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sufficiently large and general set of points in a projective
space
Damian Maingi
To cite this version:
Damian Maingi. On the Minimal Resolution Conjecture for an ideal of sufficiently large and general set of points in a projective space. Algebraic Geometry [math.AG]. Universite Nice Sophia Antipolis, 2010. English. �tel-01295042�
´
Ecole Doctorale Sciences Fondamentales et Appliqu´ees
TH`
ESE
pour obtenir le titre de Docteur en Sciences
de l’Universit´e de Nice-Sophia Antipolis Sp´ecialit´e:
MATH´EMATIQUES
pr´esent´ee et soutenue par
Damian M MAINGI
Sur la conjecture de la r´esolution minimale de l’ideal d’un arrangement g´en´eral d’un grand nombre de points dans un espace projectif
Th`ese dirig´ee par Charles WALTER
Rapporteurs: Alessandra SARTI
James ALEXANDER
soutenue au laboratoire J.A. Dieudonn´e le 21 Septembre 2010, devant le Jury compos´e de:
Alessandra SARTI Professeur Universit´e de Poitiers Rapporteur Andr´e HIRSCHOWITZ Professeur Universit´e de Nice-Sophia Antipolis Examinateur Christian PESKINE Professeur Universit´e Pierre et Marie Curie Pr´esident du Jury Charles WALTER Professeur Universit´e de Nice-Sophia Antipolis Directeur de th´ese Laura COSTA Professeur Universitat de Barcelona Examinateur
points in a Projective space
Damian M MAINGI
Je remercie Dieu de m’avoir donn´e la bonne sant´e et la capacit´e de voyager `a France pour faire cette these.
Je remercie mes parents et toute ma famille specialement Romana et Irene et tous mes amis pour leur support.
J’exprime ma r´econnaissance au Professeur Charles Walter qui a accept´e de diriger cette th`ese dans des conditions difficiles de voyager entre France et Kenya par exemple. Ses conseils et ses encouragements m’ont permis de mener `a bien ce travail. Sa disponibilit´e, sa patience et sa gentillesse m’ont apport´e un soutien indispensable. Je lui en suis tr`es r´econnaissant.
Je remercie le Professeur Michel Walschimdt parce que c’est lui qui m’a recommend´e d’aller faire une these en France et aussi pour avoir obtenu un directeur de these pour moi.
Je remercie le Professeur Andr´e Galligo pour avoir accept´e d’ˆetre mon directeur de th`ese pour quelques semaines et aussi pour son avis pendant mon sejour en France.
Je remercie l’ambassade de la France au Kenya de m’avoir permis de poursuivre ce doctorat en m’accordant une bourse couvrant toutes mes d´epenses en France. Mes remerciements sp´eciaux vont `a son excellence, Mme Elizabeth Barbier pour que continuer soutienne ce programme. Je n’oublierai pas de remercier M. Cyrille Le Deaut de son appui et son remplacement Docteur Severine Fogel et plus de mercis `a Lorys Gratien, Stephanie et `a Ann B Gallo pour leurs aides incessantes `a Nairobi `a l’ambassade de la France.
Je voudrais exprimer ma gratitude sinc`ere `a Professeur Andr´e Hirschowitz pour sa patience, sa gentillesse et pour son souci me con¸cernant pendant les nombreuses rencontres nous avons eu, pour avoir arregl´e les choses que cette th`ese ˆetre realiser et pour avoir accept´e d’ˆetre un examinateur.
Je remercie vivement Professeur Alessandra Sarti pour l’acceptation `a ˆetre rapporteur et d’ˆetre venu dans la soutenance de cette th`ese.
Je remercie Professeur Christian Peskine d’avoir accept´e la tˆache ingrate d’ˆetre le president de ce jury. Je remercie ce dernier d’avoir accept´e d’ˆetre membre du jury avec peu de temps.
Je remercie aussi vivement Professeur Laura Costa pour l’acceptation `a ˆetre examinateur et d’ˆetre venu dans la soutenance de cette th`ese.
Je remercie Professeur Philippe Maisonobe de pr´evoir tous mes besoins pendant mon s´ejour en France.
Je remercie mes colleagues de l’Universit´e de Nairobi, particuli`erement, Dr Jamen Were, le directeur de l’Ecole de Mathematiques de l’Universite de Nairobi, mon grand professeur et ami Claudio Achola, je remercie Anne Wang’ombe, Professeur Ramadhas R pour avoir sugg´ere de faire une these immediatement apres faire ma matrise.
son souci continu pour tout ce que je faisais, particulirement en ce qui concerne apprendre le fran¸cais, l’adaptation ici `a Nice et tout. Je remercie aussi Janine Lachkar, Stephanie Ferero, et Marie-Christine Bermond, toujours l`a, surtout durant les moments difficiles.
Ma grˆace sinc`ere `a Jean-Marc Lacroix, il ´etait toujours l`a quand j’ai eu besoin de son aide, installent un nouveau ordinateur et bureau pour moi chaque nouvelle ann´ee. Je remercie Monsieur Jean-Paul Prad´ere pour le tirage de cette th`ese.
Enfin je remercie mes amis de l’Universit´e de Nairobi, le departement de Chemie: Flora surtout, et les autres, toujours l`a les dernier jours avant la soutenance.
1. INTRODUCTION
1.1. State of the art . . . 1
1.2. The Crux of the thesis . . . 3
2. METHODS 2.1. Preliminaries . . . 6
2.2. Notation . . . 7
2.3. M´ethodes d’Horace 2.3.1. M´ethodes d’Horace simple . . . 9
2.3.2. The Vectorial Methods . . . 9
2.3.3. Plane Divisorial . . . 12
2.3.4. Hypercritical m`ethode d’Horace . . . 13
3. MAXIMAL RANK FOR P3 3.1. Introduction . . . 15
3.2. Statements for the inductive methods . . . 16
3.3. The General Hypotheses . . . 19
3.4. The Main Theorem . . . 19
3.5. The Methods used for P3 3.5.1. The inductive steps . . . 20
3.5.2. The initial cases . . . 25
3.6. Other lemmas and corollaries . . . 28
4. MAXIMAL RANK FOR P4 4.1. Introduction . . . 31
4.2. Statements for the the inductive steps . . . 32
4.3. The General Hypotheses, the Goals and the Main Theorem . . . 34
4.4. The Methods used for P4 4.4.1. The initial cases . . . 36
4.4.2. The Inductive steps . . . 41
4.5. Other Lemmas and Corollaries . . . 46
1. INTRODUCTION
1.1. State of the art. The Minimal Resolution Conjecture (MRC) was first explicitly for-mulated by A Lorenzini in her PhD thesis [18] and later in [17]. It deals with the question of the form of the minimal free resolution for ideals of general points in projective spaces i.e. for a general set of points {P1, . . . , Ps} ∈ Pn, with s ≥ n + 1, then the homogeneous ideal of
the sub-scheme of the union of these points, IS ⊂ R = k[x0, . . . , xn], k an algebraically closed
field and R the homogeneous coordinate ring of Pn, has the following expected form:
0 −−−−→ Fn−1 −−−−→ · · · −−−−→ Fp −−−−→ · · · −−−−→ F0 −−−−→ IS −−−−→ 0,
Fp = R(−d − p)ap−1L R(−d − p − 1)bp,
d being the smallest integer satifying s ≤ h0(Pn,OPn(d)), with
ap = max{0, h0(Pn, Ωp+1Pn (d + p + 1)) − rk(Ωp+1Pn )s}, bp = max{0, rk(Ωp+1Pn )s − h0(Pn, Ω p+1 Pn (d + p + 1))}, and d + n − 1 n < s ≤d + n n .
Specifically, the Minimal Resolution Conjecture says that for all p whereby p is a non-negative integer such that 0 ≤ p ≤ n, we should have apbp= 0 i.e. there are no strange or ghost terms
for the resolution of the Ideal, IS as above.
When p = 0, we would have ap−1 = d+nn − s and when p = n then bp = s − d+n−1n
since d + n − 1 n < s ≤d + n n .
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 evaluation map below is of maximal rank i.e. it is surjective or injective or both; see [13].
H0 Pn, Ωp+1Pn (l) −→
s
M
i=1
Ωp+1Pn (l)|Pi.
The MRC for a general set of s points in Pn over the years has had a lot of attention and still does. It is has been verified and thus is known to be true for Pn where n ≤ 3. Several works have been done in that light see [9] for full references and discussion here we only cite
several that are of interest to us i.e. in line with what we shall do in this work.
The MRC was verified for P2by F Gaeta[11], while E Ballico [2] and together with A Geramita in [4] verified it for P3. C Walter [19] tackled the minimal free resolution for P4 in which his work yields many values but misses out the most difficult values.
Specifically, C Walter gave bounds for the dimension of H0 for which the homogeneous ideal IS of s general points in P4 does not satisfy the MRC (i.e. apbp 6= 0 for some p). He proved
that a set S of s positive integers in P4 whose homogeneous ideal does not satisfy the MRC has asymptotic density 0. In this thesis I prove that ap = 0 or bp = 0 for p = 0 which inturn
implies that apbp = 0. See the sequence below:
· · · −−−−→ R(−d − 2)b1L R(−d − 1)a0 −−−−→ R(−d − 1)b0L R(−d)(d+44 )−s −−−−→ IS −→ 0
We shall make use of a version of the m´ethode d’Horace using elementary transformations of vector bundles which was introduced by A Hirschowitz in letter to R Hartshorne in 1984 (which was never published) where he showed maximal rank for the case of 28 points in P3 i.e. the map below is of maximal rank.
