Terracini Loci: Dimension and Description of Its Components

: We study the Terracini loci of an irreducible variety X embedded in a projective space: non-emptiness, dimensions and the geometry of their maximal dimension’s irreducible components. These loci were studied because they describe where the differential of an important geometric map drops rank. Our best results are if X is either a Veronese embedding of a projective space of arbitrary dimension (the set-up for the additive decomposition of homogeneous polynomials) or a Segre–Veronese embedding of a multiprojective space (the set-up for partially symmetric tensors). For an arbitrary X , we give several examples in which all Terracini loci are empty, several criteria for non-emptiness and examples with the maximal defect possible a priori of an element of a minimal Terracini locus. We raise a few open questions.


Introduction
To compute the dimension of the set of all complex tensors with a prescribed format and a prescribed tensor rank, usually one uses algebro-geometric tools.The same tools are used for computing the dimension of objects of fixed rank (e.g., for partially symmetric tensors or for additive decompositions of homogenous polynomials of prescribed degree and number of variables).This integer, i.e., the dimension of the set Σ we are interested in, is often computed by the dimension of the tangent space of Σ at a "general" point [1].The data are an embedded variety X ⊂ P r and we fix a positive integer x.We consider points q ∈ P r in the linear span of x points of X.For the equivalence classes (up to the multiplication by a non-zero scalar) of non-zero tensors the integer x would be the tensor rank, the variety X would be a multiprojective space Y and the dimensions of the factors of Y give the format of the tensor.The same Y (but with a different embedding and r) works for partially symmetric tensors.For the additive decomposition of degree d forms in n + 1 variables the variety X is the order d Veronese embedding of P n and r = −1 + ( n+d n ).We explain now the set-up and the main results and open questions of our paper.
Let X ⊂ P r be an integral and non-degenerate variety.Set n := dim X.For any closed subscheme Z of P r , let Z denote the linear span of Z, i.e., the intersection of all hyperplanes H containing Z, with the convention Z = P r if no hyperplane contains Z.For any q ∈ P r , the X-rank of q is the minimal cardinality of a finite set S ⊂ X such that q ∈ S .Fix a positive integer x.The closure (in the Zariski or Euclidean topology) σ x (X) of the set of all q ∈ P r of X-rank ≤ x is called the x-secant variety of S. For any p ∈ X reg , i.e., for any smooth point of X, let (2p, X) or just 2p denote the closed subscheme of X with (I p ) 2 as its ideal sheaf.Let S(X reg , x) denote the set of all S ⊂ X reg with cardinality x.For any S(X reg , x), let (2S, X) (or just 2S) denote the union of all 2p, p ∈ S. We have deg(2S) = x(n + 1) and, hence, dim 2S ≤ min{r, x(n + 1) − 1}.By the Terracini lemma ( [2] (Cor.1.11)) dim σ x (X) = dim 2S for a general S ∈ S(X reg , x).In this paper (as in [3]), we consider dim 2S for non-general S ∈ S reg and call the x-Terracini locus T(X; x) of X the set of all S ∈ S(X reg , x) such that dim 2S < min{r, x(n + 1) − 1}.As in [3], we say that S ∈ T(X; x) is minimal and write S ∈ T(X; x) if S / ∈ T(X; #S ) for any S S. In this paper, we also study the geometric properties of the minimal Terracini loci.Set A(X) := ∪ x>0 T(X; x) and A 0 (X) = ∪ x>0 T(X; x).There are only finitely many nonempty addenda in the union appearing in the definition of A(X) (part (c) of Proposition 1).By a theorem of Chevalley ( [4] (Ex.II.3.22)), the set A(X) is a finite union of quasi-projective varieties.Hence, α(X) := dim A(X) is well-defined if we use the convention α(X) = −∞ if A(X) = ∅.Moreover, the maximal integer t(X) such that T(X; t(X)) = ∅ is welldefined if we use the convention t(X) = −∞ if A(X) = ∅.In the definition of A(X) 0 , infinitely many x may occur ( [3] (Th. 1.1(iii))).By the definition of T(X; x) , we have T(X; x) ⊆ T(X; x) and T(X; x) = T(X; x) if T(X; y) = ∅ for all y < x, i.e., if x is the first integer such that T(X; x) = ∅.In particular, A(X) = ∅ if and only if A 0 (X) = ∅.This is our only use of the full Terracini locus.For all finite sets, S ⊂ X reg such that 2S = P r the non-negative integer δ(S) := (n + 1)#S − 1 − dim 2S is called the defect of S. For any S ∈ A(X) we have 0 < δ(S) ≤ n + 1 (Remark 8).For any x such that T(X; x) = ∅, let δ(X, x) be the maximum of all δ(S), S ∈ T(X; x) .Let δ(X) denote the maximum of all δ(S) for all S ∈ A(X), with the convention δ(X) = −∞ if A(X) = ∅.For any integer n ≥ 1, we give an example of an n dimensional embedded variety X such that δ(X) = n + 1 (Example 3).
In Sections 4 and 5, we study these notions for the Veronese embeddings of projective spaces.Our main results are for the case n = 2. Fix positive integers n and d.Let v d : P n → P r , r = −1 + ( n+d n ), denote the order d Veronese embedding of P n .For any positive integer x, let T(n, d; x) (resp.T(n, d; x) ) denote the set of all S ∈ S( ) is a locally closed subset of a finite union of locally closed subsets of S(P n , x) and, hence, it is a quasi-projective variety (quite often reducible).We call α(n, d) the maximal dimension of an irreducible component of A(n, d) with the convention We call t(n, d) the maximal integer x such that T(n, d; x) = ∅ with the convention In Section 4, we prove the following theorems.
Theorem 1. Fix integers n ≥ 2 and d ≥ 3, then there is a positive real number c n,d such that Note that n ( n+d n )/(n + 1) ∼ d n /n!, up to terms of order d n−1 .For fixed n and large d, we may take As a corollary of Theorem 2, we prove the following result.In Section 2, we fix the notation and provide some observations.Section 3 considers arbitrary embedded varieties X ⊂ P r .We provide some examples with A(X) = ∅ and several criteria for the non-emptiness of A(X).Among these criteria, we single out Proposition 3 and Corollary 2.
Sections 4 and 5 are devoted to the Veronese embeddings.In Section 4, we prove Theorems 1 and 2. In Section 5, we provide complete descriptions of T(2, d; x) for many x.
Section 6 contains an extension of Theorem 1 to the case of Segre-Veronese embeddings of multiprojective spaces (Theorem 4), i.e., the embedded variety X whose X-rank is equal to the partially symmetric rank of partially symmetric tensors.
I want to thank the referees for their useful observations.

