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Original Articles

Recovery from Power Sums

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Abstract

We study the problem of recovering a collection of n numbers from the evaluation of m power sums. This yields a system of polynomial equations, which can be underconstrained (m < n), square (m = n), or overconstrained (m > n). Fibers and images of power sum maps are explored in all three regimes, and in settings that range from complex and projective to real and positive. This involves surprising deviations from the Bézout bound, and the recovery of vectors from length measurements by p-norms.

1 Introduction

This article offers a case study in solving systems of polynomial equations. Our model setting reflects applications of nonlinear algebra in engineering, notably in signal processing [Citation15], sparse recovery [Citation7], and low rank recovery [Citation8]. Suppose there is a secret list of complex numbers z1,z2,,zn. Our task is to find them. Measurements are made by evaluating the m power sums cj=i=1nziaj, where A={a1,a2,,am} is a set of m distinct positive integers. Our aim is to recover the multiset z={z1,,zn} from the vector c=(c1,,cm).

To model this problem, for any given pair (n,A), we consider the polynomial map(1) ϕA,C : CnCm,where   ϕj = x1aj+x2aj++xnajfor  j=1,2,,m.(1)

We are interested in the image and the fibers of the map ϕA,C. The study of these complex algebraic varieties addresses the following questions: Is recovery possible? Is recovery unique? This problem is especially interesting when z1,z2,,zn are real, or even positive. Hence, we also study the maps ϕA,R and ϕA,0 that are obtained by restricting ϕA,C to Rn and R0n, respectively. For any of these, we study the following system of m equations in n unknowns:(2) ϕA,(x)  =  c.(2)

There are three different regimes. If m > n then (2) is overconstrained and has no solution, unless c=ϕA,C(z) for some zCn, and we anticipate unique recovery of {z1,,zn}. If m = n then (2) is expected to have finitely many solutions, at most the Bézout number a1a2an. If m < n then the solutions to (2) form a variety of expected dimension nm.

Example 1

(n = 3). We illustrate the three regimes. Consider the multiset z={6,8,13}. We first allow m = 4 measurements, with A={2,5,7,8}. Then the system (2) equals(3) x12+x22+x32=269,x15+x25+x35=411837,x17+x27+x37=65125605,x18+x28+x38=834187553.(3)

For the lexicographic term order with x1>x2>x3, we compute the reduced Gröbner basis{ x1+x2+x327 ,  x22+x2x3+x3227(x2+x3)+230 ,  (x36)(x38)(x313) }.

This is a 0-dimensional radical ideal, having six zeros, so z={6,8,13} is recovered uniquely.

We next take m = 3 with A={2,5,7}. Here, we solve the first three equations in (3). This square system has 66 complex solutions, four less than the Bézout number 70=2×5×7. Finally, we allow only m = 2 measurements, with A={2,5}. The first two equations in (3) define a curve of degree 10 = 2 × 5 in C3. Its closure in P3 is a singular curve of genus 14.

Remark 2.

In applications, noise in the data is a concern. This makes our problem interesting even for A={1,2,,m}. However, from the perspectives of algebraic geometry and exact computations, the dense case is solved. The power sums reveal the elementary symmetric functions, via Newton’s identities. Our recovery problem amounts to finding the roots of a polynomial of degree n in one variable. For related work see [Citation1]. The recent article [Citation15] studies a more general version of the dense problem: The authors consider the composition γ of the map ϕA with a linear map T:CmCn where mn and A={1,,m}. Their Theorem 1 states that, if T is generic, then γ1(γ(x)) is finite for every xCm.

In this paper, A is any set of m distinct positive integers. Our presentation is organized as follows. In Section 2 we show that, for mn, the fiber of ϕA,C above a generic point in Cm has the expected dimension nm. For m > n we expect the recovery of complex multisets from power sums to be unique when gcd(a1,,am)=1. This is stated in Conjecture 6. In Section 3, we study the case m = n. We propose a formula for the number of solutions of (2). This number is generally less than the Bézout number a1a2an. For instance, in Example 1, the drop is from 70 to 66. We shall explain this. This issue is closely related to the question, put forth in [Citation3], for which sets A the power sums form a regular sequence. We present supporting evidence for the conjectures made in [Citation3] and we offer generalizations.

In Section 4 we turn to the images of the power sum maps ϕA,C,ϕA,R, and ϕA,0. The image of ϕA,C is constructible and has the expected dimension min(m,n), but it is generally not closed in Cm. In the overconstrained case (m > n), we study the degree and equations of the closure of the image. For instance, the image of ϕA,C:C3C4 in Example 1 is defined by a polynomial of degree 45 with 304 terms. The image of the real map ϕA,R is semialgebraic in Rm. It is closed if some ai is even. Moreover, the orthant R0n is mapped to a closed subset of R0m. It is a challenging task to find a semi-algebraic description of the image. We take first steps by exploring its algebraic boundary. Delineating the real image involves the ramification locus in Cn and its image in Cm, which is the branch locus of ϕA,C.

In Section 5, we examine our problem over the positive real numbers. Here, the recovery from power sums is equivalent to recovery from length measurements by various p-norms. This enables a better understanding of the map ϕA,0. We prove that recovery is unique in the square case n = m, see Proposition 24. The image of ϕA,0 is expressed as a compact subset in the probability simplex Δm1. Theorem 27 characterizes the structure of this set.