H0 P3, ΩP3(5) −→ 28 M i=1 ΩP3(5)|P i
We use the m´ethode d’Horace to prove MRC for P3 first, see [10], even though it is already been done using other methods and then we proceed to apply them for the case of P4. This method is after the strategy of the title character in Corneille’s Horace based on a story in Livy here applied to “kill” points i.e. the vanishing of H0(TS ⊗ ΩP4(d + 1)) for
certain significant values of S by induction on d i.e. choosing a hyperplane Pn−1⊂ Pn and
running the points to Pn−1 for those significant values of S and by induction on d this stops after a while.
M Ida used the m´ethode d’Horace to get the number of generators for the ideal of a generic union of lines in P3 see [14]. F Lauze used the m´ethode d’Horace to tackle the MRC for P2,
P3 [16] and Pn [15, 16] getting the highest syzygies of IS, an−2 and bn−2 in all Pn, dually
he found the minimal generators of the canonical module ωS = Extn−1R (IS, R(−n − 1)) =
Extn
R(R/IS, R(−n − 1)).
This m´ethode d’Horace has been quite useful and many articles have been done on its use in other areas for example; it was used to study the dimension of the spaces of hypersurfaces in Pn with with singularities imposed using “fat points” (zero-dimensional schemes containing nilpotents) by A Hirschowitz and Alexander in [1]. E Ballico and C Fontanari in [3] used it for error-correcting codes.
The MRC for a general set of points in Pn for n ≥ 5, it has been verified for many cases for instance for n + 1, n + 2, n + 3, n + 4, and n+22 − n points in Pn see [7], [13], [17] just to mention but a few. Most notably, A Hirschowitz and C Simpson [14] where they proved that the MRC holds if the number of points is very large compared to the dimension
of the projective space i.e. s is very large compared to n (remember Pn!) large compared to r.
So far so good but then in 1993, F. Schreyer announced that he had found several cases where computer calculations suggested that MRC predicted the wrong resolution i.e 11 points in P6 and 12 points in P7. M Boij[6] in his thesis provides evidence and also S Beck and M Kreuzer in [5].
D Eisenbud and S Popescu in [8] showed that the MRC fails for a general set of s points in Pn where n ≥ 6, n 6= 9 and s = n + b(3 +√(8n + 1))/2c and later on together with F Schreyer and C Walter in [9] they gave an infinite family of counter examples, one in each Pn i.e. when n ≥ 6, n 6= 9, they showed that the minimal resolution conjecture (MRC) fails for a general set of s points in Pn for almost (1/2)√n values of s using exterior algebra methods.
1.2. The Crux of the thesis. Let k be an algebraically closed field, Pn be a projec-tive space over k and R = k[x0, x1, ..., xn] be the homogeneous coordinate ring of Pn. If
S = {P1, P2, ..., Ps} is a general set of s points in Pn, then the ideal, IS has the expected
minimal resolution of the form below for P4:
R(−d − 3)b2 R(−d − 2)b1 R(−d − 1)b0 L −−−−→ L −−−−→ L −−−−→ IS −−−−→ 0 R(−d − 2)a1 R(−d − 1)a0 R(−d)(d+44 )−s x R(−d − 4)s−(d+34 ) L R(−d − 3)a2 x 0 .
If ai = 0 or bi = 0 for i = 0, 1, 2 then the MRC holds for P4 and thus aibi = 0. F. Lauze in
his PhD thesis [16], proved that a2 = 0 or b2 = 0 and in this thesis we would like to show
that a0= 0 or b0 = 0.
We shall do so by proving that the map below is of maximal rank.
H0 P4, ΩP4(d + 1)
−−−−→ Ls
i=1ΩP4(d + 1)|P i
For this consider the exact sequence
Here, W = H0(OPn(1)), the set of linear forms and k[x0, x1, ..., xn] = Sym(W )
Tensoring the sequence above with TS(d) gives
0 −−−−→ TS⊗ ΩPn(d + 1) −−−−→ W ⊗TS(d) −−−−→ TS(d + 1) −−−−→ 0
Now taking global sections we get;
0 // H0(TS⊗ ΩPn(d + 1)) // W ⊗ Id // Id+1
H1(TS⊗ ΩPn(d + 1))
0
Thus H1(TS ⊗ ΩPn(d + 1)) = Id+1/W · Id, corresponds to the minimal generators of IS
of degree d + 1, and its dimension is b0 i.e. h1(TS⊗ ΩPn(d + 1)) = b0.
Similarly, H0(TS ⊗ ΩPn(d + 1)) is the space of linear relations among the generators of
degree d, whose dimension is a0 i.e. h0(TS⊗ ΩPn(d + 1)) = a0.
Now consider the exact sequence
0 −−−−→ TS −−−−→ OP4 −−−−→ OS −−−−→ 0
Tensoring it by ΩP4(d + 1) gives;
0 −−−−→ TS⊗ ΩP4(d + 1) −−−−→ ΩP4(d + 1) −−−−→ ΩP4(d + 1)|S −−−−→ 0
and now taking global sections yields
0 // H0(TS⊗ ΩP4(d + 1)) // H0(ΩP4(d + 1)) µ // H0 (ΩP4(d + 1)|S) H1(TS⊗ ΩP4(d + 1)) 0 We will prove that µ is of maximal rank for a general set S of s points in P4.
As result, if µ is injective then its kernel is null i.e. a0 = h0(TS⊗ ΩP4(d + 1)) = 0 and
the cokernel is not null that is b0 = h1(TS ⊗ ΩP4(d + 1)) = 4s − 1
6d(d + 2)(d + 3)(d + 4)
as expected. On other hand, if µ is surjective then we have the cokernel of µ being null i.e. 4
b0= h1(TS⊗ΩP4(d+1)) = 0 and the kernel of µ is not null that is, a0= h0(TS⊗ΩP4(d+1)) =
1
6d(d + 2)(d + 3)(d + 4) − 4s. If µ happens to be both injective and surjective i.e. bijective
then the kernel and cokernel are null that is, a0 = b0 = 0.
Since H1(T (m)) = 0 for m ≥ d because the points impose independent conditions on the hypersurfaces of degree d thanks to their being general, thus the sequence above is of finite length.
The consequence is that the homogeneous ideal IS⊂ k[X0, X1, X2, X3, X4] has the expected
number 16d(d + 2)(d + 3)(d + 4) − 4m
+ of minimal generators of degree d + 1 and the
ex-pected number 16d(d + 2)(d + 3)(d + 4) − 4m− of minimal relations of degree d + 1, where (x)+= max(x, 0) and (x)−= max(−x, 0), see Theorem 4.3 the central theorem. We prove
it by first getting bijectivity for a particular number of points and then deduce maximal rank for any general number of points within the sensible limits based on the bounds for the number of points for generality.
2. METHODS
2.1. Preliminaries. Here we start by giving the maximal rank hypotheses or statements (the so called Enonces) as in [13] by Hirschowitz and Simpson.
Let X be a smooth projective variety and X0 non-singular divisor of X. Let F be a lo-cally free sheaf on X and
0 −−−−→ F00 −−−−→ F|X0 −−−−→ F0 −−−−→ 0
be an exact sequence of locally free sheaves on X0. The kernelE of F −→ F0 is a locally free sheaf on X and we have another exact sequence of locally free sheaves on X0
0 −−−−→ F0(−X0) −−−−→ E|X0 −−−−→ F00 −−−−→ 0
and as well exact sequences of coherent sheaves on X
0 −−−−→ E −−−−→ F −−−−→ F0 −−−−→ 0 and
0 −−−−→ F (−X) −−−−→ E −−−−→ F00 −−−−→ 0.
Hypothesis 2.1. R(F , F0, y; a, b, c)
Let y, a, b and c be non-negative integers. The Hypothesis R(F , F0, y; a, b, c) asserts that there exists a points, U1, . . . , Ua, and b points, V1, . . . , Vb ∈ X0 such that for the quotients
F0
Ui −−−−→ Ai −−−−→ 0,
FVi −−−−→ Bi −−−−→ 0
there exists the points W1, . . . , Wc such that for the quotients
FWi −−−−→ Ci −−−−→ 0
with the kernel in ker(FWi −−−−→ F
0
Wi) then for a non-negative integer z, there exists y
points, Y1, . . . , Yy in X and z points Z1, . . . , Zz in X0 such that the map below is bijective.
H0 X,F −−−−→ Lai=1Ai⊕Lbi=1Bi⊕Lci=1Ci⊕Lyi=1F0Yi⊕
Lz
i=1FZi
Hypothesis 2.2. RD(F , F0, y; a, b, c)
Let y, a, b and c be non-negative integers. The Hypothesis R(F , F0, y; a, b, c) asserts that there exists a points, U1, . . . , Ua, and b points, V1, . . . , Vb ∈ X0 such that for the quotients
F0
Ui −−−−→ Ai −−−−→ 0,
FVi −−−−→ Bi −−−−→ 0
there exists the points W1, . . . , Wc such that for the quotients
γ(Y ) :FWi −−−−→ Ci(Y ) −−−−→ 0
with the kernel in ker(FWi −−−−→ F
0
Wi) then for a non-negative integer z, there exists y
points, Y1, . . . , Yy in X and z points Z1, . . . , Zz in X0 such that the map below is bijective.
H0 X,F −−−−→ Lai=1Ai⊕ Lb i=1Bi⊕ Lc i=1Ci(Y1. . . Yy) ⊕ Ly i=1F 0 Yi⊕ Lz i=1FZi Hypothesis 2.3. RD(E , F00, y0; a0, b0, c0)
Let y0, a0, b0 and c0 be non-negative integers. The Hypothesis R(E , F00, y0; a0, b0, c0) asserts that there exists a0 points, U1, . . . , Ua0, and b0 points, V1, . . . , Vb0 ∈ X0 such that for the quotients
F00
Ui −−−−→ Ai −−−−→ 0,
EVi −−−−→ Bi −−−−→ 0
there exists the points W1, . . . , Wc0 such that for the quotients
γ(Y ) :EWi −−−−→ Ci(Y ) −−−−→ 0
with the kernel in ker(EWi −−−−→ F
00
Wi) then for a non-negative integer z
0, there exists y0
points, Y1, . . . , Yy0 in X and z0 points Z1, . . . , Zz0 in X0 such that the map below is bijective.