Preliminaries
Remark 2. Let X be an integral projective variety.Set n := dim X.For each positive integer x, we have dim S(X reg , x) = nx, and S(X reg , x) is irreducible.Thus, each algebraic subset T ⊆ S(X reg , x) has dimension at most nx and dim T = nx if and only if T contains a non-empty open subset of S(X reg , x).
Remark 3. Let Z ⊂ P r be a zero-dimensional scheme.We have dim Z ≤ min{r, deg(Z) − 1}.
For any W ⊂ Z, we have dim If Z is linearly independent, then all subschemes of Z are linearly independent.
Let W be any projective variety, and let D ⊂ W be any effective Cartier divisor of W. For any zero-dimensional scheme Z ⊂ W, let Res D (Z) denote the residual scheme of Z with respect to D, i.e., the closed subscheme of W with I W : I D as its ideal sheaf.

We have Res
is exact and often called the residual exact sequence of Z with respect to D.

Arbitrary Embedded Varieties
Remark 4. We explain here the difference between T for an arbitrary X and the same symbol for the Veronese embedding of P n or the Segre-Veronese embedding of a multiprojective space Y).Elements of T(n, d; x) are subsets of P n , not v d (P n ), and an element of T(n, d; x) in the finite set S has the additional condition (often called concision) that no proper projective subspace of P n contains S. A similar restriction is required for Segre-Veronese embeddings of a multiprojective space.In [3], the following cases have ).In the case n = 1, we have A(v d (P 1 )) = ∅.In the other case we have T(v d (P n ); x) = ∅ for some x (for d = 2 take x = 2, for n = d = 3 take x = 3 with S formed by 3 collinear points).
Remark 5. Recall that T(X; x) ⊇ T(X; x) , and that if x is the minimum integer such that T(X; x) = ∅, then T(X; x) = T(X; x) .Thus, A(X) = ∅ if and only if A 0 (X) = ∅.If n = 1, we quoted a case with A 0 (X) = ∅ (the rational normal curve).There are many other examples of smooth embedded curves with A(X) = ∅.For instance, A(X) = ∅ if X is a linearly normal embedding of a smooth curve of genus g > 0 and degree d ≥ 5g.Curves are never secant defective and so part (a) of Proposition 6 only applies to some varieties of dimension n ≥ 2.
Now we explain a few cases with n ≥ 2 and A(X) = ∅.Remark 6. Obviously, T(X; x) = ∅ for all x if X = P r .Now assume dim X = r − 1.We have ∪ x>0 T(X; x) = ∅ if and only if T(X; 2) = ∅, and this is the case if and only if there is a hyperplane H tangent to 2 distinct points of X reg .For instance, if r ≥ 3 and d = 2, we have T(X; 2) = ∅ if and only if X is singular.Call v ∈ {0, . . ., r − 3} the dimension of the vertex E of the quadric cone X ⊂ P r .Fix a general (r − 1 − v)-dimensional linear subspace L ⊂ P r and set X 1 := L ∩ X.Note that E ∩ L = ∅, X 1 is a smooth quadric hypersurface of L, and X is the cone with vertex E and X 1 as a basis.A set S ∈ S(X reg , 2) is an element of T(X; 2) if and only if there is p ∈ X 1 such that S is contained in the linear span of E and p.Thus, T(X; Proposition 1.Let X ⊂ P r be an integral and non-degenerate projective variety.Set n := dim X.(a) Assume that X is secant defective, and call s the minimal integer ≥ 2 such that dim σ s (X) ≤ (n + 1)s − 2.Then, T(X; x) = ∅ for all x > s, T(X; s) contains a non-empty open subset of S(X, s), α(X) = ns and A(X) has a unique irreducible component of dimension ns.
Fix any x > s and any B ∈ S(X, x).Take any A ⊂ B such that #A = s.The set A and Observation 1 show that S / ∈ T(X; x) .The other statements of (a) are proved as in part (b) of the proof of Theorem 2. Now we prove part (b).Since dim X = n, we have α(X) ≤ nt(X).Fix an integer x > (r + 1)/(n + 1) .Take any S ∈ S(X, x) and any A ⊂ S such that #A = (r + 1)/(n + 1) .Either 2A = P r or dim 2A ≤ (n + 1)#A − 2. Thus, S / ∈ T(X; x) .Since in (a) we have s ≤ (r + 1)/(n + 1) , part (c) follows from parts (a) and (b).Proposition 2. Let X ⊂ P r be a dual defective variety in the sense of [5].Call t > 0 the dual defect of X, i.e., let 1 + t be the codimension of the dual variety X ∨ ⊂ P r∨ .Set n := dim X.Then, T(X; 2) = ∅ and dim T(X; 2) = n + t.
Proof.Since we are in characteristic 0 and X has dual defect t, a general tangent hyperplane H ∈ X ∨ is tangent to X reg along a t-dimensional linear space L. Since L ∩ X reg = ∅ and any S ⊂ L ∩ X reg with #S = 2 is an element of T(X; 2) , L ∩ X reg gives a 2t-dimensional subvariety of T(X; x) .The set of all such L is parametrized by a variety of dimension n − t.If S ⊂ X reg is an element of T(X; 2) , then it must be contained in some L ∩ X reg .We have dim S(X reg ∩ L, 2) = 2t.Proposition 3. Let X ⊂ P r be an integral and non-degenerate variety.Assume one of the following two conditions: (i) n ≥ 2 and there is a hyperplane H ⊂ P r such that X ∩ H has a multiple component not contained in Sing(X); (ii) n ≥ 3 and there is a hyperplane H such that X ∩ H has at least two irreducible components whose intersection is not contained in Sing(X).