2 Fibers

Consider the map ϕ=ϕA,C from Cn to Cm whose coordinates are ϕj=i=1nxiaj. In this section we examine the fibers of ϕ and we show that they have the expected dimension. We conclude with a discussion concerning the uniqueness of recovery in the case m=n+1.

Given a point c=(c1,,cm) in Cm, the defining ideal of the fiber ϕ1(c) equalsIc  =   ϕ1(x)c1,,ϕm(x)cm     C[x1,,xn].

Our recovery problem amounts to computing the variety V(Ic)=ϕ1(c) defined by Ic in Cn.

Proposition 3.

Assume mn. Then the following hold:

  1. The map ϕ is dominant, i.e., the image of ϕ is dense in Cm.

  2. For generic c, the ideal Ic is radical, and its variety V(Ic) has dimension n – m.

Proof.

The fiber of ϕ above a point c is the variety V(Ic)Cn. By [14, Lemma 054Z], the fibers of ϕ are generically reduced. This implies that the ideal Ic is radical for all points c outside a proper closed subset of Cm. The Jacobian of the map ϕ is the m × n matrix(4) I   =    (ϕjxi)1jm1in  =    (ajxiaj1)1jm1in.(4)

Up to multiplication by a positive integer, each m × m minor of this matrix is the product of a Vandermonde determinant and a Schur polynomial; see (8). In particular, none of these minors of I is identically zero. Thus, the Jacobian matrix I has rank m over the field C(x1,,xm). By [9, I.11.4], this implies that the polynomials ϕ1,,ϕm are algebraically independent over C. From this we conclude that the associated ring homomorphism ϕ* : C[y1,,ym]C[x1,,xn],  yiϕi(x) is injective. Hence, our map ϕ is dominant, by [14, Lemma 0CC1]. The statement in (i) that the image is dense refers either to the Zariski topology or to the classical topology. Both have the same closure in this situation, by [10, Corollary 4.20].

It now follows from [11, Theorem 9.9 (b)] that, for all points c outside a proper Zariski closed subset of Cm, the fiber ϕ1(c) has dimension nm. This finishes the proof. □

The condition that the point c is generic is crucial in Proposition 3. The following example shows that the fiber dimension can jump up for special points cCm.

Example 4

(n = 3). Let m = 3 and A={3,5,7}. The generic fiber of the map ϕ consists of 60 points in C3. Interestingly, that number would increase to 66 if the number 3 in our set A were replaced with the number 2, as we saw in Example 1. Now, consider the fiber over c=(0,0,0). It is defined by the homogeneous ideal I0=x1a+x2a+x3a : aA. This ideal defines three lines of multiplicity three, with an embedded point at the origin. The radical of this ideal equals  x1+x2,x3x1+x3,x2x2+x3,x1.

Let us assume m > n, so we are in the overconstrained case. The following statement is derived from the m = n case in Proposition 3, namely by adding additional constraints:

Corollary 5.

For m > n, the fiber of ϕ above a generic point in Cm is empty. The closure of the image of ϕ is an irreducible variety of dimension n in Cm. The same holds over R.

Describing the image of ϕ will be our topic in Sections 4 and 5. A generic point c in that image can be created easily, namely by setting c=ϕ(z) where z=(z1,,zn) is any generic point in Cn. We are interested in the fiber ϕ1(c) over such a point c. By construction, that fiber is nonempty: it contains all n! points that are obtained from z by permuting coordinates. For the remainder of this section, assume gcd(a1,,am)=1. Then we conjecture that there are no other points in that fiber. This would mean that the set {z1,,zn} can be recovered uniquely from any m of its power sums, provided mn+1.

Conjecture 6. The recovery of a set of n complex numbers from n + 1 power sums with coprime powers is unique. To be precise, for m=n+1, the map ϕ is generically injective. This means that, for generic points zCn, the fiber ϕ1(ϕ(z)) coincides with the set of n! coordinate permutations of z.

We are also interested in the following more general conjecture. Let τ=(τ1,,τn) be in R>0n and consider the map ψ:CnCm, where ψj=i=1nτixiaj. Let Stab(τ) be the subgroup of the symmetric group Sn consisting of all coordinate permutations that fix τ.

Conjecture 7. For generic points zCn, the fiber ψ1(ψ(z)) is precisely the set of all coordinate permutations of z. The cardinality of this set is equal to |Stab(τ)|.

By computing Gröbner bases, we confirmed Conjectures 6 and 7 for a range of small cases: Conjecture 6 holds for n = 4 and aAa52, and for n = 5 and aAa49. Conjecture 7 holds for (n,m)=(2,3) and aAa100, for (n,m)=(3,4) and aAa69, and for (n,m)=(4,5) and aAa49. In all cases we took random integers 1τi100.

3 Square Systems

We here fix n = m, so we study the square case. By Proposition 3, our system (2) has finitely many solutions in Cn. Our aim is to find their number. We study this for n = 2 (Proposition 10) and n = 3 (Conjecture 14). This generalizes a conjecture of Conca, Krattenthaler and Watanabe [3, Conjecture 2.10]. We conclude with a discussion of the general case n4.

Our point of departure is a result which links Proposition 3 with Bézout’s Theorem.

Proposition 8.

For general measurements cCn, the square system (2) has finitely many complex solutions xCn. The number of these solutions is bounded above by a1a2an.

