Asymptotic Euler-Maclaurin formula for semi-rational polyhedra

Asymptotic Euler-Maclaurin formula for semi-rational polyhedra

Common work with Nicole Berline.

Semi-rational polyhedra

V real vector space of dimensiond; Λlattice inV, dual latticeΛ^∗⊂V^∗; η₁, η₂, . . . , η_N elements ofΛ^∗; h₁,h₂, . . . ,h_N realnumbers.

Then the polyhedron

p(h) ={x ∈V;hη_i,xi ≤h_i} is called semi rational.

Examples :

a≤x ≤b: any interval with real end points.

x ≥0,y ≥0,x +2y ≤√ 2

Riemann Sums over Semi-rational Polyhedra

Letp(h)be a semi rational polytope of dimensiond. Consider the sum of values of a smooth functiontest on the integral points ofp:

hF(p(h),Λ),testi= X

x∈Λ∩p(h)

test(x).

Whentest is a polynomial, we (Baldoni, Berline, Koeppe, De Loera, Vergne) have given ”formulae” forhF(p(h),Λ),testi depending of the real parameterhin a ”semi-quasi-polynomial”

way. Here we show that we can give asymptotic formulae for test smooth of the same kind.

Asymptotic Riemann Sums over Semi-rational Polyhedra

Consider a real parametert. Assumepis a semi-rational polyhedron, of dimensiond, andtest be a smooth function with compact support. Define the Riemann sum :

F(t) = 1 t^d

X

x∈tp∩Λ

test(x/t)

It is clear that whent → ∞,F(t)tends toR

ptest. Thus we want to evaluate at what rate the Riemann sum converges to the integral.

The Euler-MacLaurin formula in dimension 1

Letabe a real number. Let

c={x ∈R;x ≥a}.

Let{t} ∈[0,1[be the fractional part oft ∈R. Lettest be a smooth function with compact support,B_k(x)the Bernoulli polynomial,B_k =B_k(0)the Bernoulli number.

Then X

x≥ta,x∈Z

1

ttest(x/t) = Z

x≥a

test(x)dx−

k−1

X

j=1

B_j({−ta})

t^j test^(j−1)(a)

+1

t^k(−1)^k−1 Z ∞

a

(B_k({tx})−B_k)test^(k)(x)dx.

Riemann sum minus the integral

So we obtain an asymptotic formula in the form

infty

X

j=1

f_j(t)1 t^j

wheref_j(t)are polynomial functions of the functiont → {−at}.

Furthermore, an explicit formula for the rest is given as an integral of a derivative oftest of orderk against a bounded and continuous function oftx overc= [a,∞[.

Ifa=p/qis rational, the functionf_j(t)is a periodic function oft with periodq.

Ifaandtare integers, thenf_j(t) =−B_j is just the Bernoulli number. So we obtain an asymptotic formula in ¹_tj.

Semi-quasi polynomials

A functionf(t)oft ∈Rwill be called semi-quasi polynomial if f(t)can be expressed asP({c₁t},{c₂t}, . . . ,{c_Kt}), a

polynomial function of a number of functionst → {c_jt}whereca

are real numbers. If the numbersc_j =p_j/q_j are rationals, then {c_jt}is a periodic function of periodq_j, so a semi-quasi polynomial is a periodic function oft mod some periodQ.

Form of the asymptotic for semi-rational polytope

Theorem•Letpbe a semi-rational polytope, then when t→ ∞,t real,

F(t) = 1 t^d

X

x∈tp∩Λ

test(x/t) is equivalent to

Z

p

test+

∞

X

j=1

f_j(t)

t^j hD_j,testi

wheref_j(t)are semi-quasi polynomial functions oftand hD_j,testiareintegrals of derivatives of test on faces ofp.

•Ifpis rational,f_j(t)are periodic functions oft

•Ifpis with integral vertices,f_j(t)are constants.

Comments

We have explicit forms of theD_j, but unfortunately no nice formula for the rest, meaning as a distribution supported onp.

For the standard simplex, there are some formulae with rest obtained in numerical analysis (clearly an important problem in numerical analysis).

Arrangement of hyperplanes

In the case wherepis a polytope with integral vertices, two asymptotic formulae were previously obtained : one by Guillemin-Sternberg (extending Khovanskii-Pukhlikov formula for polynomials), one by Tatsuya Tate (extending

Berline-Vergne local Euler-MacLaurin formula for polynomials).

Our work was motivated by a recent question of Le

Floch-Pelayo : How to see directly that these two formulae are the same ?.

Easy to do via renormalization of rational functions with poles on an arrangement of hyperplanes.

