Edoardo Ballico ON THE POSTULATION OF DISJOINT RATIONAL

REND. SEM. MAT.
UNIVERS. POLITECN. TORINO
Vol. 44°, 2 (1986)
Edoardo Ballico
ON THE POSTULATION OF DISJOINT RATIONAL CURVES
IN A PROJECTIVE SPACE
Riassunto. Sia C C P 3 una curva; C si dice di rango massimo se per ogni k&*l Vapplicazione di restrizione rc(k): H°(W3, 0(k))r+H°(C, 0c(k)) e iniettiva o surgettiva.
In questo lav or o si dimostra che per ogni scelta degli interi positivi s, dif..., ds esiste una curva C di rango massimo, C unione di s curve razionali disgiunte di grado
d\,...,ds
eccetto che nei seguenti 5 cash 1) s = 2, d^— d2 = 2; 2) 5 = 2,
di = 4, d2 = 2;
3) s = 3, dt = d2 = d3 = 2;
4) s = 3, dx = 4, d2 = d3 = 2;
5) s — 4, di = d2 — d$ = ^4 = 2.
Introduction.
In [6] and [8] Hartshorne and Hirschowitz gave a new method to prove
that the postulation of many curves in 1PW is the best possible one. We say
that a closed subscheme C C Wn has maximal rank if for every integer k > 1
the restriction map rc(k):H°(Wn> 0T (k))—>H°(C, 0c(k)) has maximal
rank as a map of vector spaces i.e. if it is injective or surjective. Consider in
particular a curve C with 0 C (1) not special; if C has maximal rank, then
C is contained in a hypersurfece of degree k if and only if (n
) > X (^c (^))*
In [6J Hartshorne and Hirschowitz proved that for all integers n> 3,
d > 1, the union of d general lines in IP„ has maximal rank.
Here we work over an algebraically closed field.
Here we prove the following theorem.
Classificazioneper soggetto AMS (MOS, 1980): Primary 14H45, Secondary 14N05
208
THEOREM l. Fix positive integers s, dlf..., ds; exclude the following cases:
1) s = 2, dx — d2 = 2; 2) s = 2, dj = 4, d2 = 2; 3) s = 3, d± = d2 — d^ — 2;
4) 5= 3, ^ i = 4 , d2 = ^ 3 = 2; 5) 5 = 4, d1=d2=d3=d^
= 2.
Then the union in JP3 o / 5 disjoint rational curves of degree d1,...,ds
general position has maximal rank.
in
The exceptional cases really occur. Indeed the union in P 3 of 2 disjoint
conies is contained in a reducible quadric. The union of any rational curve of
degree 4 and any disjoint conic (or the union of 3 conies) is contained in a reducible cubic surface. The union of a rational quartic and two disjoint conies
(or the union of 4 conies) is contained in a reducible quartic surface.
In a joint paper with Ph. Ellia [3] it was proved the corresponding result
in F„ for every n > 4 (without any exceptional case). Essentially for numerical reasons the case of IP3 considered here is harder and longer.
We use in an essential way reducible curves and curves with nilpotents,
not only for the inductive steps as in [6], [8], [1], but also to construct curves
with the connected components of degrees exactly dlf...,ds.
Consider a
chain T of lines L j , . . . , ! ^ , d = dx 4- ... 4- ds, in IP3 with L z nL ; =? fc 0 if
and only | i — / ' | < 1 . Consider the union Y of T and 5—1 nilpotents
X(Pr)> rr=l,... ,s— 1, where Pr is a singular point of T and x(^V) is the
first infinitesimal neighbourhood of Pr in 1P3. Then Y is a limit of a
family of s disjoint rational curves C,-, 1 = 1,... ,5. Choosing carefully the
points Pr we may assume deg C(- = dj for every i. If T has good postulation and Sing (T) lies in a smooth quadric, we may hope to prove that Y has
good postulation using the method introduced in [6] and [8]. Even this trick
of "adding nilpotents" was used in [6] and [8]. The main technical improvement with respect to [6] is in the intersection with the quadric (see § 3). In
general we have simply push the methods of [6] forward until we have obtained theorem 1.
Of course, one can try to obtain similar results for union of rational and
elliptic curves, and so on. Nonsense. As far as we know from the quotations
°f [1] ([7] and [10]) this type of results can be used mainly for the construction ot specific examples, in which the degrees are fixed and rather low. If
anybody really needs such an example, probably he will be able to prove its
existence by hinself; I hope that the tricks used in this paper will be useful.
The general picture which arises from this paper is that such examples
exist (if the degrees are big with respect to the genus) except in a relatively
small number of cases. These exceptional cases arise from geometrical reasons,
although the proofs break down for numerical reasons; the numerical reasons
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may help to detect the exceptional cases. Furthermore in many cases in the
proof of theorem 1 we prove the existence of curves of maximal rank and
whose irreducible components are lines. The euristic rule seems to be that if
you are not able to prove the existence of reducible curves with fixed degree,
genus and maximal rank, then probably there is no smooth curve with the
same degree and genus and with maximal rank. For an application of this
"rule", see [4]: no hyperelliptic curve of degree 7 and genus 3 in 1P3 has
maximal rank.
In the appendix we prove the existence of many curves not of maximal
rank. In particular we prove the following proposition:
PROPOSITION. Fix integers d,n with n +
Kd<n2/6.
a) There exists a union Y of d disjoint lines with h1(F3, J>Y{n - 1)) =£ 0,
h'iWs, JY(n)) = 0.
b) / / n > 4 and d^n
deg Y = d.
+ 2, the same holds for a smooth rational curve Y,
Note that if d<(n- l) 2 /6, then Y has not maximal rank.
For the smoothness in b) we have to use results, ideas and proofs of [2].
We prove a) also in IPN, JV>4, and we obtain that for every d we find a
union Y of d disjoint lines in IP^ such that the first integer k with
rY(k) surjective is "arbitrary": the only constraints are the Castelnuovo?s,
type theorem proved in [5] and Riemann-Roch.
I want to thank S. Greco (who detected an error in an early version of
3.5) and Ph. Ellia: a good definition is worth many theorems.
1. Notations.
Unless otherwise stated (for example in the appendix) every scheme is
contained in 1P3. For every subscheme Y of X, J>Y x *s t n e ideal sheaf of
Y in X; we will write J?Y if X — IP3 or (but only in § 3) if X is a smooth
quadric surface Q. We will write H\J>Y(ri)), h\J>Y(n)) instead of H%(IP3,
J>Y(ri)) or £*(IP3, J'yin)). If Y is a closed subscheme of P w , let rYW (k):
H°(1PM, 0YJPn(k))—+H°(Y, 0Y(k)) be the restriction map. We write rY(k)
for n = 3. We say that Y has maximal rank if for every integer k > 1 the
restriction map rY jp (k) has maximal rank as a map of vector spaces i.e. it is
injective or surjective.
A tree in IP3 is a connected, reduced curve Z in IP3 with only ordi-
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nary double points, arithmetic genus 0 and such that every irreducible component of Z is a line. In [11] (or see [8]) it is proved that a tree T of degree
d is contained in the closure Z{d) in Hilb(JP3) of the set of smooth rational
curves in P 3 of degree d. Note that Z{d) is irreducible. For positive integers 5, dx,..., ds, Z(s, dX)... ,ds) denotes the closure in Hilb(P 3 ) of the
set of smooth curves in P 3 with s connected components Q, I = 1,... ,5
deg Q = ^ , each C,- rational. Obviously Z(s, dx,... ,ds) is irreducible. The
set of trees of degree d in P 3 is not irreducible if d > 4, because two trees
of the same degree may have different shapes.
A bamboo is a tree T such that any line in T intersects at most two
other irreducible components of T. Equivalently a tree T of degree d is a
bamboo if and only if we may find an ordering Lx,..., L^ of the lines in T
such that Li O Lj ¥=0 if and only if | / - ; | < 1. For positive integers s,dx,
...,ds the set B(s, dx,... ,ds) of curves in P 3 with s connected components Cj, i = l , . . . ,5, deg Ci = di, each Q a bamboo, is irreducible.
Now fix a tree T. Since T is connected and rational, we may order the
lines Lx,...,Ld
of T in such a way that for every i with 2 < z < d there
is a unique index |3(T) < i such that Li O L ^ 0 with & < i if and only if
k=P(i). Fix a map j3:{2,...,d}—• {1,... , r f - 1 } with P(i)<i for every
f. Let B(fi) (or B(d,(3)) be the set of all trees T of degree d in IP3 such
that there exists an ordering of the lines Lx,..., Lj of T such that Li C\Lk^z0
for & < i if and only if k = j3(z). We call j3 the type of the tree T. It is easy
to check the irreducibility of B(P). Note that a tree may have different types
(corresponding to different ordering of its lines). However B(P) O B(P')¥=0
if and only if J5(j3) = B(f$'). Similarly we may speak about the type of a
disjoint union of s trees. If j3 = (Jix,... ff5s) we will write B(P) for the set
of curves with s trees of types Plf..., & as connected components. Take a
curve K whose connected components are trees-, a line L C y is said to be
a final line of Y if it intersects at most another irreducible component of Y.
For a point P in P 3 , xiP) is its first infinitesimal neighbourhood in
1P3. By definition %(P) has J>2p as ideal sheaf. Fix two distinct lines L,D
in 1P3 intersecting at a point P. Then L U D U x ( P ) is a degeneration of a
family of pairs of disjoint lines.
