The role of the cotangent bundle in resolving ideals of fat points in the plane
aa r X i v : . [ m a t h . AG ] J un The role of the cotangent bundle in resolving ideals of fat points inthe plane.
Alessandro Gimigliano Brian Harbourne Monica Id`aNovember 1, 2018
Acknowledgments: We thank GNSAGA, MUR and the University of Bologna, which supportedvisits to Bologna by the second author, who also thanks the NSA and NSF for supporting hisresearch.
Abstract
We study the connection between the generation of a fat point scheme supported at generalpoints in P and the behaviour of the cotangent bundle with respect to some rational curvesparticularly relevant for the scheme. We put forward two conjectures, giving examples andpartial results in support of them. In this paper we are concerned with minimal free graded resolutions of fat point ideals in P .Given general points P , . . . , P n ∈ P (which, unless we say something explicit to the contrary, willalways be assumed to be general), and nonnegative integers m , . . . , m n , let I ( Z ) denote the ideal I ( P ) m ∩ · · · I ( P n ) m n of R = K [ P ] = K [ x , x , x ] (where K is any algebraically closed field andwhere I ( P i ) is the ideal generated by all forms that vanish at P i ). We refer to I ( Z ) as a fat point ideal, and if Z is the subscheme defined by I ( Z ), we use m P + · · · + m n P n or Z ( m , . . . , m n ) todenote the scheme Z , and I Z for its sheaf of ideals, so that, in particular, I ( Z ) k = H ( P , I Z ( k )).In order to understand better the geometry of Z as a subscheme of P , the first thing thatcomes to mind is to see how many curves of given degree k contain Z , that is, have singularities ofmultiplicity at least m , . . . , m n at the given points P , . . . , P n ; in other words, we want to determinethe dimension, as a K -vector space, of the homogeneous component I ( Z ) k of I ( Z ).The Hilbert function h Z of I ( Z ), h Z ( k ) := dim K ( I ( Z ) k ), is not known in general, even if ithas been determined for many choices of Z . For example, it is known for all Z with n ≤ n if m = · · · = m n ≤
20 ([Hi1], [CCMO]), and for any n if m i ≤ K is the complex numbers.) It is also knownfor many additional cases. Let us say that the sequence of multiplicities m i (and by extension Z )is uniform if m = · · · = m n ≥ n ≥
9. Then [E] determines h Z for all k , as long as Z isuniform, if n is a square, extending results of [HHF]. The paper [HR] determines h Z in many otheruniform cases.All of these results are consistent with a well known conjecture by means of which one canexplicitly write down the function h Z given the multiplicities m i . Various equivalent versions ofthis conjecture have been given (see [S], [Ha4], [G], [Hi2], [Ha1]). We will refer to them collectivelyas the SHGH Conjecture. 1et us say that a fat point subscheme Z is quasi-uniform if n ≥ m = · · · = m ≥ m ≥· · · ≥ m n ≥
0. Thus uniform implies quasi-uniform. As shown in [HHF], assuming the SHGHConjecture, then h Z ( k ) = max(0 , (cid:0) k +22 (cid:1) − P i (cid:0) m i +12 (cid:1) ) holds for all k for a quasi-uniform Z . Sincethere are (cid:0) k +22 (cid:1) forms of degree k and since the requirement for a form to vanish to order m i ata point P i imposes (cid:0) m i +12 (cid:1) conditions, the SHGH Conjecture in this situation just says that theconditions imposed by the points are independent as long as h Z ( k ) > Z , that is, determining the minimal free graded resolution0 → M → M → I ( Z ) → I ( Z ). Here M and M are free R -modules of the form M = ⊕ k R t k [ − k ] and M = ⊕ k R s k [ − k ]. If h Z is known and if the graded Betti numbers t k are known,then the values of s k are easy to determine from the exact sequence above.We are hence interested in the graded Betti numbers t k . It is not hard to see that t k is thedimension of the cokernel of the map µ k − ( Z ) : I ( Z ) k − ⊗ R → I ( Z ) k , where R denotes the K -vector space spanned in R by linear forms and µ k − is the map induced by multiplication ofelements of I ( Z ) k − by linear forms. This paper is a reflection about the geometric obstacles tothe rank maximality of the maps µ k . Let us denote by Ω the cotangent bundle of P , and by p : X → P the blow up at the points P i . We first translate the problem of determining the rankof the maps µ k ( Z ) into two equivalent postulation problems for Z , one in P and the other in X :determine, for each k , the rank of the restriction map(a) ρ k = ρ k ( Z ) : H (Ω( k + 1)) → H (Ω( k + 1) | Z ); or(b) η k = η k ( Z ) : H ( p ∗ Ω( k + 1)) → H ( p ∗ Ω( k + 1) | p − Z ).We show that the point of view (a) gives some information about the failure of this rank maximalitydue to superfluous conditions imposed by Z to the restriction of Ω to some curves; but in fact thisis not enough, and the right point of view is (b), since it is then possible to take into account thesplitting of p ∗ Ω on the normalization of the appropriate rational curves, and this allows to countproperly the superfluous conditions imposed by Z to the restriction of Ω to each curve.Hence, by studying several examples and proving certain results (e.g. 5.3), we arrive at twoconjectures about the failure of the rank maximality of µ k , one when µ k is expected to be surjectiveand the other when injectivity is expected. The idea is the same in the two cases but the expectedsurjective case is much easier to formulate, and this is why we keep them distinct; in both casesthe obstruction to rank maximality is described by the presence of particular rational curves whoseintersection with Z is “too high”.Notice that a similar line of thought leads to the SHGH Conjecture: in fact, determining h ( P , I Z ( k )) amounts to computing the rank of the restriction map r k : H ( P , O P ( k )) → H ( Z, O Z ). The SHGH Conjecture says that failure of r k to have maximal rank is completelyaccounted for by the occurrence of curves C ⊂ P whose strict transform ˜ C ⊂ X is an exceptionaldivisor (i.e., a smooth rational curve of self-intersection − C ∩ Z is too big with respect to O ( k ) | C (or, expressing things on the blow up, such thatthe inverse image ˜ Z of Z meets ˜ C in too many points with respect to kL | ˜ C , which here just meansthat ˜ C · F < −