H0 X,E −−−−→ Lai=10 Ai⊕Lb 0 i=1Bi⊕ Lc0 i=1Ci(Y1. . . Y 0 y) ⊕ Ly0 i=1F 00 Yi⊕ Lz0 i=1EZi
2.2. Notation. Since we are talking about the MRC for projective spaces and the m´ethode d’Horace then we set
X = Pn X0 = Pn−1 F = ΩPn F0 = Ω Pn−1 E = O⊕n Pn(−2) F00 =O Pn−1(−1)
The exact sequences of the elementary transformations after twisting by d + 1 are:
0 0 y y ΩPn(d) ΩPn(d) y y 0 −−−−→ OPn(d − 1)⊕n −−−−→ ΩPn(d + 1) −−−−→ ΩPn−1(d + 1) −−−−→ 0 y y 0 −−−−→ OPn−1(d) −−−−→ ΩPn|Pn−1(d + 1) −−−−→ ΩPn−1(d + 1) −−−−→ 0 y y 0 0
From which we have the Hypotheses:
H0Ω,n(d + 1; α, β, γ) = H(ΩPn(d + 1), ΩPn−1(d + 1), α, β, γ) and
H0O,n(d − 1; ρ, σ, τ ) = H(OPn(d − 1)⊕n,OPn−1(d); ρ, σ, τ ) and
H00O,n(d − 1; ρ, σ, τ ) = H(OPn(d − 1)⊕n,OPn−1(d); ρ, σ, τ )
Our method is to prove inductively certain statements HΩ,n(d + 1) and HO,n(d − 1). The
exact statements for n = 3 and n = 4 are found in Chapters 3 and 4 respectively, but roughly speaking we have:
Hypothesis 2.4. H0Ω,n(d + 1; α, β, γ)
The Hypothesis H0Ω,n(d + 1; α, β, γ) asserts that for non-negative integers α, β γ and ε sat-isfying the conditions:
0 ≤ γ ≤ 1, 1 ≤ ε ≤ n − 2,
nα + (n − 1)β + εγ = h0(Ω
Pn(d + 1)), and
(n − 1)β + εγ ≤ h0(ΩPn−1(d + 1)) having for γ = 1 a quotient Γ0 then the map
η : H0 Pn, ΩPn(d + 1) −→ α M i=1 ΩPn(d + 1)|A i ⊕ β M j=1 ΩPn−1(d + 1)|B j⊕ Γ 0 |C is bijective with h0(ΩPn(d + 1)) = d d+n
d+1 and for α general points A1. . . Aα ∈ P
n, β + 1
general points B1. . . Bβ, C ∈ Pn−1 .
Hypothesis 2.5. HΩ,n(d + 1)
The Hypothesis HΩ,n(d + 1) asserts that H0Ω,n(d + 1; α, β, γ) is true for all α, β and γ
satis-fying the conditions above.
Hypothesis 2.6. H0O,n(d − 1; ρ, σ, τ )
The Hypothesis H0O,n(d − 1; ρ, σ, τ ) asserts that for non-negative integers ρ, σ, τ and θ satis-fying the conditions:
0 ≤ τ ≤ 1, 2 ≤ θ ≤ n − 1,
nρ + σ + θτ = h0(OPn(d − 1)⊕n), and
σ + θτ ≤ h0(OPn−1(d)) having for τ = 1 a quotient Γ then the map
φ : H0 Pn,OPn(d − 1)⊕n −→ ρ M i=1 OPn(d − 1)⊕n |Ri⊕ σ M j=1 OPn−1(d)|S j⊕ Γ(S)|T is bijective with h0(OPn(d − 1)⊕n) = n d+n−1
d−1 and for ρ general points R1. . . Rρ∈ Pn, σ + 1
general points S1. . . Sσ, T ∈ Pn−1 .
Note that the quotient Γ depends on S = {S1, · · · , Sσ} so if we have a quotient at T we
write Γ(S)|T.
Hypothesis 2.7. HO,n(d − 1)
The Hypothesis HO,n(d − 1) asserts that H0O,n(d − 1; ρ, σ, τ ) is true for any ρ, σ, and τ sat-isfying the conditions above.
Hypothesis 2.8. H00O,n(d − 1; ρ, σ, τ )
A variant version of the Hypothesis H0O,n(d − 1; ρ, σ, τ ) with Γ independent of Γ0 takes the form H00O,n(d−1; ρ, σ, τ ) and it makes the same assertion as the Hypothesis H0O,n(d−1; ρ, σ, τ ) the only difference being quotient dependency.
2.3. M´ethodes d’Horace. We explain the m´ethodes d’Horace we use in their basic form i.e. without some extra complications appearing in the actual inductions.
(1) Simple
(2) Vectorial Methods 1 and 2
(3) Plane divisorial and
(4) The Hypercritical m´ethodes d’Horace.
2.3.1. M´ethode d’Horace simple. [15] lemme 1
Lemma 2.9. Suppose we have a bijective morphism of vector spaces γ : H0(X0,F0) −−−−→ L
and that we have H1(X,E ) = 0. Let µ : H0(X,F ) −−−−→ L be a morphism of vector spaces.
Then for H0(X,F ) −→ M ⊕ L to be of maximal rank it suffices that H0(X,E ) −→ M is of maximal rank i.e.
0 // H0(X,E ) // α H0(X,F ) // β H0(X0,F0) // γ ∼ = 0 0 // M // M L L // L // 0
In some of our cases γ will be at times an evaluation map of sections of a line bundle at a general set of points, so bijectivity will not be an issue. In other cases maximal rank property will follow by previous work on Pn−1. Therefore, if α is of maximal rank then β will be of maximal rank by the short 5-lemma.
2.3.2. The Vectorial Methods. At the back of our minds we have a general set S of s points in Pn and would like to prove maximal rank for
H0(ΩPn(d + 1))
µ
−−−−→ H0(Ω
Pn(d + 1)|S)
We have used two vectorial methods which we simply named, Vectorial Methods 1 and 2. Vectorial method 1 includes the differential method in order to handle fractions of points.
(i) Vectorial Method 1
The Hypothesis H0O,n(d − 1; ρ, σ, τ ) implies H0Ω,n(d + 1; α, β, γ) provided that the condi-tions in Hypothesis 2.4 and 2.6 are satisfied.
Consider the exact sequence;
0 // OPn(d − 1)⊕n // ΩPn(d + 1) // ΩPn−1(d + 1) // 0
and let B and C be general subsets of Pn−1. We specialize A to R ∪ S ∪ T with R a general set of ρ points in Pn and S and T sets of σ and τ general points in Pn−1. To run points to
Pn−1, consider the map, γ : H0(ΩPn−1(d + 1)) −→ H0(ΩPn−1(d + 1)| B) ⊕ Γ
0
|C, if the number
of points we have satisfy h0(ΩPn−1(d + 1)) then γ is bijective, if not then we specialize 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 x x H0(Ω Pn−1(d + 1)) γ −−−−→∼ = H 0(Ω Pn−1(d + 1)|{B∪S}) ⊕ Γ0| C⊕ Γ|T x x H0(ΩPn(d + 1)) −−−−→ Hβ 0(ΩPn(d + 1)|R∪S∪T =A) ⊕ H0(ΩPn−1(d + 1)|B) ⊕ Γ0| C x x H0(OPn(d − 1)⊕n) −−−−→α H0(OPn(d − 1)⊕n |R) ⊕ H0(OPn−1(d)|S) ⊕ Γ| T x x 0 0
Thus the map γ is bijective and hence if α is bijective then β is bijective as well and this gives H0O,n(d − 1; ρ, σ, τ ) implies H0Ω,n(d + 1; α, β, γ).
There are other cases where we will need fractions of points that is when h0(X0,F0) = h0(Pn−1, Ωn−1P (d + 1)) is not divisible by rk(F0) = n − 1 for instance when dealing with P3 to P2, dim 1 will be a fraction, P4 to P3 we will have fractions and what can we do like in this case to get dimension 2 or 1 and so on? We use the next method to handle such situations.
Differential m´ethode d’Horace([13] lemme 1)
Lemma 2.10. Suppose we are given a surjective morphism of vector spaces,
λ : H0(Pn−1, Ω
Pn−1(d + 1)) // L
and suppose there exists a point Z0∈ Pn−1 such that
H0(Pn−1, ΩPn−1(d + 1)) // L ⊕ ΩPn−1(d + 1)|Z0
and suppose H1(Pn,OPn(d − 1)⊕n) = 0. Then there exists a quotient OPn(d − 1)⊕n
|Z0 −→ D(λ)
with kernel contained in ΩPn−1(d)|Z0 of dimension dim(D(λ)) = rk(ΩPn(d + 1)) − dim(ker λ)
having the following property. Let µ : H0(Pn, ΩPn(d + 1)) −→ M be a morphism of
vec-tor spaces then there exists Z ∈ Pn−1 such that if H0(Pn,OPn(d − 1)⊕n) −→ ML D(λ) is
of maximal rank then H0(Pn, ΩPn(d + 1)) −→ M ⊕ LL ΩPn(d + 1)|Z is also of maximal rank.
The sequences below aid in explanation, although the actual reasoning of Hirschowitz and Simpson in [13] is somewhat more complicated.
0 // H0(OPn(d − 1)⊕n) // α H0(ΩPn(d + 1)) // β H0(ΩPn−1(d + 1)) // γ 0 0 // M L D(λ) // M L L L ΩPn(d + 1)|Z // L L D0(λ)|Z // 0
The map γ is bijective by inductive hypothesis on the dimension of ΩPn−1(d + 1) and thus
the bijectivity of α implies that of β by the short 5-lemma.