Fix a smooth quadric surface Q, Q = Pj x P j . Let F be a closed subscheme of P 3 . By the residual scheme ResQ(Y) of Y with respect to Q
we mean the closed subscheme of P 3 with the kernel of the natural map
Ojp " ^ Horn {J'y, OQ) as ideal sheaf. If L U D are lines contained in Q and
P = LCiD, then P = Res Q (I U D U x(P)). If 5 C Q and i? is a positive
divisor of Q, we may consider the residual scheme (in Q) of S with respect
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to R, with ideal sheaf Ker(0g—• Horn (J S , 0R)). A double point of Q is
a scheme T C Q with r re( j a point and h°(T, 0T) = 2. A double point of
Q is said to be regular if it is not contained in a line contained in Q. A triple
point of Q is the first infinitesimal neighbourhood \Q (?) m Q °f a point
P of Q, i.e. it is x(P) ^ (?• If £>£) are lines intersecting transversally Q
and P = (L D D) G Q, then (L U D) H g is the union of two simple points
and one double point while Q O (L U D U x(P)) contains a triple point with
support P.
Take Y in Z(s, d 1? ..., ds)\ for simplicity we will assume d : = dj + ...
+ ds>4.
Let 7 be the set of integers k>2
Note that if kEI,
then
fe+lG/
such that (k*3)>s
because if kei,k>2,
(k + 3)>d.
+ kd.
The
minimal integer in 7 is said to be the critical value of Y. Y has maximal
rank if and only if rY (n — 1) is injective and rY(n) is surjective (Castelnuovo's lemma [9] p. 99), where n is the critical value of Y.
We define the integers r(n), q(n),a(n), b{n) by the relations:
nr(n) + l+q(n)
(n +i)a(n)
= (n + 3) ,
+ b(n) = (n+J)
0<q(n)<n-l;
,
0<b(n)<n.
A union Y of a(n) disjoint lines has maximal rank if and only if Ker (rY(n — 1))
= 0 and dim Ker(r y (w)) = bin). A connected rational curve X of degree
r(n) has maximal rank if and only if Ker (rx (n — 1)) = 0 and dim Ker (rx (n)) =
= q(n). We have a{n) = [{n + 3){n + 2)/6], b(n) = 0 if 72 = 0 , l m o d 3 ,
£(3* + 2) = * + l . We have r(6k +1) = 6k2 + 8* + 3, q{6k+l) = 0\
r(6k + 2) = 6k2 + 10* + 4 , #(6* +2) = 3* + 1 ; r(6* + 3) = 6k2 + 12* + 6,
q(6k + 3) = 2* 4-1; r(6* + 4) = 6* 2 + 14* + 8, q(6k + 4) = 3* + 2;
r(6* + 5) = 6* 2 4 - 1 6 * + 11, 4(6* + 5) = 0; r ( 6 * + 6) = 6* 2 + 18* + 13,
q(6k + 6) = 5k + 5.
2. Definition of H(w).
In this section we define an assertion H(n) and we prove (lemma 2.1)
that if H(u) is true for all u>m, then theorem 1 is true for all data s,
d\,... ,ds with critical value at least m + 1.
H(n): a) Fix s,dlt'...,ds
with critical value n. Then there exists
du..., ds) with h°(JY(n))
= C\l)-n(dx
+ ... +
ds)-s.
YEZ(s,
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b) Fix s,dlf...,ds
with critical value n + l and (
) + w>s +
.+ n(d1 + ... + ds); then there exists K G Z C s , ^ , . . . , ^ ) .such that
ib°( y ( « » - = 0 .
LEMMA 2.1. 7/ H(u) holds for every u>m, then theorem 1 is true for every
datum s, dx,..., ds with critical value at least m 4- 1.
Proof. Take a datum s,d1,...,ds
with critical value & > w + l . We may
find r < s , r , < ^ for z < r such that H ( & - 1 ) , part b), applies to r,r,-.
Hence we find Y in Z(r, r 1 ? ..., rr) with h°(J>Y(k — 1)) = 0. Adding to 7
d?1 + . . . + ^ s - : r i — ... — rr suitable lines we find that a general element W of
Z(s, rfj,..., ds) has•••^°(«/w(fe - 1)) = 0. By H(k), part a), a general element
Z of Z(5, rfj,..., ds) . has rz{k) surjective. The lemma follows from the irreducibility of Z{s,dx,... ,ds), semicontinuity and Castelnuovo's lemma
(19] p. 99). •
We say that H(n) holds for a datum s, d^ with critical value n (resp.
n +1) if part a) (resp. part b)) of H(n) holds for the datum 5, d\.
Remark 2.2. Fix c, rx,..., rc with critical value n. We may find s, dlf..., ds
with s > c and ^ > r,- for i < c, critical value n and such that
(n+^)-n{dx
+ ... + r f s ) - 5 < »
(1)
If you prove that there exists F G Z f x , ^ , . . , , ^ ) with rY(n) surjective and
every irreducible component of degree 1, then you may find T C F , T E
Z(c,r1,... ,rc), with rT(n) surjective. Hence H(n), part a), holds for the
datum c, r,-.
We will always prove H(n), part a), in this way. We will call a datum
s, di for f/(w) maximal if either it satisfies (1) or it has critical value n + 1 .
3. Intersection with a smooth quadric.
In this section we want to control the postulation of the intersection of
a disjoint union of bamboos (or of trees of fixed type) with a smooth quadric
Q. From [1] lemma 6.2 it follows by induction on d := dx + ... + ds that for
every choice of positive integers s,di (with dx > 2 if 5 = 1) and natural
numbers a,b there is a union Y of s disjoint bamboos of degrees di with
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h°(Q, / y n ( ? ( f l ^ ) ) = m a x ( 0 , (a-\-l)(b-\-l)-2d)\
by semicontinuitythis
holds for general F, Q.
But this result is not sufficient for our purpose. Indeed we want to show
that the same result is true when Q contains every singular point of Y but,
with this only restriction, Y and Q are more or less general. This is a very
strong restriction and we cannot simply rewrite [1] §6 with suitable modifications.
We fix a smooth quadric surface Q. In this section we write J'y instead
of J>Y Q for every closed subscheme Y of Q. Recall that a double point
of Q is said to be regular if it is not contained in any line contained in Q.
We will use many times (even without mention it) the following trivial
lemma.
LEMMA 3.1. Let S C Q be a closed subscheme with finite length u and a,b
integers with b > 1. Assume h°(Q, Js(a,b)) — {a 4- 1) (b 4- 1) - u and
h°(Q, «>s(tf, £~" 1)) = max(0, (a + l)b-u).
Take a line LOQ of type
(0,1) and disjoint from S. Put x := m'm(a + l, (a 4- \)(b 4- 1) — u). Then
there exists W C L, card(W) = #, such that h°(Q,
J>s\jW(a>b))=
= (a + l)(b +
l)-u-x.
Proof
If not, we have h°(Q, Js(a, b - 1)) = h°(Q, J>SKJL{a,b)) >
> (a + 1) (b + 1 ) - u ~x, contradiction.
•
LEMMA 3.2. Fix natural numbers a, b,d with b > 3,2 d < {a -I-1) (b 4-1).
Then there exists a subset W of Q, card (W) = d, such that, if J is any
union of d regular double points with support W, h°(Q, Jy(a,b)) =
= (a + l)(b + 1) — 2d, Furthermore we may assume that there exists an
ordering Px,..., Pd of the points of W such that the line spanned by Pi and
P,-_ i is not contained in Q for every i <with 1 (a + 1) (b 4-1) - 1,
because if E,F are 0-dimensional schemes with E D F, the restriction map
H°(E, 0E) —+ //°(F, 0F) is surjective. Let d = (a + l)k 4- e, k,e integers,
0 < £ < # . We have k> 2, since b > 3 and, if b — 3, (a + l)(b + 1) is even.
We take £ + 1 lines L1,...,Lk + i of type (0,1) on Q. We consider a
subset W = {?!,...,P d } C Q with the following properties:
(i) P{ €Lk
+
i if and only if i = (k + 1) / with ;' < e;
(ii) PjELj, /'<&, if and only if either i=j + (k + l)s,
= (k + l)e+j + ks,
0<s<a-e-l.
s<>e, or z =
214
We want to show that a general choice of W satisfying (i), (ii) will work.
Take any union J of d regular double points with support W. Any form
/ G H ° ( Q , <0j{a,b)) vanishes on Lif i<*k, and the quotient g of f by the
equation of Li U ... U Lk vanishes on every point of W H (L1 U ... U L&)
and on the double points with support in L^ + 1 . Hence g2 divides /.Note
that for any union / of e < # regular double points with support on L^ + 1 ,
h°(Q, JI(a,l))=-2(a + l)-2e and b°(Q, J>j(a,0)) = max(0, a 4- 1 -2e).
Indeed the first equality follows adding a + l—e regular double points with
support on L^ + 1 . For the second equality use induction on e\ if 2e < # + 1,
we can find e lines Rj of type (1,0), ; = 1, ...,£, such that the union T
of 2Rj, / = 1,... , e - 1 , and Re containes e—\ double points on Z,£ + 1
and a point P of L^ +1 but no regular double point with support on P. Now
the first part of the lemma is proved.
The second part of the lemma follows from the construction.
•
From lemma 3.2 it follows at once the following corollary.
COROLLARY 3.3. Fix non negative integers a,b with & > 3 and positive
integers s,dl,...,ds.
We may find YGB(s,dl,...
,ds) with Sing(Y)CQ f
d i m ( y n Q ) = 0 and b*(Q, SYnQ(a,b))
= max(0, (a + l)(b + l)-2d,
with d = dx 4- ... 4- ds.
Proof Any final line of Y intersects Q in a point of Sing(F) and a point
which may be a general point of Q. •
In the statement of 3.3 fix a subset S C Sing(y). By semicontinuity we
may deform Y to Y' E B(s, dx,... ,ds) and S to S'CSing(y') such that
*°(G, ^ y ' n e ( « ^ ) ) = max(0,(« + l)(^ + l ) - 2 r f ) and (Sing(F') O Q) red =
= S'.
DEFINITION. Take E G B(s, dly... ,ds) with d i m ( £ n Q ) = 0. We say that
Q has k ( 0 < / J < 5 ~ 1 ) good secants Lj, ; = 1, ...,&, to £ if E U Lj is
disjoint union of s — & bamboos. Equivalently, we can order the connected
components Cx, i = l , . . . , s , of £ in such a way that Ly intersects Cy and
Cy+i, every L;- intersect a different final line of E.
