## 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 $$T\in \mathbb {C}^{n_1+1} \otimes \cdots \otimes \mathbb {C}^{n_k+1}$$ can be decomposed in a unique way as a sum of pure tensors:

\begin{aligned} T=\sum _{i=1}^r v_{1,i}\otimes \cdots \otimes v_{k,i}, \end{aligned}

with $$v_{j,i}\in \mathbb {C}^{n_j+1}$$, for $$j=1, \ldots , k$$. Of course the minimum r realizing the above expression is a crucial value and it is called the rank of T.

From the applied point of view, identifiability in tensor decomposition arises naturally in numerous areas, we quote as examples Phylogenetics, Quantum Physics, Complexity Theory and Signal Processing (cf. e.g. [3, 19, 21, 25, 30, 38,39,40,41, 44, 49, 51, 52]).

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  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  in .

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 $$\mathbb {C}$$) 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 $$\mathcal {S}(Y,T)$$ 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 $$\mathcal {S}(Y,q)$$ 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 $$T\in (\mathbb {C}^2)^{\otimes 4}$$, 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 $$2\times 2$$ matrix. Our main Theorem 7.1 states that a rank-3 tensor T is identifiable except if

1. 1.

T is a $$3\times 3$$ matrix and $$\dim (\mathcal {S}(Y,T))=6$$;

2. 2.

there exist $$v_1,v_2,v_3\in \mathbb {C}^2$$ s.t. $$T\in \mathbb {C}^2\otimes v_2 \otimes v_3+ v_1 \otimes \mathbb {C}^2 \otimes v_3+ v_1 \otimes v_2\otimes \mathbb {C}^2$$ and $$\dim (\mathcal {S}(Y,T))\ge 2$$;

3. 3.

$$T\in (\mathbb {C}^2)^{\otimes 4}$$ and $$\dim (\mathcal {S}(Y,T))\ge 1$$;

4. 4.

$$T\in \mathbb {C}^3\otimes \mathbb {C}^2\otimes \mathbb {C}^2$$ and it is constructed as in Example 3.6. In this case, $$\dim (\mathcal {S}(Y,T))=3$$;

5. 5.

$$T\in \mathbb {C}^3\otimes \mathbb {C}^2\otimes \mathbb {C}^2$$ and it is constructed as in Example 3.7. In this case, $$\mathcal {S}(Y,T)$$ contains two different four-dimensional families;

6. 6.

$$T \in \mathbb {C}^{m_1}\otimes \mathbb {C}^{m_2}\otimes (\mathbb {C}^2)^{k-2}$$, where $$k\ge 3$$ and $$m_1,m_2\in \{2,3 \}$$. In this case $$\dim (\mathcal {S}(Y,T))\ge 2$$ and T is constructed as in Proposition 3.10. If $$m_1+m_2+k \ge 6$$ then $$\dim (\mathcal {S}(Y,T))=2$$.

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.

## Preliminaries and Notation

We will always work over an algebraically closed field $$\mathbb {K}$$ of characteristic 0.

### Definition 1.1

Let $$X \subset \mathbb {P}^N$$ be a non-degenerate projective variety, the X-rank $$r_{X}(q)$$ of a point $$q \in \langle X \rangle$$ is the minimal cardinality of a finite set $$S\subset X$$ such that $$q \in \langle S \rangle$$.

### Notation 1.2

Let $$A\subset \mathbb {P}^N$$ be any subset. With an abuse of notation we denote by $$\langle A \rangle$$ the projective space spanned by A.

Let $$V_1, \ldots , V_k$$ be vectors spaces of dimension $$n_1+1, \ldots , n_k+1$$, respectively, the Segre variety is the image of the following embedding:

\begin{aligned}&\nu : \mathbb {P}(V_1)\times \cdots \times \mathbb {P}(V_k) \rightarrow \mathbb {P}(V_1 \otimes \cdots \otimes V_k)\\&\quad ([v_1], \ldots , [v_k])\mapsto [v_1 \otimes \cdots \otimes v_k]. \end{aligned}

### Notation 1.3

We denote by Y the multiprojective space

\begin{aligned} Y:{=}\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k} \end{aligned}

and by X the image of Y via Segre embedding, i.e. $$X=\nu (Y)$$.

We denote the projection on the ith factor as

\begin{aligned} \pi _i: Y \longrightarrow \mathbb {P}^{n_i} . \end{aligned}

The space corresponding to forget the ith factor in the multiprojective space Y is denoted by $$Y_i$$:

\begin{aligned} Y_i:{=} \mathbb {P}^{n_1}\times \cdots \times \widehat{\mathbb {P}^{n_i}}\times \cdots \times \mathbb {P}^{n_k}. \end{aligned}

With $$\nu _i : Y_i \longrightarrow \mathbb {P}^{N'}$$ we denote the corresponding Segre embedding, in particular $$X_i:{=}\nu (Y_i)$$.

The projection on all the factors of Y but the ith one is denoted with $$\eta _i$$:

\begin{aligned} \eta _i : Y \longrightarrow Y_i . \end{aligned}

Obviously all fibers of $$\eta _i$$ are isomorphic to $$\mathbb {P}^{n_i}$$.

### Definition 1.4

For any $$q \in \mathbb {P}^N$$, $$\mathcal {S}(Y,q)$$ denotes the set of all subsets $$A \subset Y$$ such that $$\sharp (A) =r_X(q)$$ and $$q \in \langle \nu (A) \rangle$$ and we will say that if $$A \in \mathcal {S}(Y,q)$$, then A evinces the rank of q. Moreover, we say that $$q\in \langle X \rangle$$ is identifiable if $$\sharp \mathcal {S}(Y,q) =1$$.

### Notation 1.5

Sometimes we will also use the following multi-index notations: for $$1 \le i \le k$$, $$\varepsilon _i=(0,\dots ,0,1,0,\dots ,0)$$, where the only 1 is in the ith place and $$\hat{\varepsilon _i}$$ which is a k-tuple with all one’s but the ith place, which is filled by 0, i.e. $$\widehat{\varepsilon _i}=(1, \dots , 1,0,1, \dots ,1)$$.

### Definition 1.6

The rth secant variety of X is $$\sigma _r(X):{=} \overline{\bigcup _{p_, \ldots , p_r\in X} \langle p_1 , \ldots , p_r}\rangle$$ where the closure is the the Zariski closure. The set of points of X-rank equal to r is sometime denoted as $$\sigma _r^0(X)$$. If $$\dim \sigma _r(X)< \min \{rn+r-1, \dim \langle X \rangle \}$$, the variety X is said to be r-defective, otherwise X is r-regular. If X is r-defective, the difference $$\delta _r= \min \{rn+r-1, \dim \langle X \rangle \}- \dim \sigma _r(X)$$ is called the rth secant defect of X.

We will often use the so called Concision/Autarky property (cf. [48, Prop. 3.1.3.1] [9, Lemma 2.4]) that we recall here.

### Lemma 1.7

(Concision/Autarky) For any $$q \in \mathbb {P}(V_1 \otimes \cdots \otimes V_k)$$, there is a unique minimal multiprojective space $$Y'\simeq \mathbb {P}^{n'_1}\times \cdots \times \mathbb {P}^{n'_k} \subseteq Y\simeq \mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$ with $$n'_i \le n_i$$, $$i=1 , \ldots , k$$ such that $$\mathcal {S}(Y,q)=\mathcal {S}(Y',q)$$.

### Definition 1.8

(concise Segre) Given a point $$q \in \mathbb {P}^N$$, we will call concise Segre the variety $$X_q:{=}\nu (Y')$$ where $$Y' \subseteq Y$$ is the minimal multiprojective space $$Y'\subseteq Y$$ such that $$q \in \langle \nu (Y') \rangle$$ as in Concision/Autarky Lemma 1.7.

### Remark 1.9

The minimal $$Y'$$ defining the concise Segre of a point q can be obtained as follows. Fix any $$A \in \mathcal {S}(Y,q)$$, set $$A_i:{=} \pi _i(A) \subset \mathbb {P}^{n_i}$$, $$i=1,\dots , k$$, where the $$\pi _i$$s are the projections on the ith factor of Notation 1.3. Each $$\langle A_i \rangle \subseteq \mathbb {P}^{n_i}$$ is a well-defined projective subspace of dimension at most $$\min \{ n_i, r_X(q)-1 \}$$. By Concision/Autarky we have $$Y' = \prod _{i=1}^k \langle A_i \rangle$$. In particular q does not depend on the ith factor of Y if and only if for one $$A \in \mathcal {S}(Y,q)$$ the set $$\pi _i(A)$$ is a single point.

### Remark 1.10

Let $$q \in \mathbb {P}^N$$ and consider $$A \in \mathcal {S}(Y,q)$$. We claim that there is no line $$L \subset X$$ such that $$\sharp (L \cap \nu (A) )\ge 2$$. Obviously if $$\sharp (L \cap \nu (A) )>2$$ 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 $$\mathcal {S}(Y,q)$$ have. So assume that there exists a line $$L\subset X$$ such that $$\sharp (L \cap \nu (A) )=2$$; let $$u,v \in A$$ be the preimages of those points, i.e. $$u\ne v$$ and $$\langle \nu (u), \nu (v) \rangle = L$$. Then $$r_X(q) > 2$$ because if $$r_X(q)=2$$ then we would have $$q \in L \subset X$$, so the rank of q will be 1. Let $$E= A{\setminus } \{ u,v \}$$. Then we will have that $$q \in \langle \nu (E) \cup L \rangle$$, so we can find a point $$o \in L$$ such that $$q \in \langle \nu (E) \cup \{ o \} \rangle$$, which would imply $$r_X(q)< \sharp A$$.

### A Very Useful Lemma

Let X be a non degenerate irreducible projective variety embedded in $$\mathbb {P}^N$$ via an ample line bundle $$\mathcal {L}$$. Let $$Z\subset X$$ be a zero-dimensional scheme and let $$D\subset \mathbb {P}^N$$ be a fixed hyperplane, i.e. $$D\in \vert \mathcal {L}\vert$$. 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 $$\mathcal {I}_{Z}:\mathcal {I}_D$$. The ideal sheaf $$\mathcal {I}_{D\cap Z,D}\otimes \mathcal {L}$$ 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:

\begin{aligned} 0 \rightarrow \mathcal {I}_{Res_D(Z)}\otimes \mathcal {L}(-D) \rightarrow \mathcal {I}_Z \otimes \mathcal {L} \rightarrow \mathcal {I}_{D\cap Z,D}\otimes \mathcal {L} \rightarrow 0. \end{aligned}

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 $$X \subseteq \mathbb {P}^n$$ be an irreducible variety embedded by the complete linear system associated with $$\mathcal L = \mathcal O_X(1)$$. Let $$p\in \mathbb {P}^n$$ and let AB be zero-dimensional schemes in X such that $$p \in \langle A \rangle$$, $$p \in \langle B \rangle$$ and there are no $$A' \subsetneq A$$ and $$B' \subsetneq B$$ with $$p \in \langle A'\rangle$$ or $$p \in \langle B' \rangle$$. Suppose $$h^1(\mathcal {I}_{B}(1)) = 0$$. Let $$C \subseteq \mathbb {P}^n$$ be an effective Cartier divisor such that $$\mathrm {Res}_C(A) \cap \mathrm {Res}_C(B) = \emptyset$$. If $$h^1 (X, \mathcal {I}_{\mathrm {Res_C(A \cup B)}} (1) (-C)) = 0$$ then $$A \cup B \subseteq C$$.

We rephrase it in terms of sets of points of multiprojective spaces embedded via $$\vert \mathcal {O}(1,\dots ,1) \vert$$.

Let $$k\ge 2$$, let $$Y=\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$ such that $$X:{=}\nu (Y)\subset \mathbb {P}^N$$, where $$N=\prod (n_i+1)-1$$. Let $$q \in \mathbb {P}^N$$ be a point of X–rank r and let $$A,B\in \mathcal {S}(Y,q)$$ be sets of points evincing the rank of q and write $$S:{=}A\cup B$$. In this setting, the irreducible variety X considered in Lemma 1.11 is the Segre variety. The residual scheme $$Res_C(S)$$ is therefore $$S{\setminus } (S\cap C)$$. The assumption $$h^1(\mathcal {I}_B(1))=0$$ of [13, Lemma 2.4], in the setting of Segre varieties becomes $$h^1(\mathcal {I}_B(1,\dots ,1))=0$$, which means that the points of $$\nu (B)$$ 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.

