope to show, in this lecture, that calculating the volume of a convex rational polytope or calculating the number of points with integral coordinates inside this polytope are similar problems (both difficult). Convex polytopes arise in many fields of mathematics: symplectic geometry, representation theory, algebraic geometry. Computing volumes is important. It is also important to study integral points in rational polytopes: they are the integral solutions of systems of linear inequations with integral coefficients. For example, solving the equation 5x +10y +20z = 105 with x,y,z positive integral 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 number of solutions is 36). Similarly regulating flows in networks is reduced to linear inequalities. Notice that already the problem of finding if there exists an integral point in a rational polytope is non-trivial (a polytope is called rational if its vertices have rational coordinates). H. Lenstra ([?]) showed in 1983 that there is, for a fixed dimension n, a polynomial time algorithm checking if a rational polytope P in Rn contains an integral point: |P ∩ Zn|= 6 ∅, and A. Barvinok ([?]) showed in 1994 that there is, for a fixed dimension n, a polynomial time algorithm giving the number |P ∩ Zn| of integral points in P . I shall explain here explicit and very similar formulae for both the volume and the number of integral points in rational convex polytopes. The formula for the volume is deduced in a straightforward way from the inversion formula of Jeffrey-Kirwan ([?]) for the Laplace transform. The formula for the number of points given in ([?]) follows from a multidimensional residue theorem, deduced from the one dimensional residue theorem with the help of a result due to A. Szenes ([?]) on separation of variables. From these formulae, the relations obtained by Khovanskii-Pukhlikov ([?]) and more generally by Brion-Vergne ([?]), Cappell-Shaneson ([?]), Guillemin ([?]) between volumes and the number of integral points will become clear. The residue formulae for volume and number of points have a theoretical interest: for example, the periodic-polynomial dependence of the formula in terms of the inequalities defining the polytope is clear from the given formulae. They can also be used for the computation of these quantities, at least for network polytopes. This is work in progress with Velleda Baldoni and Jesus de Loera.

3 c

0 b a

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 Rn defined by linear inequations. A third way of representing a convex polytopes is as a set of positive solutions to linear equations: i.e. the first positive quadrant intersected with a 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 the inequations x y z x ≥ 0, y ≥ 0,z ≥ 0 and + + ≤ 1 a b c or is also clearly isomorphic to the convex polytope in R4 described by x y z x ≥ 0, y ≥ 0,z ≥ 0,h ≥ 0 and + + + h =1. a b c

A convex polytope is called integral if its vertices have integral coordinates. A convex polytope is called rational if its vertices have rational coordinates. Our topic of discussion here is the calculation of the volume of convex polytopes, together with the calculation of the number of points with integral coordinates in an integral convex polytope P .

4 k 8 7 6 5 4 3 2 1 0 k 0 1 2 3 4 5 6 7 8 2 Z2 (k+1)(k+2) Volume(k∆) = k /2 |(k∆ ∩ )| = 2 k =8

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 kn vol(k∆) = . n! The number of integral points in k∆ is given (as we shall see shortly) by (k + 1)(k + 2) ··· (k + n) p (k)= . n n!

We refer to Section ?? for the relation between points in k∆ and a basis of vector spaces V (k) attached to P . xn Thus we see that the discrete analogue to the monomial n! giving the (x+1)(x+2)···(x+n) volume of k∆ for x = k is the polynomial pn(x) = n! which gives the number of integral points in k∆ for x = k. This polynomial pn(x) xn has same leading term n! , but it has the advantage that it takes integral values on all integers, and any polynomial which takes integral values on all integers is a linear combination with integral coeﬃcients of such polynomials pn(x).

5 More generally, to any integral convex polytope P is associated a polynomial 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 n iP (k) = cardinal(kP ∩ Z ), k =0, 1, 2,... is given by a polynomial expression in k called the Ehrhart polynomial. The n 1 n−1 Z ﬁrst two leading terms of this polynomial are vol(P )k + 2 vol (δP )k +··· . 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 δP of the polytope P . The volume of a face F of the boundary δP is computed using a normalized measure built from the Lebesgue measure on the affine space AF spanned by the face. The normalization is done in such a way that n the fundamental domain of the lattice Z ∩ 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, the first 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 of the 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→ f(ξ) Zn ξ∈kPX∩ n+m is a polynomial function of k. The leading term of this polynomial is k P f(x)dx. We also note here the important reciprocity law for the Ehrhart polyno-R 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 0 n iP 0 (k) = cardinal(kP ∩ Z ) to be the number of integral elements in the interior of kP . Then n iP 0 (k)=(−1) iP (−k).

6 0 k |(0, k) ∩ Z| = k − 1

0 k |[0, k] ∩ Z| = k +1

0 k

lenght [0, k]= k

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 than 0 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 dilated polytope kP is a polynomial in k, with leading term kn vol(P ) is far from being obvious. Let us give an example involving dilatations in just one direction, where we shall see that even the asymptotics between the volume and the number of integral points is only true under dilations in all directions. Example: The hanging pyramid. + 3 Let m ∈ R . The hanging pyramid Pm is the convex hull in R of the vertices 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, but there are no integral points in Pm other than its 4 vertices. Whatever the

7 3 Vol(Pm)= m/6 |Pm ∩ Z | =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 pyramid projects 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 4 vertices. The asymptotics in m of the volume and the number of integral points are therefore very diﬀerent in this example. Before going to the proof of Ehrhart’s theorem, let us have a look at the formulae relating the number of points in an integral convex polytope and its volume in dimension 1 and 2:

In dimension 1: Let us consider the polytope given by the interval P = [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 a convex polytope with integral vertices dilated by k is given by: k i (k)= k2 vol(P )+ (number of integral points on the boundary of P ) +1. P 2 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 convex polytope in unions and differences of integral simplices. Thus we consider (n+1) points in Zn and the convex polytope given by the convex hull of these points. Ehrhart gave a formula for the number of integral points contained in this set. n Let us start with the standard simplex ∆ in R , 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 the number p(n, k) of solutions in non-negative integers (x0, x1,...,xn) of the equation 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 (k + 1)(k + 2) ··· (k + n) p(n, k)= . n!

