MINIMALITY AND UNIQUENESS FOR DECOMPOSITIONS OF SPECIFIC TERNARY FORMS

ELENA ANGELINI AND LUCA CHIANTINI

Abstract. The paper deals with the computation of the rank and the identifiability of a specific ternary form. Often, one knows some short Waring decomposition of a given form, and the problem is to determine whether the decomposition is minimal and unique. We show how the analysis of the Hilbert-Burch matrix of the set of points representing the decomposition can solve this problem in the case of ternary forms. Moreover, when the decomposition is not unique, we show how the procedure of liaison can provide alternative, maybe shorter, decompositions. We give an explicit algorithm that tests our criterion of minimality for the case of ternary forms of degree 9. This is the first numerical case in which a new phenomenon appears: the span of 18 general powers of linear forms contains points of (subgeneric) rank 18, but it also contains points whose rank is 17, due to the existence of a second shorter decomposition which is completely different from the given one.

1. Introduction

The paper deals with homogeneous polynomials (forms) F, which will be also seen as symmetric tensors, over the complex field ℂ. Our analysis starts with a given Waring expression of F in terms of powers of linear forms

F = λ_{1}L_{1}^{d} + ⋯ + λ_{r}L_{r}^{d}

(d is the degree of F). Our target is to determine the minimality and the uniqueness of the expression, up to a scalar multiplication of the L_{i}’s. Indeed, in our setting minimality is a consequence of uniqueness. So, when uniqueness holds, r is the (Waring) rank of F.

The determination of the rank of a given form, as well as the the uniqueness of a Waring expression, is relevant in several aspects of tensor theory related to physics, signal processing, statistics, chemistry, artificial intelligence, etc. For some examples of these relations, taken from the huge literature on the subject, let us mention the papers [2], [3], [6], [34].

Our analysis, besides its theoretical interest, will produce the following concrete application. There are several situations in which one knows some Waring expression of a specific form F. For instance, this happens when F splits in a sum of blocks whose rank is well known, or when one applies to F some heuristic algorithm, e.g. based on local regression. In these cases, however, the question about the minimality of the expression remains open. We will see in examples (see Section 5.2) that a local analysis of the Waring expression cannot guarantee its minimality, except when the length r is rather small (as, e.g., in the Kruskal’s criterion). We will produce a method which, in principle, can solve the minimality problem, and then guarantee that r is the rank of F, for all Waring expressions of ternary forms. The consequent algorithm that one can construct depends on the degree d. In the last section, we produce a concrete example of the algorithm, for ternary forms of degree 9.

2000 Mathematics Subject Classification. 14N07, 14J70, 14C20, 14N05, 15A69, 15A72.

This work was supported by the National Group for Algebraic and Geometric Structures, and their Applications (GNSAGA – INdAM) and by the Italian PRIN 2015 - Geometry of Algebraic Varieties (B16J15002000005).

We attack the problem with tools of projective algebraic geometry. Thus, we start with the polynomial ring R = ℂ[x_{0}, …, x_{n}] and we indicate with R_{d} the homogeneous piece of degree d. Any linear form L determines a point (which, by abuse of notation, we still indicate with L) in the projective space ℙ^{n} over the linear space R_{1} of forms of degree 1. In this notation, F corresponds to a point in the projective space over R_{d}. We identify ℙ(R_{d}), of (projective) dimension N = \binom{n+d}{d} − 1, with ℙ(S^{d}(R_{1})). The map v_{d} : ℙ^{n} → ℙ^{N} which sends a linear form L to its d-th power is universally known as the d-th Veronese map.

A Waring expression F = λ_{1}L_{1}^{d} + ⋯ + λ_{r}L_{r}^{d} of F determines a subset A = {L_{1}, …, L_{r}} ⊂ ℙ^{n} such that F sits in the span of v_{d}(A). In our notation, we will say that A is a decomposition of length r of F. The decomposition A is nonredundant if F does not lie in the span of v_{d}(A′), for any proper subset A′ ⊂ A. The decomposition A is minimal (resp. unique) if there are no other subsets B ⊂ ℙ^{n}, of length r′ < r (resp. r′ ≤ r) such that F belongs to the span of v_{d}(B). So, when A is minimal, its length is the (Waring) rank of F.

There are methods that, in some setting, can prove that a decomposition A = {L_{1}, …, L_{r}} of F is unique, and also minimal. For instance, one can take several flattenings of the tensors associated to F, L_{1}^{d}, …, L_{r}^{d} and use them to prove the uniqueness, as in [21], [22]. Alternatively, one can study the catalecticant map α defined by some partial derivatives of F, and the intersection of the image of α with a suitable Veronese variety, as in [29], [30]. The most famous method to prove uniqueness goes back to Kruskal, [28]. Kruskal used the Kruskal’s rank of the matrix whose entries are the coefficients of the L_{i}’s (i.e. the coordinates of the L_{i}’s, as points in ℙ^{n}) to produce an inequality which, if satisfied, guarantees that the decomposition is unique, and minimal. Kruskal’s criterion has been refined in [15], by taking into account the matrices defined by powers of the L_{i}’s.

All the aforementioned methods have a common problem: they will not provide an answer when the degree d and the length r grow.

From the point of view of projective geometry, it is clear that the Waring rank is constant in a Zariski open subset of ℙ^{N}, i.e. it is constant outside a subset defined by algebraic equations: a small subset, in any reasonable metric. Denote with r_{g} the generic value of the rank, which holds in a dense subset of ℙ^{N}. Then, it is almost straightforward that a decomposition cannot be unique when r > r_{g} (too many parameters). When r = r_{g} generic uniqueness can hold, but it is a quite rare phenomenon, completely classified (see [24]). On the contrary, for subgeneric values r < r_{g}, uniqueness holds for a minimal Waring expression of a sufficiently general form F, except for a short list of cases (see [16]). On the other hand, for a specific decomposition of a specific form F, whose rank is unknown, all the known methods can guarantee its uniqueness or minimality only when the length r is much smaller than r_{g} (see Section 3 of [4], for a concrete bound).

The analysis that we propose overcomes this difficulty, and will guarantee the minimality, and also the uniqueness, of a given decomposition, in principle for all r < r_{g}. We will give a concrete example for forms of degree 9 in three variables.

The theoretical situation can be summarized as follows. Geometrically, the Waring expression F = λ_{1}L_{1}^{d} + ⋯ + λ_{r}L_{r}^{d} presents F as a point of the (linear) span of the set v_{d}(A). When r is big, even if the L_{i}’s are general, it turns out that the span of v_{d}(A) contains both points F for which the decomposition A is unique, and points F′ = μ_{1}L_{1}^{d} + ⋯ + μ_{r}L_{r}^{d} for which uniqueness, and also minimality, do not hold (though A is still a non-redundant decomposition of F′). Examples of this situation are described in Section 3 below. It may happen indeed (Case 2 of Proposition 3.4) that F′ is non-redundantly spanned by v_{d}(A), yet there exists another decomposition B of F′, completely different from A and with less summands. Thus, methods based on the geometrical analysis of A alone (as the Kruskal’s or the catalecticant approaches) cannot distinguish between forms F, F′ in the span of v_{d}(A), so they will fail to guarantee e.g. minimality, when the span contains points with different behavior. The answer can be obtained only by analyzing A and the coefficients λ_{i}’s of the Waring expression.

Our analysis follows the guidelines introduced in [4]. We attack the problem with a set of tools typical for the study of the geometry of finite sets in projective spaces: Hilbert-Burch matrices, the Cayley-Bacharach property, and liaison. An extensive illustration of the interplay between decompositions and the geometry of finite sets can be found in the book [27]. Papers [10], [8], [9] are based on the study of Hilbert functions of points. What is really new in the present paper, as well as in [4], is the observation that given a finite set A ⊂ ℙ^{2}, the possible alternative (maybe even shorter) decompositions of any ternary form F in the span of v_{d}(A) can be recovered geometrically, by playing with the Hilbert-Burch matrix of A and liaison. Thus, we can characterize the (algebraic) subset Θ of the span of v_{d}(A) consisting of forms for which A is non-redundant, but yet they have an alternative decomposition of length r′ < r. Moreover, we produce concrete algorithms which guarantee that a form F does not lie in the ‘bad’ set Θ.

In the paper, we illustrate in details the procedure in the case of ternary forms of degree 9. This is the first numerical case in which the full list of phenomena that one can theoretically predict for forms in the span of a fixed set v_{d}(A) effectively appears. So the full strength of the method can be illustrated. In particular, in Case 2 below we will show how one can use a decomposition which is “wrong” (i.e. non-minimal) to obtain a minimal decomposition.

For ternary forms of degree 9 the generic rank is r_{g} = 19, see [1]. We consider the highest sub-generic value r = 18. We show that for any generic choice of a subset A of 18 points in ℙ^{2} (the projective space of linear forms in three variables) the span Λ of v_{9}(A) contains: (i) points F for which the decomposition A is minimal and unique, (ii) points F′ for which A in non-redundant, but there exists an alternative decomposition B ≠ A, of length 18, (iii) points F′′ for which A is non-redundant, but there exists an alternative decomposition of length 17 (so their rank is smaller than 18). Moreover, we prove that the (closure of) the set of points satisfying (ii) is a hypersurface of Λ, and it is a birational image of a Grassmannian of lines (see Theorem 4.6 below). We also determine properties of the closure Θ of the set of points satisfying (iii), which is a subvariety of Λ.

Finally, we produce an algorithm to test if a given form F = λ_{1}L_{1}^{9} + ⋯ + λ_{18}L_{18}^{9} in the span Λ of v_{9}({L_{1}, …, L_{18}}) lies in the ‘bad’ locus Θ defined above, i.e. its rank is smaller than 18. The algorithm analyses both A and the coefficients λ_{i}’s of the Waring expression. When the answer provided by the algorithm is negative, we can conclude that A is a minimal decomposition of F, so that F has rank 18. Similar algorithms for detecting the uniqueness of the decomposition can be easily produced, though they need more parameters. Finally, when a second decomposition of length r′ ≤ r exists, we show how we can recover the new decomposition from the known one A (see the end of the examples in Section 5.2).

The paper is structured as follows. In Sect. 2 we introduce our setting and recall the main tools, coming from classical algebraic geometry, that characterize our analysis. In particular, we mention the Hilbert function and the Cayley-Bacharach property for finite sets in the projective space, the notion of Hilbert-Burch matrix and liaison theory for sets of points in the projective plane. Sect. 3 is the core of the paper: we develop our analysis for ternary forms of degree 9, dealing with ranks that are outside the range of applicability of the celebrated Kruskal’s criterion. Sect. 4 disposes on the case in which two decompositions of a ternary nonic intersect. Finally, Sect. 5 is devoted to the effective algorithm we developed according to the criterion of minimality (and uniqueness) obtained in Sect. 3. Several numerical examples are presented. Acknowledgments. This work was supported by the National Group for Algebraic and Geometric Structures, and their Applications (GNSAGA – INdAM) and by the Italian PRIN 2015 - Geometry of Algebraic Varieties (B16J15002000005).

2. Preliminaries

2.1. Notation. For any finite set A we denote with ℓ(A) the cardinality. For d, n ∈ ℕ, let ℂ^{n+1} be the space of linear forms in x_{0}, …, x_{n}, so that S^{d}ℂ^{n+1} is the space of forms of degree d in n + 1 variables over ℂ. Every form T ∈ S^{d}ℂ^{n+1} defines an element of ℙ(S^{d}ℂ^{n+1}) ≅ ℙ^{N} (N = \binom{n+d}{d} − 1), which, by abuse, we still denote by T. We denote with v_{d} : ℙ^{n} → ℙ^{N} the Veronese embedding of ℙ^{n} of degree d, which is given by

v_{d}([a_{0}x_{0} + … + a_{n}x_{n}]) = [(a_{0}x_{0} + … + a_{n}x_{n})^{d}].

For any subset Z ⊂ ℙ^{N} we denote with ⟨Z⟩ the linear span of Z. For instance, if A = {P_{1}, …, P_{r}} ⊂ ℙ^{n} is a finite set, ⟨v_{d}(A)⟩ is the linear space in ℙ^{N} spanned by the points v_{d}(P_{1}), …, v_{d}(P_{r}).

With the above notations we give the following definitions.

Definition 2.1. Let A ⊂ ℙ^{n} be a finite set, and T ∈ S^{d}ℂ^{n+1} a form of degree d. We say that A computes T if T ∈ ⟨v_{d}(A)⟩. We say that a set A which computes T is non-redundant if there are no proper

subsets A′ of A such that A′ computes T.

We say that a set A which computes T is minimal if there are no sets B, with ℓ(B) < ℓ(A), such that B computes T.

If A computes T and it is minimal, the cardinality ℓ(A) is called the (Waring) rank of T. In this case we say that A computes the rank of T.

Definition 2.2. A form T is identifiable if there exists a unique set A that computes the rank of T.

2.2. Kruskal’s criterion for symmetric tensors. One of the most used criteria for detecting the identifiability of a form T is here described, with some extensions, in geometric terms and adapted to the case of forms (= symmetric tensors). We start with a definition which is the geometric analogue of the Kruskal’s rank of a matrix.

Definition 2.3. The d-th Kruskal’s rank of a finite set A ⊂ ℙ^{n} is

k_{d}(A) = max {k | dim⟨v_{d}(A′)⟩ = ℓ(A′) − 1 for all A′ ⊂ A with ℓ(A′) ≤ k}.

In other words, k_{d}(A) is the maximal k such that the image under v_{d} of any subset of A of cardinality at most k is linearly independent.

Remark 2.4. For any d, it holds that k_{d}(A) ≤ min{N + 1, ℓ(A)}.

Moreover, if k_{d}(A) = min{N + 1, ℓ(A)} (i.e. k_{d}(A) is maximal), then for all A′ ⊂ A the d-th Kruskal’s rank k_{d}(A′) is also maximal.

If A is sufficiently general, then k_{d}(A) = min{N +1, ℓ(A)} i.e. the d-th Kruskal’s rank is maximal (see e.g. Lemma 4.4 of [15]).

The Kruskal’s rank is fundamental in the statement of the reshaped Kruskal’s criterion, see [15], which says that T has rank ℓ(A) and it is identifiable when A ⊂ ℙ^{n} is a non-redundant set computing T ∈ ℙ(S^{d}ℂ^{n+1}), provided that ℓ(A) ≤ (k_{d_{1}}(A) + k_{d_{2}}(A) + k_{d_{3}}(A) − 2)/2 for some partition d = d_{1} + d_{2} + d_{3} of d.

In the case of ternary forms, the reshaped Kruskal’s criterion has been recently extended in [4], [7], [32], see Theorem 2.11 below.

2.3. The Hilbert function of finite sets and its difference. For any choice of representatives for the points of a finite Z of length ℓ, one can consider the evaluation of forms of fixed degree d at the chosen representatives, which determines a map ev(d) : S^{d}ℂ^{n+1} → ℂ^{ℓ}. The rank of the map does not depend on the choice of the representatives.