1, where F = kL − m E − · · · − m n E n , L is the pullback to X of a general line in P and E i is the exceptional locus obtained by blowing up the point P i ).Unfortunately, things are quite complicated when studying the postulation with respect to arank 2 vector bundle; for example, as said above, we have to take into consideration the splittingof p ∗ Ω on the normalization of a rational plane curve, which is not known in general (see 2.1), andis actually an interesting problem per se . In our examples we have made use, when necessary, of aMacaulay 2 script which allows us to compute splitting types (see Section A2.3 of [GHI]).2he use of the cotangent bundle in problems concerning the generation of homogeneous idealsof subschemes of a projective space was introduced by A.Hirschowitz, and used for the first timefor curves in P (see [I1]).Our conjectures assume that the fat point scheme Z postulates well in the degree k we areconsidering, i.e. that h ( I Z ( k )) = 0; but notice that, assuming the SHGH Conjecture, we canalways reduce ourselves to considering fat points Z with good postulation, and for these we needto study only the map µ α , where α is the initial degree of I ( Z ) (see 2.3).Here is what is currently known about resolution of fat point ideals in P . For uniform ([Ha2])or quasi-uniform ([HHF]) Z , it is conjectured that the maps µ k have maximal rank for all k . Werefer to these as the Uniform Resolution and Quasi-Uniform Resolution Conjectures. The UniformResolution Conjecture has been proved for m = 1 ([GM]), m = 2 ([I2]) and m = 3 ([GI]); moregenerally, if m i ≤ i , and the length of Z is sufficiently high, [BI] determines the graded Bettinumbers in all degrees. Verifications of the Quasi-Uniform Resolution Conjecture in some cases weregiven in [HHF], under the assumption of the SHGH Conjecture. Some outright verifications weregiven by [HR]. By applying the results of [E] to results of [HHF], it also follows that the UniformResolution Conjecture holds for all m not too small, as long as n is an even square. Finally, theBetti numbers are known for all Z with n ≤ n ≤ We now establish some terminology and notations and recall some basic concepts.By curve we will mean a 1-dimensional scheme without embedded components.The surface obtained from P by blowing up general points P i is always denoted by X , p : X → P is the morphism given by blowing up the points, E i is the exceptional curve obtained byblowing up the point P i and L is the divisorial inverse image under p of a line in P . We will alsouse L and E i to denote the linear equivalence class of the given divisor, in which case the divisorclass group Cl( X ) is the free abelian group on the basis L, E , . . . , E n . The intersection form on X is such that the basis elements are orthogonal with − L = E i = − i .Given a divisor F on X , we will use F to denote its divisor class and sometimes even the sheaf O X ( F ), and we will for convenience write H ( F ) for H ( X, O X ( F )). For each F , there is a naturalmultiplication map µ F : H ( F ) ⊗ H ( L ) → H ( F + L ).If Z = m P + · · · + m n P n is a fat point scheme, it is clear that, under the correspondence of H ( P , I Z ( k )) with H ( X, kL − P m i E i ), the map µ k ( Z ) : H ( I Z ( k )) ⊗ H ( O P (1)) → H ( I Z ( k + 1))is just the map µ tL − P m i E i .Given a curve C ⊂ P , we denote the multiplicity of C at P i by m ( C ) P i = r i , and ˜ C = dL − P r i E i will denote its strict transform. Note that d is just the degree of C . If C ⊂ P is anintegral curve such that ˜ C ⊂ X is smooth and rational, we write O ˜ C ( k ) instead of O P ( k ). We recallthat ˜ C is an exceptional divisor (of the first kind) in X if ˜ C = dL − P r i E i is smooth and rationalwith − C = d − P r i , which by the adjunction formula implies − K X · ˜ C = − d + P r i ,since K X = − L + E + · · · + E n . 3et Y be a smooth projective variety, D a divisor and A a subscheme of Y ; the residual scheme A ′ = res D A is the subscheme of Y whose sheaf of ideals I res D A is given by the exact sequence:0 → I res D A ( − D ) → I A → I A ∩ D,D →
0, where I A ∩ D,D is the sheaf of ideals on D defining thescheme-theoretic intersection of A and D as a subscheme of D .If Z = m P + · · · + m n P n in P is a fat point scheme, and C is a plane curve whose propertransform is ˜ C = dL − P r i E i , the residual sequence tensored by O P ( k ) becomes: 0 → I Z ′ ( k − d ) →I Z ( k ) → I Z ∩ C,C ( k ) →
0, where Z ′ = res C Z has homogeneous ideal ( I ( Z ) : I ( C )).Now if we set F k ( Z ) = kL − m E − · · · − m n E n , we have F k ( Z ) − ˜ C = ( k − d ) L − P ( m i − r i ) E i and its cohomology is the cohomology of a fat point scheme provided that m i − r i ≥ i ;more precisely, F k ( Z ) − ˜ C = F k − d ( Z ′ ) if r i ≤ m i . Thus divisors corresponding to residuals are easyto compute.Setting Ω = Ω P , recall the Euler sequence on P :0 → Ω(1) → O P ⊗ H ( O P (1)) → O P (1) → . Now let C ⊂ P be a degree d integral curve and assume ˜ C ⊂ X smooth and rational. Sincethe Euler sequence is a sequence of vector bundles, its pullback to X restricted to ˜ C is still exact,and gives 0 → p ∗ Ω(1) | ˜ C → O ˜ C ⊗ H ( O X ( L )) → O ˜ C ( d ) → . ( ∗ )In the following we set p ∗ Ω(1) | ˜ C ∼ = O ˜ C ( − a C ) ⊕ O ˜ C ( − b C ) , where we always assume a C ≤ b C ; looking at the Chern classes in ( ∗ ) gives a C + b C = d . We willsay that the splitting type of C or ˜ C is ( a C , b C ) and the splitting gap is b C − a C .In some cases we can immediately determine the splitting type. Suppose that m is the maximumvalue of m ( C ) P i . See [As] or [F1], [F2] for the proof of the following lemma: Lemma 2.1
