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

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 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 functionsw(k) ofk, wherek 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 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 can associate 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 in k with leading termk^dimM/2vol(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 holonomytaround 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 polytopeP determines a toric manifoldM of dimension 2 dimP 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 k^dimPvol(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.

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 dis- crete parameter (the integerk), 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, vol- ume 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.

Lecture 1

Residue formulae for Bernoulli polynomials and

Verlinde sums

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 Rk

0 x^mdx= ^k_m+1^m+1.

Form= 0, we have:

0⁰+ 1⁰+ 2⁰+· · ·+k⁰ =k+ 1.

Form= 1, we have:

0 + 1 + 2 + 3 +· · ·+k= k²+k 2 . Form= 2, we have:

k³ k² k

Form= 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 following theorem shows:

Theorem 1 (Jakob BERNOULLI -(1654-1705)) For any positive integerm, the sum S_m(k) =Pk

a=0a^m of the m^th-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 Bernoulli polynomials.

Given a real numbert, the Bernoulli polynomialsB(m, t) are defined by the following relation:

z e^tz e^z−1 =

∞

X

m=0

B(m, t)z^m m!.

This means, we expand z_ez^etz−1 in a Taylor series atz = 0. The coefficient of ^z_m!^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,

· · ·

It follows immediately from the definition of the Bernoulli polynomials that

d

dtB(p, t) = pB(p−1, t), Z 1

0

B(p, t)dt = 0 ifp >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 e^z −1 =

∞

X

m=0

Bm

z^m m!. Since

1

1−e^−z = 1

e^z−1+ 1, we have P∞

m=0Bmz^m

m! +z = P∞

m=0Bm(−z)^m

m! so that the Bernoulli numbers are equal to 0 formodd, except whenm= 1 in which case we haveB1 =−¹₂.

Now remark that 1

(m+ 1)!(B(m+ 1, t+ 1)−B(m+ 1, t)) is the coefficient of z^m+1 in the Taylor series of

z(e^(t+1)z

e^z−1 − e^tz

e^z−1) = ze^tz(e^z−1)

e^z−1 =ze^tz. Thus we obtain:

B(m+ 1, t+ 1)−B(m+ 1, t) = (m+ 1)t^m. We write this equality for t= 0,1, . . . , k:

B(m+ 1,1)−B(m+ 1,0) = (m+ 1)0^m B(m+ 1,2)−B(m+ 1,1) = (m+ 1)1^m

· · ·

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

Adding up these equations, we obtain

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

As

ze^(t+1)z

e^z−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 formula in 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 a polynomial. The sum over all non-zero integers n ∈Z, n6= 0:

B(φ)(t) =X

n6=0

φ(2iπn)e^2iπnt

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

n6=0 1

n^α converges forαsufficiently large. It defines a periodic function oft. It always converges as a generalized function oft, which is smooth when t /∈Z. For 0< t <1 andpsufficiently large, the residue formula in the plane yields:

B(φ)(t) = residuex=0(φ(x) e^xt 1−e^x).

(For −1< t <0, the formula is B(φ)(t) = −residue_x=0(φ(x)_1−e^ext−x).) In particular, t 7→ B(φ)(t) is given by a polynomial formula whenever 0< t <1.

From the definition ofB(p, t), it follows that B(p, t) =−p! residue_x=0(x^−p e^xt

1−e^x)

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

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

B(p, t)

p! =−residue_x=0(x^−p e^xt

1−e^x) =−X

n6=0

e^2iπnt (2iπn)^p.

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 < t < 1. If p ≥ 2, the sum in the right hand side is absolutely convergent and the formula is valid for all 0≤t ≤1.

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

B2g = −(2g)!X

n6=0

1 (2iπn)^2g

= 2(−1)^g+1(2g)!(2π)^−2gζ(2g) where ζ(k) =P∞

n=1n^−k is the zeta function at point k.

Forg = 1, as B₂ = ¹₆, we have

∞

X

n=1

1 n² = π²

6 .

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

F(z) = 1 (1−z)^p.

Given a positiveinteger t, let us consider the expression:

W(p, t)(k) = X

ω^k=1, ω6=1

ω^tF(ω) = X

1≤n≤(k−1)

e^2iπnt/k (1−e^2iπn/k)^p. It is very similar to the formula for 0< t < k:

k^pB(p, t/k)

p! =−X

n6=0

e^2iπnt/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 −^B_p!^p.