LEMMA 3.4. Fwc positive integers a,b,dx,..., ds with & > 4; for every index
i, i — 1,... ,s, ta&£ <w integer ex, 0 < e,- < 3. Fwc a system of lines on Q, say
the lines of type (0,1). There exists Z EB(s, d1)... ,ds) such that:
215
(i) the singular points Sing(Z) of Z are contained in Q, dim (Z Pi Q) = 0;
(ii) let Cj, i = 1,... ,5, be the connected components of Z and L{, L\ the
final lines of C, (Li - Lj if d{ = 1); we take points P,GL,*nQ,
P\ E L\ n (2, P.,Pj £ Sing (Z) ; if d{ = 1 w* assam* P, =£Pj; w* assume
that Q contains s-1 good secants of type (0,1) to Q; let S be the
union of the d~s := ^ 4- ... 4- ds -:s double points of Z 0 Q and of
Pi, Pj for every index ij.with d{ = 1 or 2, dj - 2 or 3; flte»
(iii) h°(Q,
s
( ^ » ) = max(0,(a + l)(Z7 + l)-length(5)).
Proo/. The main difference with respect to 3.3 is given by the good secants.
Use lemma 3.1 at most s - 1 times. If we fix lines of type (1,0) we need
b> 3. Alternatively, repeat the proof of 3.2 inverting the role of the two
systems of lines on Q, •
Here is a key lemma
LEMMA 3.5. Fix integers a,b,t with 0<a <b,t>0.
Then there exists
the union T of t triple points on Q in general position such that
b°(Q, ST(a,b)) = mBx(0y (a + l)(b 4 - 1 ) - It) if one of the following conditions is satisfied:
(i) a = 1 or a > 3;
(ii) "a = 2 and b is odd-,
(iii)"a = "2, b is even and It ¥=(a + l)(b 4-1).
Proof. Note that we may assume \(a + l)(b +
l)-.3t\<2.
Assume a = 1. We take the union T of t triple points, no two of them
with support contained in a line contained in Q. Any curve of type (l,fc)
containing T contains the union U of the t lines of type (0,1) containing
T red . The residual scheme of T with respect to U is Tred i.e. t general
points of Q.
Now assume a = 2. First assume b odd; we may assume t = b + 1. •
Take 2 lines R,L of type (1,0) and consider the union W of (&4-l)/2
general triple points with support on R and (b + l)/2 general triple points
with support on L. The residual scheme of W with respect to R L is
Wted, which is not contained in any form of type (0,&). Now assume a - 2
and b even. If t = b + 1, the union of any b + 1 triple points is contained
in an unreduced curve of type (2,b) on Q, with support of type (1, b/2).
If £ £ + 2 we put on i? (fc + 2)/2 triple points, on L (b+2)/2 triple
216
points and win. Now assume t b. We take til general triple points with
support on L, (t + l)/2 general triple points with support on JR, b + \ - til
general simple points on L, b + l~ {t + l)/2 general simple points on R
and win. The proof given for a = 2, b odd works for every a,b such that a
is odd or b is odd. In particular it works if a = 3 and b = 3 or 4.
Now assume tf>3, b~>3 and a+b>8,
hence t~>a+h. We use
induction on a. Note that if h = (a + 1) (b + 1) - 3 t , then h+a+b —
= (<z - 1) (b - 1 ) - 3(t~a -b). Let / be the union of t~a~b
general
triple points with b°(Q, ^(a - 2, b - 2)) = a +b + h. Take a general smooth
curve E of type (2,2).
Claim. For the union A of a+b general points .of E, h°(Q,
^\jA(a-2,
b - 2)) = max(/?,0). If not, for the union D of a+b-l
general points
of E,H°(Q, J'mcia - 2, b - 2)) is not empty and has E in its base locus.
Hence b°(Q, ^â - 4, b - 4)) =£ 0. If min(<z,£)=3, this is obviously false.
If a > 5 and & > 5 the inequality gives a contradiction for general T
(using also case (iii) if b — 6. If, say, # = 4 the contradiction comes from
3t~> 5(b + \)~ 2, hence t> b — 3. We have proved the claim. For a general
positive divisor 5 on E, 25 is not linearly equivalent to QE(a,b). Hence
every curve of type (a,b) containing 2S with card (S) = a + b contains
E. Let T be the union of / and oi a + b general triple points with support
contained in Q. The residual scheme of T to E is the union of / and a+b
general points of E. The lemma follows from the claim.
•
REMARK 3.6. Note that 3.5 is false if a = 0 (or b = 0). Indeed for the
general union T of £ triple points, b°(Q, JT(0,b)) = max (0, b + 1 - It).
LEMMA 3.7. Fix natural numbers a,b,d,s,k,r
with. a>\, b>2,
s>l,
d>3,k<d-s,
2k+r<2d. Assume eithe a>3,b>3
or 3k<(a + l)(b + l).
Fix a union Y of s disjoint trees, say YEB((3), with d i m ( K n Q ) = 0 and
degY^d.
Fix k+r points Plt...,Pk,
Au...,Ar
of YC\Q with Pj G
6 Sing (7), .Ay £ Sing (F) for every i,j. We may deform Y to Y'EB(f$)
with a deformation which sends Pj into P\ G Y' Pi Q, Aj into Aj G Y' O Q,
and such that:
(i) sing(y')ng = ?;,;..,?;> dim(r'ne) = o;
(ii) let Z be the union of x e 0P/), 1<*'<&, and (Y'\{P[,... ,P'k, A[,
...,A'r})PiQ; then we have h°(Q, Jz(a,b)) = max(0, (a + l)(b + 1 ) -2d + r-k).
217
Fix t<:S; let Q, l < / < £ , be some of the connected components of Y
and Cj the corresponding components of Y'; assume that Q contains t—\
good secants of type (0,1) Lj to Y with Lj linking Cj and Cj+i. Then
we may find Y' as above and such that Q contains t—\ good secants Lj
of type (0,1) to Y' with L'j intersecting Cj and Cj+i if we assume
(Lli>...VLt-l)nQC{Al,\..,Ar}.
Proof. By lemma 3.5 we may assume that the union J of \n (î),
lî^k
imposes min(3&, (a 4- 1) (b -f 1)) conditions to V : = H°(Q, 0Q(a,b)). Then
from the points P, we will build Y' £ £ ( 0 ) using [1] lemma 6.2 in such a
way that the conditions (i) and (ii) are satisfied. Recall what we need about
[1] lemma 6.2. Take a line D intersecting transversally Q and a subset
TCQ
of length t with h°(Q, JT(a,b)) = max (0, (a + 1) (b + 1) - t),
Z>°(Q, ^ r , ( « - l , fc-l))<max(0, (* +l)(fc + l ) - * - 2 ) where T' is the
maximal subscheme of T with T' U (D O Q) D T; there exists a line L intersecting D and with h°(Q, t#r{J(Lr[Q)(a,b)) = m3x(0, (a+l)(b + l)-'
—t—2)î note that the hypothesis excludes the case t — 2, a = b = 1, 7 =
We start with a line R through one of the Pj, j <k : JU (R HQ)
gives no trouble since R n (Q\{P;}) may be a general point of Q. Then we
use [1] lemma 6.2; if a connected component of degree c of Y is bounded
by P| and P^ we add in this way c — 1 lines i?i,..., Rc-\ > for Kc we take
a general line through P,- intersecting i? w , m<zC~- 1, with m uniquely
determined by the type j8 of y. We have to control the postulation of the
point Rc n (Q\{Pi}) which may vary freely in a curve of type (1,1); we use
the proof of lemma 3.1 or of [1] lemma 6.2.
The last condition of the lemma gives no trouble. Indeed we may obtain
final lines D; of C\,D\ of Cj +1 such that there is a good secant line A of
type (0,1) to D U D'. We have excluded R n (D U D') from the postulation, while the other 2 points of (D UD') O (Q\R) may be general points
of Q: through any point of P 3 there is a line intersecting two given skew
lines.
•
4. The hypothesis Bin — 2) and a proof of H(n), n — 6k + 4 , 6k + 5, 6k -1-6
for most data. We will not prove H(n) by induction using H(n — 2) as any
reader of [6] suspects. Instead we will prove some assertion B(m) about the
general postulation of certain configurations of bamboos. For n = 6k+4,
6k + 5, 6k + 6 we prove in this section that if B(n - 2) is true, then H(n)
218
holds for a datum s, d{ with at most 3n indices i with d( = 2 (see lemmas
4.1, 4.2, 4.3 for precise statements). In the next section we will introduce a
technical hypothesis B\n) and, using it, we will prove by induction the
hypotheses B(m) we need.
B(n) if n = 2,3,4 or 6 mod (6); There exists (Z,Q) such that:
1) Z
is the union of (q{n)+\)
disjoint bamboos, degZ = r(n)
and
2) Q is a smooth quadric containing Sing(Z) and q(n) disjoint good secants to Z; dim (Z HQ) = 0.
JB(ra) if n = 1 or 5 mod (6). There exists (Z,Q) such that:
1) Z
is the union of
n + 1 disjoint bamboos, degZ = r ( w ) ~ l
and
2) Q is a smooth quadric containing Sing(Z), n disjoint good secants to
Z; d i m ( Z n g ) = 0.
From now on in this section we fix a datum s, d± which is maximal for
some H(m) and with critical value n, n — m or m 4- 1. Let c be the number of indices / with d\ = 1; let / be the set of indices with d\ > 3; we put
M = [(4-l)/2].
LEMMA 4.1. B(6^ + 3) implies that H(6k 4- 5) holds for the maximal datum
s, dj with critical value n if k > 0 and the following condition is satisfied:
1) c + yLiyi>2k
+ l if n = 6k + 5 , c+ ,Eiyi>2k
+ 2 if n = 6k + 6 .
