ivic divisor problem

8/6/2019 Ivic Divisor Problem

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CEJM 2(4) 2004 494508

On the Riemann zeta-function and the divisorproblem

Aleksandar Ivic Katedra Matematike RGF-a,

Universiteta u Beogradu,

usina 7, 11000 Beograd, Serbia (Yugoslavia)

Received 14 April 2004; accepted 19 July 2004

Abstract: Let ( x) denote the error term in the Dirichlet divisor problem, and E (T ) the errorterm in the asymptotic formula for the mean square of | ( 12 + it )|. If E (t) = E (t) 2 (t/ 2)with (x) = ( x) + 2(2 x) 12 (4 x), then we obtain

T

0 (E

(t))4

dt T 16/ 9+

.

We also show how our method of proof yields the bound

R

r =1

t r + G

t r G | (12 + it )|2 dt

4

T 2+ G 2 + RG 4T ,

where T 1/ 5+ G T, T < t 1 < < t R 2T, tr +1 t r 5G (r = 1 , . . . , R 1).c Central European Science Journals. All rights reserved.

Keywords: Dirichlet divisor problem, Riemann zeta-function, mean square and twelfth moment of | ( 12 + it )|, mean fourth power of E (t)MSC (2000): 11N37, 11M06

1 Introduction and statement of results

Let, as usual,( x) =

n x

d(n) x(log x + 2 1) 14 , (1)

E-mail: [email protected], [email protected]


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A. Ivic / Central European Journal of Mathematics 2(4) 2004 494508 495

and

E (T ) = T

0 | (12 + it )|2 dt T log

T 2

+ 2 1 , (2)where d(n) is the number of divisors of n, = (1) = 0 .577215. . . is Eulers constant.Thus ( x) denotes the error term in the classical Dirichlet divisor problem, and E (T ) isthe error term in the mean square formula for | ( 12 + it )|. An interesting analogy betweend(n) and | ( 12 + it )|2 was pointed out by F.V. Atkinson [1] more than sixty years ago. Inhis famous paper [2], Atkinson continued his research and established an explicit formulafor E (T ) (see also the authors monographs [7, Chapter 15] and [8, Chapter 2]). The mostsignicant terms in this formula, up to the factor ( 1)n , are similar to those in Voronoisformula (see [7, Chapter 3]) for ( x). More precisely, in [13] M. Jutila showed that E (T )should be actually compared to 2 (T/ (2)), where

(x) := ( x) + 2(2 x) 12 (4 x). (3)Then the arithmetic interpretation of (x) (see T. Meurman [16]) is

12

n 4x

(1)n d(n) = x(log x + 2 1) + (x). (4)

We have the explicit, truncated formula (see e.g., [7] or [18])

( x) =1

2x1

4

n N d(n)n

3

4 cos(4nx 14 ) + O(x1

2 + N 1

2 ) (2 N x). (5)

One also has (see [7, eq. (15.68)]), for 2N x, (x) =

12x

14

n N

(1)n d(n)n34 cos(4nx 14 ) + O(x

12 + N

12 ), (6)

which is completely analogous to (5).

M. Jutila, in his works [13] and [14], investigated both the local and global behaviourof

E (t) := E (t) 2 t

2.

He proved the mean square bound

T + H

T H (E (t))2 dt HT 1/ 3 log3 T + T 1+ (1 H T ), (7)

which in particular yields

T

0(E (t))2 dt T 4/ 3 log3 T. (8)

Here and later denotes positive constants which are arbitrarily small, but are not nec-essarily the same at each occurrence. The bound (8) shows that, on the average, E (t)


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496 A. Ivic / Central European Journal of Mathematics 2(4) 2004 494508

is of the order t1/ 6 log3/ 2 t, while both E (x) and ( x) are of the order x1/ 4. Thisfollows from the mean square formulas (see e.g., [8])

T

0 2(x) dx = (6 2) 1

n =1

d2(n)n 3/ 2T 3/ 2 + O(T log4 T ), (9)

and

T

0E 2(x) dx = 23 (2)

1/ 2

n=1

d2(n)n 3/ 2T 3/ 2 + O(T log4 T ). (10)

The mean square formulas (9) and (10) also imply that the inequalities < 1/ 4 and < 1/ 4 cannot hold, where and are, respectively, the inma of the numbers a and bfor which the bounds

( x) xa , E (x) xb (11)

hold. For upper bounds on , see e.g., M.N. Huxley [5]. Classical conjectures are that = = 1 / 4 holds, although this is notoriously difficult to prove. M. Jutila [13] succeededin showing the conditional estimates: if the conjectural = 1 / 4 holds, then this impliesthat 3/ 10. Conversely, = 1 / 4 implies that (x) x+ holds with 3/ 10.Although one expects the maximal orders of ( x) and (x) to be approximately of thesame order of magnitude, this does seem difficult to establish.

In what concerns the formulas involving higher moments of ( x) and E (t), we referthe reader to the authors works [6], [7] and [10] and D.R. Heath-Brown [4]. In particular,

note that [10] contains proofs of

T

0E 3(t) dt = 164

T 2

0( (t))3 dt + O(T 5/ 3 log3/ 2 T ),

T

0E 4(t) dt = 325

T 2

0( (t))4 dt + O(T 23/ 12 log3/ 2 T ). (12)

In a recent work by P. Sargos and the author [12], the asymptotic formulas of K.-M.Tsang [19] for the cube and the fourth moment of ( x) were sharpened to

X1 3(x) dx = BX 7/ 4 + O(X + ) (B > 0) (13)and

X

1 4(x) dx = CX 2 + O(X + ) (C > 0) (14)

with = 75 , =2312 . This improves on the values =

4728 , =

4523 , obtained in [19].

Moreover, (13) and (14) remain valid if ( x) is replaced by (x), since their proofs usednothing more besides (5) and the bound d(n) n . Hence from (12) and the analoguesof (13)(14) for (x), we infer then that

T

0E 3(t) dt = B1T 7/ 4 + O(T 5/ 3 log3/ 2 T ) (B1 > 0),

T

0E 4(t) dt = C 1T 2 + O(T 23/ 12+ ) (C 1 > 0). (15)


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The main aim of this paper is to provide an estimate for the upper bound of the fourthmoment of E (t), which is the rst non-trivial upper bound for a higher moment of E (t).The result is the following

Theorem 1.1. We have T

0(E (t))4 dt T 16/ 9+ . (16)

Note that the bounds (8) and (16) do not seem to imply each other. For the proof of (16) we shall need several lemmas, which will be given in Section 2. The proof of Theorem 1.1 will be given in Section 3. Finally, in Section 4, it will be shown how themethod of proof of Theorem 1.1 can give a proof of

Theorem 1.2. Let T 1/ 5+

G T, T < t 1 < < t R 2T, tr +1 t r 5G (r =1, , R 1). Then R

r =1

t r + G

t r G | (12 + it )|2 dt

4

T 2+ G 2 + RG 4T . (17)

The bound in (17) easily gives the well-known bound (see Section 4)

T

0 | ( 12 + it )

|12 dt T 2+ , (18)

due to D.R. Heath-Brown [3] (who had log 17 T instead of the T -factor). It is still es-sentially the sharpest result concerning high moments of | ( 12 + it )|. General sums of zeta-integrals over short intervals, analogous to the one appearing in (17), were treatedby the author in [9].

2 The necessary lemmas

Lemma 2.1. (O. RobertP. Sargos [17]). Let k 2 be a xed integer and > 0 be given.Then the number of integers n1, n2, n 3, n 4 such that N < n 1, n 2, n3, n4 2N and

|n1/k1 + n

1/k2 n

1/k3 n

1/k4 | < N 1/k

is, for any given > 0, N (N 4 + N 2). (19)

Lemma 2.2. Let 1 G T/ log T . Then we have

E (T ) 2

G

0E (T + u) e u

2 /G 2 du + O(G log T ), (20)

and E (T )

2G

0E (T u) e u

2 /G 2 du + O(G log T ). (21)


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Proof of Lemma 2.2 . The proofs of (20) and (21) are analogous, so only the formerwill be treated in detail. From (2) we have, for 0 u T ,

0

T + uT | ( 12 + it )|2 dt = ( T + u) log T + u2 + 2 1 (22)

T log T 2 + 2 1 + E (T + u) E (T ). (23)This gives

E (T ) E (T + u) + O(u log T ),hence

G log T

0E (T ) e u

2 /G 2 du G log T

0(E (T + u) + O(u log T )) e u

2 /G 2 du.

The proof of (20) is completed when we extend the integration to [0 , ) making a smallerror, and recall that

0 e

u 2 /G 2 du = 12G,

0 ue u 2 /G 2 du = 12 G.

Lemma 2.3. Let 1 G T . Then we have

T 2

=2

G

0

T 2

u2

e u2 /G 2 du + O(GT ). (24)

Proof of Lemma 2.3. Both the cases of the + and sign in (24) are treated analogously.For example, we have12G T 2

0

T 2 +

u2 e

u 2 /G 2 du=

0

(T ) T 2 + u2 e u2 /G 2 du

= G log T

0 T

2 ) ( T 2 + u2 e u2 /G 2 du + O(1)

G log T

0 2 T n 2 (T + u)

(1)n d(n) + O((1 + |u|)log T ) du G2T ,

where we used (4) and d(n) n. This establishes (24).

The next lemma is F.V. Atkinsons classical explicit formula for E (T ) (see [2], [7] or[8]).

