The Euler–Maclaurin formula for simple integral polytopes

Abstract

We give a Euler–Maclaurin formula with remainder for the sum of a smooth function on the integral points in a simple integral lattice polytope. Our proof uses elementary methods.

The Classical Formulas in Modern Garb

Let f be a function of class C^2k+1 on the real line. The classical Euler–Maclaurin formula computes the sum of the values of f over the integer points in an interval with integer end points in terms of the integral of f over variations of that interval and a remainder term that involves the integral of f^(2k+1) times a periodic function over the interval. Explicitly, let Td be the formal power series

1

so that Td(−S) is the exponential generating function for the Bernoulli numbers b_k,

(We follow the notation of ref. 1.) Let

so that

Because b_2k+1 = 0 for k > 0, the only difference between Td(S) and L(S) is the absence of the linear term in L(S). We also have

because this last power series can be written as

We will denote the truncation of the power series L(S) (which involves only even powers of S) at the even integer 2k by L^2k(S). Then, setting D = ∂/∂x,

is a differential operator with constant coefficients involving only even-order derivatives. In particular, if g is a function with 2k continuous derivatives, then

2

If f is any function on the real line and a < b are integers, we set

If f is a function of compact support, we extend this notation to the half ray [a, ∞) by writing

The classical Euler–Maclaurin summation formula with remainder for a function f of class C^2k+1 can be written as

3

where

and

with

where B_2k+1 is the (2k + 1)th Bernoulli polynomial. Indeed, other than minor changes in notation, this is formula 298 in ref. 2. If f is a polynomial, this becomes an exact formula when 2k + 1 is greater than the degree of f. Notice that Eq. 3, when applied to a C^2k+1 function of compact support, says that

4

for either choice of +/−, where

and

Indeed, we need only choose b so large that the support of f is contained in [a, b − 1] and then apply Eqs. 3 and 2. Conversely, if we know Eq. 4 for functions of compact support, then we can deduce Eq. 3. Indeed, multiply f by a smooth function of compact support, which is one in a neighborhood of [a, b], and observe that

5

when applied to a function of compact support.

A slight variant on Eq. 3 is

6

which can be derived from Eq. 3 by adding (1/2)(f(a) + f(b)) to both sides. This formula also obviously becomes exact when applied to a polynomial and k is sufficiently large.

The early references to the Euler–Maclaurin formula are Euler (3, 4) and Maclaurin (5), although apparently Poisson (6) was the first to consider the remainder.

Some Recent Results

A polytope in ℝⁿ is called integral if its vertices are in the lattice ℤⁿ; it is called simple if exactly n edges emanate from each vertex; and it is called regular if, additionally, the edges emanating from each vertex lie along lines that are generated by a ℤ basis of the lattice ℤⁿ.

Kantor and Khovanskii (7–9) generalized the classical exact Euler–Maclaurin formula for polynomials to higher dimensional convex polytopes Δ, which are integral and regular. In fact, their formula is a generalization of the exact version of Eq. 6: Let d denote the number of facets (faces of codimension one) of the polytope Δ, and let Δ(h₁, … , h_d) denote the polytope obtained from Δ by expanding outward a distance h_i in the direction of the ith facet. More precisely, if the polytope is described by the inequalities

7

where the u_i are primitive elements of the dual lattice, then Δ(h) is defined as the polytope given by

8

Their formula then asserts that for any polynomial p

9

When applied to the constant polynomial p ≡ 1, this formula computes the number of lattice points in Δ in terms of the volumes of “dilations” of Δ. Pommersheim (10) proved a similar formula for tetrahedra that are not necessarily regular. The Kantor–Khovanskii formula was generalized to polytopes that are integral and simple by Cappell and Shaneson (refs. 11–13 and S. E. Cappell and J. L. Shaneson, personal communication), and subsequently by Guillemin (14) and by Brion and Vergne (15). Cappell and Shaneson used their theory of characteristic classes of singular varieties and had the key idea of using the operator L as we do here; Guillemin used the Kawasaki–Riemann–Roch formula on symplectic toric orbifolds; and Brion and Vergne used Fourier analysis. All these generalizations involve “corrections” when the simple polytope is not regular. They are all exact formulas. Cappell and Shaneson (personal communication) have also investigated the problem of deriving a formula with remainder. The approach we suggest here is elementary.