Proof.Assume n ≥ 2, and the hyperplane H ⊂ P r such that X ∩ H has a multiple component not contained in Sing(X).Call T the reduction of this multiple component.By construction (2S, X) ⊂ H for all finite sets S ⊂ T ∩ X reg .Thus, if #S 0, then S ∈ T(X; #S).Take the minimal integer x such that a general S ⊂ T gives an element of T(X; x) and hence (since each subsets of X are general element of T) S ∈ T(X; x) .
With the assumption of (ii), take as T an irreducible component of the intersection of two different irreducible components of X ∩ H.
As an example in which we may apply Proposition 3, we define the following corollary.
Corollary 2. Let X ⊂ P r be an integral and non-degenerate variety.We have A(X) = ∅ if X is a linearly normal embedding of a complete linear system |L| with L very ample and one of the following conditions is satisfied: (i) n ≥ 2 and L ∼ = R ⊗2 with R ample and spanned; (ii) n ≥ 3 and L ∼ = R ⊗ M with R and M ample and spanned.
Proof.First assume n ≥ 2 and L ∼ = R ⊗2 with R ample and spanned.Fix a general D ∈ |R|.
Since R is spanned, D Sing(X).Take H := 2D to apply part (i) of Proposition 3. Now assume n ≥ 3 and L ∼ = R ⊗ M with R and M ample and spanned.Take a general D 1 ∈ |R| and a general D 2 ∈ |M|.Since R and M are ample and spanned dim Proposition 4. Let X ⊆ P r and Y ⊆ P s be integral and non-degenerate varieties.Let ν : P r × P s → P rs+s+r denote the Segre embedding of P r × P s .Set W With no loss of generality, we may assume r ≤ s and hence and hence A(W) = ∅.Now assume c + r ≥ rs + r + s, i.e., (r + 2)/2 ≥ rs + s.The latter inequality is always false because s ≥ 2. Take a subset Σ ⊆ S(X reg , x).We say that Σ is a covering family if ∪ S∈Σ S is Zariski dense in X, i.e., it contains a non-empty open subset of X reg .Assume that Σ is a covering family and take a general p ∈ X red .Let Σ p denote the set of all S ∈ Σ such that p ∈ S. Let L Σ,p denote the connected component containing p of the closure in X reg of ∪ S∈Σ p S. The non-negative integer (Σ) := dim L Σ,p only depends on Σ and it is called the local covering dimension of Σ.
Example 1.Take X as in Proposition 2, i.e., assume X is dual defective with dual defect t > 0. The family T(X; 2) is a covering family with t as its local covering dimension.
Example 2. Take X as in part (a) of Proposition 1, i.e., assume X secant defective with s the first integer such that dim σ s (X) < ns − 1. Obviously, s ≥ 2. We saw that T(X; s) contains a non-empty open subset of S(X reg , x).Thus, T(X; s) is a covering family.Since s ≥ 2, T(X; s) has local covering dimension n.Now assume s = 2 and take any n-dimensional W ⊂ P r such that T(W; 2) is a covering family of local dimension 2. We get that T(W; 2) contains a general element of S(X reg , 2) and hence dim σ 2 (W) < 2n − 1 by the Terracini lemma.Theorem 3. Let X ⊂ P r be an integral and non-degenerate variety.Set n := dim X. Assume the existence of a variety W ⊂ X such that W ∩ Sing(X) = ∅ and the linear span W of W has dimension y < r.Set m := dim W. If (n + 1) (y + 2)/(m + 1) ≤ r, then A(X) = ∅ and the minimal integer w such that T(X; w) = ∅ satisfies w ≤ (y + 2)/(m + 1) .
Proof.For a very large integer a, we have deg((2A, W)) > m + 1 for all sets A ⊂ X reg such that #A = a and hence (2A, W) is not linearly independent.Let z be the minimal integer such that dim (2A, W) ≤ z(m + 1) − 2 for a general A ⊂ W reg such that #A = z.If the non-degenerate embedded variety W ⊂ W is secant defective, then z is the minimal integer such that dim σ z (W) ≤ z(m + 1) − 1.If X is not secant defective, then z = (y + 2)/(m + 1) .Since W Sing(X) and A is general in W, A ⊂ X reg .Remark 3 gives dim 2A ≤ z(n + 1) − 2. Since z ≤ (y + 2)/(m + 1) and (n + 1) (y + 2)/(m + 1) ≤ r, z(n + 1) ≤ r.Thus, A ∈ T(X; w) ⊆ A 0 (X).Hence, A(X) = ∅ and the minimal integer w such that T(X; w) = ∅ satisfies w ≤ z.Proposition 5. Let X ⊂ P r be an integral and non-degenerate variety.Set n := dim X. Assume the existence of a variety W ⊂ X such that W Sing(X) and the linear span W of W has dimension y < r.Set m := dim W. Assume that W is secant defective in W and call s the minimal integer such that dim σ s (W) ≤ s(m + 1) − 2. If (n + 1)s ≤ r, then A(X) = ∅ and the minimal integer w such that T(X; w) = ∅ satisfies w ≤ s.
Proof.Take a general A ∈ S(W, s).Since W Sing(X) and A is general in W, A ⊂ X reg .We conclude as in the proof of Theorem 3.