9 Proof. The result is obvious for k = −1, −2,..., −n, both sides of the formula being equal to 0. Let us set k ≥ 0 and consider the generating function: ∞ −kt Zk(t)= p(n, k)e , t> 0. Xk=0 −kt We shall recover p(n, k) as the coeﬃcient of e in Zk(t). We have

−t(x0+x1+···+xn) Zk(t) = e Z+ n+1 (x0,...,xnX)∈( ) n ∞ = e−txi i=0 x =0 Y Xi 1 = . (1 − e−t)n+1

∞ k 1 1 ∞ k+n k Diﬀerentiating k=0 z = 1−z , we get (1−z)n+1 = k=0 n z and hence

P ∞ P 1 k + n = e−kt (1 − e−t)n+1 n Xk=0 k+n (k+1)(k+2)···(k+n) with n = n! . Corollary 5

(k + 1)(k + 2) ··· (k + n) i (k) = cardinal(k∆ ∩ Zn)= . ∆ n!

Notice that i∆(k) is indeed a polynomial in k with constant term 1 and 1 n 1 (n+1) n−1 leading terms n! k + 2 (n−1)! k + ··· as announced in Ehrhart’s theorem 1 since vol(∆) = n! and the boundary of ∆ is the union of (n + 1) standard 1 simplices, each of volume (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 n Theorem 6 Let P be a simplex in R with integral vertices α0,α1,...,αn. The number of integer points in kP is

n k − r + n i (k)= cardinal((P, r)) P n r=0 X where for an integer r between 0 and n

n n n+1 n (P, r)= u ∈ [0, 1[ | ui = r, uiαi ∈ Z ( i=0 i=0 ) X X 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 have integer coordinates. An element of kP is an element of the convex hull of the 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 point w can be written in an unique way as:

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

n n with ki non-negative integers, 0 ≤ ui < 1, and i=0 ki + i=0 ui = k. As n n the elements αi are in Z , the point w is in Z if and only if P P n u0α0 + u1α1 + ··· + unαn ∈ Z .

n n Notice that the number i=0 ui = k− i=0 ki is a non-negative integer. This integer r is less or equal to n, as the value (n +1) for r would compel all ui P P 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 satisfy k0 + 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 and k ≥ 0, the integer k − r is greater or equal to −n and it follows from what we have proven before that the number of solutions of the equation

k0 + k1 + ··· + kn = k − r k−r+n is n , for any k ≥ 0. We therefore obtain Ehrhart’s formula: n k − r + n i (k)= cardinal((P, r)) . P n r=0 X This expression is polynomial in k.

The sets (P, r) are not very easy to determine. Furthermore, the decomposition of P in integral simplices is also not immediate. So the formula above has mainly a theoretical interest. We shall give later on more speciﬁc result on the Ehrhart polynomial.

Let P be a rational convex polytope: the vertices of P have rational coordinates, instead of integral coordinates. Ehrhart analyzed more generally n the behavior in k of the function k 7→ iP (k) = cardinal(kP ∩ Z ) and proved that this function is given by a“periodic-polynomial”, i.e. is a polynomial function in k with coeﬃcients periodic functions of k. In other words, there exists 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 < 1 d. For example, if P := [0, 2 ], then, with d = 2, iP (2k) = k + 1, while iP (2k +1) = k + 1. These two formulae can be assembled together to give k 1 k the periodic-polynomial function iP (k)= 2 +1 − 4 (1 − (−1) ). For a given rational polytope P , it is not easy to determine the smallest value p of the integer d with the property that functions k 7→ iP (dk + j) are polynomials in k. Of course, when P has integral vertices, we just proved that p = 1. Similarly, if q is an integer such that, for all vertices s of P , qs ∈ Zn, then p divides q. It may happen that p is strictly smaller than q. Example. Stanley’s pyramid. Consider the following example given by Stanley ([?]): let P be the convex polytope in R3 with vertices O = 1 1 (0, 0, 0), A = (1, 0, 0),B = (0, 1, 0), D = (1, 1, 0) and C = ( 2 , 0, 2 ). Then P is not integral, however (k + 1)(k + 2)(k + 3) i (k)= P 3! 12 C

D 0

1/2

A B

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 the formula that we used: k + n i (k)= . p n 2 Example 2. Let P be the triangle in R 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: k +2 k +1 i (k)= + =(k + 1)2. P 2 2 (The volume of P is 1.) 3 Example 3. Let P be the simplex in R 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: k +3 k +2 k +1 i (k) = +4 + P 3 3 3 = (k + 1)3.

(The volume of P is 1.) 4 Example 4. Let P the simplex in R 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 4 1 i (k)=(k + 1)(k + 2)(k2 + k + ). P 3 2 (The volume of P is 1.)