Definition 2.5. Let Z be a finite set in ℙ^{n}. The Hilbert function of Z is the map h_{Z} : ℤ ⟶ ℕ such that h_{Z}(j) = 0, for j < 0, h_{Z}(j) = rank(ev(j)), for j ≥ 0.

In the Veronese map v_{d} : ℙ^{n} → ℙ^{N}, the pull-back of linear forms in ℙ^{N} are forms of degree d in ℙ^{n}. Hence, the rank of ev(d) is equal to the dimension of the space of forms of degree 1 in ℙ^{N} mod the ideal of v_{d}(Z). Geometrically, this means that the rank of ev(d) corresponds to dim(⟨v_{d}(Z)⟩) + 1.

We will often use the first difference of the Hilbert function Dh_{Z}, given by

Dh_{Z}(j) = h_{Z}(j) − h_{Z}(j − 1), j ∈ ℤ.

We collect some useful elementary properties of h_{Z} and Dh_{Z}. They are well know and contained (sparsely) in the literature, see e.g. [27], [31]. A summary can be found in [14].

Remark 2.6. (i) Dh_{Z}(j) ≥ 0 for all j; h_{Z}(j) = Dh_{Z}(j) = 0 for j < 0; h_{Z}(0) =

Dh_{Z}(0) = 1. (ii) h_{Z}(j) = ℓ(Z) for all j ≫ 0, so that Dh_{Z}(j) = 0 for j ≫ 0 and ∑ Dh_{Z}(j) =

ℓ(Z). (iii) If Z′ ⊂ Z, then for all j ∈ ℤ we have h_{Z′}(j) ≤ h_{Z}(j) and Dh_{Z′}(j) ≤ Dh_{Z}(j).

Proposition 2.7. For any i > 0 such that Dh_{Z}(i) ≤ i, it holds

Dh_{Z}(i) ≥ Dh_{Z}(i + 1),

i.e. the function Dh_{Z} becomes non-increasing from i on. Therefore, if Dh_{Z}(i) = 0, then Dh_{Z}(j) = 0 for any j ≥ i.

The following proposition is a straightforward application of the Grassmann formula in projective spaces.

Proposition 2.8. Let A, B ⊂ ℙ^{n} be disjoint finite sets such that both v_{d}(A) and v_{d}(B) are linearly independent. Set Z = A ∪ B. For any d ∈ ℕ,

dim(⟨v_{d}(A)⟩ ∩ ⟨v_{d}(B)⟩) = ℓ(Z) − h_{Z}(d) − 1.

As pointed out e.g. in [4], it follows that when A, B are two disjoint sets that compute a form T, then ⟨v_{d}(A)⟩ ∩ ⟨v_{d}(B)⟩ is non-empty, thus the union Z = A ∪ B satisfies h_{Z}(d) < ℓ(Z). It follows:

Proposition 2.9. Let T ∈ S^{d}ℂ^{n+1} and let A, B ⊂ ℙ^{n} be disjoint non-redundant finite sets computing T. Pose Z = A ∪ B ⊂ ℙ^{n}. Then Dh_{Z}(d + 1) > 0.

The following result, which we will use several times, is a consequence of a theorem proved by E. Davis in [19], for finite subsets of the plane, and then generalized to points in any projective space in [12].

Proposition 2.10. Let T ∈ S^{d}ℂ^{3} and let A, B ⊂ ℙ^{2} be disjoint, non-redundant finite sets computing T. Pose Z = A ∪ B. Then there are no indices j ≤ d such that 0 < Dh_{Z}(j) = Dh_{Z}(j + 1) < j.

Proof. See [4] Proposition 2.20. □

In the case of ternary forms, in [4], [7], [32] the authors show that one can use an analysis of the Hilbert function in order to extend the reshaped Kruskal’s criterion, as follows.

Theorem 2.11. Let T ∈ ℙ(S^{d}ℂ^{3}) and let A ⊂ ℙ^{2} be a non-redundant set computing T. Then T is identifiable of rank r if one of the following holds:

• d = 2m is even, k_{m−1}(A) = min\{\binom{m+1}{2}, r\}, h_{A}(m) = r ≤ \binom{m+2}{2} − 2;

• d = 2m+1 is odd, k_{m}(A) = min\{\binom{m+2}{2}, r\}, h_{A}(m+1) = r ≤ \binom{m+2}{2}+⌊\frac{m}{2}⌋.

2.4. The Cayley-Bacarach property.

Definition 2.12. A finite set Z ⊂ ℙ^{n} satisfies the Cayley-Bacharach property in degree i, abbreviated as CB(i), if for all P ∈ Z, it holds that every form of degree i vanishing at Z \ {P } also vanishes at P.

The main consequence of the Cayley-Bacharach property on the Hilbert function of a set Z is contained in the following two results, proven in [5], by means of a deep result in [25].

Theorem 2.13. If a finite set Z ⊂ ℙ^{n} satisfies CB(i), then for any j such that 0 ≤ j ≤ i + 1 we have

Dh_{Z}(0) + Dh_{Z}(1) + ⋯ + Dh_{Z}(j) ≤ Dh_{Z}(i + 1 − j) + ⋯ + Dh_{Z}(i + 1).

Corollary 2.14. Let T ∈ S^{d}ℂ^{3} and let A ⊂ ℙ^{2} be a non-redundant finite set computing T. Let B ⊂ ℙ^{2} be another non-redundant finite set computing T and assume A∩B = ∅. Then Z = A∪B satisfies the Cayley-Bacharach property CB(d).

2.5. The Hilbert-Burch matrix of a set of points in ℙ^{2}. Let A ⊂ ℙ^{2} be a finite set and call I_{A} the homogeneous ideal of A, in the polynomial ring R = ℂ[x_{0}, x_{1}, x_{2}]. Then there exists an exact sequence of graded R-modules (called a minimal resolution for I_{A}), as follows:

(1)  0 ⟶ F_{1} ⟶^{M} F_{0} ⟶ I_{A} ⟶ 0

where F_{1}, F_{0} are free graded R-modules, i.e. we have

F_{0} = ⊕_{i=1}^{s}R(−d_{i}) F_{1} = ⊕_{j=1}^{s−1}R(−e_{j})

(R(i) is the graded module R with the degrees shifted by i. The shift is needed in order to get that the maps in the sequence are graded homomorphisms). The map F_{0} ⟶ I_{A} sends the standard basis of F_{0} to a set of minimal generators for I_{A}, and the numbers d_{i}’s correspond to the degrees of the chosen minimal generators. The map F_{1} ⟶ F_{0} is given by a matrix of forms M, which describes the first syzygies of A, i.e. the relations between the minimal generators.

Definition 2.15. The matrix of forms M which determines the map F_{1} ⟶ F_{0} is called a Hilbert-Burch matrix for A. The degree matrix of M is called the degree Hilbert-Burch matrix of A.

Theorem 2.16. (Hilbert-Burch Theorem, see [17]) The Hilbert-Burch matrix M depends on a choice of a set of minimal generators for the ideal I_{A}, but the degree Hilbert-Burch matrix depends only on A. The minimal generators which determine the surjection F_{0} ⟶ I_{A} are precisely the (s − 1) × (s − 1) minors of the matrix M (taken with the corresponding sign).

Example 2.17. A set of points Z ⊂ ℙ^{2} is complete intersection if there are two plane curves F, G such that Z = F ∩ G. In this case, the homogeneous ideal I_{Z} is generated by the forms F, G which define the two curves. If d_{1}, d_{2} denote the degrees of F, G respectively, then a minimal resolution of I_{Z} looks like

0 ⟶ R(−d_{1} − d_{2}) ⟶^{M} R(−d_{1}) ⊕ R(−d_{2}) ⟶ I_{Z} ⟶ 0

and the Hilbert-Burch matrix M is given by M = (\begin{matrix}G \\ −F \end{matrix} ).

2.6. Linked sets and mapping cone. Let A, B ⊂ ℙ^{2} be two sets of points such that Z = A ∪ B is complete intersection of two curves F and G, of degrees respectively d_{1} and d_{2}. In this case, we say that A and B are linked by (or also that B is the residue of A in) a complete intersection of type (d_{1}, d_{2}).

When the two sets are linked, a Hilbert-Burch matrix of B can be found from a Hilbert-Burch matrix of A via the mapping cone procedure (for more details, we refer to [23], [33], and Proposition 5.2.10 of [31]).

The homogeneous ideal I_{B} is given by

I_{B} = I_{Z} : I_{A} = {f ∈ R : fI_{A} ⊂ I_{Z}}.

The inclusion I_{Z} ⊂ I_{A} determines on the resolutions of I_{A} and I_{Z} a commutative diagram:

(2) \begin{aligned}0 & ⟶ & R(−d_{1} − d_{2}) & ⟶^{M′} & R(−d_{1}) ⊕ R(−d_{2}) & ⟶ & I_{Z} & ⟶ & 0 \\ & & ↓ ϕ′ & & ↓ ϕ & & ↓ \\ 0 & ⟶ & F_{1} & ⟶^{M} & F_{0} & ⟶ & I_{A} & ⟶ & 0 \end{aligned}

where the rightmost vertical map is the inclusion.

Then I_{B} is the image of the dual map R(d_{1}) ⊕ R(d_{2}) ⊕ F_{1}^{∨} ⟶^{(ϕ′,M′)^{∨}} R(d_{1} + d_{2}), twisted by −d_{1} − d_{2}. A resolution of I_{B} is given by:

0 → F_{0}^{∨}(−d_{1} − d_{2}) ⟶^{(ϕ,M)^{∨}} R(−d_{2}) ⊕ R(−d_{1}) ⊕ F_{1}^{∨}(−d_{1} − d_{2}) ⟶^{(ϕ′,M′)^{∨}} I_{B} → 0.

Notice that the resolution of I_{B} obtained by the mapping cone procedure is not necessarily minimal, in the sense that some summands of the map F_{0}^{∨}(−d_{1} −d_{2}) → R(−d_{2}) ⊕ R(−d_{1}) ⊕ F_{1}^{∨}(−d_{1} − d_{2}) could be factored out.

Example 2.18. Let A be a general set of three points in ℙ^{2}. The ideal I_{A} is generated by three quadrics, and a minimal resolution is given by

0 ⟶ R(−3)^{2} ⟶^{M} R(−2)^{3} ⟶ I_{A} ⟶ 0,

where the Hilbert-Burch matrix M is a 3 × 2 matrix of linear forms ℓ_{ij}. If we take a general quadric F and a general cubic G containing A, we get a linkage between A and another set B of three points in the plane. Diagram (2) looks like

0 ⟶ R(−5) ⟶ R(−2) ⊕ R(−3) ⟶ I_{Z} ⟶ 0
↓ ϕ′  ↓ ϕ  ↓
0 ⟶ R(−3)^{2} ⟶^{M}  R(−2)^{3}  ⟶ I_{A} ⟶ 0

and the mapping cone gives the resolution

0 → R(−3)^{3} ⟶^{(ϕ,M)^{∨}} R(−3) ⊕ R(−2) ⊕ R(−2)^{2} ⟶ I_{B} → 0.

The matrix of (ϕ, M)^{∨} is obtained as follows. Since the minimal generators of I_{A} are the minors of M, F, G have a representation as determinants of matrices as follows

F = det (\begin{matrix}ℓ_{11} & ℓ_{12} & c_{13} \\ ℓ_{21} & ℓ_{22} & c_{23} \\ ℓ_{31} & ℓ_{32} & c_{33} \end{matrix}) G = det (\begin{matrix}ℓ_{11} & ℓ_{12} & h_{13} \\ ℓ_{21} & ℓ_{22} & h_{23} \\ ℓ_{31} & ℓ_{32} & h_{33} \end{matrix})

for some choice of constants c_{i3} and linear forms h_{i3}. Thus the matrix of (ϕ, M)^{∨} is given by

(\begin{matrix}c_{13} & c_{23} & c_{33} \\ h_{13} & h_{23} & h_{33} \\ ℓ_{11} & ℓ_{21} & ℓ_{31} \\ ℓ_{12} & ℓ_{22} & ℓ_{32} \end{matrix}) .

This resolution is not minimal, because we can choose the resolution of I_{A} so that F is the first generator. In this case the row (c_{13}, c_{23}, c_{33}) is equal to (1, 0, 0) and can be factored out. So, we can drop the cubic generator of I_{B}. The resolution becomes

0 → R(−3)^{2} ⟶^{N} R(−2) ⊕ R(−2)^{2} ⟶ I_{B} → 0

and the Hilbert-Burch matrix N is:

N = (\begin{matrix}h_{23} & h_{33} \\ ℓ_{21} & ℓ_{31} \\ ℓ_{22} & ℓ_{32} \end{matrix}) .

2.7. Grassmannian Varieties. Let r, s ∈ ℕ such that r ≤ s and let 𝔾(r, s) be the Grassmannian variety of r-dimensional linear spaces in ℂ^{s}. 𝔾(r, s) is an algebraic subvariety of a projective space by means of the Plücker embedding

(3)  i : 𝔾(r, s) ↪ ℙ(Λ^{r}ℂ^{s})

defined by

i(⟨v_{1}, …, v_{r}⟩) = [v_{1} ∧ … ∧ v_{r}]

where v_{1}, …, v_{r} are r linearly independent vectors of ℂ^{s}. The homogeneous coordinates of 𝔾(r, s) on ℙ^{\binom{s}{r}−1} ≅ ℙ(Λ^{r}ℂ^{s}) are called Plücker coordinates. We recall that

dim 𝔾(r, s) = r(s − r).

Let W ∈ 𝔾(r, s). We can associate to W the r × s matrix with complex entries M_{W} whose rows contain the coordinates of a basis {v_{1}, …, v_{r}} of W. We notice that this representation is not unique: if we multiply by an element U ∈ GL(r, ℂ), the new matrix represents the same point of 𝔾(r, s). The Plücker coordinates of 𝔾(r, s) are the minors of size r of M_{W}, which are simply multiplied by a number when we substitute M_{W} with UM_{W}.

The embedding i : 𝔾(r, s) ↪ ℙ(Λ^{r}ℂ^{s}) is defined by a divisor h_{r,s} of 𝔾(r, s). According to Lemma 11.1 of [20], the divisor h_{r,s} ⊂ 𝔾(r, s) associated to i is

h_{r,s} = {Λ ∈ 𝔾(r, s)| dim(Λ ∩ L̃) > 0}

where L̃ ≅ ℂ^{s−r−1} is a fixed subspace of ℂ^{s}.

3. Forms of degree 9 in three variables

In this section, let us assume that n = 2, d = 9 and let T ∈ S^{9}ℂ^{3}. Thus T is a form of degree 9 in three variables, which is associated to a curve of degree 9 in ℙ^{2}.