We have min ( m, d − m ) ≤ a C ≤ d − m , and d = a C + b C . Note that the splitting type is completely determined if d − m ≤ m + 1, and it is (min( m, d − m ) , max( m, d − m )). When d − m > m + 1 it is not known in general what the splitting type is,but it can be computed fairly efficiently; see Section A 2.3 in [GHI].If f : A → B is a linear map between vector spaces, we say that f is exp-onto (i.e., expected to beonto), resp. exp-inj (i.e., expected to be injective), if dim A ≥ dim B , resp. dim A ≤ dim B . The expected dimension for the cokernel of f is defined to be exp-dim cok( f ) := max(0 , dim B − dim A ).So, for example, exp-dim cok( µ k ( Z )) = max(0 , h ( I Z ( k + 1)) − h ( I Z ( k ))) . We say that a fat point scheme Z has good postulation in degree k , if the map r k is of maximalrank, i.e. if h ( I Z ( k )) h ( I Z ( k )) = 0. We say that Z has good postulation if the maps r k havemaximal rank for all k , and we say that Z is minimally generated if the maps µ k all have maximalrank (i.e., Z is minimally generated if µ k is onto when it is exp-onto and injective when it is exp-inj).A few additional notions will be useful. Given a 0-dimensional scheme Y , we denote by l ( Y )the length of Y ; hence l ( m P + · · · + m n P n ) = P i (cid:0) m i +12 (cid:1) .We define α = α ( Z ) to be the least k such that h ( I Z ( k )) is positive, and we define τ = τ ( Z )to be the least k such that h ( I Z ( k )) = 0.Recall that if h ( I Z ( k )) = 0 then h ( I Z ( t )) = 0 for t ≥ k , and µ t ( Z ) is surjective for t ≥ k + 1, bythe Castelnuovo-Mumford lemma [Mu2]. 4 emark 2.2 Let Z be a fat points subscheme of P (supported at general points). Then α − ≤ τ .If Z has good postulation, then α − ≤ τ ≤ α .In fact, α − ≤ τ follows by taking cohomology of 0 → I Z ( k ) → O P ( k ) → O Z →
0. Goodpostulation gives h ( I Z ( k )) h ( I Z ( k )) = 0, which implies τ ≤ α . Remark 2.3
Since µ k ( Z ) (being the 0-map) is trivially injective for all k < α and it is surjectivefor k ≥ τ + 1, we need only consider µ k in degrees k (if any) with α ≤ k ≤ τ .If Z has good postulation, then either τ = α −
1, and the Betti numbers for I ( Z ) are completelydetermined, or τ = α , in which case we need only consider µ α ; if µ α is exp-onto, then Z is minimallygenerated if and only if µ α is surjective, while if µ α is exp-inj, Z is minimally generated if and onlyif µ α is injective.Now drop the good postulation assumption, and take any Z ; if k ≥ α , assuming the SHGHConjecture it is always possible (and easy to do explicitly, by factoring out the fixed part of H ( I Z ( k )); see [GH]) to replace k and Z by a k ′ and Z ′ (supported at the same points) such thatthe kernels of µ k ( Z ) and µ k ′ ( Z ′ ) have the same dimension, but such that Z ′ has good postulationin degree k ′ . Thus (assuming the SHGH Conjecture) we can reduce to considering only fat points Z with good postulation and with α = τ , and for these we need to study only the map µ α .The forthcoming Remark 2.4 and Lemma 2.5 will be useful in the next section: Remark 2.4
Let C be a curve of degree d in P ; then the exact sequence 0 → O P ( t − d ) →O P ( t ) → O P ( t ) | C → h ( O P ( t ) | C ) = (cid:0) t +22 (cid:1) for 0 ≤ t ≤ d − h ( O P ( t ) | C ) = (2 td + 3 d − d ) for t ≥ d .The same exact sequence twisted by Ω and the cohomology of the cotangent bundle (see for example[OSS]): h ( P , Ω( k )) = h ( P , Ω( − k )) = ( k − k ≥
10 if k ≤ , h ( P , Ω( k )) = ( k = 01 if k = 0together give: h (Ω( t ) | C ) = ( ( t − t + 1) if 1 ≤ t ≤ d − d (2 t − d ) if t ≥ d − , h (Ω( t ) | C ) = 0 for t ≥ max(1 , d − . Lemma 2.5
Let C be a plane curve having a singularity of multiplicity r at a point P , and let Z be the m -fat point supported at P ; then l ( Z ∩ C ) = (cid:0) m +12 (cid:1) − (cid:0) m − r +12 (cid:1) . Proof.
Let x, y be local coordinates at P , A m := K [ x, y ] / ( x, y ) m the coordinate ring of Z , f = 0a local equation for C , where f has initial degree r , and ( ¯ f ) := ( f ) A m ; then, l ( Z ∩ C ) is thedimension of the K -vector space A m /( ¯ f ). If r ≥ m , ¯ f = 0, so dim K A m /( ¯ f ) = (cid:0) m +12 (cid:1) (which wasalready obvious since Z ⊂ C ). If r < m , it is easy to prove that the vector space ( ¯ f ) has dimension (cid:0) m +1 − r (cid:1) using an appropriate induction. 5 Various equivalent postulation problems
In this and in the following sections k will always denote a positive integer, and Z , as usual, a fatpoint subscheme supported at general points of P .In this section we are going to translate the problem of determining the rank of the maps µ k ( Z ) : I ( Z ) k ⊗ R → I ( Z ) k +1 into three different, but closely related, postulation problems. By postulation problem we mean the computation of the rank of a restriction map H ( F ) → H ( F | Y )with F a vector bundle and Y a subscheme of a given scheme. One of these approaches, i.e. thetranslation into a postulation problem in the 3-fold P (Ω) with respect to a rank 1 bundle, is herebecause we find it intrinsically interesting, altough we’ll use it only to understand the geometry ofcertain examples. The other two approaches will lead to conjectures 6.1 and 6.7.We now define the three restriction maps in which we are interested: ρ k = ρ k ( Z ), ψ k = ψ k ( Z ), η k = η k ( Z ).The multiplication map µ k = µ k ( Z ) comes from considering the Euler sequence twisted by I Z ( k ) and taking cohomology:(1 ∗ ) 0 → H (Ω( k +1) ⊗I Z ) → H ( I Z ( k )) ⊗ H ( O P (1)) µ k → H ( I Z ( k +1)) → H (Ω( k +1) ⊗I Z ) → H ( I Z ( k )) ⊗ H ( O P (1)) → . . . In the forthcoming Lemma 3.1 we compare this to the cohomology sequence obtained by re-stricting Ω to Z :(2 ∗ ) 0 → H (Ω( k +1) ⊗I Z ) → H (Ω( k +1)) ρ k → H (Ω( k +1) | Z ) → H (Ω( k +1) ⊗I Z ) → H (Ω( k +1)) = 0Now consider the projective bundle π : P (Ω) → P with the invertible sheaf E t = O P (Ω) (1) ⊗ π ∗ O P ( t ) . We set T = π − ( Z ) ⊂ P (Ω) . By [Ht] Ex. III.8.1, III.8.3 and III.8.4, R i π ∗ O P (Ω) (1) = 0 for i >
0, hence R i π ∗ E t ∼ = R i π ∗ O P (Ω) (1) ⊗O P ( t ) = 0 for i >
0, so that H i (Ω( t )) ∼ = H i ( E t ) for all i ≥
0; in particular, H ( E k +1 ) = 0 for any k ≥
0. Taking ψ k to be the canonical restriction map, we have the exact sequence:(3 ∗ ) 0 → H ( E k +1 ⊗ I T ) → H ( E k +1 ) ψ k → H ( E k +1 | T ) → H ( E k +1 ⊗ I T ) → . We will also work in the blow up p : X → P . Set˜ Z = X i ≥ m i E i ⊂ X and consider the exact sequence:(4 ∗ ) 0 → H ( p ∗ Ω( k +1) ⊗I ˜ Z ) → H ( p ∗ Ω( k +1)) η k → H ( p ∗ Ω( k +1) | ˜ Z ) → H ( p ∗ Ω( k +1) ⊗I ˜ Z ) → H ( p ∗ Ω( k + 1)) = 0 for the following reason: R i p ∗ O X = 0 for i > p ∗ O X ∼ = O P hence,by [Ht] III.8.3, R i p ∗ p ∗ Ω( k + 1) ∼ = R i p ∗ ( O X ⊗ p ∗ Ω( k + 1)) ∼ = R i p ∗ O X ⊗ p ∗ Ω( k + 1) = 0 for i > p ∗ p ∗ Ω( k +1) ∼ = Ω( k +1), and so by [Ht] ex.III.8.1 H i ( p ∗ Ω( k +1)) ∼ = H i ( p ∗ p ∗ Ω( k +1)) = H i (Ω( k +1))for all i ≥