Example 5: The hanging pyramid Pm. Let Pm be the simplex with vertices 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 1 1 i (k)= (k + 1)(k + 2)(k +3)+(m − 1) (k − 1)(k)(k + 1) Pm 6 6 m k = k3 + k2 + (12 − m) +1. 6 6 m 12−m Notice on this last formula that we indeed have iPm (1) = 6 +1+ 6 +1 = 4. But we see here that the coeﬃcient of k in the Ehrhart polynomial is a negative number, when m is large enough. In particular, the coeﬃcients of Ehrhart polynomials are not necessarily positive.

14 mk3 |k(P ) ∩ Z3| = Vol(kPm)= 6 m

mk3 2 k 6 + k + (12 − m) 6 +1

Figure 5: Hanging pyramid

15 Finally, we include here a formula due to Mordell ([?]) for the number of points in the polytope P (a, b, c) deﬁned by the vertices 0, A = ae1, B = be2, C = ce3 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 in P (a, b, c) is not a rational function of a, b, c. Here is the Ehrhart polynomial for the number of points with integral coordinates contained in kP (a, b, c):

1 1 i (k)= abck3 +( (bc + ca + ab + 1))k2 P (a,b,c) 6 4 1 1 bc ca ab 1 +( (a + b + c +3)+ ( + + + ))k 4 12 a b c abc −(s(bc, a)+ s(ca, b)+ s(ab, c))k +1. Here s(p, q) is the Dedekind sum, deﬁned by

q i pi s(p, q)= (( ))(( )) q q i=1 X 1 with ((x)) = 0 if x is an integer and ((x)) = x − [x] − 2 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, the coefficients add up to 1. Thus Magic(n) is defined as the intersection of Rn2 the positive quadrant in , cut by the 2n linear equations i xij = 1 for all 1 ≤ j ≤ n and j xij = 1 for all 1 ≤ i ≤ n. Clearly the sum of coefficients in all lines is equal to the sum of coefficients on all columnsP and its value is n. Thus thisP 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 x x x x x  21 22 23 24 25 x31 x32 x33 x34 x35 x x x x x   41 42 43 44 45 x x x x x   51 52 53 54 55   with xij ≥ 0 and x11 + 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 number of integral elements in kMagic(n) is a polynomial of the form

(n−1)2 mnk + ··· +1, where mn is the volume of Magic(n). Finding iMagic(n)(k) boils down to counting magic squares: square (n × n)-matrices ﬁlled up with non-negative integers 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 1 2 4 2 .   2 1 5   We have iMagic(1)(k)=1,

iMagic(2)(k)= k +1, k +4 k +3 k +2 i (k)= + + , Magic(3) 4 4 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 computation 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 the Ehrhart polynomial for n = 9 requires 325 days of computer time on a 1GHz PC running under Linux. The leading term of the Ehrhart polynomial (the volume) requires 15 seconds for n = 7, 54 minutes for n = 8, 317 hours for n = 9. We now consider the Chan-Robbins-Yuen polytope CRYn, namely the subset of the set of (n × n) doubly stochastic matrices consisting of those with just one non-zero line above the diagonal permitted. For example, CRY5 is the set of matrices

x11 x12 0 0 0 x x x 0 0  21 22 23  x31 x32 x33 x34 0 x x x x x   41 42 43 44 45 x x x x x   51 52 53 54 55   with xij ≥ 0 and

xij =1, for all 1 ≤ j ≤ 5, i X and xij =1, for all 1 ≤ i ≤ 5. j X The dimension of this convex polytope is n(n − 1)/2. The corresponding Ehrhart polynomial is not known but the leading term is known (Chan- Robbins-Yuen ([?]) and Zeilberger ([?])). It is given by

n−2 (2i)! kn(n−1)/2 ( ) . (i + 1)!i! (n(n − 1)/2)! i=1 Y 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 difficult to compute. There are very few cases for which they are explicitly known. One can however hope to find some relations between them. A relation together with a way to compute the volume and the number of integral convex polytopes follows from Brion’s formulae ([?]), see also ([?],[?],[?]). This formula has been used by Barvinok ([?]) to give an algorithm to compute in polynomial time (when the dimension is fixed) the number 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 of variables, which leads to a direct comparison between volumes and number of points via residue formulae. Furthermore, the residue formulae seem to yield a more efficient computational tool than Brion’s formulae, at least for transportation polytopes. Michel Brion provided a formula for integral of exponential functions over convex polytopes or for sums of exponentials over integral points. They generalize the following formulae:

• For any a, b ∈ R b eay eby exydx = − + y y Za which follows by integration.

