SESHADRI CONSTANTS AT VERY GENERAL POINTS

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1 TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 357, Number 8, Pages S Article electronically published on December 8, 004 SESHADRI CONSTANTS AT VERY GENERAL POINTS MICHAEL NAKAMAYE Abstract We study the local positivity of an ample line bundle at a very general point of a smooth projective variety We obtain a slight improvement of the result of Ein, Küchle, and Lazarsfeld 0 Introduction The goal of this paper is to study the Seshadri constant of an ample line bundle A at a very general point of a smooth projective variety X Definition 01 Suppose X is a smooth projective variety, x X, anda an ample line bundle on X Then we define the Seshadri constant of A at x by c 1 A C ɛx, A = inf C x mult x C ; here the infimum runs over all integral curves C X passing through x Equivalently, if π : Y X denotes the blow up of X at x with execptional divisor E, then ɛx, A =sup{r Q + π A re isample} r The Seshadri constant ɛx, A measureshowmanyjetsna separates at asymptotically as n In the case when X is a surface, it is known [EL] that ɛ, A 1 Meanwhile, Ein, Küchle, and Lazarsfeld [EKL] have established a lower bound in arbitrary dimension ɛ, A 1 dim X The factor of dim X appearing in the general result is a function of the gap argument used in the proof The same gap argument is also responsible for the presumably extra factor of dim X in the known results for global generation of adjoint bundles see [S] for example Our goal in this paper is to make some progress toward obtaining better lower bounds for ɛ, A Ultimately, however, one would like a bound which does not depend on the dimension of X, and so from our point of view the interest of this work lies more in the methods used than in the explicit results The basic idea employed here, namely that a singular Seshadri exceptional subvariety influences the dimension count for sections with specified jets, was already presented in [N] The counting is difficult and we have tried to find a compromise between computational Received by the editors September, 003 and, in revised form, January 8, Mathematics Subject Classification Primary 14C0 385 c 004 American Mathematical Society

2 386 MICHAEL NAKAMAYE complexity versus obtaining the best possible results Thus we have counted very carefully in the three fold case and less so in the higher-dimensional case In order to clarify our basic strategy we will go through the argument completely in the three fold case to obtain: Theorem 0 Suppose X is a smooth three fold and X a very general point Then for any ample line bundle A on X we have ɛ, A 1 The proof of Theorem 0 will use the work of Ein and Lazarsfeld [EL] on surfaces as well as the uniform bounds on symbolic powers obtained in [ELS] We will then establish the following slight quantitative improvement of the main result in [EKL] for arbitrary dimension: Theorem 03 Suppose X is a projective variety of dimension d 4 and A an ample line bundle on X Then for a very general point X we have the lower bound 3d +1 ɛ, A > 3d The proofs of Theorem 0 and Theorem 03 follow [EKL] line for line, our only innovation coming in counting jets The fundamental observation is that if ɛ, A is small, this puts restrictions on h 0 X, na m nα /mnα+1 for various values of α The smaller ɛ, A becomes, the greater these restrictions are In [EKL] the lower bound on the multiplicity which can be imposed at comes from the Riemann Roch theorem We improve upon this by considering the above-mentioned obstructions, which translates into a better lower bound for ɛ, A The main difficulty in establishing the lower bound ɛ, A 1 d 1 in general is that the set of points where the bound ɛ, A 1 d from [EKL] fails may contain divisors once d 3 The reader only interested in the method employed and not the details of counting should skip to after going through Lemma 13 Finally, we would like to point out the similarity between the counting methods used here and those employed by Faltings and Wüstholz [FW] to reprove and extend the Schmidt subspace theorem In particular, the measure theoretic aspect of [FW] is very closely related to our counting of jets The asymptotic dimensions h 0 X, na m nα /mnα+1,as, can be used to define a measure µ on [0,mA], with ma defined in Here µa, b would measure the asymptotic cost of raising the multiplicity at from a to b Notation and conventions All varieties considered will be defined over the complex numbers C Apointx of an irreducible variety X will be called very general if x belongs to the complement of countably many closed, proper subvarieties If x X, thenm x O X is the maximal ideal sheaf of x Suppose V X is an irreducible subvariety and s H 0 X, L Then ord V s is the order of vanishing of s along V If L is a line bundle BSL denotes the stable base locus of L, thatis, BSL ={x X : sx = 0 for all s H 0 X, nl for all n>0} If s H 0 X, L, then Zs X denotes the zero scheme of the section s