Let A = {P_{1}, …, P_{r}} ⊂ ℙ^{2} be a finite set that computes T. Our target is to determine conditions on T such that A is the unique set that computes the rank of T.

We will assume in this section that A is sufficiently general, in a very precise sense. We assume indeed that:

(i) A is non-redundant; (ii) k_{4}(A) = min{15, r}; (iii) h_{A}(5) = r.

It is a standard fact indeed that when r ≤ 21, all A in a Zariski open subset of (ℙ^{2})^{r} satisfy the previous conditions.

Remark 3.1. For a general T ∈ S^{9}ℂ^{3}, according to the Alexander-Hirschowitz Theorem [1], the rank is 19, and a simple parameter count shows that identifiability cannot hold (see e.g. [24]). On the other hand, if r ≤ 17 and A satisfies the previous conditions, then, by Theorem 2.11, T is identifiable of rank r.

Therefore, we will assume in this section that r = 18, hence h_{A}(5) = 18 and k_{4}(A) = 15. The last equality means that for all subsets A′ of 15 points in A we have h_{A′}(4) = dim⟨v_{4}(A′)⟩ + 1 = 15. Thus A′ lies in no quartic curves. It follows that A itself is contained in no quartic curves, hence h_{A}(j) = dim(S^{j}ℂ^{3}) for j = 1, 2, 3, 4. Since h_{A}(5) = r = 18, then h_{A}(j) = 18 for all j ≥ 5. The values of h_{A} and its difference can be summarized in the following table:

(4)  \begin{matrix}j & 0 & 1 & 2 & 3 & 4 & 5 & 6 & … \\ h_{A}(j) & 1 & 3 & 6 & 10 & 15 & 18 & 18 & … \\ Dh_{A}(j) & 1 & 2 & 3 & 4 & 5 & 3 & 0 & … \end{matrix} .

In Sect. 5 we produce an algorithm which, starting with the coordinates of the points of A, tests whether or not A satisfies the generality conditions (i) - (iii). Conditions (i) - (iii) determine the shape of the degree Hilbert-Burch matrix of A.

Proposition 3.2. Assume A satisfies conditions (i) - (iii). Then there exist three forms Q_{1}, Q_{2}, Q_{3} ∈ S^{5}ℂ^{3} of degree 5 and one form S ∈ S^{6}ℂ^{3} of degree 6 such that the ideal I_{A} is minimally generated by Q_{1}, Q_{2}, Q_{3}, S. The resolution of I_{A}, determined by this choice of generators, has the form

0 ⟶ R(−7)^{3} ⟶^{M} R(−5)^{3} ⊕ R(−6) ⟶ I_{A} ⟶ 0.

Proof. Since the space (I_{A})_{5} of quintics through A is three-dimensional, a basis Q_{1}, Q_{2}, Q_{3} for (I_{A})_{5} determines three minimal generators for I_{A}. Since h_{A}(5) = 18, by the Castelnuovo-Mumford Regularity Theorem (see [13]) I_{A} is generated in degree 6. If we multiply the Q_{i}’s by 3 independent linear forms, which generate S^{1}ℂ^{3}, we obtain a subspace Λ_{6} of dimension 9 in (I_{A})_{6}. (I_{A})_{6} has dimension 10, since h_{A}(6) = 18. It follows that by taking S ∈ (I_{A})_{6} \Λ_{6}, we get a minimal set of 4 generators for I_{A}, which thus determines a surjection α : R(−5)^{3} ⊕ R(−6) ⟶ I_{A}. The graded piece (I_{A})_{7} has dimension dim(R_{7}) − h_{A}(7) = 36 − 18 = 18. By multiplying each Q_{i} by a basis for S^{2}ℂ^{3} and S by a basis for S^{1}ℂ^{3}, we get 21 elements in (I_{A})_{7}. Thus we have at least 3 independent relations of degree 7 among Q_{1}, Q_{2}, Q_{3}, S. Since 3 = 4−1, by the Hilbert-Burch Theorem 2.16, the independent relations of degree 7 are exactly 3 and the kernel of the map α is R^{3}(−7). □

Now, assume that B = {P′_{1}, …, P′_{ℓ(B)}} ⊂ ℙ^{2} is another finite set computing T such that

(i) ℓ(B) ≤ 18; (ii) B is non-redundant

and set Z = A ∪ B ⊂ ℙ^{2}.

We assume, for the rest of the section, that the intersection A ∩ B is empty.

We will analyze the case in which the intersection is non-empty in the next section. Observe that, by Corollary 2.14, the last assumption: implies that Z satisfies the Cayley-Bacharach property CB(9).

Remark 3.3. Since Z satisfies CB(9), then, by Theorem 2.13, Remark 2.6 (iii) and (4), we get that

Dh_{Z}(6) + Dh_{Z}(7) + Dh_{Z}(8) + Dh_{Z}(9) + Dh_{Z}(10) ≥
≥ Dh_{A}(0) + Dh_{A}(1) + Dh_{A}(2) + Dh_{A}(3) + Dh_{A}(4) = 15.

Moreover, being Dh_{A}(5) = 3, then by Remark 2.6 (iii) we obtain that

(5) 36 ≥ ℓ(Z) ≥ 15+Dh_{Z}(5)+Dh_{Z}(6)+Dh_{Z}(7)+Dh_{Z}(8)+Dh_{Z}(9)+Dh_{Z}(10) ≥
≥ 18 + Dh_{Z}(6) + Dh_{Z}(7) + Dh_{Z}(8) + Dh_{Z}(9) + Dh_{Z}(10).

Therefore

(6)  Dh_{Z}(6) + Dh_{Z}(7) + Dh_{Z}(8) + Dh_{Z}(9) + Dh_{Z}(10) ≤ 18.

Proposition 3.4. Assume that A ∩ B = ∅. Then one the following cases occurs for the first difference of the Hilbert function of Z: Case 1:

\begin{matrix}j & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & … \\ Dh_{Z}(j) & 1 & 2 & 3 & 4 & 5 & 6 & 5 & 4 & 3 & 2 & 1 & 0 & ⋯ \end{matrix}

and ℓ(B) = 18, ℓ(Z) = 36; Case 2:

\begin{matrix}j & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & … \\ Dh_{Z}(j) & 1 & 2 & 3 & 4 & 5 & 5 & 5 & 4 & 3 & 2 & 1 & 0 & ⋯ \end{matrix}

and ℓ(B) = 17, ℓ(Z) = 35; Case 3:

\begin{matrix}j & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & … \\ Dh_{Z}(j) & 1 & 2 & 3 & 4 & 5 & 5 & 5 & 5 & 3 & 2 & 1 & 0 & ⋯ \end{matrix}

and ℓ(B) = 18, ℓ(Z) = 36.

Proof. Notice that since A imposes the maximal number of conditions to forms of degree j = 1, 2, 3, 4, then by Remark 2.6 iii), the same holds for Z. By Proposition 2.9, we know that Dh_{Z}(10) ≥ 1. Assume that Dh_{Z}(10) ≥ 2. Then, by Proposition 2.10, necessarily Dh_{Z}(9) ≥ Dh_{Z}(10) + 1 ≥ 3, Dh_{Z}(8) ≥ Dh_{Z}(9)+1 ≥ 4, Dh_{Z}(7) ≥ Dh_{Z}(8)+1 ≥ 5, which implies, by (6), that Dh_{Z}(6) ≤ 4, which contradicts Proposition 2.7. Therefore

Dh_{Z}(10) = 1.

Moreover, for any j ≥ 11 we get Dh_{Z}(j) = 0, and also Dh_{Z}(9) ≥ 2, otherwise we violate Proposition 2.10. Assume that Dh_{Z}(9) ≥ 3. Then, by arguing as before, Dh_{Z}(8) ≥ Dh_{Z}(9) + 1 ≥ 4, Dh_{Z}(7) ≥ Dh_{Z}(8) + 1 ≥ 5 and Dh_{Z}(6) ≤ 5. If Dh_{Z}(6) = 5, then Dh_{Z}(7) = 5 and, by the first line of (5), Dh_{Z}(5) + Dh_{Z}(8) + Dh_{Z}(9) ≤ 10, so that Dh_{Z}(5) ≤ 3, a contradiction. If Dh_{Z}(6) ≤ 4, then, by Proposition 2.7, Dh_{Z}(7) ≤ 4, a contradiction. Therefore

Dh_{Z}(9) = 2

and Dh_{Z}(8) ≥ 3, by Proposition 2.10. Assume that Dh_{Z}(8) ≥ 4. Then, with the same arguments as above, Dh_{Z}(7) ≥ 5. If Dh_{Z}(7) = 5, then Dh_{Z}(6) ≥ 5 and, being ∑_{i=0}^{10} Dh_{Z}(i) ≤ 36, then Dh_{Z}(5) ≤ 4, a contradiction; if Dh_{Z}(7) > 5, then, by(6), Dh_{Z}(6) < 6, which provides a contradiction to Proposition 2.7. As a consequence, we get that

Dh_{Z}(8) = 3

so that, by Proposition 2.10 again, Dh_{Z}(7) ≥ 4. Case 3. Assume that Dh_{Z}(7) ≥ 5. If Dh_{Z}(7) > 5, then, by Proposition 2.7, Dh_{Z}(6) > 5, moreover, by (5) and by Proposition 2.10, 36 ≥ ℓ(Z) ≥ Dh_{Z}(5) + Dh_{Z}(6) + Dh_{Z}(7) + 21. In particular it turns out to be Dh_{Z}(5) ≤ 3 < Dh_{Z}(6), a contradiction with Proposition 2.7. Hence we have

Dh_{Z}(7) = 5.

By Proposition 2.7 we know that Dh_{Z}(6) ≥ 5. In particular, if Dh_{Z}(6) > 5, then, as above, by (5) and by Proposition 2.10 one obtains Dh_{Z}(5) < 5, which violates Proposition 2.7. Thus, necessarily,

Dh_{Z}(6) = 5

which implies that

Dh_{Z}(5) = 5.

We get Case 3 of the statement. In particular, we get ℓ(Z) = 36 and ℓ(B) = 18. Cases 1, 2. Now assume that

Dh_{Z}(7) = 4.

Therefore, by Proposition 2.10, we get that Dh_{Z}(6) ≥ 5. If Dh_{Z}(6) > 5, then, by (5), Dh_{Z}(5) ≤ 5, a contradiction with Proposition 2.7. Thus

Dh_{Z}(6) = 5.

As a consequence, by (5) and Proposition 2.7 we get that either

Dh_{Z}(5) = 6,

and we are in Case 1, with ℓ(Z) = 36, ℓ(B) = 18, or

Dh_{Z}(5) = 5,

and we are in Case 2, with ℓ(Z) = 35, ℓ(B) = 17. □

We analyze in details the three cases that appear in Proposition 3.4.

3.1. Case 1. Let us assume to be in the first case of Proposition 3.4. In fact, we will prove below that this is the general case of the three, in the sense that Case 2 and Case 3 are limits of Case 1. Recall that the ideal I_{A} is generated by three quintics Q_{1}, Q_{2}, Q_{3} and one sextic S (Proposition 3.2). By the Hilbert-Burch Theorem and by Proposition 3.2, there exist c_{uv} ∈ S^{2}ℂ^{3} and ℓ_{j} ∈ S^{1}ℂ^{3}, for u, v, j ∈ {1, 2, 3}, such that A has a Hilbert-Burch matrix M:

(7)  M = (\begin{matrix}c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \\ ℓ_{1} & ℓ_{2} & ℓ_{3} \end{matrix}) .

Q_{1}, Q_{2}, Q_{3}, S coincide, respectively, with (−1)^{i} times the minor obtained by leaving out the i-th row of M, i ∈ {1, 2, 3, 4}. In other words, we have:

(8) Q_{1} = − |\begin{matrix}c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \\ ℓ_{1} & ℓ_{2} & ℓ_{3} \end{matrix}| , Q_{2} = |\begin{matrix}c_{11} & c_{12} & c_{13} \\ c_{31} & c_{32} & c_{33} \\ ℓ_{1} & ℓ_{2} & ℓ_{3} \end{matrix}| ,

Q_{3} = − |\begin{matrix}c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ ℓ_{1} & ℓ_{2} & ℓ_{3} \end{matrix}| , S = |\begin{matrix}c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \\ c_{31} & c_{32} & c_{33} \end{matrix}| .

The Hilbert function tells us that Z is contained in no quintics and it lies in two independent sextics. Moreover, we know that Z satisfies CB(9), and its Hilbert function is the same as the one of a complete intersection of type (6, 6). By the Main Theorem of [18], it follows that Z is a complete intersection of type (6, 6). So I_{Z} is generated by F, F′, where F, F′ ∈ (I_{A})_{6}. We get a commutative diagram

0 ⟶ R(−12) \underrightarrow{(\begin{matrix}F′ \\ −F \end{matrix} )}  R(−6)^{2}  ⟶ I_{Z} ⟶ 0
↓ M_{1}  ↓ M_{2}  ↓
0 ⟶ R(−7)^{3}  ⟶^{M}  R(−5)^{3} ⊕ R(−6) ⟶ I_{A} ⟶ 0

where M is described in (7), and the matrices M_{1}, M_{2} are defined by writing

F = L_{1}Q_{1} + L_{2}Q_{2} + L_{3}Q_{3} + aS F′ = L′_{1}Q_{1} + L′_{2}Q_{2} + L′_{3}Q_{3} + a′S

so that

M_{2} = (\begin{matrix}L_{1} & L′_{1} \\ L_{2} & L′_{2} \\ L_{3} & L′_{3} \\ a & a′ \end{matrix}) , M_{1} = (\begin{matrix}q_{1} \\ q_{2} \\ q_{3} \end{matrix}) ,

with q_{i} ∈ S^{5}ℂ^{3} for i ∈ {1, 2, 3}.

By the mapping cone, a resolution of I_{B} is given by

(9)  0 → R(−7)^{3} ⊕ R(−6) ⟶^{N} R(−6)^{2} ⊕ R(−5)^{3} → I_{B} → 0

where

N = (\begin{matrix}L_{1} & L_{2} & L_{3} & a \\ L′_{1} & L′_{2} & L′_{3} & a′ \\ c_{11} & c_{21} & c_{31} & ℓ_{1} \\ c_{12} & c_{22} & c_{32} & ℓ_{2} \\ c_{13} & c_{23} & c_{33} & ℓ_{3} \end{matrix}) .

and notice that the last three rows represent the transpose of M.