Lemma 2.4. Let 0 < A < A be any two xed constants such that AT < N < A T , and let N = N (T ) = T/ (2) + N/ 2 (N 2/ 4 + NT/ (2))1/ 2. Then

E (T ) = 1(T ) + 2(T ) + O(log2 T ), (25)

where 1(T ) = 2 1/ 2(T/ (2))1/ 4

n N

(1)n d(n)n 3/ 4e(T, n) cos(f (T, n)) , (26)

2(T ) = 2n N

d(n)n 1/ 2(log T/ (2n )) 1 cos(T log T/ (2n ) T + / 4), (27)


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We combine now (20) with (24) with the + sign (when E (T ) 0) or (21) with (24) withthe sign (when E (T ) 0), to obtain by the Cauchy-Schwarz inequality

(E (t))2 G 1

G log T

G log T

e u2 /G 2 (E (t + u))2 du + G2T , (36)

provided that T t 2T, T G T 5/ 12. Keeping in mind the preceding discussionwe thus have (replacing ( t + u)1/ 4 with t1/ 4 by Taylors formula, with the error absorbedby the last term in (37)) by using (6), (25), (35) and (36),

(E (t))2 G 1 G log T

G log T e u

2 /G 2 ( 23(X ; u) + 24(X, N ; u) +

25(X, N ; u)) d u

+ T 1+ N 1, (37)

where we set

3(X ; u) := t1/ 4 n X

(1)n d(n)n 3/ 4

e(t + u, n ) cos(f (t + u, n )) cos( 8n (t + u) / 4) , (38) 4(X, N ; u) := t1/ 4

X


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left-hand side of (33) is majorized by the maximum taken over |u| G log T times

2T

T (E (t))2( 24(X, N ; u) +

25(X, N ; u) +

26(X ; u) + T

1+ N 1) dt

2T

T (E (t))4 dt

2T

T 44(X, N ; u) + 45(X, N ; u) + 46(X ; u) dt

1/ 2

+ T 7/ 3+ N 1, (42)

where we used the Cauchy-Schwarz inequality for integrals and (8). Thus from (42) wehave the key bound

2T

T (E (t))4 dt max

|u | G log T 2T

T 44(X, N ; u) +

45(X, N ; u) +

46(X ; u) dt

+ T 7/ 3+ N 1. (43)

To evaluate the integrals on the right-hand side of (43) we note rst that

2T

T 44(X, N ; u) + . . . dt

5T/ 2

T/ 2(t) 44(X, N ; u) + . . . dt, (44)

where (t) is a smooth, nonnegative function supported in [ T/ 2, 5T/ 2] , such that (t) =1 when T t 2T . The integrals of 44(X, N ; u), 45(X, N ; u) and 46(X ; u) are allestimated analogously. The sums over n are divided into log T subsums of the form

K 0), where := 8(m + n k l ). (45)

Therefore, in the case of 5(X, N ; u), there remains the estimate

2T

T 45(X, N ; u) dt 1 + T

1+ max|u | G log T

supX K N

5T/ 2

T/ 2(t)

K


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Proceeding analogously as in (47), we obtain that

2T

T 46(X ; u) dt T

1+ max1 K X, | | T 1 / 2

T 1K 3(K 4K 1/ 2| |+ K 2)

T

(T 1/ 2

X 13/ 2

+ X 5

), (48)since instead of (mnkl ) 3/ 4 in (46) now we shall have (mnkl )3/ 4t 1 (see (41)).

The estimation of 4(X, N ; u) (see (39)) presents a technical problem, since the cosinescontain the function f (t, n ), and Lemma 2.1 cannot be applied directly. First we notethat, by using (28), we can expand the exponential in power series to get rid of the termsa5n5/ 2t 3/ 2 + . . . . In this process the main term will be 1, and the error terms will makea contribution which will be (for shortness we set a = 8, b = 1623 and = t + u)

max|u | G log T

supX K N

T 5T/ 2

T/ 2(t)

K


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integrations by parts over t, only the portion of the sum for which | | T 1/ 2 willremain, where is given by (45). In the remaining sum we use (49) (once with A = 14 and once with A = 34 ), noting that e

iz = cos z + i cos(z 12 ). We remark that, for|v| u0, we can use the crude estimate (u) T 1/ 6, hence for this portion the estimationwill be quite analogous to the preceding case. Next we note that

u 1

u 0 1/ 8v 1/ 4 exp( ib 1/ 4v3/ 2 8nv ) dv log T ( = t + u, |u| G log T ),

writing the integral as a sum of log T integrals over [U, U ] with u0 U < U 2U u1, and applying the second derivative test to each of these integrals. We also remarkthat the contribution of the O-term in (32) will be, by trivial estimation,

u 0T 1/ 8u 7/ 4 du T 1/ 8u 3/ 40 1

if we suppose that (50) is satised. It remains yet to deal with the integral with v > u 1in (49), when we note that

v

b 1/ 4v3/ 2 8nv T 1/ 4v1/ 2 (v > u 1),provided that C in (50) is sufficiently large. Hence by the rst derivative test

u 1(v) cos(8n ( + v)

14 ) dv

1 + T 1/ 8u 1/ 41 T 1/ 4u 1/ 21

1 + T 1/ 4K 3/ 4 1 + T 1/ 4X 3/ 4 1,

since K X T 1/ 3. Thus the contribution of the integrals on the right-hand side of (49) is log T .

Then we can proceed with the estimation as in the case of 5(X, N ; u) to obtain

2T

T 44(X, N ; u) dt T

3/ 2+ N 1/ 2 + T 2+ X 1.

Gathering together all the bounds, we see that the integral in (33) is

T T 3/ 2N 1/ 2 + T 2X 1 + T 1/ 2X 13/ 2 + X 5 + T 7/ 3+ N 1 , (51)

provided that (40) holds. Finally we choose

X = T 1/ 3 , N = T 5/ 9,

so that (40) is fullled. The above terms are then T 16/ 9+ , and the proof of Theorem1 is complete. The limit of the method is the bound T 2X 1 T 5/ 3, which wouldyield the exponent 5 / 3 + in (16). The true order of the integral in (16), and in generalthe order of the k-th moment of E (t), is elusive. This comes from the denition E (t) =E (t) 2 (t/ (2)), which makes it difficult to see how much the oscillations of thefunctions E and cancel each other.


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4 The proof of Theorem 1.2

We shall show now how the method of proof of our Theorem 1.1 may be used to yieldTheorem 1.2. Our starting point is an expression for the integral

tr +2 G

t r 2Gr (t)| ( 12 + it )|2 dt, (52)

where t r is as in the formulation of Theorem 1.2, and r C is a non-negative function

supported in [ t r 2G, t r + 2 G] that equals unity in [ t r G, t r + G]. The integral in (52)majorizes the integral

t r + G

t r G | (12 + it )|2 dt, (53)

which is of great importance in zeta-function theory (see K. Matsumoto [15] for an ex-tensive account on mean square theory involving (s)). One can treat the integral in (52)by any of the following methods.(a) Using exponential averaging (or some other smoothing like r above), namely the

Gaussian weight exp( 12 x2), in connection with the function E (T ), in view of F.V.Atkinsons well-known explicit formula (cf. Lemma 2.4). This is the approach em-ployed originally by D.R. Heath-Brown [3].

(b) One can use the Voronoi summation formula (e.g., see [8, Chapter 3]) for the explicitexpression (approximate functional equation) for

| ( 12 + it )

|2 = 1( 12 + it )

2( 12 + it ),where (s) = (s) (1 s), namely

(s) = 2 s s 1 sin( 12 s )(1 s).Voronois formula is present indirectly in Atkinsons formula, so that this approachis more direct. The effect of the smoothing function r in (53) is to shorten the sumapproximating | |2 to the range T 2 (1 G 1T ) n T 2 (T = t r ). After this nointegration is needed, and proceeding as in [7, Chapters 7-8] one obtains that theintegral in (53) equals O(GT ) plus a multiple of

t r +2 G

t r 2Gr (t)

k T 1+ G 2(1)kd(k)k 1/ 2

14

+t

2k

1/ 4

sin f (t, k ) dt, (54)

where f (t, k ) is given by (28).(c) Instead of the Voronoi summation formula one can use the (simpler) Poisson sum-

mation formula, namely

n =1

f (n) =

0f (x) dx + 2

n =1

0f (x) cos(2nx ) dx,

provided that f (x) is smooth and compactly supported in (0 , ). In [11] a sketchof this approach is given.

We begin now the derivation of (17), simplifying rst in (54) the factor (1 / 4 +t/ (2k)) 1/ 4 by Taylors formula, and then raising the expression in (54) to the fourth


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power, using Holders inequality for integrals. It follows that the sum in (17) is boundedby

RG 4T + T 1G3R

r =1

t r +2 G

t r 2Gr (t)

k T 1+ G 2(

1)kd(k)k 1/ 4 sin f (t, k )

4dt

RG 4T + T 1G3 5T/ 2

T/ 2(t)

k T 1+ G 2(1)kd(k)k 1/ 4 sin f (t, k )

4dt, (55)

where (t) is a non-negative, smooth function supported in [ T/ 2, 5T/ 2] such that (t) = 1for T t 2T , hence (m )(t) m T m . Therefore it suffices to bound the expression

I K := 5T/ 2

T/ 2(t)

K


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which is (17).It remains to show how (17) gives the twelfth moment estimate (18). Write

2T

T | ( 12 + it )|12 dt

r T +1| ( 12 + i r )|12, (57)

where for r = 1 , 2, . . . we set

| ( 12 + i r )| := maxT + r 1 t T + r | ( 12 + it )|.Let {t r,V }be the subset of { r }such that

V | ( 12 + it r,V )| 2V (r = 1 , . . . , R V ),where clearly V may be restricted to O(log T ) values of the form 2m such that log T V T 1/ 6, since ( 12 + it ) = o(t1/ 6) (see [7, Chapter 7]). Now since we have (see e.g., [8,Theorem 1.2]), for xed k

N ,

| ( 12 + it )|k log t t+ 12

t 12| ( 12 + it + iu)|k du + 1 ,

it follows that, for some points t r ([T, 2T ]) with r = 1 , . . . , R , R RV , 1 GT, t

r +1 t

r 5G,

RV V 2 R V

r =1| ( 12 + it r,V )|2

R V

r =1

log T t r,V + 12

t r,V 12| ( 12 + it )|2 dt + 1

R

r =1

log T t r + G

t r G | (12 + it )|2 dt + RV log T

log T (R )3/ 4R

r =1

t r + G

t r G | (12 + it )|2 dt

1/ 4+ RV log T

T (RV G + R3/ 4V T

1/ 2G 1/ 2),

where the estimate of Theorem 1.2 was used, with RV replacing R. If we take G = V 2T 2 ,then we obtain

R1/ 4V T 1/ 2+ G 3/ 2,

which givesRV T 2+ G 6 T 2+ V 12 .

Then the portion of the sum in (57) for which | ( 12 + i r )| T 1/ 10+ islog T max

V T 1 / 10+ RV V 12 T 2+ .


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But for values of V such that V T 1/ 10+ , the above bound easily follows from the largevalues estimate (the fourth moment) R T 1+ V 4. This shows that the integral in (57)is T 2+ , and proves (18). Note that the author [9, Corollary 1] proved the bound

R

r =1

t r + G

t r G| ( 12 + it )|4 dt

2RG 2 log8 T + T 2G 1 logC T (58)

for some C > 0, where T < t 1 < . . . < t R 2T , t r +1 t r 5G for r = 1 , . . . , R 1 and1 G T . The bound (58), which is independent of Theorem 1.2, was proved by amethod different from the one used in this work. Like (17), the bound (58) also leads tothe twelfth moment estimate (18).

References[1] F.V. Atkinson: The mean value of the zeta-function on the critical line, Quart. J.

