Ranks on the boundaries of secant varieties

In many cases (e.g. for many Segre or Segre embeddings of multiprojective spaces) we prove that a hypersurface of the $b$-secant variety of $X\subset \mathbb {P}^r$ has $X$-rank $>b$. We prove it proving that the $X$-rank of a general point of the join of $b-2$ copies of $X$ and the tangential variety of $X$ is $>b$.


Introduction
Let X ⊂ P r be an integral and non-degenerate variety defined over an algebraically closed field.For any q ∈ X the X-rank r X (q) of X is the minimal cardinality of a set S ⊂ X such that q ∈ S , where denote the linear span.For any q ∈ P r let S(X, q) denote the set of all finite subsets S ⊂ X such that q ∈ S and ♯(S) = r X (q).For any integer s > 0 let σ s (X) ⊆ P r be the s-secant variety of X, i.e. the closure of the union of all linear space S with S ⊂ X and ♯(S) = s.See [17] for many applications of X-ranks (e.g. the tensor rank) and secant varieties (a.k.a. the border rank).The algebraic set σ s (X) is an integral variety of dimension ≤ s(1 + dim X) − 1 and σ s (X) is said to be non-defective if it has dimension min{r, s(1+dim X)−1}.Every secant variety of a curve is non-defective ([3, Corollary 4]).Let τ (X) ⊆ P r be the tangential variety of X, i.e. the closure in P r of the union of all tangent spaces T p X, p ∈ X reg .The algebraic set τ (X) is an integral variety of dimension ≤ 2(dim X) and τ (X) ⊆ σ 2 (X).For any integer b ≥ 2 let τ (X, b) denote the join of one copy of τ (X) and b − 2 copies of X.If X is a curve, then dim τ (X, b) = min{r, s(1 + dim X) − 2} (use b − 2 times [3, part 2) of Proposition 1.3] and that dim τ (X) = 2) and hence τ (X, b) is a non-empty codimension 1 subset of σ b (X) if X is a curve and r > 2b.For a variety X of arbitrary dimensional usually τ (X, b) is a hypersurface of σ b (X), but this is not always true.For instance, if σ 2 (X) has not the expected dimension one expects that τ (X, b) = σ b (X) and this is the case if X is smooth ([14, Corollary 4]).A general q ∈ τ (X) has r X (q) = 2 (and hence for any b ≥ 2 a general o ∈ τ (X, b) has X-rank ≤ b) if a general tangent line to X reg meets X at another point of X, i.e. if X is tangentially degenerate in the sense of [15].It is easy to check that X is tangentially degenerate if and only if the curve X ∩ M ⊂ M is tangentially degenerate, where M is a general codimension n − 1 linear subspace of X. H. Kaji proved that in characteristic zero a smooth curve in P m , t ≥ 3, is not tangentially degenerate ( [15,Theorem 3.1]) and this is true also if the normalization map of X ∩ M is unramified ([15, Remark 3.8]) or if X ∩ M has only toric singularities ( [11]).See [16] for the state of the art (at that time) on tangentially degenerate curves and a list to the examples known in positive characteristic.
In [7] we raised the following question and gave a positive answer (in characteristic zero) when X is a curve.
Our aim is to refine this question for n := dim X > 1 and get (in some cases) a positive answer.Take a general q ∈ τ (X, b).There is o ∈ X reg , a degree 2 connected zero-dimensional scheme v with v red = {o} and p 1 , . . ., p b−2 ∈ X such that p i = p j for all i = j, p i = o for all i and q ∈ v ∪ {p 1 , . . ., p b−2 } .For a general q ∈ τ (x, b) the set (o, p 1 , . . ., p b−2 ) is general in X b−1 and v is a general tangent vector to X at o. Let Z(X, b) be the set of all degree b schemes v∪{p 1 , . . ., p b−2 } with p i = p j for all i = j and o : (ii) Is ♯(Z(X, b, q)) = 1 for a general q ∈ τ (X, b) ′ ?(iii) Is r X (q) > b for a general q ∈ τ (X, b)?If (i) and (iii) are true, then the set of all q ∈ σ b (X) with r X (q) > b has dimension b(n + 1) − 2 (i.e. over C it has Hausdorff dimension 2b(n + 1) − 4).To get a positive answer the first part of Question 1.1 for X and b it is not necessary to prove that (i) and (iii) hold and probably (ii) never will be used to prove (i) and (iii), but (ii) is a nice question, similar to ask if ♯(S(X, o)) = 1 for a general o ∈ σ b (X) (this is called the identifiability of σ b (X)).The way we prove (iii) in the next theorem we get with a very similar proof also (ii), while (i) comes for free.