Remark 3.5. Let S′ be any form of degree 6 in I_{A}. Then S′ corresponds to an element of the homogeneous piece (I_{A})_{6}, which is a 10-dimensional linear space. By the description of the generators of I_{A}, there are linear forms u_{1}, u_{2}, u_{3} and a constant c ∈ ℂ such that S′ is the determinant of the matrix:

(\begin{matrix}u_{1} & u_{2} & u_{3} & c \\ c_{11} & c_{21} & c_{31} & ℓ_{1} \\ c_{12} & c_{22} & c_{32} & ℓ_{2} \\ c_{13} & c_{23} & c_{33} & ℓ_{3}. \end{matrix})

If we pose

u_{1} = a_{0}x_{0} + a_{1}x_{1} + a_{2}x_{2}, u_{2} = a_{3}x_{0} + a_{4}x_{1} + a_{5}x_{2}, u_{3} = a_{6}x_{0} + a_{7}x_{1} + a_{8}x_{2}

then we can associate to S′ the 10 coordinates (a_{0}, …, a_{8}, c). This provides a set of coordinates of S′ in the linear space (I_{A})_{6}.

In particular, we have

(10)  F = − |\begin{matrix}L_{1} & L_{2} & L_{3} & a \\ c_{11} & c_{21} & c_{31} & ℓ_{1} \\ c_{12} & c_{22} & c_{32} & ℓ_{2} \\ c_{13} & c_{23} & c_{33} & ℓ_{3} \end{matrix}|

(11)  F′ = |\begin{matrix}L′_{1} & L′_{2} & L′_{3} & a′ \\ c_{11} & c_{21} & c_{31} & ℓ_{1} \\ c_{12} & c_{22} & c_{32} & ℓ_{2} \\ c_{13} & c_{23} & c_{33} & ℓ_{3} \end{matrix}| ,

so that, if we pose

(12) L_{1} = a_{0}x_{0} + a_{1}x_{1} + a_{2}x_{2}, L_{2} = a_{3}x_{0} + a_{4}x_{1} + a_{5}x_{2}, L_{3} = a_{6}x_{0} + a_{7}x_{1} + a_{8}x_{2} L′_{1} = a_{10}x_{0} + a_{11}x_{1} + a_{12}x_{2}, L′_{2} = a_{13}x_{0} + a_{14}x_{1} + a_{15}x_{2}, L′_{3} = a_{16}x_{0} + a_{17}x_{1} + a_{18}x_{2},

then we can associate to the pair (F, F′) the matrix

W = (\begin{matrix}a_{0} & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & a_{9} \\ a_{10} & a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} & a_{17} & a_{18} & a_{19} \end{matrix}) ,

where a_{9} = a, a_{19} = a′. The pair (F, F′) determines a two dimensional linear subspace Λ of (I_{A})_{6}, hence an element of the Grassmannian of 2-dimensional linear subspaces in a 10-dimensional space. The Plücker coordinates of Λ are precisely the 2 × 2 minors of W, in the sense of (3).

Since the residue scheme B depends only on the linear space Λ spanned by F, F′, we can summarize the analysis in the following remark.

Remark 3.6. Let T ∈ S^{9}ℂ^{3} be a form with a decomposition A ⊂ ℙ^{2} of length r = 18, which satisfies conditions (i)-(iii) above, i.e. A is non-redundant, k_{4}(A) = 15 and h_{A}(5) = 18. Assume that there exists a second decomposition B for T, disjoint from A, such that Z = A ∪ B has the difference Hilbert function of Case 1. Then there exists a 2-dimensional subspace Λ ⊂ (I_{A})_{6} such that B is linked to A by two sextics which give a basis of Λ.

We can prove that the converse holds.

Remark 3.7. Choose a general 2-dimensional subspace Λ ⊂ (I_{A})_{6} and a basis F, F′ of Λ. Since I_{A} is generated in degree 6 and Λ is general, the residue B of A in the complete intersection Z of F, F′ is a set of 18 distinct points which does not intersect A (i.e. Z is smooth, as a consequence of the classical Bertini’s Theorem). Moreover, since h_{Z}(9) = 35 < ℓ(Z) = 36, by Proposition 2.8 the linear spaces ⟨v_{9}(A)⟩ and ⟨v_{9}(B)⟩ meet in exactly one point T. Thus, T is a form of degree 9 which has two different (disjoint) decompositions of length 18.

Remark 3.7 tells us that, in Case 1, we have a (rational) map

(13)  f : 𝔾(2, 10) −−→ ⟨v_{9}(A)⟩,  f(⟨F, F′⟩) = ⟨v_{9}(A)⟩ ∩ ⟨v_{9}(Z \ A)⟩,

whose image contains tensors T with a second decomposition B of length 18, such that Z = A ∪ B has a Hilbert function as in Case 1 of Proposition 3.4. Since the Grassmannian of 2-dimensional subspaces of (I_{A})_{6} has dimension 2(10 − 2) = 16, while the span of v_{9}(A) is 17-dimensional, it turns out that a general element T ∈ ⟨v_{9}(A)⟩ cannot belong to the (closure of) the image of f.

The next goal is to analyze the map f and determine when a given T belongs to the closure of im(f).

Recall that when A, B are linked by a complete intersection, the Hilbert function of B can be derived from the Hilbert function of A. The procedure is well-known, and widely described in [31]. If Z = A ∪ B is complete intersection of two sextics, and we are in Case 1, then the Hilbert function of B satisfies

\begin{matrix}j & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & … \\ Dh_{B}(j) & 1 & 2 & 3 & 4 & 5 & 3 & 0 & 0 & ⋯ \end{matrix}

Thus, from the Hilbert functions of A and B, we know that both (I_{A})_{9} and (I_{B})_{9} determine 37-dimensional subspaces of the linear space R_{9}, which has dimension 55. By standard facts on the intersection of ideals, I_{A} ∩ I_{B} is the ideal of the union Z = A ∪ B. Thus, from the Hilbert function of Z one knows that

dim((I_{A})_{9} ∩ (I_{B})_{9}) = dim((I_{Z})_{9}) = 55 − 35 = 20.

From the Grassmann formula one computes:

dim((I_{A})_{9} + (I_{B})_{9}) = 37 + 37 − 20 = 54.

It follows that (I_{A})_{9} + (I_{B})_{9} is a hyperplane in R_{9}, thus it determines a point in the dual projective space ℙ^{∨} = ℙ(R_{9})^{∨}, of dimension 54.

Before going on, we need to make a general remark on the relation between points and forms in projective spaces, clarifying the roles of elements of ℙ^{N} and its dual. The following remark collects standard facts for the relations between projective geometry and linear algebra.

Remark 3.8. Let ℙ(V) be a projective space and let ℙ(V)^{∨} be its dual, defined as the set of hyperplanes in ℙ(V). If we fix coordinates in ℙ(V), then there are dual coordinates in ℙ(V)^{∨}, defined as follows: for any hyperplane H in ℙ(V), the coefficients of an equation H in the fixed coordinates of ℙ(V) are dual coordinates for the point [H] representing H in ℙ(V)^{∨}. Thus, for a point T ∈ ℙ(V), the coordinates of T are coefficients for an equation of the hyperplane dual to T, in the dual coordinates of ℙ(V)^{∨}. If Λ is a linear subspace of ℙ(V), the dual subspace Λ^{∨} ⊂ ℙ(V)^{∨} is the set of hyperplanes containing Λ. Thus for Λ_{1}, Λ_{2} ⊂ ℙ(V) subspaces, the intersection Λ_{1}^{∨} ∩ Λ_{2}^{∨} corresponds to the dual of the linear subspace Λ_{1} + Λ_{2}.

Coming to our situation, up to now we indicated with ℙ^{2} the projective space over the space of linear forms in the ring R = ℂ[x_{0}, x_{1}, x_{2}], i.e. ℙ^{2} = ℙ(R_{1}). There is a natural interpretation of ℙ(R_{9})^{∨} = (ℙ^{54})^{∨} as ℙ(S^{9}(R_{1}^{∨})). If T ∈ R_{9} is a form, then the coefficients of T are coordinates of T in the natural basis defined by monomials. Thus T represents an equation for the hyperplane of ℙ(R_{9})^{∨} in the dual set of coordinates. In this interpretation, forms in the span Λ of powers L_{1}^{9}, …, L_{18}^{9} of linear forms are equations for hyperplanes in ℙ(R_{9})^{∨} associated to points of Λ. It follows that, if A = {L_{1}, …, L_{18}} ⊂ ℙ^{2}, then the space ℙ((I_{A})_{9}) has a natural interpretation as the set of hyperplanes in ℙ^{54} which are dual to points of Λ.

A consequence of the previous remark is the interpretation of (I_{A})_{9} + (I_{B})_{9} expressed in the following.

Proposition 3.9. Assume that ⟨v_{9}(A)⟩∩⟨v_{9}(B)⟩ = {T }. Then the subspace (I_{A})_{9}+ (I_{B})_{9} is a hyperplane in R_{9} which corresponds to the dual of the point T, in the natural interpretation of ℙ(R_{9})^{∨} given in the notation above.

Proof. T belongs to the span of the L_{i}^{9}’s exactly when the hyperplane dual of T contains the intersection of the hyperplanes dual to the L_{i}^{9}. In the dual space, the intersection of the hyperplanes dual to the L_{i}^{9} is defined by forms in (I_{A})_{9}. The claim follows. □

The next result describes birationally the geometric locus (hypersurface) of forms of degree 9 in the span of v_{9}(A) for which a second decomposition exists. Indeed, it refers to the situation in which the Hilbert function of A ∪ B is as in Case 1, but we will prove (see Proposition 3.21 and Proposition 3.24) that the remaining two cases lie in the border of Case 1.

Theorem 3.10. Let f : 𝔾(2, 10) −−→ ⟨v_{9}(A)⟩ be the rational map defined in (13) and let D ⊂ 𝔾(2, 10) be the pull-back of a hyperplane in f. Then D is a divisor in the class h_{2,10}. Therefore f is defined by a linear subseries of the complete linear series defined by h_{2,10}. Hence f is a linear projection of the Plücker embedding of 𝔾(2, 10).

Proof. Our aim is to show that, if H is a hyperplane of ℙ^{54} not containing ⟨v_{9}(A)⟩, then f^{−1}(H) is linearly equivalent to the divisor h_{2,10}. We will do that for a special hyperplane H.

In the dual space (ℙ^{54})^{∨}, H corresponds to a point [H]. Therefore, we want to describe for which elements W ∈ 𝔾(2, 10) the hyperplane H_{f(W)} associated to the point f(W) contains the point [H]. As in Remark 3.8, an equation for H, whose coefficients are a set of representatives for the dual point [H], determines a form of degree 9 in R, which we call G_{H}. By construction,

H_{f(W)} = ℙ((I_{A})_{9} + (I_{B(W)})_{9})

where B(W) = {P′_{1}(W), …, P′_{18}(W)} ⊂ ℙ^{2} is such that Z(W) = A ∪ B(W) is the complete intersection of type (6, 6) associated to W. Thus

(14) [H] ∈ H_{f(W)} if and only if

there exist G_{A} ∈ (I_{A})_{9}, G_{B(W)} ∈ (I_{B(W)})_{9} such that G_{H} = G_{A} + G_{B(W)}.

The condition now becomes purely algebraic. Take a set Y of representatives of the points of A, and call ρ̄ : R → ℂ^{18} the evaluation map of forms on the set Y. The natural inclusion of I_{Z(W)} in I_{B(W)} gives rise to an exact sequence

(15)  0 → (I_{Z(W)})_{9} ⟶ (I_{B(W)})_{9} ⟶^{ρ} ℂ^{18}

where the rightmost map ρ = ρ̄_{|(I_{B(W)})_{9}} is not surjective.

Claim 3.11. The following are equivalent:

(i) there exist G_{A} ∈ (I_{A})_{9}, G_{B(W)} ∈ (I_{B(W)})_{9} such that G_{H} = G_{A} + G_{B(W)}, (ii) ρ(G_{H}) belongs to ρ((I_{B(W)})_{9}).

Proof. If (i) holds, then

ρ(G_{H}) = ρ(G_{A} + G_{B(W)}) = ρ(G_{A}) + ρ(G_{B(W)}) = ρ(G_{B(W)})

which implies (ii). On the other side, assume (ii) so that there exists G_{B(W)} ∈ (I_{B(W)})_{9} such that ρ(G_{B(W)}) = ρ(G_{H}). Therefore ρ(G_{H} − G_{B(W)}) = 0, so that G_{H} − G_{B(W)} belongs to (I_{A})_{9} and (i) holds. □

By combining (14) with Claim 3.11, our aim turns out to be the following: given a form G_{H} of R_{9}, we want to describe the elements W ∈ 𝔾(2, 10) such that ρ(G_{H}) belongs to ρ((I_{B(W)})_{9}).

Claim 3.12. For any W ∈ 𝔾(2, 10) set Z′(W) = Z(W) \ {P_{17}, P_{18}}. Then for any form G of degree 9 in I_{Z′(W)}, the residue ρ(G) is fixed, up to scalar multiplication.

Proof. We apply Proposition 5.2.10 of [31] (Mapping cone) to the commutative diagram (see Section 2.6)

0 ⟶ R(−3) \underrightarrow{(\begin{matrix}−c \\ ℓ \end{matrix})} R(−1) ⊕ R(−2) ⟶ I_{\{P_{17},P_{18}\}} ⟶ 0
↑  ↑  ↑
0 ⟶ R(−12) \overrightarrow{(\begin{matrix}−F′ \\ F \end{matrix})}  R(−6)^{2}  ⟶ I_{Z(W)} ⟶ 0

where c, ℓ are generators for the ideal of {P_{17}, P_{18}}. We get that I_{Z′(W)} admits a resolution of the form

(16)  0 → R(−11) ⊕ R(−10) \underrightarrow{(\begin{matrix}ϕ_{1} & ψ_{1} \\ ϕ_{2} & ψ_{2} \\ −c & ℓ \end{matrix})} R(−6)^{2} ⊕ R(−9) → I_{Z′(W)} → 0

where ϕ_{1}, ϕ_{2} ∈ S^{5}ℂ^{3}, ψ_{1}, ψ_{2} ∈ S^{4}ℂ^{3} are defined by the central vertical map of the diagram. The ideal I_{Z′(W)} is thus generated by F, F′, G_{0} with

(17)  F = |\begin{matrix}ϕ_{1} & ψ_{1} \\ −c & ℓ \end{matrix}| = ℓϕ_{1} + cψ_{1}

(18)  F′ = |\begin{matrix}ϕ_{2} & ψ_{2} \\ −c & ℓ \end{matrix}| = ℓϕ_{2} + cψ_{2}

(19)  G_{0} = |\begin{matrix}ϕ_{1} & ψ_{1} \\ ϕ_{2} & ψ_{2} \end{matrix}| = ϕ_{1}ψ_{2} − ϕ_{2}ψ_{1}.

Then for any G ∈ (I_{Z′(W)})_{9} we get G = U_{1}F_{1} +U_{2}F_{2} +qG_{0}, q ∈ ℂ, where U_{1}, U_{2} are suitable cubics. Since F_{1}, F_{2} vanish at Z(W), then the residue ρ(G) is a multiple of ρ(G_{0}). □

In particular, for W ∈ 𝔾(2, 10), we can find scalars α_{W} , β_{W} such that the residue ρ(G) is a scalar multiple of (0, …, 0, α_{W} , β_{W}) for all G ∈ (I_{Z′(W)})_{9}.

