696
Views
0
CrossRef citations to date
0
Altmetric
Articles

Polynomial interpolation and residue currents

ORCID Icon
Pages 1772-1781 | Received 18 Feb 2021, Accepted 23 Feb 2021, Published online: 15 Mar 2021

Abstract

We show that a global holomorphic section of O(d) restricted to a closed complex subspace XPn has an interpolant if and only if it satisfies a set of moment conditions that involves a residue current associated with a locally free resolution of OX. When X is a finite set of points in CnPn this can be interpreted as a set of linear conditions that a function on X has to satisfy in order to have a polynomial interpolant of degree at most d.

AMS SUBJECT CLASSIFICATIONS:

1. Introduction

Let i:XCn be a subvariety or complex subspace whose underlying space, Xred, is a finite set of points {p0,,pr}Cn. Let g be a holomorphic function on X, i.e. a global holomorphic section of OX, and let GC[ζ1,,ζn] be a polynomial. We say that G interpolates g if the pull-back of G to X equals g, i.e. iG=g.

If X is reduced, then a holomorphic function g on X is just a function from Xred={p0,,pr} to C, and we have that G interpolates g if G(pj)=g(pj) for each j=0,,r. In the univariate case this is referred to as Lagrange interpolation. If X is not reduced, then at each point pj, G also has to satisfy some conditions on its derivatives. In the univariate case this is referred to as Hermite interpolation, see Example 4.3.

The motivating question for this note is the following. What are the necessary and sufficient conditions on g for the existence of an interpolant of degree at most d?

Let AX denote the vector space of holomorphic functions on X, i.e. AX=H0(Cn,OX). Since the set of holomorphic functions on X that have an interpolant of degree at most d is a linear subspace of AX, we have that a function gAX has an interpolant of degree at most d if and only if it satisfies a finite set of linear conditions. In this note we will show how these linear conditions can be explicitly realized as a set of moment conditions that involves a so-called residue current associated with a locally free resolution of OX.

Recall that since Xred is a finite set of points, X can be viewed as a closed complex subspace of Pn, and we have that polynomials of degree at most d on Cn naturally correspond to global holomorphic sections of the line bundle O(d)Pn via d-homogenization. This motivates the following more general notion of interpolation that we shall consider in this note. Let i:XPn be a closed complex subspace of arbitrary dimension. Let Φ and φ be global holomorphic sections of O(d) and OX(d)=iO(d), respectively. We say that Φ interpolates φ if iΦ=φ.

From a minimal graded free resolution of the homogeneous coordinate ring of X, SX, we obtain a locally free resolution of OX of the form (1) 0O(En)fnf2O(E1)f1OPnOX0(1) where Ek=O()βk,, see [Citation1] and [Citation2, Section 6]. The βk, are referred to as the graded Betti numbers of SX. We equip the Ek with the natural Hermitian metrics. In [Citation2], Andersson and Wulcan showed that with (Equation1), one can associate a residue current R that generalizes the classical Coleff–Herrera product [Citation3], see Section 2. It can be written as R=k,Rk,, where each Rk, is an O()βk,-valued (0,k)-current. In [Citation4], the same authors proved a result which as a special case gives a cohomological condition in terms of the current R for when Φ interpolates φ. In Section 3 we will show that in our setting this condition amounts to the following set of moment conditions.

Theorem 1.1

Let XPn be a closed complex subspace, and let R be the residue current associated with  (Equation1). Moreover, let ω be a nonvanishing holomorphic O(n+1)-valued n-form. A global holomorphic section φ of OX(d) has an interpolant if and only if for each ℓ it holds that (2) PnRn,φhω=0(2) for all global holomorphic sections h of O(dn1).

Recall that the interpolation degree of X is defined as inf{d:all global holomorphic sections of OX(d) has an interpolant}. In particular, if Xred is a finite set of points in Cn, then the interpolation degree of X is the smallest number d such that any gAX has an interpolant of degree at most d. Define (3) tk(SX)=sup{:βk,0}.(3) (We use the convention that the supremum of the empty set is .) As a consequence of Theorem 1.1 we get the following bound of the interpolation degree.

Corollary 1.2

Let XPn be a closed complex subspace with homogeneous coordinate ring SX. The interpolation degree of X is less than or equal to tn(SX)n.