3 SESHADRI CONSTANTS AT VERY GENERAL POINTS 387 If α R, then α denotes the round down of α, that is the largest integer less than or equal to α Similarly, α denotes the round up of α, the smallest integer greater than or equal to α 1 The three fold case Before proving Theorem 0 we review the strategy of [EKL] Suppose then that X is a smooth projective variety of dimension d and A an ample line bundle on X Furthermore, let be a very general point of X Ein, Küchle, and Lazarsfeld study the linear series ka m kα for various values of α Roughly speaking, the argument of [EKL] goes as follows Suppose that ɛ, A < 1 d and let C be a curve with 11 A C mult C < 1 d Moreover, we assume that C is chosen from a flat family F X T defined over a smooth affine variety T of dimension d with a quasi finite map φ : T X, where A C t mult φt C t < 1, t T d Consider, then, for k sufficiently divisible, the linear series 1 ka m kα, kα Z If kα > kɛ, A, then by 11 the curve C is in the base locus of this linear series Using the fact that C moves in the family F in order to differentiate in the parameter direction, Ein, Küchle, and Lazarsfeld then show that any divisor D ka m kα vanishes along C to order at least kα kɛ, A In particular, taking α>ɛ, A in 1, we see that if D ka m kα,thend vanishes to order greater than kɛ, A alongc and hence vanishes along all curves in C t F with φt C The next step in the argument is to show that a subfamily of the curves {C t } φt C, defined over a constructible subset W φ 1 C, sweep out an irreducible surface S X The argument is then iterated and the base locus of ka m kα, α > rɛ, A, is shown to contain an irreducible subvariety of dimension r, sweptoutbyanr 1- dimensional subfamily of F After iterating this argument d times, a contradiction is reached because the linear series ka m kdɛ,a+1 is forced to be empty but by hypothesis dɛ, A < 1 and a simple argument using the Riemann Roch theorem yields the desired contradiction Fundamental to the argument of [EKL] is the following differentiation result: Lemma 13 [EKL] Proposition 3 Suppose X is a very general point and let W X be an irreducible subvariety Let π : Y X be the blow up of with exceptional divisor E and let W is the strict transform of W in Y Write αw = inf { W BSπ A βe} β Q

4 388 MICHAEL NAKAMAYE Suppose γ>αw and 0 s H 0 X, na m nγ Then ord W s nγ αw n +1 To obtain this version from [EKL], Proposition 3, let Γ X T be the graph of φ : T X Letp 1 : X T X and p : X T T denote the projections to the first and second factors respectively Consider BS p 1 ka I kαw +1 Γ By hypothesis for all k>0these subschemes contain an irreducible component Z k X T so that Z k π 1 t W for any t with φt = As k one obtains a fixed subscheme Z X T with W Z t,wherez t = Z π 1 t According to [EKL], Proposition 3, any section σ H 0 X, p 1kA I kγ Γ must vanishto orderat leastkγ kαw +1 alongz Indeed, if not, after differentiating kγ αw +1 times we obtain σ H 0 X, p kαw +1 1kA IΓ not vanishing along Z; this is a contradiction Lemma 13 is the translation of this statement for the family Z to the fibres Z t Note that when applying Lemma 13 we often will assume, for simplicity, that ord W s nγ αw Indeed, for the asymptotic estimate on jets, the round down and 1 are irrelevant Proof of Theorem 0 Suppose that Theorem 0 were false and ɛ, A < 1 Thus through a very general point X there is a curve C with A C / mult C < 1 Choosing a suitable family of such curves F X T as above, we claim that there is an open set U X such that for each x U there is an irreducible curve C x satisfying A C x = p and mult C x =q with p/q < 1/ If ɛ, A =p/q and there is a curve C through with mult C =q and A C = p, then this is satisfied If there were a Seshadri exceptional surface S at, that is, a surface with deg A S mult S = ɛ, A, then an immediate contradiction is obtained using Lemma 13: choose p/q < γ < 1and0 σ H 0 X, na m nγ /m nγ+1 According to Lemma 13 ord S σ nγ pn/q +1 Since mult S, this is not possible for n 0 Since there must either be a Seshadri exceptional curve or a Seshadri exceptional surface when ɛ, A < 1 we must have a Seshadri excptional curve C through as desired The goal of the proof is to estimate h 0 X, na m 3pn q 14 n 3 We will show that this it is positive and then we have a contradiction from 3p+αn [EKL] which shows that the linear series na m q is empty for any α>0 Let π : Y X be the blow up of X at with exceptional divisor E Choose a