Fix two non-zero scalars α, β ∈ ℂ. Let v_{α,β} = (v_{1}, …, v_{18}) ∈ ℂ^{18} be defined by

v_{j} = \begin{cases}0 j ∈ \{1, …, 16\} \\ α j = 17 \\ β j = 18 \end{cases}

Since the evaluation map ρ̄ surjects in degree 9, we can find H such that the form G_{H} satisfies ρ(G_{H}) = v_{α,β}. We compute f^{−1}(H) for this hyperplane H. Notice that H cannot contain ⟨v_{9}(A)⟩, for G_{H} does not vanish at P_{17}, P_{18}. By Claim 3.11, W belongs to f^{−1}(H) if and only if there exists G ∈ (I_{B(W)})_{9} with ρ(G) = ρ(G_{H}). Such a form G vanishes at the points P_{1}, …, P_{16}, thus it belongs to (I_{Z′(W)})_{9}. We obtain by Claim 3.12 that W ∈ f^{−1}(H) if and only if

(20)  det (\begin{matrix}α & β \\ α_{W} & β_{W} \end{matrix} ) = 0.

Now we compare the two expressions that we have for F, from (10) and (17). We get that

ℓϕ_{1} + cψ_{1} = L_{1}Q_{1} + L_{2}Q_{2} + L_{3}Q_{3} + aS,

but Q_{1}, Q_{2}, Q_{3}, S vanish at P_{17}, P_{18}, thus they belong to the ideal spanned by ℓ, c. In particular, we can write Q_{i} = M_{i}c + N_{i}ℓ and S = M̄c + N̄ℓ. Since ℓ, c give a complete intersection, they have only trivial syzygies, and hence we conclude that there exists a form U of degree 3 such that

ϕ_{1} = Uc + L_{1}N_{1} + L_{2}N_{2} + L_{3}N_{3} + aN̄
ψ_{1} = −Uℓ + L_{1}M_{1} + L_{2}M_{2} + L_{3}M_{3} + aM̄.

Similarly, if we compare the two expressions that we have for F′, from (11) and (18), we get that

ϕ_{2} = U′c + L′_{1}N_{1} + L′_{2}N_{2} + L′_{3}N_{3} + a′N̄
ψ_{2} = −U′ℓ + L′_{1}M_{1} + L′_{2}M_{2} + L′_{3}M_{3} + a′M̄.

for some form U′ of degree 3. Now put the previous expressions in the formula (19) for G_{0}. When we compute the residue of G_{0} at Z(W), the contribution of U, U′ disappears, since c, ℓ vanish at the points P_{17}, P_{18}. Moreover the M_{i}, N_{i}, M̄, N̄ are fixed and do not depend on the choice of W. It follows that (α_{W} , β_{W}) is an expression in terms of the 2 × 2 determinants of the coefficients of the L_{i}’s and a and the coefficients of the L′_{i}’s and a′. That is: (α_{W} , β_{W}) is a linear expression in the Plücker coordinates of W. This concludes the proof of the theorem. □

Of course, we do not know which projection of the Grassmannian determines the rational map f. It is likely that the projection is highly special.

Next, we want to show that general elements in the image of f have exactly two decompositions.

Proposition 3.13. The map f : 𝔾(2, 10) −−→ ⟨v_{9}(A)⟩ defined in (13) is birational onto the image.

Proof. We aim to show that, for some P ∈ im(f) ⊂ ⟨v_{9}(A)⟩, the set f^{−1}(P) is finite and has degree 1. We proceed via a computational approach in Macaulay2 [26] (over a finite field, but then the proof holds also over ℂ). For a detailed description of our procedure, we refer to the ancillary file nonics3.txt. We start by selecting a finite set A = {P_{1}, …, P_{18}} ⊂ ℙ^{2}, whose elements have random coefficients. Then, we construct the Hilbert-Burch matrix of A and we fix 2 forms F, F′ of degree 6 in I_{A}, so that one is not the multiple of the other. This is equivalent to a choice of 6 linear forms L_{1}, L_{2}, L_{3}, L′_{1}, L′_{2}, L′_{3} and 2 scalars a, a′ (since the choice is general, we may assume that ∂_{x_{0}}L_{1} = 1, ∂_{x_{0}}L′_{1} = 0, a = 0, a′ = 1). We get a residual set B_{F,F′}, whose ideal admits a resolution as in (9). By means of (13), we compute f(F, F′) and we pose P = f(F, F′). Let (p_{0}, …, p_{54}) be a representative vector for P.

In order to obtain f^{−1}(P), in the first and second row of the Hilbert-Burch matrix N of B_{F,F′} we change L_{j} (resp. L′_{j}) with L_{j} = a_{3j−3}x_{0} + a_{3j−2}x_{1} + a_{3j−1}x_{2} (resp. with L′_{j} = a_{3j+7}x_{0} + a_{3j+8}x_{1} + a_{3j+9}x_{2}), for j ∈ {1, 2, 3} (with a_{0} = 1 and a_{10} = 0) and we consider the 55 × 54 matrix MFix′′ whose columns provide a set of generators for (I_{A})_{9} + (I_{B_{F,F′}})_{9}. Notice that MFix′′ is divided in 2 blocks: the first 37 columns have integer entries, while in the last 17 the entries depend linearly on the 2x2 minors of the matrix

W = (\begin{matrix}1 & a_{1} & a_{2} & a_{3} & a_{4} & a_{5} & a_{6} & a_{7} & a_{8} & 0 \\ 0 & a_{11} & a_{12} & a_{13} & a_{14} & a_{15} & a_{16} & a_{17} & a_{18} & 1 \end{matrix}) ,

i.e. on the Plücker coordinates of W in the sense of (3). According to the previous remark, write MF ix′′ = (A_{1}|A_{2}). Therefore

f^{−1}(P) = {(a_{1}, …, a_{8}, a_{11}, …, a_{18}) ∈ ℂ^{16} | (p_{0}, …, p_{54}) · MF ix′′ = 0_{1×55}}.

Since (p_{0}, …, p_{54}) · A_{1} = 0_{1×37} provide trivial conditions, then

(21) f^{−1}(P) = {(a_{1}, …, a_{8}, a_{11}, …, a_{18}) ∈ ℂ^{16} | (p_{0}, …, p_{54}) · A_{2} = 0_{1×17}}.

We notice that the 17 equations appearing in (21) provide a linear system in the Plücker coordinates of W. Our computations in Macaulay2 show that f^{−1}(P) has dimension 0 and degree 1, which concludes the proof. □

Proposition 3.13 implies the following:

Remark 3.14. If T ∈ S^{9}ℂ^{3} of rank 18 is a general point in im(f), i.e. a general unidentifiable nonic of rank 18, then there exist exactly two finite sets computing the rank of T.

In other words, the space ⟨v_{9}(A)⟩ contains a variety Θ, which is the closure of a linear projection of 𝔾(2, 10), whose general points consist of forms in S^{9}ℂ^{3} of rank 18, that admit two finite sets computing the rank.

Remark 3.15. Given T ∈ S^{9}ℂ^{3} of rank 18 with a non-redundant finite set A = {P_{1}, …, P_{18}} ⊂ ℙ^{2} computing it, such that k_{4}(A) = 15 and h_{A}(5) = 18, by following the proof of Proposition 3.13, in principle one can develop a criterion that establishes the uniqueness of such an A.

The corresponding algorithm is explained in more details in Sect. 5.

3.2. Case 2. Let us assume to be in the second case of Proposition 3.4. This is the unique case in which the new decomposition has less than 18 summands, so that the rank of T is ≤ 17. Indeed we will show (see Corollary 4.2 below) that in this case the rank is exactly 17. In Section 5 we will provide a code which can exclude that a given form T falls in this case.

Look at the sequence

0 → (I_{Z})_{s} → R_{s} → ℂ^{35}.

For s = 5, we get dim(I_{Z})_{5} = dim R_{5} − h_{Z}(5) = 21 − 20 = 1, so that Z is contained in a unique quintic Q. For s = 7 one gets in the same way dim(I_{Z})_{7} > 6, so there exists G ∈ (I_{Z})_{7} which is not a multiple of Q. Since Z satisfies CB(9) and the Hilbert function of Z is the same as the Hilbert function of a complete intersection of type (5, 7), by the Main Theorem of [18] Z is a complete intersection of type (5, 7). In particular, I_{Z} = (Q, G), with Q ∈ S^{5}ℂ^{3} and G ∈ S^{7}ℂ^{3} and a minimal resolution of I_{Z} is given by

0 → R(−12) \underrightarrow{(\begin{matrix}−G \\ Q \end{matrix})} R(−5) ⊕ R(−7) → I_{Z} → 0.

Again, fix three quintics Q_{1}, Q_{2}, Q_{3} and one sextic S that generate the ideal of A, so that a minimal resolution of I_{A}, with Hilbert-Burch matrix M, is as described in Proposition 3.2.

Remark 3.16. For any choice of the quintic generators Q_{1}, Q_{2}, Q_{3}, the ideal I_{A} coincides with the ideal generated by the Q_{i}’s in degree 7. Indeed, it follows from the description of the Hilbert-Burch matrix M that, for a general choice of the quintic forms, there are no relations of degree 7 involving only Q_{1}, Q_{2}, Q_{3}. In other words, there are no non-trivial quadrics q_{1}, q_{2}, q_{3} such that ∑ q_{i}Q_{i} is the zero polynomial. It follows that the ideal J generated by Q_{1}, Q_{2}, Q_{3} satisfies dim J_{7} = 18, which is exactly the dimension of (I_{A})_{7}, as computed from the Hilbert function.

Remark 3.17. In the space ℙ^{20} = ℙ(R_{5}) which parameterizes quintic forms (up to scalar multiplication) the generators Q_{1}, Q_{2}, Q_{3} of I_{A} determine a plane ℙ^{2} = Π. Fix a general quintic Q that belongs to the plane, so that in particular Q belongs to I_{A}. Call J the ideal generated by Q. The quotient (I_{A})_{7}/J_{7} is a vector space of dimension 12. The variety E, consisting of all complete intersections Z of type (5, 7) containing A, is a ℙ((I_{A})_{7}/J_{7})-bundle, i.e. a ℙ^{11}-bundle, over an open subset of Π. The bundle structure is given by the map which sends Z = S ∩ Q to Q ∈ Π. Then E has dimension 13. Sets B of 17 points of ℙ^{2} linked to A by a complete intersection of type (5, 7) are parameterized by E. As in Remark 3.7, we get that a general such set B determines one point of the span of v_{9}(A) with a second decomposition of length 17. Hence there exists a rational map

(22)  f′ : E −−→ ⟨v_{9}(A)⟩  f′(Z) = ⟨v_{9}(A)⟩ ∩ ⟨v_{9}(Z \ A)⟩,

such that the closure Θ′ of the image of f′ (of dimension at most 13) is the closure of the locus of forms T ∈ ⟨v_{9}(A)⟩ for which A is non-redundant but the rank is 17.

We will provide in Sect. 5 an algorithm which guarantees that a given form T does not lie in the subvariety Θ′, so it has rank 18. In order to produce the algorithm, we need a description of the form T which is the intersection of ⟨v_{9}(A)⟩ and ⟨v_{9}(B)⟩, where B is linked to A by a complete intersection Q ∩ G of type (5, 7).

Remark 3.18. Fix a quintic Q ∈ I_{A}, which can be written as

Q = a_{1}Q_{1} + a_{2}Q_{2} + a_{3}Q_{3},

for a choice of the scalars a_{i}’s. Notice that if a_{3} ≠ 0 (resp. a_{2} ≠ 0. a_{1} ≠ 0), then the ideal generated by Q_{1}, Q_{2}, Q_{3}, Q coincides with the ideal generated by Q_{1}, Q_{2}, Q (resp. Q_{1}, Q, Q_{3} or Q, Q_{2}, Q_{3}). By Remark 3.16, a general septic G ∈ I_{A} can be written as

G = q_{1}Q_{1} + q_{2}Q_{2} + q_{3}Q_{3},

for a choice of conics q_{1}, q_{2}, q_{3}. Let B = B(Q, G) be the residue of A in the complete intersection of Q and G. As in case 1, from the Hilbert functions of A, B and Z = A ∪ B, we know that (I_{A})_{9}, resp. (I_{B})_{9}, determines a 37-dimensional, resp. a 38-dimensional, subspace of the linear space R_{9}, which has dimension 55. Moreover dim((I_{A})_{9} + (I_{B})_{9}) = 54, so that (I_{A})_{9} + (I_{B})_{9} is a hyperplane in R_{9}. As in proposition 3.9, the hyperplane (I_{A})_{9} + (I_{B})_{9} is dual to the point T of intersection between ⟨v_{9}(A)⟩ and ⟨v_{9}(B)⟩.

As in case 1, the mapping cone procedure provides an effective way of computing (I_{B})_{9}, thus also the sum (I_{A})_{9}+(I_{B})_{9}. Namely, the ideal I_{B} is defined by the minors of the Hilbert-Burch matrix M, to which we add the columns that define Q, G. We get the matrix:

(23)  M′ = (\begin{matrix}a_{1} & a_{2} & a_{3} & 0 \\ q_{1} & q_{2} & q_{3} & 0 \\ c_{11} & c_{21} & c_{31} & ℓ_{1} \\ c_{12} & c_{22} & c_{32} & ℓ_{2} \\ c_{13} & c_{23} & c_{33} & ℓ_{3} \end{matrix}) .

In particular, B depends on the choice of forms in the first two rows

(\begin{matrix}a_{1} & a_{2} & a_{3} & 0 \\ q_{1} & q_{2} & q_{3} & 0 \end{matrix}) .

Theorem 3.19. The map f′ : E −−→ ⟨v_{9}(A)⟩ defined in (22) is birational onto the image.

Proof. As in Proposition 3.13, we proceed via a computational approach in Macaulay2 [26]. For a detailed description of our procedure, we refer to the ancillary file

nonics1.txt.

