International Mathematical Forum, Vol. 6, 2011, no. 8, 389 - 398Full Tex

Dieudonne, Laboratoir´e J. A.

Maingi, Damian

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International Mathematical Forum, Vol. 6, 2011, no. 8, 389 - 398Maximal Rank for ΩPnDamian M. MaingiThe School of MathematicsUniversity of NairobiP.O. Box 30197 00100 Nairobi, Keniadmaingi@uonbi.ac.ke&Laboratoiré J A DieudonneUniversité de Nice–Sophia Antipolis06108 Nice Cedex 02 Francedmaingi@unice.frAbstract. Let k an algebraically closed field and R the homogeneous coordinate ringof Pn and ΩPn the cotangent bundle of Pn. In this paper I prove that for a given set S ofs general points in Pn then the evaluation map H0(Pn, ΩPn(l)) −→ ⊕si=1 ΩPn(l)|Pi is ofmaximal rank. Implying that a0 = 0 or b0 = 0 so that a0b0 = 0 as conjectured by AnnaLorenzini [4, 5] see below· · · −−−→ R(−d − 2)b1 ⊕R(−d − 1)a0 −−−→ R(−d − 1)b0 ⊕R(−d)(d+nn )−s −−−→ IS −→ 0Mathematics Subject Classification: 13D02, 16E05Keywords: Elementary Transformations, Cotangent Vector Bundle1. introductionFor a general set of points {P1, . . . , Ps} ∈ Pn, with s ≥ n + 1, then the homogeneous idealof the sub-scheme of the union of these points, IS ⊂ R = k[x0, . . . , xn], k an algebraicallyclosed field and R the homogeneous coordinate ring of Pn, has the following expectedform:0 −−−→ Fn−1 −−−→ · · · −−−→ Fp −−−→ · · · −−−→ F0 −−−→ IS −−−→ 0,390 D. M. MaingiFp = R(−d − p)ap−1⊕R(−d − p − 1)bp,d being the smallest integer satifying s ≤ h0(Pn, OPn(d)), withap = max{0, h0(Pn, Ωp+1Pn (d + p + 1)) − rk(Ωp+1Pn )s},bp = max{0, rk(Ωp+1Pn )s − h0(Pn, Ωp+1Pn (d + p + 1))}, and(d + n − 1n)< s ≤(d + nn).The problem can be reduced to showing the following; for all 0 ≤ p ≤ n − 1 and non-negative integer l then existence of the above resolution is the same as saying the evalu-ation map below is of maximal rank i.e. it is surjective or injective or both; see [1].H0(Pn, Ωp+1Pn (l)) −→s⊕i=1Ωp+1Pn (l)|Pi.For this consider the exact sequence0 −−−→ ΩPn(1) −−−→ W ⊗ OPn −−−→ OPn(1) −−−→ 0Here, W = H0(OPn(1)), the set of linear forms and k[x0, x1, ..., xn] = Sym(W )Tensoring the sequence above with TS(d) gives0 −−−→ TS ⊗ ΩPn(d + 1) −−−→ W ⊗ TS(d) −−−→ TS(d + 1) −−−→ 0Now taking global sections we get;0  H0(TS ⊗ ΩPn(d + 1))  W ⊗ Id  Id+1H1(TS ⊗ ΩPn(d + 1))0Thus H1(TS ⊗ΩPn(d + 1)) = Id+1/W · Id, corresponds to the minimal generators of IS ofdegree d + 1, and its dimension is b0 i.e. h1(TS ⊗ ΩPn(d + 1)) = b0.Maximal Rank for ΩPn 391Similarly, H0(TS ⊗ ΩPn(d + 1)) is the space of linear relations among the generators ofdegree d, whose dimension is a0 i.e. h0(TS ⊗ ΩPn(d + 1)) = a0.Now consider the exact sequence0 −−−→ TS −−−→ OPn −−−→ OS −−−→ 0Tensoring it by ΩPn(d + 1) gives;0 −−−→ TS ⊗ ΩPn(d + 1) −−−→ ΩPn(d + 1) −−−→ ΩPn(d + 1)|S −−−→ 0and now taking global sections yields0  H0(TS ⊗ ΩPn(d + 1))  H0(ΩPn(d + 1))μ H0(ΩPn(d + 1)|S)H1(TS ⊗ ΩPn(d + 1))0We will prove