Identifiability of rank-3 tensors

Rank-2 and rank-3 tensors are almost all identifiable with only few exceptions. We classify them all together with the dimensions and the structures of all the sets evincing the rank.


Introduction
Identifiability of tensors is one of the most active research areas both in pure mathematics and in applications. The core of the problem is being able to understand if a given tensor T ∈ C n1+1 ⊗ · · · ⊗ C n k +1 can be decomposed in a unique way as a sum of pure tensors: with v j,i ∈ C nj +1 , for j = 1, . . . , k. Of course the minimum r realizing the above expression is a crucial value and it is called the rank of T .
From the pure mathematical point of view, being able to understand if a tensor is identifiable is a very elegant problem that goes back to Kruskal [45] and finds more modern contributions with the language of Algebraic Geometry and Multilinear Algebra in eg. [31,32,34,36,35,22,23,42,50,26,4,5,12,11,18]. Except for very few contributions [45,38,46] which work for certain specific classes of given tensors, all the others regards the identifiability of generic tenors of certain rank. From the computational point of view, as far as we know, the unique algorithm dealing with the identifiability of any given tensor is a numerical one developed in Bertini [14] in [20].
Dealing with generic tensors of given rank r brings the problem into the setting of secant varieties of Segre varieties (cf. Definition 1.6), namely the closure (either Zariski or Euclidean closures can be used for this definition if working over C) of the set of tensors of rank smaller or equal than r. Knowing if a generic tensor of certain rank is identifiable gives an indication regarding the behaviour of specific tensors of the same rank. Namely, the dimension of the set S(Y, T ) of rank-1 tensors computing the rank of a specific tensor T (cf. Definition 1.4) can only be bigger or equal than the dimension of S(Y, q) where q is a generic tensor of rank equal to the rank of T (this will be explained in Remark 3.2 for the specific case of rank-3 tensors T ∈ (C 2 ) ⊗4 , but it is a well known general fact for which we refer [43, Cap II, Ex 3.22, part (b)]). Since the cases in which generic tensors of fixed rank are not-identifiable are rare (cf. eg. [18,26,42,47,22,27,28,29,34]), the knowledge of generic tensors' behaviour doesn't help all the applied problems where the ken of a specific tensor modeling certain precise samples is required.
In the present manuscript we present a systematic study of the identifiability of a given tensor starting with those of ranks 2 and 3. We give a complete classification of these first cases: we describe the structures and the dimensions of all the sets evincing the rank. In terms of generic tensors of rank either 2 or 3, everything was already well known form [1,8,27,29,34,36,45,37]. What it was missing was the complete classification for all the tensors of those ranks.
The paper is organized as follows. After the preliminary Section 1 where we introduce the notation and the main ingredients needed for the set up, we can immediately show the identifiability of rank-2 tensors in Section 2. In Section 3 we explain in details all the examples where the nonidentifiability of a rank-3 tensor arises. In Sections 5 and 6 we show that all the examples of the previous section are the only possible exceptions to non-identifiability of a rank-3 tensor. Section 7 is actually devoted to collect all the information needed (but actually already proved at that stage) to conclude the proof of our main Theorem 7.1. Notation 1.3. We denote by Y the multiprojective space Y := P n1 × · · · × P n k and by X the image of Y via Segre embedding, i.e. X = ν(Y ). We denote the projection on the i-th factor as The space corresponding to forget the i-th factor in the multiprojective space Y is denoted by Y i : The projection on all the factors of Y but the i-th one is denoted with η i : Obviously all fibers of η i are isomorphic to P ni . Definition 1.4. For any q ∈ P N , S(Y, q) denotes the set of all subsets A ⊂ Y such that ♯(A) = r X (q) and q ∈ ν(A) and we will say that if A ∈ S(Y, q), then A evinces the rank of q. Moreover we say that q ∈ X is identifiable if ♯S(Y, q) = 1. Notation 1.5. Sometimes we will also use the following multi-index notations: for 1 ≤ i ≤ k, ε i = (0, . . . , 0, 1, 0, . . . , 0), where the only 1 is in the i-th place andε i which is a k-uple with all one's but the i-th place, which is filled by 0, i.e. ε i = (1, . . . , 1, 0, 1, . . . , 1). Definition 1.6. The r-th secant variety of X is σ r (X) := p,...,pr∈X p 1 , . . . , p r where the closure is the the Zariski closure. The open part of the r-th secant variety of X is sometime denoted as σ 0 r (X) := {q ∈ X | r X (q) = r}. If dim σ r (X) < min{rn + r − 1, dim X }, the variety X is said to be r-defective, otherwise X is r-regular. If X is r-defective, the difference δ r = min{rn + r − 1, dim X } − dim σ r (X) is called the r-th secant defect of X.