Select a finite set A = {P_{1}, …, P_{18}} ⊂ ℙ^{2}, and construct the Hilbert-Burch matrix M of A. The choice of a quintic form Q and a septic form G in I_{A}, not multiple of Q, is equivalent to a choice of scalars a_{1}, a_{2}, a_{3} and quadrics q_{1}, q_{2}, q_{3}. Assume first that a_{1} ≠ 0, so that, after rescaling, we may assume a_{1} = 1. In this case notice that Q_{1} is generated by Q_{2}, Q_{3}, Q. Thus the residue B = B(Q, G) of A in the complete intersection Q ∩ G is also the residual in the complete intersection of Q ∩ G′ where G′ = (q_{2} − a_{2}q_{1})Q_{2} + (q_{3} − a_{3}q_{1})Q_{3} = q′_{2}Q_{2} + q′_{3}Q_{3}. I.e., we may assume q_{1} = 0. In particular B = B(Q, G) depends on the choice of a_{2}, a_{3} and the 12 coefficients a_{4}, …, a_{15} of the quadrics q_{2}, q_{3}. Now we can compute how many choices of parameters a_{2}, … a_{15} determine a given hyperplane P = (I_{A})_{9} + (I_{B(Q_{0},G_{0})})_{9} for a general choice of the forms Q_{0} ∈ (I_{A})_{5}, G_{0} ∈ (I_{A})_{7}. This can be done by Macaulay2. We choose a nonic P corresponding to a hyperplane (I_{A})_{9} + (I_{B(Q_{0},G_{0})})_{9} where Q_{0}, G_{0} are determined randomly, with no coefficients a_{1}, …, a_{15} equal to 0. The coefficients (p_{0}, …, p_{54}) of P in (ℙ^{54})^{∨} can be easily computed via the mapping cone. Take the 55 × 54 matrix N whose columns provide a set of generators for (I_{A})_{9} + (I_{B(Q,G)})_{9}, for Q, G general. Then, in the subset of the net Π of Q_{1}, Q_{2} and Q_{3} in which a_{1} ≠ 0

(24)  f′^{−1}(P) = {(a_{2}, …, a_{15}) ∈ ℂ^{14} | (p_{0}, …, p_{54}) · N = 0_{1×54}}.

In practice, in order to simplify the computation, observe that in an open set of the variety E defined in Remark 3.17, we may also assume that one among a_{4}, …, a_{15} is equal to 1. Thus, in the open set, we get a parameterization of E with 13 parameters, corresponding to the dimension of E. By varying the coefficients among a_{1}, a_{2}, a_{3} and among a_{4}, …, a_{15} which are set equal to 1, we obtain a complete scan of f^{′−1}(P). Our computations in Macaulay2 show that, in any case, f^{′−1}(P) has dimension 0 and degree 1, i.e. it is a point. This concludes the proof. □

In the rest of the section, we will prove that Case 2 is a degeneration of Case 1. We keep all the previous notation.

Proposition 3.20. Let P_{0} ∈ {G = 0} be a general point. Then Z ∪ {P_{0}} is a limit for t = 0 of a family {Z̃_{t}} (over a small disc), with Z̃_{t} complete intersection of type (6, 6) containing A for all t ≠ 0.

Proof. The proof is direct. Let ℓ_{1}, ℓ_{2} be general generators of I_{P_{0}}. Since P_{0} ∈ {G = 0} then there are forms E_{1}, E_{2} of degree 6 such that G = E_{1}ℓ_{1} + E_{2}ℓ_{2}. Since G is irreducible, for t ∈ ℂ general the forms ℓ_{1}Q + tE_{1}, ℓ_{2}Q + tE_{2} determine a complete intersection Z̃_{t} of type (6, 6). By taking the flat limit for t = 0, the ideal of Z̃_{t} degenerates to the ideal generated by ℓ_{1}Q, ℓ_{2}Q, G, which is the ideal of Z ∪ {P_{0}}.

It remains to prove that we can assume A ⊂ Z̃_{t}. Let A_{t} be the subset of Z̃_{t} which degenerates to A (note that A_{t} is well defined since t moves in a small disc). Since the ideal of A is generated in degree 6, the same holds for the ideal of A_{t}. Moreover (I_{A})_{6} and (I_{A_{t}})_{6} have the same dimension. Thus the space of sextic curves containing A_{t} degenerates to the space of sextic curves containing A. It follows that the set of complete intersections of type (6, 6) containing A_{t} degenerates to the set of complete intersections of type (6, 6) containing A. In particular, Z belongs to the closure of the set of complete intersections of type (6, 6) containing A. □

Proposition 3.21. Let T be a form of degree 9 with decompositions A, B such that Z = A ∪ B has Hilbert function as in Case 2. Then there exists a family of tensors T_{t}, over a small disc Δ, such that T_{0} = T, T_{t} belongs to span of v_{9}(A) for all t ∈ Δ and for t ≠ 0 there exists a second decomposition B_{t} of T_{t}, which degenerates to B, with A ∪ B_{t} complete intersection of type (6, 6).

Proof. The claim is an immediate consequence of Proposition 3.20, because for Z_{t} belonging to both Case 1 and Case 2 we have ℓ(Z_{t}) − h_{Z_{t}}(9) = 1, thus the spans of v_{9}(A) and v_{9}(B_{t}) intersect in one point, for all t ∈ Δ. □

It follows that nonic forms with two decompositions A, B such that A ∪ B is as in Case 2 are degeneration of forms with two decompositions A, B_{t} with A ∪ B_{t} as in Case 1.

3.3. Case 3. Let us consider the third case of Proposition 3.4. We will prove again that the tensor T is the limit of a family tensors T_{t} ∈ v_{9}(A) with two decompositions whose union is a complete intersection of two sextic curves.

As in the previous situation, Proposition 2.10 implies that Z is contained in a plane quintic. Moreover, the evaluation map determines exact sequences

0 → (I_{Z})_{s} → R_{s} → ℂ^{36}.

Looking at the cases s ∈ {5, 8, 9, 12}, we get that Z is contained in a unique quintic Q, and there exists a pencil of plane curves of degree 8 containing Z and not containing Q. Fix two curves O_{i} ∈ S^{8}ℂ^{3}, i = 1, 2 such that Z ⊂ Q ∩ O_{1} ∩ O_{2}.

Lemma 3.22. The set Z as above is limit for t = 0 of a family {Z_{t}}, containing A, such that for t ≠ 0 the set Z_{t} belongs to an irreducible quintic Q containing A.

Proof. If the unique quintic Q that contains Z is irreducible, then the claim is obvious because irreducibility is locally open in families. Assume then that Q is reducible. In any case, the Hilbert function of Z coincides with the Hilbert polynomial in degree 10, thus, by the Castelnuovo-Mumford regularity Theorem (see [13]), the ideal of Z is generated in degree 11. Link Z first with a complete intersection of type (5, 11), then link the residue with a complete intersection of type (5, 7). By using twice the mapping cone (see Sect. 2.6), one realizes that the final residue W is a complete intersection of 2 quartics.

There exists a family {W_{t}} of sets of 16 points, with W_{0} = W, whose generic element is a complete intersection of two quartics and it is contained in an irreducible quintic curve which contains A.

Indeed, if X, X′ are the equations of two quartics whose intersection is W, then Q = zX + z′X′, where z, z′ are two linear forms which intersect in some point P ∈ Q. Now, move Q in a family of quintics {Q_{t}}, containing A, whose general element is irreducible. Move P in a family of points P_{t} such that P_{t} ∈ Q_{t}. The linear forms z, z′ generalize to two families of linear forms {z_{t}}, {z′_{t}} such that z_{t}, z′_{t} define P_{t}. Then, there are two families of quartics {X_{t}}, {X′_{t}} such that Q_{t} = z_{t}X_{t} + z′_{t}X′_{t} and the families specialize to X, X′ for t = 0. So, just take W_{t} to be the intersection of X_{t}, X′_{t}.

Once the existence of the family {W_{t}} is established, the existence of {Z_{t}} follows immediately by linking back the general element of {W_{t}}, first with a complete intersection of type (5, 7), and then with a complete intersection of type (5, 11). □

Thus, since we want to find Z as a limit, we may assume that Q is irreducible. The residue X of Z in the complete intersection Q ∩ O_{1} is a scheme of length 4, whose Hilbert function can be computed from the Hilbert functions of Z and the complete intersection (see Section 2.6). It follows that X is contained in a line. Thus X is complete intersection of a line ℓ and a quartic q. The mapping cone implies thus that the homogeneous ideal of Z is generated by Q, O_{1}, O_{2}, and its minimal resolution is given by

(25)  0 → R(−12) ⊕ R(−9) ⟶^{M} R(−5) ⊕ R(−8)^{2} → I_{Z} → 0

where

(26)  M = (\begin{matrix}S & q_{1} \\ q_{2} & ℓ_{1} \\ q & ℓ \end{matrix})

denotes the Hilbert-Burch matrix of I_{Z}.

Note that S ∈ S^{7}ℂ^{3}, q_{i} ∈ S^{4}ℂ^{3} for i ∈ {1, 2}, ℓ_{1} ∈ S^{1}C^{3}, and Q, O_{1}, O_{2} coincide, respectively, with (−1)^{i} times the minor obtained by leaving out the i-th row of M, i ∈ {1, 2, 3}.

Let W be the residue of Z in the intersection O_{1}, O_{2}. The fact that O_{1}, O_{2} intersect properly, for a general choice of the two forms of degree 8 in the ideal of Z, follows from the resolution above. By the mapping cone, W is a set of length 28, complete intersection of S and q_{1}. Thus, the residue of W in the intersection of q_{1} and O_{1} is the set X.

Move the set X in O_{1}, to a general set X_{t} of 4 points, q_{1} moves to a quartic containing X_{t}. Taking residues, we obtain a family of sets {W_{t}} of length 28 in O_{1} such that W_{0} = W. For t ≠ 0 the set W_{t} is the residue of a general set of 4 points, which is complete intersection of 2 quadrics. Thus the resolution of the ideal of W_{t} is:

0 → R(−10)^{2} → R(−4) ⊕ R(−8)^{2} → I_{W_{t}} → 0.

The ideal of W_{t} has two minimal generators in degree 8, one of which is O_{1} (fixed).

Lemma 3.23. We can move X to X_{t} so that the second generator O′_{t} of degree 8 moves to O_{2} as t goes to 0.

Proof. By the mapping cone, for t general the residue of X_{t} in a complete intersection (4, 8) is contained in a pencil of curves of degree 8. The limit of this pencil determines a pencil in the 2-dimensional space of curves of degree 8 through W. We need to prove that we can choose the family {X_{t}} so that the limit contains O_{2}. But this is clear for all the limits contain O_{1} and the limit changes if we vary the family. □

Notice that, for t general, the residue of W_{t} with respect to O_{1} ∩O′_{t} is a complete intersection of two sextics. It follows that Z is limit of a family of 36 points, whose general element is a complete intersection of two sextics.

Collecting all the previous lemmas, we get:

Proposition 3.24. In Case 3, the form T is the limit of a family tensors T_{t} ∈ v_{9}(A) with two decompositions whose union is a complete intersection of two sextic curves.

Proof. We know that Z is limit of a family {Z_{t}} whose general element is a complete intersection of 2 sextics. We can conclude that Z is limit of a family {Z_{t}} as above, whose general element contains A, because the family of complete intersections of type (6, 6) containing Z is irreducible, and it is the limit of the set of complete intersections of type (6, 6) containing a general set of 18 points. □

4. The case of non-empty intersection

In this section we assume that A satisfies the genericity conditions (i)-(iii), but we drop the assumption that A ∩ B is empty. We will see that the case can be characterized in terms of the Case 2 of the previous section.

By arguing as in the proof of Claim 4.2 of [4], we have the following:

Proposition 4.1. If A∩B ≠ ∅, then the cardinality of B is 18 and the intersection contains only one point.

Proof. Assume A ∩ B = {P_{1}, …, P_{s}}, with 1 ≤ s < 18. Fix coordinates T_{1}, …, T_{18} for the points of v_{9}(A) and coordinates T′_{s+1}, …, T′_{ℓ(B)} for the points of v_{9}(B \ A). Then there is a choice of scalars such that

T = a_{1}T_{1} + … + a_{18}T_{18} = b_{1}T_{1} + … + b_{s}T_{s} + b_{s+1}T′_{s+1} + … + b_{ℓ(B)}T′_{ℓ(B)}

Define

T_{0} = (a_{1} − b_{1})T_{1} + … + (a_{s} − b_{s})T_{s} + a_{s+1}T_{s+1} + … + a_{18}T_{18} =
= b_{s+1}T′_{s+1} + … + b_{ℓ(B)}T′_{ℓ(B)}.

T_{0} is an element of S^{9}ℂ^{3} admitting two disjoint decompositions: A and B_{0} = B \A. Notice that, since

(27)  T = T_{0} + b_{1}T_{1} + … + b_{s}T_{s},

necessarily B_{0} is non-redundant for T_{0}, otherwise B is redundant for T. Denote by A′ ⊂ A a non-redundant decomposition of T_{0}. Notice that P_{i} ∈ A′ for all i ∈ {s + 1, …, 18}, otherwise, by (27), A is a redundant decomposition for T. It turns out that T_{0} has two non-redundant decompositions, A′ and B_{0}, with ℓ(A′) ≤ 18 and ℓ(B_{0}) = ℓ(B) − s ≤ 18 − s ≤ ℓ(A′). Since A satisfies properties (ii) and (iii), then, by Remark 2.4, k_{4}(A′) = min{15, ℓ(A′)} and h_{A′}(5) = ℓ(A′) ≤ 18, and so, by Theorem 2.11, B_{0} cannot exist, unless ℓ(A′) = 18, i.e. A = A′. Thus also A is non-redundant for T_{0}.

The tensor T_{0} has two non-redundant decompositions A, B_{0}, with ℓ(B_{0}) ≤ ℓ(B)− s, and A ∩ B_{0} = ∅. Now assume that either s ≥ 2 or ℓ(B) < 18, hence ℓ(B_{0}) ≤ 16. Thus Z, which is also equal to A ∪ B_{0}, is a set of at most 34 points which, from Corollary 2.14, must satisfy the property CB(9). It follows that:

ℓ(Z) ≥ ∑_{i=0}^{4} Dh_{A}(i)+Dh_{Z}(5)+∑_{i=6}^{10} Dh_{Z}(i) ≥ 2 ∑_{i=0}^{4} Dh_{A}(i)+Dh_{Z}(5) = 30+Dh_{Z}(5).

It follows Dh_{Z}(5) ≤ 4. Thus, by Proposition 2.7, Dh_{Z}(i) ≤ 4 for all i > 4. Since Dh_{Z}(10) > 0, the difference Hilbert function Dh_{Z} cannot be strictly decreasing from 5 to 10. This contradicts Proposition 2.10.

□

Corollary 4.2. Let T ∈ S^{9}ℂ^{3} be a form with a decomposition A ⊂ ℙ^{2} of length r = 18, which satisfies conditions (i)-(iii) above, i.e. A is non-redundant, k_{4}(A) = 15 and h_{A}(5) = 18. Then the rank of T is either 17 or 18.