Math. Oxford , Vol. 10, (1939), pp. 122128.

[2] F.V. Atkinson: The mean value of the Riemann zeta-function, Acta Math. , Vol. 81,(1949), pp. 353376.

[3] D.R. Heath-Brown: The twelfth power moment of the Riemann zeta-function,Quart. J. Math. (Oxford) , Vol. 29, (1978), pp. 443462.

[4] D.R. Heath-Brown: The distribution of moments in the Dirichlet divisor problems,Acta Arith. , Vol. 60, (1992), pp. 389415.

[5] M.N. Huxley: Area, Lattice Points and Exponential Sums , Oxford SciencePublications, Clarendon Press, Oxford, 1996.

[6] A. Ivic: Large values of the error term in the divisor problem, Invent. Math. , Vol.71, (1983), pp. 513520.

[7] A. Ivic: The Riemann zeta-function , John Wiley & Sons, New York, 1985.

[8] A. Ivic: The mean values of the Riemann zeta-function , LNs 82, Tata Inst. of

Fundamental Research, Bombay (distr. by Springer Verlag, Berlin etc.), 1991.[9] A. Ivic: Power moments of the Riemann zeta-function over short intervals, Arch.

Mat. , Vol. 62, (1994), pp. 418424.

[10]A. Ivic: On some problems involving the mean square of | ( 12 + it )|, Bull. CXVI Acad.Serbe, Classe des Sciences mathematiques , Vol. 23, (1998), pp. 7176.[11]A. Ivic: Sums of squares of | ( 12 + it )| over short intervals, Max-Planck-Institut f ur Mathematik, Preprint Series , Vol. 52, (2002), pp. 12.[12]A. Ivic and P. Sargos: On the higher moments of the error term in the divisor problem ,to appear.

[13]M. Jutila: Riemanns zeta-function and the divisor problem, Arkiv Mat., Vol. 21,(1983), pp. 7596; ibid. Vol. 31, (1993), pp. 6170.

[14]M. Jutila: On a formula of Atkinson, In: Proc. Coll. Soc. J. BolyaiBudapest81 ,Vol. 34, North-Holland, Amsterdam, 1984, pp. 807823.
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[15]K. Matsumoto: Recent developments in the mean square theory of the Riemann zetaand other zeta-functions, In: Number Theory , Birkhauser, Basel, 2000, pp. 241286.

[16]T. Meurman: A generalization of Atkinsons formula to L-functions, Acta Arith. ,Vol. 47, (1986), pp. 351370.

[17]O. Robert and P. Sargos: Three-dimensional exponential sums with monomials, J.reine angew. Math. (in print).

[18]E.C. Titchmarsh: The theory of the Riemann zeta-function , 2nd Ed., UniversityPress, Oxford, 1986.

[19]K.-M. Tsang: Higher power moments of ( x), E (t) and P (x), Proc. London Math.Soc.(3) , Vol. 65, (1992), pp. 6584.


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CEJM 2(4) 2004 509515

On the Apostol-Bernoulli Polynomials

Qiu-Ming Luo

Department of Mathematics,Jiaozuo University,

Jiaozuo City, Henan 454003, The Peoples Republic of China

Received 19 July 2001; accepted 23 May 2003

Abstract: In the present paper, we obtain two new formulas of the Apostol-Bernoullipolynomials (see On the Lerch Zeta function. Pacic J. Math., 1 (1951), 161167 .), using theGaussian hypergeometric functions and Hurwitz Zeta functions respectively, and give certainspecial cases and applications.c Central European Science Journals. All rights reserved.

Keywords: Bernoulli numbers, Bernoulli polynomials, Apostol-Bernoulli numbers, Apostol-Bernoulli polynomials, Gaussian hypergeometric functions, Stirling numbers of the second kind,Hurwitz Zeta functions, Lerch functional equation MSC (2000): Primary: 11B68; Secondary: 33C05, 11M35, 30E20

1 Introduction

An analogue of the classical Bernoulli polynomials were dened by T. M. Apostol (see[1]) when he studied the Lipschitz-Lerch Zeta functions. We call this polynomials theApostol-Bernoulli polynomials. First we rewrite Apostols denitions below

Denition 1.1. Apostol-Bernoulli polynomials B n (x, ) are dened by means of thegenerating function (see [1, p.165 (3.1)] or [4, p.83])

zexz

ez

1=

n =0

B n (x, )zn

n!, (|z + ln | < 2) (1)

setting = 1 in (1), Bn (x) = B n (x, 1) are classical Bernoulli polynomials.

[email protected]


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510 Q.-M. Luo / Central European Journal of Mathematics 2(4) 2004 509515

Denition 1.2. Apostol-Bernoulli numbers B n () := B n (0, ) are dened by means of the generating function

z

ez

1=

n =0

B n ()zn

n!, (|z + ln | < 2) (2)

setting = 1 in (2), Bn = B n (1) are classical Bernoulli nunbers.

T.M.Apostol not only gave elementary properties of polynomials B n (x, ) in [1], butalso obtained the recursion formula for the set of numbers B n () using the Stirling num-bers of the second kind (see [1, p. 166 (3.7)]) as follows

Bn() = n

n 1

k=1

(

1)k k!k (

1) 1 k S (n

1, k), (n

N0; () > 0, = 1) (3)

where S (n, k ) denote the Stirling numbers of the second kind which are dened by meansof the following expansion (see [3, p.207, Theorem B])

xn =n

k =0

xk

k!S (n, k ). (4)

By applying binomial series expansion and Leibnizs rule, we rst obtain the repre-sentation of the polynomials B n (x, ) involving the Gaussian hypergeometric functions,and thereout deduce Apostols formula (3); afterward we prove Theorem 3.1 using Lerchfunctional equation with related Hurwitz Zeta function. Furthermore we show that themain result in [9, p.1529, Theorem A] is only a special case of Theorem 3.1.

2 Apostol-Bernoulli Polynomials and GaussianHypergeometric functions

Theorem 2.1. If n is a positive integer and () > 0, = 1 are complex numbers, thenwe have

B n (x, ) = nn 1

l=0

n 1l

l( 1) l 1l

j =0

( 1) jl

j j l(x + j )n l 1

F [l n + 1 , l; l + 1; j/ (x + j )] (5)

where F [a, b; c; z] denotes Gaussian hypergeometric functions dened by (cf. [5, p.44 (4)])

F [a, b; c; z] :=

n =0

(a)n (b)n(c)n

zn

n!, |z| < 1 (6)

where ()0 = 1 , ()n = ( + 1) ( + n 1) = ( + n )( ) , (n 1).


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Q.-M. Luo / Central European Journal of Mathematics 2(4) 2004 509515 511

Proof 2.2. We differentiate both sides of (1) with respect to the variable z. ApplyingLeibnizs rule yields

B n (x, ) = D nzzexz

e z 1z =0

, Dz =ddz

.

=( 1) 1n

k=0

nk

kxn k D k 1z

1(ez 1) + 1

1

z =0.

(7)

Since binomial series expansion

(1 + w) 1 =

l=0

( w)l , |w| < 1 (8)

setting w =

1 (ez

1), we have

B n (x, ) = ( 1) 1n

k=0

nk

kx n kk 1

l=0

1

l

D k 1z {(ez 1)l}| z =0 . (9)

By the denition of Stirling numbers of the second kind (see [5, p.58 (15)])

(ez 1)l = l!

r = l

S (r, l )zr

r !(10)

yields

B n (x, ) =n

k=1

nk

kk 1

l=0

( 1)l l( 1) l 1l!S (k 1, l)xn k . (11)

We change sum order of k and l, and using the formula below (see [5, p.58 (20)])

S (n, k ) =1k!

k

j =0( 1)

k j k j j

n

(12)

we obtain

B n (x, ) = nn 1

l=0

l( 1) l 1xn k l 1l

j =0

( 1) jl

j j l

n l 1

k=0

n 1n k 2

jx

k

. (13)

Applying (6) to (13) readily yields

B n (x, ) = nn 1

l=0

n 1l

l( 1) l 1xn k l 1

l

j =0

( 1) jl

j j l .F [l n + 1 , 1; l + 1; j/x ] (14)


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Finally, we apply the known transformation [10, 15.3.4]

F [a, b; c; z] = (1 z) a F [a, c b; c; z/ (z 1)],

and (2.2) immediately obtain (5).

Remark 2.3. H .M. Srivastava and P. G Todorov considered earlier the formula of thegeneralized Bernoulli polynomials (see [6, p.510 (3)]), for = 1, which is a complemen-tarity of our result (5), for = 1, as follows

Bn (x) = nk =0nk

k!(2k)!

k

j =0

( 1) jk j

j 2k (x + j )n k

F [k n, k 1; 2k + 1; j/ (x + j )]. (15)

Remark 2.4. We will also apply the representation (5) in order to derive an interestingspecial case considered by T. M. Aspotol in (3).

By the well-known formula [10, 15.1.20]

F [a, b; c; 1] =(c)(c a b)(c a)(c b)

(c = 0 , 1, 2, . . . , (c a b) > 0),

upon setting a = l n + 1 , b = l, and c = l + 1 yields

F [l n + 1 , l; l + 1; 1] =n 1

l

1

, (0 l n). (16)

In view of (16), the special case of our formula (5) when x = 0 gives Apostols represen-tation (3).

Remark 2.5. The formula of classical Bernoulli numbers considered by H. W. Gould in[7, p.49, Eq.(17)] is also a complementarity of the Apostols formula (3), for = 1

Bn

=n

k =0

( 1)kn + 1

n k

n + k

k

1

S (n + k, k). (17)

Remark 2.6. There is a relationship between Apostol-Bernoulli polynomials B n (x, )and Stirling numbers of the second kind:

B n (x, ) =n

k =1

knk

k 1

j =0

( 1) j j ( 1) j 1 j !S (k 1, j )xn k , ( = 1). (18)

Remark 2.7. Recently, Luo also obtained the relation between the classical Bernoullipolynomials and the Stirling numbers of the second kind [8], which is a complementarityof (18), for = 1:

Bn (x) =n

k =0

n k

s =0

ns + k

k + s + 1s

s + 2 kk

1

S (s + 2 k, k)xn s k . (19)


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3 Apostol-Bernoulli polynomials and the Hurwitz ZetaFunction

It is well-known that Hurwitz-Lerch Zeta functions are dened by the innite series (see

[5, p.121, (1)])

(z,a,s ) =

k =0

zk

(a + k)s, (20)

(a C \ Z 0 ; s C , when |z| < 1; (s) > 1, when |z| = 1) .