0. 6 emma 3.1 If Z has good postulation in degrees k and k + 1 , then µ k is injective, resp. surjective,if and only if ρ k is injective, resp. surjective. Moreover, if h ( I Z ( k )) = 0 , thenexp-dim cok µ k = exp-dim cok ρ k = max (0 , l ( Z ) − k ( k + 2)) , andcok µ k = cok ρ k = H (Ω( k + 1) ⊗ I Z ) . Proof. If h ( I Z ( k )) = 0, µ k is injective, that is, H (Ω( k + 1) ⊗ I Z ) = 0, so also ρ k is injective. If h ( I Z ( k )) = 0, we have that cok µ k = H (Ω( k + 1) ⊗ I Z ) = cok ρ k , and ker µ k = H (Ω( k + 1) ⊗I Z ) = ker ρ k , so that the difference between the dimension of the domain and the dimension of thecodomain is the same: h ( I Z ( k )) h ( O P (1)) − h ( I Z ( k + 1)) = 3( (cid:0) k +22 (cid:1) − l ( Z )) − ( (cid:0) k +32 (cid:1) − l ( Z )) = k ( k + 2) − l ( Z ) = h (Ω( k + 1)) − h (Ω( k + 1) | Z ). Lemma 3.2
The following conditions are equivalent: (i) ρ k is injective, resp. surjective; (ii) ψ k is injective, resp. surjective; (iii) η k is injective, resp. surjective. Proof. i ) ⇔ ii ): one has π ∗ ( E t ) ∼ = Ω( t ), and (see for example [I3], 2.1) π ∗ ( E t | T ) ∼ = Ω( t ) | Z , π ∗ ( E t ⊗ I T ) ∼ = Ω( t ) ⊗ I Z ; hence H ( E t ) ∼ = H (Ω( t )), H ( E t ⊗ I T ) ∼ = H (Ω( t ) ⊗ I Z ), H ( E t | T ) ∼ = H (Ω( t ) | Z ) .i ) ⇔ iii ): One has p ∗ O X ∼ = O P ; by Prop. 2.3 of [AH], one has also: p ∗ O ˜ Z ∼ = O Z ; hence itfollows (see for example the proof of Lemma 2.3 in [I3], taking into account that p − ( Z ) = ˜ Z ) that p ∗ I ˜ Z ∼ = I Z . By the projection formula we get p ∗ ( p ∗ Ω( k + 1)) ∼ = Ω( k + 1), so p ∗ ( p ∗ Ω( k + 1) | ˜ Z ) ∼ =Ω( k + 1) | Z and p ∗ ( p ∗ Ω( k + 1) ⊗ I ˜ Z ) ∼ = Ω( k + 1) ⊗ I Z . Hence the dimensions of the first three vectorspaces in (4 ∗ ) and in (2 ∗ ) are the same, so we conclude that ρ k is of maximal rank if and only if η k is. Now we are interested in studying the behaviour of the restriction of Ω( k + 1) to a curve in P .This will help us in the study of ρ k and hence (see Section 3) of µ k . In what follows C will be acurve of degree d in P . Definition 4.1
We denote by β = β C,Z,k : H (Ω( k + 1) | C ) → H (Ω( k + 1) | C ∩ Z )the restriction map. We also set γ ( C, Z, k ) := exp-dim cok β C,Z,k = max { , l ( Z ∩ C ) − h (Ω( k + 1) | C ) } . If m ( C ) P i = r i ≤ m i + 1, by Lemma 2.5 l ( Z ∩ C ) = P ( r i m i − (cid:0) r i (cid:1) ). So by Remark 2.4, we find for k + 2 ≥ d and r i ≤ m i + 1 γ ( C, Z, k ) = max { , X ( r i m i − r i ! ) − d (2 k + 2 − d ) } . roposition 4.2 Assume h ( I Z ( k )) = 0 . If C ⊂ P is a curve of degree d ≤ k + 2 , thendim cok µ k ≥ dim cok β C,Z,k . In particular, if there exists a (not necessarily integral) curve C of degree d ≤ k + 2 such thatdim cok β C,Z,k > exp-dim cok µ k , then µ k is not of maximal rank.Proof . Since h ( I Z ( k )) = 0, Z has good postulation in degree k and k + 1, so, by Lemma 3.1,dim cok µ k = dim cok ρ k = h ( I Z ⊗ Ω( k + 1)).Now set t := k + 1 and consider the commutative diagram:0 0 0 ↓ ↓ ↓ → I res C Z ⊗ Ω( t − d ) → I Z ⊗ Ω( t ) → I Z ∩ C,C ⊗ Ω( t ) → ↓ ↓ ↓ → Ω( t − d ) → Ω( t ) → O C ⊗ Ω( t ) → ↓ ↓ ↓ → O res C Z ⊗ Ω( t − d ) → O Z ⊗ Ω( t ) → O Z ∩ C ⊗ Ω( t ) → ↓ ↓ ↓ ↓ ↓ ↓ → H ( I res C Z ⊗ Ω( t − d )) → H ( I Z ⊗ Ω( t )) → H ( I Z ∩ C,C ⊗ Ω( t )) ǫ →↓ ↓ ↓ → H (Ω( t − d )) → H (Ω( t )) → H (Ω( t ) | C ) → ↓ ↓ ρ k ↓ β C,Z,k → H ( O ⊕ res C Z ) → H ( O ⊕ Z ) → H ( O ⊕ Z ∩ C ) → ↓ ↓ ↓ ǫ → H ( I res C Z ⊗ Ω( t − d )) → H ( I Z ⊗ Ω( t )) → H ( I Z ∩ C,C ⊗ Ω( t )) → ↓ ↓ ↓ where h (Ω( t ) | C ) = 0 and h ( I res C Z ⊗ Ω( t − d )) = 0 since h (Ω( t − d )) = 0 (this because d ≤ t + 1).One has that dim cok µ k = dim cok ρ k = h ( I Z ⊗ Ω( t )) ≥ h ( I Z ∩ C,C ⊗ Ω( t )) = dim cok β C,Z,k .As a consequence we have a first criterion to find schemes Z for which µ k fails to have maximalrank: Corollary 4.3
Assume h ( I Z ( k )) = 0 . If there exists a (not necessarily integral) curve C of degree d ≤ k + 2 such that γ ( C, Z, k ) > if µ k is exp-onto, or γ ( C, Z, k ) > l ( Z ) − k ( k + 2) if µ k is exp-inj,then µ k is not of maximal rank.Proof . This follows directly by Proposition 4.2, since dim cok β C,Z,k ≥ γ ( C, Z, k ).8 xample 4.4
Let Z = Z (3 , , , µ ( Z ) does not have maximal rank. To see this directly,let C be the line through P and P . Then Z has good postulation (see [Ha5] or [Ha3] for calculatingthe Hilbert function), l ( Z ) = 11, h ( I Z (3)) = 0, h ( I Z (4)) = 4, h ( I Z (5)) = 10, and α = τ = 4.Thus I ( Z ) is generated in degrees at most 5, but since C is in the base locus of H ( I Z (4)) but thezero locus of the whole ideal is just Z , there must be a generator of degree 5, so the map µ ( Z ) isnot surjective, and since it is exp-onto it is hence not of maximal rank (see 2.3).Alternatively, note that Corollary 4.3 applies: µ is exp-onto but γ ( C, Z, > | C ∼ = O C (3) ⊕ O C (4), we see h (Ω(5) | C ) = 9 <
10 = 2 l ( Z ∩ C ). In other words, Z ∩ C imposes one superfluous condition to the sections of Ω(5) | C . But a point of P imposes 2condition to a rank 2 bundle; so if we wish to understand what’s going on geometrically we haveto move to P (Ω). Here (cf. Lemma 3.2) we have to check the dimension of the space of globalsections of E = O P (Ω) (1) ⊗ π ∗ O P (5) vanishing on the 1-dimensional scheme T = π − ( Z ), or,equivalently, on the 0-dimensional scheme T ′ := ( T ∩ P ( E )) ∪ ( T ∩ P ( G )), where E ⊕ G is a localtrivialization of Ω. (In fact, since E is O P (1) on the fibers, the inverse image of a point π − ( P )can be replaced by two generic points in the fiber. For the non reduced case, and for furtherdetails, see [I1], [I3], [GI].) Hence we are looking at the postulation with respect to the invertiblesheaf E of the 0-dimensional scheme T ′ in P (Ω); since Ω(5) | C ∼ = O C (3) ⊕ O C (4), we have a curve D := P ( O C (3)) ⊂ P (Ω) | C ⊂ P (Ω) and T ′ ∩ D has length 5, while E | D ∼ = O D (3) (see [Ht] V.2.6).It is now clear that it is possible to find a subscheme of T ′ of length l ( T ′ ) − T ′ to E , that is, T ′ does not postulate well with respect to E . Example 4.5