We now define the homogenized system (HS) to be the system (2), where cj is replaced by cjx0aj. Note that (HS) has its solutions in Pn. What we are interested in for our recovery problem are the solutions that do not lie in the hyperplane at infinity {x0=0}. Next, we define the system at infinity (SI) to be (2) with c = 0. The solutions of (SI) are in Pn1. The cone over that projective scheme is the zero fiber of the map ϕ. We will use the notations (HS) and (SI) both for the systems of equations and the projective schemes defined by them.

Remark 9.

The scheme (HS) is in general not the projective closure of the affine part defined by (2), as it can contain higher-dimensional components. For example, set n=m=4, and let A consist of four odd coprime integers. The variety in C4 defined by the system (2) is zero-dimensional by Lemma 3. However, the scheme (HS) is not zero-dimensional in P4, since it contains the lines defined by xi=xj,xk=xl,x0=0 for {i,j,k,l}={1,2,3,4}.

The solutions of (HS) that lie in the hyperplane {x0=0} are precisely the solutions to (SI). However, the multiplicities are different. If the variety (SI) in Pn1 is finite, then the number of solutions to (2) in Cn equals a1a2an minus the total length of (HS) along (SI). For n = 2, this observation fully determines the number of solutions to (2) in terms of A.

Proposition 10

(n = 2). Assume a1<a2. For generic (c1,c2)C2, the number of common solutions in C2 to the equations x1a1+x2a1=c1 and x1a2+x2a2=c2 equals  a1(a2gcd(a1,a2))  if both a1/gcd(a1,a2) and a2/gcd(a1,a2) are odd. It equals the Bézout number a1a2 otherwise.

Proof.

First assume gcd(a1,a2)=1. The binary forms x1a1+x2a1 and x1a2+x2a2 are relatively prime, unless both a1 and a2 are odd, so x1+x2 divides both forms. In the former case, (SI) has no solutions, so the number of solutions to (2) equals the Bézout number a1a2. If a1 and a2 are odd, then (SI) ={x1+x2=0} defines the point (1:1) on the line P1, corresponding to the point P=(1:1:0) of the scheme (HS). The multiplicity of (HS) at P can be computed locally in the chart {x20} by setting x2=1. It is the multiplicity at the point (0, 1) of the affine scheme in C2 defined by the ideal  I =  x1a1x0a11, x1a2x0a21 .

Write mP=x0,x11 for the maximal ideal of P=(0,1) in the local ring OP of the curve V(x1a1x0a11)C2. In OP we have x11=x1a11u=x0a1u, where u is a unit. In fact, u is a certain product of cyclotomic polynomials in x1. Therefore, x0 is a uniformizer, i.e., mP=x0, and x11 is contained in mPa1mPa1+1. From this we concludex1a2x0a21  =  (((x11)+1)a2x0a21)  =  i=1a2(a2i)(x11)ix0a2    mPa1mPa1+1.

Hence, x1a2x0a21 vanishes to order a1 at P. We conclude that the multiplicity of (HS) in P is a1. Therefore, the system (2) has a1(a21)=a1(a2gcd(a1,a2)) solutions in C2.

Finally, suppose that a1 and a2 are not relatively prime, and set g=gcd(a1,a2). We replace x1, x2 by x1g,x2g, and we apply our previous analysis to the two equations(5) (x1g)a1/g+(x2g)a1/g = c1and(x1g)a2/g+(x2g)a2/g = c2.(5)

The system (5) has solutions at infinity if and only if a1/g and a2/g are both odd. In that case, we have (SI) ={x1g+x2g=0}, which defines the g points (ζi:1)i=1,,g in P1, where ζ is a primitive gth root of –1. Each of the corresponding points in (HS) has multiplicity a1. This can be computed analogously to what we did for P in the argument above. □

We turn to n = 3, and we assume gcd(a1,a2,a3)=1. Our problem is now much harder. It is unknown when (SI) has any solutions in P2. No solutions means that the power sums ϕ1,ϕ2,ϕ3 form a regular sequence. Conca, Krattenthaler and Watanabe [3, Conjecture 2.10] suggest that this holds if and only if a1a2a30 mod 6; we call this the CKW conjecture. They prove the “only if” part in [3, Lemma 2.8]. Another proof for this part is given by the next lemma. Set Ap={a1 mod p,a2 mod p,a3 mod p} for p = 2, 3. Thus, A2{0,1} and A3{0,1,2}. We assumed Ap={0} for p = 2, 3. Let ζ be a primitive cube root of unity.

Lemma 11.

The points (1:1:0),(1:0:1), and (0:1:1) are in (SI) if and only if 0A2, and the points (1:ζ:ζ2) and (1:ζ2:ζ) are in (SI) if and only if 0A3.

Proof.

If n is a prime, ξ is a primitive nth root of unity, and a is a multiple of n, then the power sum x1a+x2a++xna does not vanish at (1,ξ,,ξn1), but rather it evaluates to n. We obtain the assertion by specializing to n = 2 and n = 3. □

The CKW conjecture states that (SI) has no solutions when 0A2A3. It is shown in [Citation3, Theorem 2.11] that this holds if {1,n}A with 2n7, or if {2,3}A. The proof rests on the expression of power sums in terms of elementary symmetric polynomials.

In what follows we present conjectures that imply the CKW conjecture. We begin with a converse to Lemma 11. Theorems 13 and 15 verify all conjectures for some new cases.

Conjecture 12. We have (SI){(1:1:0),(1:0:1),(0:1:1),(1:ζ:ζ2),(1:ζ2:ζ)}.

This generalizes [3, Conjecture 2.10] since the five possibilities for points on (SI) do not occur if 0A2A3. We show some new cases of the conjecture using computational tools.