5 SESHADRI CONSTANTS AT VERY GENERAL POINTS 389 rational number α and a large positive integer n with nα Z Thenwehave h 0 X, na h 0 X, na m αn 15 = = αn 1 αn 1 h 0 X, na m k h 0 X, na m k+1 h 0 Y,π na ke h 0 Y,π na k +1E We have E P and using 15 and the exact sequence 0 H 0 Y,π na k +1E H 0 Y,π na ke H 0 E,π na ke we find h 0 X, na h 0 αn 1 X, na m αn = h 0 16 Y P, Ok, where h 0 Y P, Ok denotes the dimension of the subspace of H 0 P, Ok coming via restriction from H 0 Y,π na ke Our goal, then, is for each value of k, to bound h 0 Y P, Ok from above We next define critical numbers where the base locus of ka m kα is forced to jump for numerical reasons: α 1 = p q, α 3 = p q There is also a more subtle jumping value between α 1 and α 3,atleastforq sufficiently large, for which we require an extra definition Let Z PT X = P denote the zero dimensional subscheme of degree q given by T C Then one can define a Seshadri constant associated to Z as follows Suppose ψ : Y P is a birational map with Y smooth and ψ 1 I Z =O Y E Then ɛz, O1 = sup{r Q + : ψ O1 re isnef} r Then, as we will see below, the base locus of ka m kα is forced to contain a surface as soon as α>α,whereα satisfies α p/q = ɛ T C, O PTX1 α Note that a surface could enter the base locus of ka m kα for α<α The numbers we chose are the worst case scenario, the case where the linear series ka generates the most possible jets at If it generates fewer jets, the numbers in the argument only improve We note here that the reader not interested in the counting details can skip the analysis involving α Indeed, when we prove Theorem 0 below there is enough room in the estimates so that the key result is Lemma 11 which applies to the jet analysis once we have exceded α 3 We included a more complete analysis both in order to reveal the subtleties involved in counting and because in other cases the more detailed analysis may be required By Lemma 13 we know that any section of ka m kβ for β>α1 must vanish along C to multiplicity at least kβ α 1 Once β>α we claim that the base locus

6 390 MICHAEL NAKAMAYE of ka m kβ must contain a surface S which passes through Indeed,ifnot,then choose s 1,s ka m kβ so that T Zs 1 and T Zs 1 meet properly inside T X By Lemma 13 we have mult C s 1 kβ α 1 andmult C s kβ α 1 By Theorem A of [ELS], we have f 1,f I kβ α/ C,whereI C is the ideal sheaf of C and f 1 and f are local equations for s 1 and s Let π : Y X be the blow up of X at with exceptional divisor E LetD 1 = π Zs 1 kβe E and D = π Zs kβe E We have E P and D 1,D are curves of degree kβ meeting properly along T C, each with multiplicity at least kβ α 1 / along T C Considering the pencil of divisors spanned by D 1 and D shows that ψ Okβ kβ α 1 / E is nef where ψ : Y P is a resolution of I Z as above It follows that ɛz, O1 kβ α 1/ kβ > α p/q α, contradicting the definition of α Using Lemma 13 again, we conclude that there is a surface S X such that for β>α any divisor D ka m kβ must vanish along S to order at least kβ α Finally, let S be the surface swept out by {C x } x Z with Z φ 1 C the constructible subset considered above By Lemma 13 any divisor D ka m kβ must vanish along S to order at least kβ p q We are now prepared to bound h 0 Y P, Ok from above, using the information about the order of vanishing of each section of HY 0 P, Ok along T C, T S, and T S We will divide the estimate into four cases: 0 k nα 1, nα 1 <k nα, nα <k nα 3, nα 3 <k 3np q We assume for simplicity that α i Q and nα i Z For those α i which are irrational, it suffices in the argument below to replace nα i by nα Notealsothat if α >α 3, one simply einates the third interval, replacing α by α 3 in the second interval For small values of k one expects na to generate all k jets and the estimate is h 0 Y P, Ok k + 17, 0 k nα 1 Next, for nα 1 <k nα any section σ HY 0 P, Ok vanishes to order at least k nα 1 / along T C PT X giving the estimate 18 h 0 Y P, Ok k + k nα1 / +1 q + ok, nα 1 <k nα This is established in Lemma 113 below Next suppose nα < k nα 3 Let σ H 0 X, na m k For n suitably divisible, write Zσ =as + S, with mult S =nα