We prove the following result.
Theorem 1.2.Take b ≥ 2. Let X ⊂ P r , be a an integral and non-degenerate variety, which is non-singular in codimension 1. Set n := dim X. Assume O X (1) = L ⊗ R and the existence of base point free linear spaces , the morphisms u V : X → P v−1 and u W : X → P w−1 , are birational onto their images, that the closures of their images X V and X W have singular locus of dimension ≤ n − 1, and that dim σ 2 (X V ) = 2n + 1. Assume that the image of the multiplication map If u v and u W are embeddings the assumptions on the singularities of X V and X W are satisfied if and only if X is non-singular in codimension 1.
We apply Theorem 1.2 to the case of certain Segre-Veronese embeddings of multiprojective spaces (see Example 2.6), but since we assumed that both L and R are birationally very ample, we cannot useTheorem 1.2 for the most important case: tensors, i.e. the Segre embedding of a multiprojective space.For tensors we prove the following result.
Theorem 1.3.Let X ⊂ P r , r + 1 = s i=1 (n i + 1), be the Segre embedding of the multiprojective space P n1 × • • • × P ns .Fix an integer b ≥ 2 and assume the existence of a decomposition E ⊔ F = {1, . . ., s} such that i∈E (n i + 1) > b + 3 + i∈E n i and i∈E (n i + 1) The assumptions of Theorem 1.3 imply ♯(E) ≥ 2 and ♯(F ) ≥ 2 and hence they exclude the case s = 2, 3.The exclusion of the case s = 2 is not a fault of our too restrictive assumptions.If s = 2 every q ∈ τ (X) \ X has X-rank 2 ( [8], [12,Proposition 1.1]) and hence a general q ∈ τ (X, b) has rank at most b.The paper [10]  For a better description of the X-ranks of σ 3 (X) for s = 3 see [12].In this case τ (X) is not contained in the singular locus of σ 2 (X) ( [12,Theorem 1.3].We expect that the same holds for τ (X, b) for certain very positively embedded X.For the case b = 2, see [19].
We work over an algebraically closed field K with characteristic zero.

Proof of Theorem 1.2
For any integer b > 0 let A(X, b) denote the set of all subsets of X with cardinality b.For any zero-dimensional scheme Z ⊂ X and any effective Cartier divisor D of X the residual scheme of Z with respect to D is the closed subscheme of X with I Z : I D as its ideal sheaf.We have For any line bundle L on X we have an exact sequence (the residual sequence of I Z ⊗ L with respect to D): For any L ∈ Pic(X), any linear subspace V ⊆ H 0 (X, L) and any zero-dimensional scheme For any integral variety M and any o ∈ M reg let (2o, M ) be the first infinitesimal neighborhood of o in M , i.e. the closed subscheme of M with (I o,M ) 2 as its ideal sheaf.Lemma 2.1.Let X P r , be an integral and non-degenerate variety, which is scheme-theoretically cut out by quadrics.Then X is not tangentially degenerate.
Proof.Take a general q ∈ X reg and a general line L ⊂ P r tangent to X reg at q and assume that (L ∩ X) red contains a point o = q.Since the connected component of L ∩ X containing q contains the divisor 2q of L and X is scheme-theoretically cut out by quadrics, we have L ⊂ X.Since L is general, we get τ (X) ⊆ X and so τ (X) = X.Let M ⊂ P r be a general linear space with codimension n − 1.The scheme X ∩ M is an integral curve spanning M and we get τ (X ∩ M ) = X ∩ M , contradicting the assumption X P r .
Remark 2.2.The homogeneous ideal of a Segre-Veronese variety X ⊂ P r is generated by the 2 × 2 minors of flattenings ([17, Theorem 6.10.6.5]) and in particular (unless X = P r ) it is not tangentially degenerate by Lemma 2.1.Just to know that X is scheme-theoretically cut out by quadrics (to be able to apply Lemma 2.1) is easier, since this is easily seen to be true if it is true for the Segre embedding of X.
Lemma 2.3.Let X ⊂ P r , r ≥ 3 + n, be an integral and non-degenerate n-dimensional variety, which is non-singular in codimension 1.Let L ⊂ P 3 be a general tangent line to X reg .Let ℓ L : P r \ L → P r−2 denote the linear projection from L. Then ℓ L|X\X∩L is birational onto its image.