It is not at all obvious from its definition that the functionk 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 degree term −k^{p B}_p!^p. If p is odd and p 6= 1, this polynomial is of degree less or equal to p−1.

For allp≥1, we have the residue formula:

W(p, t)(k) = −kresiduex=0

e^tx (1−e^x)^p

1

(1−e^−kx)dx

.

Proof. Consider the 1-formk_(1−z)^zt p

z^k (1−z^k)

dz

z. From the conditions 0≤t≤p, and k ≥1, it follows that this form has no residues at 0 and∞. Poles of the factor _(1−z¹k) arise at z = e^2iπn/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 of residues of this 1-form is equal to 0, we obtain from the residue theorem

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

z^t (1−z)^p

z^k (1−z^k)

dz z

.

A change of variable z =e^x in the residue yields:

W(p, t)(k) = −kresiduex=0

e^tx (1−e^x)^p

1

(1−e^−kx)dx

.

This last expression depends onkvia the Laurent expression of _1−e^k−kx.Recall from the definition of the Bernoulli numbers that

kx e^kx−1 =

∞

X

r=0

Br

(kx)^r r! .

Hence

W(p, t)(k) = −

∞

X

r=0

BrRr(p, t)k^r r!

where

Rr(p, t) = (−1)^rresiduex=0

e^tx

(1−e^x)^px^r−1dx

so thatRr(p, t) vanishes for larger. In fact, as _(1−e¹x)^p has a pole at 0 of order p, we see that Rr(p, t) vanishes forr > p, so that W(p, t)(k) is a polynomial of degree less or equal than p. More precisely, it is of degree pfor even pand of degree less or equal to p−1 for odd p, p 6= 1. For p even or p = 1, the highest degree term is−B_p^k_p!^p. Ifp is odd,p6= 1, the term of degreepink is equal to 0 asBp = 0, while the term of degree (p−1) is−(p/2−t)Bp−1 k^p−1

(p−1)!.

5 A relation between Bernoulli series and trigono- metric sums

Define:

V(q, k) =

k−1

X

n=1

1

4^q(sin(πn/k))^2q.

As we will see later on, V(q, k) is a special case of a Verlinde sum.

We haveV(q, k) = (−1)^qW(2q, q)(k). Indeed (−1)^qW(2q, q)(k) = (−1)^q X

1≤n≤(k−1)

e^2iπnq/k (1−e^2iπn/k)^2q

=

k−1

X

n=1

1

(1−e^2iπn/k)^q(1−e^−2iπn/k)^q

=

k−1

X

n=1

1

4^q(sin(πn/k))^2q.

Thus, from Theorem 4, we obtain that k7→V(q, k) is a polynomial ink, of degree 2q. Notice that V(q, k) is a sum of positive real numbers so that

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

(−1)^q+1 B2q

(2q)!k^2q.

(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 Taylor series at the origin of the function of x

A(q, x) : =ˆ (x/2)^2q (sinh(x/2))^2q

= 1− q

12x²+ ( q

1440 + q²

288)x⁴+· · · .

Substitutex= _∂t^∂ in ˆA(q, x), and consider ˆA(q, ∂) as a series of differential operators in powers of∂ := _∂t^∂. The action of ˆA(q, ∂) on a polynomial function of t is well defined.

Theorem 5

V(q, k) = (−1)^(q+1)k^2q

2q!( ˆA(q, ∂/k)·B(2q, t))|_t=0.

From this expression, we see again that the highest degree term of the polynomial function k7→V(q, k) is (−1)^q+1B2qk^2q

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 expres- sion arises when calculating the integral R

Mgch(L^k−2) ˆA(Mg)(where ch is the Chern character andA(Mˆ g)is theAˆgenus). This evaluates the dimension of the space of so-called generalized theta functions: the dimension of the space of holomorphic sections of the holomorphic line bundle L^k−2 over M_g. We shall come back to this analogy later.)

Proof. How to prove this theorem: Consider the residue formula for the Bernoulli polynomial

B(2q, t) = −(2q!) residue_x=0

x^−2q e^xt (1−e^x)

.

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

(−1)^q+1k^2q

2q!A(q, ∂/k)B(2q, t) = (−1)ˆ ^qk^2qresiduex=0

A(q, x/k)xˆ ^−2q e^xt (1−e^x)

= residuex=0

1

(1−e^x/k)^q(1−e^−x/k)^q e^xt (1−e^x)

.

Thus at t= 0, we obtain (−1)^q+1k^2q

2q!