### Lemma 1.13

Let $$k\ge 2$$ and consider $$Y=\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$, where all $$n_i\ge 1$$. Let $$q\in \mathbb {P}^N$$, $$A,B\in \mathcal {S}(Y,q)$$ be two different subsets evincing the rank of q and write $$S=A\cup B$$. Let $$D \in \vert \mathcal {O}_Y(\varepsilon ) \vert$$ be an effective Cartier divisor such that $$A\cap B \subset D$$, where $$\varepsilon =\sum _{i \in I} \varepsilon _i$$ for some $$I\subset \{ 1,\dots , k\}$$ as introduced in Notation 1.5. If $$h^1(\mathcal {I}_{S{\setminus } S\cap D}(\hat{\varepsilon }))=0$$ then $$S\subset D$$.

The above lemma gives a sufficient condition so that the whole $$S=A\cup B$$ is contained in a given divisor D of the variety X. If AB are two disjoint distinct sets evincing the rank of a tensor q of X–rank 3 the assumption that $$A\cap B \subset D$$ 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 $$X_q=\nu \left( \mathbb {P}^1_i\right) ^{\times k}$$. Therefore, for the rest of this section, we will focus our attention to Segre varieties of products of $$\mathbb {P}^1$$s.

### Remark 2.1

If the concise Segre $$X_q$$ of a tensor $$q\in \sigma _2(X)$$ is a $$\nu ( \mathbb {P}^{1}\times \mathbb {P}^{1})$$, then $$\sigma _2(X_q)$$ parameterizes the $$2 \times 2$$ 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 $$(\mathbb {P}^1)^{\times k}$$ with $$k\ge 3$$.

### Definition 2.2

The variety $$\tau (X)$$ is the tangent developable of a projective variety X, i.e. $$\tau (X)$$ is defined by the union of all tangent spaces to X.

Recall that a tensor $$q\in \tau (X) {\setminus } X$$ 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]).

### Proposition 2.3

Let $$q \in \sigma _2^0(X)$$. Then $$| \mathcal {S}(Y,q)|> 1$$ if and only if the concise Segre $$X_q$$ of q is $$X_q=\nu (\mathbb {P}^1 \times \mathbb {P}^1)$$.

### Proof

We only need to check the case of $$k \ge 4$$ 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 $$k\ge 4$$.

Since X is cut out by quadrics, then if a line $$L \subset \mathbb {P}^N$$ is such that $$\deg (L \cap X)>2$$ then $$L \subset X$$ and the points of L have X-rank 1. Let $$A,B \in \mathcal {S}(Y,q)$$, either $$\langle A \rangle =\langle B \rangle$$ or $$\langle A \rangle \cap \langle B \rangle =\{q\}$$. In fact, in the first case $$A=B$$ since $$r_X(q)=2$$ and therefore $$\langle A\rangle$$ is not contained in X; moreover, X is cut out by quadrics. In the second case $$A\ne B$$. We can, therefore, assume that $$A,B \in \mathcal {S}(Y,q)$$ are two disjoint sets: $$A=\{a, a' \}$$, $$B=\{ b, b' \}$$, where $$a=(a_1, \dots , a_k)$$, $$a'=(a_1',\dots , a_k')$$ and $$b=(b_1,\dots ,b_k)$$, $$b'=(b_1',\dots ,b_k')$$. Since $$a \ne a'$$, we may assume that at least one of their coordinates is different. Actually we can assume that all the $$a_i\ne a_i'$$, otherwise, by the concision property, one could consider one factor less. The same considerations hold for B.

Now suppose that there exists an index $$i\in \{1 ,\dots ,k \}$$ such that $$\{a_i,a_i' \} \ne \{ b_i, b_i'\}$$ and let such an index be $$i=1$$: $$\{a_1,a_1' \} \ne \{ b_1, b_1'\}$$.

Now we proceed by induction on k. Let $$\eta _k$$, $$\nu _k$$, and $$X_k$$ be as in Notation 1.3. Let $$\tilde{q}=(q_1,\dots ,q_{k-1})$$ be the projection $$\eta _k(q)$$, then $$\eta _k(A)\ne \eta _k(B)$$ and $$\emptyset \ne \langle \nu _k(\eta _k(A)) \rangle \cap \langle \nu _k(\eta _k(B)) \rangle \supset \{ \tilde{q} \}$$ because $$\{ q\}\subset \langle \nu (A) \rangle \cap \langle \nu (B) \rangle$$. So $$r_{X_k}(\tilde{q})= 2$$ and $$| \mathcal {S}(Y_k,\tilde{q})|\ge 2$$, 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 $$i=, 1\ldots , k$$ we have that $$\{ a_i,a'_i\}=\{b_i,b'_i \}$$.

Without loss of generality assume that $$a_1=b_1$$ and $$a'_1=b'_1$$, moreover up to permutation there exists an index $$e \in \{1, \dots , k-1 \}$$ such that $$b_i=a_i$$ and consequently $$b'_i=a'_i$$ for $$1 \le i \le e$$ and $$b_i=a'_i$$ and $$b'_i=a_i$$ for $$e+1 \le i \le k$$. Eventually by exchanging the role of the first e elements with the others, we have that $$k-e \ge 2$$ because by assumption $$k \ge 4$$. Let $$H \in |\mathcal {O}_Y(0,\dots ,0,1) |$$ be the only element containing $$a'$$, $$H= \mathbb {P}^1 \times \cdots \times \mathbb {P}^1 \times \{ a'_k\} \cong (\mathbb {P}^1)^{\times k-1}$$; then $${\text {Res}}_H(A\cup B)=\{ a',b'\}$$ and since $$k-e\ge 2$$ we have that $$\eta _k(a')\ne \eta _k(b')$$, i.e. $$h^1(\mathcal {I}_{{\text {Res}}_H(A \cup B)}(1,\dots ,1,0))=0$$. By Lemma 1.13, we get $$a'=b'$$ which contradicts the fact that $$A \cap B=\emptyset$$.

### Corollary 2.4

Let q be any rank-2 tensor. If q is not-identifiable, then there is a bijection between $$\mathcal {S}(Y,q)$$ and $$\mathbb {P}^2 {\setminus } L$$, where $$L\subset \mathbb {P}^2$$ is a projective line, $$q \in \tau (X)$$ and L parametrizes the set of all degree 2 connected subschemes V of Y such that $$q \in \langle \nu (V) \rangle$$.

### 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, $$X\subset \mathbb {P}^3$$ is a quadric surface. Denote by $$H_q \subset \mathbb {P}^3$$ the polar plane of X with respect to q. Since $$q \notin X$$, we have that $$q \notin H_q$$ and the intersection $$X \cap H_q =\{ p \in X \; |\; T_pX \ni q \}$$ is a smooth conic. Remark also that by definition a point $$o \in X$$ is such that $$q \in T_oX$$ if and only if $$o \in X \cap H_q \subset \tau (X)$$.

Fix $$o \in H_q$$, then

• if $$o \notin X$$ the line given by $$\langle o,q \rangle$$ is not tangent to X and when considering the intersection $$\langle o,q\rangle \cap X$$, it is given by two points $$p_1, p_2 \notin \{o,q\}$$ such that $$\{ p_1, p_2\} \in \mathcal {S}(Y,q)$$;

• if $$o \in X$$, i.e. $$o \in X \cap H_q$$, then the line $$\langle o,q \rangle$$ is tangent to X.

Consider $$\Pi _q= \{ \text {lines } L \subset \mathbb {P}^3 \text { passing through } q\} \cong \mathbb {P}^2$$ and consider the following isomorphism $$\varphi : H_q \longrightarrow \Pi _q$$ defined by $$p \mapsto \langle p,q \rangle$$. Clearly, $$\varphi (X \cap H_q)$$ is a smooth conic $$\mathcal {C}$$ of $$\Pi _q$$. Moreover, one can notice that $$\Pi _q {\setminus } \varphi (X \cap H_q) \cong \mathbb {P}^2 {\setminus } \mathcal {C}$$ 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 $$q \in \mathbb {P}^N$$ 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 $$X_q= \nu (\mathbb {P}^{n_1} \times \cdots \times \mathbb {P}^{n_k} )$$, with $$n_1, \dots ,n_k \in \{1,2 \}$$.

First of all let us remark that the matrix case is highly not-identifiable even for the rank-3 case.

### Remark 3.1

In the case of two factors (i.e. $$k=2$$), a rank-3 tensor q is a $$3\times 3$$ matrix of full rank. The dimension of the concise Segre X of $$3\times 3$$ matrices is 4 and $$\dim (\sigma _3(X))=\min \{\dim (\mathbb {P}^8), 3\dim (X)+2 \}=\min \{8,14\}=8$$. Thus, $$\dim \mathcal {S}(Y,q)= 14-8=6$$ for all $$q \in \mathbb {P}^8$$ of rank 3.

Consider now the third secant variety of the Segre embedding of $$Y= \mathbb {P}^{n_1} \times \cdots \times \mathbb {P}^{n_k}$$, where $$n_i\in \{1,2 \}$$, the following Examples 3.6 and 3.7 and Proposition 3.10 provides instances of not-identifiability that we will show to be essentially the only classes of not-identifiable rank-3 tensors in $$\mathbb {C}^{n_1+1}\otimes \cdots \otimes \mathbb {C}^{n_k+1}$$ (cases (4), (5) and (6), respectively, of our main Theorem 7.1) more than the well-known ones (matrix case, points on tangential variety of $$\nu ((\mathbb {P}^1)^{\times 3})$$, and elements of the defective $$\sigma _3(\nu ((\mathbb {P}^1)^{\times 4}))$$—items (1), (2) and (3) respectively of Theorem 7.1).

In the following remark we explain the behaviour on $$\sigma _3((\mathbb {P}^{1})^{\times 4})$$.

### Remark 3.2

It has been shown in  (cf. also [27, 29]) that the third secant variety of a Segre variety X is never defective unless either $$X=\nu (\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1)$$ or $$X=\nu (\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^a)$$, with $$a\ge 3$$.

The case in which q is a rank-3 tensor in $$\langle \nu (\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^a) \rangle$$ with $$a\ge 3$$ 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 $$X=\nu (\mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1 \times \mathbb {P}^1)$$ and $$q\in \langle X \rangle$$ can also be easily handled. The fact that $$\dim (\sigma _3(X))$$ is strictly smaller than the expected dimension proves that the generic element of $$\sigma _3(X)$$ has an infinite number of rank-3 decompositions. By definition of dimension there is no element of $$\sigma _3(X)$$ s.t. its tangent space has dimension equal to the expected one: $$\dim (T_q(\sigma _3(X)))\le \dim \sigma _3(X)$$ for all $$q\in \sigma _3(X)$$. This does not exclude the existence of certain special rank-3 tensors q such that $$\dim (T_q(\sigma _3(X)))=\dim (T_{q'}(AbSec_3(X)))< 14$$ where $$AbSec_3(X):{=}\{(p; p_1, p_2,p_3)\in \mathbb {P}^{15} \times X^{\times 3} \, | \, p \in \langle p_1, p_2, p_3 \rangle \}$$ 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 $$\sigma _3^0(X)$$ 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 $$\mathbb {P}^2$$. Let $$q\in \sigma _3^0(\nu (Y))$$, with $$\nu (Y)$$ the concise Segre of q and let $$A,B \in \mathcal {S}(Y,q)$$ be two disjoint subsets evincing the rank of q. By Autarky $$\langle \pi _1(A)\rangle = \langle \pi _1(B)\rangle = \mathbb {P}^2$$; moreover when considering the restrictions of the projections $$\pi _{1\vert A}$$ and $$\pi _{1\vert B}$$ to the subsets A and B, respectively; they are both injective and both $$\pi _1(A)$$ and $$\pi _1(B)$$ contain linearly independent points.

### Remark 3.4

Consider $$Y= \mathbb {P}^2 \times \mathbb {P}^1 \times \mathbb {P}^1$$ and an irreducible divisor $$G \in |\mathcal {O}_Y(0,1,1) |$$. Then $$\sigma _2(\nu (G))\subsetneq \sigma _3(\nu (G)) =\langle \nu (G) \rangle = \mathbb {P}^8$$. Indeed G is nothing else than the Segre–Veronese variety () of $$\mathbb {P}^2\times \mathbb {P}^1$$ embedded in bi-degree (1,2), i.e. $$G\cong \mathbb {P}^2 \times \mathbb {P}^1$$, $$\mathcal {O}_Y(1,1,1)_{|_{G}}\cong \mathcal {O}_{\mathbb {P}^2 \times \mathbb {P}^1}(1,2)$$ and $$\mathcal {O}_Y(1,0,0)\cong \mathcal {O}_Y(1,1,1)(-G)$$. The classification of the dimensions of secant varieties of such a Segre–Veronese can be found in [10, 15, 17, 33].