Setting z = e2iz gives the Lipschitz-Lerch Zeta function (z,a,s ) (see [1, p.161])

(z,a,s ) =

k=0

e2kiz

(a + k)s, (21)

(a C \ Z 0 ; z R \ Z , (s) > 0; z Z , (s) > 1).

When s = n is negative integer, the Lipschitz-Lerch Zeta function (z,a,s ) was evalu-ated by T. M. Apostol using polynomials B n (x, ) (see [1, p.164]):

(z,a, n) = B n +1 (a, e 2iz )

n + 1. (22)

If we set z = 0 in (22), then we have the formula (see [2, p.264, Theorem 12.13])

(a, n) = Bn +1 (a)n + 1

(23)

where (a, s ) are Hurwitz Zeta functions dened by (a, s ) := k=01

(a + k)s(see [2,

p.249]) and Bn (x) are classical Bernoulli polynomials.Further, if we set a = 0 in (23), then we have the known formula (see [2, p.266,

Theorem 12.16]):

( n) = Bn +1

n + 1(24)

where (s) are Riemann Zeta functions dened by (s) := k =01ks

(see [2, p.249]) andBn := Bn (0) are classical Bernoulli numbers.

In this section, we will apply the Lerch functional equation to obtain the representa-tion of Apostol-Bernoulli polynomials B n (x, ) at rational points x =

pq

. Clearly, it is a

generalized form of [9, p.1529, Theorem A].

Theorem 3.1. For n N \ { 1}; pZ , qN , z C we have

B n pq

, e2iz = n!

(2q)n

q

j =1

z + j 1

q, n exp

n2

2(z + j 1) p

qi

+q

j =1

j z

q, n exp

n2

+2( j z) p

qi . (25)


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Proof 3.2. In view of the Lipschitz-Lerch Zeta functions (z,a,s ) (21), and using theknown series identity

k=1

f (k) =q

j =1

k =0

f (qk + j ), (qN ) (26)

we obtain that

(z,a,s ) = q sq

j =1

a + j 1

q, s exp(2(kq + j 1)iz ), (k N 0). (27)

Setting z =pq

yields

p

q,a ,s = q s

q

j =1

a + j 1

q, s exp

2( j 1) pi

q, ( pZ , qN ). (28)

On the other hand, by the Lerch functional equation (see [5, p.125, (29)]), for 0 < z 0 and B(r / a, w) is a beta function.

Distribution (5) is a special limit case of the Bessel distribution investigated by T.Srodka (1973). It was also analyzed by J. Seweryn (1986) and by W. Ogi nski (1979) and

was applied in reliability theory.Let us consider the GNBD (4) that depends on cy :

P (x; n,cy) =n

n + xn + x

x(cy)x (1 cy)n + x x , x = 0 , 1, 2, . . . , (6)

where 0 < cy < 1, n = 1, 2, . . . , cy < 1, 1, and Y is a random variable withGBD (5).

Theorem 3.3. The probability function of the compound distribution

GNBD Y

GBD

is given by the formula

P GB (x) = D

k=0

( c)kn + x x

k

(bw)ka B

x + r + ka

, w , (7)

where

D =n

n + x

n + x

xcx (bw)

xa

B(r/a,w ),

x = 0 , 1, 2, . . . ; a, b, w, r > 0, where cy < 1, 0 < cy < 1, n > 0, 1.

Proof. From formulas (3), (5) and (6) we have

P GB (x) = aD 1

(bw)1 /a

0

yx+ r 1 1 ya

bw

w 1

(1 cy)n + x xdy

= aD 1

k=0

( c)kn + x x

k

(bw)1 /a

0

yx+ r + k 1 1 ya

bw

w 1

dy,

where

D1 =n

n + x

n + x

xcx

(bw)r/a B(r/a, w ).


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T. Gerstenkorn / Central European Journal of Mathematics 2(4) 2004 527537 531

Substituting ya /bw = t, we get

P GB (x) = D

k=0

( c)kn + x x

k(bw)

ka

1

0

tx + r + k

a 1(1 t)w 1dt,

x = 0 , 1, 2, . . . , a, b, w, r > 0, n > 0, 1, 0 < c 1

(bw)1/a

and, from the denition of the beta function, we obtain (7). In the special case whenn, Nwe have a simpler formula given by

P GB (x) = Dn + x x

k=0

( c)kn + x x

k

(bw)ka B

x + r + ka

, w . (8)

Case (ii) of (4) the case of = 0 is discussed next in Section 3.2.

3.2 Special cases

In the case when = 0 (the binomial distribution) the proof of Theorem 3.3 is similarto, but even simpler than, the proof given above. More specically, we get

P GB (x) = D

n x

k=0

( c)kn x

k(bw)

ka B

x + r + ka

, w

where

D =

n

xcx (bw)

xa

B(r/a,w ).

Here we get a compound of the binomial distribution with the generalized beta dis-tribution:

binomial Y

generalized beta ( = 0)

P 0GB (x) =

n

xcx (bw)

xa

B(r/a,w )

n x

k=0

( c)kn x

k

(bw)ka B

x + r + ka

, w , (9)

x = 0 , 1, 2, . . . , n .

T. Gerstenkorn (1982, p. 90, (6)) obtained the same result. Moreover, Gerstenkorndetermined some special cases of this compoundcompounds of the binomial distributionwith the beta distribution, generalized gamma, exponential, arcsine, Erlang, 2, gamma,


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532 T. Gerstenkorn / Central European Journal of Mathematics 2(4) 2004 527537

Rayleigh, Maxwell, , normal truncated, and Weibull. In the Gerstenkorn paper, factorialmoments and ordinary moments of order l of the distributions are also determined.

From (8) we also have the compound of the generalized negative binomial with thebeta distribution

GNBD Y beta ( b = 1/ w, a = 1)

P B(x) =n ck

n + x

n + x

x

B(r, w )

k=0

( 1)kn + x x

kB(x + r + k, w).

It can be demonstrated that

k=0

( 1)kn + x x

kB(x + r + k, w) = B(x + r, w + n + x x)

holds. Hence, we get

P B(x) =n ck

n + xn + x

x

B(x + r, w + n + x x)B(r, w )

, (10)

a result that, for = 1 and c =1, yieldsnegative binomial

Y beta

P 1B(x) =n + x 1

x

B(x + r, w + n)B(r, w )

(11)

or, after using the formula

n [x,

1] = n(n + 1)( n + 2) . . . (n + x 1),

P 1B(x) =n [x, 1]r [x, 1]w[n, 1]

x!(r + w)[x+ n, 1].

Furthermore, a consequence of Theorem 3.3 is for = 1, n = 1 , 2, . . . the compounddistribution

Pascal binomial Y

generalized beta

P 1GB (x) =n + x 1

x

cx (bw) xaB(r/a,w )

n

k=0

( c)kn

k(bw)

ka B

x + r + ka

, w . (12)

Some special compounds, as mentioned before for binomialgeneralized beta, can beobtained for distribution (12).


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4 Factorial moments and ordinary (crude) moments of a com-pound of the negative binomial distribution with the gener-alized beta distribution

Denition 4.1. Let X y and X be random variables with distribution functions F (x|y)and H (x), respectively (see (1)), and let parameter y have distribution G(y). Then, whenone keeps in mind the formula for the so-called factorial polynomial

x[l] = x(x 1)(x 2) . . . (x (l 1)),

m[l] = E (X [l]) =

E (X [l]y )dG(y) (13)

is called a factorial moment of order l of the variable X with compound distribution (1).

Relation (13) will be symbolized as follows:

E (X [l]y ) Y G(y). (14)

Theorem 4.2. The factorial moment of order l of the compound distribution negativebinomial

Y generalized beta is given by the formula

mNB GB[l] = n[l, 1]

cl

(bw)l

a

B(r/a, w )

k=0

lk

( c)k(bw) ka B l + r + ka , w (15)

Proof. The factorial moment of order l of the negative binomial distribution is given bythe formula

mNB[l] (n, p) = pq

l

n [l, 1], 0 < p < 1, q = 1 p

(W. Dyczka (1973, p. 223, (63))). Consequently, by (14), the factorial moment of this

order of distribution (12) is, if we let p = cy, the following:mNB GB[l] = m

NB[l] (n,cy) Y GB (y; a,b,r,w )

=an [l, 1]

(bw)r/a B(r/a,w )

(bw)1 /a

0

cy1 cy

l

yr 1 1 ya

bw

w 1

dy

=an [l, 1]cl

(bw)r/a B(r/a,w )

k=0

l

k( c)k

(bw)1 /a

0

yl+ r + k 1 1 ya

bw

w 1

dy.

Substituting ya /bw = t, we get

m [l] =n [l, 1]cl(bw)

la

B(r/a,w )

k=0

l

k( c)k(bw)

ka

1

0

tl+ r + k

a 1(1 t)w 1dt,


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a result that yields (15).

A special case of Theorem 4.2 is the following theorem.

Theorem 4.3. The factorial moment of order l of the compound distribution negativebinomial Y

beta ( b = 1 /w, a = 1 , c = 1) is given by the formula (for w > l)

mNB B[l] = n[l, 1] B(l + r, w l)

B(r, w )(16)

or by the formula

mNB B[l] =n [l, 1]r [l, 1]

(w 1)[l]. (17)

Proof. In this case we have

mNB B[l] =n [l, 1]

B(r, w )

k=0

l

k( 1)kB(l + r + k, w).

It can be demonstrated that

k=0

lk

( 1)kB(l + r + k, w) = B(l + r, w l),

thus giving formula (16). Formula (17) is then evident.Further special cases can easily be obtained by taking account of the remarks included

in Section 3.

The ordinary (crude) moments of the compound distributions under consideration are

obtained by using the formulam l =

l

k=0

S lkm [k],

where S lk stands for the so-called Stirling numbers of the second kind. Bohlmann (1913)seems to be the rst to give this formula; the tables for these numbers can be found, forinstance, in A. Kaufmann (1968) or in J. Lukasiewicz and M. Warmus (1956).

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CEJM 2(4) 2004 538560

Zero-dimensional subschemes of ruled varieties

Edoardo Ballico 1, Cristiano Bocci 2, Claudio Fontanari 1

1 Department of Mathematics,University of Trento,

38050 Povo (TN), Italy 2 Department of Mathematics,University of Milano,

I-20133 Milano, Italy

Received 22 September 2003; accepted 23 August 2004

Abstract: Here we study zero-dimensional subschemes of ruled varieties, mainly Hirzebruchsurfaces and rational normal scrolls, by applying the Horace method and the Terracini method.c Central European Science Journals. All rights reserved.