The existence of λ surjective means we have some points in Pn−1 for that but the (whole as opposed to fractions) points are not enough to satify h0(ΩPn−1(d + 1)) i.e. bijectivity
so specialize an extra point, Z, Pn to Pn−1 for which LL D(λ)
|Z is injective, exceeding
the dimension, h0(ΩPn−1(d + 1)) required and thus the quotient D is such that it factors into
D(λ)|Z ∼=OPn−1L D|Z0 and thus D0|Z is the dimension we required to satisfy h0(ΩPn−1(d + 1))
i.e for γ’s bijectivity.
The sequences for the quotient are as follows: 0 0 dim n − 1 ΩPn−1(d)|Z // // _ D0|Z _ dim n − 3 (n − 2) dim n OPn(d − 1)⊕n |Z // // D|Z ∼=OPn−1|Z⊕ D0|Z dim n − 2 (n − 1) dim 1 OPn−1(d)|Z OP3(d)|Z dim 1 0 0
(ii) Vectorial Method 2
Consider the exact sequence;
0 // ΩPn(d) // OPn(d − 1)⊕n // OPn−1(d) // 0
and let S and T general sets of σ and τ points in Pn−1, specialize R to A ∪ B, where A is 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 number of
points we have are enough to satisfy h0(OPn(d)) then γ bijective, if not then we specialize 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 points we construct a diagram of exact sequences as follows;
0 0 H0(OPn−1(d)) OO γ // H 0(O Pn−1(d)|{S∪T ∪B}) OO H0(OPn(d − 1)⊕n) OO // H0(O Pn(d − 1)⊕n |R) ⊕ H0(OPn−1(d)|S) ⊕ ΓT OO H0(ΩPn(d)) OO // H0(Ω Pn(d + 1)|A) ⊕ H0(ΩPn−1(d + 1)|B⊕ ΓC OO 0 OO 0 OO
The map γ is bijective giving the Hypothesis H0Ω,n(d; α, β, γ) implies H0O,n(d − 1; ρ, σ, τ ). When the number of points we have in Pn−1 are few relative to d we use the plane divisorial method in preference to this method.
2.3.3. Plane Divisorial. The Hypothesis HO,n(d − 2; ρ0, σ, τ ) implies HO,n(d − 1; ρ, σ, τ ).
Let R be a general set of ρ points in Pn, S and T be general sets of σ and τ points in Pn−1 such that they are fewer relative to d (i.e. when Vectorial Method 2 fails). We
choose a hyperplane H ⊂ Pn disjoint from S and T with H ∼= Pn−1 and specialize ρ0 points from Pn to H (i.e. R0 is the set we have after specializing from R in Pn) so that H0(H,OH(d − 1)⊕n) −→ H0(OH(d − 1)⊕n|R0) is bijective that is set ρ − ρ0 = h0(OPn−1(d−1)) and
so taking global sections for the sequence
0 −−−−→ OPn(d − 2)⊕n −−−−→ OPn(d − 1)⊕n −−−−→ OH(d − 1)⊕n −−−−→ 0
we construct a diagram of exact sequences: 0 0 x x H0(H,OH(d − 1)⊕n) α −−−−→ ∼ = H 0(O H(d − 1)⊕n|R0) x x H0(Pn,OPn(d − 1)⊕n) −−−−→β H0(OPn(d − 1)⊕n |R ⊕ H0(OPn−1(d)|S) ⊕ Γ|T x x H0(Pn,OPn(d − 2)⊕n) −−−−→ Hγ 0(OPn(d − 2)⊕n |R0 ⊕ H0(OPn−1(d − 1)|S ⊕ Γ|T x x 0 0
Hence if α is bijective then β is bijective and this gives the Hypothesis HO,n(d − 2; ρ0, σ, τ ) implies HO,n(d − 1; ρ, σ, τ ).
2.3.4. Hypercritical m`ethode d’Horace.
Lemma 2.11. Consider H0O,n(d − 1; s1, s2, 0) with d ≥ 1, s1, and s2 being non-negative
integers that satisfy: ns1+ s2 = h0(OPn(d − 1)⊕n) and s2 ≤ h0(OPn−1(d)). Now suppose
that the H0(ΩPn(d)) −→ H0(ΩPn(d)|S
1) is injective and H
0(O
Pn(d − 1)⊕n) −→ H0(OPn(d −
1)⊕n|S1) is surjective with a general S1⊆ Pnthen the Hypothesis H0O,n(d − 1; s1, s2, 0) is true.
This Lemma is for when we have no quotient.
Proof. From the Hypothesis H0O,n(d − 1; s1, s2, 0) we have a set S1 of s1 general points in Pn
and a set S2 of s2 general points in Pn−1.
Consider the exact sequence:
0 −−−−→ ΩPn(d) −−−−→ OPn(d − 1)⊕n −−−−→ OPn−1(d) −−−−→ 0
0 ker φ // H0(O Pn−1(d)) 0 // H0(ΩPn(d)) _ // H0(O Pn(d − 1)⊕n) // φ H0(OPn−1(d)) // 0 0 // H0(ΩPn(d)|S 1) // H 0(O Pn(d − 1)⊕n|S 1) // 0 0
That is ker φ maps injectively on a subspace V ⊆ H0(OPn−1(d)) i.e.
ker φ γ // α )) S S S S S S S S S S S S S S S S S S S H0(OPn−1(d)) V( 55 k k k k k k k k k k k k k k k k β H0(OPn−1(d)|S2) H0(OPn−1(d)|S2)
The Hypothesis H0O,n(d − 1; s1, s2, 0) asserts that s2 = dim V for S2 ⊆ Pn−1 general, then α
is bijective and since β is bijective since OPn−1(d) is a line bundle also, since V depends only
on S1 but not S2 then γ has no choice but to be bijective thus
H0 OPn(d − 1)⊕n −→ s1 M i=1 OPn(d − 1)⊕n |Ri⊕ s2 M j=1 OPn−1(d)|S j
is bijective and the Hypothesis H0O,n(d − 1; s1, s2, 0) is true.
Lemma 2.12. Consider H0O,n(d − 1; s1, s2, 1) where d ≥ 1, s1, s2 and 2 ≤ θ ≤ n − 1 are
non-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 2.11 i.e. H0(ΩPn(d)) −→ H0(ΩPn(d)|S
1) is injective
and H0(OPn(d − 1)⊕n) −→ H0(OPn(d − 1)⊕n|S
1) is surjective then the Hypothesis H
00 O,n(d −
1; s1, s2, 1) is true.
Proof. The proof is identical to the previous Lemma since this was the Hypothesis with a quotient OPn(d − 1)⊕n−→ Γ with Γ not depending on the Sjs.
3. MAXIMAL RANK FOR P
3
Here the MRC for P3 is verified using the methods of Horace that will be used in P4 even though it has been verified before using other methods. The results have been published in [10], while here a complete outline is given.
3.1. Introduction. Let k be an algebraically closed field, P3 be a projective space over k and R = k[X0, X1, X2, X3] be the homogeneous coordinate ring of P3. If S = {P1, P2, ..., Ps}
is a general set of s points in P3, with s ≥ 4, then the ideal, IS has the expected minimal
resolution if the map
µ : H0 P3, ΩP3(d + 1) −→ s M i=1 ΩP3(d + 1)|P i
is of maximal rank. We wish to prove that µ is of maximal rank and as a consequence we have the following theorem.
Theorem 3.1. Suppose we have a general set S, of s points in P3, s ≥ 4 such that the map µ : H0 P3, ΩP3(d + 1) −→ Lsi=1ΩP3(d + 1)|S
i is of maximal rank then the homogeneous
ideal IS ⊂ k[X0, X1, X2, X3] has 12d(d + 2)(d + 3) − 3m
+ number of minimal generators of
degree d + 1 and 12d(d + 2)(d + 3) − 3m− number of minimal relations of degree d + 1, where (x)+= max(x, 0) and (x)−= max(−x, 0).
3.1.1. Elementary transformation and Quotient Sequences. We fix a hyperplane P2⊂ P3. Then there is an elementary transformations of vector bundles on P3 along the divisor P2 consisting of the following the exact sequences after twisting by d + 1:
0 0 y y ΩP3(d) ΩP3(d) y y 0 −−−−→ OP3(d − 1)⊕3 −−−−→ ΩP3(d + 1) −−−−→ ΩP2(d + 1) −−−−→ 0 y y 0 −−−−→ OP2(d) −−−−→ ΩP3|P2(d + 1) −−−−→ ΩP2(d + 1) −−−−→ 0 y y 0 0
Suppose d ≥ 0 then h0(ΩP3(d + 1)) = 12d(d + 2)(d + 3), h0(ΩP2(d + 1)) = d(d + 2) and
h0(OP3(d − 1)⊕3) = 12(d + 2)(d + 1)d.
bar means the starting case, 1 overline bar meant the second case (inductive step) so l, third l if we were to have such and so on.
In the hypotheses below we will be using quotients of fibresOP3(d − 1)⊕3|W of the forms:
0 0 dim 2 ΩP2(d)|T // // _ D|T0 _ dim 1 dim 3 OP3(d − 1)⊕3|W // // D|W ∼=OP2(d)|W ⊕ D0 |T dim 2 dim 1 OP2(d)|W OP2(d)|W dim 1 0 0
Here t = w implying T is the same point as W since in the vectorial method 2, the map H0(OPn−1(d)) −→ H0(OPn−1(d)|{Z∪X} is an evaluation sections of a line bundle at
points. Meaning its bijectivity follows by specializing enough points to satify the dimension of H0(OPn−1(d)) and so our quotient decomposes i.e. D|W ∼= D
0
|W ⊕OP3|T and this occurs
in the case H0Ω,3(d; r, s, t) implies H0O,3(d − 1; u, v, w) below.