We will often use the so called Concision/Autarky property (cf. [ Lemma 1.7 (Concision/Autarky). For any q ∈ P(V 1 ⊗ · · · ⊗ V k ), there is a unique minimal multiprojective space Y ′ ≃ P n ′ 1 × · · · × P n ′ k ⊆ Y ≃ P n1 × · · · × P n k with n ′ i ≤ n i , i = 1, . . . , k such that S(Y, q) = S(Y ′ , q). Definition 1.8. (concise Segre) Given a point q ∈ P N , we will call concise Segre the variety Remark 1.9. The minimal Y ′ defining the concise Segre of a point q can be obtained as follows. Fix any A ∈ S(Y, q), set A i := π i (A) ⊂ P ni , i = 1, . . . , k, where the π i 's are the projections on the i-th factor of Notation 1.3. Each A i ⊆ P ni is a well-defined projective subspace of dimension at most min{n i , r X (q) − 1}. By Concision/Autarky we have Y ′ = k i=1 A i . In particular q does not depend on the i-th factor of Y if and only if for one A ∈ S(Y, q) the set π i (A) is a single point. Remark 1.10. Let q ∈ P N and consider A ∈ S(Y, q). We claim that there is no line L ⊂ X such that ♯(L ∩ ν(A)) ≥ 2. Obviously if ♯(L ∩ ν(A)) > 2 we would have at least 3 points that evince the rank of q on a line, which is a contradiction with the linearly independence property that sets in S(Y, q) have. So assume that there exists a line L ⊂ X such that ♯(L ∩ ν(A)) = 2; let u, v ∈ A be the preimages of those points, i.e. u = v and ν(u), ν(v) = L. Then r X (q) > 2 because if r X (q) = 2 then we would have q ∈ L ⊂ X, so the rank of q will be 1. Let E = A \ {u, v}. Then we will have that q ∈ ν(E) ∪ L , so we can find a point o ∈ L such that q ∈ ν(E) ∪ {o} , which would imply r X (q) < ♯A.

Identifiability on the 2-nd secant variety
In this section we study and completely determine the identifiability of points on the second secant variety of a Segre variety.
By Remark 1.9, the concise Segre of a border rank-2 tensor q is X q = ν P 1 i ×k . Therefore for the rest of this section we will focus our attention to Segre varieties of products of P 1 's.
Remark 2.1. If the concise Segre X q of a tensor q ∈ σ 2 (X) is a ν(P 1 × P 1 ), then σ 2 (X q ) parameterizes the 2 × 2 matrices of rank at most 2 for which it is trivial to see that they can be written as sum of two rank-1 matrices in an infinite number of ways.
For the rest of this section we will therefore focus on Segre varieties of (P 1 ) ×k with k ≥ 3.
Definition 2.2. The variety τ (X) is the tangent developable of a projective variety X, i.e. τ (X) is defined by the union of all tangent spaces to X.
Recall that a tensor q ∈ τ (X) \ X has rank equal to 2 if and only if the conciesd Segre X q of q is a two-factors Segre, moreover it is not-identifiable for any number of factors (cf. eg. [8,Remark 3]). Proposition 2.3. Let q ∈ σ 0 2 (X). Then |S(Y, q)| > 1 if and only if the concise Segre X q of q is X q = ν(P 1 × P 1 ).
Proof. We only need to check the case of k ≥ 4 since k = 2, 3 are classically known. The case of matrix is obviously not-identifiable (cf. Remark 2.1), while the identifiabily in the case k = 3 is classically attributed to Segre and it is also among the so called Kruskal range (cf. [45], [36,Thm. 4.6], [34,Thm. 1.2]), see also [37, line 7 of page 484]. We assume therefore that k ≥ 4.