Another similar example is given by Z = Z (4 , , , , µ ( Z ) is again exp-onto and fails to have maximal rank. To see this, let C be the conic through the 5 points P i . Again Z has good postulation ([Ha5], [Ha3]), and we have l ( Z ) = 31, h ( I Z (6)) = 0, h ( I Z (7)) = 5, h ( I Z (8)) = 14, α = τ = 7. As before, C is in the base locus of I ( Z ) , so while the map µ isexp-onto it is not surjective, hence does not have maximal rank. Alternatively, again Corollary 4.3applies: γ ( C, Z, >
0. In more detail, Z does not postulate well with respect to Ω(8) (i.e., thenumber of sections of Ω(8) vanishing on Z is greater than the length of Z would lead us to expect),since h (Ω(8) | C ) = 28 < ·
15 = 2 l ( Z ∩ C ). In other words, Z ∩ C imposes 2 superfluous conditionson the sections of Ω(8) | C ∼ = O P (13) ⊕ (here we have used the fact that Ω | C ∼ = O P ( − ⊕ ).If we work in P (Ω) (cf. Lemma 3.2), we have to consider the postulation with respect to E of the 1-dimensional scheme T = π − ( Z ), or, as in the previous example, of the 0-dimensionalscheme T ′ := ( T ∩ P ( E )) ∪ ( T ∩ P ( G )), E ⊕ G again being a local trivialization of Ω. SinceΩ(8) | C ∼ = O P (13) ⊕ , we have two curves, D and D , both contained in P (Ω) | C , where T ′ ∩ D i haslength 15, while E | D i ∼ = O P (13). It is hence possible to find a subscheme of T ′ of length l ( T ′ ) − T ′ on E ; i.e., T ′ does not postulate well with respect to E . Thesetwo superfluous conditions give a contribution of 2 to the cokernel. Example 4.6
The map µ for Z = 3 P + 3 P + 3 P fails to have maximal rank; Z postulates well([Hi1]), l ( Z ) = 18, h ( I Z (4)) = 0, h ( I Z (5)) = 3, h ( I Z (6)) = 10, α = τ = 5, and µ is exp-inj.But µ is not injective; actually, if L ij is the line through P i and P j , and C is the union of L , L and L , the cubic C is a fixed component for | I ( Z ) | , hence the three generators of I ( Z ) are ofthe form CF i , i = 1 , ,
3, where F , F and F are the three conics which generate I ( P + P + P ).Since h ( I P + P + P (3)) = 7, the dimension of the image of µ is also 7, i.e. I ( Z ) needs 3 generatorsin degree 6, not just one.Alternatively, note that Corollary 4.3 applies: γ ( C, Z,
5) = 3 > l ( Z ) − C is a triangle, hence reducible with aritmethic genus 1. What happens here is that Ω(6) | L ij ∼ = O L ij (4) ⊕ O L ij (5), so that Z ∩ L ij imposes one superfluous condition to the sections of Ω(5) | L ij L ij ; one of these superfluous conditions wouldn’t bother the rankmaximality of µ , which is expected to be injective with a 1-dimensional cokernel; the other twoconditions give a contribution of 2 to the cokernel. If we reinterpret the situation in P (Ω), we haveto consider the postulation with respect to E of a certain 0-dimensional scheme T ′ , analogously towhat happens in the previous examples; here there are three curves D ij := P ( O L ij (4)) such that E | D ij ∼ = O D ij (4), while T ′ ∩ D ij has length 6; T ′ does not postulate well with respect to E . Noticeanyway that the reducible curve D ∪ D ∪ D causing troubles is now the union of three disjointsmooth rational curves, since a point P in D ij is the point in P (Ω | L ij ) representing the tangentdirection of L ij at P .These first three examples are easy to treat by taking into account the occurrence of fixedcomponents. The next example (as well as example 5.4) shows that this is not always the case. Example 4.7 If Z = 9 P + · · · + 9 P , the map µ fails to have maximal rank. Again Z postulateswell ([Ha2], [Ha5], [Ha3]), l ( Z ) = 315, h ( I Z (23)) = 0, h ( I Z (24)) = 10, h ( I Z (25)) = 36, α = τ = 24, and µ is exp-inj. But | I ( Z ) | is fixed component free and µ is not injective; if itwere, dim Im µ would be 30, but in fact it is 29 ([Ha2]). Once more, Corollary 4.3 applies: γ ( C, Z, > l ( Z ) − C := P C i , where C i is a cubic with m ( C i ) P j = 1 for i = j ,2 for i = j . Again, C is not irreducible. What happens here is that the superfluous conditionsimposed by Z ∩ C on Ω(25) | C are more than the expected dimension for the cokernel of ρ , since2 l ( Z ∩ C ) − h (Ω(25) | C ) = 616 −
609 = 7 > l ( Z ) − ρ , and hence µ , arenot injective by Corollary 4.3. (Notice that taking into account just one of the curves C i is notenough: in fact, h (Ω(25) | C i ) = 141 < l ( Z ∩ C i ) = 142; but this only says that dim cok ρ ≥ In examples 4.4, 4.5, 4.6 and 4.7, failure of µ k ( Z ) to have maximal rank was related to Z imposingtoo many conditions on the global sections of Ω( k + 1) | C , and we checked it just by a dimensioncount, i.e. the expected dimension γ ( C, Z, k ) of the cokernel of β C,Z,k was too big. But Ω( k + 1) | C is a rank two vector bundle, so it can happen that the dimension of the cokernel is bigger thanits expected dimension. This of course cannot occur with a rank one bundle on P , since if A isa 0-dimensional scheme on P , the cokernel of the restriction map H ( O P ( t )) → H ( O A ) alwayshas the expected dimension.Instead if we consider for example the restriction map H ( O P ⊕ O P (2)) → H ( O ⊕ A ) where A is the union of two points, then the expected dimension of the cokernel is 0 but the actual dimensionis 1; A imposes 1 condition too many on H ( O P ). This is possible because the splitting gap of O P ⊕ O P (2) is 2. In the previous examples this behaviour did not arise: in examples 4.4 and 4.5, C is a line or a smooth conic with splitting gap 1, respectively 0; in example 4.7, C i is a singularcubic, so we don’t look at Ω( k + 1) | C i , but we can look at the splitting of the pull-back of Ω( k + 1)on ˜ C i , and we find that the splitting gap is 1.The forthcoming example 5.4, instead, illustrates a situation such that µ k ( Z ) is exp-onto, butthere exists a curve C with splitting gap 2, and dim cok β C,Z,k >
0, so µ k ( Z ) is not onto althoughexp-dim cok β C,Z,k = γ ( C, Z, k ) = 0. So it seems evident that, if we want to formulate a conjectureabout the rank maximality of µ k ( Z ), it is necessary to take into consideration the splitting type,and to consider the real cokernel of the maps β C,Z,k ; this is what we are going to do next.