• For a, b integers

b ay by uy e e e = y + −y . Z 1 − e 1 − e u=Xa,u∈ This last result follows from writing integers between a and b as differences of integers strictly greater than b and integers greater or equal than a and summing arithmetic progressions. The first integral formula generalizes from the interval polytope [a, b] to a general convex polytope P ⊂ Rn. 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) s s s that at each vertex s ∈ V(P ) start n edges a1, a2, . . . , an of the polytope P : s Rn vectors ai are elements of such that near s the convex polytope P is the

19 Figure 6: Non-generic pyramid versus generic pyramids

n s set of points of the forms s+ i=1 tiai , with ti ≥ 0(ti small). In other words, n R+ s the tangent cone at s to P is the affine cone s + i=1 ai . P nRn R s s s We identify the exterior product Λ with , and |a1 ∧a2 ∧···∧an| is, by P s definition, the absolute value of the determinant of the (n×n)-matrix with ai s as column vectors. Elements ai are defined here only up to proportionality. Then, we have:

s s hs,yi n hx,yi |a1 ∧···∧ an|e (−1) e dx = s s . P ha1, yi···han, yi Z s∈VX(P ) Notice that this formula does not depend of the choice of the length of s the edges ai passing through the vertex s. With n = 1, setting P = [a, b], then there are 2 vertices a and b. We can a b choose a1 = b − a, a1 = a − b, and the above formula yields back

b |(b − a)|eay |(a − b)|eay eay eby exy = − + = − + . (b − a)y (a − b)y y y Za In particular, Brion’s formula gives a formula for vol(P ) knowing the vertices s and the edges through s.

s s n n |a1 ∧···∧ an|hs, yi vol(P )=(−1) s s ha1, yi···han, yi s∈VX(P ) 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 is an integral convex polytope. Furthermore, we assume that at each vertex s, s Zn s s we can choose integral vectors ai ∈ such that |a1 ∧···∧ an| = 1, and such n R+ s that, as before, the tangent cone at s to P is the aﬃne cone s + i=1 ai . We call such a convex polytope a Delzant polytope. An integral convex polytope is rarely a Delzant polytope: P 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

hs,yi hξ,yi e e = s s . ha1,yi han,yi Zn (1 − e ) ··· (1 − e ) ξ∈XP ∩ s∈VX(P )

Example. Brion’s formula to integrate an exponential over the standard simplex ∆ dilated by k in R2 is

1 eky1 eky2 ey1x1+y2x2 dx dx = + + , 1 2 y y (−y )(y − y ) (−y )(y − y ) Zk∆ 1 2 1 2 1 2 1 2 while the formula to sum up an exponential on all the integral points (p1, p2) in k∆ is

1 eky1 eky2 ey1p1+y2p2 = + + . (1 − ey1 )(1 − ey2 ) (1 − e−y1 )(1 − ey2−y1 ) (1 − e−y2 )(1 − ey1−y2 ) (p1,pX2)∈k∆

6 Partition polytopes

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

21 k

0 k

Figure 7: Simplex and edges through vertices

Let P and Q be two convex polytopes in Rn. The Minkowski sum is deﬁned as P + Q = {x + y| x ∈ P, y ∈ Q}. The following theorem, proved in ([?]), is a generalization of Ehrhart’s theorem.

n Theorem 8 • Let P1, P2,...,Pr be convex polytopes in R . Then the function

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

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

• Assume P1, P2,...,Pr are integral convex polytopes. Then the function on (Z+)r given by

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

is a polynomial function of (k1, k2,...,kr). We shall study more general families of convex polytopes.

22 Let Φ = [α1,α2,...,αN ] n be a sequence of N vectors in R (elements αi may not be all distinct). We assume that all vectors αi lie strictly on the same side of a hyperplane, so N Rn that Φ generates an acute cone C(Φ) = { i=1 tiαi| ti ≥ 0} in (an acute cone is a cone which does not contain any straight line). P N Let {w1,w2,...,wN } be the canonical basis of R . We deﬁne a linear map from RN to Rn by:

AΦ(x1, x2,...,xN )= xiαi. i=1 X

We may write the linear map A := AΦ asa(n×N)-matrix A with column vectors the vectors αi. For a ∈ C(Φ) deﬁne

−1 RN PΦ(a)= AΦ (a) ∩ + , 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 a convex polytope called partition polytope. Example. Transportation polytopes 2n Let R with canonical basis e1,...,en,f1,...,fn and let Φ be the set of n2 vectors Φ=[(ei + fj)| 1 ≤ i, j ≤ n]. Then A(xij)= xij(ei + fj).

The convex polytope PΦ(a1e1 +a2e2X+···+anen +b1f1 +b2f2 +···+bnfn) can be identified to the set of (n×n)-matrices (xij) with non-negative coefficients and with given sums of coefficients in each row and given sums of coefficients of each column. Indeed the vector equation xij(ei + fj)=(a1e1 + a2e2 + ···+anen +b1f1 +b2f2 +···+bnfn) is equivalent to the series of 2n equations: P n

xij = ai, j=1 X 23 n

xij = bj. i=1 X n n Of course, to get a solution, we need that i=1 ai = j=1 bj. When a1 = a2 = ··· = an = b1 = b2 = ··· = bn = 1, we recover the poly- P P tope Magic(n). For general (ai, bj) (such that ai = bj), this polytope is called a transportation polytope, and is very important in studying ﬂow in networks. P P

n If Φ is a sequence of vectors in Z , then PΦ(a) is a rational convex polytope. Example: Let A = (6, 10, 15), then A(x1, x2, x3)=6x1 + 10x2 + 15x3.