A(q, ∂/k)B(2q, t)ˆ

|_t=0

= residuex=0

1

(1−e^x/k)^q(1−e^−x/k)^q 1 (1−e^x)

. The change of variables x7→ −kx leads to

(−1)^q+1k^2q 2q!

A(q, ∂/k)B(2q, t)ˆ

|_t=0

= −kresiduex=0

1

(1−e^x)^q(1−e^−x)^q 1 (1−e^−kx)

= −(−1)^qkresiduex=0

e^qx (1−e^x)^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+1k^2q 2q!

A(q, ∂/k)B(2q, t)ˆ

|_t=0 = V(q, k)

=

k−1

X

n=1

1

(1−e^2iπn/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:

Indeed applying formally ˆA(q, ∂/k) to the sum−P

n6=0 e^2iπnt

(2iπn)^2q expressing the Bernoulli polynomial _2q!¹ B(2q, t), we would obtain:

(−1)^(q+1)k^2q 2q!

A(q, ∂/k)B(2q, t)ˆ

|t=0 = (−1)^qk^2qX

n6=0

A(q,ˆ 2iπn/k) 1 (2iπn)^2q

= X

n6=0

1

(1−e^2iπn/k)^q(1−e^−2iπn/k)^q. This last expression is highly divergent, for at least two reasons: first, for n 6= 0 multiple ofk, the term to add to the sum is equal to∞, second, all the terms in the arithmetic progressionn+kjgive the same summand. However, it gives a hint of why this operator ˆA(q, ∂/k) occurs in the comparison. The

”renormalized” sum consists in restricting the sum to 0 < n < k, a set of representatives 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

X

n=0

1

sin((n+ 1)π/(k+ 2))^2q takes positive integral values on integers k. Notice that for k = 0, we have Ver(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 these functions take integral values on integers.

6 Preliminaries on semi-simple Lie algebras

The Verlinde sums we saw corresponded to the Lie group SL(2,C) with Lie algebrasl(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, Yi= 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, Yi = 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 algebrah is a maximal abelian subalgebra of gand 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.

Forsl(2), we write an element t of h as t=

t1 0 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

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

Let S = {α₁, . . . , α_r} be a simple system of roots; we set H_i to be the unique element of h such that λ(Hi) = _hα²

i,αiihλ, αii, for any λ ∈ h^∗. 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 ⊂ h^∗_R 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, ifhλ, αi 6= 0 for allα ∈∆. 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 representationU ofgin 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 allH∈h}

is not reduced to zero. We have V = P

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

IfV is an irreducible finite-dimensional representation ofg, then V has a unique highest weight λ (all other weightsλ^′ of the representation V satisfy

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

D:={λ∈h^∗_R| λ(H_i) 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 λ as its highest weight. LetUλ 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

B(p,g)(t) = X

λ∈Preg

e^(2iπλ,t) (Q

α∈∆⁺2iπhα, λi)^p.

This series converges for psufficiently 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. OnhR/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:

B(p, t1) = −B(p, sl(2))(t) =−X

n6=0

e^2iπt1n (2iπn)^p. Here

t ∈hR=

t₁ 0 0 −t1

with t1 ∈]0,1[.

Similarly to the case ofsl(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.

Consider the example of sl(3): we get B(q, sl(3))(t₁, t₂) = X

n16=0,n26=0,n1+n26=0

e^{2iπ(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 0

0 t2 0

0 0 −(t₁+t₂)



.

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

(−1)^q X

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

e^{2iπ(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 P

n6=0 e^2iπnt (2iπn)^p

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

(−1)^qB(q, su(3))(t₁, t₂) =

−residue_x=0

residuey=0

e^{{t1}x−{t2}y} x^qy^q(x+y)^q

1

(1−e^x)(1−e^−y)

+residuex=0

residuey=0

e^{{t1+t2}x+{t2}y} x^qy^q(x+y)^q

1

(1−e^x)(1−e^y)

. Here {t} denotest−[t] where [t] is the integral part oft.

For example, we haveB(2, su(3))(0,0) =−₃₀₂₄₀¹ .

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

X

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

e^{2iπ(t1n−t2m)}

(2iπn)^q(2iπm)^q(2iπ(n+m))^q and the second residue should lead to the sum

X

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

e^{2iπ(t1+t2)n+(t2)(n+m)} (2iπn)^q(2iπm)^q(2iπ(n+m))^q

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: LetGbe the compact simply connected group whose Lie algebra is a compact form of g. Up to some normalization, forp= 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 V(q, k) =

k−1

X

n=1

1

4^q(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 associated cone of dominant integral weights. Let ρ= ¹₂P

α∈∆⁺α.