that μ is of maximal rank for a general set S of s points in Pn.As result, if μ is injective then its kernel is null i.e. a0 = h0(TS ⊗ ΩPn(d + 1)) = 0 andthe cokernel is not null that is b0 = h1(TS ⊗ΩPn(d+1)) as expected. On other hand, if μis surjective then we have the cokernel of μ being null i.e. b0 = h1(TS ⊗ ΩPn(d + 1)) = 0and the kernel of μ is not null that is, a0 = h0(TS ⊗ ΩPn(d + 1)).2. preliminariesWe use the statements (the so called Enonces) as in [1] by Hirschowitz and Simpson whichF Lauze used in [2] to proof maximal rank for TPn .Let X a smooth projective variety and X ′ non-singular divisor of X. Let F be a locallyfree sheaf on X and0 −−−→ F′′ −−−→ F|X′ −−−→ F′ −−−→ 0be a exact sequence of locally free sheaves on X ′. The kernel E of F −→ F′ is a locallyfree sheaf on X and we have another exact sequence of locally free sheaves on X ′0 −−−→ F′(−X ′) −−−→ E|X′ −−−→ F′′ −−−→ 0and as well exact sequences of coherent sheaves on X0 −−−→ E −−−→ F −−−→ F′ −−−→ 0392 D. M. Maingiand0 −−−→ F(−X) −−−→ E −−−→ F′′ −−−→ 0.We have the following hypotheses:R(F, F′, y; a, b, c)RD(F, F′, y; a, b, c)RD(E, F′′, y′; a′, b′, c′)2.1. Notation. Set X = Pn, X ′ = Pn−1, F = ΩPn , F′ = ΩPn−1 , E = O⊕nPn(−2), F′′ =OPn−1(−1).The exact sequences of the elementary transformations after twisting by d + 1 are:0 0⏐⏐	⏐⏐	ΩPn(d) ΩPn(d)⏐⏐	 ⏐⏐	0 −−−→ OPn(d − 1)⊕n −−−→ ΩPn(d + 1) −−−→ ΩPn−1(d + 1) −−−→ 0⏐⏐	 ⏐⏐	 ∥∥∥0 −−−→ OPn−1(d) −−−→ ΩPn |Pn−1(d + 1) −−−→ ΩPn−1(d + 1) −−−→ 0⏐⏐	⏐⏐	0 0From which we have the hypotheses:H ′Ω,n(d + 1; α, β, γ) = H(ΩPn(d + 1), ΩPn−1(d + 1), α, β, γ) andH ′ ,n(d − 1; ρ, σ, τ) = H(OPn(d − 1)⊕n, OPn−1(d); ρ, σ, τ) andH ′′ ,n(d − 1; ρ, σ, τ) = H(OPn(d − 1)⊕n, OPn−1(d); ρ, σ, τ).For the plane divisorial, with H ⊆ Pn a hyperplane isomorphic to Pn−1 we shall utilizethe sequence;0 −−−→ OPn(d − 2)⊕n −−−→ OPn(d − 1)⊕n −−−→ OH(d − 1)⊕n −−−→ 0..Hypothesis 2.1. H ′Ω,n(d + 1; α, β, γ)The hypothesis H ′Ω,n(d + 1; α, β, γ) asserts that for non-negative integers α, β, γ and εsatisfying the conditions:0 ≤ γ ≤ 1, and 1 ≤ ε ≤ n − 2,nα + n − 1β + εγ = h0(ΩPn(d + 1)), andMaximal Rank for ΩPn 393(n − 1)β + εγ ≤ h0(ΩPn−1(d + 1)) having for γ = 1 a quotient Γ′ then the mapη : H0(Pn, ΩPn(d + 1)) −→α⊕i=1ΩPn(d + 1)|Ai ⊕β⊕j=1ΩPn−1(d + 1)|Bj ⊕ Γ′|Cis bijective with h0(ΩPn(d + 1)) = d(d+nd+1)and for α general points A1 . . . Aα ∈ Pn, β + 1general points B1 . . . Bβ, C ∈ Pn−1.Hypothesis 2.2. HΩ,n(d + 1)The hypothesis HΩ,n(d + 1) asserts that H′Ω,n(d + 1; α, β, γ) is true for all α, β and γsatisfying the conditions above.Hypothesis 2.3. H ′ ,n(d − 1; ρ, σ, τ)The hypothesis H ′ ,n(d − 1; ρ, σ, τ) asserts that for non-negative integers ρ, σ, τ and θsatisfying the conditions:0 ≤ τ ≤ 1 and 2 ≤ θ ≤ n − 1,nρ + σ + θτ = h0(OPn(d − 1)⊕n), andσ + θτ ≤ h0(OPn−1(d)) having for τ = 1 a quotient Γ then the mapφ : H0(Pn, OPn(d − 1)⊕n) −→ρ⊕i=1OPn(d − 1)⊕n|Ri ⊕σ⊕j=1OPn−1(d)|Sj ⊕ Γ(S)|Tis bijective