Proof We take s,dj satisfying 1). Take (Z,Q) satisfying J5(6& + 3). We
want to add to Z u := dl + ... + ds — 6k2 — 12 k — 6 disjoint lines in Q.
Note that u>2k + l because u >a(6k + 5)-r(6k 4- 3) = 3k + 3. Put
v \— min(w — 2k — 1, c). We add in Qu disjoint lines L,*, l < f < w , such
that L,-, l < / ' < 2 ^ 4 - l , are the good secants to Z and exactly v of the
lines Li do not intersect Z. Let W be the union of Z and the L,-. We
want to add to W s — v—1 nilpotents x(P) f° r suitable PESing(W) in
such a way that the union Y of W and the \(P) is m B{s, dX)... ,ds) and
r y (6& + 5) is surjective.
Consider the abstract situation. We have a curve Z, deg Z = 6k2 4-I- 12& 4-6, disjoint union of 2& + 2 bamboos. We can add 2& + 1 lines
Aj such that the union of Z and the Aj is a connected bamboo. We add v
lines not intersecting Z, w~v — 2k —\ suitable lines intersecting Z at most
219
at a point and s - v - 1 nilpotents. In this way we may obtain a curve in
B(s, ^ i , . . . , ds). However in out concrete situation we have a big constraint.
Suppose you want to link a line Lj to a line A of Z,A not a final line of
Z. In our construction Lj is contained in the quadric Q and the points of
A C\ Q are singular points of Z. Suppose B C Z,B a line with A O B =£ 0
and that you want to link a line L ; , ; < 2 k 4- 1, to A C) B. A U 5 U Lj has
arithmetic genus 0 if it spans JP3 but to separate the connected components
we cannot add a nilpotent at A H B. At most we may put a nilpotent at the
2 points of Q Pi ((^4 U B)\(A D £)); in this way we obtain a rational curve
of degree at least 3. For example if d\ — 2 for each i and,, u > 2 k 4- 1, this
construction fails. This construction works for any d\ exactly if condition
1) is satisfied because q(6k + 5) = 0.
Now we prove that we may assume rY>(n) surjective where (Z'} Y') is
a suitable deformation of (Z,F). First we deform Z to Z' such that P' G
G Sing(Z') fl Q if and only if P' is a deformation of PG Sing(Z) and
X(P) G 7; furthermore if \(P) £ ^ with P^Lj, j < 2& + 1, we assume that
the corresponding connected components of Y' are not linked by a good
secant in Q; in this way we will apply the lemmas in § 3.
The case of H(6k + 5), part b) is similar and omitted. Fix a general
subset SCQ
with card (5) = (
,
) — (6k + 5)(d1 + ... + ds)~s;
we have
to prove that we may assume h°( ^Y'us(6k
+ 5)) = 0. Take / G
GH°(J>Y'\jS(6k + 5)) and assume that the restriction g of f to Q vanishes. Then / is divided by the equation q of Q-, by £(6& + 3) we obtain
//g = 0. We have g = 0 by 3.6 and an identity about binomial coefficients
which explains the Euler characteristic of the projective spaces. •
Note that a(6k+4)-r(6k
+ 2) = 3 * + . 3 > 3 * + l, r(6k+4) ~r(6k + 2) =
= 4fc+4, a(6k+6)-r(6k+4)
= 3&+4 and r(6k +4) ~r(6k + 2) =4fc + 5.
Hence we obtain exactly in the same way the following two lemmas.
LEMMA 4.2. B(6k + 2) implies that H(6k 4-4) holds for the maximal
datum s,dj with critical value n if k~>2 and the following condition is
satisfied:
1)
c + yLiyi>k + l if n = 6k+*
, c 4- 2,-y,- >k + 2 if n = 6& 4- 5 .
LEMMA 4.3. #(6& +4) implies that H(6k + 6) holds for the maximal datum
s, dj with critical value n if &>2 tfftd the following condition is satisfied:
1)
c + 2 ^ > * + l i/ rc = 6&+6 , c + X{yi>k
4-2 i/ w = 6k + l .
220
5. Proof of B(n) for all n.
In this section we will prove the hypothesis B(n) for all w > 1. We
need two technical hypothesis B'(6k + 1), B'(6k + 6).
B'(6k + 1). There exists (Z,A,B,Q)
such that:
1) Z is the union of 6k+ 2 disjoint bamboos, deg Z = 6k2 + 8 & + 2 and
b°(Jz(6k + l)) = 0; Z = AUB where A has 4*4-1 connected components and B has 2 k 4-1 connected components;
2) Q is a smooth quadric containing Sing(Z); dim(Zn<2) = 0; Q contains 4k good secants of type (1,0) to A and 2k good secants of type
(0,1) to B.
B'(6k+6).
There exists (Z,A,B,Q)
such that:
1) Z is the union of 5k+6 disjoint bamboos, deg Z = 6k2 + 18& 4- 13 and
h°(J;z(6k + 6)) = 0; Z = AUB where A has 2&+2 connected components and B has 3&+4 connected components;
2) Q is a smooth quadric containing Sing(Z), 2k + 1 good secants of type
(1,0) to ^4 and 3& + 3 good secants of type (0,1) to B\ dim (Z H Q) =
= 0.
Note that B(l) and B'(l) are trivially true. For B{2) take the union
Y of a tree of degree 3 and a disjoint line; Y is not contained in any quadric;
check that it cannot be contained in a smooth quadric or in a reducible quadric or in a quadric cone. In lemmas 5.1,..., 5.8 we will prove the following
chains of implications: for all k > 0, B'(6k + 1) => B(6k + 3) => £(6fc 4- 5) -*
=>B'(6k+7), B(6k + 5) =* £(6*4-7), B(6k+2)=* B(6k+4) =* B(6k+6),
B(6k+4) => J B ' ( 6 ^ + 6 ) =* 5 ( 6 * + 8 ) .
Thus at the end of the section we will have proved B{n) for all n > l
and B'(n) if w = 0,1 mod (6). The explicit values of r(n), q(n) given in § 1
explain the numbers which will appear in the proof of most lemmas.
LEMMA 5.1. B'{6k + l) implies B{6k + 1) for all
k>0.
Proof We take {Z,A,B,Q) given by B(6k + l). In Q we take the 4k
good secants D,«, i = \,... ,4k to A. The union of ^4 and the Dj has two
final lines not contained in Q, say Llf'L2.
Let Aj, j= 1,2, be a line of
type (1,0) in Q intersecting L ; . Let Ri,R2 be two lines of type (1,0)
in Q, each fy intersecting a different final line of B, none of them inter-
221
secting a good secant to B. Let Y be the union of Z, D,-, i — 1,...,4&,
Aj,Rj, / = 1,2. We want to prove that h°(^Y(6k 4- 3)) = 0. Take / G
G H ° ( t / F ( 6 ^ 4-3)). Assume that the restriction g of f to Q vanishes.
Then / is divided by the equation q of Q; by B'(6k 4- l)f/q = 0, hence
/ = 0. Since g vanishes on 4&+4 lines of type (1,0) it is sufficient to
prove that the union of 6 k2 + 8k double points and the 12 & + 4 - 8& - 4
simple points of Z C\ Q not on these lines gives independent conditions for
forms of type ( 2 & - 1 , 6&4-3) (if k > 0); if & = 0, g = 0 because it
vanishes on 4 lines of type (1,0). For k>0 we use lemma 3.4.
Now we show how to deform Y to a curve satisfying B(6k 4- 3). We
move the lines of A,Dj, Ax ,A2 in such a way that the configuration W
obtained has no irreducible component in Q but every singular point of W
is contained in Q. Then we take a point P in a final line of W and a point
P' ERX such that the line of type (0,1) containing P contains P \ We
deform Rx with a family of lines through P'. We can deform B U RlU R2
to a curve W' with lines i?,' corresponding to i?,- but not contained in Q,
with S i n g ( W ) c Q , P' G flj and Q contains 2& good secants to W\ By
semicontinuity we may assume rw\j w'(6k + 3) bijective.
•
The proofs of the other implications are very similar. Hence in the
proofs of lemmas 5.2,...,5.8 (and of most lemmas in this paper) we will
give only the construction corresponding to the first third of the proof of
lemma 5.1. We will leave to the reader the duty of checking that the other
parts of the proof of lemma 5.1 work also for the other lemmas.
LEMMA 5.2. B(6k 4- 3) implies B(6k + 5) for every
k>0.
Proof. Take (Y, Q) satisfying B(6k + 3). Suppose for examples that Q has
2&4-1 good secants of type (1,0) to Y. Let Z be the union of Y and
4&+4 lines D,- of type (0,1) on Q, no D; intersecting Y. For a deformation Z' of Z, (Z',Q) satisfies B(6k + 5). In particular we have to move
the lines D,- outside Q but mantaining fixed one of their point, to obtain
2k + 14-(4&4-4) good secants of type (1,0).
•
LEMMA 5.3. B(6k + 5) implies B'(6k +7) for every
k>0.
Proof Take (F, Q) satisfying 5(6^4-5), with Q containing 6& + 5 good
secants of type (1,0) to Y, say D,-. Let T be the union of Y and 4& + 6
lines Rj, j = 1,... ,4k + 6 , of type (1,0) on Q with Rj = Dj for ; =
= 1,..., 2k. Let Q, i = l , . . . ,6k + 6 , be the connected components of Y;
222
we assume that Dj links Q and C, + 1 for 1 < z < 6k + 5. We assume that
R2k+i intersects a final, line of Cx not intersecting D1,R2k+2 intersects
a final line of C 6 ^ +5 not intersecting D 6 & +4 , that i?2&+3 a n d ^2^+4 m _
tersect different final lines of C 6 ^ +6 and that Rj does not intersect F for
; = 2k + 5,...,4& + 6. Let ^4' be the union of Q for z<6& + 5 and # ;
for / < 2 & + 2. Let 22' be the union of C6^ + 6 and Rj for / ' > 2 & + 2 .