### Proposition 3.5

For the Segre embedding of $$Y=\mathbb {P}^2 \times \mathbb {P}^1 \times \mathbb {P}^1$$ fix $$G_1 \in |\mathcal {O}_Y(0,1,0) |$$ and $$G_2 \in |\mathcal {O}_Y(0,0,1) |$$ and define $$G:{=} G_1 \cup G_2$$ to be their union. We have that for $$\{i,j\}=\{1,2\}$$, $$\dim \langle \nu (G_i)\rangle =5$$, $$\dim \langle \nu (G)\rangle =8$$, $$\sigma _2(\nu (G_i))=\langle \nu (G_i) \rangle$$ and $$\langle \nu (G) \rangle$$ is the join of $$\sigma _2(\nu (G_i))$$ and $$\nu (G_j)$$.

### Proof

First of all remark that, for $$i=1,2$$, $$G_i\cong \mathbb {P}^2 \times \mathbb {P}^1$$, $$\mathcal {O}_Y(1,1,1)_{|_{G_i}} \cong \mathcal {O}_{\mathbb {P}^2\times \mathbb {P}^1}(1,1)$$ and G is a reducible element of $$|\mathcal {O}_Y(0,1,1) |$$. With an analogous computation of the one in Remark 3.4 one sees that $$\dim \langle \nu (G) \rangle =8$$ and $$\sigma _2(\nu (G_i))=\langle \nu (G_i) \rangle$$. It remains to show that $$\langle \nu (G) \rangle = \mathcal {J}$$, where $$\mathcal {J}$$ denotes the join of $$\sigma _2(\nu (G_i))$$ and $$\nu (G_j)$$ with $$\{i,j\}=\{1,2\}$$. We remark that since $$\sigma _2 (\nu (G))= \mathbb {P}^5$$, then $$\mathcal {J}=\mathrm {Join}(\mathbb {P}^3,\nu (G_1),\nu (G_2))$$. To show that $$\mathcal {J}=\mathbb {P}^8$$ it is sufficient to see that $$\dim (\sigma _2(\nu (G_i)\cap \nu (G_j)))=1$$ and this is a straightforward computation since the elements of $$\nu (G_1)$$ are tensors with a second factor fixed, while the elements of $$\nu (G_2)$$ have the third factor fixed, and to have the equality between an element of $$\sigma _2(\nu (G_1))$$ and an element of $$\nu (G_2)$$ it is sufficient to impose two linear independent conditions and, therefore, since $$\dim (\nu (G_2))=3$$ we have that the intersection has dimension 1.

### Example 3.6

Take $$Y=\mathbb {P}^2 \times \mathbb {P}^1 \times \mathbb {P}^1$$, consider the Segre embedding on the last two factors and take a hyperplane section which intersects $$\nu (\mathbb {P}^1 \times \mathbb {P}^1)$$ in a conic $$\mathcal {C }$$, then take a point $$q \in \langle \nu (\mathbb {P}^2 \times \mathcal {C} )\rangle$$. Such a construction is equivalent to consider an irreducible divisor $$G \in |\mathcal {O}_Y(0,1,1) |$$, so $$G\cong \mathbb {P}^2 \times \mathbb {P}^1$$ embedded via $$\mathcal {O}(1,2)$$, then $$\dim \sigma _2(\nu (G))=7$$, thus $$\sigma _2(\nu (G)) \subsetneq \langle \nu (G) \rangle \simeq \mathbb {P}^8$$. As a direct consequence we get that a general point $$q \in \langle \nu (G) \rangle$$ has $$\nu (G)$$-rank 3 and it is not-identifiable because of the not-identifiability of the points on $$\langle \mathcal {C} \rangle$$ and by [43, Cap II, Ex 3.22, part (b)]. Thus, $$\dim (\mathcal {S}(G,q))= 3$$.

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.

### Example 3.7

Fix $$Y=\mathbb {P}^2 \times \mathbb {P}^1 \times \mathbb {P}^1$$. Consider $$G_1 \in |\mathcal {O}_Y(0,0,1)|$$, $$G_2 \in |\mathcal {O}_Y(0,1,0)|$$ and call $$G=G_1 \cup G_2$$ which is a reducible element of $$|\mathcal {O}_Y(0,1,1)|$$. By Proposition 3.5, $$\dim \langle \nu (G) \rangle =8$$; moreover, by a dimension count, we have $$\langle \nu (G_i)\rangle = \sigma _2(G_i)$$, for $$i=1,2$$, both having dimension 5. By Proposition 3.5, we also have that $$\langle \nu (G) \rangle =\mathcal {J}_1=\mathcal {J}_2$$, where $$\mathcal {J}_1 =\text {Join}( \sigma _2(\nu (G_1)), \nu (G_2))$$ and $$\mathcal {J}_2= \text {Join}( \sigma _2(\nu (G_2)) , \nu (G_1))$$. A general $$q \in \langle \nu (G)\rangle$$ has rank 3 and for the subsets evincing its rank we have a 4-dimensional family of sets A such that $$\sharp (A)=3$$, $$\sharp (A\cap G_1)=2$$, $$\sharp (A \cap G_2)=1$$, $$A\cap G_1 \cap G_2=\emptyset$$ and $$q \in \langle \nu (A)\rangle$$. Such a family has dimension 4 since $$G_1$$ is a non defective threefold in $$\mathbb {P}^5$$; therefore, there exists a 2-dimensional family of sets of cardinality 2 in $$G_1$$ spanning a general point of $$\mathbb {P}^5$$; moreover, q sits in a 2-dimensional family of lines joining points of $$G_1$$ and $$G_2$$. Analogously, by looking at q as an element of $$\mathcal {J}_2$$, we get the existence of a 4-dimensional family of sets B such that $$\sharp (B)=3$$, $$\sharp (B\cap G_2)=2$$, $$\sharp (B \cap G_1)=1,$$ $$A\cap G_1 \cap G_2=\emptyset$$ and $$q \in \langle \nu (B) \rangle$$. So we proved that $$\mathcal {S}(G,q)$$ contains at least two dimensional families of solution. Thus, $$\dim \mathcal {S}(G,q)\ge 4$$.

### Proposition 3.8

Let $$q\in \sigma _3^0(\nu (\mathbb {P}^{2}\times \mathbb {P}^{1}\times \mathbb {P}^{1}))$$ and suppose that there exist $$A,B\in \mathcal {S}(Y,q)$$ s.t. $$\sharp (A\cup B)=6$$. Then there exists a unique $$G\in |\mathcal {O}_Y(0,1,1)|$$ containing $$S=A\cup B$$. For such a G we have that $$\mathcal {S}(Y,q)=\mathcal {S}(G,q)$$.

### Proof

Call $$S:{=}A\cup B$$, by Remark 3.3, both $$\pi _{1\vert A}$$ and $$\pi _{1\vert B}$$ are injective and both $$\pi _1(A)$$ and $$\pi _1(B)$$ are sets containing linearly independent points. So $$h^1(\mathcal {I}_A(1,0,0))=h^1(\mathcal {I}_B(1,0,0))=0$$. Now $$h^0(\mathcal {O}_Y(0,1,1))= 4$$, so there exists $$G \in \vert \mathcal {O}_Y(0,1,1) \vert$$ containing B. Moreover, $$S{\setminus } S\cap G\subseteq A$$ but since $$h^1(\mathcal {I}_{A}(1,0,0))=0$$ we have that $$S\subset G$$. This holds for any $$G \in \vert \mathcal {I}_B(0,1,1)\vert$$, so $$\langle \nu _1(\eta _1(A))\rangle \subset \langle \nu _1(\eta _1(B))\rangle$$. The same holds exchanging the roles of A and B; thus, $$\langle \nu _1(\eta _1(A))\rangle = \langle \nu _1(\eta _1(B))\rangle$$.

Assume G is irreducible, then B contains three linearly independent points on G, so the points of B are uniquely determined by G.

Assume G is reducible, i.e. $$G=G_1 \cup G_2$$, with $$G_1 \in \vert \mathcal {O}_Y(0,1,0) \vert$$ and $$G_2 \in \vert \mathcal {O}_Y(0,0,1) \vert$$. Remark that, by Autarky, it does not exist any $$E \in \mathcal {S}(Y,q)$$ which is all contained in $$G_i$$, for $$i=1,2$$, because G is a multiprojective subspace of Y.

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 $$G=G_1 \cup G_2$$ that contains them.

### Corollary 3.9

If $$q\in \sigma _3^0(\nu (\mathbb {P}^{2}\times \mathbb {P}^{1}\times \mathbb {P}^{1}))$$ is such that there exist two disjoint sets $$A, B\in \mathcal {S}(Y,q)$$, then q can be either as in Example 3.6 and $$\dim (\mathcal {S}(Y,q))= 3$$ or as in Example 3.7 and $$\dim (\mathcal {S}(Y,q))= 4$$.

### Proof

This is a direct consequence of the uniqueness of the $$G \in |\mathcal {O}_Y(0,1,1)|$$ s.t. $$\mathcal {S}(Y,q)=\mathcal {S}(G,q)$$ in Proposition 3.8.

### Proposition 3.10

Let $$Y':{=}\mathbb {P}^1\times \mathbb {P}^1\times \{ u_3\} \times \cdots \times \{ u_k\}$$ be a proper subset of $$Y=\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$, $$k\ge 2$$. Take $$q' \in \langle \nu (Y') \rangle {\setminus } \nu (Y')$$, $$A\in \mathcal {S}(Y',q')$$ and $$p\in Y {\setminus } Y'$$. Assume that Y is the minimal multiprojective space containing $$A\cup \{ p\}$$ and take $$q \in \langle \{q',\nu (p) \} \rangle {\setminus } \{q' , \nu (p) \}$$.

1. 1.

$$\sum _{i=1}^k n_i\ge 3$$, $$n_1,n_2\le 2$$, $$n_3, \ldots , n_k\le 1$$ and if $$k\ge 3$$ then $$r_{\nu (Y)}(q)>1$$;

2. 2.

If $$k\ge 3$$ and $$\sum _{i=1}^k n_i\ge 4$$ then $$r_{\nu (Y)}(q)=3$$ and $$\mathcal {S}(Y,q)=\{ \{p\} \cup A\}_{A\in \mathcal {S}(Y',q')}$$.

3. 3.

$$\nu (Y)$$ is the concise Segre of q.

### Proof

First of all, remark that $$r_{\nu (Y)}(q)>1,$$ otherwise there exists $$o\in Y$$ s.t. $$q =\nu (o)$$ and $$q' \in \langle \nu (\{o,p\})\rangle$$. Since $$r_{\nu (Y)}(q') =2$$, we would have $$\{o,p\}\in \mathcal {S} (Y,q')$$ and by Autarky we get $$\{o,p\}\subset Y'$$, contradicting the assumption $$p\notin Y'$$.

The fact that $$n_1+\cdots +n_k\ge 3$$ is obvious from the fact that $$p \notin Y'$$ so $$Y\ne Y'$$.

Since $$q'$$ is a $$2 \times 2$$ matrix of rank 2, $$\dim \mathcal {S}(Y',q')=2$$ and $$Y'$$ is the minimal multiprojective subspace of Y containing A, the minimal multiprojective subspace containing $$Y'\cup \{ p\}$$ is Y. So since $$\mathbb {P}^{n_i}=\langle \pi _i(Y'\cup \{ p \} )\rangle$$, we get $$1\le n_i \le 2$$ 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 $$\mathcal {S}(Y,q)$$ is of type $$A\cup \{p\}$$ with $$A\in \mathcal {S}(Y',q')$$, then Autarky and the fact that Y is the minimal multiprojective space containing $$A\cup \{p\}$$ will imply that $$\nu (Y)$$ is the concise Segre of q. So let us prove item 2.

The proof is by induction on the number of factors. Step (A) is the basis of induction for the case in which Y has at least one factor of projective dimension 2 ($$k=3$$), Step (B) is the basis of induction for the case in which all the factors of Y have projective dimension 1 ($$k=4$$), Steps (C) and (D) are the induction processes of Step (B) and Step (A), respectively.