It can be shown by purely algebraic means that the interpolation degree of X is in fact equal to tn(SX)n, see e.g. [Citation5, Corollary 1.6]. If Xred consists of a finite set of points, then it can be shown that tn(SX)n is equal to the Castelnuovo–Mumford regularity of SX, see [Citation1, Exercise 4E.5], and the statement in this case is Theorem 4.1 in [Citation1].

In Section 4 we will consider the case when Xred is a finite set of points in Cn. In this case the corresponding versions of Theorem 1.1 and Corollary 1.2 first appeared in [Citation6]. We will consider some examples where we explicitly write down the conditions for when gAX has an interpolant of degree at most d. In particular, we will obtain the precise conditions for when the Hermite interpolation problem has a solution.

2. Residue currents

Let f be a holomorphic function in an open set in Cn. Let ξ be a smooth (n,n)-form with compact support. In [Citation7], using Hironaka's desingularization theorem, Herrera and Lieberman proved that the limit (4) limϵ0|f|>ϵξf(4) exists. Thus (Equation4) defines a current known as the principal value current, which is denoted by [1/f]. The residue current Rf of f is the (0,1)-current ¯[1/f]. It is easy to see that Rf has its support on V(f)=f1(0), and that it satisfies the following duality principle: A holomorphic function Φ belongs to the ideal (f) if and only if RfΦ=0.

Example 2.1

Let ζ0C. We have that the action of ¯[1/(ζζ0)] on a test form ξ(ζ)dζ is given by (5) ¯1ζζ0,ξ(ζ)dζ=2πiξ(ζ0).(5)

2.1. Residue currents associated with generically exact complexes

We will now consider a generalization of the above construction due to Andersson and Wulcan. Consider a generically exact complex of Hermitian holomorphic vector bundles over a complex manifold Y of dimension n, (6) 0Enfnf2E1f1E00,(6) i.e. a complex that is exact outside an analytic variety ZY of positive codimension. The vector bundle E=kEk has a natural superbundle structure, i.e. a Z2-grading, E=E+E, where E+=kE2k and E=kE2k+1, which we shall refer to as the subspaces of even and odd elements, respectively. This induces a Z2-grading on the sheaf of E-valued currents C(E); if ωξ is an E-valued current, where ω is a current and ξ is a smooth section of E, then the degree of ωξ is the sum of the degree of ξ and the current degree of ω modulo 2.

We say that an endomorphism on E is even (resp. odd) if it preserves (resp. switches) the degree. If α is a smooth section of EndE, then it defines a map on C(E) via α(ωξ)=(1)(degα)(degω)ωα(ξ), where ω is a current and ξ is a smooth section of E. In particular, the map f=k=1nfk defines an odd map on C(E). We define an odd map on C(E), =f¯, which, since f and ¯ anti-commute, satisfies 2=0. The map ∇ extends to an odd map on C(EndE) via Leibniz's rule, (αξ)=(α)ξ+(1)degααξ. In [Citation2], Andersson and Wulcan constructed EndE-valued currents U=U=k+1Uk, and R=R=k+1Rk, where Uk and Rk are Hom(E,Ek)-valued currents of bidegree (0,k1) and (0,k), respectively, which satisfy (7) U=idER,R=0.(7) The current R is referred to as the residue current associated with (Equation6) and it has its support on Z.

Suppose that the complex of locally free sheaves corresponding to (Equation6), (8) 0O(En)fnf2O(E1)f1O(E0),(8) is exact. When the Ek are equipped with Hermitian metrics, we shall refer to (Equation8) as a Hermitian resolution of the sheaf O(E0)/imf1. In this case it holds that R=0 if 1, and henceforth we shall write Rk for Rk0. Moreover, we have that R satisfies the following properties:

Duality principle: A holomorphic section Φ of E0 belongs to imf1 if and only if RΦ=0.

Dimension principle: If codimZ>k, then Rk=0.