Definition 5.1
Let C be a curve of degree d in P , such that its strict transform ˜ C = dL − P r i E i is smooth and rational in the surface X obtained by blowing up the points P i . Given a positive10nteger k and taking cohomology of the exact sequence 0 → p ∗ Ω( k + 1) | ˜ C ⊗ I ˜ C ∩ ˜ Z → p ∗ Ω( k + 1) | ˜ C → p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z →
0, where ˜ Z = P m i E i , we get the restriction map θ = θ C,Z,k : H ( p ∗ Ω( k + 1) | ˜ C ) → H ( p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z ) . In order to measure the superabundance of conditions imposed by ˜ C ∩ ˜ Z on the sections of p ∗ Ω( k +1) | ˜ C we also set δ ( C, Z, k ) = dim cok θ C,Z,k . Writing a and b for a C and b C , we have p ∗ Ω( k + 1) | ˜ C ∼ = O ˜ C ( − a + dk ) ⊕ O ˜ C ( − b + dk ). Moreover,˜ C · ˜ Z = P r i m i .Since b ≤ d and by assumption k ≥
1, we have dk − b ≥ h ( p ∗ Ω( k + 1) | ˜ C ) = 0. Hence δ ( C, Z, k ) = h ( p ∗ Ω( k + 1) | ˜ C ⊗ I ˜ C ∩ ˜ Z ) = h ( O P ( − a + dk − P r i m i ) ⊕ O P ( − b + dk − P r i m i )) =max(0 , l ( ˜ Z ∩ ˜ C ) − h ( O ˜ C ( − a + dk )) + max(0 , l ( ˜ Z ∩ ˜ C ) − h ( O ˜ C ( − b + dk )), so that finally δ ( C, Z, k ) = max(0 , X r i m i − dk + a −
1) + max(0 , X r i m i − dk + b − . In certain cases, δ is nothing more than γ : Theorem 5.2
Let C ⊂ P be a curve whose strict transform ˜ C = dL − P r i E i is smooth andrational in X , and assume d ≤ k + 2 and r i − ≤ m i for all i . Thencok β C,Z,k ∼ = cok θ C,Z,k hence δ ( C, Z, k ) ≥ γ ( C, Z, k ) , with equality if and only if cok β C,Z,k has its expected dimension (thisoccurs, for example, if P r i m i − dk + a − ≥ ).Proof . The maps θ = θ C,Z,k and ¯ θ = ¯ θ C,Z,k : H ( p ∗ p ∗ Ω( k + 1) | ˜ C ) → H ( p ∗ p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z ) are themaps on cohomology coming from the exact sequences 0 → p ∗ Ω( k + 1) | ˜ C ⊗ I ˜ C ∩ ˜ Z → p ∗ Ω( k + 1) | ˜ C → p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z → p ∗ , and it is clear that cok ¯ θ ∼ = cok θ. We also have an exact sequence 0 → O C → p ∗ O ˜ C → S →
0, where S = ⊕ P ∈ Sing ( C ) ˜ O P / O P and ˜ O P denotes the integral closure of O P . Letting δ P be the length l ( ˜ O P / O P ), one has (by [Ht],Ex. IV.1.8 and Cor. V.3.7) p a ( C ) = p a ( ˜ C ) + P P ∈ Sing ( C ) δ P . But 0 = p a ( ˜ C ) = (cid:0) d − (cid:1) − P (cid:0) r i (cid:1) and p a ( C ) = (cid:0) d − (cid:1) , so P P ∈ Sing ( C ) δ P = P (cid:0) r i (cid:1) , hence l ( S ) = P (cid:0) r i (cid:1) .There is a natural map 0 → O C ∩ Z → p ∗ O ˜ C ∩ ˜ Z ; let us denote the cokernel by S ′ . Since r i ≤ m i + 1for all i , by lemma 2.5 we have l ( S ′ ) = l ( ˜ C ∩ ˜ Z ) − l ( C ∩ Z ) = P r i m i − P ( r i m i − (cid:0) r i (cid:1) ) = P (cid:0) r i (cid:1) .Now consider the diagram0 → O C → p ∗ O ˜ C → S → ↓ ↓ → O C ∩ Z → p ∗ O ˜ C ∩ ˜ Z → S ′ → ↓ ↓ R p ∗ I ˜ C ∩ ˜ Z, ˜ C There is a map
S → S ′ making the diagram commute, and it has to be surjective since R p ∗ I ˜ C ∩ ˜ Z, ˜ C = 0 by [Ht] III.11.2. Hence it is a surjective map between sheaves supported atpoints and of the same lenght, so we conclude S ′ ∼ = S , which gives us the exact sequence 0 →O C ∩ Z → p ∗ O ˜ C ∩ ˜ Z → S ′ →
0. 11ensoring this and the exact sequence at the beginning of the proof by Ω( k + 1), taking intoaccount that p ∗ O ˜ C ⊗ Ω( k + 1) ∼ = p ∗ p ∗ Ω( k + 1) | ˜ C and p ∗ O ˜ C ∩ ˜ Z ⊗ Ω( k + 1) ∼ = p ∗ p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z (cf.the projection formula, [Ht] III.8.3), recalling that H (Ω( k + 1) | C ) = 0 for k + 2 ≥ d (see Remark2.4), and finally writing β = β C,Z,k , we get0 → H (Ω( k + 1) | C ) → H ( p ∗ p ∗ Ω( k + 1) | ˜ C ) → H ( S ⊕ ) → ↓ β ↓ ¯ θ ↓ ∼ = → H (Ω( k + 1) | C ∩ Z ) → H ( p ∗ p ∗ Ω( k + 1) | ˜ C ∩ ˜ Z ) → H ( S ⊕ ) → θ ∼ = cok β .The inequality δ ( C, Z, k ) ≥ γ ( C, Z, k ) is now clear, since δ is the dimension of cok θ C,Z,k , while γ is merely the expected dimension of cok β C,Z,k . For the rest, assuming P r i m i − dk + b − ≥ P r i m i − dk + a − ≥ h ( p ∗ Ω( k + 1) | ˜ C ) = h ( p ∗ p ∗ Ω( k + 1) | ˜ C ) = h (Ω( k + 1) | C + 2 P (cid:0) r i (cid:1) ,we have δ ( C, Z, k ) = 2 l ( ˜ Z ∩ ˜ C ) − h ( p ∗ Ω( k + 1) | ˜ C ) = 2( l ( ˜ Z ∩ ˜ C ) − P (cid:0) r i (cid:1) ) − h (Ω( k + 1) | C =2 l ( Z ∩ C ) − h (Ω( k + 1) | C ) = γ ( C, Z, k ). Corollary 5.3
Assume h ( I Z ( k )) = 0 and moreover that there exists an integral curve C ⊂ P such that ˜ C = dL − P r i E i is smooth and rational in X with d ≤ k + 2 , r i − ≤ m i for all i and δ ( C, Z, k ) > exp-dim cok µ k ( Z ) . Then µ k ( Z ) is not of maximal rank.Proof . We have exp-dim cok µ k < δ ( C, Z, k ) = dim cok θ C,Z,k = dim cok β C,Z,k and we concludeby Proposition 4.2.We now show how to use this last result. A significant difference here with the three previousexamples is that the splitting gap for (any irreducible component of) C was 0 or 1 previously; inExample 5.4 it is 2. Example 5.4