Theorem 13.

Conjecture 12 holds for all a1<a2<a3 with a1+a2+a3300.

Proof.

Let P=(α:β:γ) be a point on (SI), corresponding to P=(α:β:γ:0) on (HS). After permuting coordinates, we may assume α0. Then P is in the affine chart C2 of P2 given by x=x2/x1 and y=x3/x1. The restriction of (SI) to that plane C2 is defined by(6) xa1+ya1+1 = xa2+ya2+1 = xa3+ya3+1  =  0.(6)

Conjecture 12 states that the number of solutions to the system (6) is 0, 2 or 4, as follows:

We verified the counts in the second column for all a1<a2<a3 with a1+a2+a3300. We did this using the Gröbner basis implementation in the computer algebra system magma. The same would be doable with other tools for bivariate equations. □

Conjecture 14. For n = 3 and gcd(a1,a2,a3)=1, the following holds for the system (2):

If 0A3, then we have #Solutions = {a1a2a3 if A2={1,0};a1a2a33a1a2 if A2={1}.

If A3={1} or {2}, then we have #Solutions = {a1a2a34a1 if A2={1,0};a1a2a34a13a1a2 if A2={1}.

If A3={1,2}, then we have   #Solutions = {a1a2a32iA if A2={1,0};a1a2a32iA3a1a2 if A2={1}.

Here iA is the index of nilpotency of the zero-divisor x0 in the homogeneous system (HS).

At present we do not have a simple formula for the number iA in all cases. Computationally, it can be found from the homogeneous ideal I=f1,f2,f3 that is generated by  fj=x1aj+x2aj+x3ajcjx0aj  for j = 1, 2, 3. Using ideal quotients, the definition is as follows:(I:x0iA1)    (I:x0iA)  =  (I:x0iA+1).

From our computations it seems that iA is always either a1 or a2 or 2a1.

Theorem 15.

Conjecture 14 holds for all a1<a2<a3 with a320 or a1+a2+a340.

Proof.

Our approach is to compute the multiplicity in (HS) for each point that is known (by Theorem 13) to lie in (SI). We conjecture that these multiplicities are as follows:

  1. If A2={1}, then the point (1:1:0:0) has multiplicity a1a2 in (HS);

  2. if A3={1,2}, then the point (1:ζ:ζ2:0) has multiplicity iA in (HS);

  3. if |A3|=1, then 2iA=2a1 and this is the multiplicity of (1:ζ:ζ2:0) in (HS).

These claims imply Conjecture 14, by our previous analysis. Indeed, if 0A2A3, then (SI) is empty and the number of solutions to (2) is the Bézout number a1a2a3. Otherwise, we need to subtract the multiplicities above, according to the various cases. Here the number in (i) is multiplied by 3 since the S3-orbit of (1:1:0:0) has three points, and the numbers in (ii) and (iii) are multiplied by 2 since the S3-orbit of (1:ζ:ζ2:0) has two points.

For our computations, we fix P{(1:1:0:0),(1:ζ:ζ2:0)}, we focus on the affine chart C3={x1=1}, and we consider the ideal I=f1,f2,f3 in the local ring OP,C3. The quotient V=OP,A3/I is a vector space over C, and its dimension is the multiplicity of (HS) at P. We computed this dimension for all values of a1,a2,a3 in the stated range, and we verified that (i), (ii), and (iii) are satisfied. This was done using Gröbner bases in magma. □

Extending Conjecture 14 to n4 seems out of reach at the moment, for two reasons. First of all, the conditions on A for (SI) to have no solutions are less simple. For n = 4 with gcd(a1,a2,a3,a4)=1, Conca, Krattenthaler and Watanabe [3, Conjecture 2.15] state three conditions on A under which (SI) has no solutions. They show that all three conditions are necessary. We verified their conjecture using Gröbner bases in magma for a1+a2+a3+a4100. Secondly, in the event that (SI) does have solutions, it is not at all obvious what these should be. In general, they are not given only by points whose coordinates are roots of unity, as was the case for n = 3. This happens already for n = 4 as the following example shows:

Example 16.

Set n = 4 and A={2,4,9,10}. The system (2) has 576 solutions which is 144 less than the Bézout number 720. The scheme (SI) in P3 which is defined by the ideal x12+x22+x32+x42,x14+x24+x34+x44,x19+x29+x39+x49,x110+x210+x310+x410 contains 72 distinct points. The minimal polynomial of each of the coordinates of the points in (SI) has degree 36. Every root of this polynomial occurs in each coordinate in exactly two points.

4 Images

We now study the images of the power sum maps ϕA,C, ϕA,R, and ϕA,0. The recovery problem (2) has a solution if and only if the measurement vector c lies in that image. We know from Chevalley’s Theorem [10, Theorem 4.19] that im(ϕA,C) is a constructible subset of Cm. Over the real numbers, the Tarski-Seidenberg Theorem [10, Theorem 4.17] tells us that im(ϕA,R) is a semialgebraic subset of Rm and im(ϕA,0) is a semialgebraic subset of R0m. It follows from Proposition 3 that, for each of these images, the dimension equals min(n,m).

We first examine whether the images are closed. We use the classical topology on Rm or Cm. This makes sense not just over R, but also over C, since the Zariski closure of the image of any complex polynomial map coincides with its classical closure [10, Corollary 4.20].

Proposition 17.