with h0(OPn(d − 1)⊕n) = n(d+n−1d−1)and for ρ general points R1 . . . Rρ ∈ Pn,σ + 1 general points S1 . . . Sσ, T ∈ Pn−1.Hypothesis 2.4. H ,n(d − 1)The hypothesis H ,n(d − 1) asserts that H ′ ,n(d − 1; ρ, σ, τ) is true for any ρ, σ, and τsatisfying the conditions above.Hypothesis 2.5. H ′′ ,n(d − 1; ρ, σ, τ)A variant version of the hypothesis H ′ ,n(d − 1; ρ, σ, τ) with Γ independent of Γ′ takesthe form H ′′ ,n(d− 1; ρ, σ, τ) and it makes the same assertion as the hypothesis H ′ ,n(d−1; ρ, σ, τ) the only difference being quotient dependency.3. the methods of horaceMéthode d’Horace simple[3] lemme 1Lemma 3.1. Suppose we have a bijective morphism of vector spaces γ : H0(X ′, F′) −−−→ Land that we have H1(X, E) = 0. Let μ : H0(X, F) −−−→ L be a morphism of vectorspaces. Then for H0(X, F) −→ M⊕L to be of maximal rank it suffices that H0(X, E) −→M is of maximal rank.394 D. M. MaingiDifferential méthode d’Horace([1] lemme 1)Lemma 3.2. Suppose we are given a surjective morphism of vector spaces,λ : H0(Pn−1, ΩPn−1(d + 1))  L and suppose there exists a point Z′ ∈ Pn−1 such thatH0(Pn−1, ΩPn−1(d + 1))   L ⊕ ΩPn−1(d + 1)|Z ′ and suppose H1(Pn, OPn(d − 1)⊕n) =0. Then there exists a quotient OPn(d−1)⊕n|Z ′ −→ D(λ) with kernel contained in ΩPn−1(d)|Z ′of dimension dim(D(λ)) = rk(ΩPn(d + 1)) − dim(kerλ) having the following property.Let μ : H0(Pn, ΩPn(d + 1)) −→ M be a morphism of vector spaces then there existsZ ∈ Pn−1 such that if H0(Pn, OPn(d − 1)⊕n) −→ M⊕D(λ) is of maximal rank thenH0(Pn, ΩPn(d + 1)) −→ M ⊕ L⊕ΩPn(d + 1)|Z is also of maximal rank.The sequences for the quotient are as follows:00dim n − 1 ΩPn−1(d)|Z   D′|Z dim n − 3 (n − 2)dim n OPn(d − 1)⊕n|Z  D|Z ∼= OPn−1 |Z ⊕ D′|Zdim n − 2 (n − 1)dim 1 OPn−1(d)|ZOP3(d)|Zdim 10 03.1. The Vectorial Methods.Lemma 3.3. Vectorial Method 1Let α, β, γ, d and ε be non-negative integers satisfying the conditions of Hypothesis 2.1and ρ, σ, τ and θ non-negative integers satisfying the conditions of Hypothesis 2.3 thenthe Hypothesis H ′ ,n(d − 1; ρ, σ, τ) implies H ′Ω,n(d + 1; α, β, γ).Proof. Consider the exact sequence;0  OPn(d − 1)⊕n  ΩPn(d + 1)  ΩPn−1(d + 1)  0and let B and C be general subsets of Pn−1. We specialize A to R ∪ S ∪ T with R ageneral set of ρ points in Pn and S and T sets of σ and τ general points in Pn−1. To runpoints to Pn−1, consider the map, γ : H0(ΩPn−1(d + 1)) −→ H0(ΩPn−1(d + 1)|B) ⊕ Γ′|C , ifMaximal Rank for ΩPn 395the number of points we have satisfy h0(ΩPn−1(d + 1)) then γ is bijective, if not then wespecialize as many more points as we need to Pn−1 in order for γ to become bijective.Taking global sections for the exact sequence above and evaluating we construct;0 0⏐⏐⏐⏐H0(ΩPn−1(d + 1))γ−−−→∼= H0(ΩPn−1(d + 1)|{B∪S}) ⊕ Γ′|C ⊕ Γ|T⏐⏐ ⏐⏐H0(ΩPn(d + 1))β−−−→ H0(ΩPn(d + 1)|R∪S∪T=A) ⊕ H0(ΩPn−1(d + 1)|B) ⊕ Γ′|C⏐⏐⏐⏐H0(OPn(d − 1)⊕n) α−−−→ H0(OPn(d − 1)⊕n|R ) ⊕ H0(OPn−1(d)|S) ⊕ Γ|T⏐⏐⏐⏐0 0From the above diagram of exact