Let Z = 4 P + · · · + 4 P + P ; then µ ( Z ) fails to have maximal rank (see [FHH]).Note that Z has good postulation and I ( Z ) is fixed component free (apply [Ha5] or [Ha3]). Wehave l ( Z ) = 71, h ( I Z (10)) = 0, h ( I Z (11)) = 7, h ( I Z (12)) = 20, α = τ = 11, hence µ α isexp-onto. The map µ ( Z ) is not surjective. This can be attributed to to the existence of a rationalcurve C of degree 8 with r i := m ( C ) P i = 3 for 0 ≤ i ≤
7, and r = 1; ˜ C ⊂ X is a smoothrational curve of self-intersection ˜ C = 0. This time we cannot read failure of maximal rank onthe sections of Ω, since h (Ω(12) | C ) = 128 = 2 l ( Z ∩ C ); i.e., γ ( C, Z,
11) = 0. Instead, the splittinggap for C is 2, since (see the proof of Lemma 12 of [FHH]) p ∗ (Ω(1)) | ˜ C ∼ = O ˜ C ( − ⊕ O ˜ C ( − p ∗ (Ω(12)) | ˜ C ∼ = O ˜ C (85) ⊕ O ˜ C (83). The scheme ˜ Z := P m i E i intersects ˜ C in a 0-dimensional schemeof length P r i m i = 85, so ˜ Z ∩ ˜ C is too much for O P (83) (and not enough for O P (85)); that is,the cohomology of the exact sequence 0 → p ∗ Ω(12) | ˜ C ⊗ I ˜ C ∩ ˜ Z → p ∗ Ω(12) | ˜ C → p ∗ Ω(12) | ˜ C ∩ ˜ Z → → H ( O ˜ C ⊕ O ˜ C ( − → H ( p ∗ Ω(12) | ˜ C ) θ → H ( p ∗ Ω(12) | ˜ C ∩ ˜ Z ) → H ( O ˜ C ⊕ O ˜ C ( − → θ is not of maximal rank. (This cannot happen if the splitting gap is 0 or 1.) Since δ ( C, Z,
11) =1 >
0, we see by Corollary 5.3 that µ ( Z ) fails to have maximal rank.The previous examples might lead one to think that the curves C that need to be taken intoconsideration are the ones with C ≤
0. The following example shows that this is not the case.12 xample 5.5
Let Z = 15( P + . . . + P ) + 13( P + P ) + 9 P + 2( P + . . . + P ); then l ( Z ) = 719, h ( I Z (37)) = 22, h ( I Z (38)) = 61, so that µ ( Z ) is exp-onto but in fact it does not have maximalrank; precisely, dim cok( µ ) = 1; this has been computed with Macaulay 2 ([GS]).Now consider a curve C whose strict transform ˜ C is an irreducible curve in the linear system | L − E + . . . + E ) − E + E ) − E − E + . . . + E ) | (such a C exists, since ˜ C = 2 D with D a Cremona transform of a line, hence the linear system above contains Cremona transformsof conics). One has ˜ C = 4. The splitting type for ˜ C is (14 ,
20) (it is possible to compute itwith the script in [GHI]); then (5.1) δ ( C, Z,
37) = 1. Here too we cannot work in P ; in fact, γ ( C, Z,
37) = 0.
In each of our examples above, failure of µ k ( Z ) to be surjective is accompanied by δ ( C, Z, k ) > Conjecture 6.1
Let Z = P m i P i be a fat point scheme in P (for general points P i ), with h ( I Z ( k )) = 0 . Say µ k ( Z ) is exp-onto. Then µ k ( Z ) fails to be surjective if and only if thereexists an integral curve C ⊂ P whose strict transform ˜ C = dL − P r i E i is smooth and rational in X , with d ≤ k + 2 , r i ≤ m i + 1 and δ ( C, Z, k ) > . Remark 6.2
In fact, in every example we have found for which µ k ( Z ) fails to have maximal rank,we have h ( kL − P m i E i − ˜ C ) >
0, and hence d ≤ k .The “if” part of Conjecture 6.1 is true, and is Corollary 5.3. Here are some counterexamples to the“if” part of conjecture 6.1 with d > k + 2 and r i > m i + 1 for some i : Z = P , k = 1, ˜ C = 4 L − E − E − · · · − E ; Z = P + · · · + P , k = 2, ˜ C = 5 L − E − E + E + E ) − ( E + · · · + E ); Z = P + · · · + P , k = 3, ˜ C = 8 L − E + · · · + E ) − E ; Z = 2 P + 2 P + P + · · · + P , k = 4, ˜ C = 7 L − E − E − E + · · · + E ).The problem in each case is, in some sense, that C is too big.In the case when µ k ( Z ) is exp-inj the situation is more complicated. We have already seen inExamples 4.6, 4.7 that the curve C needs not be irreducible; the following example shows that itcan also be nonreduced. Example 6.3
Let Z = 60( P + . . . + P ); then l ( Z ) = 14640, h ( I Z (169)) = 0, h ( I Z (170)) = 66, h ( I Z (171)) = 238, α = τ = 170, so that µ ( Z ) is exp-inj but in fact it does not have maximalrank (see [Ha2]); precisely, exp-dim cok( µ ) = 40, while the actual dimension is 48.Let C j be a sextic with r j,i = m ( C j ) P i = 2 for i = j and r j,j = m ( C j ) P j = 3, j = 1 , . . . ,
8. Thesplitting type for C i is (3 ,
3) by 2.1 ; then (5.1) δ ( C j , Z, , P i r j,i − · −
1) =4. In order to take into account the contribution of each C j , we do as in Example 4.7 and we consider C = P C j , but this is still not enough since 8 · < exp-dim cok( µ ).So we go on: since res C j Z = 57 P j + 58 P i = j P i , we find δ ( C j , res C j Z, −
6) = 2. If we add upthe contribution not only of Z but also of res C j Z for all the C j ’s, we then find dim cok( µ ) ≥ δ ( C j , res C j ( res C j Z ) , −
12) = 0.Notice that since the splitting type for C i is balanced, we can work directly in P ; it is easy tocheck that γ (2 C, Z, γ , but this is not always thecase for injectivity too. In fact, in the following example bijectivity is expected, and γ = 0, while δ = 1. Example 6.4
Let Z = 11( P + . . . + P ) + 5 P + 2 P ; then l ( Z ) = 480, h ( I Z (30)) = 16, h ( I Z (30)) = 0, h ( I Z (31)) = 48, so that µ ( Z ) is exp-bijective but in fact it does not have maxi-mal rank. To see this, it is enough to apply Corollary 5.3 with ˜ C = 19 L − E + . . . + E ) − E − E .The splitting type for C is (8 ,