Proof.Since we are in characteristic zero, it is sufficient to prove that ℓ L|X\X∩L is generically injective, i.e. that for a general q ∈ X the plane L ∪ {q} intersects X only in q and the set (X ∩ L) red .If n = 1, then this is true by [7,Lemma 2.5].Now assume n > 1 and that the lemma is true for varieties of dimension < n.Since n > 1, for a general hyperplane M ⊂ P n the scheme X ∩ M is an integral variety non-singular in codimension 1 and spanning M .Since dim X ∩ M > 1, some tangent line of X reg is contained in M and it is tangent to (X ∩ M ) reg .Since L is a general tangent line of X reg , we get that for a general hyperplane H ⊃ L the scheme X ∩ H is integral and spans M .Since L is a general tangent line, the set X ∩ L is finite.Take p ∈ X reg ∩ L such that L ⊂ T p X. Since dim T p X > 1 a general H ⊃ L does not contain T p X, i.e.X ∩ H is smooth at p. We move L among the tangent lines of (X ∩ H) reg and apply the inductive assumption to X ∩ H.We get that for a general q ∈ X ∩ H the plane L ∪ {q} intersects X ∩ H (and hence X) only in q and the set (X ∩ L) red .Moving H among the hyperplanes containing L we get the lemma.Lemma 2.4.Fix an integer b ≥ 2. Let X ⊂ P r , r ≥ 4 + n, be an integral and non-degenerate n-dimensional variety which is non-singular in codimension 1 and take a general Z ∈ Z(X, b).
Remark 2.5.Let X ∈ P r be an integral and non-degenerate variety.Set n := dim X.Let τ (X) ⊆ P r be the tangential variety of X.In characteristic zero if τ (X) = P r we have X ⊆ Sing(τ (X)).For a general x ∈ τ (X) there is o ∈ X reg and a line L ⊆ T o X with x ∈ L \ {o}.The tangent space of τ (X) is constant at all points of τ (X) reg ∩ L. L is uniquely determined by a degree 2 zerodimensional scheme v ⊂ M such that v red = {o}.Let Z(o, v) denote the following zero-dimensional scheme of X (and hence of P r ) with Z(o, v) red = {o} and deg(Z(o, v)) = 2n + 1.It is sufficient to define the ideal J of Z(o, v) in the local ring O X,o .Since O X,o is an n-dimensional regular local ring, there is a regular system of parameters x 1 , . . ., x r such that x 2 1 , x 2 , . . ., x n generate the ideal sheaf of v in P r .Take as J the ideal generated by all x i x j x k , i, j, k ∈ {1, . . ., , n} and all x 1 x i , i = 1, . . ., x n .Now we check that this definition depends only on X, o and v, but not on the choice of x 1 , . . ., x r .Let µ be the maximal ideal of O X,o .Take another regular system of parameters y 1 , . . ., y n of O X,o with y 2 1 , y 2 , . . ., y n generating the ideal sheaf of v in X.Since O X,o is regular, the completion ÔX,o of O X,o with respect to its maximal ideal is isomorphic to K[[x 1 , . . ., x n ]].In K[[x 1 , . . ., x n ]] we have y i = L i + M i , with M i a power series with no constant and no linear term, L 1 , . . ., L n linear forms in x 1 , . . ., x n with invertible Jacobian with respect to x 1 , . . ., x n and there is a non-zero constant c such that x 1 −cy 1 ∈ (x 2 , . . ., x n )+µ 2 .Thus y 1 , . . ., y n gives the same ideal.We have T x τ (X) ⊃ Z(o, v).Now assume that the scheme Z(o, v) is linearly independent in P r , i.e. that dim Proof of Theorem 1.2: Taking a linear projection we reduce to prove the theorem when the map Set Z := ∪ o∈S (2o, X).By [3, Corollary 1.10] it is sufficient to prove that h 1 (I Z (1)) = 0. We use induction on the integer b, starting the induction here with the obvious case b = 1.Fix o ∈ S and set S ′ := S \ {o} and B := ∪ o∈S ′ (2o, X).By the inductive assumption we may assume h 1 (I B (1)) = 0. Thus it is sufficient to prove that (2o, X) gives n + 1 independent conditions to H 0 (I B (1)).Since we may take o general after fixing S ′ , o is not in the base locus of H 0 (I B (1)).Take N ⊆ T o X with 0 ≤ dim N ≤ n and maximal giving independent conditions to H 0 (I B∪{o} (1)) and let N ′ ⊆ (2o, X) the corresponding zero-dimensional scheme with deg(N To get a contradiction it is sufficient to prove that H 0 (I N ′′ (1)) H 0 (I N ′ (1)).Since S is general, we may assume S ∩ I V = S ∩ I W = ∅.Since u V and u W are embedding at o, we have dim V (−N ′′ ) = dim V (−N ′ ) − 1.Since we may take S ′ general after fixing o, we have dim Using V ⊗ W we see that H 0 (O P r (1)) separates the 2-jets of X at o and in particular h 1 (I Z ′ (1)) = 0, concluding the proof of the case b = 2.We proved also that (3o, X) is linearly independent in P r , where (3o, X) is the closed subscheme of X with (I o,X ) 3 as its ideal sheaf.Now assume b > 2 and that the last assertion is true for the integer b − 1, i.e. assume that the zero-dimensional scheme