7 SESHADRI CONSTANTS AT VERY GENERAL POINTS 391 This is possible since σ must vanish to order at least k nα along the surface S Let ρ : H 0 X, na as m nα H 0 P, Onα be the restriction homomorphism Then by definition we have h 0 Y P, Ok = dimimageρ Using the construction in [EKL], 38, we see that there exists an irreducible subvariety V X T such that S = V X t forsomet with φt = Inparticular, since is a very general point it follows that there is a surface S algebraically equivalent to S, not containing, namelys = V X ξ for a general point ξ T Chooser sufficiently large so that ra+ bs S is very ample for all b>0 Choose D ra + as S so that D does not contain and let E = D + as Then tensoring by E gives an injection ρ E : H 0 X, na as m nα which preserves multiplicity at We conclude that H 0 X, n + ra m nα 19 h 0 Y P, Ok h 0 X, n + ra m nα /m nα+1, nα <k nα 3 Finally suppose nα 3 <k 3pn q Suppose that mult σ =k, σ H 0 X, na We know from Lemma 13 that mult S σ k nα 3 Since, according to Lemma 11 below mult C S 3, we can write Zσ =as + S with mult S =k 3k nα 3 Arguing as in the previous case then gives 110 h 0 Y P, Ok h 0 X, n + ra m 3nα3 k /m 3nα3 k+1, nα3 <k 3pn q We are now prepared to evaluate the it 14 using 16 We assume to begin with that q 5 In particular this means that ɛ T C, O PTX1 < 1/ and this guarantees that α <α 3 We divide the sum into four ranges of k determined by our critical numbers α 1,α,α 3 First,by17, Next, using 18 we see that nα k=nα 1+1 nα 1 h 0 Y P, Ok h 0 Y P, Ok n 3 α3 1 6 n 3 α3 α3 1 6 qα α 1 3 4

8 39 MICHAEL NAKAMAYE Using 18 and 19 gives nα 3 k=nα +1 h 0 Y P, Ok n 3 nα 3 k=nα +1 α 3 α h 0 X, n + ra m n+r nα n+r /m α qα α 1 8 n 3 nα n+r n+r +1 Finally, for 3np/q k=nα 3+1 h0 Y P, Ok, using 110 and arguing as in the above case we can remove the r which is fixed But the sum 3np/q k=nα 3+1 h 0 X, na m 3nα3 k /m 3nα3 k+1 is simply every other term of the sum nα 3 h0 Y P, Ok, and since our upper bound for h 0 Y P, Ok varies as a piecewise polynomial, this gives 3np/q k=nα 3+1 h 0 P, Ok n 3 α 3 6 qα α 1 3 +α 3 α 4 1 Combining all of the above estimates gives 111 3np/q 3 3 h 0 P, Ok n 3 α q α α 1 8 α 3 6 qα α 1 3 α +α 3 α 4 q α α 1 8 α 3 α 6 +α 3 α Note that in the last inequality we have omitted two of the negative or defect terms which were obtained by the detailed analysis above The reason for this is that for q sufficiently large, the estimate 111 turns out to be sufficient to establish Theorem 0 while small values of q can be dealt with by hand We included all of the counting details nonetheless, as this is the technical heart of this paper In order to compute the upper bound in 111, we need to know the value of α The Seshadri constant ɛ T C, O PTX1 is, however, very difficult to compute and so we look at the worst case scenario In particular, the bound in + O1 and then decreases The bound in 19 then repeats the last value for the bound in 18 and then when one reaches 110 the values start to decrease The O1 term will have no effect on the asymptotic 18 increases until x = np q 4

