ON THE X-RANK OF A POINTS OF THE TANGENT DEVELOPABLE OF A CURVE IN A PROJECTIVE SPACE

Let X ⊂ Pn be a smooth curve. For any P ∈ Pn the X-rank of P is the minimal cardinality of a set S ⊂ X such that P ∈ 〈S〉, where 〈 〉 denote the linear span. Let τ(X) ⊂ Pn be the tangent developable of X. We compute upper bounds for the X-rank of all P ∈ τ(X) or of the general P ∈ τ(X), mainly if X is a canonically embedded curve. To do that we define some invariants for the pair (X,OX (1)) and compute them if X is canonically embedded and either X is a smooth plane curve or it has general moduli. AMS Subject Classification: 14N05, 14H52


Introduction
Fix an integral and non-degenerate variety X ⊂ P n .For any P ∈ P n the X-rank r X (P ) of P is the minimal cardinality of a subset S ⊂ X such that P ∈ S , where denote the linear span ( [16], [8], [15]).Now assume that X is a smooth curve and that n ≥ 3. Let τ (X) ⊂ P n be the tangent developable of X, i.e. the unions of all tangent lines T O X.Quite often the maximal among the integers r X (P ) is achieved by a point of τ (X) \ X (e.g. this is the case when X is a rational normal curve by a theorem of Sylvester's ( [16,Theorem 4.1], [8,Theorem 23], but it appears quite often elsewhere ( [4])).In this note we give some criteria to compute upper bounds for points of τ (X) \ X when X is a smooth curve, usually when X is the canonical model of a non-hyperelliptic smooth curve.There are also some easy and well-known lower bound and sometimes we get in this way a good picture).As a byproduct of these definitions we prove the following result.
Corollary 1.Let C ⊂ P 2 , d ≥ 4, be a smooth pane curve of degree d.Let X ⊂ P g−1 , g = (d − 1)(d − 2)/2, be the canonical model of C and φ : C → X the canonical map Then r X (P ) ≤ 2d − 5 for all P ∈ τ (X) and r X (P ) = d − 2 for a general P ∈ τ (X) and r X (P ) = d − 3 for some P ∈ τ (X).If d ≥ 5 and O ∈ C is such that the tangent line to C at O is neither a flex nor a multitangent, then r X (P ) = d − 3 for at least one and at most d − 2 points of T φ(O) (X) \ {φ(O)} and r X (P ) = d − 2 for all other points of T φ(O) (X) \ {φ(O)}.We have r X (P ) ≥ 2d − 6 for at least one P ∈ τ (X).
In section 2 we study the lowest integers r X (P ), P ∈ τ (X) \ X.In section 3 we study r X (P ) ∈ τ (X) (see Proposition 6 for curves with general moduli, Propositions 2, 3 and 5 for an arbitrary curve and Theorem 1 for a result giving, with a little effort, Corollary 1).
We work over an algebraically closed field K such that char(K) = 0.
2. The Lowest Ranks of Points of τ (X) \ X Notation 1.Let C ⊂ P n be a smooth, connected and non-degenerate curve.Let β(C) be the maximal integer such that every zero-dimensional subscheme of C with degree at most β(C) is linearly independent.
Let X be a smooth curve of genus g ≥ 3. The gonality gon(X) of X is the minimal degree of a line bundle L on X with h 0 (X, L) ≥ 2. Let Gon(X) denote the set of all line bundles L on X with degree gon(X) and h 0 (X, L) ≥ 2. By the definition of gonality we have Gon(X) = ∅.For each L ∈ Gon(X) we have h 0 (X, L) = 2 and X is base point free.The set Gon(X) is a closed subset of Pic gon(X) (X) and dim(Gon(X)) ≤ 1 ( [11]).If X is not hyperelliptic and Fix L ∈ Gon(X).Since L is spanned and h 0 (X, L) = 2, the complete linear system |L| induces a degree gon(X) morphism h L : X → P 1 .Let R(L) be the set of all O ∈ X such that h 0 (X, L(−2O)) > 0. Since L has no base points, we have Easy examples (standard cyclic coverings with degree k and large g) show that we may have R ′ (L) = ∅ for all L ∈ Gon(X).In particular we may have ∆(X) 1 = ∅.Proposition 1. Assume gon(X) ≥ 6.We have r X (P ) ≥ gon(X) − 2 for all P ∈ τ (X) \ X. Fix O ∈ X.