Let $$E\in \mathcal {S}(Y,q)$$, if we will show that $$E\supset \{ p \}$$ and that there exists $$B \in \mathcal {S}(Y',q')$$, such that $$E=B\cup \{p \}$$, we will be done. Assume that there is no $$B\in \mathcal {S}(Y',q')$$ such that $$E=B\cup \{p \}$$. Fix any $$A\in \mathcal {S}(Y',q')$$ and set $$S:{=}A\cup \{ p \} \cup E$$.

1. (A)

[Case $$k=3$$, $${n_1=2}$$, $${n_2=n_3=1}$$] First assume $$p\in E$$ and set $$E':{=} E{\setminus } \{p\}$$ and $$F= A\cup E'$$. Since $$\cap _{B\in \mathcal {S} (Y,q')}\eta _3(B)=\emptyset$$, taking another $$A\in \mathcal {S} (Y,q')$$ if necessary we may assume $$\eta _3(A)\cap \eta _3(E') =\emptyset$$. Set $$\{D\}:{=} |\mathcal {I} _p(0,0,1)|$$. By Lemma 1.13, we have $$h^1(\mathcal {I} _{S{\setminus } S\cap D}(1,1,0)) >0$$ and hence (since $$\sharp F\le 4$$) $$h^0(\mathcal {I} _{S{\setminus } S\cap D}(1,1,0))\ge 3$$. This must be true for all $$A\in \mathcal {S} (Y',q')$$ and hence we have $$h^0(Y_3,\mathcal {I} _{\eta _3(Y')\cup \eta _3(E')}(1,1)) \ge 3$$. Since $$\eta _3(Y')\in \vert \mathcal {O}_{Y_3}(1,1) \vert$$ we have $$h^0(Y_3,\mathcal {I}_{\eta _3(Y')}(1,1)) =1$$, contradicting the previous inequality.

From now on suppose $$p\notin E$$. As above we may assume $$\eta _3(A)\cap \eta _3(E) =\emptyset$$.

Fix $$o\in E$$. Since $$h^0(\mathcal {O} _Y(1,1,0)) =6$$ and $$\sharp A \cup \{ p \} \cup \{ o\}= 4$$ there is $$G\in |\mathcal {O} _Y(1,1,0)|$$ containing $$A\cup \{p\} \cup \{o\}$$. Assume for the moment $$S\nsubseteq G$$, i.e. $$E\nsubseteq G$$. We have $$h^1(\mathcal {I} _{S{\setminus } S\cap G}(0,0,1)) >0$$; thus, $$\sharp E\ge 3$$. Since $$\sharp E \le 3$$, we get $$\sharp E =3$$ (and hence q has rank 3 and $$\nu (Y)$$ is the concise Segre containing q), $$S{\setminus } S\cap G =E{\setminus } \{o\}$$ and $$\sharp \pi _3(E{\setminus } \{o\}) =1$$. Taking a different $$o\in E$$ we get $$\sharp \pi _3(E) =1$$, i.e. $$\nu (Y)$$ is not the concise Segre of q, a contradiction.

Now assume $$S\subset G$$. Since this must be true for all $$G\in |\mathcal {I} _{A\cup \{p,o\}}(1,1,0)|$$, we get $$| \mathcal {I} _{A\cup \{p,o\}}(1,1,0)| \supseteq |\mathcal {I} _{\{p\}\cup E}(1,1,0)| \ne \emptyset$$. Note that $$\eta _3(Y')\in \vert \mathcal {O} _{Y_3}(1,1)\vert$$ and hence $$h^0(Y_3,\mathcal {I} _{\eta _3(Y')}(1,1)) =1$$. Since $$n_1=2$$ and Y is the minimal multiprojective space containing q, we have $$\eta _3( p)\notin \eta _3(Y')$$. Thus, $$h^0(Y_3,\mathcal {I} _{\eta _3(Y')\cup \{\eta _3( p)\}}(1,1)) =0$$, a contradiction since $$|\mathcal {I} _{A\cup \{p,o\}}(1,1,0)|\ne \emptyset$$.

2. (B)

[Case $$k=4$$, $${n_1=n_2=n_3=n_4=1}$$] Fix $$G\in |\mathcal {O} _Y(0,0,1,1)|$$ containing E. Assume $$S\nsubseteq G$$. Since $$S{\setminus } E=A\cup \{ p\}$$, by Lemma 1.13, we have $$h^1(\mathcal {I} _{A \cup \{ p\} }(1,1,0,0)) >0$$. Call $$p'$$ the projection of p via $$Y\rightarrow Y'$$. Since $$\mathcal {O} _{\mathbb {P}^1\times \mathbb {P}^1}(1,1)$$ is very ample we get that either $$p'\in A$$ or that $$\sharp (\pi _i(A\cup \{p'\})) =1$$ for some $$i\in \{1,2\}$$. The second possibility is excluded, because $$\sharp (\pi _1(A))=\sharp (\pi _2(A)) =2$$ for any $$A\in \mathcal {S} (Y',q')$$. The first possibility is excluded taking instead of A another general $$A_1\in \mathcal {S} (Y',q')$$. Now assume $$S\subset G$$. We get $$A\subset G$$. This is ruled out taking another $$A\in \mathcal {S} (Y',q')$$ since a general $$a\in Y'$$ is contained in some $$B\in \mathcal {S} (Y',q')$$. Thus, we would have that $$Y'\subset G$$ which is a contradiction.

3. (C)

[Case $$k\ge 5$$, $${n_i=1}$$ for all is] We exclude this case by induction on k, the base case $$k=4$$ being excluded in (B). Fix $$o\in \mathbb {P}^1{\setminus } \{p_k,u_k\}$$, set $$M:{=} \pi _k^{-1}(o)$$, i.e. $$M=(\mathbb {P}^1)^{\times k-1}\times \{ o\}$$ and call $$\Lambda :{=} \langle \nu (M)\rangle$$. Note that $$(Y'\cup \{p\})\cap M=\emptyset$$. Denote by $$r=2^k-1$$ and define $$r':{=} \dim \Lambda = 2^{k-1}-1$$.

Consider the following linear projection form $$\Lambda$$:

\begin{aligned} \ell : \mathbb {P}^r{\setminus } \Lambda \rightarrow \mathbb {P}^{r'}. \end{aligned}
(3.1)

Note that $$\nu (Y) \cap \Lambda = \nu (Y_k)\times \{o\}$$ and that $$\ell _{|\nu (Y){\setminus } M} =$$ $$\nu _k(\eta _k(Y{\setminus } M))$$. We identify $$\mathbb {P}^{r'}$$ with the target projective space of $$Y_k$$. Since $$(Y'\cup \{p\})\cap M = \emptyset$$, $$\ell$$ is well defined on $$Y'\cup \{p \}$$ and it acts as the composition of $$\eta _k$$ and the Segre embedding.

By the inductive assumption $$\mathcal {S} (Y_k,\ell (q)) =\{ B \cup \eta _k({p}) \}_{B\in \mathcal {S} (\eta _k(Y'),\eta _k(q'))}$$. Thus, for any $$E\in \mathcal {S} (Y,q)$$ there is $$B\in \mathcal {S}(Y',q')$$ such that $$\eta _k(E) =\eta _k(B\cup \{p\})$$. Since $$\eta _{k|E}$$ is injective by Remark 1.10 and $$\mathcal {S} (Y,q)\supseteq \{B\cup \{p\}\}_{B\in \mathcal {S} (Y',q')}$$, we get $$\mathcal {S} (Y,q) =\{B\cup \{p\}\}_{B\in \mathcal {S} (Y',q')}$$.

4. (D)

[Case $$k\ge 3$$, $${n_1=2}$$, $$n_1+\cdots +n_k \ge 5$$] If only one of the factors is a $$\mathbb {P}^2$$ we use Step (A) as base of the induction and then we construct a projection similar to (3.1). Indeed $$Y=\mathbb {P}^2\times (\mathbb {P}^1)^{k-1}$$, where $$k\ge 4$$. Fix $$o \in \mathbb {P}^1 {\setminus } \{ p_k, u_k\}$$, set $$M:{=}\pi _k^{-1}(o)$$ and define $$\Lambda :{=} \langle \nu (M)\rangle$$. Denote by $$r=3\cdot 2^{k-1}-1$$ and by $$r'=\dim \Lambda :{=}3\cdot 2^{k-2}-1$$. We consider the linear projection $$\ell : \mathbb {P}^r {\setminus } \Lambda \rightarrow \mathbb {P}^{r'}$$ which acts as the composition of $$\eta _k$$ and the Segre embedding. By the inductive assumption $$\mathcal {S} (Y_k,\ell (q)) =\{ B \cup \eta _k({p}) \}_{B\in \mathcal {S} (\eta _k(Y'),\eta _k(q'))}$$. Thus, for any $$E\in \mathcal {S} (Y,q)$$ there is $$B\in \mathcal {S}(Y',q')$$ such that $$\eta _k(E) =\eta _k(B\cup \{p\})$$. Since $$\eta _{k|E}$$ is injective by Remark 1.10 and $$\mathcal {S} (Y,q)\supseteq \{B\cup \{p\}\}_{B\in \mathcal {S} (Y',q')}$$, we get $$\mathcal {S} (Y,q) =\{B\cup \{p\}\}_{B\in \mathcal {S} (Y',q')}$$.

Now assume also $$n_2=2$$, so that we must have $$k\ge 3$$. Let $$Y=\mathbb {P}^2\times \mathbb {P}^2\times (\mathbb {P}^1)^{k-2}$$ and fix $$o\in \mathbb {P}^2{\setminus } \pi _2(Y')$$. Set $$M:{=} \pi _2^{-1}(o)$$, and $$\Lambda :{=} \langle \nu (M)\rangle$$. Then $$r=9\cdot 2^{k-2}-1$$, $$\dim \Lambda =9\cdot 2^{k-3}-1$$. Let $$r':=9\cdot 2^{k-3}-1$$ and consider the linear projection $$\ell : \mathbb {P}^r{\setminus } \Lambda \rightarrow \mathbb {P}^{r'}$$ from $$\Lambda$$ which acts on $$\nu (Y)$$ as the composition of the Segre embedding and the map $$\mathbb {P}^2\times \mathbb {P}^2\times (\mathbb {P}^1)^{k-2}{\setminus } \mathbb {P}^2\times \{o\}\times (\mathbb {P}^1)^{k-2} \rightarrow \mathbb {P}^2\times (\mathbb {P}^1)^{k-1}$$, which is the linear projection $$\mathbb {P}^2{\setminus } \{o\}\rightarrow \mathbb {P}^1$$ on the second factor and the identity on any other factor. Since $$(Y'\cup \{p\})\cap M =\emptyset$$, $$\ell (q)$$ is well defined. We conclude since we already proved the statement in the case where only one of the factors is a $$\mathbb {P}^2$$.

## 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.

### Lemma 4.1

Let q be a not-identifiable tensor and let A and B two distinct sets evincing the rank of q. Define $$S:=A \cup B$$. If $$\sharp (S) \ge 5$$ and $$\dim \langle \nu (S) \rangle =2$$, then the rank of q cannot be 3.

### Proof

Assume the existence of such a rank-3 tensor q with 2 distinct decompositions A and B s.t. $$\sharp (A \cup B) \ge 5$$. The plane $$\langle \nu (S) \rangle$$ contains at least five not-collinear points. Note that $$\langle \nu (S) \rangle \not \subseteq X$$, otherwise also $$q\in X$$ which contradicts $$r_X(q)=3$$. So $$\langle \nu (S) \rangle \cap X$$ contains a conic $$\mathcal {C}$$. Either if it is reduced or not, the two secant variety of $$\mathcal {C}$$ fills $$\langle \nu (S) \rangle =\mathbb {P}^2$$. So $$r_X(q) \le 2$$, which is an absurd.

### Lemma 4.2

Let q be a not-identifiable rank-3 tensor and let $$A,B \in \mathcal {S}(Y,q)$$ be distinct. Then $$\sharp (A \cap B) \le 1$$.