Keywords: Ruled varieties, Hirzebruch surfaces, fat points, Horace lemma, Grassmann defectivevarietiesMSC (2000): 14N05

1 Introduction

Let Z Pr

be a closed subscheme. For any integer t, let Z,t,r be the restriction mapH 0(P r , OP r (t)) H 0(Z, OZ (t)). We say that Z has maximal rank if for every integert > 0 the linear map Z,t,r has maximal rank, i.e. it is either injective or surjective. Weare going to investigate this property for a zero-dimensional Z .A very classical motivation for such an interest is provided by the following

Denition 1.1. Let X P r be an integral nondegenerate projective variety of dimensionn . The h-secant variety Sech (X ) of X is the Zariski closure of the set { pP r : p lies in

This research is part of the T.A.S.C.A. project of I.N.d.A.M., supported by P.A.T. (Trento) andM.I.U.R. (Italy). E-mail: [email protected] E-mail: [email protected] E-mail: [email protected]


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the span of h + 1 independent points of X }. The expected dimension of Sech (X ) is

expdim Sech (X ) = min {(n + 1)( h + 1) 1, r }

and X is h-defective with h-defect h (X ) if

h (X ) = expdim Sec h (X ) dim Sech (X ) > 0.

A basic tool for understanding defective varieties is the classical Lemma of Terracini(see [33] for the original version and [18] or [1] for a modern proof), which says that, forP 1, . . . , P h +1 general points and P general in their span, one has

T P (Sech (X )) < T P 1 (X ), . . . , T P h +1 (X ) >

with equality holding in characteristic zero. Therefore, at least over the eld C , X is not h-defective if and only if the restriction map

Z, 1,r : H 0(P r , OP r (1)) H 0(Z, OZ (1))

has maximal rank, whereZ := h +1i=1 2P i X.

More generally, for Z := h +1i=1 m i P i X with m i 2, the restriction map Z, 1,r fails tohave maximal rank if and only if the union of the ( m i 1)-osculating space at P i fori = 1 , . . . , n spans a linear space of dimension less than the expected one. In other words,the linear system of hyperplane sections having a contact of order m i at P i has dimensiongreater than the expected one, i.e. it is a special linear system.

Here we focus on ruled varieties, by applying as an essential tool the so-called Horacemethod. We recall its set-up in Section 2, while in Section 3 we collect the main results of the present paper. In particular, Horace method provides a quick proof of the followingwell-known

Proposition 1.2. Fix integers x > 0, a, b 3. Let Z F 0 be a general union of x doublepoints of F 0. Then the restriction map Z,a,b : H 0(F 0, OF 0 (ah + bF )) H 0(Z, OZ (ah +bF )) has maximal rank.

We stress that the above result is sharp: indeed, it is easy to verify that homogeneousspecial systems arise when we consider the value a = 2. This fact is true not onlyfor F 0, but for F e with e 0. For example, consider the system L := |2h + (2 d +2e)F 2d+ e+1i=1 2P i | on F e , e 0; we have h0(F e , OF e (2h + (2 d + 2 e)F ) = 6 d + 3 e + 3,hence the virtual dimension of L is 1 then we would expect that L is empty. Sinceh0(F e , OF e (h +( d+ e)F ) = 2 d+ e+2, there is a unique curve C in the system |h +( d+ e)F |passing through P i , i = 1 , . . . , 2d + e + 1: its double 2 C is a divisor on F e of type2h + (2 d + 2 e)F with 2d + e + 1 double points, hence 2 C L and L is special.

We also remark that, at least in characteristic zero, Proposition 1.2 may be provedby using [5], Lemma 4.2, or [5], Proposition 4.1, [16], Lemma 3, [32], Lemma 2.2, and


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certain reducible curves with negative intersection with K F e . From this point of viewthe cases e = 0 , 1 are quite easier, because in this case K F e is ample. However, Horacemethod allows us to handle also fat points of higher multiplicity. For instance, we obtainthe following

Theorem 1.3. Fix integers b 4, a 11, s > 0 and m i , 1 i s, such that 1 m i 3for every i. Fix s general points P i F 0, 1 i s, and set Z := m i P i . Then therestriction map Z,a,b : H 0(F 0, OF 0 (ah + bF )) H 0(Z, OZ (ah + bF )) has maximal rank.

We do not have a unied way to handle all cases with 3 a 10 left open byTheorem 1.3, hence we omit the few cases with low a which can be easily checked by hand.Indeed, the paper [26] by Antonio Laface contains the complete list of all homogeneousspecial systems with multiplicity two or three. We point out that such a list gives someinformation also in the non-homogeneous case, just by applying the Horace method (seeProposition 3.6).

For F e , e > 0, the proof of Theorem 1.3 gives a similar, but weaker, statement(Theorem 3.11); roughly speaking, we have to replace b with b ea in our assumptions.

Our work suggests several open questions, which are collected in Section 4. In par-ticular, Question 4.4 should be viewed as the analogue for Hirzebruch surfaces of thecelebrated Segre and Harbourne-Hirschowitz conjectures for the projective plane.

Next, in Section 5 we consider higher dimensional scrolls and we apply the samemethods to the study of a zero-dimensional subscheme which is curvilinear, i.e. its Zariskitangent space at a reduced point has dimension at most one.

Finally, in Section 6 we turn to another geometrically meaningful kind of defectivity,the so-called Grassmann defectivity, which is dened as follows.

Denition 1.4. Let X P r be an integral nondegenerate projective variety of dimensionn dened over the eld C . The (k, h)-Grassmann secant variety Seck,h (X ) of X is theZariski closure of the set {l G (k, r ) : l lies in the span of h + 1 independent points of

X }. The expected dimension of Seck,h (X ) isexpdim Seck,h (X ) = min {(h + 1) n + ( k + 1)( h k), (k + 1)( r k)}

and X is (k, h)-defective with (k, h)-defect k,h (X ) if

k,h (X ) = expdim Sec k,h (X ) dim Seck,h (X ) > 0.

This property was classically investigated by Alessandro Terracini in his beautifulpaper [34], going back to 1915. Recent advances in the subject are due to Ciro Ciliberto,

Luca Chiantini, and Marc Coppens ([10], [11], [12], and [17]); see also the papers [19],[21], and [20], all three strongly inspired by the pioneering work of Terracini.Here we give a new application of the same method in order to answer the following

Question 1.5. (Ciliberto, [15]) Can a smooth projective variety be both defective andGrassmann defective?


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After describing a natural generalization of Terracinis approach, we derive explicitnumerical conditions in order to be Grassmann defective for a projective variety. A closerinspection to the case of rational normal scrolls makes it easy to produce innite seriesof smooth varieties (even surfaces!) being both defective and Grassmann-defective, for

instance:

Example 1.6. For every h 4 the smooth rational scroll

X (1, 2h + 2) P 2h +4

is both h-defective and (1 , h)-defective.

We work over an algebraically closed eld K of characteristic zero. Indeed, our proof

of Proposition 1.2 is characteristic free, but in the proof of Theorem 1.3 and in all relatedproofs we need char(K ) = 0 because we quote [13]. We point out that in the case p := char( K ) > 0 it would be sufficient to assume p > max{a, 4}, but we stress that wedo not know any characteristic free proof of Theorem 1.3.

2 Preliminaries

Let Y be any algebraic variety, Z Y a closed subscheme and D Y an effective Cartier

divisor of Y . For any P Y let 2P or 2{P, Y } be the closed subscheme of Y with I 2P,Y as

its ideal sheaf. Thus, if P Y reg , then length(2 {P, Y }) = 1+dim( Y ). The same notationwill be used when P is an arbitrary integral closed subvariety of Y . The residual schemeResD (Z ) of Z with respect to D is the closed subscheme of Y with Hom ( I D,Y , I Z,Y )as the ideal sheaf. Thus Res D (Z ) Z . If P D reg (and hence P Y reg ) we have2{P, Y } D = 2 {P, D } and Res D (2{P, Y }) = {P } (with its reduced structure).

For any L Pic( Y ) we have an exact sequence on Y :

0 I ResD

(Z ) ,Y L( D) I Z,Y L I Z D,D (L |D ) 0 (1)

From (1) we obtain the following very elementary form of Horace Lemma.

Lemma 2.1. For any L Pic( Y ) we have h0(Y, I Z,Y L) h0(Y, I ResD (Z ) ,Y L( D))+h0(D, I Z D,D (L |D )) and h1(Y, I Z,Y L) h1(Y, I ResD (Z ) ,Y L( D))+ h

1(D, I Z D,D (L |D )).

The next remark will be called the double residue trick.

Remark 2.2. Fix a projective variety W , a closed subscheme T of W , a line bundleL Pic( W ), an effective Cartier divisor D on W , a point P D reg with P /T red . LetZ W be a zero-dimensional scheme such that Z red = {P }. Set Z 1 := Z . For anyinteger i 2 dene inductively the scheme Z i by the formula Z i = Res D (Z i 1). HenceZ i Z i 1 with strict inequality if Z i 1 = . Set a i := length( Z i D). The non-increasing


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sequence (a1, . . . ) will be called the associated sequence of Z with respect to D. We havelength( Z ) = i 1 a i . Fix an integer i 1. By [4], Lemma 2.3 (see in particular Fig. 1 atp. 308) there is a scheme U Z whose associated sequence is obtained from the associatedsequence (a1, . . . ) of Z omitting the term a i and a subscheme E D with E red = {P }

and length( E ) = a i such that to check h1(W, I Z T L) = 0 (resp. h

0(W, I Z T L) = 0) it

is sufficient to prove h1(D, I (T D )E,D (L |D )) = h1(W, I ResD (T )U L( D)) = 0 (resp.h0(D, I (T D )E,D (L |D )) = h0(W, I ResD (T )U L( D)) = 0).If a1 = 3, a2 = 2, a3 = 1, a j = 0 for j > 3 and i = 3, then we will call the (1 , 3, 2)-trick this application of [4], Lemma 2.3.

Z = -(1,3,2)-trick

E =

U =

If a1 = 3, a2 = 2, a1 = 1, a j = 0 for j > 3 and i = 2, then we will call the (2 , 3, 1)-trick this application of [4], Lemma 2.3.

Z = -(2,3,1)-trick

E =

U =

If a1 = 2, a1 = 1, a j = 0 for j > 2 and i = 2, then we will call the (1 , 2)-trick thisapplication of [4], Lemma 2.3.