. , n} and schemes
(c) In this step we prove that ♯(Z(X, b, q)) = 1 for a general q ∈ τ (X, b).Fix a general q ∈ τ (X, b) and assume ♯(Z(X, b)) > 1 and so there are Z, A ∈ Z(X, b) with Z = A. Since dim τ (X, q) = b(n + 1) − 2, a dimensional count shows that Z(X, b, q) is finite for a general q ∈ τ (X, b).Hence we may assume that Z(X, b, q) is finite.A dimensional count gives that Z and A are general in Z(X, b), but of course we do not assume any generality for Z ∪ A. In particular we may assume Z ∩ (I Let D ⊂ X be the hypersurface whose equation is a general element of V (−Z).Let E denote the residual scheme Res D (Z ∪ A) of Z ∪ A with respect to the effective Cartier divisor D ⊂ X.Since Z ⊂ D, we have E = Res D (A).Thus E is a closed subscheme of A and E = ∅ if and only if A ⊂ D. Note that each element of Z(X, b) has only finitely many subschemes.Since dim τ (X, b) = b(n + 1) − 2 and q is general in τ (X, b), we have q / ∈ Z ′ for any Z ′ Z and q / ∈ A ′ for any A ′ A. Since q ∈ Z ∩ A , A = Z, q / ∈ Z ′ for any Z ′ Z and q / ∈ A ′ for any A ′ A, we have h 1 (P r , I Z∪A (1)) > 0. Since u V (X) is not singular in codimension 1 and it is embedded in a projective space of dimension ≥ n+2, u V (X) is not tangentially degenerate ([15, Theorem 3.1]).By Lemma 2.4 applied to X V ⊂ P v−1 the scheme u V (Z) is the scheme-theoretic theoretical base locus of (d) In this step we prove that r X (q) > b for a general q ∈ τ (X, b).Since dim τ (X, b) > dim σ b−1 (X) by step (b), we have r X (q) = b.Take Z ∈ Z(X, b, q) and S ∈ S(X, q).Since S ∈ S(X, q), there is no S ′ S with q ∈ S ′ .Since dim τ (X, b) = b(n + 1) − 2, we may assume that Z is general in Z(X, b) and that q / ∈ Z ′ for any Z ′ Z. Since Z is not reduced, we have Z = S. Hence h 1 (I Z∪S (1)) > 0. As in step (c) we see that Z is the intersection of the open set X \ I V with the scheme-theoretic base locus of V (−Z).Fix a general q ∈ τ (X, b) and assume r X (q) ≤ b.Take Z ∈ Z(X, b, q).If we have S with S ∩ (I V ∪ I W ) = ∅ and dim W (−S) = w − b, then we may apply verbatim the proof in step (c) with S instead of A. If S ∩ (I V ∪ I W ) = ∅ and dim V (−S) = v − b, then we may apply the proof in step (c) taking (W, Z) instead of (V, Z) and (V, S) instead of (W, A).Call τ τ non-empty open subset of τ (X, b) such that for each q ∈ τ τ we have r X (q) = b and q ∈ Z with Z sufficiently general in Z(X, b) ) and (X W , u W (Z)) satisfy the thesis of Lemma 2.4).The set S(X, q) is constructible and hence it makes sense to speak about the irreducible component of S(X, q) and of their dimension.Let σ b (X) ′ denote the set of all a ∈ σ b (X) such that there is a finite set B ⊂ X with ♯(B) = b, a ∈ B and a / ∈ B ′ for any B ′ B. The set σ b (X) ′ is constructible (it is the image of a an open subset of the abstract join of b copies of X).Hence τ := τ τ ∩ σ b (X) ′ is constructible.By assumption τ contains a non-empty open subset of τ (X, b) and hence it is irreducible and of dimension b(n + 1) − 2. Let Γ b ⊆ X (b) be the set of all S ∈ S(X, q) with q ∈ τ .Since dim τ = n(b + 1) − 2, we have dim Γ b ≥ nb − 1.If dim Γ b = nb, then for a general q ∈ τ we may take as S a general subset of X with cardinality b, concluding the proof in this case.Thus we may assume that each irreducible component of the constructible set Γ b has dimension nb − 1.Thus there is a non-empty open subset τ ′ of τ such that S(X, q) is finite for all q ∈ τ ′ .Restricting τ ′ if necessary we may assume that the positive integer ♯(S(X, q)) is the same for all q ∈ τ ′ .Let X (b) denote the symmetric product of b copies of X and let m : ) and in the base locus of W (−S).This is impossible, since X is embedded in P r and (by our reduction at the beginning of the proof) the image of the map ρ : , be the Segre-Veronese embedding of the multiprojective space and either s ≥ 3 or s = 2 and (c 1 , c 2 ) = (1, 1) or s = 1 and c 1 ≥ 3. We claim that dim τ (X, b) = b(n + 1) − 2, dim σ b (X) = b(n + 1) − 1 and r X (q) > b and ♯(Z(X, b, q)) = 1 for a general q ∈ τ (X, b).By Remark 2.2 to apply Theorem 2.6 it is sufficient to observe that the variety σ 2 (X V ) has dimension 2n + 1, where X V is the Segre-Veronese embedding of X by the complete linear system |O X (c 1 , . . ., c s )|, by [1, Theorem 4.2].