Remark 7.
Many examples are easy using Theorem 3 or a small adaptation of its proof.For instance, if X contains a line L Sing(X), it is sufficient to assume r > 2n to obtain T(X; 2) = ∅ because for any S ⊂ L \ L ∩ Sing(X) such that #S = 2, we have deg(2S, L) = 4 and dim (2S, L) = 1.
0 and S ∈ T(X; x), then δ(S) increases with x.Thus, all cases of the embedded variety, quite frequently, with T(X; x) for infinitely many x have the property that +∞ is the supremum of all δ(S), S ∈ A(X) 0 .For n = 1, this phenomenon does not occur because T(X; Example 3. Fix an integer n ≥ 1 and let X ⊂ P n+1 be a smooth hypersurface such that there is a hyperplane H ⊂ P n+1 tangent to X at at least two points.Fix p, q ∈ X such that p = q and H is tangent at each point of S := {p, q}.Note that H = 2p = 2q and hence H = 2S .We have S ∈ T(X; 2) (Remark 6).Since H = 2S , dim H = n and deg(2S) = 2n + 2, we have δ(S) = n + 1.Hence, δ(X) = n + 1.
(a) Let V d,x denote the Severi variety of all integral and nodal curves C ⊂ P 2 of degree d and geometric genus (d − 1)(d − 2)/2 − x, i.e., all integral nodal curves C ⊂ P 2 of degree d and with exactly x nodes ( [6]).Fix a general C ∈ V d,ρ and set S := Sing(C).Thus, #S = ρ.Since S ⊆ Sing(C), h 0 (I 2S (d)) > 0. Thus, h 1 (I 2S (d)) > 0. The Bézout Theorem gives S = P 2 .Thus, to prove that S ∈ T(2, d; ρ) , it is sufficient to prove that h Claim 1: The monodromy group of nodes of V d,ρ is the full symmetric group S ρ .
Proof of Claim 1.Let V d,ρ denote the closure of V d,ρ in the Hilbert scheme of P 2 .We use the first two lemmas in [6].We have ) branches, each of them corresponding to a subset of Sing(D) with cardinality ρ (the prescribed nodes).Thus, the monodromy group of the nodes of V d,0 is the full symmetric group.
By Claim 1, to get a contradiction, we may assume that h 1 (I 2S (d)) > 0 for all S ⊂ S, such that #S = ρ − 1.We saw that the set of all S ⊂ S, S = Sing(C), C ∈ V d,ρ , #S = ρ − 1, depends on 2(ρ − 1) parameters and, hence, a general such that S satisfies h , where A is a general element of S(P 2 , ρ − 1).Thus, h To conclude the proof it is sufficient to prove that h Observation 2: Fix a positive integer t.Since each Y i , i > 0, is a complete intersection, for every zero-dimensional scheme Z ⊂ Y i , we have h Proof of Claim 1.It is sufficient to prove that h 1 (Y i , I Y i ∩2S (d)) = 0 (Observation 2).Note that S is the residual scheme of Y i ∩ 2S with respect to the divisor Y i+1 of Y i .Since Y i+1 is the complete intersection of Y i and X i+1 , the residual exact sequence Recall that Y 0 = P n , and that h 1 (Y n−1 , I 2S ,Y n−1 (d)) = 0 (Observation 1).Thus, applying n − 1 times Claim 1, we get h 1 (I 2S (d)) = 0. Hence, S is minimal.Remark 12.In the definition of T(n, d; x) , each S ∈ T(n, d; x) is assumed to span P n .Dropping this condition (i.e., considering T(v d (P 2 ), P r ), r = (d 2 + 3d)/2, v d : P 2 → P r the order d Veronese embedding) would not change Theorem 2 and Corollary 1.The case n = 1 ([3] (Lemma 3.4)) shows that sometimes A(X) = ∅.