The constructible set  im(ϕA,C) is generally not closed in Cm. The semialgebraic set  im(ϕA,R) is closed in  Rm when  0A2, but it is generally not closed otherwise. Finally, the semialgebraic set  im(ϕA,0) is always closed in R0m.

Proof.

Let m=n=2,a1=1 and a23 odd. Fix any c=(c1,c2)C2, with c1=0 and c2=0. Then (2) has no complex solution because ϕ1(x) divides ϕ2(x). Thus, c does not belong to the image of ϕA,C or ϕA,R. However, c is in the closure of  im(ϕA,C) because that closure is C2 by Proposition 3 (i). The same counterexample works over the real numbers. Let us now set c1=ϵ for ϵ>0 very small and solve ϕ1(x)=ϵ by setting x2=ϵx1. This substitution in ϕ2(x)=c2 gives a polynomial equation in one variable x1 of odd degree a2. Such an equation always has a real solution x1(ϵ). The image of the point (x1(ϵ),ϵx1(ϵ)) converges to c in R2 as ϵ0 from which we conclude that c lies in the closure of im(ϕA,R).

We are left with the cases where the image is closed. First suppose 0A2. Then there is an even element in A, say ai. Let cRm be in the closure of  im(ϕA,R). There exists a sequence {x(l)}l0 of points in Rn such that ϕA,R(x(l)) converges to c as l. Since ai is even, we have ϕi(x(l))=||x(l)||aiai, which converges to ci0 as l. Hence, the sequence {x(l)}l0 is bounded in the norm ||·||ai. There is a subsequence that converges to some point x() in Rn. Since the power sum map ϕA,R is continuous, the image of x() is equal to c. Therefore, cim(ϕA,R), and we conclude that the real image is closed. Finally, take A arbitrary and consider the nonnegative power map ϕA,0. Let cR0m be in the closure of  im(ϕA,0) and let {x(l)}l0 be a sequence of nonnegative points whose images ϕA,0(x(l)) converge to c as l. Now, for any index i, the norm ||x(l)||ai=ϕi(x(l))1/ai is bounded, so there exists a convergent subsequence of {x(l)}l0. Let x()R0n be its limit. Again, by the continuity of the power map, we have ϕA,0(x())=c. From this we conclude that  im(ϕA,0) is closed. □

Example 18.

Set n=m=2 and A={1,3}. The image of ϕA,R is the nonclosed set{(0,0)}    { cR2 : (c1<0  and  c134c2)   or   (c1>0  and  c134c2) }.

On the other hand, the image of the map restricted to the nonnegative orthant is closed:(7)  im(ϕA,0)  =  { cR02 : c2c134c2 }.(7)

In Section 5 we generalize this description of the image of ϕA,0 to other power sum maps.

We next examine our images through the lens of algebraic geometry. Let c1,,cm be variables with degree(ci)=ai. These are coordinates on the weighted projective space WPm1 with weights given by A. We regard ϕ=ϕA,C as a rational map from Pn1 to WPm1. The following features of the image will be characterized in Theorem 21: (i) For m=n+1, the closure of the image im(ϕ) is an irreducible hypersurface in WPm1. We give a formula for its degree, which is the weighted degree of its defining polynomial in the unknowns c1,,cm. (ii) For mn, we describe the positive branch locus of the map ϕ. This is a hypersurface in Cm. By reasoning as in the proof of [8, Theorem 3.13], this hypersurface represents the algebraic boundary of the image of ϕA,0.

To study the branch locus of ϕ, we start with the ramification locus R. This consists of points in Cn where ϕ is not smooth [2, Section 2.2, Proposition 8]. Set ϕj=i=1nxiaj and μ=min{n,m}. Let IC[x1,,xn] be the ideal generated by the μ×μ minors of the Jacobian matrix I as in (4); this is an ideal of height |mn|+1. Its variety R=V(I) is the set of points where I has rank less than μ. Each maximal minor of I, up to multiplication with a positive integer, has the form(8) (xi1xi2xiμ)ai11·1j<kμ​​​​(xijxik) · S(xi1,xi2,,xiμ),(8) for some 1i1<<iμn. The last factor is a Schur polynomial.

In the square case m = n, the variety R is a reducible hypersurface in Cn, given by the vanishing of one polynomial (8). Write g=gcd(a11,,am1). By [4, Theorem 3.1], the Schur polynomial S is either constant, which happens when (ai1)/g=i1 for 1in, or it is irreducible. Let R be the closure in Cn of R(ijV(xixj)V(xk)). Thus, R is the nontrivial component in the ramification locus. Our discussion implies the following:

Proposition 19.

Assume m = n. The ramification variety R is either empty, in which case we have (ai1)/g=i1 for 1in, or it is an irreducible hypersurface of degree i=1n(ai1)(n2)n(a11).

Example 20.

For A={3,6,7} and n = 3, the ideal I is principal. Its generator factors asx12x22x32(x1x2)(x1x3)(x2x3)(x12x22+x12x2x3+x1x22x3+x12x32+x1x2x32+x22x32).

The variety R is the quartic surface in C3 defined by the Schur polynomial in the last factor.

In the overdetermined regime (m > n), there are partial results by Fröberg and Shapiro [Citation5], who study the closure of R(ijV(xixj)) in Cn which we denote by R. Notably, it remains an open problem to find the dimension of R. Assuming a1=g=1, the first interesting case is n = 3, m = 5. Proving the dimension of R to be the expected one is equivalent to showing that three complete homogeneous polynomials form a regular sequence. This brings us back to Conca, Krattenthaler and Watanabe [3, Conjecture 2.17].