The polytope PΦ(k) is the convex hull of its 3 vertices k k k ( , 0, 0), (0, , 0), (0, 0, ). 6 10 15

If A is surjective, then the polytopes PΦ(a) are of dimension N − n, for a in the interior of C(Φ). The polytopes PΦ(a) are contained in aﬃne spaces parallel to E := Ker A. If we denote by f1,f2, ..., fN the restrictions of the N N linear forms x1, x2,...,xN on R to the subspace E, then for u ∈ R , the polytope PΦ(Au) is isomorphic to the polytope in E = Ker A described by the inequations Q(u) := {y ∈ E| hfi, yi + ui ≥ 0}. Indeed, if y ∈ Q(u), the point u + y lies in P (Au). Some of the inequalities above might be irrelevant. If Au = Av, polytopes Q(u) and Q(v) are just translations of each other in the space Ker A by the vector u − v. Consider the polytope PΦ(A(u + hwk)) = PΦ(Au + hαk). Then the polytope Q(u + hwk) is isomorphic to the polytope PΦ(Au + hαk) and is deﬁned by the same inequations as the polytope Q(u) except that one of the inequalities hfi, yi + ui ≥ 0 has been replaced for i = k by the inequality hfk, yi + (uk + h) ≥ 0. In particular all polytopes PΦ(a + h), when a is generic 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 deformations for the matrix:

1001 A = . 0111 2 The space Ker A is isomorphic to R with basis b1 = (w3 − w2), b2 = (w4 − w1 − w2). We write y ∈ Ker A as y = y1b1 + y2b2. To describe the family Q(u), we may vary u in a supplementary subspace to Ker A. We choose 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 this example).

• 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 u1 − u2

A u2 − u1

C A u2

0 B 0 B u2 u2

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 irrelevant and does not produce a face of Q(u).

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

Let us, following Gelfand-Kapranov-Zelevinski ([?]), decompose the cone C(Φ) as a union of “chambers”. By deﬁnition, a chamber is a connected component of the open subset of C(Φ) obtained by removing the boundaries of all the cones C(σ) spanned by subsets σ of Φ forming a basis of V ∗. In the case of the matrix 1001 A = 0111

26 e2 − e3

1 2 (e1 − e3)

1 e1 − e2 2 e2 1 3 e1

1 e3 2 e2

Figure 9: Chambers for A3

then α1 = e1 = α3, α2 = e2 and α4 = e1 + e2. The cone C(Φ) is the ﬁrst quadrant, but to deﬁne chambers, we have to remove the half-lines + + + R 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 polytope PΦ(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 wall n of a chamber. If {P1, P2,...,Pr} is a set of convex polytopes in R , then the family of polytopes obtained by taking their Minkowski sums

{t1P1 + ··· + trPr} with ti ≥ 0 can be embedded as a subset of a family of partition polytopes PΦ(a) where a varies in the closure of a chamber C of C(Φ). Figure 9 is the drawing of chambers for the matrix

100101 A := 010111   001011 For a general matrix A, it is diﬃcult to describe chambers of the cone C(Φ). The program PUNTOS available on the homepage of Jesus de Loera

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

7 Generating functions

The method we used to compute the volume as well as number of points of a rational convex polytopes is based on generating functions. We are in the setting of partition polytopes, with a surjective map A : N n N R → R . We denote by (w1,w2,...,wN ) the canonical basis of R , by αk = A(wk) and by Φ := [α1,α2,...,αN ]. We assume that all vectors αk lies strictly on an open half-space delimited by an hyperplane. Then the dual cone to the cone C(Φ) generated by Φ is an open cone: it consists of all n vectors z ∈ R such that hz,αki > 0, for all 1 ≤ k ≤ N. For a ∈ C(Φ), let

N i(a) = cardinal(PΦ(a) ∩ Z ) and v(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

1 v(a)e−ha,zida = , hα,zi ZC(Φ) α∈Φ 1 −ha,zi Q i(a)e = −hα,zi . Zn α∈Φ(1 − e ) a∈CX(Φ)∩ Q Proof. Let z in the dual cone to C(Φ) and let us compute

F (z) := e−hA(x),zidx. RN Z + N We ﬁrst write A(x)= i=1 xiαi. The integral reads P

28 −hP xiαi,zi F (z) = e dx1dx2 ··· dxN RN Z + N −xihαi,zi = e dxi R+ i=1 Y Z N 1 = . hα ,zi i=1 i Y On the other hand, we can use Fubini formula. We ﬁrst integrate over the x such that A(x)= a, then we integrate on a. Thus

F (z)= e−hA(x),zidx da. RN Za∈C(Φ) Z{x∈ + | A(x)=a} ! RN The set {x ∈ + | A(x) = a} is our partition polytope PΦ(a) and the −hA(x),zi −ha,zi integral over this set of e is e vol(PΦ(a)). We thus obtain the ﬁrst formula of the proposition. The second formula arises in the same way by calculating in two diﬀerent −hA(x),zi ways the sum ZN e . x∈ +

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

P (z) f(z) := , zM

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

f(z)= v(h)e−hzdh Zh≥0 for z > 0. I claim that v is given by the residue formula

hx v(h) = residuex=0(f(x)e ),

M up which is obvious to check. Indeed, f(z) is of the form p=1 zp , and we may p−1 directly check the formula for 1 = h dh. zp h≥0 (p−1)! P hx The dependance of v(h) in h is viaR the Taylor series of e at x = 0. The function f(x) has a pole at x = 0 of order M, thus we need to take the Taylor development of ehx only up to order M. In particular, it is clear that v(h) is a 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 P (z) F (z) := , (1 − z)M where P (z) is a polynomial in z. We assume that F (z) tends to 0 when z tends to ∞. Expand the function F (z) as a Taylor series at the origin. Then there exists a polynomial function i(k) such that

∞ F (z)= i(k)zk Xk=0 for |z| < 1. I claim that i is given by the residue formula dz i(k)= −residue (F (z)z−k ). z=1 z Indeed, integrating on a small circle near z = 0 the Taylor expansion of F , we obtain dz i(k) = residue (F (z)z−k ). z=0 z

30 −k dz From our assumption, it follows that the 1-form F (z)z z has no residue at ∞. Thus we obtain, from the residue theorem on P1(C), that for any k ≥ 0, dz i(k)= −residue (F (z)z−k ). z=1 z Writing z = e−x, we also obtain

−x kx i(k) = residuex=0(F (e )e ), which is strikingly similar to the formula

hx v(h) = residuex=0(f(x)e ).