Examples in the Plane
For any zero-dimensional scheme Z ⊂ P 2 , Z = ∅, let τ(Z) denote the last integer t such that h Remark 13.Let Z ⊂ P 2 be a non-empty zero-dimensional scheme.Set z := deg(Z) and d := τ(Z).Assume z ≤ 3d.Then either there is a line L such that deg(L ∩ Z) ≥ d + 2 or there is a conic D such that deg(Z ∩ D) ≥ 2d + 2 or z = 3d, and Z is the complete intersection of a plane cubic and a degree d plane curve ([9] (Remarques at p. 116)).
Remark 14.Let W ⊂ P 2 be a zero-dimensional scheme.Fix an integer t ≥ −2.Since W is zerodimensional, h 1 (W, O W (t)) = 0. Hence, the long cohomology exact sequence of the exact sequence We recall the following result [3]  The aim of this section is to give a complete classification of all S ∈ T(2, d; x) for even d and x ∈ {3d/2, 3d/2 + 1} and for all odd d and x = (3d + 1)/2.From the complete description, we obtain dim T(2, d; x) for all d and x considered in Propositions 7-9 (Remark 16).
We need the following observation, which contains the definition of the critical scheme ([3] ( §2)).
We prove the following three propositions, which give some restrictions on x, S and the critical schemes of S. See the three subsections for the existence parts in all cases listed in the proposition.Thus, the conditions in Propositions 7-9 are necessary and sufficient conditions on x, S and the critical schemes of S.
We exclude the case of three lines through a common point belonging to S because at least one of these lines would contain at least 1 + d/2 points of S. The minimality of S also implies that C has no multiple components.Thus, C is reducible and there is a line ), the long cohomology exact sequence of (4) gives h Since S is minimal and , we get S ⊂ D 1 , a contradiction.Thus, every critical scheme is contained in C, and it is a complete intersection.If S ⊂ C reg , then 2S ∩ C is the unique critical scheme of S. Now assume S C reg .First, assume that C is integral and, hence, has a unique singular point, p.By assumption p ∈ S, set S := S \ {p}.The scheme Z is the union of the degree 3d − 2 scheme Z 1 := Z ∩ C reg and a degree 2 scheme Proof.First, assume deg(Z) ≤ 3d.By Remark 13 and the minimality of S, the scheme Z is the complete intersection of a unique cubic C and a degree d plane curve T and, in particular, deg(Z) = 3d.Thus, exactly one of the connected components of Z has degree 1. Assume d ≥ 9.For each line L, we have #(S ∩ L) gives that C contains every critical scheme of S. As in the proof of Proposition 7, we get that S has a unique critical scheme.Now assume deg(Z) = 3d + 1.Thus, all connected components of Z have degree 2. Recall that h 1 (I Z (d)) = 1 and h 1 (I Z (d + 1)) = 0 (Remark 15), i.e., τ(Z) = d ≥ 4 − 3 + deg(Z)/4 (here we use that d ≥ 7).We use the case s = 4 of [9] (Cor.2).Since deg(Z) = 4d, Ref. [9] (Cor.2) gives the existence of an integer t ∈ {1, 2, 3} and a degree t plane curve C such that Since S is minimal, (5) excludes the cases t = 1, 2. Thus, deg for all lines L by the minimality of S. Thus, (4) with C instead of D 1 , which gives that every critical scheme of S is contained in C. If S ⊂ C reg , then we obtain the uniqueness of the critical scheme.The case of a singular C is solved as in the proof of Proposition 7. Proposition 9. Assume d ≥ 6 and even.Fix S ∈ T(2, d; 3d/2 + 1) , and let Z be a critical scheme for S.Then, Z is contained in an integral plane cubic C and one of the following cases occurs: A necessary condition for being critical is that no Z ⊂ Z such that deg(Z ) = 3d is the complete intersection of C and a degree d plane curve.Now we prove the uniqueness of the cubic curve.Assume S ⊂ C with C a cubic and C = C .Since #S > 9, C and C have a common component, J.We obtain a contradiction and hence the uniqueness of the critical scheme as in the proof of Proposition 7.
In the next subsections, we show the existence parts for Propositions 7-9 with a description of the sets S ∈ T(2, d; x) and their critical schemes.We show here that in many cases S is not the complete intersection of C and a degree d/2 curve.Take an integral plane cubic.The multiplication by 2 is a surjective map µ : Pic 0 (C) → Pic 0 (C) between the connected one-dimensional algebraic group Pic 0 (C) and itself.Set G := ker(µ).The group G has cardinality 4 if C is smooth, cardinality 2 if C is nodal and cardinality 1 if C is cuspidal.Thus, if C is cuspidal, S is a complete intersection, while if S is either smooth or nodal, we may take Take any such R. Since deg(R) ≥ 2, |R| is base free and dim |R| = 3d/2 > 1.Thus, there are ∞ 3d/2−1 reduced sets A ∈ |R| such that A ⊂ C reg .Each such A is an element of T(2, d; 3d/2) but not a complete intersection.Remark 17. Assume d odd and d ≥ 11.We proved that T(2, d; (3d + 1)/2) = T 1 ∪ T 2 with T 1 ∩ T 2 = ∅ and T 1 (resp.T 2 ) formed by the set of all S ∈ T(2, d; (3d + 1)/2) with a critical scheme of degree 3d (resp.3d + 1).Since d ≥ 11, each minimal S has a unique critical scheme Z.Let Z i , i = 1, 2, denote the set of all criticals associated with some S ∈ T i .Since each S ∈ T i has a unique critical scheme, T i and Z i are isomorphic as algebraic sets.
(a) In this step, we describe T 1 .The examples given before this remark give a family F ⊆ T 1 of dimension dim |O P 2 (3)| + dim |R| = 9 + 3(d − 1)/2 − 1 = 8 + (3d − 1)/2 because if C is smooth (resp.with a node, resp.with a cusp), the possible R's are 4 (resp.2, resp.1).Take an integral and singular C ∈ |O P 2 (3)|.Set {p} := Sing(C).No S ∈ T 1 may have p as a connected component of its critical scheme Z because Z would not be a complete intersection of C and another curve.Now assume that the cubic C is reducible.Since Z is a complete intersection, the degree 1 connected component is contained in C reg .Thus, dim T 1 = 8 + (3d − 1)/2.
(b) Now we describe T 2 .Fix an integral C ∈ |O C (3)|.To obtain an element of T 2 , it is sufficient to take S ⊂ C reg such that #S = (3d + 1)/2 and then consider as critical scheme Z for S the only scheme Z ⊂ C such that Z red = S and each connected component of Z has degree 2; we only need to exclude a lower dimensional family of sets S which would give an element of T 1 .We find that the subset of T 2 obtained in this way is an irreducible quasi-projective variety of dimension (3d + 1)/2 + 9. Now assume C integral and singular.Call p the singular point of C and assume p ∈ S.There is a one-dimensional family of degree two connected schemes v ⊂ C such that v red = p.We get a lower-dimensional family.Now assume C is reducible.Any line L (resp.smooth conic) contains at most (d + 1)/2 (resp.d + 1) points of S. Thus, S ⊂ C reg and we see that this family has a lower dimension.Thus, dim T 2 = (3d + 1)/2 + 9.

