Minimal and Primitive Terracini Loci of a Four-Dimensional Projective Space

: We study two quite different types of Terracini loci for the order d -Veronese embedding of an n -dimensional projective space: the minimal one and the primitive one (defined in this paper). The main result is that if n = 4, d ≥ 19 and x ≤ 2 d , no subset with x points is a minimal Terracini set. We give examples that show that the result is sharp. We raise several open questions.


Introduction
Let X ⊂ P r be an integral and nondegenerate variety defined over an algebraically closed field K with characteristic zero.For any positive integer x, let S(X reg , x) denote the set of all S ⊂ X reg such that #S = x.For any p ∈ X reg , let 2p denote the closed subscheme of X with (I p ) 2 as its ideal sheaf.For each S ∈ S(X reg , x), set 2S := ∪ p∈S 2p.We have deg(2S) = x(dim X + 1).By the Terracini lemma ( [1] Cor. 1.10), the zero-dimensional scheme 2S is a key player for the study of the secant varieties of X.
Let T (X, x) denote the set of all S ∈ S(X reg , x) such that ⟨2S⟩ ̸ = P r and dim⟨2S⟩ ≤ deg(2S) − 2 = x(dim X + 1) − 2, where ⟨ ⟩ denotes the linear span.The set T (X, x) is called a Terracini set of the embedding X ⊂ P r or the x-Terracini locus of X.These Terracini loci are the sets at which a certain morphism connected to the x-secant variety ramifies (see Section 6 for motivations).The integer x(dim X + 1) − dim⟨2S⟩ + 1 is the dimension of the kernel of the differential of this map at a point associated with S ( [1] Cor. 1.10).Knowing that S / ∈ T (X, x) guarantees that S is an isolated solution for the X-rank problem of a sufficiently general p ∈ ⟨S⟩ (Section 6).In previous works, it was clear that these sets T (X, x) (which are algebraic subsets for the Zariski topology) have a rich geometry ( [2]).Sometimes T (X, x) = ∅ for all x > 0, e.g., if X is a rational normal curve ([2] Th. 1.1(i)).Assuming T (X, x) ̸ = ∅ for some x, the minimal integer x, t(X, min), such that T (X, x) ̸ = ∅ is certainly important (see Remark 1 for an easy lower bound on it).
But not all Terracini sets are "equally interesting".Take S ∈ T (X, x).Quite often, there are integers y > x and S 1 ∈ S(X reg , y − x) such that S ∩ S 1 = ∅ and S ∪ S 1 ∈ T (X, y).Indeed, the condition "dim⟨2(S ∪ S 1 )⟩ ≤ deg(2(S ∪ S 1 )) − 2 = y(dim X + 1) − 2" is true by the corresponding inequality for 2S and we only need ⟨2(S ∪ S 1 )⟩ ̸ = P r .This is true for all p ∈ X reg \ S if dim⟨2S⟩ ≤ r − dim X − 2, and this inequality is very often satisfied in interesting ranges of integers x.In many cases, from a single S ∈ T (X, x), we obtain larger sets that give T (X, y) ̸ = ∅ for all y > x ([2] Th. 1.1(iii)).Take S ∈ T (X, x).We say that S is minimal if dim⟨2S ′ ⟩ = deg(2S ′ ) − 1 for all S ′ ⊊ S, i.e., if no proper subset of S is Terracini.Let T (X, x) ′ denote the set of all minimal S ∈ T (X, x).It is easy to check that T (X, x) ′ = ∅ for all x ≫ 0 (Remark 4).Thus, minimal Terracini sets are more important.Of course, elements of T (X, t(X, min)) are minimal.Now assume that X is the image v d (P n ), n > 1, of the d-Veronese embedding v d : P n → P r , r = ( n+d n ) − 1 of P n .We take finite sets in P n instead of v d (P n ).For any positive integer x, let T 1 (n, d; x) denote the set of all S ∈ S(P n , x) such that v d (S) ∈ T (X, x).Fix a line L ⊂ P n and take S ⊂ L such that #S = 1 + ⌈d/2⌉.Since h 0 (O P 1 (d)) = d + 1, dim⟨2ν d (S)⟩ ≤ deg(2S) − 2. Since n > 1 and d ≥ 2, (1 + ⌈d/2⌉)(n + 1) ≤ ( n+d n ).Thus, S ∈ T 1 (n, d; 1 + ⌈d/2⌉).This set S is not very interesting; it lies in a tiny part of P n , and this (many points in a tiny part) is the only reason to be an element of T 1 (n, d; x).Let T (n, d; x) denote the set of S ∈ T 1 (n, d; x) such that ⟨S⟩ = P n , i.e., P n is the minimal projective space containing S. A similar notation and assumption is available for Segre embeddings of multiprojective spaces.Take S ∈ T (n, d; x).We say that S is minimally Terracini and write S ∈ T (n, d; x) ′ if dim⟨ν d (S ′ )⟩ = (n + 1)#S − 1 for all S ′ ⊊ S ( [2]) .Note that ν d (T (n, d; x) ′ ) = T (X, x) ′ ∩ ν d (T (n, d; x)).