Now assume mn. Then R contains all linear spaces defined by nm+1 independent equations of the form xi = xj or xk = 0. We call these positive ramification components. This name is justified as follows. The Schur polynomial S in (8) has positive coefficients, and therefore cannot vanish at nonzero points in the nonnegative orthant R0n. Hence, only these components contribute to the ramification locus of the positive map ϕA,0. A positive branch hypersurface is any irreducible hypersurface in weighted projective space in WPm1 that is the closure of the image of a positive ramification component under the power sum map ϕ.

Theorem 21.

The following hypersurfaces in WPm1 are relevant for the image of our map.

  1. If m=n+1, then the image of ϕ is an irreducible hypersurface in WPm1 whose weighted degree is at most (a1a2am)/(m1)!. If this ratio is an integer, then this bound can be attained. Specifically, if m = 3 and a1a2a3 is even, then it can be attained.

  2. If mn, then the weighted degree of any positive branch hypersurface of ϕ is at most the Bézout number a1a2am.

Proof.

Let mn. The restriction of ϕ to any positive ramification component is a rational map from Pm2 to WPm1. After renaming the xi if needed, we can write its coordinates as(9) i=1m1τixiaj   for j=1,,m,(9) where τ1,,τm1 are positive integers. Let Hτ denote the image of this map in WPm1. This also covers the case m=n+1 in (i) since ϕ has coordinates as in (9) with τ1==τm1=1. Hence, all hypersurfaces in (i) and (ii) have the form Hτ. Our aim is to compute their degrees.

Fix positive integers τ1,,τm1 and set zj=cj1/aj. Consider the projective space P2m2 with coordinates x1,,xm1,z1,,zm. Let Z denote the variety in P2m2 defined by the homogeneous polynomials i=1m1τixiajzjaj, for j=1,,m. By the same reasoning as in Proposition 3, this variety is irreducible and it is a complete intersection of degree a1a2am.

We consider the image of Z under the coordinate projection(10) π:P2m2 Pm1, (x1::xm1:z1::zm)(z1::zm).(10)

The closure of π(Z) is essentially the hypersurface Hτ we care about, but it lives in Pm1. Its degree in Pm1 with coordinates (z1::zm) coincides with the degree of Hτ in WPm1 with coordinates (c1::cm). Indeed, these two hypersurfaces have the same defining polynomial, up to the substitution cj=zjaj. The Refined Bézout Theorem [6, 12.3] implies(11) deg(π(Z))    deg(Z)deg(π|Z)  =  a1a2amdeg(π|Z),(11) where equality holds if π has no base locus. This immediately proves (ii).

We proceed with proving (i). The degree of π|Z is the size of its generic fiber. This equals the size of the generic fiber of the map given by (10). Conjecture 7 states that the size of the generic fiber of π is the size of the stabilizer of τ=(τ1,,τm1) in the symmetric group Sm1. In particular, it would follow that the generic fiber is a single point if and only if the τi are all distinct, and it consists of (m1)! points if and only if the τi are identical. We do not know yet whether this conjecture holds. But, in any case, the number |Stab(τ)| furnishes a lower bound for the size of a generic fiber and thus for deg(π).

Since the hypersurface im(ϕ) in (i) equals Hτ for τ1==τm1=1Zτ1==τm1=1, we conclude from (11) that its weighted degree is at most (a1a2am)/(m1)!. Equality can only hold when the base locus of π on the variety V(ϕj(x)zjaj : j=1,,m) is empty. This happens precisely when the system at infinity (SI) is the empty set. A necessary condition for this to happen is that (m1)! divides the Bézout number a1a2am. One checks that this is also sufficient when m = 3: we saw in Proposition 10, that (SI) is empty when a1a2 is even. □

Example 22

(m = 3). Suppose gcd(a1,a2,a3)=1 and B=a1a2a3 is even. If n = 2, then the image of ϕ is a curve of expected degree B/2 in WP2. If n3, then every positive branch curve has expected degree B/2 or B. For instance, if n = 4, then the ramification component {x1=0,x3=x4} should give a branch curve of degree B, while {x1=x2,x3=x4} should give a branch curve of degree B/2. We shall see pictures of such curves in the next section.

Example 23

(m = 4). If n = 3 and 6 divides B=a1a2a3a4, then we expect the image of ϕ to have weighted degree B/6. This would follow from the conjectures in Sections 2 and 3. The positive branch surfaces for n4 should have degrees B/6,B/2 or B. If 6 does not divide B, then the weighted degrees of the image and branch surfaces in WP3 are determined by the base loci. This takes us back to Conjecture 14. To be very explicit, let A={2,5,7,8} as in Example 1. Here B/6=560/6=93.333. The image of our map ϕ:P2WP3 is defined by a homogeneous polynomial of weighted degree 90 with 304 terms, namely9c1451050c141c43724c140c22+22400c139c2c331000c138c32+1966899200c1c411+1258815488c22c410.

By contrast, consider A={2,5,7,9}. Now, B/6=105 is an integer, and this equals the weighted degree of the image surface. Its defining polynomial has 388 terms, and it looks like59049c135c27459270c134c26c359049c135c35+255150c133c26c4++6350400c2c34c48324000c36c47.