d even and x
We prove that case (i) of Proposition 9 occurs.Fix an integral cubic plane curve C and p, q ∈ C reg , p = q.Set L Remark 18.The family just constructed has pure dimension dim |O P 2 (3)| + 2 + dim |R| = 10 + (3d − 1)/2.Now assume that C is singular with Sing(C) = {p}.The family with p as one of the degree 1 connected components of the critical scheme has a lower dimension.
and assume d i ≥ 2(n − 1) − n i for all i.Let c be the maximal integer such that d i ≥ c(n − 1) − n i for all i.Set Y := P n 1 × • • • P n s and r := −1 + ∏ k i=1 ( n i +d i n i ).Let X ⊂ P r be the image of the Segre-Veronese embedding of multidegree (d 1 , . . ., d r ) of Y.There is a positive real number c (only depending on k, n 1 , . . ., n k and d 1 , . . ., d k and for fixed n 1 , . . ., n k growing of order n in min{d 1 , . . ., In the last part of the proof of Theorem 4, we give an explicit value of c in terms of k, n 1 , . . ., n k and d 1 , . . ., d k .From this value, one can play, keeping some of the integers n 1 , . . . ,n k , d 1 , . . . ,d k and sending the other to infinity.The easiest case is when n Observation 1: The Künneth formula gives h i (O Y (t 1 , . . ., t k )) = 0 for all i ≥ 1 if t h ≥ −n h for all h = 1, . . ., k.
If so, what is the order of increasing of lim inf k α(X k ) and lim sup k α(X k )? Give lower and upper bounds for α(X k ), k ≥ 3. The same for t(X k ).