### Proof

Suppose, by contradiction, that A and B have 2 distinct points in common and call the set of these two points E. Let $$A=E \cup \{u\}$$ and $$B=E \cup \{v\}$$. Since the rank of q is 3, $$q \notin \langle \nu (E) \rangle$$, but since by definition $$q\in \langle \nu (A) \rangle \cap \langle \nu (B) \rangle$$ we have that $$\langle \nu (E)\rangle \subsetneq \langle \nu (A) \rangle \cap \langle \nu (B) \rangle$$. Clearly $$\langle \nu (E)\rangle$$ is a line; therefore, $$\dim \langle \nu (A) \rangle \cap \langle \nu (B) \rangle > 1$$, but $$\langle \nu (A) \rangle$$ and $$\langle \nu (B) \rangle$$ are both planes, so $$\langle \nu (A) \rangle =\langle \nu (B) \rangle$$. In the plane $$\langle \nu (A) \rangle$$, we have two different lines: $$\nu (E)$$ and $$\langle \nu (u),\nu (v) \rangle$$, which mutually intersect in at most a point $$q'$$. Remark that $$q' \notin X$$ because otherwise the line $$\langle \nu (E)\rangle$$ would have at least 3 points of rank 1 and so we would have $$\langle \nu (E)\rangle \subset X$$, contradicting Remark 1.10. So $$r_X(q')=2$$ and $$\sharp \mathcal {S}(Y,q')\ge 2$$, by Proposition 2.1 we get that actually $$q' \in \langle \nu (Y')\rangle$$, where $$Y'=\mathbb {P}^1 \times \mathbb {P}^1$$. But also $$E, \{ u,v\} \subset Y'$$, so $$q \in \langle \nu (Y')\rangle$$, 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 $$A \cup B$$ 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 $$\sharp A\cup B =5$$ in Sect. 5 from the case $$\sharp A\cup B=6$$ 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 $$E\subset Y\simeq \mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$ of cardinality at most 3 does not impose independent conditions to multilinear forms over $$Y_i\simeq \mathbb {P}^{n_1}\times \cdots \times \hat{\mathbb {P}^{n_i}} \times \cdots \times \mathbb {P}^{n_k}$$, $$i=1, \ldots , k$$ (i.e. $$h^1(\mathcal {I}_E(\hat{\varepsilon }_i)) >0$$) if and only if one of the following cases occurs:

1. 1.

$$\sharp (E) =3$$ and there is $$j\in \{1,\dots ,k\}{\setminus } \{i\}$$ such that $$\sharp (\pi _h(E)) =1$$ for all $$h\notin \{i,j\}$$;

2. 2.

there are $$u, v \in E$$ such that $$u\ne v$$ and $$\eta _i(u)=\eta _i(v)$$.

### Proof

The fact that both items 1. and 2. imply that $$h^1(\mathcal {I}_E(\hat{\varepsilon }_i)) >0$$ is obvious. Let us describe the other implication.

By definition $$H^0(\mathcal {O}_Y(\hat{\varepsilon }_i)) \cong H^0(\mathcal {O}_{Y_i}(1,\dots ,1)$$, and $$\mathcal {O} _Y(\hat{\varepsilon }_i)$$ is not a very ample line bundle. So we cannot be sure about the injectivity of the restriction $$\eta _{i|E}$$ of $$\eta _i$$ to the finite set E.

If $$\eta _{i|E}$$ is not injective one immediately gets that $$h^1(\mathcal {I}_E(\hat{\varepsilon }_i) )>0$$. Moreover, if $$\eta _{i|E}$$ is not injective it means that there are 2 distinct points of E, say u and v which are mapped by $$\eta _i$$ onto the same point, i.e. we are in item 2. of this lemma.

Now assume that $$\eta _{i|E}$$ is injective (i.e. we are not in item 2.). This implies that $$\sharp E = \sharp \eta _i(E)$$. We have by hypothesis that $$h^1(\mathcal {I}_E(\hat{\varepsilon }_i))>0$$. Since by definition $$h^1(\mathcal {I}_E(\hat{\varepsilon }_i)) =h^1(Y_i, \mathcal {I}_{\eta _i(E)}(1,\dots ,1))$$ we have that $$\eta _i(E)$$ does not impose independent conditions to the multilinear forms over $$Y_i$$; therefore, $$\sharp (\eta _i(E))\ge 3$$ which clearly implies that $$\sharp (\eta _i(E))=3$$ since by hypothesis the cardinality of E is at most 3. Now $$\eta _i(E)$$ is a set of 3 distinct points on $$Y_i$$ which does not impose independent conditions to the multilinear forms over $$Y_i$$, and $$\mathcal {O}_{Y_i}(1,\dots ,1)$$ is very ample, therefore the 3 points of $$\eta _i(E)$$ must be mapped to collinear points by the Segre embedding $$\nu _i$$ of $$Y_i$$. Hence, by the structure of the Segre variety $$\nu _i(Y_i)$$, we get that $$\langle \nu _i(\eta _i(E))\rangle \subseteq \nu _i(Y_i)$$ and there is $$j\in \{1,\dots ,k\}{\setminus } \{i\}$$ such that $$\sharp (\pi _h(\eta _i(E))) =1$$ for all $$h\notin \{i,j\}$$. Since $$h\ne i$$, we have $$\pi _h(\eta _i(E)) =\pi _h(E)$$.

## 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 $$\sharp A\cup B$$ can only be either 5 or 6. This section is fully devoted to the case in which $$\sharp A\cup B=5$$, i.e. A and B share only one point and call it p:

\begin{aligned} S:= A\cup B, \; \; \sharp S=5, \; \; A\cap B =\{p\} \hbox { and } A'=A {\setminus } \{ p \}, \; \;B'=B{\setminus } \{ p \}. \end{aligned}
(5.1)

The matrix case is well known, therefore we will always assume that q is an order-$$k\ge 3$$ tensor, i.e. $$q\in \langle \nu (Y) \rangle$$ with $$Y=\prod _{i=1}^k\mathbb {P}^{n_i}$$ and $$k\ge 3$$.

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 $$\mathbb {P}^1$$s only (see Proposition 5.2).

This will completely cover the cases of not-identifiable rank-3 tensors with the condition (5.1) since, by Remark 1.9, the concise Segre of a rank-3 point q is $$X_q= \nu (\mathbb {P}^{n_1} \times \cdots \times \mathbb {P}^{n_k} )$$, with $$n_1, \dots ,n_k \in \{1,2 \}$$.

### 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. $$Y=\mathbb {P}^2\times \mathbb {P}^{n_2}\times \cdots \times \mathbb {P}^{n_k}$$ with $$n_i\in \{1,2\}$$ for $$i=1, \ldots , k$$ and $$k\ge 3$$. Let $$q\in \sigma _3^0(\nu (Y))$$, with $$\nu (Y)$$ the concise Segre of q. If there exist two sets $$A,B\in \mathcal {S}( Y,q)$$ evincing the rank of q such that $$\sharp A\cap B=1$$ then q is as in Proposition 3.10.

### Proof

Consider a divisor $$M\in \vert \mathcal {O}_Y(\varepsilon _1) \vert$$ containing $$A'=A{\setminus } \{ p\}$$. By Concision/Autharky $$S \nsubseteq M$$, so, by Lemma 1.13, either $$h^1(\mathcal {I}_{S {\setminus } S \cap M}(\hat{\varepsilon }_1))>0$$ or $$p \notin M$$ and $$A'\cup B'\subset M$$. We study separately the two cases.

1. 1.

First assume $$h^1(\mathcal {I}_{S {\setminus } S \cap M}(\hat{\varepsilon }_1))>0$$.

The divisor M contains $$A'$$ by definition so $$\sharp (S {\setminus } S\cap M) \le 3$$; moreover, if we define $$Y_1:=\mathbb {P}^{n_2}\times \cdots \times \mathbb {P}^{n_k}$$ with $$n_i=1,2$$ for $$i=2, \ldots , k$$, we have that $$\mathcal {O}_{Y_1}(1,\ldots , 1)$$ is very ample; therefore, we can apply Lemma 4.4 and say that one of the following occurs:

1. (i)

$$\sharp (S{\setminus } S\cap M)=3$$ and there exists a projection $$\pi _i$$, with $$i\in \{2, \ldots ,k \}$$ such that $$\sharp (\pi _i(S{\setminus } S\cap M))=1$$;

2. (ii)

There exist $$u,v \in (S {\setminus } S\cap M)$$ such that $$u\ne v$$ and $$\eta _1(u)=\eta _1(v).$$

We remark that case (ii) implies that $$\pi _i(u)=\pi _i(v)$$ for all $$i >1$$. Since M contains $$A'$$, we have that $$S {\setminus } S \cap M =\{u,v\} \subseteq B$$, we can exclude case (ii) thanks to Remark 1.10.

So only case (i) is possible. Since $$\sharp (S{\setminus } S\cap M)=3$$ we have that $$S {\setminus } S \cap M=B$$ and there exist an index $$i \in \{ 2,\ldots , k \}$$ such that $$\sharp \pi _i(S{\setminus } S\cap M)=1$$. The fact that there is $$i\in \{2,\ldots , k\}$$ such that $$\sharp (\pi _i(B)) =1$$, means that B only depends by $$k-1$$ factors, contradicting Autarky.

2. 2.

Now assume $$A'\cup B'\subset M$$.

Let $$Y''$$ be the minimal multiprojective space contained in M and containing $$A'\cup B'$$. Since $$q\in \langle \langle \nu (Y'')\rangle \cup \{p\})\rangle$$ and $$p\notin Y''$$, there is a unique $$o\in \langle \nu (Y'')\rangle$$ such that $$q\in \langle \{\nu ({p}),o\}\rangle$$. Since $$\langle \nu (A)\rangle$$ (resp. $$\langle \nu (B)\rangle$$) is a plane containing $$\nu ({p})$$ and q, there is a unique $$o_1\in \langle \nu (A')\rangle$$ (resp. $$o_2\in \langle \nu (B')\rangle$$) such that $$q\in \langle \{\nu ({p}),o_1\}\rangle$$ (resp. $$q\in \langle \{\nu ({p}),o_2\}\rangle$$). The uniqueness of o gives $$o=o_1=o_2$$. Since $$o_1=o_2$$, we get a tensor of rank 2 with $$A'$$ and $$B'$$ as solutions. Thus, q is as described in Proposition 3.10.

### Proposition 5.2

Let $$Y=(\mathbb {P}^1)^{\times k}$$ with $$k\ge 3$$ and let $$q\in \sigma _3^0(\nu (Y))$$ be such that there exist two different sets $$A,B\in \mathcal {S}(Y,q)$$ with the property $$\sharp (A\cup B)=5$$, where $$\nu (Y)$$ is the concise Segre of q. Then k can only be either 3 or 4. If $$k=3$$ then q belongs to a tangent space of $$\nu ((\mathbb {P}^1)^{\times 3})$$ and $$\dim (\mathcal {S}(Y,q))\ge 2$$. If $$k=4$$ then $$\dim (\mathcal {S}(Y,q))\ge 1$$.

### Proof

If $$k=3$$ then the only rank-3 tensors in $$\langle \nu (\mathbb {P}^1)^{\times 3})\rangle$$ are those belonging to the the tangential variety of the Segre variety (cf. [9, 24]) for which $$\dim (\mathcal {S}(Y,q))\ge 2$$ (cf. [1, 8, 27, 28]).

The case $$k=4$$ is covered by Remark 3.2.

Assume $$k>4$$ and write $$Y=\prod _{i=1}^k\mathbb {P}^1_i$$. Let $$S=A\cup B$$ as in (5.1).

We build a recursive set of divisors to being able to cover the whole set S as follows. Let $$o_i\in \mathbb {P}^1_i$$, $$i=2,3,4$$ be such that:

1st divisor:

$$\pi _4^{-1}(o_4)\cap S \ne \emptyset$$ and call $$M_4:=\pi _4^{-1}(o_4)$$;

2nd divisor:

$$\pi _3^{-1}(o_3)\cap (S {\setminus } (S\cap M_4)) \ne \emptyset$$ and call $$M_3:=\pi _3^{-1}(o_3)$$.

3rd divisor:

If $$M_3\cup M_4$$ already covers the whole S (i.e. $$S\subset M_3\cup M_4$$), set $$M_2$$ to be any divisor $$M_2\in |\mathcal {O} _Y(\hat{\varepsilon }_2)|$$.

4th divisor:

Otherwise, if $$S\nsubseteq M_3\cup M_4$$, choose $$o_2\in \mathbb {P}^1_2$$ such that $$\pi _2^{-1}(o_2)\cap (S {\setminus } S\cap (M_3\cup M_4))\ne \emptyset$$ and set $$M_2:=\pi _2^{-1}(o_2)$$.

Now it may happen that either $$S\subset M_2\cup M_3 \cup M_4$$ or not. We study those two cases in (a) and (b), respectively.