The function F (e−x) has a pole of order M at x = 0. So it is clear that i(k) is a polynomial in k of degree less or equal to M − 1. A similar multidimensional residue formula is used in the ﬁnal formulae of Theorems ?? and ??. The same idea of moving a residue from a contour near 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 crucial tool 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 M N Q . Here M,N,Q are positive integers. Expand (1−z1) (1−z2) (1−z1z2)

k1 k2 F (z1,z2)= v(k1, k2)z1 z2 k1≥X0,k2≥0 in Taylor series. Then there exists two polynomial functions v1(a1, a2) and v2(a1, a2) such that v1(a, a)= v2(a, a) and such that

v(k1, k2)= v1(k1, k2) if k1 ≥ k2, while v(k1, k2)= v2(k1, k2) if k1 ≤ k2. Indeed, we have, from the Cauchy theorem,

1 1 dz dz v(k , k )=( )2 z−k1 z−k2 1 2 1 2 2iπ (1 − z )M (1 − z )N (1 − z z )Q 1 2 z z Z|z1|=ǫ1,|z2|=ǫ2 1 2 1 2 1 2 31 whatever small non-zero real numbers ǫ1, ǫ2 we choose. It is tempting, as in the one-dimensional case, to use the other pole z1 = z2 = 1 to compute this integral. When choosing exponential coordinates e−x1 ,e−x2 near this pole, we are led to consider the 2-form

1 k1x1 k2x2 e e dx1dx2. (1 − e−x1 )M (1 − e−x2 )N (1 − e−(x1+x2))Q In a neighborhood of (0, 0), the function 1 J(x1, x2)= (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} is holomorphic 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 1 v (a , a )=( )2 J(x , x )ea1x1+a2x2 dx dx 1 1 2 2iπ 1 2 1 2 ZC(ǫ1,ǫ2) with ǫ2 > ǫ1 > 0, while 1 v (a , a )=( )2 J(x , x )ea1x1+a2x2 dx dx , 2 1 2 2iπ 1 2 1 2 ZC(ǫ1,ǫ2) with ǫ1 > ǫ2 > 0. In other words, the analogue of the residue formula for

−x kx i(k) = residuex=0(F (e )e ), for the Taylor series of F in one variable is replaced by the two formulae (both obviously polynomial in k1, k2.)

−x1 −x2 k1x1+k2x2 v1(k1, k2) = residuex2=0residuex1=0(F (e ,e )e ),

−x1 −x2 k1x1+k2x2 v2(k1, k2) = residuex1=0residuex2=0(F (e ,e )e ).

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 a hyperplane {α =0} in V . Let R∆ be the ring of rational functions on VC with poles on the hyperplanes {α =0}. If F ∈ R∆, we have, for z ∈ VC

P (z) F (z)= nα α∈∆hα,zi for some polynomial P and non-negativeQ 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 of 1 k+1 the function k+1 z except when k = −1. Thus there is just a particular 1 function z which has no primitive. We can write: 1 C[z,z−1]= C ⊕ ∂ C[z,z−1]. z z Let us come back to a general hyperplane arrangement and let us give a −1 description of R∆ which generalizes this decomposition of C[z,z ]. Let σ be a subset of ∆, such that elements of σ form a basis of V ∗. We will say that σ is a basic subset of ∆. Setting

σ = {αi1 ,αi2 ,...,αin }, we deﬁne the ”simple fraction” 1 fσ(z)= hαi1 ,zihαi2 ,zi···hαin ,zi and the space S∆ generated by simple fractions:

S∆ = Cfσ, σ X where σ runs over all basic subsets of ∆.

33 1 2 n Choose a basis {e ,e ,...,e } of V . The partial derivative ∂i acts on R∆. We denote n

∂R∆ = ∂iR∆. i=1 X 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 by the 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 the set + ∆ = {z1,z2, (z1 + z2)} 1 2 + − of linear forms on V := {z = z1e + z2e }. Let ∆ = ∆ ∪ ∆ . There ∗ + are 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 between the 3 corresponding simple fractions: 1 1 1 = + . z1z2 z1(z1 + z2) 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 SAn is obtained as follows: set 1 fw(z)= (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 SAn has 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 whenever F is a sum of derivatives. For instance, the residue of the following function z3 1 1 1 − + (z1 − z2)(z1 − z3)(z2 − z3)z1z2z3 (z1 − z2)(z2 − z3)z3 (z1 − z3)(z3 − z2)z2 vanishes. Indeed, we can verify that this function is equal to:

(z1 − 2z3) (z1 − 2z2) −∂2 + ∂3 . z3(z1 − z3)(z2 − z3) z2(z1 − z2)(z3 − z2)

The total residue vanishes on homogeneous functions of R∆ which are of degree m, whenever m =6 −n. Thus we can extend the total residue to P (z) functions F (z)= nα , where P is a holomorphic function deﬁned near Qα∈∆hα,zi 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→ hhC,φii on S∆, deﬁned by hhC, | det σ|fσii =1 if C ⊂ C(σ) hhC, | det σ|fσii =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 hhC,φii is diﬃcult in general. The linear form hhC,φii can be realized as an integration on a cycle in the space VC \∪α∈∆{α =0} depending of the chamber C.