3.
Let X n,k ⊂ P r , n ≥ 2, k ≥ 2 and r = (n + 1) k − 1, denote the Segre embedding of (P n ) k .Give lower and upper bound on α(X n,k ) and t(X n,k ) for fixed k for n → +∞.

Methods
There are no experimental data and no part of a proof is completed numerically.All results are given with full proofs.

Conclusions
Let X ⊂ P r be a smooth variety.For each positive integer x let S(X, x) be the set of all subsets of X formed by x.The Terracini loci A(X) of X are subsets of ∪ x>0 S(X, x) (the element of ∪ x>0 S(X, x) at which a certain differential drop rank), which are the finite union of algebraic varieties.For some X they are empty.We study their non-emptiness, the maximal dimension of their irreducible components and the maximal and minimal x such that S(X, x) ∩ A(X) = ∅.Our best results are if X is either a Veronese embedding of a projective space P n , all n > 0 (the set-up for the additive decomposition of homogeneous polynomials) or a Segre-Veronese embedding of an arbitrary multiprojective space (the set-up for partially symmetric tensors).For the Veronese embedding of the plane, we describe exactly the Terracini loci for low x values.For an arbitrary X, we provide several examples in which all Terracini loci are empty, several criteria for non-emptiness and examples with the maximal defect a priori possible of an element of a Terracini locus.We list four open questions.

Proposition 7 . 3 )
Assume d ≥ 6 and even.Fix S ∈ T(2, d; 3d/2) , and let Z be a critical scheme for S.Then, deg(Z) = 3d, and Z is the complete intersection of a reduced plane cubic C and a curve of degree d.If C is reducible, then S ⊂ C reg and any line L ⊂ C contains d/2 points of S. If d ≥ 8, then S has a unique critical scheme, and S is contained in a unique cubic C. Proof.Since deg(Z) ≤ 3d, Remark 13 gives that either Z is the complete intersection of a cubic C and a degree d curve or there is a line L such that deg(L ∩ Z) ≥ d + 2 or there is a conic D such that deg(Z ∩ D) ≥ 2d + 2. The existence of the line L is excluded, because we would have #(S ∩ L) ≥ d/2 + 1 and hence S would not be minimal.The existence of the conic D is excluded because we would have #(S ∩ D) ≥ d + 1 and hence S would not be minimal.The minimality of S also gives that C has no multiple components because S ⊂ C red and we excluded the cases where C red is a line or a conic.Since d ≥ 4, C is the unique cubic containing the complete intersection Z. Set S 1 := S \ C ∩ Sing(C).Consider the residual exact sequence of C: 0 → I S 1 (d − 3) → I 2S (d) → I 2S∩C,C (d) → 0 (Assume h 1 (I S 1 (d − 3)) > 0. Since d ≥ 6, we have #(S 1 ) ≤ #S = 3d/2 ≤ 3(d − 3) − 1.By Remark 13, either there is a line R such that #(S 1 ∩ R) ≥ d − 1 (excluded by the minimality of S because 2d − 2 ≥ d + 2), or there is a conic D such that #(D ∩ S 1 ) ≥ 2(d − 3) + 2, excluded because the minimality of S gives #(S ∩ D) ≤ d + 1 and d ≥ 6).Thus, h 1 where C = L ∪ D and D is a conic, perhaps reducible.Since #S = 3d/2 and #(L ∩ D ∩ S) ≥ 2, either #(S ∩ L) > d/2 or #(S ∩ D) > d.Thus, S is not minimal, a contradiction.Now assume d ≥ 8, i.e., assume #S > 9. Take another critical scheme Z and call C the unique plane cubic containing Z .Assume C = C .Since Z red = S (Remark 15), we would have S ⊆ C ∩ C .Since #S > 9, C and C would have in common an irreducible component, D 1 , with x := deg(D 1 ) ∈ {1, 2}.Write C = D 1 ∪ D, C := D 1 ∪ D with deg(D) = deg(D ) = 3 − x and deg(D ∩ D ) = (3 − x) 2 .Consider the residual exact sequence of D 1 :