The methods used always work for many points in P3 and d but for small values of d it gets tricky to handle the initial cases and so we work them out explicitly.
3.2. Statements for the inductive methods.
Hypothesis H0Ω,3(d + 1; r, s, t). There exists R1, . . . , Rr ∈ P3, S1, . . . , Ss ∈ P2, and when
t = 1, a quotient ΩP2|T D0|T of dimension 1 for T ∈ P2 such that the restriction map (1)
is bijective. H0 P3, ΩP3(d + 1) −→ r M i=1 ΩP3(d + 1)|R i⊕ s M j=1 ΩP2(d + 1)|S j⊕ D 0 |T (1)
Hypothesis H0O,3(d − 1; u, v, w). For D : (P2)v −→ Gr(1, ΩP2|W) ⊆ Gr(2,O⊕3
P3|W), and
g = 1, there exists U1, . . . , Uu ∈ P3, V1, . . . , Vv, W ∈ P2 such that the restriction map (2) is
bijective. H0 P3,OP3(d − 1)⊕3 −→ u M i=1 OP3(d − 1)⊕3|U i⊕ v M j=1 OP2(d)|V j⊕ D(V )|W (2) 16
Hypothesis H00O,3(d − 1; u, v, w). For D : (P2)v −→ Gr(1, ΩP2|W) ⊆ Gr(2,O⊕3
P3|W), and
g = 1, there exists U1, . . . , Uu ∈ P3, V1, . . . , Vv, W ∈ P2 such that the restriction map (3) is
bijective. H0 P3,OP3(d − 1)⊕3 −→ u M i=1 OP3(d − 1)⊕3|U i ⊕ v M j=1 OP2(d)|V j⊕ D|W (3)
Making use of the above hypotheses we define two lemmas and give their proofs. Lemma 3.2. (a) If H0Ω,3(d + 1; r, s, t) is true, then we have
2s + t ≤ h0 ΩP2(d + 1) = d(d + 2), (4a) 2s + t ≡ h0 ΩP3(d + 1) (mod 3), (4b) r = 1 3 h 0 Ω P3(d + 1) − 2s − t (4c)
(b) If d ≥ 0, s ≥ 0, and 0 ≤ t ≤ 1 are non-negative integers verifying (4a) and (4b), then the r defined by (4c) satisfies r ≥ d+23 ≥ 0.
Proof. (a) Suppose H0Ω,3(d + 1; r, s, t) is true then we have the following exact sequences:
0 0 x x H0(P2, ΩP2(d + 1)) β −−−−→ surj Ls i=1ΩP2|S i⊕ D 0 |T γ x θ x H0(P3, ΩP3(d + 1)) α −−−−→∼ = Lr i=1ΩP3|R i⊕ Ls i=1ΩP2|S i⊕ D 0 |T δ x φ x H0(P3,O P3(d − 1)⊕3) λ −−−−→ inj Lr i=1ΩP3|R i x x 0 0
From (1) one can construct the exact sequences above from which γ and θ are surjective (while δ and φ are injective) with α being bijective and in addition the maps β ◦ γ and θ ◦ α commute i.e. β ◦ γ = θ ◦ α and so β has no choice but to be surjective thus we attain 2s + t ≤ h0 ΩP2(d + 1) which is (4a).
From the fact that α is bijective we have that 3r + 2s + t = h0(ΩP3(d + 1)) and making
r the subject gives (4c) as was required above i.e. r = 13 h0 ΩP3(d + 1) − 2s − t.
From (4c) and since r is a non-negative integer then h0 ΩP3(d + 1) − 2s − t ≡ 0 (mod 3) i.e
2s + t ≡ h0 ΩP3(d + 1)) (mod 3) which is (4b).
For (b), we have by (4a) and (4c) r ≥ 13(h0 ΩP3(d + 1) − h0 ΩP2(d + 1)) = 13h0(OP3(d −
Lemma 3.3. (a) If H0O,3(d − 1; u, v, w) is true, then we have
v+w ≤ h0(OP2(d)) and w ≤ h0(ΩP3(d)) (5a)
v + 2w ≡ 0 (mod 3) (5b)
u = 13(h0 OP3(d−1)⊕3)−v−2w) (5c)
(b) If d ≥ 1, v ≥ 0, and 0 ≤ w ≤ 1 are non-negative integers verifying (5a) and (5b), then the u defined by (5c) satisfies u ≥ 0.
Proof. (a) Suppose H0O,3(d − 1; u, v, w) is true then we have the following exact sequences:
0 0 x x H0(P2,OP2(d)) β −−−−→ surj Lv i=1OP2|V i⊕OP2|W γ x θ x H0(P3,OP3(d − 1)⊕3 α −−−−→ ∼ = Lu i=1O ⊕3 P3|U i ⊕ Lv i=1OP2|V i⊕ D|W δ x φ x H0(P3, ΩP3(d)) −−−−→λ inj Lu i=1O ⊕3 P3|U i⊕ D 0 |W x x 0 0
From (2) we can construct the exact sequences above from which γ and θ are surjective (while δ and φ are injective) and α is bijective and in addition the maps β ◦ γ and θ ◦ α commute so β has no choice but to be surjective thus we get v + w ≤ h0 OP2(d). Also β is
an evaluation map of fibres of a line bundle and so D|W of dimension 2 is decomposed into
OP2|W ⊕ D0|W thus w ≤ h0(ΩP3(d)) for all d ≥ 0 thus we attain (5a).
Since α is bijective we have that 3u + v + 2w = h0(O
P3(d − 1)⊕3) thus which on making
u subject gives u = 13 h0 OP3(d − 1)⊕3 − v − 2w which is (5c)
From (5c) we have 3u, h0 OP3(d − 1)⊕3 are ≡ 0 (mod 3) then v + 2w ≡ 0 (mod 3) which
gives us v + 2w ≡ 0 (mod 3) which is (5b).
For (b), when d ≥ 2 by (5a) and (5c) we have u ≥ 13(h0(ΩP3(d)) − w) ≥ 1. When d = 1, we
have 0 ≤ w ≤ h0(ΩP3(1)) = 0 and u ≥ 13(h0(ΩP3(1)) − w) = 0.
Hence u ≥ 0 for d ≥ 1.
18
3.3. The General Hypotheses.
Hypothesis HΩ,3(d + 1). For all s ≥ 0, all 0 ≤ t ≤ 1 and r verifying (4a), (4b), and (4c),
the hypothesis H0Ω,3(d + 1; r, s, t) is true.
Hypothesis HO,3(d − 1). For all v ≥ 0, all 0 ≤ w ≤ 1 and u verifying (5a), (5b), and (5c), the hypothesis H0O,3(d − 1; u, v, w) is true.
Goal. To prove HΩ,3(d + 1) for d ≥ 2 and HO,3(d − 1) for d ≥ 1.
The goal follows from the initial cases and the inductive steps and it suffices because we have the following theorem which deals with maximal rank property for P3
3.4. The Main Theorem.
Theorem 3.4. Suppose HΩ,3(d + 1) is true. Then for any non-negative integer m, there
exists a set S = {R1, R2, . . . , Rm} of m points in P3 such that the evaluation map
H0 ΩP3(d + 1) −→ m M i=1 ΩP3|R i is of maximal rank. Proof. Set r = b13h0(ΩP3(d + 1))c
(a) If h0(ΩP3(d + 1)) ≡ 0 (mod 3) then r is the critical number of fibres of dimension 3
required for our map
H0 ΩP3(d + 1) −→ r M i=1 ΩP3|R i to be bijective.
(i) If m = r then we have bijectivity i.e. we have the same number of points as the critical number, our map will biject, i.e. then since α is bijective and ι the iden-tity then it follows that β is bijective, i.e.
H0 ΩP3(d + 1) β // αUUUUUUUUU** U U U U U U U U Lm i=1ΩP3|P i Ln i=1ΩP3|P i ⊕ Lr i=n+1ΩP3|P i ι OO
(ii) If m < r, i.e. we have less points than the critical number of points, our map will surject, i.e. since α is bijective and γ surjective then it follows that β is surjective, i.e.
H0 ΩP3(d + 1) β // // α ∼ =UUUUUUUU** U U U U U U U U U Lm i=1ΩP3|P i Lm i=1ΩP3|P i ⊕ Lr i=m+1ΩP3|P i γ OOOO
(iii) If m > r, then we have more points than the critical number, our map will inject, i.e. then since α is injective (and surjective) and γ surjective then it follows that β is injective. H0 ΩP3(d + 1) α ∼ = // w βUUUUUUUU** U U U U U U U U U Lr i=1ΩP3|P i Lr i=1ΩP3|P i ⊕ Lm i=r+1ΩP3|P i γ OO
(b) If h0(ΩP3(d + 1)) 6≡ 0 (mod 3) since h0(ΩP3(d + 1)) = 12d(d + 2)(d + 3) then we have
h0(ΩP3(d+1)) ≡ 2 (mod 3) and so the r points in P3and one point in P2 make up the
critical number for bijectivity of the map H0 ΩP3(d + 1) −→ Lri=1ΩP3|R
i ⊕ ΩP2|S
and from which we now have;
(i) if m = r + 1 then we have injectivity since the number of points we have are more than the critical number of points. That is, we have injectivity for the map H0 ΩP3(d + 1) −→ Lri=1ΩP3|R
i⊕ ΩP3|S.
(ii) If m > r + 1 then we have injectivity since the number of points are many more i.e. the map H0 ΩP3(d + 1) −→ Lri=1ΩP3|R
i⊕ ΩP3|S ⊕
Lm
i=r+2ΩP3|R
i is injective.
(iii) if m < r + 1 then we have fewer points than the critical number and so our map H0 Ω
P3(d + 1) −→ Lmi=1ΩP3|R
i will surject.