Since X is cut out by quadrics, then if a line L ⊂ P N is such that deg(L ∩ X) > 2 then L ⊂ X and the points of L have X-rank 1. Let A, B ∈ S(Y, q), either A = B or A ∩ B = {q}. In fact, in the first case A = B since r X (q) = 2 and therefore A is not containd in X and X is cut out by quadrics. In the second case A = B. We can therefore assume that A, B ∈ S(Y, q) are two disjoint sets: . Since a = a ′ , we may assume that at least one of their coordinates is different. Actually we can assume that all the a i = a ′ i , otherwise, by the concision property, one could consider one factor less. The same considerations hold for B. Now suppose that there exists an Let η k , ν k , and X k be as in Notation 1.3. Letq = (q 1 , . . . , q k−1 ) be the projection η k (q), then η k (A) = η k (B) and ∅ = ν k (η k (A)) ∩ ν k (η k (B)) ⊃ {q} because {q} ⊂ ν(A) ∩ ν(B) . So r X k (q) = 2 and |S(Y k ,q)| ≥ 2, which is a contradiction because X k is a concise Segre of k − 1 factor (where k > 3) and a point of it cannot have more than a decomposition. Thus for all i =, 1 . . . , k Without loss of generality assume that a 1 = b 1 and a ′ 1 = b ′ 1 , moreover up to permutation there exists an index e ∈ {1, . . . , k − 1} such that b i = a i and consequently b ′ i = a ′ i for 1 ≤ i ≤ e and b i = a ′ i and b ′ i = a i for e + 1 ≤ i ≤ k. Eventually by exchanging the role of the first e elements with the others, we have that k − e ≥ 2 because by assumption k ≥ 4. Let H ∈ |O Y (0, . . . , 0, 1)| be the only element containing a ′ , Corollary 2.4. Let q be any rank-2 tensor. If q is not-identifiable, then there is a bijection between S(Y, q) and P 2 \ L, where L ⊂ P 2 is a projective line, q ∈ τ (X) and L parametrizes the set of all degree 2 connected subschemes V of Y such that q ∈ ν(V ) .
Proof. It suffices to work with a Segre variety of 2 factors only because by Proposition 2.3 it is the only not-identifiable case in rank-2. Thus X ⊂ P 3 is a quadric surface. Denote by H q ⊂ P 3 the polar plane of X with respect to q. Since q / ∈ X we have that q / ∈ H q and the intersection ∈ X the line given by o, q is not tangent to X and when considering the intersection o, q ∩ X, it is given by two points Consider Π q = {lines L ⊂ P 3 passing through q} ∼ = P 2 and consider the following isomorphism ϕ :

Examples of not-identifiable rank-tensors
The purpose of this section is to explain in details the phenomena behind the not-identifiable rank-3 tensors. In the main Theorem 7.1 they will turn out to be the unique cases of not-identifiability for a rank-3 tensor. From now on we always consider q ∈ P N such that r X (q) = 3, therefore, by Remark 1.9, we may assume that q is an order-k tensor with at most 3 entries in each mode, i.e. the concise Segre of q is X q = ν(P n1 × · · · × P n k ), with n 1 , . . . , n k ∈ {1, 2}.
First of all let us remark that the matrix case is highly not-identifiable even for the rank-3 case.
In the following remark we explain the behaviour on σ 3 ((P 1 ) ×4 ).

Remark 3.2.
It has been shown in [1] (cf. also [27,29]) that the third secant variety of a Segre variety X is never defective unless either X = ν( The case in which q is a rank-3 tensor in ν(P 1 ×P 1 ×P a ) with a ≥ 3 corresponds to a not-concise tensor (cf. Remark 1.9) therefore it won't play a role in our further discussion.