Then (Z,^4,J5,Q) satisfies £'(6&+7), where (Z,^4,5) is a suitable deformation of (A'UB',A',Bf).
•
LEMMA 5.4. B(6k + 5) implies B(6k 4-1) for every
k>0.
Proof. Take (F, Q) satisfying £(6& + 5) with <2 containing 6k+ 5 good
secants to Y of type (1,0). Let T be the union of Y and 4k+6 lines
i?z-, i = 1,... ,4&+6, of type (0,1) on Q with R4k + 5, R4k + 6 not intersecting Y and each Rj, / < 4k + 4, intersecting a different final line of F.
•
LEMMA 5.5. B(6k + 2) implies B (6k +4) for every
k>0.
Proof Take (Y,Q) satisfying B(6k+2), with Q containing 3& + 1 good
secants of type (1,0) to F. Take 4&+4 lines Dj, i = 1,... ,4k + 4 , of
type (0,1) on Q, with D 4 ^ + 4 not intersecting F and each Dj, j <
<4& + 3, intersecting a different final line of F. A deformation of the
union of F and the Dj will work.
•
LEMMA 5.6. B(6k + 4) implies B(6k + 6) /or every k > 0.
Proof. Take (F, Q) satisfying £(6&+6), with Q containing 3& + 1 good
secants of type (1,0) to F. Take 4& + 5 lines D,-, /' = 1,... ,4k 4- 5, of
type (0,1) on Q; we assume that each Dj, ; < 2 & + 2 , intersects a different final line of F, while Y n Dj = 0 for z>2& + 3. A suitable deformation of the union of F and the Dj, 1 0.
Proof. Take (Y,Q) satisfying B(6k4-4) with Q containing 3k + 2 good
secants Dj, i = 1,..., 3k + 2, of type (1,0) to F. Let Q, i = l , . . . ,
3 & + 3, be the connected components of F and assume that Dj intersects
Cj and Cj+1 for every i. Consider in Q 4k + 5 lines Rj, j — 1,... ,4k 4 5,
223
of type (1,0) with fl2,-i and fl2, intersecting different final lines of Q
for K i < & + 1, fly not intersecting Y for j>2k+2.
Let A' be the
union of the Q for & 4 2 < i < 3 & + 3 and B' the union of C, for 1 0.
Proo/. Take (Z,A,fl,Q) satisfying B'(6k + 6 ) . Consider 4&+7 lines fly,
/ = 1,... ,4k + 7, of type (1,0) with fly good secant to A for / < 2k 4- 1
and every fl,, i>2k + l, intersecting a different final line of B.
•
6. C O - 2 ) and //(«) for most data, n-6k
4-7, 6k + 9.
In this section we give partial results about H(6k + 7) and H(6& + 9)
for every & > 0. These results are similar to (but weaker than) the result
proved in § 4, § 5 for H(m), m = 6k 4 4, 6k 4- 5, 6k 4- 6. In the next section
we will obtain similar results for H(6k 4 8) using a slightly different path.
Here we need the following assertions C(6k 4- 5), C(6k'.4- 7), & > 0.
C(6& + 5), & > 0 . There exists (Z, fl, (1 <i<
2& 4- 3), Q) such that:
1) Z is a bamboo df degree 6&2 + 14& + 9 and fl,- C Z, with R1}R2
lines of Z;
final
2) Q is a smooth quadric containing Sing(Z) and satisfying the following
condition 3);
3) for any choice of 2&+2 singular points Pj, j = 1,... ,2k + 2, with
Pj $ Rj for every /,/, there exists a line Ly through Pj such that for
the union Y of Z and every Lj we have £°( t ^ K (6& 4- 5)) = 0.
Note that Y has arithmetic genus 0 if for every /, / = 1,..., 2k + 2,
the lines of Y containing fly span P 3 . In this case Y * is a deformation of
a family of trees of degree r(6k + 5): move every Lj along one of the lines
of Z containing Py.
C(6k+7),
k>0.
There exists (Z, fl; (1 < / ' < 2& 4 5), Q) such that:
1) Z is a bamboo of degree 6k2 4-14& + 11 and fl,- C Z, with fl2*+4 and
fl2& + 5 final lines of Z, fl x intersecting R2k+4, R2 intersecting R2k + s\
2) Q is a smooth quadric and the following condition 3) holds:
224
3) fix 6k + 6 singular points Pj, 1 < / < 6 H 6 , of Z with Pj $ R{ for
every *',/; then there exist 6k+6 lines Ly, l < / < 6 & + 6 , with Pj E Lj
such that /b°(,/y(6& 4-7)) = 0, where y is the union of Z and the Ly;
furthermore we may assume that Q contains every singular point of Z
except eventually the P,, 1 0; hence we
have proved C(6&4-5) and C(6k 4-7) by the results in §5. Then we use
C(6& + 5) and C(6k+7)
to obtain respectively that H(6k+1)
and
H(6k 4- 9) hold for many data (see lemmas 6.3, 6.4 for precise statements).
LEMMA 6.1. B(6k + 3) implies C(6k + 5) for every
k>0.
Proof Take (Y, Q) satisfying £(6&4-3). Let Riy 3<i<2k+3,
be the
good secants to Y contained in Q, say of type (1,0). Let Rj, / = .1,2, be
lines of type (1,0) intersecting the two final lines of Y not intersecting the
good secants of Y. Let Z be the union of Y and the Rj, j = 1,... ,2k 4- 3.
(Z,Ri}Q)
satisfies 1), 2) of C(6& + 5). Fix 2&+2 points Pj} ; = 1,...,
2& + 2, of Sing (Z) with Py £ #, for every /',/ (that is to say Py G Sing(F));
we add lines Lj, l < ; < 2 & + 2 , of type (1,0) on Q with Pj E Lj. Then
we copy the last two parts of the proof of lemma 5.1. Note that for any form
g of type (6k + 5, 6k 4- 5) on Q, divisible by the equation q of the union
of the Rj and the Lj and vanishing on F H Q , g/q vanishes on every Pj.
Hence the postulation of Y H Q is controlled by 3.3 and 3.4 (the points Pj
can be general points of Q). •
LEMMA 6.2. C(6k 4- 5) implies C(6k 4- 7) /or m?ry & > 0.
Proof Take (Z, #,• (1 < / ' < 2k + 3), Q) satisfying C(6k + 5). By semicontinuity we may assume that Q contains no connected component of Z. Fix
6k + 6 singular points Pj of Z not on the'i?,-, hence in Q. Take lines Ly,
l < / < 2 & + 2, satisfying 3) of C(6k 4- 5). By semicontinuity now we may
assume Pj£Q, hence L;- £ Q, for l < / < 2 & + 2 . Then we add 4&+6
lines Lj, j = 2& 4- 3,..., 6k + 8, of type (1,0) on Q with PjELj and
conclude as in 5.1. Note however that the union of Z and the Lj, 1 < / <
< 2& + 2, is not a bamboo; we need the notion of type of a tree to work in
an irreducible variety B(f3). These trees are very similar to bamboos, at least
for large k, and it would be sufficient to have 3.3 and 3.4 for bamboos and
then use [1] lemma 6.2 for the remaining lines. •
225
We fix a maximal datum s,d{ for H(m), m = 6k + 7 or 6k + 9 , with
critical value w, n = m or m + 1. We put d = i x + ... + ds. Let c be the
number of indices i with dj = 1. Put y,- = [(d,- - l)/2].
LEMMA 6.3. H(6k + 7) holds for the maximal datum s,d{ with critical
value n if k > 0 and the following conditions are satisfied:
1) c + 2 j y j > 6 H 8 ^ » = 6 H 7 ) c + 2 i j / < > 6 H 9 ^ = 6H8;
2)
d-s>4k+4.
Proof. We consider the case n = 6& + 7. Note that 3& + 5 < d - r(6& + 5) <
< 4 f c + 6 . We start from {Z,Ri}Q) satisfying C(6k + 5). Then in Q we add
d~r(6k + 5) lines D, of type (1,0) and a few nilpotents x(^r)> l ^ r <
< s — 1, P y ^ Q H Sing(Z), P'r i^Pj for every r,j. We obtain something in
Z(s,dl)... ,ds) and win. By semicontinuity we may take instead of Z\j U Lj,
F ; == Z\j U Ly with Lj intersecting Z at a smooth point of Z contained in
an irreducible component of Z containing Pj. If d — 5 > 4& + 4 we are able
to find enough curves of degree > 2 containing Lj and the necessary nilpotents in Q (used to "separate different connected components"). •
In the same way we obtain the following lemma.
LEMMA 6.4. H(6k+9)
holds for the maximal datum s,d^ with critical
value n if the following conditions are satisfied:
1) c + Eiyi.> 10k + 15 if n = 6k + 9 , c + 2 f ^ > 1 0 * + 16 if n = 6&+10;
2) d - s > 12fc + 12.
Note that we will use 6.3 and 6.4 with much weaker bounds both on 1)
and 2) (except for low k). This will follow from the statements in § 8, 9, 10
which cover more than the cases omitted in 6.3, 6.4.
We will leave to the reader the duty to repeat, when necessary, the
relevant part of the proof of 5.1 needed in the proofs of the lemmas of the
remaining sections.
7. D(6k + 6) and the proof of H(6k + S) for most data.
In this section we will prove that H(6k + 8) is true for many maximal
data s,d{ with critical value n = 6k+8 or 6&+9, in particular if for at
most In indices i we have d\ = 2 and we have not too much indices i
226
with di = 1 (see 7.2 for a precise statement). But now the proofs are more
involved than in § 6. .
We need another technical hypothesis D(6k 4-6), & > 0.
D(6k+6),
k>0.