9 SESHADRI CONSTANTS AT VERY GENERAL POINTS 393 estimate and thus the worst case to cosider is α = np q 4 With this value of α we need to assume that q 9 in order to guarantee that α <α 3 We find then, using 111, 3np/q h 0 P, Ok n 3 3 p 3 pq 8 p + 6q 4 3 qq 4 q 4 3p 3 q 4 + 9p3 q 8 qq 4 3 = 1 6 One checks that when q 10 and p/q < 1/ then 1 3p 3 6 q 4 + 9p3 q 8 qq 4 3 < 1 6 It follows from 16 that when q 10, h 0 X, na m 3pn q n 3 > 0, and this concludes the proof of Theorem 0 when q 10 If q < 10, then there are only four possibilities which are not einated by [EKL], namely p/q = /5,p/q = 3/7,p/q = 3/8, and p/q = 4/9 We outline here how to einate the cases p/q =3/7 andp/q =4/9which are the most difficult of the four The counting here goes as follows For 0 k np/q we use 17 For np/q < k np/q, we use the estimate in 8 below Finally for np/q < k 3np/q, we use 110 This gives, in the case where p/q =3/7, For p/q =4/9 we find 3np/q 3np/q h 0 P, Ok n h 0 P, Ok n Lemma 11 Suppose C satisfies A C mult C = p q < 1 Let S be the surface swept out by {C x } x Z φ 1 C Then mult C S 3 Proof of Lemma 11 To see why S must be singular along C note that for a general point ξ φz C there is a curve C ξ S such that A C ξ mult C ξ = p q < 1 If S were smooth at a general point ξ C, then it would follow that ɛξ,a S < 1 and this is impossible since Ein and Lazarsfeld [EL] have established that on a

10 394 MICHAEL NAKAMAYE smooth surface the set of points where the Seshadri constant can be less than one is at most countable In order to refine this argument, let π : X X be an embedded resolution of S Forξ C general let C ξ be the strict transform of C ξ in X and write π 1 ξ C ξ = {x 1,,x r } Suppose moreover that ψ : C C ξ is a desingularization with ψ 1 π 1 ξ = {y 1,,y s } Choose a linear series D on X with D sufficiently positive so that if E D is a general member through ξ, thenix j, C ξ Ẽ : X =mult xj C ξ for 1 j r: this is possible by [F], 145 Then by [F], 145 and 7117, we have s r q = iξ,c ξ D : X = ord yj ψ π D = mult xi Cξ j=1 Now we have s =mult C S and thus if s = we find that for i =1ori = π A C ξ mult xi C ξ < 1 Let S X be the resolution of S The curves C ξ move in a one-parameter family along the surface S and thus {x S : ɛx, π A < 1} is not countable, violating the main result of [EL] Note that in [EL] Ein and Lazarsfeld state the main result for ample line bundles but the proof holds unchanged for a big and nef line bundle Indeed, the only point where [EL] uses ampleness is to show that curves with bounded degree relative to the appropriate ample bundle A move in finitely many families, but this also holds more generally when A is big and nef Lemma 113 Suppose mult C =q and nα 1 <k nα Then h 0 Y P, Ok k + k nα1 +1/ q + ok, nα 1 <k nα Proof of Lemma 113 Let A = O P 1, where P = PT X and let Z P be the projectivized tangent cone of C at By definition of ɛz, A, given δ>0so that ɛz, A δ Q for all n>0 sufficiently large and divisible, the evaluation map H 0 P, On H 0 P, On O P /I nɛz,a δ Z is surjective According to [F], Example 434, lo X /IZ r qr = + Or In particular, for nα 1 <k nα we see that that Lemma 113 holds since any section of HY 0 P, Ok vanishes to order at least k nα 1 / along Z Counting jets in the general case In addition to the computational complexity involved in estimating h 0 X, na m k for different values of k, the central difficulty in the higher-dimensional case is Lemma 11 which uses the fact that on a surface X the set {x X : ɛx, A < 1} is countable for an ample line bundle A In particular, it is critical for Lemma 11 i=1

11 SESHADRI CONSTANTS AT VERY GENERAL POINTS 395 that this set contains no divisor and this is not known in higher dimension In order to prove Theorem 03, we begin by recalling a key definition from [N] Definition 1 For an ample Q divisor A on a smooth surface X we let ma = sup{mult D} D DivX Q effective} D A Here denotes numerical equivalence The importance of Definition 1 lies in the following simple result: Lemma Suppose X is a projective variety of dimension d and A an ample line bundle on X If ɛ, A < ma, d then given δ>0 there exists an irreducible proper subvariety Y X, ofdimension at least one, such that ɛξ,a Y <ɛ, A+δ Here ξ is a very general point of Y and A Y is the restriction of A to Y Proof of Lemma Suppose, to the contrary, that for some δ>0 ɛξ,a Y ɛ, A+δ for every irreducible Y X As above, following [EKL], 34, given ɛ>0thereisa a family of curves F X T with 3 A C t α = <ɛ, A+ɛ, t T mult φt C t This gives a chain of subvarieties 4 V 1 V V d, where V 1 = C for a very general point and V i+1 = CV i in the notation of [EKL], Lemma 351 In particular, V i+1 is obtained by adjoining curves in the family F By [EKL], Lemma 351, each V i is irreducible According to Lemma 13 any section s H 0 X, na m nα+ vanishes along C to order at least nα + 1 and hence vanishes along V Proceding inductively using Lemma 13 we find that if s H 0 X, na m dnα+d,then 5 s V d =0 By hypothesis, ma >dɛ, A and thus, shrinking ɛ in 3 if necessary, we can assume that s is not indentically zero in 5 and thus dimv d d 1 It follows from4thatforsome1 r d 1wemusthave V r = V r+1 In particular for a general, hence smooth, point ξ V r, we find a curve C ξ V r with mult ξ C ξ = α A C ξ Hence ɛξ,a V r α<ɛ, A+δ Thus we can take Y = V r and this proves Lemma