Z = -(1,2)-trickE =U =

Now assume that we want to prove h i (W, I B T L) = 0 ( i = 0 or i = 1) for a general zero-dimensional scheme B with B red = {P 1} and with associated sequence (3 , 1, 0, . . . ),where P 1 is a general point of a sufficiently general irreducible Cartier divisor D1; inour set-up, D1 will be a general ber of the ruling. Set M := Res D 1 (B). To prove thesought-for vanishing it is sufficient to prove h i (D1, I (T D 1 ){P },D 1(L |D 1 )) = h i (W, I T M L( D1)) = 0. This application of [4], Lemma 2.3, will be called the (1 , 3)-trick . Andso on for other associated sequences.

Example 2.3. Fix P D such that D is smooth at P , hence Y is smooth at P . LetZ Y be a zero-dimensional scheme such that Z red = {P }. Assume Z curvilinear, i.e.assume that the Zariski tangent space of Z at P has dimension at most one. There isa germ C at P of a smooth curve containing Z . Let be the intersection multiplicityof C with D at P . Set a := length( Z ). Hence Z is the divisor aP of C and henceOZ = OC / OC ( aP ). Set b := [a/ ] 0 and c = a b. We have a i (Z ) = for

1 i b, ab+1 (Z ) = c and a i (Z ) = 0 for all i b + 2.

Remark 2.4. Let Y be an irreducible variety, L Pic( Y ) and V H 0(Y, L) a nite-dimensional vector space. Assume V = {0}. For a general P Y we have V H 0(Y, I {P }L) = V and hence dim( V H 0(Y, I {P } L)) = dim( V ) 1. In several proofs, we are


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going to use this obvious remark without any further mention. By this observation, inall the statements of this paper we could assume m i 2 for every i without any lossof generality.

3 Ruled surfaces

Let F e , e 0, be the Hirzebruch surface with invariant e, i.e. such that e is the minimalself-intersection of a section of the ruling of F e . Hence F 0 = P 1 P 1 is isomorphic to asmooth quadric surface and F 1 is isomorphic to the blowing-up of P 2 at one point. Wex a ruling f : F e P 1; f is unique if and only if e > 0. We have Pic( F e) = Z2 and wetake, as a basis of Pic( F e), a section h of f with h2 = e and a class, F , of the ruling f .Thus h F = 1 and F 2 = 0. We will use both the additive and the multiplicative notationfor line bundles and divisors on F e . We have F e = 2h (2 + e)F , and it is easy tocheck h0(F e , OF e (ah + bF )) = 0 if a < 0. Moreover, by the projection formula we havef (OF e (ah + bF )) =

ai=0 OP 1 (b ie) for every a 0. Thus h0(F e , OF e (ah + bF )) = 0 if

a 0 and b < 0, h0(F e , OF e (ah + bF )) =t 1i=0 (b ie +1) if 0 (t 1)e b < te for some

integer t with 0 < t a, h0(F e , OF e (ah + bF )) =ai=0 (b ie+1) = (2 b+2 ae)(a +1) / 2 if

a 0 and b ae 1 and h1(F e , OF e (ah + bF )) = 0 if a 0 and b ae 1. Now we will seehow to translate any statement concerning the postulation of a zero-dimensional schemeZ F e in a statement concerning an interpolation problem for suitable polynomials (seeRemark 3.1 for the case e = 0 and Remark 3.2 for the case e > 0).

Remark 3.1. Fix integers 0 and 0. Since F 0 = P 1 P 1, the vector spaceH 0(F 0, OF e (h + F )) may be parametrized by the set of all bihomogeneous polynomialsin the variables x0, x1, y0, y1 whose monomials have degree with respect to x0, x1 anddegree with respect to y0, y1.

Remark 3.2. Fix integers e > 0, a 0 and b. Take variables z0, z1 and w and assign the

weight one to z0 and z1 and the weight e to w. Since f (OF e (ah + bF )) =ai=0 OP 1 (b ie),the vector space H 0(F e , OF e (ah + bF )) corresponds to the vector space of all weighted

homogeneous polynomials in z0, z1 and w with b as total weight degree and with degree a as a polynomial in w.

Remark 3.3. Fix integers e 0, a 0, b ea and t 0. Fix A R F e with R |F |,and A a zero-dimensional scheme such that length( A) = t. Since h1(F e , OF e (ah + ( b 1)F )) = 0, the restriction map : H 0(F e , OF e (ah + bF )) H 0(R, OR (a)) is surjective.We have h0(F e , I A (ah + bF )) = h0(F e , OF e (ah + bF )) min{t, a +1 }, hence h1(F e , I A (ah +bF )) = 0 if and only if t a + 1.

Lemma 3.4. Let Z F 0 be the general union of 5 (resp. 6) double points. Thenthe restriction map H 0(F 0, OF 0 (3h + 3 F )) H 0(Z, OZ (3h + 3 F )) is surjective (resp.injective).


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Proof 3.5. We will do only the surjectivity part, the injectivity part being similar andeasier. Let T F 0 be a general element of |h + 2 F |. Hence T = P 1. Notice thatT (3h+3 F ) = 9 and h1(F 0, OF 0 (2h+ F )) = 0. Hence the restriction map H 0(F 0, OF 0 (3h+3F )) H 0(T, OT (3h + 3 F )) is surjective. Let Z F 0 and union of 5 double points with

Z red T . Thus length( Z T ) = 10. By Horace Lemma we have hi

(F 0, I Z (3h + 3 F )) =h i (F 0, I Z red (2h + F )), i = 0 , 1. Since there is a unique C |2h + F | containing Z red , weare done.

Proof of Proposition 1.2. If 3x (a +1)( b+1) 1 we need to prove the surjectivity of Z,a,b . If 3x > (a +1)( b+ 1) 1 we need to prove the injectivity of Z,a,b . We will only dothe case 3x (a +1)( b+ 1) 1, the case 3x > (a +1)( b+ 1) 1 requiring only notationalmodications. Hence we assume 3 x (a + 1)( b+ 1) 1. Just taking a suitable union of more double points, it is sufficient to prove the case ( a+1)( b+1) 3 3x (a+1)( b+1) 1. We will assume that these inequalities are satised. If a = b = 3, then x = 5 and theresult follows from Lemma 3.4. Hence we may assume a + b 7 and, say, b a. Suppose,for the moment, b 5. Thus, by induction on the integer a + b, we may assume that theresult is true for the pair ( a, b 1) and the pair ( a, b 2), b 5. Fix D |F |. First assumea odd. Let W F 0 be a general union of x (a +1) / 2 double points of F 0. Let Z F 0 bea general union of (a+1) / 2 double points of F 0 with (Z )red D. Thus Res D (Z ) = ( Z )redand length( Z D) = a + 1 = h0(D, OD (ah + bF )). By Horace Lemma 2.1 it is sufficientto show h1(F 0, I W (Z ) red (ah + ( b 1)F )) = 0. By the inductive assumption and theinequality 3( x (a + 1) / 2) (a + 1) b 1 we have h1(F 0, I W (ah + ( b 1)F )) = 0. Noticethat ( a + 1) / 2 + 3(x (a + 1) / 2) = 3 x (a + 1)( b + 1) + ( a + 1) b. Hence to showthat h1(F 0, I W (Z ) red (ah + ( b 1)F )) = 0 and the generality of ( Z )red it is sufficient toshow that h0(F 0, I W D (ah + ( b 1)F )) h0(F 0, I W (ah + ( b 1)F )) (a + 1) / 2. Sinceh0(F 0, I W D (ah +( b 1)F )) = h0(F 0, I W (ah +( b 2)F )), it is sufficient to use the inductiveassumption. Indeed, in all cases with 3 x (a + 1)( b + 1) 3, the inductive assumptionimplies h0(F 0, I W (ah +( b 2)F )) = 0. Now assume a even. Let A F 0 be a general unionof x a/ 2 1 double points of F 0. Take a general union B of a/ 2 double points of F 0 withsupport on D and a general P D. Apply the double residue trick (Remark 2.2) withrespect to P . In order to conclude, it is sufficient to prove h1(F 0, I AB red 2{P,D }(ah + ( b1)F )) = 0. Indeed, we have h1(F 0, I 2{P,D }(ah +( b 1)F )) = 0 because a > 0 and b 1 > 0,and h1(F 0, I B red 2{P,D }(ah + ( b 1)F )) = 0 because h1(F 0, I 2{P,D }(ah + ( b 1)F )) = 0,h0(F 0, I 2{P,D }(ah + ( b 1)F )) card( B red ) and B red is general in D. Hence we see thath1(F 0, I AB red 2{P,D }(ah + ( b 1)F )) = 0 because h1(F 0, I B red 2{P,D }(ah + ( b 1)F )) = 0,h0(F 0, I B red 2{P,D }(ah + ( b 1)F )) = ( a + 1) b card( B red ) 3 card( A) and A is generalin F 0.

Consider now the case b = 4. We can restrict our attention to 3 a b (in fact, if a > b = 4 then we can apply the previous argument replacing a by b and b by a). Thusthe surjectivity of the map Z,a,b is given by h1(F 0, I Z (ah + 4 F )) = 0 where Z = xi=1 2P iwith 5a + 2 3x 5a + 4 and a 4. By the same argument of Lemma 3.4 or by adirect computation (for example with a computer algebra system) we see that the systems


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I 6i =1 2P i (3h + 4 F ) and I 8i =1 2P i (4h + 4 F ) are non-special and the claim follows.

Proof of Theorem 1.3. Let Z = s 3i=1 3P i s 3 + s 2i= s 3 +1 2P i be a zero-dimensional scheme of

fat points, where s3 and s2 are respectively the number of triple points and double points.

By Laface we can easily treat the cases s2 = 0 or s3 = 0. In fact, if s2 = 0, then Z is theunion of triple points and I Z (ah + bF ) is never special if a, b 4. When s3 = 0, Z is theunion of double points. In this situation I Z (ah + bF ) is never special if a, b 3.

Notice that we will never be able use induction on the parameters appearing in thestatement of Theorem 1.3. We will just try to control the postulation of a certain zero-dimensional scheme in nitely many steps; if at one step, we cannot go on, then wefailed. If all the steps can be done and is exhausted (resp. the line bundle OF 0 (ah + bF )is exhausted), then we obtain h1(F 0, I (ah + bF )) = 0 (resp. h0(F 0, I (ah + bF )) = 0).

By semicontinuity we will obtain Theorem 1.3 for certain data s, m i , 1 i s, relatedto . Even more, instead of is some of the steps we will use a virtual scheme and applyRemark 2.2.