5 Recovery from p-norms

Focusing on the positive region, we now investigate the properties of the map ϕA,0. The key fact to be used throughout is that the power sum of degree p represents the p-norm:(12) ||x||p  =  ( i=1nxip )1/p for all  x=(x1,,xn)R0n.(12)

Hence, our recovery problem for nonnegative vectors xR0n is equivalent to recovery of x from values of the p-norms ||·||p, where p runs over a prespecified set A of positive integers. We are interested in existence and uniqueness of vectors with given p-norms for pA.

Let us begin with the basic identifiability question: How many different p-norms are needed to reconstruct a vector in R0n from their values up to permuting the n coordinates? Conjecture 6 together with (12) would imply that n + 1 different norms suffice. On the other hand, it follows from Proposition 3 that at least n different p-norms are necessary. But are these n measurements already sufficient? We start by showing that this is indeed the case.

Proposition 24.

For m = n, recovery from p-norms is always unique. Given any set A of n positive integers, the map ϕA,0:R0nR0n is injective up to permuting coordinates.

Proof.

Write ϕ=ϕA,0. We proceed by induction on n. For n = 1, the map ϕ is obviously injective as it is strictly increasing. Thus, we have unique recovery for n = 1. Let us now prove the statement for arbitrary n. Our argument is based on the calculus fact that a differentiable function from a real interval to R is injective if its derivative has constant sign.

Consider the cone of decreasing vectors, C={xRn:x1>x2>>xn0}. Let X1, X2 be two arbitrary distinct points from this cone. Our claim states that they map to two different points under the map ϕ, i.e., that ϕ is injective on C. Let L be the line segment from X1 to X2 and consider the restriction ϕ|L:RRn of ϕ to L that is now a function in one variable. Its derivative is given by the product of the Jacobian matrix of ϕ, which we denote by Iϕ, evaluated at L, and the vector (X2X1). First notice that if X1,n=X2,n=0, then we are in the case where the induction hypothesis applies. Let us now w.l.o.g. assume X2,n>0. Then Iϕ is an n × n matrix whose determinant is of the form (8).

The Schur polynomial S does not vanish on L∖{X1}, and neither do the linear factors. Hence, the coordinates of the vector Iϕ·(X2X1) do not vanish at any point on L∖{X1}. Each coordinate is a function of constant sign on the whole segment L. This shows that ϕ is injective on the line L. As X1 and X2 were chosen to be arbitrary points, we conclude that ϕ is injective on the whole cone C. For a much more general version of this argument, we refer to the equivalence of conditions (inj) and (jac) in [12, Theorem 1.4]. □

The proof above is not algorithmic. It does not tell us how to invert ϕ. Our current method of choice for recovery is solving the equations using numerical algebraic geometry.

The next goal is to characterize the semialgebraic set  im(ϕA,0) inside the nonnegative orthant R0m. Starting with m = 2, we first present a generalization of the formula (7).

Proposition 25.

Set m=2n and A={a1<a2}. Then the nonnegative image equals (13)  im(ϕA,0)  =  { cR02  :  c2a1  c1a2  na2a1c2a1 }.(13)

Proof.

At any point xR0n, our map evaluates the norms ||x||aj=ϕj(x)1/aj for i = 1, 2. The first norm is larger than or equal to the second one: ||x||a1||x||a2. They agree at coordinate points. Their ratio is maximal at e=(1,1,,1). This gives the inequalities1    ||x||a1||x||a2    ||e||a1||e||a2 = n1/a1n1/a2.

All values in this range are obtained by some point xR0n. We now raise both sides to the power a1a2 and thereafter we clear denominators. This gives the inequalities in (13). □

The proof of Proposition 25 suggests that the study of the nonnegative image  im(ϕA,0)  can be simplified by replacing the power sum map by the normalized map into the simplex(14) ψA : R0n  Δm1 :  x  1j=1m||x||aj·( ||x||a1,||x||a2,,||x||am).(14)

Here, Δm1={uR0m : u1+u2++um=1} is the standard probability simplex. If we know the image of this map then that of the power sum map can be recovered as follows:(15)  im(ϕA,0)  =  { cR0m  :  1j=1mcj1/aj( c11/a1, c21/a2,, cm1/am )  im(ψA) }.(15)

We next consider the case m = 3. For every n3, the image is a nonconvex region in the triangle Δ2. These regions get larger as n increases. We illustrate this for an example.

Example 26.

Set A={2,3,4}. For n3, the image of the norm map ψA into the triangle Δ2 is an n-gon with curvy boundary edges that lies inside the subtriangle {c1>c2>c3}. The edges and diagonals of this n-gon are the following (n2) curvy segments for 1i<jn:Bij  =  ψA({xR0n : x1==xixi+1==xj, and xk=0 for k>j}).

The Zariski closure of Bij is an irreducible curve. There are n/2 distinct sets Bij with i = j. For ij, the Zariski closures of Bij and Bji,j are the same. Hence, we obtain(n2)n22+n2  =  n24distinct complex algebraic curves as Zariski closures of these curvy segments.

For n = 3, there are two distinct branch curves: one curve of degree 12, given by the segment B12, and one of degree 24, given by the two segments B13 and B23. For n = 6, shows the curvy hexagon im(ψA). Its 15 curvy segments form nine distinct branch curves, six of degree 24 and three of degree 12. The latter are given by B12,B24,B36. The curvy segment B12 is red in both pictures. For n = 2, we have B12=im(ψA). For n3, the curvy segment B12 is one of the n boundary edges of im(ψA). The Zariski closure of the curvy segment B12 is the branch curve {c1124c16c264c212+12c12c26c343c14c282c312=0}.