In Section 5, we consider the following set: T (n, d; x).Let T (n, d; x) be the set of all S ∈ T (n, d; x) such that h 1 (I 2A (d)) = 0 for all A ⊊ S such that ⟨A⟩ = P n .We say that T (n, d; x) is the primitive Terracini loci of the Veronese variety v d (P n ).Obviously, In our opinion, the minimal Terracini locus is the most important one, and in [2] it was shown how different T (n, d; x) ′ and T (n, d; x) are.In the second part of this paper (Section 5), we show how different the primitive Terracini locus is with respect to the other ones.By the semicontinuity theorem for cohomology, the sets T (n, d; x) ′ , T (n, d; x) and T (n, d; x) are locally closed subsets of the set S(P n , x) of all subsets of P n of cardinality x.In particular, it makes sense to speak about their dimension.
Remark 1.Take an integral and nondegenerate variety X ⊂ P r .Recall the positive integer t(X, min), which is a key invariant of the embedding X ⊂ P r , and that every S ∈ T (X, min) is minimal.Now assume X = v d (P n ), n ≥ 2, d ≥ 4. With this general definition (as discussed in [2]), we would obtain t(X, min) = ⌈(d + 2)/2⌉ for all n > 1. Obviously, this is not interesting, since in the applications, to see the difficulty of a problem, it is better to consider only multivariate polynomials which are "concise", i.e., such that there is no linear change in coordinates making the polynomials depending on a smaller number of variables.With this restriction, Theorem 1, Remark 8 and Proposition 2 solve the minimal x-problem for n = 4.For n = 2, 3, the problem was solved in [2].The case n = 2 was easy, while the proof of the case n = 3 was longer.In our opinion, the proof of Theorem 1 is as short as possible with our technology.We stress that these tools are useful for other problems (see the discussion of Question 5 at the end of Section 2).
A similar restriction should be added in the study of other important (for the applications) embedded varieties, e.g., the multiprojective spaces (which are associated with tensors and partially symmetric tensors) and Grassmannians (which are associated with antisymmetric tensors).For tensors and partially symmetric tensors, it is equivalent to study exactly the concise tensors.
We recall the following conjecture ( [2] In Section 2, we give the key definition (critical scheme) used in [2] and give several results (and a question) on the Hilbert function of a zero-dimensional scheme Z ⊂ P n .
Section 3 is devoted to the proof of Theorem 1.An outline of the proof is presented at the beginning of the section.
In Section 4, we consider the range x > 2d: the range of the last four questions.
In Section 5, we consider T (n, d; x).We give conditions on n, d and x in order to have T (n, d; x) ̸ = ∅ and other conditions implying T (n, d; x) = ∅.We classify the sets T (n, d; x) if x < 3d/2 (Theorem 3).The main difference between T (n, d; x) and T (n In Section 6, we give the main motivations for the study of the Terracini loci.Several tools used here and in [2] (in particular zero-dimensional schemes, not just finite sets) are useful for other topics, e.g., the description of evaluation codes and the computation of their minimum distance and higher Hamming weights ( [3,4]).
The author thanks the referees for several helpful suggestions.

Preliminary Results
For any sequence {w i }, i ≥ 1 of non-negative integers, we say that {w i } is weakly decreasing if w i ≥ w i+1 for all i ≥ 1.
A rational normal curve C ⊂ P r is an integral and nondegenerate curve of degree deg(C) = r.All rational normal curves of P r are smooth and rational and they are the nondegenerate curves of P r of minimal degree.