Step 1: First of all we study the case b = 4 and a arbitrary. We consider the case withs3 > 0 and s2 > 0 and we may assume 5( a + 1) 2 6s3 + 3 s2 5(a + 1) + 2 (allother cases are reduced to this one). We assume s3 2. We x a ber A and twodifferent points P, Q A. We use 3P and the (2 , 3, 1)-trick with respect to Q. Sincelength( Z A) = 3 + 2 we have h i ( I Z A (4F )) = 0, for i = 0 , 1. Thus h i ( I Z (ah + 4 F )) =

hi( I ResA (Z )((a 1)h + 4 F )). The residual scheme Z := Res A (Z ) has s3 1 triple

points, s2 double points, another double points 2 P and a virtual scheme Q with layers(3, 1). Moreover, we observe that P and Q are general in A. Since on A we haveagain a virtual scheme of length 2 + 3 = 5, then h i ( I Z A (4F )) = 0, for i = 0 , 1 andh i ( I Z (ah + 4 F )) = h i ( I Z ((a 2)h + 4 F )). Z := Res A Z consists of s3 2 triple points,s2 double points and 2 simple points.

If s3 is an even number we apply the previous argument s3 times and we obtainh i ( I Z (ah + 4 F )) = h i ( I Z ( s 3 ) ((a s3)h + 4 F )), where Z (s 3 ) = N M with N union of s2

double points and M union of s3 simple points. One big inductive step consists in thereduction of an assertion on |(a x)h+4 F |, where x is an even integer with 0 x s3 2,to an assertion on |(a x 2)h + 4 F | by making two smaller inductive steps, rst from|(a x)h+4 F | to |(a x 1)h+4 F | and then from |(a x 1)h+4 F | to |(a x 2)h+4 F |,each time using A |h| . In this part of the proof, for every big inductive step we take adifferent Ai general in |h| , 1 i s3/ 2. Since the points in N are general, from [26] itfollows that h1(F 0, I N ((a s3)h +4 F )) = 0; we also have h0(F 0, I N ((a s3)h +4 F )) s2.This inequality would be sufficient to obtain h1(F 0, I N M ((a s3)h+4 F )) = 0 if the pointsin M were general. Unluckily this is not the case, but the only restriction on M is theexistence of s3/ 2 general |h |, 1 i s3/ 2, such that card( Ai M ) = 2 for all i; inparticular, the points in M i := M Ai are general in Ai . Dene M i := i j =1 M j and setM 0 := . We claim that if h1(F 0, I N M i ((a s3)h +4 F )) = 0 for some i {0, . . . , s 3/ 2},then also h1(F 0, I N M i +1 ((a s3)h + 4 F )) = 0. Indeed, the claim follows from theinequality h0(F 0, I N M i ((a s3)h + 4 F )( Ai+1 )) 2 and the fact that M i+1 is general


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in Ai+1 . From this claim we inductively obtain h1(F 0, I Z ( s 3 ) ((a s3)h + 4 F )) = 0, asdesired.

If s3 is an odd number we apply the previous argument s3 1 times. Then we applyan Horace method using the last triple point and a double point. Again, the length on

A is 3 + 2 = 5 and we obtain hi

( I Z (ah + 4F )) = hi

( I Z ( s 3 ) ((a s3)h + 4F )), where Z (s

3)

consists of s2 double points and s3 simple points.Finally, when s3 = 1 we just apply an Horace method using the triple point and a

double point obtaining again h i ( I Z (ah + 4 F )) = h i ( I Z (1) ((a 1)h + 4 F )), where Z (1)

consists of s2 double points and 1 simple points.If a s3 then Z has maximal rank for bidegree ( a, 4). Moreover, by the assumption

6s3 + 3 s2 5(a + 1) + 2, we see that a < s 3 only if a 10.

Step 2: Fix an effective divisor R |F | and P 1 R. We assume the existence of

an integer k such that 1 k s and integers j i , 1 i k, 1 j i s, j 1 = 1, j i = j r if i = r , such that ki=1 m j i = a + 1. Choose P j i R, with P j i = P j r if i = r . Choose general P j F 0 for j = { j1, . . . , j k }. Set Z 1 := ki=1 m j i P j i and Z 2 = j = { j 1 ,...,j k }m j P j . By semicontinuity it is sufficient to show h1(F 0, I Z 1Z 2 (ah + bF )) = 0.Set Z := ki=1 (m j i 1)P j i Z 2. Thus Z = Res R (Z 1 Z 2). By Horace Lemma we have

h i (F 0, I Z 1Z 2 (ah + bF )) = hi (F 0, I Z (ah + ( b 1)F )) i = 0 , 1

and hence we may continue taking again R as Cartier divisor and trying to insert someof the points P j , j = { j1, . . . , j k }, into R. Notice that the set ( Z 2)red is general in F 0, butthat the points P j 1 , . . . , P j k are not general in F 0.

Step 3: Now assume that there are no such integers 1 k s, j i , 1 i k, 1 j i s, j1 = 1, j i = j r if i = j , such that

ki=1 m j i = a + 1. We take the integers k and j i so that

ki=1 m j i a +1 and

ki=1 m j i is maximal under this restriction. We claim that we may

apply the double residue trick (Remark 2.2) with respect to many P j with j = { j1, . . . , j k }so that the total sum of the new virtual intersection with R is again a +1. If k = s, thenwe cannot do that, but in this subcase we immediately get h1(F 0, I Z (ah + bF )) = 0 byHorace Lemma and the assumption m i 3 b + 1.

If k < s by the maximality of ki=1 m j i and the assumption m i 3 for all i we have

a 1 k

i=1

m j i a.

If outside R we still have a fat point with multiplicity three, then we may alwaysincrease by one (case ki=1 m j i = a) or by two (case

ki=1 m j i = a 1) as we want the

length of the restriction to R of a subscheme by using Remark 2.2 and respectively the(1, 3, 2)-trick and the (2 , 3, 1)-trick.

If outside R we have a double point, then we use the (1 , 2)-trick to increase by onethe length on R and we specialize it to a double point 2 Q, Q general in R, to increase bytwo the length on R. In each step we apply the double residue trick only at one pointbecause m i 3. Furthermore, at each step we need to consider at most one scheme


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having support on R and coming from a double residue trick of some previous steps,because m i 3 for every i (and, indeed, if there is any such scheme, then it comes fromthe previous step).

More in detail we obtain the following situations. Let k3 and k2 be respectively the

number of triple points and double points such that k = k2 + k3. Case ki=1 m j i = a 1

i) If (k3 = s3) or (k3 s3 and k2 < s 2) then we specialize a double point Q toR. The residual scheme consists of s3 k3 triple points, s2 + k3 k2 1 doublepoints, k2 + 1 simple points.

ii) If (k3 < s 3 and s2 = k2) then we use the (2 , 3, 1)-trick on a triple point Q. Theresidual scheme consists of s3 k3 1 triple points, s2 + k3 k2 double points,k2 simple points and a virtual scheme with layers (3 , 1).

Casek

i=1 m j i = aiii) If (k3 = s3) or (k3 s3 and k2 < s 2) then we use the (1 , 2)-trick on a doublepoint Q. The residual scheme consists of s3 k3 triple points, s2 + k3 k2 1double points, k2 simple points and a virtual scheme with layers (2).

iv) If (k3 < s 3 and s2 = k2) then we use the (1 , 3, 2)-trick on a triple point Q. Theresidual scheme consists of s3 k3 1 triple points, s2 + k3 k2 double points,k2 simple points and a virtual scheme with layers (3 , 2).

Step 4: We use the set-up of Remark 2.2. If we apply the (1 , 3)-trick or the (1 , 3, 2)-

trick (resp., (1 , 2)-trick, (2 , 3)-trick, or (1 , 2)-trick) at the next step (i.e. for the integerb := b 1) we obtain a connected virtual scheme whose intersection with R has lengthlarger by two (resp., by one) than the one of the original scheme. If P R and m > 0,ResR (mP ) = ( m 1)P (with the convention 0 P = ), hence length( R ResR (mP )) =m 1 = length( R mP ) 1. Since in our set-up we may always take m 3 anda + 1 4 3, we are always sure that at the next step we have a virtual scheme whoseintersection with R has length at most a + 1. It follows that we can always apply thegeneral machinery of Step 3 also with the integer b < b.

Step 5: After b 4 steps we arrive at b = 4. We are going to mimic the proof of the case b = 4 given in Step 1 with suitable modications. In the present case wehave a scheme Z = Z 1 Z 2 Z 3 with Z 1 union of general fat points with multiplicity1, 2 or 3; Z 2 union of general fat points with multiplicity 1 or 2, supported by pointsof R; Z 3 either the empty set or the disjoint union of at most 2 schemes each of whichbelong to the following list: (2 , 0, . . .); (3, 1, 0, . . .); (3, 0, . . .); (3, 2, 0, . . .), where we havespecied the length of the intersection with R obtained via residual schemes. Furthermore,we have length( Z 2 Z 3) 2a 1 and length( R (Z 2 Z 3)) a; notice also thatlength( A (Z 2 Z 3)) 2 for any A |h| and length( A (Z 2 Z 3)) = 2 if and only if atthe point A R the scheme Z 2Z 3 is either a double point or a point of type (3 , 1, 0, . . .)with respect to R.

If length( A (Z 2Z 3)) = 2, then we insert a triple point on A. Let U be the union of all Ai |h | such that length( Ai (Z 2Z 3)) = 2. Since outside U (Z 2Z 3)) the schemeZ 2Z 3 (not just its reduction) is contained in R, it follows that E := Res U (Z 2Z 3) R.


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We claim that if h1(F 0, I S (eh + 4 F )) = 0 for a suitable scheme S Z 1 Z 2 Z 3 anda suitable e a, then h1(F 0, I S E (eh + 4 F )) = 0. Indeed, by [13] applied to the linearsystem induced by H 0(F 0, I S (eh + 4 F )) on R, for general E red we have h0(F 0, I S E (eh +4F )) = max {0, h0(F 0, I S (eh + 4 F )) h0(F 0, I S (eh + 4 F )( R))}. Hence in order to

check h1(F 0, I S E (eh + 4 F )) = 0 it is sufficient to prove that h

0(F 0, I S (eh + 4 F )( R))

h0(F 0, I S (eh + 4 F )) length( E ). Since OF 0 (eh + 4 F )( R) = OF 0 ((e 1)h + 4 F ), weget the above inequality by a standard inductive argument (we refer to the proof of Proposition 1.2 and to Step 1 for more details in a similar situation).