Proof of Theorem 1.3
In this section X = P n1 × • • • × P ns .For any i ∈ {1, . . ., s} let π i : X → P ni denote the projection onto the i-th factor of X. Set O X (ε i ) := π * i (O P n i (1)).For any E ⊆ {1, . . ., s} set O X (E) := ⊗ i∈E O X (ε i ) ∈ Pic(X) and let π E : X → i∈E P ni denote the projection onto the factors of X with label in E.
Proof of Theorem 1.3:There are many papers, which could be used to see that dim σ b (X) = b(n + 1) − 1 ([2], [13], and if n i = n for all i, [18] (case s = 3) and [4], any s); this is also a consequence of [10,Corollary 4.15], which implies that S(X, o) is finite for a general o ∈ σ b (X).Take a general q ∈ τ (X, b).Thus there is Z ∈ Z(X, b) with q general in Z .Since q is general in τ (X, b), Z is general in Z(X, b) and q is general in Z .In particular q / ∈ Z ′ for any Z ′ Z.Take an integer c ≤ b and assume the existence of W ∈ (Z(X, c) ∪ A(X, c)) with q ∈ W and q / ∈ W ′ for any W ′ W and W = Z.Since q ∈ ( Z ∩ W \ Z ∩ W ), we have h 1 (I Z∪W (1)) > 0.
(a) Fix a decomposition E ⊔ F with i∈E (n i + 1) > b and i∈F (n i + 1) > b.In this step we prove that c = b and that h 1 (I W (E)) = 0 and H 0 (I W (E)) = H 0 (I Z (E)); note that this would also imply that π (a3) Exchanging the role of E and F we also get h 1 (I W (F )) = 0, that π F |W is an embedding and that H 0 (I W (F )) = H 0 (I Z (F )).
(b) Take E and F as in step (a).Since Z is general in Z(X, b), the scheme π E (Z) is general in Z(X E , b).Since H 0 (I Z (E)) = H 0 (I W (E)), Lemma 2.4 applied to X E gives π E (W ) ⊆ π E (Z).Since π E|W is an embedding, we first get π E (W ) = π E (Z) and then that W ∈ Z(X, q) and W / ∈ A(X, q).This is sufficient to see that r X (q) > b.By step (a3) we also get π F (W ) = π F (Z). Hence π i (W ) = π i (Z) for all i ∈ {1, . . ., s}.This is not enough to say that W has only finitely many possibilities (obviously W red has only finitely many possibilities) and so to prove that dim τ (X, b) = b(n + 1) − 2 we need to work more.Fix again a general q ∈ τ (X, b) and assume that dim τ (X, b) < b(n + 1) − 2, i.e. assume that Z(X, b, q) is infinite.The set Z(X, b, q) is constructible and hence it makes sense to speak about the dimensions of the irreducible components of Z(X, b, q).Since dim τ (X, b) < b(n + 1) − 2, each of the irreducible components of Z(X, b, q) has positive dimension.Let Γ be the irreducible component of Z(X, b, q) containing Z.A general U ∈ Γ may be considered as a general element of Z(X, b) and hence we may apply Lemma 2.4 for X V and u V (U ) and for X W and u W (U ).Since there are only finitely many sets W red , W ∈ Γ, for a general U ∈ Γ \ {Z} we have U red = Z red and so deg(U ∩ Z) = b − 1.Since q ∈ ( Z ∩ U \ Z ∩ U ), we get U = Z and hence U ⊂ Z , contradicting Lemma 2.4.