Letθ be the highest root. Let us also introduce the set (called an alcove):

Ak:=

λ∈ D| 2hλ, θi hθ, θi ≤k

(the definition of Ak does not depend of the choice of the scalar product.) Leth be the dual Coxeter number h = 2^hρ,θi_hθ,θi + 1. When g =sl(n), then h=n.

Consider the scalar product <·,·> such that< θ, θ >= 2.

Letq be an integer. The Verlinde sum is

Ver(q,g)(k) = c^q_G(k+h)^rq X

u∈Ak

1 Q

α∈∆⁺(1−e2iπ<α,u+ρ>

k+h )^q(1−e⁻2iπ<α,u+ρ>

k+h )^q

= c^q_G(k+h)^rq X

u∈Ak

1 Q

α∈∆⁺(2 sin(π<α,u+ρ>

k+h ))^2q.

Herec_G is an explicit constant depending onG. It is such that Ver(1,g)(0) = 1.

Wheng=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.

Forg=sl(3), the above sum is

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

× X

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 arising as the space of generalized theta functions which is the space of holomor- phic sections of the holomorphic line bundleL over the manifold M(G, g) = M(G, g,0)) of a Riemann surface of genusg =q+ 1 with central chargek. 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 differ- ential operators ˆA(q, ∂/(k+h)) to the Witten series. The form of the wanted operator ˆA(q, ∂/(k +h)) can be guessed from the same intuitive argument of “differentiating” under the sum sign. The correct argument follows imme- diately from the explicit residue formula. It allows to compute the integral R

M(G,g)ch(L^k) ˆA(M(G, g)) (ch is the Chern character and ˆA(M(G, g)) is the Aˆgenus), at least when k is sufficiently large.

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 = 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,

whereC_r, B_r, D_rare the series of classical simple Lie algebras andG₂, 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).

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 u1, u2 ∈G, we denote [u1, u2] =u1u2u⁻¹₁ u⁻¹₂ .

The manifoldM(G, g, t) is defined to be:

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

g

Y

i=1

[ui, vi] =e^2iπt}/T.

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 (u^′₁, v^′₁, . . . , u^′_g, v^′_g) if there exists t∈T such thatu^′_i =tuit⁻¹ and v_i^′ =tvit⁻¹.

• The main reference on the geometric properties of the manifoldM(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 mod- uli space of stable bundles. J. Differential Geom. 1992, pp 131-149.

In the general case, the volume ofM(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 manifoldM(G, g, t) together with the com- putation 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.

• The Verlinde formula was conjectured by:

E. Verlinde: Fusion rules and modular transformations in 2D-conformal field theory. Nuclear Physics B1988 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 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 onk) 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).

Lecture 2

Residue formulae for volumes and number of

integral points in convex rational polytopes

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 inter- est in this topic comes from my efforts to understand the relations between symplectic manifolds and group representations, the existence of such rela- tions 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 modelQ(M, ω) forM. It is important to understand the relation between both quantities. The dimension q(k) of the vector space Q(M, kω) is a polynomial ink, and the volume of the man- ifoldM is the limit ofk^−dimM/2q(k) whenk tends to∞. The full expression for the dimension ofQ(M, kω) is the content of the Riemann-Roch theorem, which expresses this integerq(k) as an integral overM. 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 ofP 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 al- gebraic 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 sym- plectic 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 coor- dinates inside this polytope are similar problems (both difficult). Convex polytopes arise in many fields of mathematics: symplectic geometry, repre- sentation 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 withx, y, z positive

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 ratio- nal 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 for- mula of Jeffrey-Kirwan ([?]) for the Laplace transform. The formula for the number of points given in ([?]) follows from a multidimensional residue the- orem, 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.

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 inRⁿ, or equivalently a compact set ofRⁿ defined by linear inequa- tions. 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≥0, y ≥0, z≥0 and x 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 and x

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 convex polytopes, together with the calculation of the number of points with integral

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 polytopek∆ 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 basis of vector spaces V(k) attached to P.

Thus we see that the discrete analogue to the monomial ^x_n!ⁿ giving the volume of k∆ for x = k is the polynomial pn(x) = (x+1)(x+2)···(x+n)

n! which

gives the number of integral points in k∆ for x=k. This polynomialpn(x) has same leading term ^x_n!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).