For any projective scheme M, any effective Cartier divisor D of M and any zerodimensional scheme Z ⊂ M of the residual scheme Res D (Z) of Z with respect to D is the zero-dimensional subscheme of M with I Z,M : I D,M as its ideal sheaf.We have Res Often, we will say that (1) is the residual exact sequence of D without mentioning Z and L. Let Z red denote the reduction in Z, i.e., the set of all p ∈ M such that {p} ⊆ Z.The set Z red is finite, and #S ≤ deg(Z) and Z red and Z have the same number of connected components.
A key definition used in [2] is the notion of a critical scheme.
Definition 1.Take a finite set S ⊂ P n such that h Remark 2. Take a finite set S ⊂ P n such that h 1 (I 2S (d)) > 0. There is a critical scheme of S ( [5,6] and [2] Using Remark 2, we obtain the following lower bound for the integer t(X, min): Lemma 1.Let X ⊂ P r be an integral and nondegenerate variety.Let γ(X) be the minimal degree of a zero-dimensional scheme Z ⊂ X reg such that dim⟨Z⟩ ≤ deg(Z) − 2 and ⟨Z⟩ ̸ = P r , with the convention γ(X) = ∞ if there is no such Z (e.g., for a rational normal curve).Then, t(X, min) ≥ ⌈γ(X)/2⌉ if γ(X) is finite while T (X, x) = ∅ for all x if γ(X) = ∞.
Remark 3. Often, it is easy to compute γ(X).For instance, if X ⊂ P r is the image of the d-Veronese embedding of P n , d ≥ 3, then γ(X) = d + 2.
Remark 4. Let X ⊂ P r be an integral and nondegenerate variety.Set n := dim X.By the Terracini lemma ([1] Cor.1.10), we have T (X, x) ′ = ∅ for all x > ⌈(r + 1)/(n + 1)⌉ (see [2] Prop.3.5) for the case of the Veronese varieties.Now assume that X is secant-defective and let k be the first integer such that the k-secant variety has dimensions of at most k(n + 1) − 2. In this case, T (X, k) ′ contains a general S ∈ S(X, k) by the Terracini lemma ([1] Cor.1.10 (b)).Thus, in this case, k is the maximal integer y such that T (X, y) ′ ̸ = ∅.By the semicontinuity theorem for cohomology, we have T (X, k) = S(X reg , k).
Remark 5. Fix positive integers d, z and a zero-dimensional scheme Z ⊂ P 2 such that deg(Z) = z and h 1 (I Z (d)) > 0.
(a) Assume z ≤ 3d.Then, Z is in one of the following cases ( [7] Rem.(i) at.p. 116): (i) There is a line L such that deg(L ∩ Z) ≥ d + 2; (ii) There is a conic D such that deg(D ∩ Z) ≥ 2d + 2; (iii) z = 3d and Z is the complete intersection of a plane curve of degree three and a plane curve of degree d.
(b) Assume z ≤ 4d − 4 and z ≥ 16.Then, either Z is as in one of the cases (i), (ii) or (iii) of part (a) or there is a plane cubic C such that deg(Z ∩ C) ≥ 3d + 1 or z = 4d − 4 and Z is the complete intersection of a plane curve of degree four and a plane curve of degree d − 1 The following result is well-known for finite sets, but we need it for certain very mild zero-dimensional schemes: Proposition 1. Assume d ≥ 12. Let Z ⊂ P 3 be a zero-dimensional scheme such that ⟨Z⟩ = P 3 , its connected components have degree ≤ 2, z := deg(Z) ≤ 3d + 3 and no line contains at least ⌈(d + 2)/2⌉ + 1 points of Z red .Assume h 1 (I Z (d)) > 0.Then, one of the following cases occurs: > 0, then we use Remark 5. Thus, we may assume h we use Remark 5 for the integer d − 1.In both cases, we are in one of the cases (i), (ii) or (iii) of Proposition 1 for the integer d − 1, and all cases are contained in a plane.Hence, in this case, we have deg(Res then we are in case (i) for the integer d.Thus, we may assume deg > 0, then we conclude using Remark 5 (more precisely, d is odd and the conic is singular).Thus, we may assume h 1 (I M∩Z (d)) = 0.The residual exact sequence of M gives h 1 (I Res M (Z) (d − 1)) > 0. As above, we obtain the existence of a line R such that deg we allow the case R = L, but it does not occur by the assumption on Z red ), then we are in case (ii) with a singular conic because L ⊂ M and deg(R Assume for the moment that J, R and L are pairwise disjoint.Let Q ′ be the only quadric containing J ∪ L ∪ R. The quadric Q is smooth and J, L and R are in the same ruling of Q ′ , say the ruling Taking the plane ⟨L ∪ J⟩, we obtain a contradiction, unless J ∪ R ∪ L is a reducible rational normal curve. (b) By step (a), we are allowed to assume z 1 ≤ 4. Hence, Z is contained in no reducible quadric.