1. (a)

Here we assume that $$S\subset M_2\cup M_3 \cup M_4$$. Since $$\sharp (S)=5$$ there is at least one of the $$M_i$$s containing at least two points of S, and there are two of the $$M_i$$s whose union contains at least 4 points of S: wlog we may assume that $$\sharp (S \cap (M_3\cup M_4))\ge 4$$.

• Assume $$\sharp (S\cap (M_3\cup M_4))= 4$$. Since $$\mathcal {O} _Y(1,1,0,0,\dots )$$ is globally generated, we have that $$h^1(\mathcal {I} _{S{\setminus } S\cap (M_3\cup M_4)}(1,1,0,0,1,1,\dots )) =0$$, contradicting Lemma 1.13.

• Assume $$S\subset M_3\cup M_4$$. Therefore, there is one of the $$M_i$$s containing at least 3 points of S,  let $$\sharp (M_4\cap S) \ge 3$$. Since $$S\nsubseteq M_4$$, we get $$h^1(\mathcal {I} _{S{\setminus } S\cap M_4}(\hat{\varepsilon }_4)) >0$$ (by Lemma 1.13); hence, $$\sharp (S{\setminus } S\cap M_4) =2$$ and

\begin{aligned} S{\setminus } S\cap M_4=\{u,v\} \hbox { with } \pi _i(u)=\pi _i(v), \; \forall i\ne 4.\end{aligned}
(5.2)

Since $$h^1(\mathcal {I} _{S{\setminus } S\cap M_3}(\hat{\varepsilon }_3)) >0$$ (again by Lemma 1.13, we get that either there are $$w, z\in S{\setminus } S\cap M_3$$ such that $$w\ne z$$, $$\pi _i(w)=\pi _i(z)$$ for all $$i\ne 3$$ or $$\nu _4(\eta _4(S\cap M_4))$$ (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

\begin{aligned} w, z\in S{\setminus } S\cap M_3 \hbox { such that } w\ne z, \, \pi _i(w)=\pi _i(z) \; \forall i\ne 3. \end{aligned}
(5.3)

In (5.2) and (5.3), we have 4 distinct points uvwz such that $$\sharp (\pi _5(\{u,v,w,z\})) =1$$. Take $$M_5\in |\mathcal {O} _Y(\varepsilon _5)|$$ containing $$\{u,v,w,z\}$$. Since $$h^1(\mathcal {I} _{S{\setminus } S\cap M_5}(\hat{\varepsilon }_5)) =0$$, Autarky and Lemma 1.13 give a contradiction.

2. (b)

Assume $$S\nsubseteq M_2\cup M_3\cup M_4$$. By Lemma 1.13, we get $$h^1(\mathcal {I} _{S{\setminus } S\cap (M_2\cup M_3\cup M_4)}(1,0,0,0,1,1,\dots ))>0$$. Thus, $$\sharp (S{\setminus } (M_2\cup M_3\cup M_4)) =2$$, say $$S{\setminus } (M_2\cup M_3\cup M_4) =\{u,v\}$$ and $$\pi _i(u) =\pi _i(v)$$ for all $$i\ne 2,3,4$$. But in this case it is sufficient to change the original choice of $$o_4$$ and take as $$o_4$$ the point $$\pi _4(u)$$ and the the new divisor $$M_4$$ will contain 2 points of S, i.e. uv therefore we are able to get new divisors $$M_2$$, $$M_3$$ with the same construction as above leading to the case $$S\subset M_2\cup M_3\cup M_4$$ excluded in step (a).

## 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 $$\sharp A\cup B$$ can only be either 5 or 6. This section is fully devoted to the case in which $$\sharp A\cup B=6$$, i.e. A and B are disjoint:

\begin{aligned} S:= A\cup B, \; \; \sharp S=6, \; \; A:=\{a_1,a_2,a_3\}, B:=\{b_1,b_2,b_3\} \; \; A\cap B =\emptyset . \end{aligned}
(6.1)

First of all, let us show that if q is a rank-3 tensor whose concise Segre $$\nu (Y)$$ has at least two factors of projective dimension 2, it never happens that in $$\mathcal {S}(Y,q)$$ there are two disjoint sets.

### Remark 6.1

Let $$Y=(\mathbb {P}^{2})^{\times k_1}\times (\mathbb {P}^{1})^{\times k_2}$$ and $$S\subset Y$$ a set of 6 distinct points. Consider $$I\subseteq \{k_1+1, \ldots , k_1+k_2\}$$ and $$\varepsilon :=\sum _{i\in I}\varepsilon _i$$. Suppose there exists a divisor $$M\in |\mathcal {O}_Y(\varepsilon )|$$ intersecting S in 4 points. Call $$\{u,v\}:=S{\setminus } (S\cap M)$$. In this setting one can apply Lemma 1.13 and get that $$h^1(\mathcal {I}_{\{u,v\}}(\widehat{\varepsilon } ))>0$$ (where $$\widehat{\varepsilon }$$ is a $$(k_1+k_2)$$-tuple with 0s in position of the indices appearing in $$\varepsilon$$ of I and 1s everywhere else) and $$\pi _h(u)=\pi _h(v)$$ for any $$h\in \{1,\ldots , k_1+k_2\}{\setminus } I$$.

### Proposition 6.2

Let Y be a multiprojective space with at least three factors and at least two of them of projective dimension 2, i.e. $$Y=\mathbb {P}^2\times \mathbb {P}^2\times \mathbb {P}^{n_3}\times \cdots \times \mathbb {P}^{n_k}$$ with $$n_i\in \{ 1,2\}$$ for $$i=1,\dots ,k$$ and $$k\ge 3$$. Let $$q\in \sigma _3^0(\nu (Y))$$, with $$\nu (Y)$$ the concise Segre of q. If $$A,B\in \mathcal {S}( Y,q)$$ evince the rank of q, then A and B cannot be disjoint.

### Proof

The proof is by absurd: assume that there exist $$A,B\in \mathcal {S}( Y,q)$$ with $$A\cap B= \emptyset$$. By Remark 3.3, we have that $$\langle \pi _i(A)\rangle = \langle \pi _i(B)\rangle =\mathbb {P}^2$$ for $$i=1,2$$. Fix $$W\in |\mathcal {I}_B(\varepsilon _2+\varepsilon _3)|$$ (it exists, because $$h^0(\mathcal {O}_Y(\varepsilon _2+\varepsilon _3)) = h^0(\mathbb {P}^2\times \mathbb {P}^{n_3}, \mathcal {O}_{\mathbb {P}^2\times \mathbb {P}^{n_3}}(1,1)) = 3(n_3+1)>4$$). Since $$\pi _{1|A}$$ is injective, we have $$h^1(\mathcal {I} _A(\varepsilon _1))=0$$. Thus, $$S\subset W$$ by Lemma 1.13. In this way, we have shown that

\begin{aligned} \hbox {any divisor } D \in \vert \mathcal {O}_Y(\varepsilon _2+\varepsilon _3) \vert \hbox { containing } B \hbox { contains also }A . \end{aligned}
(*)

Claim 6.2.1 $$\pi _3(a_i)=\pi _3(b_i)$$ where $$a_i,b_i$$ are as in (6.1), for $$i=1,2,3$$.

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 $$\pi _j(a_i)=\pi _j(b_i)$$ for $$i=1,2,3$$ and for $$j=1, \ldots , k$$ 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 $$J_3\subset \mathbb {P}^{n_3}$$ containing $$\pi _3({b_i})$$, (where the $$b_i$$s are as in (6.1), $$i=1,2,3$$) 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 $$M_3:= \pi _3^{-1}(J_3)$$. Take a line

\begin{aligned} L_2\subset \mathbb {P}^2 \hbox { containing }\{\pi _2(b_j),\pi _2(b_k)\} \hbox { with } j,k\ne i \hbox { and set } M_2:= \pi _2^{-1}(L_2).\qquad \end{aligned}
(**)

We have $$B\subset M_2\cup M_3\in |\mathcal {O} _Y(\varepsilon _2+\varepsilon _3)|$$. Thus, from (*), we get that $$M_2\cup M_3$$ contains also A. Since $$A\nsubseteq M_2$$ by Autarky, there is $$a\in A\cap M_3$$, i.e. there is $$a\in A$$ such that

\begin{aligned} \pi _3(a) =\pi _3({b_i}) \end{aligned}
(6.2)

(in fact if $$n_3=1$$ it is trivial, if $$n_3=2$$ then we have already remarked that $$\pi _3(b_i)$$ is the only point of $$J_3$$ belonging to $$\pi _3(Y)$$). Since $$\pi _{i\vert A}$$ is injective for $$i=1,2$$ (cf. Remark 3.3), the points of A projecting on $$\pi _3(b_i)$$ are different for different is except if there are $$b_i\ne b_j$$ such that $$\pi _3(b_i)=\pi _3(b_j)$$. Suppose that this is the case. By Lemma 4.4 we get that $$\sharp (S{\setminus } S\cap M_3)=2$$. Thus, if for $$i\ne j$$ $$\pi _3(b_i)=\pi _3(b_j)$$ there are 2 points of A and 2 points of B in $$M_3$$, i.e. $$\sharp (S\cap M_3)= 4$$. Suppose that $$S\cap M_3=\{ a_3,b_3,a_2,b_2\}$$. By [13, Lemmas 2.4 and 2.5] (also [7, Lemma 5.1, item (b)]) $$h^1(\mathcal {I}_{S{\setminus } S\cap M_3}(\hat{\varepsilon }_3))>0$$, i.e. $$\pi _i(a_1)=\pi _i(b_1)$$ for all $$i\ne 3$$. This is a contradiction since we already know that $$\pi _3(a_2)=\pi _3(b_2)$$ and we would have $$a_2=b_2$$, which contradicts the assumption that $$A\cap B=\emptyset .$$

Therefore, the points $$a\in A$$ of (6.2) are all different for different choices of is. So we may assume that $$\pi _3(a_i)=\pi _3(b_i)$$ for $$i=1,2,3$$ and the $$\pi _3(b_i)\ne \pi _3(b_j)$$ for $$i\ne j$$.

The argument of the proof of Claim 6.2.1 can be repeated verbatim for all the others $$\pi _j$$s with the only caution that when we do the case $$j=2$$ we have to use a line $$L_1\subset \mathbb {P}^2$$ containing $$\{\pi _1(b_j),\pi _1(b_k)\}$$ with $$j,k\ne i$$ and set $$M_1:= \pi _1^{-1}(L_1)$$ instead of $$M_2$$ and $$L_2$$ in (**). Moreover, (*) clearly holds if we replace the $$\varepsilon _2$$ with $$\varepsilon _1$$ and $$\varepsilon _3$$ with $$\varepsilon _j$$ for any $$j=3, \ldots , k$$. As already highlighted this concludes the proves since $$\pi _j(a_i)=\pi _j(b_i)$$ for $$i=1,2,3$$ and for $$j=1, \ldots , k$$ which is a contradiction with A and B being distinct.

This shows that under the assumption (6.1), we can exclude the case where the Segre variety has at least two factors of projective dimension 2.

Let us focus on the four-factor case.

### Proposition 6.3

Let $$Y=\mathbb {P}^2\times \mathbb {P}^1\times \mathbb {P}^1 \times \mathbb {P}^{1}$$. Let $$q\in \sigma _3^0(\nu (Y))$$, with $$\nu (Y)$$ the concise Segre of q. There do not exist two disjoint sets $$A,B \in \mathcal {S}(Y,q)$$ evincing the rank of q.

### Proof

Assume by contradiction that there exist two disjoint sets $$A,B \in \mathcal {S}(Y,q)$$ evincing the rank of q and, moreover, assume that no $$\eta _{i\vert S}$$ is injective, for $$i=2,3,4$$.

By Remark 1.10, for each $$i=2,3,4$$ there exists $$a \in A$$, $$b \in B$$ such that $$\eta _i(a)=\eta _i(b)$$. Fix $$H:=\pi _1^{-1}(L)$$, where $$L\subset \mathbb {P}^2$$ is a line containing $$\pi _1(a_1)$$ and $$\pi _1(a_2)$$, where $$a_1,a_2\in A$$. Since we assumed that no $$\eta _{i\vert S}$$ is injective, then there exist $$b_1,b_2 \in B$$ such that $$\pi _1(a_i)=\pi _1(b_i)$$, for $$i=1,2$$. Thus, $$H\supset \{a_1,a_2,b_1,b_2 \}$$ and by Autarky $$S \not \subset H$$, so there is at least an element of S out of H, e.g. $$a_3\in S{\setminus } \{a_1,a_2,b_1,b_2 \}$$. Thus, we have $$h^1(\mathcal {I}_{S{\setminus } S\cap H}(0,1,1,1))=0$$ contradicting Lemma 1.13. So there exists at least one integer $$h\in \{2,\dots ,4\}$$ such that $$\eta _{h\vert S}$$ is injective.