Proof.We have β(X) = gon(X) − 1 ≥ 5. Since β(X) ≥ 4, for each P ∈ τ (X) there is a unique zero-dimensional scheme O ∈ X with P ∈ T O X. Fix P ∈ τ (X) \ X and take the unique O ∈ X such that P ∈ T O X, i.e. such that P ∈ 2O .Fix S ⊂ X evincing r X (P ), i.e. fix S ⊂ X with ♯(S) = r X (P ) and P ∈ S .By [7, Lemma 1] we have h ∈ S, we have 2O + S = 2O ∪ S, i.e. 2O + S is the minimal zero-dimensional scheme containing both 2O and S. Since 2O + S is a divisor of a line bundle evincing the gonality of X, Grassmann's formula gives that 2O ∩ S is a single point, P L .We get W = ∅ and the existence of a surjection M(X, O) → W. Hence ♯(W) ≤ µ(X, O).Fix any P ′ ∈ W, say associated to L ∈ ∆(X) 1 , and take the only S ⊂ X such that 2O + S ∈ |L|.

We have O /
∈ S and we saw that {P ′ } = 2O ∩ S .Fix any P ∈ T O X. Since P ′ = O, we have P ∈ {O} ∪ S and hence r X (P ) ≤ gon(X) − 1.

The Rank of a General Point of the Tangent Developable
Let C be a smooth and connected projective curve of genus g.We recall the following definitions ([10], [5], [6]).For each integer r ≥ 1 the r-gonality gon(C, r) of C is the minimal integer d such that there is a degree d line bundle L on C with h 0 (C, L) ≥ r + 1 (note that h 0 (C, L) = r + 1 and that L has no base points for each such L).The integer gon(C, 1) is the gonality gon(C) of C. For every integer r ≥ 2 the r-birational gonality birgon(C, r) is the minimal integer d such that there is a degree d line bundle L on C with h 0 (C, L) ≥ r + 1 and the rational map induced by |L| is birational onto its image (notice that h 0 (C, L) = r + 1 and L has no base points for each such L).For every integer r ≥ 3 the r-embedding gonality embgon(C, r) is the minimal integer d such that there is a degree d line bundle L on C with h 0 (C, L) ≥ r + 1 and the rational map induced by |L| is an embedding.
If C has general moduli and 1 ≤ r ≤ g − 2, then gon(C, r) = ⌈rg/(r − 1)⌉ + r − 1 and the set W r d (C) of line bundles on C evincing d := gon(C, r) has pure dimension ρ(g, r, d) Let C be a smooth and connected projective curve of genus g.   for the arithmetic genera of non-degenerate curve in P r with degree ≤ r + 1, we get that φ(C) is an elliptic curve.Hence C is a bielliptic curve, a contradiction.Now assume g = 5.We need to exclude the case deg(φ) = 3.This is excluded, because in this case φ(C) would be a smooth conic and hence C would be a trigonal curve.
Proof.It is sufficient to prove the inequality for all P ∈ τ (X) \ X. Fix O ∈ X such that P ∈ 2O \ {O}.Take a line bundle L on X evincing d(X, O).Therefore we have 2O + S ∈ |L| with S a reduced divisor and O not in the support of S. Hence 2O ∩ S is a point, P ′ , in 2O \{O}.Since P ∈ {P ′ , O}, we get r In the same way we prove the following result.Proposition 3. Let X ⊂ P g−1 , g ≥ 4, be a canonically embedded curve such that d ′ (X, 2) is defined.We have r X (P ) ≤ d ′ (X, 2) − 1 for a general P ∈ τ (X).
The proof of Proposition 2 and 3 gives the following result.Proof.The inequality d ′ (C, 2) ≤ d−3 follows from Proposition 5. We have It is often easier to compute the integer d ′ (C, 2) then the integer d(C, 2).It may also be Proposition 6.We have d ′ (X, 2) = ⌊g/2⌋ for a general curve X of genus g ≥ 7.
Question 1.In the set-up of Proposition 6 it is reasonable to conjecture that d(X, 2) = d ′ (X, 2) for a general smooth curve of genus g ≥ 7.

Remark 2 .
