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, e.g. [4,5,11,12,18,22,23,26,31,32,[34][35][36]42,50]. Except for very few contributions [38,45,46] which work for certain specific classes of given tensors, all the others regards the identifiability of generic tensors 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 tensors of given rank r brings the problem into the setting of rth 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. e.g. [18,22,[26][27][28][29]34,42,47]), the knowledge of generic tensors' behaviour does not help in 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 [1,8,27,29,34,36,37,45]. What it was missing was the complete classification for all the tensors of those ranks.
In Proposition 2.3, we show that rank-2 tensors T are always identifiable except if T is a 2 × 2 matrix. Our main Theorem 7.1 states that a rank-3 tensor T is identifiable except if 1. T is a 3 × 3 matrix and dim(S(Y, T )) = 6; 2. there exist v 1 , v 2 , v 3 ∈ C 2 s.t. T ∈ C 2 ⊗v 2 ⊗v 3 +v 1 ⊗C 2 ⊗v 3 +v 1 ⊗v 2 ⊗C 2 and dim(S(Y, T )) ≥ 2; 3. T ∈ (C 2 ) ⊗4 and dim(S(Y, T )) ≥ 1; 4. T ∈ C 3 ⊗ C 2 ⊗ C 2 and it is constructed as in Example 3.6. In this case, dim(S(Y, T )) = 3; 5. T ∈ C 3 ⊗ C 2 ⊗ C 2 and it is constructed as in Example 3.7. In this case, S(Y, T ) contains two different four-dimensional families; 6. T ∈ C m1 ⊗ C m2 ⊗ (C 2 ) k−2 , where k ≥ 3 and m 1 , m 2 ∈ {2, 3}. In this case dim(S(Y, T )) ≥ 2 and T is constructed as in Proposition 3.10. If The paper is organized as follows. After the preliminary Sect. 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 Sect. 2. In Sect. 3, we explain in details the examples where the non-identifiability of a rank-3 tensor arises. In Sects. 5 and 6, we show that 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. 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 ith 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 ith 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.

A Very Useful Lemma
Let X be a non degenerate irreducible projective variety embedded in P N via an ample line bundle L. Let Z ⊂ X be a zero-dimensional scheme and let D ⊂ P N be a fixed hyperplane, i.e. D ∈ |L|. Denote with Res D (Z) the residual scheme of Z with respect to D, i.e. the zero-dimensional scheme whose defining ideal sheaf is I Z : I D . The ideal sheaf I D∩Z,D ⊗ L represents the scheme theoretic intersection of D and Z, also called the trace of Z with respect to D. The residual exact sequence of Z with respect to D in X is the following: An extremely useful tool that will turn out to be crucial in many proofs of this paper is [7, Lemma 5.1]. We recall here the analogous statement given in [13,Lemma 2.4] in the setting of zero-dimensional schemes. Lemma 1.11. (Ballico-Bernardi-Christandl-Gesmundo) Let X ⊆ P n be an irreducible variety embedded by the complete linear system associated with L = O X (1). Let p ∈ P n and let A, B be zero-dimensional schemes in X such that p ∈ A , p ∈ B and there are no A A and B B with p ∈ A or p ∈ B . Suppose h 1 (I B (1)) = 0. Let C ⊆ P n be an effective Cartier divisor such that We rephrase it in terms of sets of points of multiprojective spaces embedded via |O(1, . . . , 1)|.
Let k ≥ 2, let Y = P n1 × · · · × P n k such that X : =ν(Y ) ⊂ P N , where N = (n i + 1) − 1. Let q ∈ P N be a point of X-rank r and let A, B ∈ S(Y, q) be sets of points evincing the rank of q and write S : =A ∪ B. In this setting, the irreducible variety X considered in Lemma 1.11 is the Segre variety. The residual scheme Res C (S) is therefore S\(S ∩ C). The assumption h 1 (I B (1)) = 0 of [13,Lemma 2.4], in the setting of Segre varieties becomes h 1 (I B (1, . . . , 1)) = 0, which means that the points of ν(B) are linearly independent and this assumption is satisfied since both A and B are sets evincing the rank of q.
With all this said we can state the specific version of [13, Lemma 2.4] and [7, Lemma 5.1] which is needed in the present paper. Notation 1.12. With an abuse of notation, when we will make cohomology computation, if the variety for which we compute the cohomology of the ideal sheaf is Y we will omit it. We will specify the variety only when it is not Y .
The above lemma gives a sufficient condition so that the whole S = A∪B is contained in a given divisor D of the variety X. If A, B are two disjoint distinct sets evincing the rank of a tensor q of X-rank 3 the assumption that A ∩ B ⊂ D is always satisfied.