Take any quadric + 1, we may apply either Remark 5 or ([2] Proposition 6.1).We obtain z 1 ≥ d, contradicting one of our assumptions. (b2 then we may apply either Remark 5 or [2] Proposition 6.1 to the scheme Z ∩ Q.Thus, we may assume deg(Z ∩ Q) ∈ {3d + 2, 3d + 3}.Taking Z ∩ Q instead of Z, we may assume Z ⊂ Q.We may also assume that Z is contained in every quadric containing at least a degree eight subscheme of Z. Thus, Z is contained in ∞ 1 quadrics.Since z 1 ≤ 4, Z is not contained in a reducible quadric.Thus, Z is contained in the complete intersection T of two quadric surfaces.Note that the restriction map Looking at the connected components of T, we reduce to the case T connected.If T is irreducible, it is sufficient to use that deg(Z) − #Z red = 1 and deg(Z) < 4d ([2] Remark 3.1).If T is reducible, it is sufficient to perform all possible decompositions of T and use the assumption on Z red (this is performed in [2] Lemma 6.4 for all z < 4d, with minor additional assumptions).Remark 6.We take the assumptions on d and z as in Proposition 1, except that now we assume Z ⊂ P 4 and ⟨Z⟩ = P 4 .The thesis of Proposition 1 is true and it is easy to prove it by using Proposition 1 and starting with a hyperplane H 1 such that deg(Z ∩ H 1 ) is maximal.Remark 7. Fix integers n, d and z such that n ≥ 2, d ≥ 4 and z < 3d.Let Z ⊂ P n be a zero-dimensional scheme such that deg(Z) = z, each connected component of Z has degree ≤ 2 and h 1 (I Z (d)) > 0. By Remark 5, we may assume n > 2 and use induction on n.As in the argument of Remark 6, by using hyperplanes instead of planes, we obtain that either there is a line Question 5: Take a large integer d.Are Proposition 1 or Remarks 6 and 7 still true for any degree z zero-dimensional scheme?Are they true at least for curvilinear zerodimensional schemes, i.e., for the zero-dimensional schemes whose connected components may be embedded in a smooth curve?
Solving the last question would be a key step for the study of the cactus rank of homogeneous polynomials ( [8][9][10]) and would also have nice consequences for computing the higher Hamming weights of certain evaluation codes ( [3,4] and several papers quoting them).When the evaluation code comes from a smooth curve, it would be sufficient to study Question 5 for curvilinear schemes because all zero-dimensional subschemes of a smooth curve are curvilinear.

Proof of Theorem 1
This section is devoted to the proof of Theorem 1.We outline its proof.Assume, by contradiction, that T (4, d; x) ′ ̸ = ∅ and fix S ∈ T (4, d; x) ′ .Let Z be a critical scheme of S (Definition 1).The proof is divided into two parts.
In step (a), we consider the case x ≤ 2d − 1.We prove this case by taking a general linear projection Z from a general point of P 4 into P 3 and applying several results of [2] to the image of Z.A key ingredient of step (a) is the definition of the set V of allowable points of projection.
In step (b), we consider the case x = 2d.We take We see that ℓ o (Z) is a flat limit of the family h t (Z) of projectively equivalent schemes.The semicontinuity theorem for cohomology gives h 0 (P 3 , I ℓ o (Z) (d)) > 0. We have h 0 (P 3 , O P 3 (d)) = ( d+33 ).Since 4x < ( d+3 3 ), h 0 (P 3 , I ℓ o (Z) (d)) > 0. Thus, S ′ ∈ T (3, d; x).Our assumptions on d and x are the ones made in [2] Theorem 1.4.Thus, S ′ / ∈ T (3, d, x) ′ .Since the critical scheme Z of S has only finitely many subschemes, there is a nonempty open subset V of U such that dim⟨ℓ o (Z ′′ )⟩ = min{3, dim⟨Z ′′ ⟩} for all o ∈ V. From now on, we assume o ∈ V. Since S ∈ T (4, d; 3) ′ , we obtain that there is scheme We exclude the case dim M ′ = 1 because the minimality of S implies that S does not contain ⌈d/2⌉ + 1 collinear points.Thus, M ′ is a plane.The definition of V gives that ⟨G⟩ is a plane and ℓ o|G : G → S ′′ is a linear isomorphism of plane subsets.Consider the residual exact sequence of M ′ in P 3 : By assumption, h 1 (I (2S ).Hence, h 0 (P 3 , I 2S ′′ ,P 3 (d)) > 0. Recall that Z ′′ is minimal and T (3, d; y) ′ = ∅ for all y ≤ 2d − 1 such that y ̸ = ⌈3d/2⌉ + 1.By [2] Theorem 1.3, there is a rational normal curve C containing Z ′′ and deg(Z ′′ ) ∈ {3d + 2, 3d + 3}.A rational normal curve C ⊂ P 3 such that #(ℓ o (Z) ∩ C) ≥ 3d + 2 must be the linear projection of a rational normal curve of a hyperplane of P 4 because Z has finitely many subschemes and we may take o outside the finitely many rational normal curves of P 4 containing a subschemes of Z of degree at least 3d + 2.