Let C ⊂ P n be a smooth curve.Take zero-dimensional schemes A, B ⊂ C. Therefore A and B are effective divisors of C. For each P ∈ C let a P (resp.b P ) be the degree of the connected component of A (resp.B).The effective divisor A+B has degree deg(A)+deg(B) (each P appars with multiplicity a P +b P in A+B).The scheme A∪B is the minimal subscheme of C containing both A and B. Each P ∈ C appears with multiplicity max{a P , b P } in A ∪ B. Therefore A + B = A ∪ B if and only if A ∩ B = ∅.
R, 2) be the maximum of all integers d(C, R, O), O ∈ C. Let d ′ (C, R, 2) be the minimal integer d such that there is a finite set A ⊂ C with d(C, R, O) defined and d ≥ d(C, R, O) for all O ∈ C \ A. Now assume that C has genus g ≥ 3 and that C is not hyperelliptic., i.e. assume that ω X is very ample.Set d(C, O) := d(C, ω C , O), d(C, 2) := d(C, ω C , 2) and d ′ (C, 2) := d ′ (C, ω C , 2) (if these integers are defined).Lemma 1.Let C be a smooth curve and let R ∈ Pic(C) be a line bundle with deg(R) ≥ 4. Fix O ∈ C and assume h 0 (R(−2O)) = h 0 (R) − 2, that R(−2O) is spanned and that the morphism φ associated to R(−2O) is birational onto its image.Then d(C, R, O) is defined and d be the morphism associated to |R|.Since φ is birational onto its image, ψ is birational onto its image.Set Y := ψ(C) and B := Sing(Y ).Set B ′ := ψ −1 (B).Notice that ψ induces an isomorphism between C \B ′ and Y \B.Since h 0 (R(−2O)) = h 0 (R) − 2, the zero-dimensional scheme Z := ψ(2O) has degree two.Hence Z is a line.Since deg(R(−2O)) ≥ 2, we have Y = Z and hence there is P ∈ Z \ Z ∩ Y .Let H ⊂ P r be a general hyperplane containing P .Since P / ∈ Y , Bertini's theorem implies that the scheme Y ∩ H is formed by deg(R) distinct points, none of them being in B ∪ {φ(O)}.Since P ′ ∈ Z \{ψ(O)}, Therefore there is a set A ⊂ C\({O}∪B ′ ) with ψ(A) = Y ∩H.Since

Proposition 4 .
Fix a linearly normal smooth curve C. If d(C, O C (1), O) is defined for all O ∈ C, then r C (P ) ≤ d(C, O C (1), 2) + 1 for all P ∈ σ 2 (C).If d ′ (C, O C (1), 2) is defined, then r C (P ) ≤ d ′ (C, O C (1), 2) for a general P ∈ τ (C).Proposition 5.We have d ′ (C, 2) ≤ birgon(C, 2) − 3 for every nonhyperelliptic smooth curve C of genus g ≥ 3.Proof.Fix L ∈ Pic(C) evincing birgon(C, 2) and let f : C → P 2 be the associated morphism.Fix a general O ∈ C. Since O is general, O is an ordinary point with respect to f in the sense of[14], i.e. h 0 (C, L(−2O)) = 1 and h 0 (C, L(−3O)) = 0. Therefore the only divisor D ∈ |L(−2O) has not O in its support.Since we are in characteristic zero, f (C) is not a strange curve, i.e. there is no o ∈ P 2 contained in every tangent line to a smooth point of C. Since O is general, we get T f (O) f (C) ∩ Sing(C) = ∅.Since a general tangent line of f (C) is tangent at a unique smooth point of f (C) and f is birational onto its image, D contains no point of C with multiplicity ≥ 2. Since C has positive genus, we have birgon(C, 2) ≥ 3. Hence the tangent line of f (C) at f (O) contains another smooth point, say f (Q), of f (C).Take L ′ := L(−Q) and S := D − Q.We have h 0 (O C (2O)) = 1, since C is not hyperelliptic.By construction we have h 0 (C, O C (2O + S)) = h 0 (L ′ ) = 2. Theorem 1.Let C ⊂ P 2 be a smooth curve of degree d ≥ 4. We have d ′ (C, 2) = d−3.If the tangent line to C at O is neither a flex nor a multitangent, then d(C, O) is defined and d(C, O) = d − 3.If the tangent line to C at O is either a flex or a multitangent, then d(C, O) is defined and d(C, O) = 2d − 6.We have d(C, 2) = 2d − 6.