Identifiability on the 2nd 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 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 concise Segre X q of q is a two-factor Segre; moreover, it is not-identifiable for any number of factors (cf. e.g. [8, Remark 3]).
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 identifiability 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 contained in X; moreover, 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 we proceed by induction on k. Let η k , ν k , and X k be as in Notation 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 we have that 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 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 • if o / ∈ 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 : Moreover, one can notice that Π q \ϕ(X ∩H q ) ∼ = P 2 \C are just the points of the first case.

Examples of Not-Identifiable Rank-3 Tensors
The purpose of this section is to explain in detail 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 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 will not 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: existence of certain special rank-3 tensors q such that dim(T q (σ 3 (X))) = dim(T q (AbSec 3 (X))) < 14 where AbSec 3 (X) : ={(p; p 1 , p 2 , p 3 ) ∈ P 15 × X ×3 | p ∈ p 1 , p 2 , p 3 } is the 3rd abstract secant of X and q is the preimage of q via the projection on the first factor. The impossibility of the existence of such a point is guaranteed by [43,Cap II,Ex 3.22,part (b)]. This proves that all the tensors of σ 0 3 (X) have an infinite number of rank-3 decompositions. Before explaining the other not-identifiable examples, we need some preliminary results.
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) contain linearly independent points.
. The classification of the dimensions of secant varieties of such a Segre-Veronese can be found in [10,15,17,33].
Proof. First of all remark that, for i = 1, 2, 1, 1) and G is a reducible element of |O Y (0, 1, 1)|. 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 3 , ν(G 1 ), ν(G 2 )). To show that J = P 8 it is sufficient to see that dim(σ 2 (ν(G i )∩ν(G j ))) = 1 and this is a straightforward computation since the elements of ν(G 1 ) are tensors with a second factor fixed, while the elements of ν(G 2 ) have the third factor fixed, and to have the equality between an element of σ 2 (ν(G 1 )) and an element of ν(G 2 ) it is sufficient to impose two linear independent conditions and, therefore, since dim(ν(G 2 )) = 3 we have that the intersection has dimension 1.
Assume G is irreducible, then B contains three linearly independent points on G, so the points of B are uniquely determined by G.
Without loss of generality, we may assume that two 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.
Since q is a 2 × 2 matrix of rank 2, dim S(Y , q ) = 2 and Y is the minimal multiprojective subspace of Y containing A, the minimal multiprojective subspace containing Y ∪ {p} is Y . So since P ni = π i (Y ∪ {p}) , we get 1 ≤ n i ≤ 2 for i = 1, 2 and n i = 1 for all i > 2. This ends item 1.
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.
The proof is by induction on the number of factors.

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 contains a conic C. Either if it is reduced or not, the two secant variety of C fills ν(S) = P 2 . So r X (q) ≤ 2, which is an absurd.

Lemma 4.2. Let q be a not-identifiable rank-3 tensor and let
Proof. Suppose, by contradiction, that A and B have 2 distinct points in common and call the set of these two points E.
. Clearly ν(E) is a line; therefore, dim ν(A) ∩ ν(B) > 1, but ν(A) and ν(B) are both planes, so ν(A) = ν(B) . 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.

.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 notidentifiable 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 Sect. 5 from the case A ∪ B = 6 in Sect. 6.
Another very useful behaviour that needs to be understood 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. By is not a very ample line bundle. So we cannot be sure about the injectivity of the restriction η i|E of η i to the finite set E.
If η i|E is not injective 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 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).
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 to being able to cover the whole set S as follows. Let o i ∈ P 1 i , i = 2, 3, 4 be such that: . Now it may happen that either S ⊂ M 2 ∪ M 3 ∪ M 4 or not. We study those two cases in (a) and (b), respectively. Since h 1 (I S\S∩M3 (ε 3 )) > 0 (again by Lemma 1.13, 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 ith factor. The latter case cannot arise because S does not depend only on the third, fourth and ith 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.
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.