First, define recursively the integers such that the preimages of points $$o\in \mathbb {P}^1$$ intersect maximally the set S:

\begin{aligned} \alpha _4:=\max \{\sharp (\pi _i^{-1}(o)\cap S)\}_{o \in \mathbb {P}^1;i=2,\ldots , 4}. \end{aligned}
(6.3)

By rearranging if necessary, we can assume that the index $$i=2, \ldots , 4$$ realizing $$\alpha _4$$, is $$i=4$$. Call $$K_4:=\pi _4^{-1}(o)$$. Then define

\begin{aligned} \alpha _3:=\max \{\sharp (\pi _i^{-1}(o)\cap (S{\setminus } (S\cap K_4)))\}_{o \in \mathbb {P}^1;i=2,3}. \end{aligned}
(6.4)

By rearranging if necessary, we can assume that the index $$i=2,3$$ realizing $$\alpha _3$$, is $$i=3$$. Call $$K_3:=\pi _3^{-1}(o)$$. Finally define

\begin{aligned} \alpha _2:=\max \{\sharp (\pi _2^{-1}(o)\cap (S{\setminus } (S\cap K_4\cup K_3)))\}_{o\in \mathbb {P}^1}. \end{aligned}
(6.5)

So if we denote by $$o_j\in \mathbb {P}^1$$, $$j=2,3,4$$ the points realizing $$\alpha _2, \alpha _3, \alpha _4$$, respectively, then we call

\begin{aligned} K_j:=\pi _j^{-1}(o_j) \hbox { for } j=2,3. \end{aligned}
(6.6)

Remark that by Autarky assumption $${1 \le \alpha _3 \le \alpha _4\le 5 }$$.

It is easy to see that $$\alpha _4$$ cannot be 5. In fact if $$\alpha _4=5$$, then $$\sharp (S{\setminus } S\cap K_4)=1$$ which implies that $$h^1(\mathcal {I}_{S{\setminus } S\cap K_4}(1,1,1,0))=0$$, which is a contradiction with Lemma 1.13.

So the possibilities for $$\alpha _3$$ and $$\alpha _4$$ are $$1\le \alpha _3 \le \alpha _4 \le 4$$.

Let us show that

Assume that $$(\alpha _2, \alpha _3, \alpha _4)=(1,1,1)$$. In such a case, the divisor $$K_2\cup K_3\cup K_4\in \vert \mathcal {O}_Y(\hat{\varepsilon }_1) \vert$$ would contain exactly 3 points of S. Moreover, if $$h^1(\mathcal {I}_{S{\setminus } (S\cap K_2\cup K_3\cup K_4)}(\varepsilon _1))>0$$ then by Lemma 4.4 we would have a contradiction with $$(\alpha _3,\alpha _4)=(1,1)$$. Therefore, if $$(\alpha _2, \alpha _3, \alpha _4)=(1,1,1)$$, we must have $$h^1(\mathcal {I}_{S{\setminus } (S\cap K_2\cup K_3\cup K_4)}(\varepsilon _1))=0$$, but this is a contradiction with Lemma 1.13. Thus, if $$\alpha _2=1$$ then $$K_3\cup K_4$$ should contain at least 3 points of S, i.e. $$\alpha _3\ge 1$$ and $$\alpha _4\ge 2$$.

Now assume that $$(\alpha _2,\alpha _3)=(1,1)$$. Then $$\pi _{3|S}$$ is injective. The idea is to build a divisor $$F\in \vert \mathcal {O}_Y(\varepsilon ) \vert$$ with $$\varepsilon =\sum _{i\in I} \varepsilon _i$$, for some finite $$I \in \{1,\dots ,k \}$$, such that $$\sharp (S{\setminus } F\cap S)=2$$ and apply Remark 6.1 to F: the existence of such an F will contradict the injectivity of $$\pi _{3\vert S}$$. Let $$H_i \in |\mathcal {O}_Y(\varepsilon _i)| \hbox { such that }H_i\cap (S{\setminus } S \cap K_4) \ne \emptyset$$ for $$i=2,3$$. The divisor F is either $$F=K_4$$, or $$F=K_4\cup H_3$$ or $$K_4 \cup H_2 \cup H_3$$ if $$\alpha _4=4,3,2$$, respectively. The case $$(\alpha _2,\alpha _3,\alpha _4)=(1,2,2)$$ can be easily excluded since $$\sharp (S{\setminus } S\cap K_2\cup K_3\cup K_4)=1$$ and by Lemma 1.13, we would have $$h^1(\mathcal {I}_{S{\setminus } S\cap (K_2\cup K_3\cup K_4)}(\varepsilon _1))>0$$, which is absurd. For the same reason, $$(\alpha _2,\alpha _3,\alpha _4)=(1,2,3)$$ is also impossible because then $$\sharp (S\cap (K_3\cup K_4))=5$$ and by Lemma 1.13 we would have $$h^1(\mathcal {I}_{S{\setminus } S\cap (K_3\cup K_4)}(1,1,0,0))>0$$, which is a contradiction. This shows $$\alpha _2\ne 1$$.

We are, therefore, left with $$\alpha _2\ne 1 < \alpha _3 \le \alpha _4 =2,3,4$$.

Suppose that $$\alpha _3=\alpha _4=2$$. With these assumption one also gets $$\alpha _2=2$$. Indeed on the one hand, we just showed that we may always take $$H\in \vert \mathcal {O}_Y(\varepsilon _2)\vert$$ such that $$\sharp (S{\setminus } S\cap (K_3 \cup K_4)) \cap H)\ne 0$$, so such a H intersects S non-trivially and $$K_2$$ is among those Hs. On the other hand, $$\alpha _2\ne 1$$ by (). So $$\sharp (S\cap K_2)=2$$. By the construction of the $$K_i$$s in (6.6) for $$i=2,3,4$$, it is easy to show that

\begin{aligned} S=\mathbin {\coprod }_{i=2}^4 S\cap K_i. \end{aligned}

So, since $$S=\mathbin {\coprod }_{i=2}^4 S\cap K_i$$ and $$\sharp (S\cap K_i)=2$$ for $$i=2,3,4$$, we can apply Remark 6.1 separately to the divisors $$K_i\cup K_j$$ with $$i\ne j$$ and get that $$h^1(\mathcal {I}_{S\cap K_i}(\varepsilon _1+\varepsilon _i))>0$$ for $$i=2,3,4$$ and so $$\pi _1(S\cap K_i)=1$$ for $$i=2,3,4$$. To get a contradiction it is sufficient to apply again Remark 6.1 to $$\pi _1^{-1}(\langle \pi _1(S\cap K_3) , \pi _1(S\cap K_2) \rangle )$$. This shows that $$\sharp (\pi _i(S\cap K_4))=1$$ for $$i=2,3,4$$. Now since also $$\sharp (\pi _1(S\cap K_4))=1$$, then $$\sharp (S\cap K_4)=1$$, which is a contradiction with the assumption $$\alpha _3=2$$.

This proves that $$1<\alpha _2 \le \alpha _3$$, and $$2<\alpha _4=3,4$$.

The case $$(\alpha _3,\alpha _4)=(2,4)$$ can be excluded using the same argument of the case $$(\alpha _2,\alpha _3,\alpha _4)=(2,2,2)$$ above applying Remark 6.1 since if $$(\alpha _3,\alpha _4)=(2,4)$$ we have that $$K_4$$ plays the role of M in the remark.

We are, therefore, left with the unique possibility of $$(\alpha _3, \alpha _4)=(3,3)$$.

Claim 6.3.1 $$\sharp (\pi _2(S\cap K_4))=1$$.

### Proof of Claim 6.3.1:

Since we are in the hypothesis $$\alpha _4=3$$, the projection of $$S\cap K_4$$ 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. Call Z the image of the projection of $$S\cap K_4$$ onto the first two factors. Since $$h^1(\mathbb {P}^2\times \mathbb {P}^1,\mathcal {I}_Z(1,1))>0$$ those points must lie on a line L when applying the Segre embedding. Moreover, from Remark 3.3 we know that $$\pi _1( A)$$ and $$\pi _1( B)$$ are sets of linearly independent points and since linear subspaces of the Segre variety are all contained in a factor, we get that $$L \subset \mathbb {P}^2$$. Thus, $$\sharp (\pi _2(S\cap K_4))=1$$ proving the claim in this case.

If the projection of $$S\cap K_4$$ onto the first two factors is made by less than 3 points, there exist at least two points, $$u,v \in S \cap K_4$$ such that they share the same image under the projection. Remark that if we consider $$E\subset S\cap K_4$$ such that $$\sharp E=2$$ and take $$T\in \vert \mathcal {I}_E(1,1,0,0) \vert$$, then $$T\supset S\cap K_4$$. Indeed if $$S\cap K_4\not \subset T$$ then we have that $$T\cup K_3$$ contains exactly five points of S, which leads to a contradiction because by Lemma 1.13 we would have $$h^1(\mathcal {I}_{S{\setminus } S\cap T \cup K_3}(\hat{\varepsilon }_3))>0$$. Therefore, also the third point of $$S\cap K_4$$ share the same image of u and v and we are done.

Using the third factor instead of the second one, one gets $$\sharp (\pi _3(K_4\cap S))=1$$ and since we assumed that $$\alpha _4$$ is reached on the fourth factor we also have $$\sharp (\pi _4(K_4\cap S))=1$$. The same argument can be applied to $$S\cap K_3$$ which leads to $$\sharp (\pi _2(K_3\cap S))= \sharp (\pi _4(K_3\cap S))=1$$. Thus, $$\sharp (\pi _i(K_4\cap S))= \sharp (\pi _i(K_3\cap S))=1$$ for all $$i>1$$ which contradicts Autarky.$$\square$$

Since the identifiability of rank-3 tensors in $$\langle \nu ((\mathbb {P}^1)^{\times 4}) \rangle$$ 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. So we will deal with $$Y=\mathbb {P}^{n_1}\times (\mathbb {P}^1)^{l}$$, with $$n_1=1,2$$ and $$l\ge 4$$.

### Lemma 6.4

Let q be a rank-3 tensor of order at least 5 and let $$\nu (Y)$$ be its concise Segre. If there exist two disjoint sets $$A,B\in \mathcal {S}(Y,q)$$ as in (6.1), then there exists at least an index $$i \in \{1,\dots , k \}$$ such that $$\eta _{i\vert S}$$ and $$\pi _{i\vert S}$$ are injective.

### Proof

[Injectivity of $$\eta _{i\vert S}$$.]

Assume that no $$\eta _{i\vert S}$$ is injective, then by Remark 1.10 for any $$i=1,\dots ,k$$ there exist an element $$a\in A$$ and an element $$b \in B$$ such that $$\pi _h(a) =\pi _h(b)$$ for any $$h\ne i$$. It is easy to check that this condition, applied to two disjoint sets of 3 points each, and at least five $$\eta _i$$s, imposes either that $$A\cap B \ne \emptyset$$ (contradiction) or that one of the two sets (either A or B) depends only on 4 factors (contradicting Autarky).

[Injectivity of $$\pi _{i\vert S}$$.]

Assume that $$\eta _{i|S}$$ is injective and that $$\pi _{i\vert S}$$ is not injective. If $$i=1$$ and the first factor of Y is $$\mathbb {P}^2$$ take $$H\in \vert \mathcal {O}_Y(\varepsilon _i) \vert$$ as the preimage of a general line that contains exactly one point of S; otherwise take $$H\in \vert \mathcal {O}_Y(\varepsilon _i) \vert$$ as the preimage of a point of S. We remark that in both cases we get that $$\sharp (\pi _i(S\cap H))=1$$. Since by Autarky $$S\not \subset H$$, by Lemma 1.13, we have that

\begin{aligned} h^1(\mathcal {I}_{S{\setminus } S\cap H}(\hat{\varepsilon }_i))>0. \end{aligned}

We distinguish different cases depending on $$\sharp (S{\setminus } S\cap H)$$.

1. 1.