3.5. The Methods used for P3.
3.5.1. The inductive steps.
Lemma 3.5. First Vectorial Method
Suppose d, r, s, t satisfy (4a), (4b), and (4c) with d ≥ 2, s ≥ 0 and 0 ≤ t ≤ 1. Write h0(ΩP2(d + 1)) − 2s − t = 2v + w with v, 0 ≤ w ≤ 1 non-negative integers. Set u = r − v − w.
If we have u ≥ 0 and v + w ≤ h0(OP2(d)), then H0O,3(d − 1; u, v, w) implies H0Ω,3(d + 1; r, s, t).
Proof. Consider the exact sequence:
0 −−−−→ OP3(d − 1)⊕3 −−−−→ ΩP3(d + 1) −−−−→ ΩP2(d + 1) −−−−→ 0
on taking global sections we have the sequence below with dimensions shown:
dimd(d+1)(d+2)2 dimd(d+2)(d+3)2 dim d(d + 2) 0 −−−−→ H0(O
P3(d − 1)⊕3) −−−−→ H0(ΩP3(d + 1)) −−−−→ H0(ΩP2(d + 1)) −−−−→ 0.
Let S and T be general sets of s and t points in P2. Let R be a specialized set given
by U ∪ V ∪ W where U is a general set of u points in P3 and V, W are general sets of v, w points in P2. Then the map
H0(ΩP2(d + 1)) −−−−→ L = Hλ 0(ΩP2(d + 1)|S) ⊕ D0|T
is surjective by Proposition 3.17 and by Lemma 3.2 (4a)and the map
H0(ΩP2(d + 1)) −−−−→ L ⊕ Hλ 0(ΩP2(d + 1)|V)
is injective and thus by the differential method i.e. Lemma 3.16, there exists a quotient OP3(d − 1)⊕3 −→ D|W(S, V ) of dimension 2 or 0 (depending on whether w = 1 or 0) with
the property that if
H0(OP3(d − 1)⊕3) −−−−→ H0(OP3(d − 1)⊕3|U) ⊕ H0(OP2(d)|V) ⊕ D|W(S, V )
is bijective then
H0(Ω
P3(d + 1)) −−−−→ H0(ΩP3(d + 1)|R) ⊕ H0(ΩP2(d + 1)|S) ⊕ D0|T
is bijective, and this is H0O,3(d − 1; u, v, w) implies H0Ω,3(d + 1; r, s, t).
Lemma 3.6. Given the same circumstances for the above Lemma, then u,v, and w satisfy: (i) v + w ≤ h0(OP2(d)) − 3 and
(ii) u ≥ 0.
Proof. (i) We have by definition that 2v + w = h0(ΩP2(d + 1)) = d(d + 2) − 2s − t.
Then 2v + w + w = d(d + 2) − 2s − t + w after adding w to both sides thus 2(v + w) = d(d + 2) − 2s − t + w thus v + w = 12(d(d + 2) − 2s − t + w) = 12(d2+ 2d + w − 2s − t) ≤ 1 2(d 2+ 2d + d − 1) = h0(O P2(d)) − 3
The “=” turns to “≤” since d + 2 ≥ w − 2s − t for 0 ≤ w ≤ 1, s ≥ 0, 0 ≤ t ≤ 1 and d ≥ 0 and hence we have v + w ≤ h0(OP2(d)) − 3 as required.
(ii) We have u is defined as u = r − v − w.
We also know that r ≥ 13(h0(OP3(d − 1)⊕3)) by Lemma 3.2 and v + w ≤ h0(OP2(d))
Now, u = r − (v + w) ≥ 13(h0(OP3(d − 1)⊕3)) − h0(OP2(d)) + 3
= h0(OP3(d − 1)) − h0(OP2(d)) + 3
= d+2d (d−3)
3 + 3 ≥ 0 for d ≥ 1
and thus u ≥ 0 as required
Lemma 3.7. Second Vectorial Method
Suppose d, u, v, w satisfy (5a), (5b), and (5c). Write h0(OP2(d)) − v − w = s, where s is a
non-negative integer. Set r = u − s and t = w. If we have r ≥ 0 and 2s + t ≤ h0(Ω P2(d))
then H0Ω,3(d; r, s, t) implies H0O,3(d − 1; u, v, w).
Proof. Let V, W be general sets of v, w points in P2 and specialize U to R ∪ S where R and S are general sets of r and s points in P3 and P2 respectively, and W = T . The map H0(OP2(d)) −→ H0(OP2(d)|V ∪W) is surjective by Lemma 3.3 and sinceOP2(d) is a line
bun-dle and so we just specialize as many points, s, as we need to P2 to attain bijectivity for the
map the map H0(OP2(d)) −→ H0(OP2(d)|V ∪W ∪S) thus we construct the diagram of exact
sequences below. For w = t = 1 we have;
0 0 x x H0(P2,OP2(d)) −−−−→α Lvi=1OP2(d)|V i⊕ Ls i=1OP2(d)|S i⊕OP2(d)|W x x H0(P3,OP3(d − 1)⊕3 β −−−−→ Lu i=1OP3(d − 1)⊕3|U i⊕ Lv i=1OP2(d)|V i⊕ D|W x x H0(P3, ΩP3(d)) γ −−−−→ Lr i=1ΩP3(d)|U i⊕ Ls i=1ΩP2(d)|S i⊕ D 0 |T x x 0 0
and for w = t = 0 we have;
0 0 x x H0(P2,OP2(d)) α −−−−→ Lv i=1OP2(d)|V i ⊕ Ls i=1OP2(d)|S i x x H0(P3,OP3(d − 1)⊕3 β −−−−→ Lu i=1OP3(d − 1)⊕3|U i⊕ Lv i=1OP2(d)|V i x x H0(P3, ΩP3(d)) γ −−−−→ Lr i=1ΩP3(d)|U i⊕ Ls i=1ΩP2(d)|S i x x 0 0
The map α is bijective since OP2(d) is a line bundle so if γ is bijective, then β is bijective,
and this gives us H0Ω,3(d; r, s, t) implies H0O,3(d − 1; u, v, w)
22
Remark 1. For H0Ω,3(d; r, s, t) to hold the r, s, t defined should satisfy (i) r ≥ 0 and
(ii) 2s + t ≤ h0(ΩP2(d)) = (d − 1)(d + 1)
(i) We have r = u − s
Recall: u = 13(h0(OP3(d − 1)⊕3) − v − 2w) and s ≤ h0(OP2(d)) with equality only when
v = w = 0. So we also have s + w ≤ h0(OP2(d)). Now for r = 13(h0(OP3(d − 1)⊕3) − v − 2w) − s ≥ 1 3(h 0(O P3(d − 1)⊕3) − v − w) − h0(OP2(d)) = 13(h0(OP3(d − 1)⊕3)) − h0(OP2(d))
= d+22 d−33 ≥ 0 for all d ≥ 3 which gives us r ≥ 0.
(ii) We know s = 12(d + 1)(d + 2) − v − w 2s = (d + 1)(d + 2) − 2v − 2w 2s = (d + 1)(d + 2) − 2v − w − w 2s = (d + 1)(d + 2) − 2v − w − t since t = w 2s + t = (d + 1)(d + 2) − 2v − w. Since 2s + t = (d + 1)(d + 2) − 2v − w, and h0(Ω P2(d)) = (d − 1)(d + 1) then 2s + t ≤ h0(ΩP2(d) when (d + 1)(d + 2) − 2v − w ≤ (d − 1)(d + 1) i.e. when 3d + 3 ≤ 2v + w.
If Lemma 3.7 above fails we use this Lemma, the Plane Divisorial i.e. H0O,3(d − 2; u0, v, w) =⇒ H0O,3(d − 1; u, v, w) should be true if
H0Ω,3(d; r, s, t) =⇒ H0O,3(d − 1; u, v, w) fails i.e. when 3d + 3 ≥ 2v + w.
What happens is that we have U1, . . . , Uu ∈ P3, V1, . . . , Vv, W ∈ P2 and d happens to be
too large for the points V1, . . . , Vv, W ∈ P2 i.e. the condition we have for the Lemma 3.7 not
to fail is that v + w/2 ≥ 12(3d + 2) when it fails we need to do twist or shift to reduce the size of d for the points we have.
Lemma 3.8. Plane Divisorial Method
Suppose d, u, v, w are non-negative integers satisfying the conditions of Lemma 3.3. Set u0 = u − h0(OP2(d − 1)) = u − 1
2d(d + 1). If one has u
0 ≥ 0, and v + w ≤ h0(O
P2(d − 1)) then
H0O,3(d − 2; u0, v, w) implies H0O,3(d − 1; u, v, w).
Proof. We choose a hyperplane, H ∼= P2 disjoint from {V1, . . . , Vv, W }, then send the
re-quired number of points, Y1, . . . , Yy, say from the U1, . . . , Uu ∈ P3 so that the map α :
H0(H,OH(d − 1)⊕3) −−−−→
Ly
i=1OH(d − 1)|Yi is bijective and then twist d − 1 by 1 and we
construct an exact sequence:
1 2(d + 1)d(d − 1) 1 2(d + 2)(d + 1)d 3 2d(d + 1) 0 −−−−→ OP3(d − 2)⊕3 −−−−→ OP3(d − 1)⊕3 −−−−→ OH(d − 1)⊕3 −−−−→ 0 (u0, v, w) (u, v, w) (y, 0, 0)
We have y = 12d(d + 1) and u0 = u − y. Thus taking global sections and evaluating at corre-sponding points we construct a diagram of exact sequences:
0 0 x x H0(H,OH(d − 1)⊕3) α −−−−→ Ly i=1OH(d − 1)⊕3|Yi x x H0(P3,OP3(d − 1)⊕3) β −−−−→ Lu i=1OP3(d − 1)⊕3|U i⊕ Lv i=1OP2(d)|V i⊕ D|W x x H0(P3,OP3(d − 2)⊕3) γ −−−−→ Lu0 i=1OP3(d − 2)⊕3|U i⊕ Lv i=1OP2(d − 1)|V i⊕ D|W x x 0 0
We have the map α bijective and thus if γ is bijective then β is bijective which gives us H0O,3(d − 2; u0, v, w) implies H0O,3(d − 1; u, v, w).