Guillemin-Sternberg formula

Letp(h)be a semi-rational polytope. Assume thatp₀=p(h₀)is a Delzant polytope with integral vertices (corresponding to a smooth toric varietyM with ample line bundleL). Consider

F(t) = 1 t^d

X

x∈tp₀∩Λ

test(x/t).

DefineTodd(z) = _1−e^z−z =P∞

j=0(−1)^jB_jz^j/j!.

We can consider the series of differential operators Todd(∂_h/k) :=

N

Y

i=1

Todd(∂_hi/k) Then : Fork integer,k → ∞, we have

F(k)≡ Todd(∂_h/k) Z

p(h)

test

!

|_h=h0

But we want a ”Concrete formula”

Pretty clear thatF(k)≡P

j 1

k^jhD_j,testiwherehD_j,fiare integrals of derivatives oftest on faces of the polytopep.

We had obtained (Berline-Vergne) exact formulae with explicit operatorsD_j when test was a polynomial. Tatsuya Tate showed (directly) that these formulae hold in the asymptotic sense. It is in fact easy to deduce any of these asymptotic formulae from Euler-MacLaurin formula in dimension one, and we can formulate a result for any semi-rational polytope.Our method is as usual the Brianchon-Gram decomposition. and

”renormalization” in the space of rational functions with poles on an arrangement of hyperplanes.

Integrals and Sums over a cone

V real vector space of dimensiond, with a latticeΛ.

ca cone, dual conec^∗ with non empty interior.

I(c)(ξ) = Z

x∈c

e^hξ,xidx,

S(c)(ξ) = X

x∈c∩Λ

e^hξ,xi.

These functions are defined whenξis the opposite of the dual conec^∗ toc. For example ifc=R≥0, andξ <0, then

I(c)(ξ) = Z

x≥0

e^ξxdx =−1 ξ.

S(c)(ξ) =X

n≥0

e^nξ= 1 1−e^ξ.

Laurent series ofS(c)

I(c)is a rational function of degree−d. Define[S(c)] =P

j=−dS(c)<sub>[`]</sub>the decomposition ofS(c)in sum of rational functions of homogenous degree`.

Examplec=R≥0e₁⊕R≥0(e₁+e₂) S(c)(z₁,z₂) = 1

(1−e^z1)

1 (1−e^z1+z2)

= 1

z₁(z₁+z₂)

− 1

2z₁ − 1 2(z₁+z₂) +1

3+ 1 12

z₁ z₂ + 1

12 z₁ (z₁+z₂) +· · ·

Boundary values

Ifλis an interior point in−c^∗(-interior of the dual cone), define limλ S(c)(iξ) = lim

→0S(c)(iξ+λ) limλ S(c)_`(iξ) = lim

→0S(c)_`(iξ+λ) Define tempered distributions onV^∗.

The Fourier transform of lim_λS(c)_`(iξ)is a derivative of a locally polynomial function supported onc.

For examplec= [0,∞]; <0, lim

1

−(iξ+) = Z ∞

0

e^iξxdx.

The distribution F(t) and its Fourier transform

We consider the conecand the tempered distribution onV defined by

hF(c)(t),testi= 1 t^d

X

x∈Λ∩c

test(x/t).

It is easy to compute the Fourier transform of this distribution This is

limλ S(c)(iξ/t)

Ift → ∞,ξ/t tends to 0, and it is tempting to use the Laurent series of the functionS(c)(ξ).

TheoremWhent → ∞, lim

λ S(c)(iξ/t)≡X

j≥0

t^−jlim

λ S(c)_j(iξ).

Explicit formula

The proof of this theorem is obvious. It is clearly additive over cones, we decompose the conecin unimodular cones and then we are reduced to dimension 1 and we useEMLformula.

Obvious then to deduce Guillemin-Sternberg formula from computing this, and using Brianchon-Gram decomposition.

We want to be more explicit.

Renormalization

We consider the arrangement of hyperplanesH=H(G)inV^∗ with equationsg_a(ξ) =0 withg_a∈G⊂V the set of generators of the conec. We denote byR_G(V^∗)the space of rational functions onV^∗ with poles contained inH(G). Then the homogeneous components ofS(c)are rational functions with poles on this arrangement of hyperplanes.

LetSbe the set of subspacesLofV generated by elements of G. If[g₁,g₂, . . . ,g_N]is a sequence of elements ofG, the function QN ¹

i=1g_j(ξ) is a function inR_G(V^∗). For a subspaceL, consider the spaceB_Lconsisting of the linear span of these functions ^{Q 1}

a∈Aga(ξ) where the sequenceg_ain the denominator generates the rational subspaceL:L=⊕Rg_j. ThenB_Lis a subspace ofR_G(V^∗).