sequences, by Inductive hypothesis on Pn−1 and Lemma3.2 the map γ is bijective and hence if α is bijective then β is bijective as well and thisgives H ′ ,n(d − 1; ρ, σ, τ) implies H ′Ω,n(d + 1; α, β, γ)Lemma 3.4. Vectorial Method 2Let ρ, σ, τ and θ non-negative integers satisfying the conditions of Hypothesis 2.3 and α,β, γ and ε be non-negative integers satisfying conditions similar to those of Hypothesis2.1 with the Hypothesis H ′Ω,n(d; α, β, γ) being the same as Hypothesis 2.1 but twisted by1, then the Hypothesis H ′Ω,n(d; α, β, γ) implies H′ ,n(d − 1; ρ, σ, τ).Proof. Consider the exact sequence;0  ΩPn(d)  OPn(d − 1)⊕n  OPn−1(d)  0and let S and T general sets of σ and τ points in Pn−1, specialize R to A ∪ B, where Ais a general set of α points in Pn and B is a general set of β points in Pn−1 with C = T .Now consider the evaluation map, γ : H0(OPn−1(d)) −→ H0(OPn−1(d)|S∪T ), if the numberof points we have are enough to satisfy h0(OPn(d)) then γ bijective, if not then we spe-cialize as many more points, β, in this case, to Pn−1 in order for γ to become bijective.Taking global sections for the exact sequence above and evaluating at corresponding pointswe construct a diagram of exact sequences as follows;396 D. M. Maingi0 0H0(OPn−1(d))~~γ H0(OPn−1(d)|{S∪T∪B})~~H0(OPn(d − 1)⊕n)~~ H0(OPn(d − 1)⊕n|R ) ⊕ H0(OPn−1(d)|S) ⊕ ΓT~~H0(ΩPn(d))~~ H0(ΩPn(d + 1)|A) ⊕ H0(ΩPn−1(d + 1)|B ⊕ ΓC~~0~~0~~The map γ is bijective giving the Hypothesis H ′Ω,n(d; α, β, γ) implies H′ ,n(d− 1; ρ, σ, τ).When the number of points we have in Pn−1 are few relative to d we use the planedivisorial method in preference to this method.Lemma 3.5. Plane DivisorialLet ρ, σ, τ and θ non-negative integers satisfying the conditions of Hypothesis 2.3 and setρ′ = ρ − h0(OPn−1(d − 1)). If ρ′ ≥ 0 and σ + τ ≤ h0(OPn−1(d − 1)) then the HypothesisH ,n(d − 2; ρ′, σ, τ) implies H ,n(d − 1; ρ, σ, τ).Proof. Let R be a general set of ρ points in Pn, S and T be general sets of σ and τ pointsin Pn−1 such that they are fewer relative to d (i.e. when Vectorial Method 2 fails). Wechoose a hyperplane H ⊂ Pn disjoint from S and T with H ∼= Pn−1 and specialize ρ′points from Pn to H (i.e. R′ is the set we have after specializing from R in Pn) so thatH0(H, OH(d−1)⊕n) −→ H0(OH(d−1)⊕n|R′) is bijective that is set ρ−ρ′ = h0(OPn−1(d−1))and so taking global sections for the sequence0 −−−→ OPn(d − 2)⊕n −−−→ OPn(d − 1)⊕n −−−→ OH(d − 1)⊕n −−−→ 0Maximal Rank for ΩPn 397we construct a diagram of exact sequences:0 0⏐⏐⏐⏐H0(H, OH(d − 1)⊕n) α−−−→∼= H0(OH(d − 1)⊕n|R′)⏐⏐⏐⏐H0(Pn, OPn(d − 1)⊕n) β−−−→ H0(OPn(d − 1)⊕n|R ⊕ H0(OPn−1(d)|S) ⊕ Γ|T⏐⏐⏐⏐H0(Pn, OPn(d − 2)⊕n) γ−−−→ H0(OPn(d − 2)⊕n|R\R′ ⊕ H0(OPn−1(d − 1)|S ⊕ Γ|T⏐⏐ ⏐⏐0 0Since α is bijective then γ bijective implies β is also bijective and this gives the HypothesisH ,n(d − 2; ρ′, σ, τ) implies H ,n(d − 1; ρ, σ, τ).3.2. Hypercritical mèthode d’Horace.Lemma 3.6. Consider H ′ ,n(d−1; s1, s2, 