35 It can be easy to describe the linear form hhC,φii for some very special cases such as

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

The space R∆ consists of rational functions of (z1,z2,...,zn) with poles on the 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 any dimension, one of the chambers of the cone C(Φ) above is 1 2 Cnice = {a1e + a2e + ··· + anen| a1 > 0, a2 > 0, . . . , an > 0}. Furthermore, in this case the linear form attached to this chamber can be written in terms of iterated residues:

hhCnice,φii = residuez1=0 ··· residuezn=0(φ(z1,z2,...,zn)). The linear forms attached to any chamber C are not too diﬃcult to compute, as we know an explicit basis of the space SAn . Example. Let us consider the matrix: 1001 0 1 A := 0 1 0 −1 1 0 .   001 0 −1 −1

  + The system Φ spanned by this matrix is the system A3 . 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 on C(ǫ1, ǫ2, ǫ3). However, the function f(z1,z2,z3) with poles on {zi = zj}, we see that the relative order on ǫ1, ǫ2, ǫ3 will lead to different cycles. Let us give two examples for two different chambers.

36 e2 − e3

1 2 (e1 − e3)

1 e1 − e2 2 e2 1 3 e1

1 e3 2 e2

Figure 10: Chambers for A3

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

1 hhC ,φii =( )3 φ(z ,z ,z )dz dz dz 1 2iπ 1 2 3 1 2 3 ZC(ǫ1,ǫ2,ǫ3)

where ǫ1 > ǫ2 > ǫ3 > 0.

• Let C2 be the the chamber spanned by e1 − e2,e1,e3. Then 1 hhC ,φii =( )3 φ(z ,z ,z )dz dz dz 2 2iπ 1 2 3 1 2 3 ZC(ǫ1,ǫ2,ǫ3) 1 3 −( ) φ(z1,z2,z3)dz1dz2dz3 2iπ ′ ′ ′ ZC(ǫ1,ǫ2,ǫ3) ′ ′ ′ where ǫ1 > ǫ2 > ǫ3 > 0 and ǫ2 > ǫ1 > ǫ3 > 0.

10 Jeﬀrey-Kirwan formula for the volume

Let Φ ⊂ V ∗ be a sequence of vectors belonging to ∆ and all lying on the same 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 proof is given in ([?]).

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

eha,zi vol(PΦ(a)) = hhC, Tres ii hα1,zi···hαN ,zi 1 ha,ziN−n = hhC, Tres ii (N − n)! hα1,zi···hαN ,zi 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 it is 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

N 1 z1 vol(CRYn)= residuez1=0 ··· residuezn=0 N! z1z2 ··· zn i

n−2 Q 1 = C N! i i=1 Y 2i! where the Ci = (i+1)!i! are the Catalan numbers.

In the general case of a convex polytope realized as PΦ(a), the algorithm to compute its volume is the following. We choose ∆+ any subset of ∆ containing Φ, and contained in a half-space. We compute (or better, we know as in the case of the system An) a basis fσ of the space S∆, indexed by a ﬁnite set F of subsets of ∆+. Then, given a generic point a ∈ V ∗, we compute 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 hhC, ·ii, as hhC, | det σ|fσii = 0 or 1 according to the fact that a ∈ C(σ) or not. Then we can realize the form hhC, ·ii as an iterated residue with respect to special orders, entirely determined by our chamber C.

38 11 Residue formula for the number of integral points in rational polytopes

Let Φ be a sequence of Zn all lying on the same side of a given hyperplane n n and spanning R . Then for a ∈ Z ∩ C(Φ), the polytope PΦ(a) is a rational polytope. Let me now state the relation between the partition function and the number of points in the rational polytope PΦ(a) and indicate some properties of the function i(a). Let us introduce a notation:

(Φ) = [0, 1]α α∈Φ X 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:

−ha,zi 1 i(a)e = −hα,zi Zn α∈Φ(1 − e ) a∈CX(Φ)∩ Q where i(a) is the cardinal of the set of solutions in non-negative integers nk of the equation a = n1α1 + n2α2 + ··· + nN αN . Then, for each chamber C of the cone C(Φ), there exists a periodic- n polynomial function iC on R 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 way by a separation of variable argument due to A. Szenes ([?]) and the residue theorem in one variable. We state it ﬁrst in the unimodular case.

39 n Let now ∆ be a set of vectors in Z 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 n linearly independent vectors. In this case the convex polytopes PΦ(a) have integral vertices, whenever a is in Zn. We get almost the same formulae for the 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:

eha,zi iΦ(a)= hhC, Tres ii. (1 − e−hα1,zi) ··· (1 − e−hαN ,zi)

Both formulae for the volume v(a) (Theorem ??) and the number of points i(a) (Theorem ??) in terms of the linear form hhC, ·ii attached to C can be proven in a completely parallel way by reducing to the 1-dimensional case (treated in Section ??). The basic formula to reduce to the 1-dimensional case is an analogue of the two formulae below: 1 1 1 = 2 − 2 , z1z2(z1 + z2) z1z2 z1 (z1 + z2)