Lemma 3.9. For a non-negative integer d ≥ 3 and the conditions (5a), (5b) and (5c) in Lemma 3.3 hold and we have 3d + 3 ≥ 2v + w then
(i) v + w ≤ h0(OP2(d − 1)) and
(ii) u0= u − 12d(d + 1) ≥ 0.
Proof. (i)We have 3d + 3 ≥ 2v + w and we need to v + w ≤ h0(OP2(d − 1)) = 1
2d(d + 1).
In 2v + w ≤ 3d + 3 if we add w to both sides and divide by 2 we have,
v + w ≤ 12(3d + 3 + w) ≤ 12d(d + 1) = h0(OP2(d − 1)) for all d ≥ 3 since for the hypothesis
H0O,3(d − 2; u0, v, w) to be true, we must have d ≥ 3 and hence v + w ≤ h0(OP2(d − 1)).
(ii) For u0 ≥ 0 by definition u0 = u − y and recall y = 1
2d(d + 1) = h 0(O
H(d − 1)), since
H ∼= P2 and that u = 13(h0(OP3(d − 1)⊕3) − v − 2w) from (4c) in Lemma 3.3.
So that u0 = 13(h0(OP3(d − 1)⊕3) − v − 2w) − h0(OH(d − 1)) = 13(h0(OP3(d − 1)⊕3) − w − (v + w)) − h0(OH(d − 1)) from (i) ≥ (13h0(OP3(d − 1)⊕3) − h0(OP2(d − 1)) − w) − h0(OH(d − 1)) = 13(h0(OP3(d − 1)⊕3) − h0(OP2(d − 1)) − 3h0(OH(d − 1)) − w). = 13(h0(OP3(d − 1)⊕3) − 4h0(OP2(d − 1)) − w) ≥ 0 for d ≥ 3. Theorem 3.10. (a) For d ≥ 2, HO,3(d − 1) implies HΩ,3(d + 1)
(b) For d ≥ 3, HΩ,3(d) and HO,3(d − 2) imply HO,3(d − 1)
Proof. (a) For any r, s, t satisfying the conditions of Lemma 3.2 we define u, v, w as in Lemma 3.5. By Lemma 3.6 they satisfy the conditions of Lemma 3.3 since HO,3(d − 1) holds by the hypothesis H0O,3(d − 1; u, v, w) so by Lemma 3.5 H0Ω,3(d + 1; r, s, t) holds and this gives us the truth of HΩ,3(d + 1).
(b) For any u, v, w satifying the conditions of Lemma 3.3 we define r, s, t as in Lemma 3.7. Then we have two possibilities: The first, that 3d + 3 ≥ 2v + w and H0Ω,3(d; r, s, t) holds so HO,3(d − 1). The second, that 3d + 3 ≥ 2v + w and H0O,3(d − 1; u0, v, w) holds by Lemma 3.8 (divisorial) and so HO,3(d − 1) holds.
3.5.2. The Initial Cases.
Lemma 3.11. The hypothesis HΩ,3(d + 1) is true when d = 2
Proof. The truth of HΩ,3(3) will follow from the truth of H0Ω,3(3; r, s, t) for all possible
non-negative integers r, s, t and when d = 2.
The conditions that r, s and t must satify are: r ≥ h0(O
P3(1)) = 4
3r + 2s + t = h0(ΩP3(3)) = 20
2s + t ≤ h0(ΩP2(3)) = 8
t = 0 or 1 and from these we have the following as all the possibilities for (r, s, t): (i) (6,1,0) i.e. when r = 6, s = 1 and t = 0.
(ii)(5,2,1) i.e. when r = 5, s = 2 and t = 1. (iii) (4,4,0) i.e. when r = 4, s = 4 and t = 0.
The number of fibres of dimensions 3, 2, and 1 are r, s, and t respectively.
(i) For H0Ω,3(3; 6, 1, 0), we invoke Lemma 3.5 with d = 2, (r, s, t) = (6, 1, 0) and (u, v, w) = (3, 3, 0).
(ii) For the hypothesis H0Ω,3(3; 5, 2, 1) we invoke Lemma 3.5 with d = 2, (r, s, t) = (5, 2, 1) and (u, v, w) = (3, 1, 1) because s = 2 ≥ 1 = dim Gr(1, ΩP2|T). The hypothesis H0Ω,3(3; 5, 2, 1)
follows from H00O,3(1; 3, 1, 1) because of lemme 5 in [13] which is the same as H00O,3(1; 3, 1, 1) save that D0 does not depend on V .
(iii) Now for H0Ω,3(3; 4, 4, 0), we invoke Lemma 3.5 with d = 2, (r, s, t) = (4, 4, 0) and (u, v, w) = (4, 0, 0).
Lemma 3.12. The hypothesis HO,3(d − 1) is true when d = 3
Proof. The truth of HO,3(d − 1) will follow from H0O,3(2; u, v, w) for all possible non-negative integers u, v, w and when d = 3.
The conditions that u, v and w must satify for d = 3 are: 3u + v + 2w = h0(OP3(2)⊕3) = 30
v + w ≤ h0(OP2(3)) = 10
w = 0 or 1.
The number of fibres of dimensions 3, 1 and 2 are u, v, and w respectively. From the conditions we have the following as all the possibilities for (u, v, w):
(i) (10,0,0) (ii) (9,1,1) (iii) (9,3,0) (iv) (8,6,0) (v) (8,4,1) (vi) (7,9,0) (vii) (7,7,1)
(i) For H0O,3(2; 10, 0, 0), we invoke Lemma 3.8 with d = 2, (u, v, w) = (10, 0, 0) and (u0, v, w) = (4, 0, 0) so that H0O,3(1; 4, 0, 0) implies H0O,3(2; 10, 0, 0).
For the hypothesis H0O,3(1; 4, 0, 0) we invoke Lemma 3.8 with d = 1, (u, v, w) = (4, 0, 0) and (u0, v, w) = (1, 0, 0) and we have H0O,3(0; 1, 0, 0) implies H0O,3(1; 4, 0, 0) and for the hy-pothesis H0O,3(0; 1, 0, 0) we proceed similarly as follows:
For 4 general points R1, R2, R3, R4 in P3 to show that H0(OP3(1)⊕3) −→L4i=1OP3(1)⊕3|Ri
is bijective we choose a hyperplane say H ⊆ P3 with R
4 outside it, we send R1, R2, R3 to H
and with the exact sequence;
3 12 9
0 −−−−→ O⊕3
P3 −−−−→ OP3(1)⊕3 −−−−→ OH(1)⊕3 −−−−→ 0
(1, 0, 0) (4, 0, 0) (3, 0, 0)
we construct the exact sequence
0 −−−−→ H0(P3,O⊕3 P3) −−−−→ H0(P3,OP3(1)⊕3) −−−−→ H0(H,OH(1)⊕3) −−−−→ 0 α y β y γ y 0 −−−−→ OP2(1)|R 4 −−−−→ L4 i=1OP3(1)⊕3|Ri −−−−→ L3i=1OH(1)⊕3|Ri −−−−→ 0
We have the map γ bijective, and α is an evaluation map of a space of constants at a point and so is bijective and thus β is bijective i.e. the hypothesis H0O,3(0; 1, 0, 0) is true and it implies H0O,3(1; 4, 0, 0).
(ii) For H0O,3(2; 9, 1, 1), we invoke Lemma 3.8 with d = 2, (u, v, w) = (9, 1, 1) and (u0, v, w) = (3, 1, 1) so that the hypothesis H00O,3(1; 3, 1, 1) implies H0O,3(2; 9, 1, 1).
For the hypothesis H00O,3(1; 3, 1, 1) we invoke Lemma 3.8 with d = 1, (u, v, w) = (3, 1, 1) and (u0, v, w) = (0, 1, 1) and we have H00O,3(0; 0, 1, 1) implies H00O,3(1; 3, 1, 1) and for the hy-pothesis H00O,3(0; 0, 1, 1) we proceed as follows:
Suppose we have 2 general points, R4, R5 in P2, D|R5 is of dimension 2 and we want to
prove that H0(O⊕3P3) −→OP3|R4⊕ D|R5 is bijective so we construct the diagram:
0 −−−−→ ker ϕ|R5 −−−−→ H0(P3,O⊕3P3) −−−−→ OP2(1)|R 5⊕ D 0 |R5 −−−−→ 0 yπ yφ 0 −−−−→ OP2(1)|R 4 −−−−→ OP2(1)|R4 ⊕ D|R5 −−−−→ OP2(1)|R5⊕ D 0 |R5 −−−−→ 0 we haveOP2(1)|R5⊕ D|R0 5 ∼ = D|R5 with dim(D 0 |R5) = 1 and dim(D|R5) = 2.
π is bijective and φ follows since the other map is an identity and thus H00O,3(0; 0, 1, 1) is true.
(iii) For H0O,3(2; 9, 3, 0), we invoke Lemma 3.8 with d = 2, (u, v, w) = (9, 3, 0) and (u0, v, w) = (3, 3, 0) and so the hypothesis H0O,3(1; 3, 3, 0) implies H0O,3(2; 9, 3, 0).