Proof. Assume there exists another non-redundant decomposition of T with length ℓ(B) ≤ 18. If A ∩ B ≠ ∅, then ℓ(B) = 18 by Proposition 4.1. If A, B are disjoint, then we are either in Case 1, or in Case 2, or in Case 3 of Proposition 3.4. In any event, ℓ(A ∪ B) ≥ 35, so that ℓ(B) ≥ 17. □

Remark 4.3. Thus, if A∩B is non-empty, there exists a unique point, say P_{1} ∈ A, which also belongs to B. Put B_{0} = B \ {P_{1}}. With the notation of the proof of Proposition 4.1, we obtain that T_{0} has two non-redundant decompositions: A and B_{0}, with ℓ(B_{0}) = 17. It follows from Proposition 3.4 applied to T_{0} that Z = A ∪ B_{0} has Hilbert function as in Case 2. Thus, the previous analysis of Case 2 shows that A and B_{0} are linked in a complete intersection of type (5, 7), and T_{0} is the unique point of intersection of the spans of v_{9}(A) and v_{9}(B_{0}) (see Remark 3.7). Moreover, the tensor T belongs to the line joining v_{9}(P_{1}) to T_{0}.

Remark 4.4. Let A, B be as in Case 1 of the previous section. We know that Z = A ∪ B is the complete intersection of two sextics F, F′, and the spans of v_{9}(A), v_{9}(B) meet in one point. Assume that F moves in a family, so that the limit curves F_{0} is still irreducible, and it is tangent to F in one point of A, say in P_{1}. Then B moves in a family of finite sets whose limit B_{0} is linked to A by a complete intersection F_{0}, F′_{0}. It is clear from the construction that A, B_{0} share the point P_{1}, so that ν_{9}(P_{1}) is the unique point of intersection of the spans of v_{9}(A), v_{9}(B_{0}).

Notice that since the ideal of A is generated in degree 6, for all sextics F containing A one can find infinitely many sextics F′_{0}, containing A and tangent to F at P_{1}.

Now, we are ready to prove that the case of non-empty intersection can also be seen as a degeneration of the previous Case 1.

Proposition 4.5. When A ∩ B ≠ ∅, the form T is the limit of a family of tensors T_{t} ∈ v_{9}(A) with two decompositions whose union is complete intersection of two sextic curves.

Proof. We know, by Proposition 4.1, that A ∩ B is one point, say P_{1}, moreover B_{0} = B \ {P_{1}} is a set of 17 points, linked to A by a complete intersection of type (5, 7). Moreover the span of v_{9}(A) and v_{9}(B_{0}) meet in a form T_{0} of rank 17 and T sits in the line joining v_{9}(P_{1}) and T_{0}.

By Case 2 of the previous section and by Proposition 3.21, we know that T_{0} is a limit of a family {T_{t}}, whose general element T_{t} is the intersection of the spans of v_{9}(A) and v_{9}(B_{t}), where {B_{t}} is a family of finite sets of cardinality 18, which are linked to A by a complete intersection of two families of sextics {F_{t}}, {F′_{t}}. The limits of the two families F_{0}, F′_{0} intersect in a common quintic Q, containing A ∪ B_{0} = Z. In particular, F_{0} = Q ∪ L, where L is a general line.

Now, take a family {G_{t}} of sextics through A, such that G_{t} is tangent to F_{t} for t general (hence for all t). Such a family exists because the ideal of A is generated in degree 6. For each a ∈ ℂ consider the family {H(a)_{t}}, where H(a)_{t} = {G_{t} + aF′_{t}}. For t, a general the curves H(a)_{t} and F′_{t} link A to a finite set B(a)_{t} of cardinality 18. By the construction of Case 1 of the previous section, the generators of the ideal I_{B(a)_{t}} depend linearly on a. Thus the hyperplane defined by I_{A} +I_{B(a)_{t}} determines a pencil of hyperplanes, when a moves. For a = 0 the hyperplane I_{A} + I_{B(a)_{t}} corresponds to the intersection T (a, t) of the spans of v_{9}(A) and v_{9}(B(0)_{t}). Since, for a = 0, H_{t} = G_{t} is tangent to F_{t} at P_{1}, the intersection is v_{9}(P_{1}). When a goes to infinity and t goes to 0, the limit of T (a, t) corresponds to the limit of the points defined by the families {F_{t}}, {F′_{t}}, hence to T_{0}. For any intermediate a, we get as a limit for t = 0 a point corresponding to hyperplanes of the pencil, thus points of the line joining v_{9}(P_{1}) and T_{0}.

□

At the end of the analysis of Case 1,2,3 and the case of non-empty intersection, we see that all the forms for which there exists a second decomposition are limits of forms described in Case 1 of the previous section. Since the image of a Grassmannian in a projection is Zariski closed, we can summarize the result of the section in the following

Theorem 4.6. Let F be a ternary form of degree 9, with a Waring expression

F = λ_{1}L_{1}^{9} + ⋯ + λ_{18}L_{18}^{9}

and call A the set of points induced by {L_{1}, …, L_{18}}, so that A is a decomposition of F. Assume that A satisfies the following genericity properties of the beginning of Section 3:

(i) A is non-redundant; (ii) k_{4}(A) = 15; (iii) h_{A}(5) = 18.

Then there exists a second decomposition B of length ≤ 18 for F only if F belongs to a fixed hypersurface Θ in the span of v_{9}(A), which is the closed image of a birational projection of the Grassmannian of lines in ℙ^{9} in its Plücker embedding.

Therefore, if F does not belong to the hypersurface Θ, then the rank of F is 18 and A is the unique decomposition of F.

Remark 4.7. (The Panforte Challenge) The bad locus Θ is thus a hypersurface in a projective space ℙ^{17}.

As in [4], Remark 4.10, geometrically Θ is composed of points in which two folds of the secant variety Sec_{18} to the Veronese variety v_{9}(ℙ^{2}) cross each other. Thus, the points of Θ are singular points of the secant variety, which is a hypersurface of ℙ^{54}. It turns out that the secant variety has a singular locus of codimension 1.

We do not know an equation, or even the degree of Θ in ℙ^{17}, not even when the points of A are general. A computer based calculation did not provide an answer, in a reasonable time.

As a challenge, the second author offers a Panforte (traditional cake of Siena) to the first who will determine an equation for Θ, for a general choice of the decomposition A.

5. The algorithm

This section is devoted to the algorithm we developed based on the criterion of minimality explained in Remark 3.17, Remark 3.18, and Theorem 3.19. Our criterion is effective in the sense of [15]: it certifies that a given decomposition is minimal on a dense subset (in the Euclidean topology) of the set of forms of degree 9 having a decomposition of length 18.

5.1. The algorithm. Fix a finite set A = {P_{1}, …, P_{18}} ⊂ ℙ^{2} and a ternary nonic T in the linear span of ν_{9}(A), i.e.

T = ∑_{i=1}^{18} λ_{i}L_{i}^{9}

where, according to Remark 3.8, L_{i} is the linear form associated with P_{i} in ℙ^{2} = ℙ(R_{1}) and λ_{i} ∈ ℂ, for i ∈ {1, …, 18}. For any i, denote by v_{i} a representative vector for P_{i} and by (t_{0}, …, t_{54}) the coefficients of T in the standard monomial basis of degree 9 in 3 variables. In order to establish that the given T has exactly rank 18, proceed as follows. First, check that the tests

1) non-redundancy test: dim⟨v_{9}(v_{1}), …, v_{9}(v_{18})⟩ = 18

2) fourth Kruskal’s rank test: k_{4}(A) = 15

3) fifth Hilbert function test: h_{5}(A) = 18 provide positive answers. If so, construct the ideal I_{A} and its Hilbert-Burch matrix M as in (7). Add to the transpose of M the rows

(\begin{matrix}a_{1} & a_{2} & a_{3} & 0 \\ q_{1} & q_{2} & q_{3} & 0 \end{matrix})

so that we get the Hilbert-Burch matrix M′ of a hypothetical non-redundant decomposition B of T with ℓ(B) = 17, as in (23). By arguing as in the proof of Theorem 3.19 and by using the same notation, assume that a_{1} = 1 and q_{1} = 0 (the cases a_{2} = 1, q_{2} = 0 and a_{3} = 1, q_{3} = 0 are similar), take the matrix N_{1} whose columns yield a set of generators for (I_{A})_{9} + (I_{B(Q,G)})_{9}, for Q, G general and compute

d_{1} = max_{4≤i≤15} dim\{(a_{2}, …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{1} = \underline{0} ∩ a_{i} = 0\}.

Similarly, in the cases a_{2} = 1, q_{2} = 0 and a_{3} = 1, q_{3} = 0 compute, respectively,

d_{2} = max_{4≤i≤15} dim\{(a_{1}, a_{3} …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{2} = \underline{0} ∩ a_{i} = 0\}

and

d_{3} = max_{4≤i≤15} dim\{(a_{1}, a_{2}, …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{3} = \underline{0} ∩ a_{i} = 0\}.

Thus, if the also the next test is successful:

4) d_{1} = d_{2} = d_{3} = −1, i.e. dim f^{′−1}(T) = −1 then f^{′−1}(T) is empty and so T has rank 18.

The algorithm has been implemented in Macaulay2, over the finite field ℤ_{31991}. The detailed procedure is contained in the ancillary file nonics2.txt.

Some examples of ternary nonics of rank 18 and 17, with a non-redundant decomposition of length 18, are presented in the following subsection.

5.2. Examples. In Macaulay2, we generated a random collection of 18 points:

A = [v_{i}]_{i=1}^{18} = [\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 2 \\ −1 & 2 & 1 \\ 1 & 2 & 3 \\ 1 & −2 & 0 \\ 2 & 1 & 4 \\ 4 & 2 & −3 \\ 1 & 5 & 1 \\ 5 & 2 & 3 \\ 6 & 2 & 3 \\ 1 & 7 & 7 \\ 1 & 7 & 3 \\ 6 & 5 & 4 \\ −7 & 2 & 3 \\ 3 & 7 & 4 \\ 2 & −5 & 1 \\ 6 & 3 & −4 \\ −7 & 6 & 6 \end{matrix}] .

By abuse of notation, define A = {P_{1}, …, P_{18}} ⊂ ℙ^{2}, where P_{i} is the projective class of v_{i} and denote by L_{i} the linear form whose coefficients are given by v_{i}.

Test 1) and test 3) show, respectively, that dim⟨v_{9}(A)⟩ = rank([v_{9}(v_{i})]_{i=1}^{18}) = 18 and h_{A}(5) = rank([v_{5}(v_{i})]_{i=1}^{18}) = 18, as required. Moreover, all the 816 subsets of 15 columns of [ν_{4}(v_{i})]_{i=1}^{18} are of rank 15, so that k_{4}(A) = 15.

A case of rank 18. Let

T_{2} = ∑_{i=1}^{18} L_{i}^{9} =

= [4283x_{0}^{9} − 14212x_{0}^{8}x_{1} + 2365x_{0}^{7}x_{1}^{2} − 11335x_{0}^{6}x_{1}^{3} + 10354x_{0}^{5}x_{1}^{4} − 7342x_{0}^{4}x_{1}^{5} + 11432x_{0}^{3}x_{1}^{6}+ −15881x_{0}^{2}x_{1}^{7} −10204x_{0}x_{1}^{8} −663x_{1}^{9} −10837x_{0}^{8}x_{2} −6573x_{0}^{7}x_{1}x_{2} +6070x_{0}^{6}x_{1}^{2}x_{2} −12124x_{0}^{5}x_{1}^{3}x_{2}+ +8455x_{0}^{4}x_{1}^{4}x_{2} − 9097x_{0}^{3}x_{1}^{5}x_{2} + 200x_{0}^{2}x_{1}^{6}x_{2} + 11563x_{0}x_{1}^{7}x_{2} + 11173x_{1}^{8}x_{2} + 2810x_{0}^{7}x_{2}^{2}+ +5187x_{0}^{6}x_{1}x_{2}^{2} −1688x_{0}^{5}x_{1}^{2}x_{2}^{2} −3089x_{0}^{4}x_{1}^{3}x_{2}^{2} +8745x_{0}^{3}x_{1}^{4}x_{2}^{2} +12508x_{0}^{2}x_{1}^{5}x_{2}^{2} +151x_{0}x_{1}^{6}x_{2}^{2}+ +11119x_{1}^{7}x_{2}^{2} +11414x_{0}^{6}x_{2}^{3} +2714x_{0}^{5}x_{1}x_{2}^{3} +11939x_{0}^{4}x_{1}^{2}x_{2}^{3} +5024x_{0}^{3}x_{1}^{3}x_{2}^{3} +10884x_{0}^{2}x_{1}^{4}x_{2}^{3}+ +8404x_{0}x_{1}^{5}x_{2}^{3} +755x_{1}^{6}x_{2}^{3} +15891x_{0}^{5}x_{2}^{4} −1013x_{0}^{4}x_{1}x_{2}^{4} −11790x_{0}^{3}x_{1}^{2}x_{2}^{4} +14982x_{0}^{2}x_{1}^{3}x_{2}^{4}+ −8411x_{0}x_{1}^{4}x_{2}^{4} −5236x_{1}^{5}x_{2}^{4} +4416x_{0}^{4}x_{2}^{5} −11481x_{0}^{3}x_{1}x_{2}^{5} +14698x_{0}^{2}x_{1}^{2}x_{2}^{5} +5309x_{0}x_{1}^{3}x_{2}^{5}+ +11614x_{1}^{4}x_{2}^{5}−9777x_{0}^{3}x_{2}^{6}−2702x_{0}^{2}x_{1}x_{2}^{6}−5846x_{0}x_{1}^{2}x_{2}^{6}−10960x_{1}^{3}x_{2}^{6}−8430x_{0}^{2}x_{2}^{7}+7085x_{0}x_{1}x_{2}^{7}+ +12763x_{1}^{2}x_{2}^{7} − 14136x_{0}x_{2}^{8} − 9808x_{1}x_{2}^{8} + 9194x_{2}^{9}].

Since tests 1), 2) and 3) are successful, T_{2} is general enough, so that our criterion applies. Moreover, test 4) provides positive answer too. Therefore we conclude that A is minimal for T_{2}, i.e. T_{2} has rank 18 (ancillary nonics2.txt).

A case of lower rank. In the same span of v_{9}(A) one can find forms for which the decomposition A is non-redundant, yet there is another decomposition of length 17.