First assume we obtain a contradiction.Now assume Z ′ = Z ′ ∩ C. Since C may depend on o, we call it C(o).The rational normal curve C(o) is unique (for a fixed o ∈ V) because deg(Z ′ ) > 6.Call C o the cone with vertex o and C(o) as its base.Since this is the only remaining case to consider for x < 2d, this would occur for all o ∈ V. We obtain that Z is contained in all two-dimensional cubic cones C o , o ∈ V. Fix o, a ∈ V such that a ̸ = o.Since C a is cut out by quadrics, C a ∩ C o is strictly contained in the complete intersection of C o and a quadric, which is a degree six scheme, counting the multiplicities of its connected components.Since deg(Z) > 16, the set ∆ := ∩ u∈V C u contains a curve, T 1 , (maybe with multiple components) with deg(T 1 ) < 6.
we obtain that T 1 has no component of degree 2 and that either it is linearly isomorphic to C(o) (and hence it must be a linear section of C o not containing o) or it is a rational normal curve of P 4 containing o. Taking a general u ∈ V instead of o, we exclude the latter case.
Take a general u ∈ V and a general There is where W 1 is the union of the connected components of Z with a point of F 1 as their reduction; 2.
There is a conic There is a plane cubic C 3 such that There is a plane cubic containing There is a rational normal curve C 5 of a hyperplane of P 4 containing We recall that (as in step (a)) for any A ⊆ Z, we have dim⟨ℓ o (Z)⟩ = min{3, dim⟨ℓ o (Z)⟩}, and hence each subscheme of Z ′ contained in a plane (resp.a line) comes from a subscheme of Z contained in a plane (resp.a line).
(b1.1) Assume the existence of W 5 .Since a rational normal curve C 5 of a hyperplane of P 4 is scheme-theoretically cut out by quadric hypersurfaces and each con- (b1.2) Assume the existence of W i with i ∈ {3, 4} and call C 3 the plane cubic containing W i .Take a general hyperplane H ′ containing ⟨C 3 ⟩.Since ⟨S⟩ = P 4 and S is minimal, We also obtain that d is odd, #(S ∩ L ′ ) = (d + 1)/2 and #(S ∩ ⟨C 3 ⟩) = (3d − 1)/2.Since ⟨S⟩ = P 4 , we obtain L ′ ∩ ⟨C 3 ⟩ = ∅.Thus, there is a hyperplane H 1 ⊃ ⟨C However, taking as M a hyperplane containing a point of S ∩ R ′ , we obtain a contradiction.Now assume the existence of R.
and we excluded this case in step (b1.3.1)) or there is a line J 1 such that deg(Res U 3 (Z) ∩ J 1 ) ≥ d + 1 (it may be an irreducible component of C 2 ).Taking a general quadric hypersurface containing L ′′ ∪ J ∪ L 1 , we obtain a contradiction.