For the hypothesis H0O,3(1; 3, 3, 0) we invoke Lemma 3.8 with d = 1, (u, v, w) = (3, 3, 0) and (u0, v, w) = (0, 3, 0) and we have H0O,3(0; 0, 3, 0) implies H0O,3(1; 3, 3, 0) and for the hy-pothesis H0O,3(0; 0, 3, 0) we proceed as follows:
Suppose we have 3 general points, R4, R5, R6 in P2 and with the sequence,
0 −−−−→ ΩP3(1) −−−−→ O⊕3
P3 −−−−→ OP2(1) −−−−→ 0
on taking global sections and we construct the following;
dim 0 dim 3 dim 3
0 −−−−→ H0(P3, Ω P3(1)) −−−−→ H0(P3,O⊕3 P3) π −−−−→ H0(P2,O P2(1)) −−−−→ 0 yφ ρ y L6 i=4OP2(1)|R i L6 i=4OP2(1)|R i
We have an identity map, ι say, ρ is bijetive, since it is an evaluation of sections of a line bundle of dimesion 3 at 3 general points and π is bijective, and φ = ρ ◦ π is thus bijective, and H0O,3(0; 0, 3, 0) holds.
(iv) For H0O,3(2; 8, 6, 0), we invoke Lemma 3.7 with d = 2, (u, v, w) = (8, 6, 0) and (r, s, t) = (4,4,0) and for the hypothesis H0Ω,3(3; 4, 4, 0) we proved above in (a)(iii).
(v) For H0O,3(2; 8, 4, 1) we prove in somewhat complicated method as follows. Take P1, P2, · · · , P7 general in P4 and Q4, R general in P3.
Then H0(ΩP3(3)) −→ L7i=1ΩP3(3)|P
i is injective and so H
0(Ω
P3(3)) −→ L8i=1ΩP3(3)|P i
will also be injective.
H0(T
is a general quadric in P3, and in particular is irreducible. Since the H0(TP1,··· ,P7,Q4,R(2)) is
1 dimensional, vanishing at Q4, R ∈ Z impose independent conditions on the 3 dimensional
H0(TP1,··· ,P7(2)). By semicontinuity there is a non-empty open subset U ⊆ Z × Z of pairs of
points (S1, S2) such that S1, S2 impose independent conditions on H0(TP1,··· ,P7(2)), i.e. such
that F1 is the only quadric form (up to multiples) vanishing at P1, P2, · · · , P7, S1, S2. Since
U is non-empty in the irreducible Z × Z, it is dense.
Now since (Q4, R) ∈ U for a general P8 ∈ Z, (Q4, P8) and (P8, R) will be in U . So P8
imposes a non-trivial condition on H0(TP1,··· ,P7(2)) meaning H
0(T
P1,··· ,P8(2)) =< F1, F2> is
2 dimensional, but by construction, the only quadratic forms vanishing at P1, · · · , P7, P8 and
at either Q4 or R are in < F1> and so vanish at both Q4 and R.
Hence F2 vanishes at P1, · · · , P8 but at neither Q4 nor R and we have:
0 // H0(ΩP3(3)) //
H0(OP3(2)⊕3) // H0(OP2(3)) // 0
0 // L8i=1ΩP3(3)|Pi⊕ D0|P13(P9, · · · , P12) // L8i=1OP3(2)⊕3|Pi
and we have H0(TP1,··· ,P8(2)) =< F1, F2 >.
(vi) For H0O,3(2; 7, 9, 0), we invoke Lemma 3.7 with d = 2, (u, v, w) = (7, 9, 0) and (r, s, t) = (6,1,0) and for the hypothesis H0Ω,3(3; 6, 1, 0) we proved in (a)(i) above.
(vii) For H0O,3(2; 7, 7, 1), we invoke Lemma 3.7 with d = 2, (u, v, w) = (7, 7, 1) and (r, s, t) = (5,2,1) and for the hypothesis H0Ω,3(3; 5, 2, 1) we proved in (a)(ii) above.
These give HΩ,3(3) and HO,3(2) and then all HΩ,3(d + 1) with d + 1 ≥ 3 and all HO,3(d − 1)
with d ≥ 3 by induction.
3.6. Other lemmas and corollaries.
Lemma 3.13. For any integer d ≥ 1, the hypothesis H0O,3(d − 1; h0(O
P3(d − 1)), 0, 0) is true.
Proof. H0O,3(d − 1; h0(OP3(d − 1)), 0, 0) is a special case of HO,3(d − 1) and thus is true as long
as the number of points u in P3that we require for the truth of H0O,3(d−1; h0(OP3(d−1)), 0, 0)
are u = 13h0(OP3(d − 1)⊕3= h0(OP3(d − 1)).
Using Lemma 3.8 we can reduce H0O,3(d − 1; h0(O
P3(d − 1)), 0, 0) to H0O,3(0; 1, 0, 0) which is
true see Lemma 3.12 (i)
Lemma 3.14. The hypotheses H0O,3(0; 0, 3, 0) and H0O,3(0; 0, 1, 1) are true.
Proof. The hypothesis H0O,3(0; 0, 3, 0) follows from Lemma 3.12 (iii) and the hypothesis H0O,3(0; 0, 1, 1) follows from Lemma 3.12 (ii).
28
A consequence of these last two lemmas is the following statement.
Corollary 3.15. Let d ≥ 1 be an integer. Then the hypothesis H0O,3(d − 1; u, v, w) holds in the following cases:
(a) u = h0(OP3(d − 1)), v = 0, w = 0.
(b) u = h0(OP3(d − 1)) − 1, v = 3, w = 0.
(c) u = h0(O
P3(d − 1)) − 1, v = 1, w = 1.
Proof. They are special cases of Lemmas 3.13 and 3.14. (a) Set u = h0(OP3(d − 1)) and we have Lemma 3.13
(b) If d=1 then it is Lemma 3.14, if d ≥ 1 and use Lemma 3.8 to reduce it to Lemma 3.14 (c) If d=1 then it is Lemma 3.14, if d ≥ 1 and use Lemma 3.8 to reduce it to Lemma 3.14
Lemma 3.16. The differential m´ethode d’Horace ([13] lemme 1)
Suppose we are given a surjective morphism of vector spaces,
λ : H0(P2, ΩP2(d + 1)) // // L
and suppose there exists a point Z0 in P2 such that
H0(P2, ΩP2(d + 1)) // L ⊕ ΩP2(d + 1)|Z0
and suppose also that H1(P3,OP3(d − 1)⊕3) = 0.
Then there exists a quotient OP3(d − 1)3|Z0 −−−−→ D with kernel contained in ΩP2(d)|Z0 of
dimension dim(D) = rank(ΩP3(d + 1)) − dim(ker λ) having the following property.
Suppose µ : H0(P3, Ω
P3(d + 1)) −−−−→ M is a morphism of vector spaces then there exists Z
in P3 such that if
H0(P3,OP3(d − 1)⊕3) −−−−→ M ⊕ D is of maximal rank then
H0(P3, Ω
P3(d + 1)) −−−−→ M ⊕ L ⊕ ΩP3(d + 1)|Z is also of maximal rank.
Proposition 3.17. For all d ≥ 1, and for all p there exists P1, . . . , Pp ∈ P2 such that the
map H0(Ω
P2(d + 1)) −→Lpi=1ΩP2|P
i of maximal rank.
Proof. For d even and hence h0(ΩP2(d + 1)) = d(d + 2) even, then we have bijectivity of
H0(ΩP2(d + 1)) −→Lpi=1ΩP2|P
i for p =
1
2d(d + 2) by the MRC property for P
2 and hence
surjective for p < 12d(d + 2) and injective for p > 12d(d + 2).
For d odd and hence h0(ΩP2(d+1)) = d(d+2) odd, we have surjectivity for p = 1
2(d(d+2)−1)
and injectivity for p =12(d(d + 2) + 1) and therefore we have bijectivity of H0(ΩP2(d + 1)) −→Lpi=1ΩP2|P
i⊕ ΩP2|Q for a general Q specialized from P
2 to P1.
Proposition 3.18. For all d ≥ 1 and for all m, there exists M1, . . . , Mm∈ P2 such that the
map H0(OP2(d − 1)⊕2) −→Lmi=1OP2(d − 1)⊕2|M
i is bijective.
Proof. Follows from truth of MRC for P2 m general set of points in P2.
4. MAXIMAL RANK FOR P
4
4.1. Introduction. Let X be a smooth projective variety, X0 be non-singular divisor of X. LetF be a locally free sheaf on X and
0 −−−−→ F00 −−−−→ F
|X0 −−−−→ F0 −−−−→ 0
be a exact sequence of locally free sheaves over X0. The kernelE of F −−−−→ F0 is a locally
free sheaf on X and we have another exact sequence
0 −−−−→ F0(−X0) −−−−→ E
|X0 −−−−→ F00 −−−−→ 0
In [12] the so called ´enonc´es dependant for us from the sequence above it takes the form RD(F , F0; y, a, b, c) we have ours in the form which is general in the sense that it takes care
of dependancy on quotients as well.
We fix a hyperplane P3⊆ P4, and consider the exact sequence
0 −−−−→ OP3(1) −−−−→ ΩP4(2)|P3 −−−−→ ΩP3(2) −−−−→ 0
then there is an elementary transformations of vector bundles on P4 along the divisor P3 consisting of the following diagram of exact sequences after twisting the sheaves in above sequences by d − 1 0 0 y y ΩP4(d) ΩP4(d) y y 0 −−−−→ OP4(d − 1)⊕4 −−−−→ ΩP4(d + 1) −−−−→ ΩP3(d + 1) −−−−→ 0 y y 0 −−−−→ OP3(d) −−−−→ ΩP4|P3(d + 1) −−−−→ ΩP3(d + 1) −−−−→ 0 y y 0 0
We use these sequences for the vectorial methods and for the plane divisorial, with H ⊆ P4 a hyperplane isomorphic to P3 we shall utilize the sequence;
0 −−−−→ OP4(d − 2)⊕4 −−−−→ OP4(d − 1)⊕4 −−−−→ OH(d − 1)⊕4 −−−−→ 0.