0) with d ≥ 1, s1, and s2 being non-negative in-tegers that satisfy: ns1 +s2 = h0(OPn(d−1)⊕n) and s2 ≤ h0(OPn−1(d)). Now suppose thatthe H0(ΩPn(d)) −→ H0(ΩPn(d)|S1) is injective and H0(OPn(d − 1)⊕n) −→ H0(OPn(d −1)⊕n|S1) is surjective with a general S1 ⊆ Pn then the Hypothesis H ′ ,n(d − 1; s1, s2, 0) istrue.This Lemma is for when we have no quotient.Proof. See [6] Lemma 1.11.Lemma 3.7. Consider H ′ ,n(d − 1; s1, s2, 1) where d ≥ 1, s1, s2 and 2 ≤ θ ≤ n − 1 arenon-negative integers such that, ns1+s2+θ = h0(OPn(d−1)⊕n) and s2+θ ≤ h0(OPn−1(d)).Under the same Hypotheses as Lemma 2.1 i.e. H0(ΩPn(d)) −→ H0(ΩPn(d)|S1) is injectiveand H0(OPn(d−1)⊕n) −→ H0(OPn(d−1)⊕n|S1) is surjective then the Hypothesis H ′′ ,n(d−1; s1, s2, 1) is true.Proof. See [6] Lemma 1.12.3.3. The Main Theorem.Theorem 3.8. Suppose HΩ,n(d+1) is true. Then for any non-negative integer m, thereexists a set, M = {P1, P2, . . . , Pm} of m points in Pn such that the evaluation map, μ, isof maximal rank.μ : H0(Pn, ΩPn(d + 1)) −→m⊕i=1ΩPn(d + 1)|Pi398 D. M. MaingiProof. (a) If h0(ΩPn(d + 1)) ≡ 0 (mod n) then r is the critical number of points neededfor bijectivity i.e. the map H0(Pn, ΩPn(d + 1)) −→ ⊕ri=1 ΩPn|Pi is bijective. Set π = 1nh0(ΩPn(d + 1))we now have the following cases:(i) if m = r then our map is bijective since we have the same number of points as thecritical number i.e. the map α is bijective and γ an identity map and so μ is bijective seebelow:H0(Pn, ΩPn(d + 1))μα⊕mi=1 ΩPn|Pi⊕ni=1 ΩPn|Pi ⊕⊕ri=n+1 ΩPn|Piγ~~(ii) if m > r i.e. we have more points than the critical number and our map is injectivei.e. since α is bijective and γ surjective then our map μ has to inject see below:H0(Pn, ΩPn)∼=α μ⊕ri=1 ΩPn|Pi⊕ri=1 ΩP4|Pi ⊕⊕mi=r+1 ΩP4|Piγ~~(iii) if m < r then we have the less points than the critical number thus our map surjectsi.e. since α is bijective and γ surjective then our map μ is surjective.References[1] A Hirschowitz and C Simpson: La résolution minimale de l’idéal d’un arrangement général d’ungrand nombre de points dans Pn. Invent. Math. 126, 467–503 (1996).[2] F Lauze: Rang maximal pour TPn . Manuscripta Math. 92, 525–543 (1997).[3] F Lauze: Sur la résolution des arrangements génèriques de points dans les espaces projectifsThèse, Université de Nice-Sophia Antipolis, Septembre 1994.[4] A Lorenzini: The Minimal Resolution Conjecture. J Algebra 156, 5–35 (1993).[5] A Lorenzini: On the Betti numbers of points in projective space. Ph.D. thesis. Queen’s University,Kingston, Ontario, 1987. J Algebra 156, 5–35 (1993).[6] D M Maingi: On the Application of the Method of Horace to get the number of generators for anideal of s general points in P4 International Journal of Algebra, Vol. 4, 10, 477 – 500 (2010), toappear.Received: September, 2010

Maximal rank for ΩPn

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Maximal rank for ΩPn

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