For every r with 3k + 4 < r <min(4& + 7, 5*+ 5),
there exists (Z,W,S,D,Q) such that:
1) Z is the union of W and two lines, each of them intersecting only one
irreducible component of W; W is the union of 5 k + 6 — r disjoint bamboos, deg Z = r(6k + 6); D is a line intersecting Z, S C D, card (5) =
= r, h°(J?zus(6k+6))
= 0;
2) Q is a smooth quadric containing D as a line of type (1,0), every singular point of W and 5k + 5~r good secants of type (0,1) to W-t furthermore we assume the following condition 3):
3) for any choice of r4-1 points P,, l < j < r + l , of Sing(W) we may
assume that D intersects Px, that S contains D H £>;, 2 < / < r + l —
— (5&+6 — r), where Dy is the line of type (0,1) in Q passing through
Pj; the other 5k + 5 -r points of 5 are the points D C\Rj, where Rj,
1 < / < 5 k + 5 — r, are the good secants.
Note that r > 5 & + 5-7% hence condition 3) of D(6k + 6) is meaningfull.
First we will prove that B(6k+4-) implies D(6k +6) (lemma 7.1) and
then that D(6k + 6) implies H(6k + 8) for most data (lemma 7.2).
LEMMA 7.1. B(6k+4)
implies D{6k + 6) for every
k>0.
Proof Put x= 3k + 3-(5k+6-r)
= r-2k-3.
Take (T, Q) satisfying
B(6k+4)
with Q containing 3k + 2 good secants to T of type (1,0).
Note that if r < 4 & + 4 we will find Z satisfying D(6k+6) and such that
its connected components are bamboos. Let Z be the union of T and
4& + 5 lines Liy K i < 4 H 5 , of type (1,0) on Q. We assume that L,is a good secant to T if i < x and that L,- intersects T for all i. We assume that Lj intersects only final lines of T if i < min (4-fe 4-5, 8& + 9 - r)
and that each connected component of TU Lx U ... U Lx intersects some
Lj with j >i at a final line; here we use 4k 4- 5 — x — (max (0, r — 4k. - 4)) >
> 3& 4- 3 - #. We conclude as in 5.1; to handle the postulation of an S as
in 3) we need the proof of 3.2.
•
Fix a maximal datum s,dj for H(6£4-8) with critical value n =
= 6k + 8 or 6&4-9. Put d = dx 4- ... + ds. Let c be the number of indices
i with dj = l. Put yi = [{di ~ l)/2].
227
LEMMA 7.2. H(6k + 8) holds for the maximal datum s,d{ with critical
value n if the following conditions are satisfied:
i) c + Xtyi > 4fc 4- 9 if n = 6k + 8 , c + S , ^ > 4& + 10 i/ « = 6fc + 9;
ii) c < s - 1 and if c = s- 1, then d1 > 3;
iii) & > 0.
Proof Put r = < / - r ( 6 * + 6 ) - l . Since d>a(6&4-8), we have 3 f c + 4 <
< r < 4 & 4 - 7 ; note that r < 4 & 4 - 6 if w = 6&4-8. We consider only the
case r < 5 & 4 - 5 , since if r > 5 & + 5 the bamboo's trick works (starting
from B(6k+6)). Take (Z,W,S,D,Q) satisfying D(6&4-6) for the integer
r. We use the notations of condition 3) of D(6k + 6). Let T be the union
of Z, the lines D,-, 2 < i < r 4 - l , x(P) for all ? G 5 , x(^y) for suitable
singular points of Z. By i), ii) we may obtain TGZ(s, dlf... ,ds) and win
as in 5.1. Note that before doing this construction we deform Z in such a
way that the only singular points of Z contained in Q are the ones on which
we put a nilpotent. Also we do not put a nilpotent on a point of a good
secant; if the construction need it we deform Z, the good secant disappears
and we may add a line intersecting Z in at most a point.
•
8. E(n),F(n)
for n even: applications to H(n + 2).
In this section we define and prove the assertions E(n) for n even,
n > 6. Then using E(n~2)
we prove that H(n), w > 8 and even, n^
^ 8 mod(6), is true for all data s,d{ with many indices i with d± = 2, for
instance if dj = 2 for at least 3n indices. Then we define and prove the
assertions F(6k+4)
and F(6k+6),
k>l.
Using F(6k+6) we prove
a similar result for H(6k + 8) for every k > 1.
£(6&+2), k>l.
There exists (Z,A,B,Q)
such that:
1) Z is union of 9£ + 4 disjoint bamboos, deg Z = 6k2 + 10fc+ 3 and
h°( Jz(6k +2)) = 0; Z = A U B (disjoint union); every connected component of 5 is a reducible conic; let u be the number of connected
components of A; if k > 2 we assume w < 3& 4-2; if & = 1 we assume
u = deg A = 7.
2) Q is a smooth quadric containing Sing(Z) and u~ 1 good secants of
type (1,0) to A.
228
E(6k + 4), k > 1. There exists (Z\ A',B\
C, Q) such that:
1) Z' is union of 9&+7 disjoint bamboos, deg Z' = 6k2 4-14&+7 and
h°(Jz,(6k+4))
= 0-} Z'=A'UB'UC
(disjoint union); C is union of
3 disjoint lines; every connected component of B' is a reducible conic;
let v be the number of connected components of A'-, we assume z;<
<3& + 3.
2) Q is a smooth quadric containing every singular point of Z' and v — 1
good secants of type (1,0) to A'.
E(6k+6),
k>0.
There exists (Z", A", B\ C\ Q) such that:
1) Z" is union of 11 £4-12 disjoint bamboos, degZ" = 6k2 4- 18& + 12
and b°(Jz»(6k+6))
= 0i Z" = A" U B" U C' (disjoint union); C' is
union of 2& + 8 disjoint lines; every connected component of B" is a
reducible conic; let w be the number of connected components of A";
we assume 2 & < w < 3 & + 3.
2) Q is a smooth quadric containing Sing(Z") and w — 1 good secants of
type (1,0) to A".
We will prove in lemmas 8.1, 8.2, 8.3 that E(6k+2) =» E(6k+4) =>
=> E(6k + 6) => E(6k + 8) for every k > 1. Then we will prove £(6),£(8)
(lemma 8.4), thus completing the proof of E(n) for every even n, n^ 6.
LEMMA 8.1. £(6&4-2) implies E(6k+4)
for every
k>\.
Proof Let (Z,A,B,Q) be given by 2i(6&4-2). By semicontinuity we may
assume that dim (Z H Q) - 0. First assume k > 2. We add in Q 4& + 4
suitable lines D,-, / = 1,... ,4k + 4 , of type (0,1) with D, intersecting Z
if and only if i < 4k 4-1. If « > 2& 4-1, every line D,-, z < 4& 4-1, intersects a different final line of A. Put B1 = B and take as ^4' the union of A
and the D,, * < 4& 4-1. Assume w < 2k. Put z> = 2& 4- 1 and let U be the
union of v~ u connected components of B. We assume that the every D,,
i < 4k 4- 1, intersects a different final line of A U £/. We put Bl = B\U and
take as A' the union of ^4, W and the D,, * < 4k 4- 1.
If & = 1, we add 3 disjoint lines and 5 lines each of them intersecting
one of the components of A. We take as B' the union of B and one of the
5 new reducible conies constructed in this way.
•
LEMMA 8.2. E(6k + 4) implies E(6k 4-6) for every
k>\.
229
Proof. Let (Z', A\ B\ C, Q) be given by £(6& + 4 ) . We add in Q 4fc + 5
lines Di, i = 1,... ,4& 4- 5, of type (0,1) with D, intersecting Z if and
only if z < 2 & . First assume v>2k.
The lines D,, z < 2 & , intersects different final lines of A'. We put B" - B'\ we take as A" the union of A'
and the D f , z < 2 & , as C' the union of C and the D,, j < 2 f c + l. Now
assume v<2k-, we put w = 2k. Let W be the union of w— v connected
components of B. We assume that each D,, z < 2 & , intersects a different
final line of A' U W and that every connected component of W intersects
one of the D,. In this way we obtain the condition about the good secants.
We put B" = B'\W and take as A" a deformation of the union of A', W, D{,
z < 2 & , and as C' a deformation of the union of C and the Dy,
j>2k-\-l.
•
LEMMA 8.3. E(6k+6)
implies E(6k + 8) for every
k>\.
Proof Let (Z", 4 " , 5", C , Q) be given by E(6k 4- 6). We add in Q 4fc + 7
lines D,-, z = 1,... ,4k 4-7, of type (1,0) with Di} i<2k~l,
suitable
good secants to A" and each Dj, j > 2k, intersecting a different line of C'.
•
LEMMA 8.4. E(6) and E(8) are true.
Proof By [6] we know that the union Z of 12 general disjoint lines in P 3
are of maximal rank. We leave to the reader to check that the proof of [6]
gives E(6) and what is necessary to prove E(8): the condition of existence
of 6 good secants. Now £ ( 8 ) can be left to the reader.
•
Now we have proved E{n) for all even n with n > 6.
Let 5, di be a maximal datum for H{m), m even. Put d = di + ... 4- ds
and let e be the number of indices / with di = 2.
LEMMA 8.5. H(6k 4-4) holds for the maximal datum s,di if e > 11 k + 15 .
(k>3).
Proof. Take (Z,A,B,Q)
satisfying £(6&4-2) with dim (Z n Q) = 0. We
add in Q d~r(6k + 2 ) - 1 lines D,- of type (1,0). We assume that each
Di intersects a different final line of Z, that among the D,- there are the
u — 1 good secants to ^4 and that for each connected component W of B
either 0 or 2 of the D,- intersect W. This is possible because
a(6k+4)~
— r(6k+2) = 3k+4-&*3k+2.
If B has £ connected components, we add
230
s — t—1 suitable nilpotents. For example if L U V is a connected component of B, we may add a line D intersecting L,a line D' intersecting L'
and x(L H L'); we obtain a degeneration of a family of pair of disjoint
conies. The bound in the statement arise from t < 9k 4- 3 and r(6k + 4) +
+ l-r(6k + 2) = 4k+6.