Assume $$\sharp (S{\setminus } S\cap H)=4$$ and call $$S':=\eta _i(S{\setminus } S\cap H)$$; let $$A' \subset S'$$ such that $$\sharp A'=2$$ and call $$B':=S'{\setminus } A'$$, so $$\sharp B'=2$$. Since $$\eta _{i \vert S}$$ is injective we have that $$h^1(Y_i,\mathcal {I}_{S'}(\hat{\varepsilon _i}))=h^1(\mathcal {I}_{S{\setminus } S\cap H}(\hat{\varepsilon }_i))>0$$. So $$\langle \nu _i(A')\rangle \cap \langle \nu _i(B') \rangle \ne \emptyset$$, which means that we have at least a point $$q' \in \langle \nu _5(Y_i)\rangle$$ of rank 2 for which $$A'$$ and $$B'$$ are different subsets evincing its rank. Thus, by Proposition 2.3, since $$\sharp \mathcal {S}(Y_i,q')>1$$, the points in $$A'$$ and $$B'$$ only depend on two factors, i.e. $$\sharp (\pi _j(S'))=1$$ for at least two indices $$j\in \{1,\dots ,k \}$$. Without loss of generality assume it happens for $$j=1,2$$. If the first factor of Y is $$\mathbb {P}^2$$, let $$M_1\in \vert \mathcal {I}_{\pi _1(S')}(\varepsilon _1) \vert$$ be the preimage of a general line containing $$\pi _1(S')$$ and let $$\{ M_2\}:=\vert \mathcal {I}_{\pi _2(S')}(\varepsilon _2) \vert$$. Otherwise let $$\{ M_j\}:=\vert \mathcal {I}_{\pi _j(S')}(\varepsilon _j) \vert$$, for $$j=1,2$$; in both cases then $$h^1(\mathcal {I}_{S{\setminus } S\cap M_j}(\hat{\varepsilon }_j))>0$$. So $$S{\setminus } S\cap M_j=S\cap H$$ and $$\sharp (\eta _j(S\cap H))=1$$, for $$j=1,2$$. If we call $$S\cap H=\{u,v \}$$, it follows that $$\eta _1(u)=\eta _1(v)$$ and $$\eta _2(u)=\eta _2(v)$$, so in particular we get that $$\pi _j(u)=\pi _j(v)$$ for any j, which is a contradiction.

2. 2.

Assume $$\sharp (S{\setminus } S\cap H)=3$$. By Proposition 4.4 there exists $$j\ne i$$ such that $$\sharp (\pi _h(S{\setminus } S\cap H))=1$$ for all $$h\ne i,j$$. For all $$h>1$$ with $$h\ne j,i$$, since $$h^0(\mathcal {O}_Y(\varepsilon _h))=2$$ we get $$h^0(\mathcal {I}_{S{\setminus } S\cap H}(\varepsilon _h))=1$$. Set $$\{ M_h \}:=\vert \mathcal {I}_{S{\setminus } S\cap H}(\varepsilon _h) \vert$$, if $$h=1$$ and the first factor is $$\mathbb {P}^2$$ take $$M_h\in \vert \mathcal {I}_{S{\setminus } S \cap H}(\varepsilon _h) \vert$$ as the preimage of a general line, otherwise call $$\{ M_h \}:=\vert \mathcal {I}_{S{\setminus } S\cap H}(\varepsilon _h) \vert$$. Since we took H such that $$\sharp \pi _i(S\cap H)=1$$, there exists at least an index $$t\ne i$$ such that $$\sharp \pi _t(S\cap H)\ge 2$$. Thus, we can find $$D\in \vert \mathcal {O}_Y(\varepsilon _t)\vert$$ containing exactly one point of $$S\cap H$$. For all $$s\ne t$$, set $$W_s:=M_s\cup D$$, so $$\sharp (S{\setminus } S\cap W_s)=2$$; we remark that $$W_j\cap S=W_s\cap S$$ for any js thus we may call $$E:=S{\setminus } S\cap W_s$$.

By Lemma 1.13, we have that $$h^1(\mathcal {I}_E((1,\ldots ,1)-\varepsilon _s - \varepsilon _t))>0$$, so $$\sharp \pi _j(E)=1$$ for all $$j\ne s,t$$. Since $$E\subset H$$, we have that $$\pi _i(E)=1$$; moreover, taking $$s=1,2,3$$, if $$t\ne j$$, we get that $$\sharp E=1$$; thus, a contradiction. It remains to study what happens when $$t=j$$, i.e. if $$\sharp (\pi _j(S\cap H))\ge 2$$. In such a case, when we let s varies in $$\{ 1,\ldots , k \}{\setminus } \{i,j\}$$, we get $$\sharp \pi _s(S\cap H)= 1$$. Thus, $$\eta _j(S\cap H)=1$$, i.e. the three points of $$S\cap H$$ actually lies on a line, which is a contradiction with Remark 1.10, because two of them are points of A or B.

3. 3.

Assume $$\sharp (S{\setminus } S\cap H)\le 2$$. Since $$h^1(\mathcal {I}_{S{\setminus } S \cap H}(\hat{\varepsilon }_i))>0$$, we get that $$\sharp (S {\setminus } S\cap H)=2$$ and that $$\sharp \eta _i(S{\setminus } S\cap H) =1$$, which is a contradiction.

With these two lemmas we can conclude the case of two disjoint sets $$A,B\in \mathcal {S}( Y,q)$$ with q of rank-3.

### Proposition 6.5

Let $$q\in \sigma _3^0(\nu (Y))$$ be a tensor of order $$k\ge 5$$ and let $$\nu (Y)$$ be its concise Segre. Then $$\mathcal {S}( Y,q)$$ does not contain two disjoint sets.

### Proof

By Lemma 6.4 there exists at least an index $$i \in \{1,\dots , k \}$$ such that $$\eta _{i\vert S}$$ is injective, from which follows that the corresponding $$\pi _{i\vert S}$$ is also injective. Now if $$\eta _{j\vert S}$$ is not injective for some $$j\ne i$$ then $$\pi _{i\vert S}$$ is not injective, which is a contradiction with the assumption that $$\eta _{i\vert S}$$ is injective. Therefore, thus $$\eta _{j|S}$$ and $$\pi _{j\vert S}$$ have to be injective for all $$j=1, \ldots , k$$.

Write $$A:=\{ a_1,a_2,a_3\}$$ and $$B:=\{ a_4,a_5,a_6 \}$$. If the first factor is a $$\mathbb {P}^2$$ take $$L_1 \in \mathbb {P}^2$$ as a general line containing $$\pi _1(a_1)$$ and define $$H_1\in \vert \mathcal {I}_{a_1}(\varepsilon _1)\vert$$ as $$H_1:=\pi _1^{-1}L_1$$. For $$i=2,\dots ,5$$ take $$\{ H_i \}:=\vert \mathcal {I}_{a_i}(\varepsilon _i) \vert$$ (this is possible since by hypothesis $$k\ge 5$$). Otherwise, for all $$i=1,\dots ,k$$ take $$\{ H_i \}:=\vert \mathcal {I}_{a_i}(\varepsilon _i) \vert$$. In both cases, since every $$\pi _{i\vert S}$$ is injective we get that $$H_1\cup \cdots \cup H_5$$ contains exactly 5 points of S. Thus, from Lemma 1.13, we get that $$h^1(\mathcal {I}_{S{\setminus } (S\cap H_1\cup \cdots \cup H_5)}(0,0,0,0,0,1,\dots ,1))>0$$ which is a contradiction since $$\sharp (S{\setminus } (S\cap H_1\cup \cdots \cup H_5))=1$$.

## Identifiability of Rank-3 Tensors

The following theorem completely characterizes the identifiability of any rank-3 tensor and it is the main theorem of the present paper.

### Theorem 7.1

Let $$Y=\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$ be the multiprojective space of the concise Segre of a rank-3 tensor q. Denote with $$\mathcal {S}(Y,q)$$, the set of all subsets of Y computing the rank of q. The rank-3 tensor q is identifiable except in the following cases:

1. 1.

q is a rank-3 matrix, in this case $$\dim (\mathcal {S}(Y,q))=6$$;

2. 2.

q belongs to a tangent space of the Segre embedding of $$Y=\mathbb {P}^1\times \mathbb {P}^1\times \mathbb {P}^1$$, in this case $$\dim (\mathcal {S}(Y,q))\ge 2$$;

3. 3.

q is an order-4 tensor of $$\sigma _3^0(Y)$$ with $$Y=\mathbb {P}^1\times \mathbb {P}^1\times \mathbb {P}^1\times \mathbb {P}^1$$, in this case $$\dim (\mathcal {S}(Y,q))\ge 1$$;

4. 4.

q is as in Example 3.6, where $$Y=\mathbb {P}^2\times \mathbb {P}^1\times \mathbb {P}^1$$, in this case $$\dim (\mathcal {S}(Y,q))=3$$;

5. 5.

q is as in Example 3.7, where $$Y=\mathbb {P}^2\times \mathbb {P}^1\times \mathbb {P}^1$$, in this case $$\mathcal {S}(Y,q)$$ contains two different four-dimensional families;

6. 6.

q is as in Proposition 3.10, where $$Y=\mathbb {P}^{n_1}\times \cdots \times \mathbb {P}^{n_k}$$ is such that $$k\ge 3$$, $$n_i\in \{1,2\}$$ for $$i=1,2$$, $$n_i=1$$ for $$i>2$$ and $$\sum _{i=1}^{k} n_i\ge 4$$. In this case, $$\dim (\mathcal {S}(Y,q))\ge 2$$ and if $$n_1+n_2+k\ge 6$$ then $$\dim (\mathcal {S}(Y,q))=2$$.

### Proof

In case 1. the point q is a rank-3 matrix; therefore, it is highly not-identifiable. See Remark 3.1 for the computation of the dimension of $$\mathcal {S}(Y,q)$$.

Case 2. is also well known: see [8, Remark 3].

Case 3. corresponds to the defective 3rd secant variety of the Segre embedding of $$Y=(\mathbb {P}^1)^{\times 4}$$ and the fact that all the elements of $$\sigma _3^0(\nu (Y))$$ are not-identifiable is shown in Remark 3.2. The fact that $$\dim (\mathcal {S}(Y,q))=1$$ for the generic rank-3 tensor depends on the fact that the 3rd defect $$\delta _3$$ of $$\nu ((\mathbb {P}^1)^{\times 4})$$ is exactly 1 (cf. ). Moreover, by [43, Cap II, Ex 3.22, part (b)], we get that for any rank 3 tensor q, the dimension $$\dim (\mathcal {S}(Y,q))\ge 1$$.

Cases 4., 5. and 6. are treated in Examples 3.6 and 3.7 and in Proposition 3.10, respectively.

All the above considerations prove that the list of cases enumerated in the statement corresponds to non-identifiable rank-3 tensors. We need to show that such a list is exhaustive. Since the matrix case is already fully covered by case 1, we only need to care about tensors of order at least 3.

First of all recall that by Remark 1.9, the concise Segre of a rank-3 tensor q is $$\nu (\mathbb {P}^{n_1} \times \cdots \times \mathbb {P}^{n_k} )$$, with $$n_1, \dots ,n_k \in \{1,2 \}$$. Then consider two distinct sets $$A,B \in \mathcal {S}(Y,q)$$. By Corollary 4.3, it can only happen that $$\sharp (A\cup B)=5,6$$.

If $$\sharp (A\cup B)=5$$, the fact that our list of not-identifiable rank-3 tensors is exhaustive is proved in Propositions 5.1 and 5.2.

If $$\sharp (A\cup B)=6$$, we can first use Proposition 6.2 to exclude the all the cases in which Y has at least two factors of dimension 2. Then we start arguing by the number of factors of Y.

If Y has 3 factors and it is the product of $$\mathbb {P}^1$$s only, then the unique tensors of rank-3 are those of the tangential variety to the Segre variety and this is case 2 of our theorem. The case of $$Y=\mathbb {P}^{2}\times \mathbb {P}^{1}\times \mathbb {P}^1$$ is completely covered by Proposition 3.8 together with Examples 3.6 and 3.7 (cf. Corollary 3.9).

If Y has 4 factors and one of them is a $$\mathbb {P}^2$$, there is Proposition 6.3 assuring that $$\mathcal {S}(Y,q)$$ does not contain two disjoint sets. If Y is a product of four $$\mathbb {P}^1$$s, we are in case 3 of our theorem.

The fact that if Y has at least 5 factors then $$\mathcal {S}(Y,q)$$ does not contain two disjoint sets is done in Proposition 6.5.