Note that the second equality in (Equation7) is equivalent to (9) f1R1=0,(9) (10) fk+1Rk+1¯Rk=0,1kn1,(10) (11) ¯Rn=0.(11) Let i:XY be a closed complex subspace with ideal sheaf IX, and suppose that OX=iOY, which we identify with OY/IX, has a Hermitian resolution of the form (12) 0O(En)fnf2O(E1)f1OYOX0,(12) cf.(Equation8) where E0 is the trivial line bundle. For the associated residue current R=R1++Rn, we can view each Rk as an Ek-valued (0,k)-current. Since imf1=IX, we have that iΦ=0 if and only if RΦ=0 by the duality principle. More generally, let LY be a holomorphic line bundle. If we equip L with a Hermitian metric, then we obtain a Hermitian resolution of iL=OXL by tensoring (Equation12) with L, and we have that R is the associated residue current with this resolution as well.

2.2. The Coleff–Herrera product

Let f=(f1,,fp):CnCp be a holomorphic mapping such that V(f)=f1(0) has codimension p. In [Citation3] Coleff and Herrera gave meaning to the product (13) μf=¯1f1¯1fp,(13) which is known as the Coleff–Herrera product. In particular, if each fj only depends on ζj, then (Equation13) is just the tensor product of the one-variable currents ¯[1/fj] described above. The current μf is ¯-closed, has support V(f), and is anti-commuting in the fj. Moreover, μf satisfies the duality principle, i.e. μfΦ=0 if and only if ΦI(f), where I(f) is the ideal sheaf generated by f.

Let HY be a holomorphic Hermitian vector bundle of rank p, and let f be a holomorphic section of the dual bundle H. Let Ek=kH, and define δk:EkEk1 as interior multiplication by f. This gives a generically exact complex (Equation6). Suppose f=f1e1++fpep in some local holomorphic frame ej for H. If codimf1(0)=p, then the corresponding complex of sheaves is a Hermitian resolution of OY/I(f) known as the Koszul complex, and it was proven in [Citation8] that the associated residue current is given by R=Rp=μfe1ep.

2.3. A comparison formula for residue currents

We have the following comparison formula for residue currents, see Theorem 1.3 and Corollary 4.7 in [Citation9]. Let XX be complex subspaces of codimension p of Y. Suppose that there exist Hermitian resolutions of length p of OX and OX, respectively, and let R and R be the associated residue currents. Moreover, suppose that there exists a map of complexes

Then Rp=ψpRp.

3. Interpolation and residue currents

Let Y be a complex manifold of dimension n, and let i:XY be a closed complex subspace. Let LY be a holomorphic line bundle, and let Φ and φ be global holomorphic sections of L and iL, respectively. We say that Φ interpolates φ if iΦ=φ.

Suppose that there exists a Hermitian resolution of OX of the form (Equation12), and let R denote the associated residue current. For each point xY there is a neighbourhood U and a holomorphic section φ~ of L such that iφ~=φ on U. We define the current Rφ on Y locally as Rφ~. This is well-defined since if φ~ is another section such that iφ~=φ, then R(φ~φ~)=0 by the duality principle.

We have the following result which follows immediately as a special case of Lemma 4.5 (ii) in [Citation4].

Lemma 3.1

Let Φ and φ be global holomorphic sections of L and iL, respectively. Then Φ interpolates φ if and only if there exists a current w such that ΦRφ=w.

In other words, φ has an interpolant if and only if there exist currents w1,,wn such that ¯wn=Rnφ, and (14) ¯wk=fk+1wk+1+Rkφ,1kn1.(14) Moreover, in this case an interpolant of φ is given by Φ=f1w1. Note that Φ is holomorphic since ¯Φ=f1¯w1=f1(f2w2+R1φ)=(f1R1)φ=0. Here the last equality follows from (Equation9).

Let us now consider interpolation on Y=Pn with respect to the line bundle L=O(d). Recall that there is a Hermitian resolution of OX of the form (Equation1). We write Rk, for the O()βk,-valued component of Rk.

If Rnφ=¯wn for some current wn, then one can successively find currents wn1,,w1 such that (Equation14) holds since, in view of (Equation10), ¯(fk+1wk+1+Rkφ)=fk+1¯wk+1+(¯Rk)φ=(fk+1Rk+1¯Rk)φ=0, and it follows from, e.g. [Citation10, Theorem 10.7] that H0,k(Pn,EkO(d))=0,1kn1. We thus have the following condition for the existence of an interpolant.