(b1.4) Assume the existence of We excluded the existence of D in step (b1.3) and the existence of D ′ in step (b1.2).Thus, ).We excluded the existence of R 2 and R 3 in steps (b1.2) and (b1.3).Thus, there is R. We use L 1 and R as we used L 1 and L 2 in the first part of step (b1.4). (b2 | for all A ⊂ S such that #A = 5.Assume for the moment that S is not in a linear general position, i.e., there is a hyperplane H containing at least five points of S. Take A ⊆ S ∩ H such that #A = 5 and take the quadric 2H.Since Z ⊂ 2H by the assumption of step (b2), we obtain S ⊂ H, a contradiction.Thus, S is in the linear general position and hence every hyperplane contains at most a degree eight subscheme of Z. Set Z 0 := Z.Let H 1 be a hyperplane such that z 1 := deg(Z ∩ H 1 ) is maximal.Set Z 1 := Res H 1 (Z).Fix an integer i ≥ 2 and assume the hyperplane H j , the integer z j and the scheme Z j for all j = 1, . . ., i − 1.Take a hyperplane H i such that

Beyond Theorem 1
We recall the definition of a reducible rational normal curve ([2] §4.1).Let T ⊂ P n be a reduced, connected and degree n curve spanning P n .If T is irreducible, then it is a rational normal curve.Now assume that T has s ≥ 2 irreducible components.Since T is connected, there is an ordering T 1 , . . ., T s of the irreducible components of T (called a good ordering) such that each For any reduced and connected curve M, let p a (M) denote its arithmetic genus, i.e., set rational normal curve in its linear span.Note that Sing(T) is the set of all points of T contained in at least two irreducible components of T.An irreducible component T i of T is said to be a final component if #(Sing(T) ∩ T i ) = 1.In any good ordering T 1 , . . ., T s of T, T 1 and T s are final components, but there may be other final components (e.g., take at T the union of n general lines though some point of P n ).By [2] Proposition 4.7, there are very strong restrictions for the existence of S ∈ T (n, d; ⌈(nd + 2)/2⌉) ′ contained in a reducible normal rational curve T: n is even, d is odd, S ⊂ T reg , all final components have an odd degree and #(S ∩ T i ) = (d i d + 1)/2 for all i.Now assume n = 4.We have d 1 + • • • + d s = 4 and hence 2 ≤ s ≤ 4. Take a reducible rational normal curve T ⊂ P 4 .To have some S ∈ T (4, d; 2d + 1) ′ with S ⊂ T, we also need d the be odd, S ⊂ T reg and #(S ∩ T i ) = d i (d + 1)/2 for all final components T i ([2] Proposition 6.1).Thus, either s = 2 and {d 1 , d 2 } = {1, 3} or 3 ≤ s ≤ 4 and all final components are lines.All s ∈ {2, 3, 4} and d 1 , . . ., d s with d 1 + • • • + d s = 4 occur for some reducible normal curve of P 4 (Remark 8 and Proposition 2).If d i = d j = 1 and #(S ∩ T i ) = #(S ∩ T j ) = (d + 1)/2 for some i ̸ = j, then T i ∩ T j = ∅, because no reducible conic contained in T contains d + 1 points of S. If we also prescribe that all final components of T have an odd integer as #(S ∩ T i ) ([2] Proposition 6.1), then we obtain the following list: With this restriction, we only have the following cases: T is nodal; and T 1 , T 2 , T 4 are final components, while #(T 3 ∩ T i ) = 1 for i = 1, 2, 4. Take one of the four cases just listed.Take S ∈ T (4, d; 2d + 1) ′ such that S ⊂ T reg and set x i := #(S ∩ T i ).Since S ⊂ T reg , we have x 1 + • • • + x 4 = 2d + 1.In case (1), we have d 1 = (3d + 1)/2 and d 2 = (d + 1)/2.In case (2), we have x 1 = x 3 = (d + 1)/2 and x 2 = d.
Proof.Obviously, ⟨S⟩ = P 4 .Since h 0 (O T (d)) = 4d + 1 and deg(2S ∩ T) = 4d + 2, h 1 (I 2S (d)) > 0. Thus, S ∈ T (4, d; 2d + 1).Fix S ′ ⊂ S such that #S = 2d and set A i := S ′ ∩ T i and y i := #(A i ).Note that y i = x i for s − 1 irreducible components of T and y i = x i − 1 for the other component of T. To prove that S is minimal, it is sufficient to prove that h 1 (I 2S ′ (d)) = 0. Since the restriction map H 0 (O P 4 (d)) → H 0 (O T (d)) is surjective, it is sufficient to prove that h 1 (T, I A,T (d)) = 0.This is performed by using s − 1 Mayer-Vietoris exact sequences.