The Weighted Lawrence Decomposition of a Simple Polytope

If Δ is a polytope, we let 1 denote the function which is zero on the exterior of Δ, is one on the interior of Δ, and on the boundary of Δ takes the values

where c(x) is the codimension of the smallest dimensional face containing x. Thus, for example, for an interval [a, b] on the line, the function 1 assigns the value one to points a < x < b, zero to points outside the interval, and 1/2 to the points a and b. We use a similar definition for a polyhedral cone.

Let Δ be a polytope in a vector space V and F a face of Δ. The tangent cone to Δ at F is

The polytope is the intersection of half-spaces,

where the H_i are given by the inequalities (Eq. 7), where d is the number of facets. The facets are then

For each face F, we let I_F ⊆ {1, … , d} encode the set of facets that contain F so that

We assume that Δ is simple, so the number of elements in I_F is equal to the codimension of F. The relation

is order (inclusion)-reversing. The tangent cone at a face F is then

For a vertex v, let α_i,v for i ∈ I_v be an edge vector at v, which does not belong to the facet labeled by i. (At the moment, these vectors are only determined up to a positive scalar. Later, when discussing integral polytopes, we will make a more specific choice of the edge vectors.) In terms of the edge vectors, the tangent cone at the vertex is

Let

be all the different codimension-one subspaces of ℝⁿ that are perpendicular to edge vectors α_j,v for Δ. (For instance, if no two edges of Δ are parallel, then the number N of such hyperplanes is equal to the number of edges of Δ.) The connected components of the complement

10

are called chambers. A vector ξ is in a chamber if and only if the pairings 〈α_i,v, ξ〉 are all nonzero. The signs of these pairings only depend on the chamber containing ξ. We choose such an element ξ, call it a “polarizing vector,” and think of it as defining the “upward” direction in ℝⁿ. We polarize the edge vectors so that they all point “down”: For each vertex v of Δ and each edge vector α_i,v emanating from v, we define the polarized edge vector to be

11

Consider the cone

By a weighted version of the Lawrence theorem (16), for any choice of polarizing vector we have

12

where #v denotes the number of edge vectors whose signs were changed by the polarizing process (Eq. 11). This theorem can be proved by elementary geometry and is illustrated for the case of a triangle in Fig. 1.

Figure 1.

The Lawrence theorem.

For any function f on ℝⁿ we define

We make a similar definition for polyhedral cones if f is a function of compact support (or a function that vanishes sufficiently rapidly at infinity along the cone). For such a function we have, as a consequence of Eq. 12,

13

The Euler–Maclaurin Formula for Regular Integral Polytopes

Let O denote the standard closed orthant in ℝⁿ, so O = ∏ ℝ_≥0. For any function f of compact support we have

Thus, if we iterate Eq. 4 we obtain a Euler–Maclaurin formula with remainder for the standard orthant and a C^2k+1 function of compact support: For any choice of the +/− signs we have

14

where D_i = (∂/∂h_i), where O(h₁, … , h_n) = {x|x_i ≥ −h_i} denotes the shifted orthant, and in the remainder R(f) is given by

15

where the sum is over all subsets I of {1, … , n} with I ≠ {1, … , n}. This remainder can be expressed as a sum of integrals, over various faces of the orthant, of periodic functions times various partial derivatives of f of order no less than 2k + 1.

A regular integral orthant C is the image of the standard orthant O via an affine transformation of the form

If n_i is the basis of (ℝⁿ)* dual to the standard basis of ℝⁿ so that O(h₁, … , h_n) is given by the inequalities 〈x, n_i〉 + h_i ≥ 0, then the image of O(h₁, … , h_n) under A is given by the inequalities

We denote this expanded orthant by C(h). If f is a C^2k+1 function of compact support and g = A*f = f ○ A, then

and thus

Thus we have a Euler–Maclaurin formula for regular orthants,

16

where

We may multiply the differential operator occurring in the first term in the above expression by any number of operators of the form L^2k(D), where the D are differentiations with respect to variables other than the h_i, because all that will remain of these operators is the constant term 1, all actual differentiations yielding zero. This remark allows us to write down a Euler–Maclaurin formula for a regular integral polytope by means of the weighted Lawrence theorem: Indeed, if Δ is a regular integral polytope and we choose a polarizing vector, then we have

where

17

Notice that both

and

do not depend on the choice of polarization, and both vanish on any function f with support that is disjoint from the polytope. Thus, the same must be true of the remainder. We have proved:

Theorem 1.