Fig. 1 The image of the norm map ψA for n=6,m=3 is a curvy hexagon in a triangle. The color coding on the left shows the progression of images for n=3,4,5,6. The color coding on the right shows the algebraic degrees 12 (red) and 24 (blue) of the curvy segments.

Fig. 1 The image of the norm map ψA for n=6,m=3 is a curvy hexagon in a triangle. The color coding on the left shows the progression of images for n=3,4,5,6. The color coding on the right shows the algebraic degrees 12 (red) and 24 (blue) of the curvy segments.

We now state a theorem which generalizes our observations in Example 26 to m4. We fix nm and A={a1<<am} as before. For 1lm and any ordered set ν=(ν1,,νl)([n]l), let Rν denote the set of vectors xR0n that satisfy xi=xi+1 if i<ν1 or νri<νr+1 for some r, and xi = 0 for all i>νl, and xixi+1 otherwise. Its image Bν=ψA(Rν) is a semialgebraic subset of dimension l1 in Δm1. Proposition 25 tells us that Bν is a curvy simplex with vertices Bν1,,Bνl. We define the type of ν to be the multiset {ν1,ν2ν1,ν3ν2,,νlνl1}. We can view τ=type(ν) as a partition with precisely l parts of an integer between l and n. Let Tn,l denote the set of such partitions τ. We use the notation Stab(τ) from Conjecture 7. In analogy to the proof of Theorem 21, we denote by Hτ the image in the simplex Δm1 of a positive ramification component of type τ.

Theorem 27.

Assume mn. The norm map ψA in (14) has the following properties:

  1. The image of ψA in Δm1 is the union of the curvy (m1)-simplices Bν where ν([n]m). The curvy facets of these curvy simplices are Bμ where μ([n]m1). Some of these curvy (m2)-simplices form the boundary of the semialgebraic set  im(ψA).

  2. Two curvy (m2)-simplices Bμ and Bμ have the same Zariski closure if  type(μ)=type(μ). Thus, the irreducible branch hypersurfaces Hτ are indexed by  τTn,m1.

Proof.

For l{1,,m} and ν([n]l), the set Rν is a convex polyhedral cone, spanned by linearly independent vectors in a linear subspace of dimension lm in Rn. By Proposition 24, the map ϕA,0 is injective on Rν. Therefore, by the transformation in (15), the map ψA is injective on Rν up to scaling. This means that the image Bν=ψA(Rν) is a curvy simplex of dimension l1 inside the probability simplex Δm1. We also conclude that the boundary of im(ψA) equals the union of the images Bν=ψA(Rν), where ν runs over a certain subset of ([n]m1). These specify the algebraic boundary of im(ϕA,0). This proves (i).

To see that part (ii) holds, we write the restriction of ϕA,0 to the cone Rμ as a polynomial function in only l distinct variables xi. The jth coordinate of that restriction has the form i=1lτi xiaj, where τ=type(μ). Different cones Rμ of the same type τ are distinguished only by the orderings of the parameters x1,x2,,xl. However, they have the same linear span in Rn. Hence, after we drop the distinguishing inequalities xi > xj, the maps are the same. In particular, their images Bμ have the same Zariski closures Hτ in the simplex Δm1. □

Example 26 illustrates Theorem 27 for m=3, where |Tn,2|=n2/4 and |Stab(τ)|{1,2}. We found it more challenging to understand the geometry of our image in higher dimensions.

Example 28

(n=8,m=4). The image of ψA in the tetrahedron is a curvy 3-polytope. It is partitioned by 56=(83) curvy triangles Bν. Their types τ identify 16 clusters: two singletons, ten triples, and four of size six. These determine 16=|T8,3| branch surfaces Hτ.

Based on computational experiments, we believe that, for all pairs mn and all exponents A, the image of ψA has the combinatorial structure of the cyclic polytope of dimension m – 1 with n vertices. In particular, the boundary is formed by the curvy (m2)-simplices Bμ where μ runs over all sequences that satisfy Gale’s Evenness Condition [16, Theorem 0.7]. This predicts that the boundary in Example 28 is subdivided into 12 curvy triangles Bμ, namely those indexed by μ{123,128,134,145,156,167,178,238,348,458,568,678}. Our belief is supported by related results for the moment curve, where A={1,2,,m}, due to Bik, Czapliński and Wageringel [Citation1]. Their figures show curvy cyclic polytopes in dimension 3.

The theory of triangulations of cyclic polytopes [Citation13] now suggests an approach to unique recovery even when m < n. Each triangulation consists of a certain subset of ([n]m). If our belief is correct, then this should induce a curvy triangulation of im(ψA). A general point c in the image is contained in a unique simplex Bν of the triangulation. There is a unique z in the locus Rν with ψA(z)=c. The assignment cz serves as a method for unique recovery.

We conclude with a natural generalization of the problem discussed in this section. Let K={K1,,Km} be a set of centrally symmetric convex bodies in Rn. Each of these defines a norm || · ||Ki on Rn. The unit ball for that norm is the convex body Ki. Consider the map (16) ψK : R0n Δm1 :  x  1j=1m||x||Kj·( ||x||K1,||x||K2,,||x||Km).(16)

Problem 29. Study the image and the fibers of the map ψK. Identify the branch loci of ψK.

Supplemental material

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Additional information

Funding

Hana Melánová was supported by Supported by START-Prize Y-966 of the Austrian Science Fund.

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