For instance, one can take

(λ_{1}, …, λ_{18}) = (10308, −9437, −13956, −12270, 2135, −4854, −2213, 1755, −13629,

7308, −8496, 2940, 11348, −12437, −6712, 4086, −823, −2818)

so that

T_{1} = ∑_{i=1}^{18} λ_{i}L_{i}^{9} =

= [9666x_{0}^{9} + 13004x_{0}^{8}x_{1} + 12463x_{0}^{7}x_{1}^{2} − 13235x_{0}^{6}x_{1}^{3} − 15442x_{0}^{5}x_{1}^{4} + 15509x_{0}^{4}x_{1}^{5} + −6311x_{0}^{3}x_{1}^{6}+ −2390x_{0}^{2}x_{1}^{7} + 547x_{0}x_{1}^{8} − 119x_{1}^{9} − 14916x_{0}^{8}x_{2} + 1822x_{0}^{7}x_{1}x_{2} − 8022x_{0}^{6}x_{1}^{2}x_{2} − 9386x_{0}^{5}x_{1}^{3}x_{2}+ −2742x_{0}^{4}x_{1}^{4}x_{2}+10541x_{0}^{3}x_{1}^{5}x_{2}+1156x_{0}^{2}x_{1}^{6}x_{2}−12023x_{0}x_{1}^{7}x_{2}+4417x_{1}^{8}x_{2}−11823x_{0}^{7}x_{3}^{2}−737x_{0}^{6}x_{1}x_{1}^{2}+ −7616x_{0}^{5}x_{1}^{2}x_{2}^{2}+11293x_{0}^{4}x_{1}^{3}x_{2}^{2}−8260x_{0}^{3}x_{1}^{4}x_{2}^{2}−9332x_{0}^{2}x_{1}^{5}x_{2}^{2}+7078x_{0}x_{1}^{6}x_{2}^{2}−4553x_{1}^{7}x_{2}^{2}−15941x_{0}^{6}x_{2}^{3}+ +4339x_{0}^{5}x_{1}x_{2}^{3} − 4251x_{0}^{4}x_{1}^{2}x_{2}^{3} + 9854x_{0}^{3}x_{1}^{3}x_{2}^{3} − 22x_{0}^{2}x_{1}^{4}x_{2}^{3} + 8408x_{0}x_{1}^{5}x_{2}^{3} + 11858x_{1}^{6}x_{2}^{3}+ −9161x_{0}^{5}x_{2}^{4}−9854x_{0}^{4}x_{1}x_{2}^{4}−13165x_{0}^{3}x_{1}^{2}x_{2}^{4}−2105x_{0}^{2}x_{1}^{3}x_{2}^{4}−8715x_{0}x_{1}^{4}x_{2}^{4}+390x_{1}^{5}x_{2}^{4}−9955x_{0}^{4}x_{2}^{5}+ −11013x_{0}^{3}x_{1}x_{2}^{5}−10651x_{0}^{2}x_{1}^{2}x_{2}^{5}−3850x_{0}x_{1}^{3}x_{2}^{5}+4029x_{1}^{4}x_{2}^{5}−11735x_{0}^{3}x_{2}^{6}−12427x_{0}^{2}x_{1}x_{2}^{6}+12255x_{0}x_{1}^{2}x_{2}^{6}+ −3686x_{1}^{3}x_{2}^{6} − 2271x_{0}^{2}x_{2}^{7} + 5939x_{0}x_{1}x_{2}^{7} − 3402x_{1}^{2}x_{2}^{7} + 13298x_{0}x_{2}^{8} + 6455x_{1}x_{2}^{8} + x_{2}^{9}].

T_{1} has been obtained as the point of intersection of the span of v_{9}(A) with v_{9}(B), where B is a set of 17 points, linked to A with two curves Q_{0} and G_{0} of degree 5 and 7 in I_{A}. In particular, Q (resp. G_{0}) is defined by the determinant of the 4 × 4 matrix obtained by adding the row R_{1} = (1, 10399, 13534, 0) (resp. the row R_{2} = (0, −633x_{0}^{2} − 11455x_{0}x_{1} + 2134x_{0}x_{2} + 11038x_{1}^{2} − 8888x_{1}x_{2} − 588x_{2}^{2}, 1927x_{0}^{2} + 4114x_{0}x_{1} + 11328x_{0}x_{2} + 13814x_{1}^{2} − 10664x_{1}x_{2} − 1749x_{2}^{2}, 0)) to the transpose of the Hilbert-Burch matrix M of A, i.e.

(a_{1}, a_{2}, …, a_{15}) = (1, 10399, 13534, −633, −11455, 2134, 11038, −8888,

−588, 1927, 4114, 11328, 13814, −10664, −1749).

Notice that, according to the notation of test 4), in this case d_{1} = d_{2} = d_{3} = 0 and

max_{4≤i≤15} deg\{(a_{2}, …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{1} = \underline{0} ∩ a_{i} = 0\} = 0,

max_{4≤i≤15} deg\{(a_{1}, a_{3} …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{2} = \underline{0} ∩ a_{i} = 0\} = 0,

max_{4≤i≤15} deg\{(a_{1}, a_{2}, …, a_{15}) ∈ ℂ^{14} | (t_{0}, …, t_{54}) · N_{3} = \underline{0} ∩ a_{i} = 0\} = 0.

Therefore, the rank of T_{1} is at most 17. Indeed, the rank is exactly 17, by Proposition 3.19 and its proof (ancillary file nonics1.txt). In order to get coordinates for the points in B, one needs to solve the polynomial

system given by the maximal minors of the 5 × 4 matrix (\begin{matrix}R_{1} \\ R_{2} \\ M^{t} \end{matrix}). This can be

achieved with Macaulay2 software system: indeed, by computing the eigenvalues and eigenvectors of certain companion matrices (ancillary file nonics4.txt) one can find the following representative vectors in ℂ^{2} for the points in B

(1, 62.6659, 29.7378)
(1, 13.368 + 38.1825 i, −19.099 + 7.53788 i)
(1, 13.368 − 38.1825 i, −19.099 − 7.53788 i)
(1, 35.333, 40.797)
(1, 14.7061, 27.8538)
(1, 10.7119, 4.95399)
(1, −0.796312, 2.23381)
(1, 1.06064 + 0.13583 i, 1.62951 − 0.563286 i)
(1, 1.06064 − 0.13583 i, 1.62951 + 0.563286 i)
(1, 0.737271, −0.0631582)
(1, −0.245331, −0.76262)
(1, −0.187307, 0.100519)
(1, −0.0870499, −0.126324)
(1, 0.00104432, 0.00164595)
(1, 0.306581 + 0.0182712 i, −0.877193 − 0.031211 i)
(1, 0.306581 − 0.0182712 i, −0.877193 + 0.031211 i)
(1, 0.390447, 0.585521).

Remark 5.1. In principle one can try to use our analysis also to determine the identifiability of a form T. In this case one starts by adding, in the previous algorithm, the following test:

5) check that the solution set of the polynomial system (t_{0}, …, t_{54})·A_{2} = 0_{1×17} introduced in (21) has dimension −1.

If the answer is negative, we can conclude that there are other solutions of the system (21), so one can guess that T has other decompositions. Unfortunately, if the answer is positive, one cannot immediately conclude the identifiability of T. Namely, test 5) checks that there exists no other set B of length 18, linked to A by a complete intersection of two sextics F, F′, such that T also sits in the span of v_{9}(B), but only when A ∩ B = ∅ and F, F′ have no common components. However, we know that there are limit cases (the case A∩B non-empty, or the case 3 of Proposition 3.4) in which B exists but the intersection F ∩ F′ is not proper. To exclude these cases one needs new ad hoc tests, which can be constructed in principle, but then the procedure becomes quite laborious. We believe that the natural way to prove identifiability is to produce equations for the locus Θ described in remark 4.7, and test the vanishing of the equation for T.

An unidentifiable case. Of course, one can use the construction of Case 1 of Proposition 3.4 to produce many examples of unidentifiable forms. Let

(λ_{1}, …, λ_{18}) = (5864, 9496, 11539, 1233, −13315, −14222, 10709, −5067, 13797,
13169, −10531, 1592, 12589, 1728, −4725, −4784, −8696, 7515)

and let

T_{3} = ∑_{i=1}^{18} a_{i}L_{i}^{9} =

= [11096x_{0}^{9} + 14876x_{0}^{8}x_{1} + 14398x_{0}^{7}x_{1}^{2} − 11088x_{0}^{6}x_{1}^{3} − 3441x_{0}^{5}x_{1}^{4} + 11138x_{0}^{4}x_{1}^{5} − 3819x_{0}^{3}x_{1}^{6}+ +6626x_{0}^{2}x_{1}^{7} −9525x_{0}x_{1}^{8} −9028x_{1}^{9} +11951x_{0}^{8}x_{2} −14433x_{0}^{7}x_{1}x_{2} +15878x_{0}^{6}x_{1}^{2}x_{2} +3683x_{0}^{5}x_{1}^{3}x_{2}+ +12902x_{0}^{4}x_{1}^{4}x_{2} + 9968x_{0}^{3}x_{1}^{5}x_{2} + 1167x_{0}^{2}x_{1}^{6}x_{2} − 1011x_{0}x_{1}^{7}x_{2} + 11114x_{1}^{8}x_{2} − 1174x_{0}^{7}x_{2}^{2}+ −10039x_{0}^{6}x_{1}x_{2}^{2} + 15571x_{0}^{5}x_{1}^{2}x_{2}^{2} − 1797x_{0}^{4}x_{1}^{3}x_{2}^{2} + 7799x_{0}^{3}x_{1}^{4}x_{2}^{2} + 3353x_{0}^{2}x_{1}^{5}x_{2}^{2} − 9008x_{0}x_{1}^{6}x_{2}^{2}+ +7892x_{1}^{7}x_{2}^{2} + 8863x_{0}^{6}x_{2}^{3} + 12538x_{0}^{5}x_{1}x_{2}3 + 4156x_{0}^{4}x_{1}^{2}x_{2}^{3} + 5014x_{0}^{3}x_{1}^{3}x_{2}^{3} + 15217x_{0}^{2}x_{1}^{4}x_{2}^{3}+ +10693x_{0}x_{1}^{5}x_{2}^{3} − 14254x_{1}^{6}x_{2}^{3} − 12480x_{0}^{5}x_{2}^{4} + 15094x_{0}^{4}x_{1}x_{2}^{4} + 11796x_{0}^{3}x_{1}^{2}x_{2}^{4} − 11496x_{0}^{2}x_{1}^{3}x_{2}^{4}+ −3087x_{0}x_{1}^{4}x_{2}^{4} − 7767x_{1}^{5}x_{2}^{4} + 1751x_{0}^{4}x_{2}^{5} + 9059x_{0}^{3}x_{1}x_{2}^{5} + 14238x_{0}^{2}x_{1}^{2}x_{2}^{5} − 640x_{0}x_{1}^{3}x_{2}^{5}+ −14792x_{1}^{4}x_{2}^{5} − 14262x_{0}^{3}x_{2}^{6} − 6895x_{0}^{2}x_{1}x_{2}^{6} + 13550x_{0}x_{1}^{2}x_{2}^{6} + 7631x_{1}^{3}x_{2}^{6} + 9523x_{0}^{2}x_{2}^{7}+ −2161x_{0}x_{1}x_{2}^{7} − 3449x_{1}^{2}x_{2}^{7} − 7220x_{0}x_{2}^{8} + 395x_{1}x_{2}^{8} + x_{2}^{9}].

As in the previous cases, tests 1), 2), and 3) are successful for T_{3}. On the other hand, in this case our computations show that the polynomial system (t_{0}, …, t_{54}) · A_{2} = 0_{1×17} defined in (21) has a solution set of dimension 0 and degree 1. In particular, T_{3} admits at least two decompositions of length 18, A and another set B. Indeed, T_{3} = f(F, F′), where F (resp. F′) is the ternary form of degree 6 defined by the determinant of the 4 × 4 matrix obtained by adding the row (x_{0} + 14307x_{1} + 13416x_{2}, 11657x_{0} + 9248x_{1} + 8324x_{2}, −13193x_{0} − 1403x_{1} + 12171x_{2}, 0) (resp. the row (7694x_{1} +12549x_{2}, −12983x_{0} +538x_{1} +11728x_{2}, 743x_{0} −12966x_{1} +12870x_{2}, 1)) to the transpose of the Hilbert-Burch matrix M of A. Therefore, T_{3} is computed by two non-redundant finite sets of length 18: A and its residual set, B = B_{F,F′}, in the complete intersection (6, 6) given by F and F′. According to our theory, test 5) fails for T_{2}. Notice that test 4) is successful for T_{3}, which means that T_{3} has rank 18 (by the procedure illustrated in the ancillary file nonics3.txt). A non-trivial solution for the previous system determines the Hilbert-Burch matrix for B. Then, coordinates for the points of B can be found by solving (e.g. with the software Macaulay2) the polynomial system given by the maximal minors of its Hilbert-Burch matrix. Indeed, by computing the eigenvalues and eigenvectors of certain companion matrices (for more details on the procedure see the ancillary file nonics5.txt), we get that the following representative vectors in ℂ^{3} for the points of B:

(1, −7.96881 + 4.74847 i, 29.737 − 8.31447 + 5.39065 i)
(1, −7.96881 − 4.74847 i, −8.31447 − 5.39065 i)
(1, −8.88473, −21.3598)
(1, −3.19251, −1.98613)
(1, 2.29572 + 0.339361 i, 2.08576 + 2.10835 i)
(1, 2.29572 − 0.339361 i, 2.08576 − 2.10835 i)
(1, 1.1725, 0.789914)
(1, −0.662147 + 0.268568 i, 1.41128 + 0.060661 i)
(1, −0.662147 − 0.268568 i, 1.41128 − 0.060661 i)
(1, −0.676455 + 0.162048 i, 0.269336 − 0.242414 i)
(1, −0.676455 − 0.162048 i, 0.269336 + 0.242414 i)
(1, 0.299266 + 0.586034 i, 0.441543 + 0.153418 i)
(1, 0.299266 − 0.586034 i, 0.441543 − 0.153418 i)
(1, −0.0000365176, −0.00000906466)
(1, 0.209511, 0.479921)
(1, 0.483517 + 0.0585949 i, −0.520821 − 0.078578 i)
(1, 0.483517 − 0.0585949 i, −0.520821 + 0.078578 i)
(1, 0.503511, 0.553533).

Remark 5.2. At the end, let us compare our methods with other algorithms for the computation of the rank of forms F. In our setting, we need to start with some Waring decomposition A (in the specific case above, with a decomposition of length 18) which is general in a mild, computable sense, i.e. it satisfies conditions i), ii), iii) at the beginning of Sect. 3. Then we determine if the decomposition is minimal and, if not, we can use A to determine a minimal decomposition B (which can only be of length 17). The time in which our algorithm returns an answer can be found in the ancillary files. A recent algorithm that computes a decomposition of general forms F is described in [11]. It turns out that, since we know A, the linear systems we deal with have a number of parameters considerably smaller than the linear systems involved in Algorithm 1 of [11]. Consequently, our algorithms run much faster, and provide an answer for values of the length of the decomposition which cannot be handled presently with the general algorithm of [11] (as the first authors of [11] confirmed in a private communication).

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Dipartimento di Ingegneria dell’Informazione e Scienze Matematiche, Università di Siena, Italy

   Email address: elena.angelini@unisi.it

Dipartimento di Ingegneria dell’Informazione e Scienze Matematiche, Università di Siena, Italy

   Email address: luca.chiantini@unisi.it