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 arevol(P)kⁿ+¹₂ volZ(δP)kⁿ⁻¹+· · ·. The constant term of the Ehrhart polynomial iP(k) is equal to 1.

HerevolZ(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 A_F spanned by the face. The normalization is done in such a way that the fundamental domain of the lattice Zⁿ∩AF has volume 1.

The Ehrhart functioniP(k) is our second example of a function ofk, the polynomial feature of which does not follow evidently from its definition, the first example (given in Lecture 1) being the Verlinde sumsk7→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→ X

ξ∈kP∩Zⁿ

f(ξ)

is a polynomial function ofk. The leading term of this polynomial isk^n+mR

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 P⁰ be the interior ofP. Define

i_P⁰(k) = cardinal(kP⁰∩Zⁿ)

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

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 numberi_P⁰(k) is the number of integers strictly greater than 0 and strictly less than k. We have

iP⁰ = (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 kⁿvol(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.

Let m ∈ R⁺. The hanging pyramid Pm is the convex hull in R³ of the vertices s₀ = (0,0,0), s₁ = (1,0,0), s₂ = (0,1,0),and s₃ = (1,1, m).

Its volume depends onm:

vol(Pm) =m/6.

Assume thatm is an integer, thenPm is an integral convex polytope, but there are no integral points in Pm other than its 4 vertices. Whatever the

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 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

which relates to the length by

iP(k) = klength(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:

i_P(k) = k²vol(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 functioniP(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 inZⁿand 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.

Let us start with the standard simplex ∆ inRⁿ, 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, . . . , x_n ≥ 0, x₁+x₂+· · ·+x_n ≤ 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

p(n, k) = (k+ 1)(k+ 2)· · ·(k+n)

n! .

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:

Zk(t) =

∞

X

k=0

p(n, k)e^−kt, t >0.

We shall recover p(n, k) as the coefficient of e^−kt inZk(t).

We have

Zk(t) = X

(x0,...,xn)∈(Z⁺)ⁿ⁺¹

e^{−t(x0+x1+···+xn)}

=

n

Y

i=0

∞

X

xi=0

e^−txi

= 1

(1−e^−t)ⁿ⁺¹. Differentiating P∞

k=0z^k = _1−z¹ , we get _(1−z)¹n+1 =P∞ k=0

k+n n

z^k and hence 1

(1−e^−t)ⁿ⁺¹ =

∞

X

k=0

k+n n

e^−kt

with ^k+n_n

= (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 _n!¹kⁿ+ ¹_2(n−1)!⁽ⁿ⁺¹⁾kⁿ⁻¹ +· · · as announced in Ehrhart’s theorem since vol(∆) = _n!¹ and the boundary of ∆ is the union of (n+ 1) standard simplices, each of volume _(n−1)!¹ .

Let us now generalize this to the case of a general simplexP in Rⁿ. We give a formula due to Ehrhart for the number of points in kP.

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

X

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

X

i=0

ui =r,

n

X

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 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) with ki non-negative integers, 0 ≤ ui < 1, and Pn

i=0ki +Pn

i=0ui = k. As the elements αi are in Zⁿ, the pointw is in Zⁿ if and only if

u0α0+u1α1+· · ·+unαn∈Zⁿ. Notice that the numberPn

i=0u_i =k−Pn

i=0k_i is a non-negative integer. This integer r is less or equal ton, 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 satisfy

k0+k1+· · ·+kn =k−r and u∈(P, r), the point

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

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 is ^k−r+n_n

, for any k≥0.

We therefore obtain Ehrhart’s formula:

iP(k) =

n

X

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 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 the behavior ink of the functionk 7→i_P(k) = cardinal(kP∩Zⁿ) 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 integerd, such that the functionk 7→i_P(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 := [0,¹₂], then, with d = 2, iP(2k) = k + 1, while i_P(2k + 1) =k+ 1. These two formulae can be assembled together to give the periodic-polynomial function iP(k) = ^k₂ + 1− ¹₄(1−(−1)^k).

For a given rational polytopeP, it is not easy to determine the smallest value pof 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 = (0,0,0), A = (1,0,0),B = (0,1,0), D = (1,1,0) and C = (¹₂,0,¹₂). Then P is not integral, however

(k+ 1)(k+ 2)(k+ 3)

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 the formula that we used:

ip(k) =

k+n n

.

Example 2. LetP be the triangle in R² with vertices α0 = 0, α1 =e1, α2 = 2e2.

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