The case in which X = ν(P 1 ×P 1 ×P 1 ×P 1 ) and q ∈ X can also be easily handled. The fact that dim(σ 3 (X)) is strictly smaller than the expected dimension proves that the generic element of σ 3 (X) has an infinite number of rank-3 decompositions. By definition of dimension there is no element of σ 3 (X) s.t. its tangent space has dimension equal to the expected one: dim(T q (σ 3 (X))) ≤ dim σ 3 (X) for all q ∈ σ 3 (X). This does not exclude the existence of certain special rank-3 tensors q such that Remark 3.3. Let Y be a multiprojective space with at least two factors where at least one of them is of projective dimension 2. By relabeling, if necessary, we can assume that the first factor is a P 2 . Let q ∈ σ 0 3 (ν(Y )), with ν(Y ) the concise Segre of q and let A, B ∈ S(Y, q) be two disjoint subsets evincing the rank of q. By Autarky π 1 (A) = π 1 (B) = P 2 ; moreover when considering the restrictions of the projections π 1|A and π 1|B to the subsets A and B respectively, they are both injective and both π 1 (A) and π 1 (B) contains linearly independent points.
. The classification of the dimensions of secant varieties of such a Segre-Veronese can be found in [15,33,17,10].
Proof. First of all remark that, for i = 1, 2, With an analogous computation of the one in Remark 3.4 one sees that dim ν(G) = 8 and σ 2 (ν(G i )) = ν(G i ) . It remains to show that ν(G) = J , where J denotes the join of σ 2 (ν(G i )) and ν(G j ) with {i, j} = {1, 2}. We remark that since σ 2 (ν(G)) = P 5 , then J = Join(P 5 , ν(G 2 )). By Terracini's Lemma for joins (cf. [2, Corollary is equal to the degree-2 part of the ideal defining a scheme T ⊂ P 3 such that T = 2q 1 + 2q 2 + 2q 3 + 2q 4 + P 1 where the q i 's are generic points of P 3 and with 2q i we denote the double fat point supported at q i . Example 3.6. Take Y = P 2 × P 1 × P 1 , consider the Segre embedding on the last two factors and take a hyperplane section which intersects ν(P 1 × P 1 ) in a conic C, then take a point q ∈ ν(P 2 × C) . Such a construction is equivalent to consider an irreducible divisor G ∈ |O Y (0, 1, 1)|, so G ∼ = P 2 × P 1 embedded via O(1, 2), then dim σ 2 (ν(G)) = 7, thus σ 2 (ν(G)) ν(G) ≃ P 8 . As a direct consequence we get that a general point q ∈ ν(G) has ν(G)-rank 3 and it is not-identifiable because of the not-identifiability of the points on C and by [43, Cap II, Ex 3.22, part (b)]. Thus dim(S(G, q)) = 3.
The following example is in the same setting of the previous one, but in this case we deal with a reducible conic and in such a case we get a 4-dimensional family of solutions.
has rank 3 and for the subsets evincing its rank we have a 4-dimensional family of sets A such that On the other hand, by looking at q as an element of J 2 , we get the existence of a 4-dimensional family of sets B such that Proof. Call S := A ∪ B, by Remark 3.3, both π 1|A and π 1|B are injective and both π 1 (A) and π 1 (B) are sets containing linearly independent points. So . The same holds exchanging the roles of A and B, thus Assume G is irreducible, then B contains three linearly independent points on G, thus they are uniquely determined by G.
Assume G is reducible, i.e. G = G 1 ∪ G 2 , with G 1 ∈ |O Y (0, 1, 0)| and G 2 ∈ |O Y (0, 0, 1)|. Remark that, by Autarky, it does not exist any E ∈ S(Y, q) which is all contained in G i , for i = 1, 2. Without loss of generality, we may assume that two general points of E lies in G 1 ; then the three points of E are uniquely determined by a reducible conic, i.e. by the reducible element G = G 1 ∪ G 2 that contains them.
Item 3 will be a consequence of item 2, in fact if the structure of the elements on S(Y, q) is of type A ∪ {p} with A ∈ S(Y ′ , q ′ ), then Autarky and the fact that Y is the minimal multiprojective space containing A ∪ {p} will imply that ν(Y ) is the concise Segre of q. So let us prove item 2.