1 1 1 = 2 + −1 . (1 − z1)(1 − z2)(1 − z1z2) (1 − z1) (1 − z2) (1 − z1)(1 − z1 )(1 − z1z2) Using the formula of Theorem ?? (together with a change of variable in residues ([?])), with V. Baldoni, we have implemented a simple Maple program for calculations of number of integral points in transportation polytopes or more generally for networks. It computes number of points in (5 × 5)- transportation polytopes in approximately 4 hours. Previous algorithm were based on Brion’s formula ([?]). As an easy consequence of the two parallel formulae (Theorem ?? and Theorem ??), 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 functions on functions on V ∗ by d ∂(α)P (h)= P (h + ǫα) . dǫ ǫ=0

40 N zi Consider the function ToddN (z1,z2,...,zN ) := i=1 1−e−zi and its Taylor series at the origin: Q 1 N Todd (z):=1+ z + ··· . N 2 i i=1 X Let Φ = [α1,α2,...,αN ] our unimodular set of vectors. We now substitute zi = ∂(αi) in this series. We obtain a series of constant coefficients differential operators that we denote by ∂(α) Todd(Φ,∂) := . 1 − e−∂(α) α∈Φ Y We can apply this series of differential operators to a polynomial function on V ∗. Theorem 16 ( Khovanskii-Pukhlikov) Let Φ be an unimodular system of vectors in Zn. 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 the series of diﬀerential operators Todd(Φ,∂) to the function v(h) given by the residue formula ehh,zi v(h) := hhC, Tres ii hα1,zi···hαN ,zi we can commute the residue and the series (the residue operates automatically by truncation of series). Thus we obtain

N hα ,zi ehh,zi Todd(Φ,∂)v(h) = hhC, Tres ( i ) ii 1 − e−hαi,zi hα ,zi···hα ,zi i=1 1 N Y ehh,zi = hhC, Tres ii. (1 − e−hα1,zi) ··· (1 − e−hαN ,zi) We recognize here the residue formula for i(a).

In the general case of a system Φ spanning Zn, but not necessarily unimodular, 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: eha,z+iγi iΦ(a)= hhC, Tres ii. −hα1,z+iγi −hαN ,z+iγi Rn Zn (1 − e ) ··· (1 − e ) γ∈ X/2π

The infinite sum γ∈Rn/2πZn means the sum over a finite set F of represen- tatives of γ, for which the function P eha,z+iγi z 7→ (1 − e−hα1,z+iγi) ··· (1 − e−hαN ,z+iγi) has a non-zero total residue. (For γ generic, it is holomorphic at zero, thus its total residue is zero. It is easy to prove that indeed the set F described above is finite.) This result implies the periodic-polynomial behavior of i(a) on the sector C − (Φ), which contains C. In particular, this formula along rays yields back Ehrhart’s theorem (Section 1). As in the case of Khovanskii-Pukhlikhov, this residue formula implies the formula of Cappell-Shaneson ([?]) and Brion-Vergne ([?]) for the partition function 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 some conditions (the polytope has to be a Delzant polytope, see Section ??), there exists a compact symplectic manifold MP of dimension 2n, an action of the group T on MP and a map from MP to P , such that the preimage of the point p ∈ P is an orbit of T . In other words, the polytope P is exactly the parameter 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 a sort of inflation of the polytope P . We refer to the lecture of Michèle Audin in the proceedings of the 5th EWM meeting held in Luminy for more details. The inflated symplectic manifold corresponding to the interval [−1, 1] is the 2-sphere S ⊂ R3 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 by rotation around the axis Oz.

42 MP P

1 1

−1 −1

Figure 11: Moment map

The inﬂated symplectic manifold corresponding to the standard simplex n ∆ ⊂ R is the projective space Pn(C). We realize Pn(C) as the space

2 2 2 iθ {(z1,z2,...,zn+1)| |z1| + ··· + |zn| + |zn+1| =1}/{z 7→ e z} with identiﬁcation of all proportional points z and eiθz in the sphere S2n+1 n+1 in C , so that Pn(C) is of dimension 2n. Rotations are given by

iθ1 iθ2 iθn iθ1 iθ2 iθn (e ,e ,...,e ) · (z1,z2,...,zn,zn+1)=(e z1,e z2,...,e zn,zn+1).

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

2 2 2 F (z1,z2,...,zn,zn+1)=(|z1| , |z2| ,..., |zn| ). 2 It is clear that the image of points in Pn(C) are in ∆, indeed x1 = |z1| ≥ 2 2 0,...,xn = |zn| ≥ 0, and x1 + x2 + ··· + xn =1 −|zn+1| ≥ 0. Furthermore two 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, 0 V (k) is the space H (Pn(C), O(k))), the point (p1, p2, ··· , pn) with pi ≥ 0 n p1 p2 pn (k−Pi=1 pi) and p1 + p2 + ··· + pn ≤ k indexing the monomial z1 z2 ··· zn zn+1 .

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

• M. BRION. Points entiers dans les polytopes convexes. Expos´e780. Seminaire Bourbaki, 1993-94. Ast´erisque 227 (1995).

• W. FULTON. Introduction to Toric Varieties. Annals of Mathematics Studies, 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 in Mathematics vol. 122.

The following review articles explain the general context of quantiﬁcation of 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 manifolds. Proc. Nat. Acad. Sci. USA 93 (1996) 25, 14238–14242.

• M. VERGNE. Quantification géométrique et réduction symplectique. Seminaire Bourbaki, mars 2001.

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