•
LEMMA 8.6. H(6k 4- 6) holds for the maximal datum s,d± if e > 11 k 4- 8
(k > 2).
Pwo/. Take (Z',A',B',C,Q)
satisfying E(6k+4). We add in Q
x:=d-r(6k+4)
lines D,- of type (1,0) (3k 4- 5 < # <4& +7). We assume that
each Dj intersects a different final line of Z\ that among the D,- there are
v— 1 good secants to A't that each line in C intersects one of the D{ and
that for every connected components W of B' either 0 or 2 of the D,intersect W. •
Now we want to prove a similar statement for H(6k 4-8), k > 0. For
this we need another assertion F(6k+6). To prove F(6k+6) we need
another assertion F(6k + 4) which in turn follows from E(6k + 2).
F(6&+4), k>l.
There exists (Z\ A',B',D,Q)
such that:
1) Z' is union of 9k + 7 disjoint bamboos, deg Z' = r(6k + 4) - 1 and
b°(Jz,(6k + 4)) = 0; Z'=A'UB'UD
(disjoint union); every connected component of D is a line, deg£)<2&; every connected component of B' has even degree, deg 5 ' + degD < 22 & + 16; let a be the
number of connected components of A'\ we assume 1 <<z < k + 2.
2) 2 is a smooth quadric containing Sing(Z') and a — 1 good secants of
type (1,0) to A'.
F(6k+6),
k>\.
There exists (Z,A,B,D,Q)
such that:
1) Z is union of 11 k 4-12 disjoint bamboos, deg Z = r(6& + 6) — 1 and
h°(^z(6k
4- 6)) = 0; Z = A U £ U D (disjoint union); every connected
component of D is a line, degD = 2^ + 5; every connected component
of B has even degree, deg B < 24 & 4-16; let b the number of connected
components of A; we assume 1 l.
good secants of
231
Proof. Take (Z,AtB,Q)
satisfying E(6k + 2) with dim(ZOQ) = 0. We
add in Q 4&+4 lines Di} i = 1,... ,4k 4 4 , of type (1,0). Put h =
= min (2& 4-1, u). We assume that exactly 3 + £ — 1 of the lines D, do not
intersect Z, that h — 1 of the D, are good secants to A and that each of
the remaining lines intersects a different final line of A (at most one) or B;
we assume that for every connected component W of B either 0 or 2 of
the D, intersects W. •
LEMMA 8.8. F(6k + 4 ) implies F(6k 4- 6) /or m?ry & > 1.
Proof. Take (Z'f ^ ' , 5', D) satisfying F(6&4 4). We add 4& + 5 lines Dit
i — 1,..., 4-k 4- 5, of type (0,1) on Q. We assume that D, does not intersect Z' if z<2&, that each Dy, j>2k, intersects a different final line of
Z\ that each line in D intersects one of the Dj and that for each connected
component W of B' either 0 or 2 of the Dj intersect W. •
LEMMA 8.9. H(6k + 8) holds for the maximal datum s,dj if e> 15 & 4-14
(k > 2).
Proof Take (Z,A,B,D,Q) satisfying F(6k+6). We add in Q d~r(6k + 6)
( > 3 & 4 6) lines D,- of type (1,0). Among the D,- we choose w - 1 good
secants to A. We assume that each line of D intersects exactly one of the
Dj, that each D,- intersects a different final line of Z, that each connected
component of B intersects 0 or 2 of the D,- and that among the D,- there
are the h — \ good secants to A. If BUD has t connected components,
we add also s — t — 1 suitable nilpotents.
•
9. E(n) and applications to H(n + 2) (odd n).
Here we repeat the work done in § 8, now for odd integers. We define
the assertions E(n) (n odd, n>7), F(6k + 3) and F(6k + 5), k>l. Then
we prove E(J) (lemma 9.6) and the following chains of implications (lemmas
9.1,...,9.5):
£(6£ +1) => E(6k + 3)^>E(6k + 5), E(6k + 1) => F(6& 4 3) =* F(6fc 4- 5) =•
=* E(6k + 7) for every & > 1, hence we have £(w) for every w odd with
n>l. Then we prove, using E(n), that H(yi + 2) is true for n odd, n>l,
for every datum s,dj such that d\ = 2 for at least, say, 3n indices /'.
232
E(6k + 1), k>l.
There exists (Z,A,B,Q)
such that:
1) Z is union of 6k + 2 bamboos, deg Z = 6k2 48ft+2 and h°(Jz(6k + l)) =
= 0; Z = AUB (disjoint union); B is union of 5ft 4 1 disjoint reducible
conies and A has ft 4- 1 connected components.
2) Q is a smooth quadric containing Sing(Z) and ft good secants of type
(1,0) to A.
E(6k 4- 3), ft > 1. There exists (Z', A\ B'f C, C', Q) such that:
1) Z' is union of 8ft + 5 disjoint bamboos, deg Z' = 6ft2 4- 12 & + 5 and
h°(Jz,(6k + 3)) = 0; Z ' = ; 4 ' U 5 ' U C U C' (disjoint union); every connected component of C and Cr is a line, deg C = ft 4 1 , deg Cl == & + 2;
^4' has & + 1 connected components; every connected component of B'
is a reducible conic.
2) Q is a smooth quadric containing Sing(Z') and 2k + 1 good secants of
type (1,0) to A UC.
£(6& + 5), * > 1 . There exists (Z,A,B,D,Q)
such that:
1) Z is union of 6k + 6 disjoint bamboos, degZ = 6&2 + 16k + 10 and
£°(J* z (6& + 3)) = 0; Z = AUBUD
(disjoint union); A is union of 3
disjoint bamboos; any connected components of B has even degree,
13& + 9 < d e g £ < 1 3 & + 10.
2) Q is a smooth quadric containing Sing(Z) and 2 good secants of type
(1,0) to A
F(6k + 3), k>l.
There exists (Z', A\ B\ C, Q) such that:
1) Z' is union of 8& + 5 disjoint bamboos, degZ' = 6k2 + 12& + 5 and
h°(J>zi(6k + 3)) = 0. Every connected component of C is a line, deg C =
= 2& + 3; B' has 3& + 1 connected components of degree 2; A' has
3& + 1 connected components; Z' = A' U Br U C (disjoint union).
2) Q is a smooth quadric containing Sing(Z'), 3k good secants of type
(1,0) to A' and 2&+2 good secants of type (1,0) to C.
F(6£ + 5), ft > 1. There exists (Z", A", 5", D, Q) such that:
1) Z" is union of 6ft + 6 disjoint bamboos, deg Z" = 6ft2 + 16 ft + 10 and
h°(Jz»(6k + 5)) = 0; Z" = 4 " U B" U D (disjoint union); 4 " is the
union of 3 disjoint bamboos; every connected component of B" is a
reducible conic, degJB" = 6ft —2; every connected component of D is
a line, degD = 3ft + 4 .
233
2) Q is a smooth quadric containing Sing(Z") and 2 good secants of type
(1,0) to A".
LEMMA 9.1. E(6k 4- 1) implies E(6k 4- 3) for every k > 1.
Proof. Take (Z,A,B,Q) satisfying E(6k + 1). We add in Q 4k + 3 suitable
lines Df, i — 1,... ,4k 4- 3, of type (0,1). We assume that D, intersects Z
if and only if z<2& and that for z<2& the lines Df- intersect different
final lines of A. We take as A' a deformation of the union of A and the
lines Dj, / < 2 & , and as C U C ' a deformation of the union of the lines
Dj for 2* + l < ; < 4 * + 3. •
LEMMA 9.2. E(6k 4- 3) implies E(6k 4- 5) for every k > 1.
Prao/. Take (Z', ^ ' , 5', C, C", 0) satisfying E(6k + 3). We add in Q 4& + 5
suitable lines D,, l < / < 4 & + 5, of type (1,0). We assume that the lines
D, for 1 < i < 2& - 1 are good secants to A' UC, for 2& < i < 3k 4-1
they intersect different final lines of C\ If £4-4 is even we assume that the
D{, 3 k 4- 2 < i < 4& 4- 5, intersects different final lines of B' in such a way
that any connected component W of B' intersects either 0 or 2 of the
Di. If & + 4 is odd we assume that the same holds for 3&4-2<z<4&.4-4
while D4£ + 5 intersects a final line of A' not intersecting any Dj with
./<2fc —1. To obtain the 2 good secants of type (1,0) to A" for £ + 4
odd we have to assume that D4^ + 5 does not intersect a final line of A' U C
intersecting a good secant to A ' U C .
•
LEMMA 9.3. E(6k 4-1) implies F(6k 4-3) /or every'*k > 1.
/Voo/. Take (Z,.<4,J3,Q) satisfying £(6&4-l). As in lemma 9.1 we add in
Q 4& 4-3 suitable lines Dif i = 1,... ,4k 4- 3, of type (0,1). Now we assume that Di intersects Z if and only if z<2& and that the D,, z<2&,
intersect final lines of different connected components of B. Let T be the
union of the connected components of B intersecting the Di. We take as
A' a deformation of the union of A,T,D,, i<2k. We put B* = B\T.
•
LEMMA 9.4. F(6k 4- 3) implies F(6k 4- 5) for every k > 1.
/Voo/. Take (Z', 4 ' , £', C, Q) satisfying F(6fc + 3). We add in Q 4k 4- 5
lines Dj, * = 1,... ,4k 4- 5, of type (1,0). We assume that D,- for i<*3k is
a good secant to A', that for 3k 4-1 < i < 4k 4-1 D, does not intersects Z'
234
and that there exists 2 connected components W, W' of Br such that each
Dj, 4k + 2<j<4k
+ 5, intersects a different final line of W U W'. We take
as A" a deformation of the union of A\ W, W', D,- for i < 3k and 4k 4- 2 <
< * < 4 * + 5; put B" = B'\(WU Wf). The good secants to A" are of type
(0,1) since they come from Dj, 4 & + 2 < / < 4 & + 5, but we may interchange the name of the two system of lines on Q. •
LEMMA 9.5. F(6k + 5) implies E (6 k 4-7) for every
k>3.