Let Δ be a regular integral polytope and f a C^2k+1 function of compact support. Then

where S(f) is given by Eq. 17. This remainder is independent of the choice of polarization and is a distribution supported on the polytope Δ.

Simple Polytopes

To give a Euler–Maclaurin formula for simple integral polytopes we need a Euler–Maclaurin formula for simple orthants. Then we can apply the weighted Lawrence theorem, Eq. 12, to go from orthants to polytopes. For a simple polytope, there are n edges emanating from each vertex and n facets meeting at each vertex. If, in addition, all the vertices of the polytope are integral, we can choose the n − 1 vectors emanating from a vertex and spanning a facet to have integer entries, and this, in turn, implies that the vector u defining the hyperplane containing this facet as the solution to 〈x, u〉 + λ = 0 can be chosen to have rational entries with respect to the dual lattice. This follows from Cramer's rule. So we will define a simple orthant to consist of an intersection of n half spaces, as in Eq. 7 with d = n, and such that the u_i are rational vectors. By clearing denominators we may assume that the u_i are actually integral, and we shall impose the additional normalization condition that the u_i are primitive elements of the dual lattice, meaning that no u_i can be expressed as a multiple of a lattice element by an integer greater than one. This fixes our choice of the u_i. If the lattice spanned by the u_i is in fact the entire dual lattice, then the orthant is regular. However, in the simple case, all we know is that the u_i span a sublattice of the dual lattice of finite index.

We want to study the Euler–Maclaurin formula for a simple orthant C, and for simplicity in notation, let us temporarily assume that the vertex of C is at the origin and that our vector space is ℝⁿ with its standard basis consisting of column vectors such that the dual space consists of row vectors. We have the basis u₁, … , u_n consisting of integral primitive row vectors. We will now choose the edge vectors α_i as the dual basis to that of the u_i, i.e., 〈α_i, u_j〉 = δ_ij and C = ∑ ℝ_≥0α_i. The lattice ∑ ℤα_i is finer than the lattice ℤⁿ; in particular, the α_i need not lie in ℤⁿ. The matrix U, the rows of which are the u_i, carries the basis α_i into the standard basis: Uα_i = e_i, and thus

the standard orthant. For any f of compact support, we have

where

Let Ξ denote the lattice spanned by the u_i, and let Γ denote the finite group

This group is the character group of the group (span_ℤα_i)/ℤⁿ, which is isomorphic to ℤⁿ/Uℤⁿ via the map U. In more detail, if γ ∈ Γ, and we choose a representative γ̃ in ℤⁿ of γ, then the function

defines a character on ℤⁿ, which is independent of the choice of representative γ̃ and is identically one on Uℤⁿ and only on Uℤⁿ. Hence, by a theorem of Frobenius on finite groups, we can conclude that the function

vanishes on ℤⁿ∖Uℤⁿ and is identically one on Uℤⁿ. Also,

Thus

We can interchange the order of summation on the right and apply the Euler–Maclaurin formula (Eq. 14) to each summand (over Γ). The nonremainder piece of each summand will be

Setting x = Uy, so dx = |det U|dy, we obtain (18)

where R is the remainder R given by (15) applied to the function

In order to convert (Eq. 18) into a form to which the Lawrence theorem applies, we must replace the operators L(D_i), which are applied to the integral of f × (an exponential) with operators applied to the integral of f (with a consequent change in the remainder). This involves two steps: (i) a modification of the Euler–Maclaurin formula in one dimension that takes the exponential into account and thus a generalization of the functions P_2k+1 and (ii) a careful dissection of the groups Γ occurring at each vertex to prove that the resulting exact terms involve only even powers of the D_i. The details will be presented elsewhere.

Finally we remark that our methods give an alternative proof of the exact version of the Euler–Maclaurin formula in the case of a simple polytope as in the work of refs. 11, 14, and 15. If f(z) = e^〈x,z〉, where z is a vector in the complexified dual space such that Re z is a polarizing vector and we choose the corresponding polarization, and if furthermore ∥z∥ is sufficiently small, then we may apply the exact version of Eq. 18 to each polarized orthant. We may also explicitly compute the integrals. The resultant formula then can be manipulated by some algebra to yield the exact results of Cappell and Shaneson, Guillemin, and Brion and Vergne. Details will also be presented elsewhere. For some related results see refs. 17–20.