11) (to compute it, use [GHI]); then 5.1 gives δ ( C, Z,
30) =max(0 , − −
1) + max(0 , − −
1) = 1.On the other hand, it is easy to check (4.1) that γ ( C, Z,
30) = 0, so Corollary 4.3 is useless here.These examples motivate the following definition:
Definition 6.5
Let C ⊂ P be a degree d curve, with m ( C ) P i = r i . The h -iterated residual schemeof Z = P m i P i with respect to C is defined inductively as follows: res C, Z := Z, res
C,h Z := res C ( res C,h − Z ) . Notice that res
C,h Z = P ( m i − hr i ) P i if m i ≥ hr i .Assume now that the strict transform ˜ C = dL − P r i E i is smooth rational with a := a C , b := b C .Let t − hd ≥
1; we define inductively the h -superabundance of C : δ h ( C, Z, t ) := δ ( C, res
C,h
Z, t − hd ) . We finally set δ ( C, Z, t ) := X h =0 ,..., [ td ] δ h ( C, Z, t ) . Now let F be as usual F = tL − P m i E i , and set A h ( C, Z, t ) = − F · ˜ C + a − h ˜ C , B h ( C, Z, t ) = − F · ˜ C + b − h ˜ C . Then B h ( C, Z, t ) ≥ A h ( C, Z, t ), and if m i ≥ hr i , t − hd ≥ δ h ( C, Z, t ) = max(0 , P r i ( m i − hr i ) − d ( t − hd )+ a − , P r i ( m i − hr i ) − d ( t − hd )+ b −
1) =max(0 , A h ( C, Z, t )) + max(0 , B h ( C, Z, t )).To understand better the connection between δ and γ , the following proposition is helpful: Proposition 6.6
Let C ⊂ P be a curve with ˜ C = dL − P r i E i smooth rational. Assume that ( p + 1) r i − ≤ m i , t + 2 ≥ d ( p + 1) , and assume also that A h ( C, Z, t ) ≥ for h = 0 , . . . , p . Then,denoting by ( p + 1) C the p th infinitesimal neighborhood of C in P , one has: X h =0 ,...,p δ h ( C, Z, t ) = γ ( ( p + 1) C, Z, t ) . Proof . First notice that, if A ( C, Z, k ) ≥
0, then using adjunction formula δ ( C, Z, k ) = A ( C, Z, k )+ B ( C, Z, k ) = 2 P ( r i m i − (cid:0) r i (cid:1) ) − d (2 k + 2 − d ) ≥
0. Hence, if k + 2 ≥ d and r i ≤ m i + 1, then γ ( C, Z, k ) = 2 P ( r i m i − (cid:0) r i (cid:1) ) − d (2 k + 2 − d ) = 2 l ( Z ∩ C ) − h (Ω( k + 1) | C ) = δ ( C, Z, k ) (see 4.1,5.2).We have res
C,h Z = P ( m i − hr i ) P i , since by assumption hr i ≤ m i , for 0 ≤ h ≤ p .So, since by assumption r i − ≤ m i − hr i , t − hd + 2 ≥ d and A h ( C, Z, t ) ≥ h = 0 , . . . , p ,we have δ h ( C, Z, t ) = δ ( C, P ( m i − hr i ) P i , t − hd ) = γ ( C, P ( m i − hr i ) P i , t − hd ) = 2 l (( P ( m i − hr i ) P i ) ∩ C ) − h (Ω( t − hd + 1) | C ) for h = 0 , . . . , p .14t is easy to check (see 2.4, 2.5 and use t − dh + 2 ≥ d and r i − ≤ m i − hr i for 0 ≤ h ≤ p , and m (( p + 1) C ) P i = ( p + 1) r i ) that P h =0 ,...,p h (Ω( t − hd + 1) | C ) = h (Ω( t + 1) | ( p +1) C ) , and P h =0 ,...,p l (( P ( m i − hr i ) P i ) ∩ C ) = l (( P m i P i ) ∩ ( p + 1) C ). So conclusion follows adding up.We are now ready to formulate a conjecture for the case where injectivity is expected. Conjecture 6.7
Let Z = P i m i P i be a fat point scheme in P (for general points P i ), with h ( I Z ( k )) = 0 . Say µ k ( Z ) is exp-inj. Then µ k ( Z ) fails to be injective if and only if there exists acurve C ⊂ P such that: ˜ C = dL − P r i E i has r i ≤ m i +1 and d ≤ k +2 ; C = P n j C j , where each C j is integral with ˜ C j smooth and rational in X and ˜ C j · ˜ C j = 0 ; and P δ ( C j , Z, k ) > l ( Z ) − k ( k + 2) . The “if” part of conjecture 6.7 is true if for example j = 1 and A h ( C, Z, t ) ≥ h = 0 , . . . , n by Proposition 6.6 and Corollary 4.3.Notice that all the results on the generation for fat point schemes (see the introduction for alist of them) are consistent with Conjectures 6.1 and 6.7.We end by proving that the SHGH Conjecture together with Conjectures 6.1 and 6.7 imply theUniform and Quasi-uniform Resolution Conjectures (for the statement of these conjectures see theIntroduction). Proposition 6.8
The SHGH Conjecture together with Conjectures 6.1 and 6.7 imply the Uniformand Quasi-uniform Resolution Conjectures.Proof . Since uniform implies quasi-uniform, let Z be a quasi-uniform point scheme, i.e. Z = m P i =1 ,..., P i − P i =10 ,...,n m i P i , n ≥ m ≥ m ≥ . . . m n ≥
0. We want to prove that, assumingthe SHGH Conjecture, Conjecture 6.1 and Conjecture 6.7, the map µ k ( Z ), or equivalently the map µ F with F = kL − m P i =1 ,..., E i − P i =10 ,...,n m i E i , is of maximal rank.We can write F = ( k − m ) L − mK X + P i ≥ ( m − m i ) E i . We can assume that h Z ( k ) > µ k ( Z ) is the zero map, hence trivially injective; since Z is quasi-uniform, the SHGHconjecture then says (see introduction) that h Z ( k ) = (cid:0) k +22 (cid:1) − (cid:0) m +12 (cid:1) − P i (cid:0) m i +12 (cid:1) . In particular (cid:0) k +22 (cid:1) − (cid:0) m +12 (cid:1) >
0, which gives k ≥ m . If k = 3 m , then n = 9, in which case F = m (3 L − E −· · · − E ), so h ( F ) = 1 and µ F has maximal rank.Now let k > m . In order to prove that µ F has maximal rank, by 6.1 and 6.7 it is enoughto prove that δ ( C, Z, k ) = 0 for each ˜ C = dL − P r i E i smooth rational in X ; since δ ( C, Z, k ) =max(0 , − F · ˜ C + a − , − F · ˜ C + b −
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