Lemma 3.2

A global holomorphic section φ of OX has an interpolant if and only if Rnφ is ¯-exact, i.e. there exists a current η such that Rnφ=¯η.

Proof

Proof of Theorem 1.1

By Serre duality we have that Rn,φ is ¯-exact if and only if PnRn,φη=0 for all global ¯-closed O(d)-valued (n,0)-forms η. Note that each such form is of the form hω for some global holomorphic section h of O(dn1). Since Rnφ is ¯-exact if and only if each component Rn,φ is, the statement follows from Lemma 3.2.

Proof

Proof of Corollary 1.2

Let dtn(SX)n, see (Equation3), and let φ be a global holomorphic section of OX(d). We have for each ℓ that PnRn,φhω=0 for all global holomorphic sections h of O(dn1). Indeed, if d+n, the only such h is the zero section, and if >d+n, then Rn,=0 since βn,=0. Therefore φ has an interpolant by Theorem 1.1.

4. Polynomial interpolation

Let us now return to the topic of polynomial interpolation. Recall that the setting is that X is a complex subspace of Cn such that Xred is a finite set of points. The aim of this section is to give some examples where we explicitly compute the residue current R associated with a Hermitian resolution of OX and write down the moment conditions that Theorem 1.1 imposes on a function gAX for the existence of an interpolant of degree at most d. We do this by identifying g with a global holomorphic section φ of OX(d) and use the fact that g has an interpolant of degree at most d if and only if φ has an interpolant. More precisely, we let [z]=[z0::zn] denote homogeneous coordinates on Pn, and we view Cn as an open complex subspace of Pn via the embedding (ζ1,,ζn)[1:ζ1::ζn]. Recall that on Cn there is a frame e for O(1) such that a global holomorphic section Φ of O(d) is given by Φ(ζ1,,ζn)=G(ζ1,,ζn)e(ζ1,,ζn)d, where G is a polynomial of degree at most d on Cn.

Throughout this section we shall let ω in Theorem 1.1 be the nonvanishing holomorphic O(n+1)-valued n-form on Pn such that ω=dζ1dζne(n+1) on CnPn.

Note that the dimension principle gives that R=Rn, and throughout this section we write R rather than Rn, for the O()βn,-valued component of Rn.

Example 4.1

Let X={(0,0),(1,0),(0,1),(1,1)}C2P2. We have that X is defined by the homogeneous ideal IX=(f1,f2), where f1=z1(z1z0) and f2=z2(z2z0). A Hermitian resolution of OX is given by the Koszul complex, see Section 2.2, where we interpret (f1,f2) as a global holomorphic section of O(2)2. Thus the associated residue current takes values in O(4), and is given by the Coleff–Herrera product, see Section 2.2, R=R4=¯1ζ1(ζ11)¯1ζ2(ζ21)e(4). By a straightforward computation, cf.(Equation4), we get R4=¯1ζ1¯1ζ2¯1ζ11¯1ζ2¯1ζ1¯1ζ21+¯1ζ11¯1ζ21e(4). By Theorem 1.1, we now get the following. Since R=0 for 5, we have that any gAX has an interpolant of degree at most 2. Moreover, g has an interpolant of degree at most 1 if and only if (Equation2) holds when =4 and h = 1. In view of (Equation5) this amounts to (15) g(0,0)g(1,0)g(0,1)+g(1,1)=0,(15) which is expected since the values of g at (0,0), (1,0), and (0,1) uniquely determines a polynomial of degree at most 1 that takes the value g(1,0)+g(0,1)g(0,0) at (1,1). Note that this gives that the interpolation degree of X is 2.

We have that g has a constant interpolant if and only if (Equation2) holds for all global holomorphic h of O(1). By linearity we only need to check h=z0,z1,z2, which amounts to (Equation15), g(1,1)g(1,0)=0, and g(1,1)g(0,1)=0. This amounts to g(0,0)=g(1,0)=g(0,1)=g(1,1) as expected.