Let  1, 1, 0)) ≥ 3. This must be true for all A ∈ S(Y ′ , q ′ ) and hence we have From now on suppose p / ∈ E. As above we may assume The second possibility is excluded, because ♯(π 1 (A)) = ♯(π 2 (A)) = 2 for any A ∈ S(Y ′ , q ′ ). The first possibility is excluded taking instead of A another general A 1 ∈ S(Y ′ , q ′ ). Now assume S ⊂ G. We get A ⊂ G. This is rule out taking another A ∈ S(Y ′ , q ′ ) since a general a ∈ Y ′ is contained in some B ∈ S(Y ′ , q ′ ). Thus we would have that Y ′ ⊂ G which is a contradiction. Consider the following linear projection ℓ : . We identify P r ′ with the target projective space of Y k . Since (Y ′ ∪ {p}) ∩ M = ∅, ℓ is well-defined on the Segre image of Y k and it acts as the composition of η k and the Segre embedding. By the inductive assumption S(Y k , ℓ(q)) = (D) [Case k ≥ 3, n 1 = 2, n 1 + · · · + n k ≥ 5] If only one of the factors is a P 2 we use Step (A) as base of the induction and then we construct a projection similar to the one used in (C). Now assume also n 2 = 2, then we must have k ≥ 3. The base case is done in the following and the inductive step is made by considering a projection similar to the one used in (C).

Lemmas
In this section we collect the basic lemmas that we will need all along the proof of the main theorem of the present paper, Theorem 7.1.
The following two lemmas describe two very basic properties that two different sets A and B evincing the rank of the same rank-3 point q have to satisfy. Proof. Assume the existence of such a rank-3 tensor q with 2 distinct decompositions A and B s.t. ♯(A ∪ B) ≥ 5. The plane ν(S) contains at least five not-collinear points. Note that ν(S) ⊆ X, otherwise also q ∈ X which contradicts r X (q) = 3. So ν(S) ∩ X is a conic and r X (q) ≤ 2 either if it is reduced or not, which is an absurd.  In the plane ν(A) we have two different lines: ν(E) and ν(u), ν(v) , which mutually intersect in at most a point q ′ . Remark that q ′ / ∈ X because otherwise the line ν(E) would have at least 3 points of rank 1 and so we would have ν(E) ⊂ X, contradicting Remark 1.10. So r X (q ′ ) = 2 and ♯S(Y, q ′ ) ≥ 2, by Proposition 2.1 we get that actually q ′ ∈ ν(Y ′ ) , where Y ′ = P 1 × P 1 . But also E, {u, v} ⊂ Y ′ , so q ∈ ν(Y ′ ) , which contradicts the fact that q has rank 3.
An immediate corollary of Lemma 4.2 is the following. Corollary 4.3. If q is a rank-3 tensor and A and B are two distinct sets evincing its rank, then the cardinality of A ∪ B can only be either 5 or 6.
This corollary turns out to be extremely useful for the proof of our main result, Theorem 7.1. We will be allowed to focus only on the structure of not-identifiable points of rank-3 with at least two decompositions A and B as in Corollary 4.3. This is the reason why we will study separately the case ♯A ∪ B = 5 in Section 5 form the case ♯A ∪ B = 6 in Section 6.
Another very useful behaviour that needs to be understood in order to study the identifiability of rank-3 tensors, is the structure of the not-independent sets of at most 3 rank-1 tensors. This is what is described by the following lemma.
Lemma 4.4. A set of points E ⊂ Y ≃ P n1 × · · · × P n k of cardinality at most 3 does not impose independent conditions to multilinear forms over Y i ≃ P n1 × · · · ×P ni × · · · × P n k , i = 1, . . . , k, (i.e. h 1 (I E (ε i )) > 0) if and only if one of the following cases occurs: Proof. The fact that both items 1. and 2. imply that h 1 (I E (ε i )) > 0 is obvious. Let us describe the other implication.
is not a very ample line bundle. Therefore the restriction η i|E of η i to the finite set E may be either injective or not injective.
In the latter case one immediately gets that h 1 (I E (ε i )) > 0. Moreover if η i|E is not injective it means that there are 2 distinct points of E, say u and v which are mapped by η i onto the same point, i.e. we are in item 2. of this lemma. Now assume that η i|E is injective (i.e. we are not in item 2.). This implies that ♯E = ♯η i (E). We have by hypothesis that h 1 (I E (ε i )) > 0. Since by definition h 1 (I E (ε i )) = h 1 (Y i , I ηi(E) (1, . . . , 1)) we have that η i (E) does not impose independent conditions to the multilinear forms over Y i , therefore ♯(η i (E)) ≥ 3 which clearly implies that ♯(η i (E)) = 3 since by hypothesis the cardinality of E is at most 3. Now η i (E) is a set of 3 distinct points on Y i which does not impose independent conditions to the multilinear forms over Y i , and O Yi (1, . . . , 1) is very ample, therefore the 3 points of η i (E) must be mapped to collinear points by the Segre embedding ν i of Y i . Hence, by the structure of the Segre variety ν i (Y i ), we get that ν i (η i (E)) ⊆ ν i (Y i ) and there is j ∈ {1, . . . , k} \ {i} such that ♯(π h (η i (E))) = 1 for all h / ∈ {i, j}. Since h = i, we have π h (η i (E)) = π h (E).