12 396 MICHAEL NAKAMAYE We will derive Theorem 03 from Lemma and the following result Lemma 6 Suppose X is a projective variety of dimension d 4 and A an ample line bundle on X Then either ma > d or ɛ, A > 1 d + 1 3d Proof of Lemma 6 The proof of Lemma 6 closely follows the method of 1, though the counting is much simpler Let π : Y X be the blow up of X at with exceptional divisor E P d 1 Choose a rational number α, and a large positive integer n so that nα Z Thenwehave,withthesamenotationasabove, h 0 X, na h 0 αn 1 X, na m αn = h 0 7 Y P d 1, Ok, where, as above, h 0 Y P d 1, Ok denotes the dimension of the subspace of H 0 P d 1, Ok coming via restriction from H 0 Y,π na ke Suppose that x P d 1 = PT X is a tangent vector to C at Then for k>ɛ, An, Lemma 13 implies h 0 Y P d 1, Ok 8 h 0 P d 1, Ok m k ɛ,an 1 x Combining 7 and 8 and taking the it as n we find h 0 X, na h 0 X, na m αn n d 1 α x d 1 max { 0, x ɛ, A d 1} d 1! 0 = αd α ɛ, A d 9 d! According to 9, if α d α ɛ, A d < 1, h then 0 X,nA m αn > 0 and hence ma >α Suppose then that ɛ, A n d 1 d + 1 3d and let α =1+ 1 3d Then we find 3d d d +1 3d +1 3d +1 d 3d 3d 3d e 1/3 e /3 < 09 Thus we see that for all d sufficiently large, ma > d if ɛ, A 1 d + 1 3d Elementary calculus suffices to show that this also holds for all d 4andthis establishes Lemma 6 Proof of Theorem 03 Suppose to the contrary that 10 ɛ, A 1 d + 1 3d By Lemma 6 ma > 1+1/3d and thus ɛ, A < ma d

13 SESHADRI CONSTANTS AT VERY GENERAL POINTS 397 By Lemma given δ>0there is a proper subvariety Y X such that for a very general point ξ Y, ɛξ,a Y 1 d + 1 3d + δ 1 But by the main theorem of [EKL], we know that ɛξ,a Y dimy 1 d 1 and this is a contradiction for δ sufficiently small Note that above in 8 we have not counted carefully: in particular, the curve C is singular at and thus the tangent space to C will be more than a single point with multiplicity one Thus the counting can be improved considerably here but we were unable to obtain a significant quantitative improvement in the final result by checking this counting more carefully References [EL] [ELS] [EKL] [F] [FW] [N] [S] L Ein, R Lazarsfeld, Seshadri constants on smooth surfaces Asterisque 18, 1993, pp MR f:14031 L Ein, R Lazarsfeld, and K Smith, Uniform bounds and symboic powers on smooth varieties, Inv Math, 144, 001, pp 41 5 MR b:13001 L Ein, O Küchle, R Lazarsfeld, Local positivity of ample line bundles, J Diff Geomertry, 4, 1995, pp MR m:14007 W Fulton, Intersection Theory, Springer, 1984 MR k:14004 G Faltings and G Wüstholz, Diophantine approximations on projective spaces, Inv math, 116, 1994, pp MR g:11068 M Nakamaye, Seshadri constants and the geometry of surfaces, J ReineAngew Math 564, 003, MR01040 Y T Siu, Effective very ampleness, Invent Math, 14, 1996, pp MR a:3036 Department of Mathematics and Statistics, University of New Mexico, Albuquerque, New Mexico address: nakamaye@mathunmedu