Substracting the polar part

Let us choose a scalar product onV. ConsiderC[L^⊥]the space of polynomial functions onL^⊥. Using our scalar product, we can embeddC[L^⊥]inC[V^∗], the space of polynomial functions on V^∗.

Proposition(De Concini+Procesi) We have the direct sum decomposition

R_G(V^∗) =⊕_L∈S(C[L^⊥]⊗B_L) Thus we writeφ∈R_G(V^∗)uniquely asφ=P

LT_L(φ)where T_L(φ)∈(C[L^⊥]⊗B_L).

The holomorphic part

ForL={0}, the corresponding termT_L(φ)is thus a polynomial function onV^∗. We denote the corresponding term by

Renorm(φ)and we call the polynomialRenorm(φ)the renormalisation of the rational functionφ. It depends of the choice of a scalar product. In one variablef(z) =P

a_izⁱ, and Renorm(f) =P

i≥0a_izⁱ. In several variables, we need a scalar product to suppress the polar part (allB_LforL6=0).

Back to Example : c = R^≥0e₁ ⊕R^≥0(e₁ +e₂)

S(c)(z₁,z₂) = 1 (1−e^z1)

1 (1−e^z1+z2)

= 1

z₁(z₁+z₂)

− 1

2z₁ − 1 2(z₁+z₂) +1

3+ 1 12

z₁ z₂ + 1

12 z₁ (z₁+z₂) +· · ·

BUT WE REWRITE 1 3 + 1

12 z₁ z₂ + 1

12 z₁ (z₁+z₂) as

1 3+ 1

12 z₁ z₂ + 1

24

z₁−z₂ (z₁+z₂)+ 1

24 and the holomorphic part ofS(c)is

renorm(S(c)(z₁,z₂)) =1/3+1/24+· · ·=3/8+· · ·

Normal cones and Renormalization

f face of the conec;<f >linear span ofp−q,p,qinf. Normal coneT(f,c)in<f >^⊥with lattice<f >^⊥∩<f >+Λ Definitionµ(f,c)(ξ) =Renorm(S(T(f,c)))(ξ). This is a holomorphic function of the variableξ ∈<f >^⊥.

MAIN THEOREM(Berline-Vergne, Paycha) S(c)(ξ) = X

faces

µ(T(f,c))(ξ)I(f)(ξ)

As a corollary

F(t) = 1 t^d

X

x∈tc∩Λ

test(x/t)

<F(t),testi ≡ X

faces

1 t^codimf

X

f

Z

f

µ(f,c)(D/k)∗test HereDare the normal derivatives to the facef.

Example : Cone based on a square

EXAMPLE : letcwith generators e₃+e₁,e₃−e₁,e₃+e₂,e₃−e₂. Then

hF(t),testi ≡ Z

c

Test

+1 t

1 2

Z

boundary(c)

Test

+1 t²(2

9 Z

edges

Test− 1 36

Z

boundary

N·Test)

whereN·test is the normal derivative.Nhere is the primitive vector pointing inward.

+1

t³((1/6)Test(0)− 1 24

Z

edges

U∗test) +O(1/t⁴).

withU =e₃−e₁for the edgee₃+e₁, etc...

For a polytope p with integral vertices

The formulae add up nicely using Brianchom-Gram decomposition.

Ifhis a smooth function onpthen : t^−d X

x∈p∩Λ/t

h(x)≡X

f

t^−codim(f) Z

f

(µ(f,p)(1/t∂x)h)

whent is an integer going to∞.

This is the formula of Tatsuya Tate.

Asymptotic EML for a semi-rational polytope

Quite clear that if we use EML in dimension 1 with cones[a,∞]

andareal, we obtain also an asymptotic formula with

coefficientsf_j(t)/t^jhD_j,hiandf_j semi-quasi polynomial andD_j explicit via the renoramization procedure.

Example : simplex dimension 2

For example, for the standard simple p={x₁≥0,x₂≥0,x₁+x₂≤1}

Here is the asymptotic series for hS(t),testi= 1 t²

X

tp∩Z²

test(x/t)

andt real.

hS(t),testi −R

ptest ≡

= 1 t(1

2 Z

f1

test+1 2

Z

f2

test + (1−2∗ {t}

2 Z

f3

test))

+1 t²(−1

12 Z

f1

∂_ytest +−1 12

Z

f2

∂_xtest) + + 1

t²(−1 12 +1

2{t} −1 2{t}²)

Z

f3

(−(∂_x+∂y)/2))∗test

+1 t²(1

4test(0,0) + (3/8−(3/4){t}+ (1/4){t}²)test(1,0)) 1

t²(3/8−(3/4){t}+ (1/4){t}²)test(0,1) +O(1/t³)

In contrast to the 1-dimensional case, we dont know how to write a nice remainder in general.