The following observation gives another difference between T (n, d; x) and T (n, d; x) ′ ([2] Lemma 2.11).Proof.Assume that the lemma fails for n and d and take S ⊂ P n with minimal cardinality for which it fails.Note that S ̸ = ∅.Fix p ∈ S and set |. Take a system of homogeneous coordinates x 0 , . . ., x n of P n .Fix f ∈ H 0 (I 2A (d)).Take o ∈ A and i ∈ {0, . . ., n}.Since f vanishes on 2o, Proof.Fix a hyperplane H ⊂ P n and p ∈ P n \ H. Take A ∈ T (n − 1, d; x − 1).Identify P n−1 with the hyperplane H and hence see A as a subset of H. Set S := A ∪ {p}.All cones with a vertex at p and singular at all points of A are singular at all points of S. Thus, h 0 (I 2S (d)) ≥ h 0 (H, I 2A,H (d)) > 0. Since h 1 (H, I 2A,H (d)) > 0 and (2A, H) ⊂ 2S, h 1 (I 2S (d)) > 0. Take E ⊊ S such that ⟨E⟩.Obviously, p ∈ E. Set B := S \ {p}.We have ⟨B⟩ = H.To prove that S ∈ T (n, d; x) (and hence to prove both assertions of Proposition 3), it is sufficient to prove that h 1 (I 2E (d)) = 0. Consider the residual exact sequence of H: Proof.Fix a hyperplane H ⊂ P n , p ∈ P n \ H and a general A ∈ S(H, x − 1).Set S := A ∪ {p}.Since x − 1 ≥ n and A is general, ⟨S⟩ = P n .Since d ≥ 4, the scheme 2S is contained in the singular locus of a degree d hypersurface, the union of 2H and a degree d − 2 hypersurface singular at p. Thus, h 0 (I 2S (d)) > 0. Since n(x − 1) > h 0 (O H (d)), Proof.Assume the existence of S ∈ T (n, d; x).Since (n + 1)(x − 1) < ( n+d n ), h 0 (I 2E (d)) > 0 for all E ⊊ S. Let A denote the set of all A ⊊ S such that h 1 (I 2A (d)) > 0. Since T (n, d; x) ′ = ∅, A ̸ = ∅.Since S ∈ T (n, d; x), ⟨A⟩ ̸ = P n for all A ∈ A. The set A is partially ordered by inclusion.If A ⊆ B ⊊ S and A ∈ A, then B ∈ A. Thus, there is B ∈ A such that #B = x − 1.Since ⟨S⟩ = P n and S ∈ T (n, d; x), H := ⟨B⟩ has dimension n − 1. Set {p} := S \ H. Lemma 2 and the residual exact sequence of H imply h 1 (H, I (2B,H),H (d)) > 0. Since T (n − 1, d; x − 1)) = 0, there is E ⊊ B such that ⟨E⟩ = H and h 1 (H, I (2E,H),H (d)) > 0. Since ⟨{p} ∪ H⟩ = P n , the set E ∪ {p} gives S / ∈ T (n, d; x), a contradiction.

Motivations
In this section, we give the original motivation for the study of Terracini loci.Just to fix the notation, we conduct it in the set-up of the Veronese embeddings of a projective space.
Fix the positive integers n, d and x and let ν d : P n → P r , r = −1 + ( n+d n ) denote the Veronese embedding.Let x 0 , . . ., x n be homogeneous coordinates.Let K[x 0 , . . ., x n ] d denote the ( n+d n )-vector space of all degree d forms in n + 1 variables, i.e., K[x 0 , . . ., x n ] d = H 0 (O P n (d)).Thus, elements of P r correspond to equivalent classes of nonzero forms.Fix f ∈ K[x 0 , . . ., x n ] d , f ̸ = 0 and let [ f ] ∈ P r denote its equivalence class.An additive decomposition of f with exactly x addenda is an equality f = ∑ x i=1 ℓ d i with each ℓ i a linear form.This decomposition is equivalent to the existence of S ∈ S(P n , x) such that [ f ] ∈ ⟨S⟩ and [ f ] / ∈ ⟨S ′ ⟩ for any S ′ ⊊ S. The set S([ f ]) of the additive decompositions of [ f ] is often called the solution set of [ f ].The set S([ f ]) ⊂ S(P n , x) has a topology, the restriction to it of the Zariski topology of S(P n , x).If h 1 (I 2S (d)) = 0, then the Terracini lemma says that S is the unique additive decomposition of [ f ] which is "near" to S, i.e., S is an isolated point of S([ f ]) for the Zariski topology.Moreover, if h 1 (I 2S (d)) = 0, then h 1 (I 2A (d)) = 0 for all A ∈ S(P n , x) that are near S in the Zariski topology, and we may recover in this way (varying A) all points of P r in a Zariski neighborhood of [ f ].Given S, it is possible to quickly check the value of h 1 (I 2S (d)) by using software (it is a linear algebra problem).In all cases with (n + 1)x < ( n+d n ), we have h 0 (I 2E (d)) > 0 for all E ∈ S(P n , x).Hence, only the value of h 1 (I 2S (d)) matters, and usually ⟨S⟩ = P n .If h 1 (I 2S (d)) > 0, it is easy to check all S ′ ⊊ S to see if S is minimal or primitive.