Two different solutions with one common point
We have seen in Corollary 4.3 that if a rank-3 tensor q is not-identifiable and A, B are two sets of points on the Segre variety computing its rank, then ♯A ∪ B can only be either 5 or 6. This section is fully devoted to the case in which ♯A ∪ B = 5, i.e. A and B share only one point and call it p: The matrix case is well known, therefore we will always assume that q is an order-k ≥ 3 tensor, i.e. q ∈ ν(Y ) with Y = k i=1 P ni and k ≥ 3. We will study separately the cases in which: • Y contains at least one factor of projective dimension 2 and all the others of dimension either 1 or 2 (Proposition 5.1); • Y is a product of P 1 's only (see Proposition 5.2).
Proposition 5.1. Let Y be the multiprojective space with at least 3 factors and at least one them of projective dimension 2, i.e. Y = P 2 × P n2 × · · · × P n k with n i ∈ {1, 2} for i = 1, . . . , k and k ≥ 3. The divisor M contains A ′ by definition so ♯(S \ S ∩ M ) ≤ 3, moreover, if we define Y 1 := P n2 × · · · × P n k with n i = 1, 2 for i = 2, . . . , k, we have that O Y1 (1, . . . , 1) is very ample, therefore we can apply Lemma 4.4 and say that one of the following occurs:
The case k = 4 is covered by Remark 3.2. Assume k > 4 and write Y = k i=1 P 1 i . Let S = A ∪ B as in (5.1). We build a recursive set of divisors in order to being able to cover the whole set S as follows. Let o i ∈ P 1 i , i = 2, 3, 4 be such that: Since h 1 (I S\S∩M3 (ε 3 )) > 0 (again by [13, Lemma 2.5]), we get that either there are w, z ∈ S \ S ∩ M 3 such that w = z, π i (w) = π i (z) for all i = 3 or ν 4 (η 4 (S ∩ M 4 )) (remind Notation 1.3) is made by 3 collinear points, say with a line corresponding to the i-th factor. The latter case cannot arise because S does not depend only on the third, fourth and i-th factor of Y . Thus there exist w, z ∈ S \ S ∩ M 3 such that w = z, π i (w) = π i (z) ∀i = 3.

Two disjoint solutions
We have seen in Corollary 4.3 that if a rank-3 tensor q is not-identifiable and A, B are two sets of points on the Segre variety computing its rank, then ♯A ∪ B can only be either 5 or 6. This section is fully devoted to the case in which ♯A ∪ B = 6, i.e. A and B are disjoint: First of all let us show that if q is a rank-3 tensor whose concise Segre ν(Y ) has at least two factors of projective dimension 2, it never happens that in S(Y, q) there are two disjoint sets.
Proof. The proof is by absurd: assume that there exist A, The proof of this claim can be repeated verbatim for all the other projections with only one caution that we will highlight in the sequel. Therefore, by repeating the argument for all the projections, we will get that π j (a i ) = π j (b i ) for i = 1, 2, 3 and for j = 1, . . . , k which is a contradiction with A and B being distinct. This will conclude the proof.