Example 4.2

Let X={(0,0),(1,0),(0,1),(0,2)}C2P2. We have that X is defined by the homogeneous ideal IX=(z1a1,z1z2,z2a2), where a1=z1z0 and a2=(z2z0)(z22z0). We have Hermitian resolutions of OP2/I(z1a1,z2a2) and OX and a map of complexes:

where the upper complex is the Koszul complex, see Section 2.2. Moreover, f1=z1a1z1z2z2a2,f2=z20a1a20z1, and ψ1=100001,ψ2=a2a1. Let R and R denote the residue currents associated with the resolutions of OX and O/I(z1a1,z2a2), respectively. We have that R=R5 takes values in O(5) and is given by the Coleff–Herrera product, see Section 2.2, R5=¯1ζ1(ζ11)¯1ζ2(ζ21)(ζ22)e(5). Thus by the comparison formula, see Section 2.3, R=ψ2R=R3R4. A straightforward computation gives that R3=¯1ζ1(ζ11)¯1ζ2e(3)=¯1ζ1¯1ζ2+¯1ζ11¯1ζ2e(3), and R4=¯1ζ1¯1ζ2(ζ21)(ζ22)e(4)=12¯1ζ1¯1ζ2¯1ζ1¯1ζ21+12¯1ζ1¯1ζ22e(4). By Theorem 1.1, we now get the following. Since R=0 for 5, we have that any gAX has an interpolant of degree at most 2. Moreover, g has an interpolant of degree at most 1 if and only if (Equation2) holds when =4 and h = 1. (Note that there is no condition involving R3 since dn1<0 in this case.) In view of (Equation5) we get the condition (16) 12g(0,0)g(0,1)+12g(0,2)=0,(16) which is expected since g has an interpolant of degree at most 1 if and only if g(0,1) is the average of g(0,0) and g(0,2). Note that this gives that the interpolation degree of X is 2.

We get that g has a constant interpolant if and only if (Equation2) holds when =4 for h=z0,z1,z2, and when =3 and h = 1. Since z1 vanishes on the support of R4, this amounts to Equation (Equation16), g(0,2)g(0,1)=0 and g(1,0)g(0,0)=0. This amounts to g(0,0)=g(1,0)=g(0,1)=g(0,2) as expected.

We end this note by considering Hermite interpolation. We refer to, e.g. [Citation11–13], and references therein for a classical survey of this topic.

Example 4.3

Let p0,,prC, and let g be a holomorphic function on the complex subspace XC defined by the ideal generated by j=0r(ζpj). Here we allow for the possibility that pi=pj for some i, j, so that X is nonreduced in general, and we denote the number of times that pj occurs by mj. We have that a polynomial G interpolates g if and only if, for each j=0,,r, G(k)(pj)=g(k)(pj),k=0,,mj1. We say that a polynomial interpolates g with respect to p0,,pk, kr, if it interpolates the pull-back of g to the complex subspace defined by the ideal generated by j=0k(ζpj). We denote the unique polynomial that interpolates g with respect to p0,,pk by H[g;p0,,pk]. The coefficient of its ζk-term is referred to as the kth divided difference of g and we denote it by g[p0,,pk]. By induction it is not difficult to see that H[g;p0,,pr](ζ)=k=0rg[p0,,pk]j=0k1(ζpj), see [Citation11, Theorem 1.8]. This is referred to as Newton's formula.

We claim that gAX has an interpolant of degree at most d if and only if (gh)[p0,,pr]=0 for all polynomials h of degree at most rd−1. Let us show how this condition follows from Theorem 1.1. Since the ideal is generated by a single element, we have that the associated residue current is given by Rr+1=¯1j=0r(ζpj)e(r1). Theorem 1.1 together with Stokes' formula gives that g has an interpolant if and only if C¯1j=0r(ζpj)ghdζ=CRH[gh;p0,,pr](ζ)j=0r(ζpj)dζ=0 for all polynomials h of degree at most rd−1, where CR is a circle of radius R0. Here we have used the fact that H[gh;p0,,pr] interpolates gh. By letting R, a direct calculation gives that the second integral is equal to 2πi(gh)[p0,,pr], and hence the claim follows.

Acknowledgments

I would like to thank Elizabeth Wulcan and Mats Andersson for helpful discussions and comments on preliminary versions of this note.

Disclosure statement

No potential conflict of interest was reported by the author(s).

References