either a multitangent or an inflexional tangent of C. We have d(C, d) ≥ d, because C has at least one flex.Fix O ∈ C and assume that the tangent line to C at O is either a flex or a multitangent.Since O C (1)(−O) has no base points, O C (2)(−2O) has no base points.Hence a general D ∈ |O C (2)(−2O)| is reduced and it does not contain O in its support.Since h i (P 2 (O P 2 (2 − d)) = 0, i = 0, 1, there is a unique conic T with T ∩ C = 2O + D (as schemes).Call Z the degree two subscheme of P 2 with 2O as its support.A general element of |I Z (2)| is a smooth conic.Since D is general, T is a general element of |I Z (2)|.Therefore T is a smooth conic.We have h 0 (C, O C (2)(−2O − S)) = h 0 (C, O C (2)(−E)) = 2, because the smoothness of the conic T implies that E is not formed by 4 collinear points.Therefore d(C, ω C , O) is defined and d(C, O) ≤ 2d − 6. Assume d(C, O) ≤ 2d − 7, i.e. d(C, O) + 2 ≤ 2d − 5. Take S evincing d(C, O).We have O / ∈ S, h 0 (C, O C (2O + S)) = 2 and O C (2O + S) has no base points.Hence d(C, O) + 2 is in the Lüroth semigroup of C. Since d(C, O) + 2 ≤ 2d − 5, S. Greco and G. Raciti proved that d(C, O) + 2 ∈ {d − 1, d} ([12], [9]).We excluded the case d(C, O) + 2 = d − 1, because the tangent line of C at O is either a flex or a multitangent and hence O / ∈ ∆(C) 1 .Now assume d(C, O) + 2 = d and set R := O C (2O + S).By assumption R is a base point free line bundle of degree d.Since T O C is either a flex or a multitangent, 2O + S / ∈ |O C (1)|.Therefore R = O C (1).Fix a general A ∈ |R|.Since h 0 (R) = 2, |ω C | = |O P 2 (d − 3)), we have h 1 (P 2 , I A (d − 3)) > 0. Since deg(A) ≤ 2(d − 3) + 1, there is a line J ⊂ P 2 such that deg(A ∩ J) ≥ d − 1 ([8,Lemma 34]).Since R has no base points, we also have h 1 (P 2 , I A ′ (d−3)) = 0 for every A ′ A (Riemann-Roch and the adjunction formula).Therefore A ⊂ J ∩ C. Since deg(A) = d, we get A ∈ |O C (1)| and hence R ∼ = O C (1), a contradiction.Proof of Corollary 1. Fix O ∈ C such that the tangent line T O C of C at O is neither a tangent line to C ad O is neither a flex of C nor a multitangent of C. We have T O C ∩ C = 2O + S with S a finite set of C with cardinality d − 2 and O / ∈ S. We have M(C, O) = {O C (1)(−Q)} Q∈S .Since C has positive genus, the line bundles O C (1)(−Q) and O C (1)(−Q ′ ) are not isomorphic if Q = Q ′ .Therefore µ(C, O) = d−2.Use part (ii) of Proposition to get that r X (P ) = d−3 for at least one and at most d−2 points of T φ(O) (X)\{φ(O)}.Fix Q ∈ S and set S ′ := S \Q.Since h 0 (O C (2O+S ′ )) = h 0 (C, O C (1)(−Q)) = 2, the set T O X ∩ S ′ is a single point, P ′ , and P ′ = O.Fix any P ∈ T φ(O) X.Since P ′ = O, we have P ∈ {P ′ , φ(O)} .Therefore P ∈ {O} ∪ S .Hence r X (P ) ≤ d − 2. Now take O such that T O C is either a flex or a multitangent of C. Theorem 5 gives the existence of P ′ ∈ T φ(O) X \ φ(O) with r X (P ′ ) = 2d − 6.Since P ′ = φ(O) as above we get r X (P ) ≤ 2d − 5 for all P C). Since any non-constant morphism f : C → P 1 has only finitely many ramification points, then dim(G 1 d ′ (C,2)+2 (C)) ≥ 1.