Proof of the Claim 6.2.1. Take a general hyperplane J 3 ⊂ P n3 containing π 3 (b i ), (where the b i 's are as in (6.4), i = 1, 2, 3) by genericity we may assume that if n 3 = 2 then J 3 is a line which does not contain any other point of that projection. Set M 3 := π −1 3 (J 3 ). Take a line (in fact if n 3 = 1 it is trivial, if n 3 = 2 then we have already remarked that π 3 (b i ) is the only point of J belonging to π 3 (Y )). Of course the points of A projecting on Suppose that this is the case. Since π i|A is injective for i = 1, 2 (cf. i.e. π i (a 1 ) = π i (b 1 ) for all i = 3. This is a contradiction since we already know that π 3 (a 2 ) = π 3 (b 2 ) and we would have a 2 = b 2 , which contradicts the assumption that A ∩ B = ∅. Therefore the points a ∈ A of (6.5) are all different for different choices of i's. So we may assume that π 3 (a i ) = π 3 (b i ) for i = 1, 2, 3 and the π 3 The argument of the proof of Claim 6.2.1 can be repeated verbatim for all the others π j 's with the only caution that when we do the case j = 2 we have to use a line L 1 ⊂ P 2 containing {π 1 (b j ), π 1 (b k )} with j, k = i and set M 1 := π −1 1 (L 1 ) instead of M 2 and L 2 in (**). Moreover (*) clearly holds if we replace the ε 2 with ε 1 and ε 3 with ε j for any j = 3, . . . , k. As already highlighted this concludes the proves since π j (a i ) = π j (b i ) for i = 1, 2, 3 and for j = 1, . . . , k which is a contradiction with A and B being distinct.
This shows that under the assumption (6.4), we can exclude the case where the Segre variety has at least two factors of projective dimension 2.
Let us focus on the 4-factors case.
, with ν(Y ) the concise Segre of q. There do not exist two disjoint sets A, B ∈ S(Y, q) evincing the rank of q.
Proof. Assume by contradiction that there exist two disjoint sets A, B ∈ S(Y, q) evincing the rank of q and moreover assume that no η i|S is injective, for i = 2, 3, 4.
Proof of Claim 6.3.1: Since we are in the hypothesis α 4 = 3, the projection of S ∩ K 4 onto the first two factors of Y is made by at most 3 points.
Suppose that such a projection is made by exactly 3 points. Since h 1 (O P 2 ×P 1 (1, 1)) > 0 those points must lie on a line L when applying the Segre embedding. Moreover from Remark 3.3 we know that π 1 (A) and π 1 (B) are sets of linearly independent points and since linear subspaces of the Segre variety are all contained in a factor, we get that L ⊂ P 2 . Thus ♯(π 2 (S ∩ K 4 )) = 1 proving the claim in this case. If the projection of S ∩ K 4 onto the first two factors is made by less than 3 points, there exist at least two points, u, v ∈ S ∩ K 4 such that they share the same image under the projection. Remark that if we consider E ⊂ S ∩ K 4 such that ♯E = 2 and take T ∈ |I E (1, 1, 0, 0)|, then T ⊃ S ∩ K 4 . Indeed if S ∩ K 4 ⊂ T then we have that T ∪ K 3 contains exactly five points of S, which leads to a contradiction because by [13, Lemmas 2.4 and 2.5] (also [7, Lemma 5.1, item (b)]) we would have h 1 (I S\S∩T ∪K3 (ε 3 )) > 0. Therefore also the third point of S ∩ K 4 share the same image of u and v and we are done.
Since the identifiability of rank-3 tensors in ν((P 1 ) ×4 ) is already fully described by Remark 3.2, we are therefore done with the order-4 tensors and we can focus on tensors of order bigger or equal than 5.
Lemma 6.4. Let q be a rank-3 tensor of order at least 5 and let ν(Y ) be its concise Segre. If there exist two disjoint sets A, B ∈ S(Y, q) as in (6.4), then there exists at least an index i ∈ {1, . . . , k} such that η i|S and π i|S are injective.
Proof. [Injectivity of η i|S .] Assume that no η i|S is injective, then by Remark 1.10 for any i = 1, . . . , k there exist an element a ∈ A and an element b ∈ B such that π h (a) = π h (b) for any h = i. It is easy to check that this condition, applied to two disjoint sets of 3 points each, and at least five η i 's, imposes either that A ∩ B = ∅ (contradiction) or that one of the two sets (either A or B) depends only on 4 factors (contradicting Autarky).
[Injectivity of π i|S .] Assume that η i|S is injective and that π i|S is not injective and take H ∈ |O Y (ε i )| such that ♯(S \ S ∩ H) = ∅. Since by Autarky S ⊂ H we have that h 1 (I S\S∩H (ε i )) > 0.
We distinguish different cases depending on ♯(S \ S ∩ H).