Proposition 8 .
each point of S lies in a unique irreducible component of C. Hence, Z is unique.Assume d ≥ 7 and odd.Fix S ∈ T(2, d; (3d + 1)/2) , and let Z be a critical scheme for S.Then, Z is contained in a unique plane cubic C and one of the following cases occurs: (i) deg(Z) = 3d, a unique connected components of Z has degree 1, and Z is the complete intersection of C and a degree d curve.If d ≥ 9, S has a unique critical scheme.(ii) deg(Z) = 3d + 1 and no Z ⊂ Z with deg(Z ) = 3d is the complete intersection of C and a degree d curve.If d ≥ 11, S ⊂ C reg and S has a unique critical scheme, 2S ∩ C.
and Z is the complete intersection of C and a degree d curve.If d ≥ 8, S has a unique critical scheme.(ii) deg(Z) = 3d + 1, Z red = S and no Z ⊂ Z is the complete intersection of C and a degree d plane curve.Moreover, C is integral.If d ≥ 8, S is contained in a unique plane cubic.If d ≥ 12, S has a unique critical scheme.Proof.Remark 15 gives τ(Z) = d.Remark 13 gives 3d ≤ deg(Z) ≤ 3d + 2. Since S is minimal, #(S ∩ L) ≤ d/2 for every line and #(S ∩ D) ≤ d for every conic.Thus, if S is contained in a plane cubic C, then C is irreducible.(a) Assume deg(Z) = 3d.Remark 13 gives that Z is the complete intersection of C and a degree d curve.If d ≥ 8, then 3d/2 + 1 ≤ −1 + 3(d − 3).(b) Assume deg(Z) ∈ {3d + 1, 3d + 2}.Since d ≥ 6, d ≥ 4 − 3 + deg(Z)/4.Since τ(Z) = d, we may use [9] (Cor.2) as in the proof of Proposition 8. We first get the existence of a cubic C such that deg(Z ∩ C) ≥ 3d and, hence, h 1

5. 1 .
d even and x = 3d/2 For simplicity, we assume d ≥ 12.Take S ∈ T(2, d; 3d/2) and a plane cubic C ⊃ S. Proposition 7 gives S ⊂ C reg and, hence, O C (S) is a line bundle R such that R ⊗2 ∼ = O C (d).
the line bundle R is non-special and base point free.Take S ∈ |R| formed by (3d − 1)/2 smooth points of C, none of them being p. Set S := S ∪ {p, q} and Z = (2S , C) ∪ {p, q}.By construction Z ∈ |O C (d)|.Since C is arithmetically Cohen-Macaulay, Z is the complete intersection of C and a degree d curve.The minimal free resolution of the complete intersection Z gives h 1 (I Z (d)) = 1.Take S S. Since S ⊂ C reg , deg(2S ∩ C) = 2#S ≤ 3d and hence h 1 (C, I 2S ∩C,C (d)) = 0 if and only if 2S is the complete intersection of C and a degree d curve.Since S is contained in the integral curve C, p a (C) = 1 and #S < 3(d − 3), h 1 (I S (d − 3)) = h 1 (C, I S ,C (d − 3)) = 0. Thus, h 1 (I 2S (d)) = 0 and hence S ∈ T(2, d; 3d/2 + 1) .

4 .
Fix integer n ≥ 1, k ≥ 2 and d ≥ 1.Let X n,k,d ⊂ P r , r = ( n+d n ) k − 1, denotethe embedding of Y := (P n ) k .Study α(X n,k,d ) and t(X n,k,d ) as a function of d and compute upper/lower bounds for their limits for d → ∞.