Conclusions
Our main results are negative (certain Terracini loci are empty), but we discuss in Section 6 how the emptiness results are used.Among the existing results, we stress the ones with the lowest possible number of points for multivariate forms of fixed degrees in a given number of variables.
We raised several open questions in the introduction and listed another one (on the Hilbert function of zero-dimensional schemes) at the end of Section 2 with a discussion of its possible applications, for instance, to evaluation codes ( [3,4]).
Funding: This research had no funding.
D (Z) ⊆ Z and deg(Z) = deg(D ∩ Z) + deg(Res D (Z)).For any line bundle L on M, we have an exact sequence 0 → I Res D (Z) ⊗ L(−D) → I Z ⊗ L → I Z∩D,D ⊗ L |D → 0 (1) we reduce to step (a) for the integer d − 2. If Z ⊂ Q, we use several residual exact sequences with respect to hyperplanes of P 4 and use Proposition 1 for the intersection of Z with these hyperplanes.Proof of Theorem 1: Set z := deg(Z).Since each connected component of Z has a degree at most of two, z ≤ 2x.(a) Assume x ≤ 2d − 1.For any o ∈ P 4 , let ℓ o : P 4 \ {o} → P 3 denote the linear projection from o. Since each connected component of Z has degree ≤ 2, there is a nonempty open subset U of P 4 \ S such that for all o ∈ U , the morphism ℓ o|Z is an embedding.By the semicontinuity theorem for cohomology restricting if necessary U , we may assume that all degree z schemes ℓ o (Z), o ∈ U , have the same Hilbert function.Fix o ∈ U and set Z ′ := ℓ o (Z) and S ′ := ℓ o (S).Since ⟨S⟩ = P 4 , we have ⟨S ′ ⟩ = P 3 .Take homogeneous coordinates x 0 , x 1 , x 2 , x 3 , x 4 of P 4 such that o = [0 : 0 : 0 : 0 : 1].For any p = [p 0 : p 1 : p 2 : p 3 : p 4 ] ̸ = o, we have ℓ o (p) = [p 0 : p 1 : p 2 : p 3 ].For any constant t ̸ = 0, define h t ∈ Aut(P 4 ) by the formula h t ([p 0 : p 1 : p 2 : p 3 : p 4 ]) = [p 0 : p 1 : p 2 : p 3 : tp 4 ].

≤ 6 .
Since ⟨S⟩ = P 4 , the residual exact sequence of ⟨T 1 ⟩ gives that S is not minimal.(b) Assume x = 2d and d ≥ 19.Fix A ⊂ S such that #A = 5 and call E the union of the connected components of Z with a point of A as their reduction.We have dim |I E (2)| ≥ 4. Take a general U ∈ |I E (2)|.(b1) Assume Z ⊈ U. Since S is minimal and Z ⊈ U, the residual exact sequence of U gives h 1 (I Res U (Z) (d − 2)) > 0. Set F := Res U (Z) red .Since F ⊆ S \ A, hence, #F ≤ 2(d − 2) − 1.Thus, we may apply part (a) and [2] Th. 1.3, 1.4, 1.5 to F. As in step (a), we take a general o ∈ P 4 and set Z ′ := ℓ o (Z) with deg(Z ′ ) = z.Since h 1 (I Res U (Z) (d − 2)) > 0, we may apply step (a) for the integer d − 2. Thus, we obtain that one of the following possibilities occurs: 1.

Remark 12 .Lemma 2 .
Fix S ∈ T (n, d; x) and let Z be a critical scheme of S. Part (b) of Theorem 3 shows that sometimes Z red ̸ = S. Fix positive integers n and d.Take a finite set S ⊂ P n .If h 1 (I 2S (d)) = 0, then h 1 (I S (d − 1)) = 0.