If length( A (Z 2 Z 3)) = 1, then the connected component C of Z 2 Z 3 withC red = A R is contained in R. Set := length( C ) and notice that 3. If = 1, thenwe insert a triple point on A and simultaneously apply the (1 , 3, 2)-trick. After this smallstep, there is a length 5 scheme on A (a length 2 connected component comes from the

triple point, a length 3 connected component comes from the (1 , 3, 2)-trick). By applyingHorace Lemma 2.1 we reduce to a case with a length 3 scheme on A (one reduced pointcoming from the triple point and one length 2 connected component coming from the(1, 3, 2)-trick). Then either we apply a (2 , 3, 1)-trick or (if no triple point is left) we inserton A a double point. Assume now = 2. In this case, we insert a triple point on Aand simultaneously apply the (1 , 2, 3)-trick. After this small step, there is a length 5scheme on A, hence by applying Horace Lemma 2.1 we can reduce to a case with on Aa connected scheme of length 1 (coming from the triple point) and a connected schemeof length 3 (coming from the (1 , 2, 3)-trick). Then we apply either the (1 , 3, 2)-trick or(if no triple point is left) the (1 , 2)-trick. Finally, in the case = 3, we insert a triplepoint on A and simultaneously apply the (1 , 2, 3)-trick. After this small step, there is alength 5 scheme on A (one reduced component coming from C , one length 2 connectedcomponent coming from the triple point and another one of length 2 coming from the(1, 2, 3)-trick). By Horace Lemma 2.1 we reduce to a case with a length 5 scheme on A(one reduced point coming from C , another reduced point coming from the triple pointand one length 3 connected component coming from the (1 , 2, 3)-trick). After applyingHorace Lemma 2.1 once again, the proof is over.

Next we classify all exceptional cases on F 0 with m i 3 for all indices i, m i = 2 forat least one index i and m i = 3 for at least one index i (see [26] for the other exceptionalcases with m i 3).

Proposition 3.6. Fix integers b 2, u > 0, c > 0. take u + c general points P i F 0,1 i u + c, and set Z := ui=1 3P i

u + c j = u +1 2P i . We have h0(F 0, I Z (2h + bF ))

h1(F 0, I Z (2h + bF )) = 0 if and only if b = 2 u + c 1 and u + c 1 0 (mod 2).

Proof 3.7. Set W := u + ci=1 2P i . Apply Horace Lemma u times with respect to the divisorsR i |OF 0 (0, 1)| such that P i R i , 1 i u. Since W = Res R 1R u (Z ), we obtain bothh0(F 0, I Z (2h+ bF )) = h0(F 0, I W (2h+( b u)F )) and h1(F 0, I Z (2h+ bF )) = h1(F 0, I W (2h+(b u)F )). Thus we restrict to analyze the homogeneous system I W (2h + ( b u)F ). By[26], this system is special if it has the form I T (2h + (2 d)F ) with T = 2d+1i=1 2P i . This can


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happen only if we require 2d = b u and 2d + 1 = u + c or equivalently, b = 2 u + c 1and u + c 1 0 (mod 2).

The proof of Theorem 1.3 may be applied to many other cases. We give, for example,

the following result.

Theorem 3.8. Fix integers x 0, c j > 0, 1 j x, b 4 + x j =1 c j a 11 +x j =1 c j ,

s > 0 and m i , 1 i s, such that 1 m i 3 for every i. Fix s + x general pointsP i , Q j F 0, 1 i s, 1 j x and set Z := si=1 m i P ix j =1 c j Q j . Then the restrictionmap Z,a,b : H 0(F 0, OF 0 (ah + bF )) H 0(Z, OZ (ah + bF )) has maximal rank.

Proof 3.9. The case x = 0 is just Theorem 1.3. Fix R |F | and take as Q j x generalpoints of R. Hence ResR (x j =1 c j Q j ) =

x j =1 (c j 1)Q j (with the convention 0 P j = ) and

length( R (x j =1 c j Q j )) = c1 + + cx . Then we repeat the proof of Theorem 1.3 takingthis ber R to apply Horace Lemma (see the details of Step 4). After at most x j =1 c jsteps we have exhausted all fat points with multiplicity 4, hence we are again in theset-up of Theorem 1.3. We never use the double residue trick with respect to any pointQ j , but only with respect to the other s points. Hence in our new situation the samebounds apply.

Our method can also be applied in several situations in which the points are not

general. In this way one obtains results for complete linear systems on a (non-general)blown - up surface. Here we give only a very easy example. The interested readermay construct in the same way thousands of other examples. Following the proof of Theorem 1.3 it is easy to construct examples with maximal rank.

Example 3.10. Fix integers e 2, b > e and y such that 1 y b. Fix y distinct pointsP i F e \ h such that f (P i ) = f (P j ) if i = j and set Z := yi=1 2P i , T :=

yi=1 f 1(f (P i )).

We have h0(F e , OF e (h+ zF )) = z+1 if 0 z e 1, while h0(F e , OF e (h+ zF )) = 2 z e+2

if z e. Now assume y > b e and 2y b. Since ResT (Z ) = y

i=1 {P i }, every m H 0(F e , I Z (h+ bf )) vanishes on T , i.e. h0(F e , I Z (h+ bF )) = h0(F e , I si =1 {P i }(h+( b y)F )) >h0(F e , OF e (h + bF )) 3y.

The proofs of Theorem 1.3 and Theorem 3.8 give the following result.

Theorem 3.11. Fix integers e > 0, s > 0, 1 i s, m i , a 11 and b such thatb 4 + ae > 0, 1 m i 3 for every i and si=1 m i (m i + 1) / 2 (b ae 1)(a + 1). Fixgeneral P i F e and set Z := si=1 m i P i . Then h1(F e , I Z (ah + bF )) = 0.

Indeed, just use |F | as a ruling and take the curves A, Ai in |h + eF | instead of |h|;notice however that the prescribed bound is certainly not sharp for e > 0.

We also obtain the following result.


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Proposition 3.12. Fix integers e > 0, s > 0, a > 0, b ae, m i , 1 i s. Set := max 1 i s {m i } and assume

si=1 m i (m i + 1) / 2 (b ea 1)(a + 1 ). Fix general

P i F e , 1 i s, and set Z := si=1 m i P i . Then h1(F e , I Z (ah + bF )) = 0.

Proof 3.13. Fix an effective divisor R |F |. We take the integers k and j i , 1 i k,so that ki=1 m j i a + 1 andki=1 m j i is maximal with this restriction. We have

a + 1 ki=1 m j i a + 1 by the denition of . Choose Q j i R, with Q j i = Q j rif i = r . Choose general O j F e for j = { j1, . . . , j k }. Set Z 1 := ki=1 m j i Q j i and Z 2 = j = { j 1 ,...,j k }m j O j . By semicontinuity it is sufficient to show h1(F 0, I Z 1Z 2 (ah + bF )) = 0.Set Z := ki=1 (m j i 1)Q j i Z 2. Thus Z = Res R (Z 1 Z 2). By Horace Lemma and theinequality ki=1 m j i a + 1 it is sufficient to prove h1(F e , I Z (ah + ( b 1)F )) = 0. Inorder to do so, we may continue taking again R as Cartier divisor and specializing ontoR the support of some of the fat points m j O j , j = { j

1, . . . , j k }. Since s

i=1m i (m i +

1)/ 2 (b ea 1)(a + 1 ), after at most b ea 1 steps we have inserted all zero-dimensional schemes. Hence with another step the claim is reduced to the vanishing of h1(F e , OF e (ah + zF )) for some z b ea 1 and the proof is over.

The same proof gives the following result.

Proposition 3.14. Fix integers s > 0, b a > 0, m i > 0, 1 i s. If s = 1, thenset := 0 and assume m1 a + 1. If s 2, then set := max 2 i s {m i} and assume

m1 a + 1. Assume si=1 m i (m i + 1) / 2 (b + 1)( a + 1 ). Fix general P i F 0,1 i s, and set Z := si=1 m i P i . Then h1(F 0, I Z (ah + bF )) = 0.

Finally we turn to more general ruled varieties. Let C be a smooth projective curveof genus g 0 and E a rank n vector bundle on C . Set X := P (E ) and call f : X C the natural projection. If L Pic( X ), then there is a unique line bundle R on C suchthat L = OX (t) f (R). Assume h1(C, E R) = 0 and set (L) the maximal integerb 0 such that h1(C, E R( bP )) = 0 (i.e. h0(C, E R( bP )) = h0(C, E R) bn)

for a general P C .The proof of Proposition 3.12 gives verbatim the following results.

Theorem 3.15. Fix a ruled surface X = P (E ) and L Pic( X ) with L = OX (a)f (R).Assume a 11 and h1(C, E R) = 0. Fix integers s > 0 and m i , 1 i s, such that1 m i 3. Choose general P i X , 1 i s, and set Z = si=1 m i P i . Assume

si=1 m i (m i + 1) / 2 (a + 1) (L). Then h1(X, I Z L) = 0.

Proposition 3.16. Fix a ruled surface X = P (E ) and L Pic( X ) with L = OX (a) f (R). Assume a > 0 and h1(C, E R) = 0. Fix integers s > 0 and m i > 0, 1 i s,and general P i X , 1 i s. Set Z := si=1 m i P i . If s = 1, set = 0 and assumem1 a + 1. If s 2, set := max 2 i s {m i }. Assume m1 a + 1 and

si=1 m i (m i + 1) / 2 (a + 1 ) (L). Then h1(X, I Z L) = 0.


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Remark 3.17. Let X P r be an integral variety and P 1, . . . , P h general points of X .By the characteristic free part of Terracinis lemma ([1], part (1) of Corollary 1.11, or[18]) if Z, 1,r has maximal rank, then X is not h-defective, the converse being in generaltrue only in characteristic zero. Hence Horace method can be used to prove the non

h-defectivity of certain projective varieties even in positive characteristic.

4 Open questions

In this section we raise two questions concerning the postulation of general unions of fatpoints on a scroll over a curve.Let C be a smooth projective curve of genus g 0 and E a rank n vector bundle on C .Set X := P (E ) and call f : X C the natural projection. Let OX (1) the tautologicalquotient line bundle on X , i.e. the only line bundle on X such that f (OX (1)) = E .The restriction of OX (1) to any ber F = P n 1 of f has degree one. We have Pic( X ) =f (Pic( C )) Z OX (1). For any R Pic( C ) and any t 0 we have f (OX (t)f (R)) =S t (E ) R. For all integers i, x with 1 i n 2 we have R i f (OX (x) f (R)) = 0.For every integer b n + 1 we have Rn 1f (OX (b) f (R)) = 0.

Remark 4.1. Fix positive integers n, m and t with n 2. If t m 1, set (n,t ,m ) :=n + m 1

n . Hence if t m 1 then the integer (n,t,m ) is the length of a fat point mP

of multiplicity m on any n-fold. Let X = P (E ) be an n-dimensional scroll over a smoothcurve and L Pic( X ) such that L |F = OF (t) for any ber F of the ruling of X . Fixany P X and let F be the ber of the

ivic divisor problem

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