Proof Take (Z", A", B", D, Q) satisfying F(6k+5). We add in Q 4k+6
lines Di} i = 1,... ,4k 4-6, of type (0,1). We choose TCB", T union of
k — 3 connected components of B". We assume that each D, for 1 < * ' <
< 3 k 4- 4 intersects a different final line of D (to obtain new conies) each
Di for 3& + 5 < z < 4 & 4 - 2 intersects a final line of a different connected
component of T (to obtain new good secant), that Dj, j = 4k 4- 3, 4k + 4
intersects different final lines of A" while Dr, r = 4k4-5, 4k + 6, does
not intersects Z"'. We take as i a deformation of the union of A",T,D\
for 3* + 3 3 , we have to prove not only E(7) but also £(13) and £(19) (see
lemma 9.7).
LEMMA 9.6. E(7) is true.
Proof. Take 5 disjoint lines Aj, i—l,...,5,
whose union Y is not contained in a surface of degree 3 ([6]). Then we take a smooth quadric Q and in
Q we take 5 lines £>,-, i = l , . . . , 5 , of type (1,0) with Di intersecting A,for *'<4, D5 not intersecting Y. We may assume h°( r (5)) = 0, where
T is the union of Y and the Dj. We deform T but not change the name of
the lines in T; we may assume dim (T H Q) = 0. Consider 6 lines i?,-, i =
= 1,... ,6, of type (1,0) in Q with i?2/+i intersecting A5-i for *'=
= 0,1,2, /?2i intersecting D6-j for i = 1,2,3. We take as Z a deformation
of the union of T, the Rif j < 6, x(^4 n £4), XW3 n ^3). •
LEMMA 9.7. £(13) flwrf £(19) are true.
Proof. We prove only £(13) because the proof of £(19) is similar. We start
with a disjoint union A of 8 conies satisfying £(7) and with 6 good secants
of type (0,1) (proof of 9.6). Then we obtain B with rB(9) bijective, B
235
union of lines, with a connected component of degree 20, 1 of degree 2 and
2 of degree 1 (and with 3 good secants of type (1,0)). Then we obtain C
with r c ( l l ) bijective, a connected component of degree 20, 1 of degree 2
and 10 of degree 1 (and with 11 good secants of type (1,0)). Then we find
something satisfying £(13).
•
Fix a maximal datum s, d\ for H(m), m = 6k + 5, 6k + 6 or 6k+ 3
and let e be the number of indices i with d\ = 2. Put d = d1 + ... + <is.
LEMMA 9.8. For k>l,
+ 7/2.
H(6k + 5) holds for the datum s,d{ if e> (15 k/2) +
Proof Take (Z,A,£,Q) satisfying £(6k + 3). Then add to Z
d-6k2— 12 & — 5 ( > 3& + 3) suitable lines D, of type (1,0). We assume that every
Di intersects Z only at final lines, that no line in Z intersects 2 of the D,,
that among the D, there are the good secants to A and that for any connected component W of B either 0 or 2 of the D,- intersect W. Then we
add suitable nilpotents.
The bound arises from d <^r(6k + 5). •
LEMMA 9.9. For every k > 1, H(6^ + 7 ) W d s /or £&e maximal datum s, dj
if 2e> Ilk + 14.
Proo/. Take (Z,^,B,C,C', Q) satisfying E(6k + 5). We add in Q d ~r(6k + 5)—l suitable lines Dj of type (1,0), each of them intersecting
a final line of Z. Among the D( we choose the k good secants to A and
2& + 3 lines, each of them intersecting a different connected component of
C U C'. We assume that for every connected component W of B, either 0
or 2 of the D; intersect W. We use tf(6& +7) <d<r(6k + 7). •
In the same way we obtain the following lemma.
LEMMA 9.10. For every k>2,
s,d{ if 2 e > 1 3 & + 12.
H(6k + 3) holds for the maximal datum
236
10. G (n) and applications to H(n 4- 2), n = 6k + 7, 6k 4- 8, 6£ 4- 9.
In this section we show that H(») is true for n = 6&+7, 6& + 8 or
6& 4- 9, for a maximal datum s, d{ such that for many indices i, d%r = 1.
The following assertion is very similar to B{6k 4- 5).
G(6& + 5), & > 0 . There exists (Z,A,£,£)) such that:
1) Z is union of 6k+ 6 disjoint bamboos, deg Z = r(6k + 6 ) - 1, and
h°(J>z(6k + 5)) = 0; Z — A^JB (disjoint union of connected components) with B union of 4&4-4 disjoint lines.
2) Q is a smooth quadric containing Sing(Z) and 2& + 1 good secants
to A.
We need also the following assertions.
G(6k+7),
k>0.
There exists (Z,A,B,Q)
such that:
1) Z is union of 6& + 8 disjoint bamboos, deg Z = r(6k 4 - 7 ) - 1, and
h°( <#z(6k 4-7)) = 0; Z = A U B (disjoint union of union of connected
components); A has 2&+2 connected components; B is disjoint
union of 2 lines and 4k 4- 4 reducible conies.
2) Q is a smooth quadric containing Sing(Z) and 2£ 4-1 good secants
to A.
G(6k+6),
k>0.
There exists (Z,A,B,Q)
such that:
1) Z is union of 5k + 6 disjoint bamboos, deg Z — r(6k + 6) and
h°(^z(6k 4-6)) = 0; Z = A U B (disjoint union of union of connected
components) and B is union of 2& + 3 disjoint lines.
2) Q is a smooth quadric containing Sing(Z) and 2£4-1 good secants to A.
LEMMA
lo.i. G (6k 4- 5) is true for every
k>0.
Proof The proof of lemma 5.2 shows that B(6k + 3) implies G(6k + 5).
LEMMA 10.2. G (6k 4- 5) implies G(6k+7)
for every
•
k>0.
Proof Take (Z,A,B,Q) satisfying G(6& + 5). We add in Q 4&4-6 lines
D^, K i < 4 H 6 , such that'each Dj, l < / < 4 & + 4 , intersects a different
connected component of £ while D4^ + 5 and D 4 ^ + 6 do not intersect Z.
237
LEMMA 10.3. G(6k +6) is true for every
k>0.
Proof. Take (Z,Q) satisfying B(6k+4)\ assume that Q contains 3&+2
good secants of type (1,0) to Z. We add to Z 4-k 4- 5 lines D,, K i <
<4& + 5, of type (0,1) in Q. We assume that D, intersects Z if and only
if i<2k+2.
•
From now on in this section we fix a maximal datum s,d{ for H(m).
Let c be the number of indices i with dj — 1. Put d — dx 4- ... 4- ds.
LEMMA 10.4. For every k>0
s, di if c>6k + 8.
H(6k + 7) holds for the maximal datum
Proof Take (Z,A,B,Q) satisfying G(6k + 5). In Q we add d-r(6k + 5) 4-1
(<4& + 10) disjoint lines Dj, with Dj good secants to i for *'<2&4-l,
intersecting a final line of A not intersecting any Dj with / < /' if i =
= 2fe + 2, 2fc 4 3, D^ not intersecting Z if i > 2fe 4- 4. Then we add a few
nilpotents.
•
In a similar way we obtain the following lemmas; note that in the omitted proof of 10.5 in (Z,A,B,Q) satisfying G(6k 4-7) we put a nilpotent
on the singular points of B.
LEMMA 10.5. For every k>0
s,di if c>10k + 16.
H(6k+9)
LEMMA 10.6. For every k>0
s,d{ if c>3k + 7.
H(6k 4 8) holds for the maximal datum
11. Proof of Hip),
holds for the maximal datum
n>8.
As we will show explicitly below, the statements proved till now show
that H{n) holds except for a finite number of n. This is the important part
of theorem 1. For completness' sake we prove H(n) also for low n and
show (in the next section) that the only exceptional cases are the ones listed
in theorem 1. In this section we prove Hin) for every n > 8 using heavily
the results proved in the previous sections, in particular in §4, 6, 7, 8, 10.
We distinguish 6 cases according to the congruence class of n mod (6).
We fix a maximal datum s, dj for H(n). Assume dj>dj if i>j. Let
238
c,e,t,d, respectively the number of indices i with d{ — 1,2,3,4. Put d =
= di + ... + ds. Let J be the set of indices i with d , > 3 . Put y,-=
= [(di~ l)/2]; put Wi = l if J,; = l,3, u>, = 0 if d , = 2, 1^ = 2 if ^ =
= 4,5 and in general w,- = [(2dj - 2)/3]. We put w(n) = x,y if in B(ra)
Z has degree x and y connected components. We will write X(n-, x,y\
c,e,t,k) for the assertion of the existence of (Z,A,B,Q) such that: (i) Z
is union of y disjoint bamboos, degZ = #; Z — A U B (disjoint union of
union of connected components of Z); (i.i) B has c connected components
of degree 1, e of degree 2, t of degree 3, k of degree 4; (iii) Q is a smooth
quadric containing Sing (Z) and y — c — e — k~ t~ 1 disjoint good secants
to A.
We will assume always s > 1 and d>s, otherwise the result is known
([6] and the proofs in [1] which works even for reducible curves).
H(6k + 5): Lemma 4.1 says that H(6k + 5), k > 1, is true if c 4- 2^y,- >
> 2k + 2. Note that 3 (c + 2,- yfi > d - 2e (and equality holds if and only
if d{ = 2 or 3 for every i). Hence for k > 1 H(6k + 5) holds for s,di if
d - 2 t f > 6 & + 10. By lemma 9.8 H(6* + 5), * > 1 , holds for s,d* if
2 e > 15£4-7. Hence it is sufficient to have d> 21 £ + 16. Since d>
>a(6k + 5) = 6k2 4-15& + 9, this is true for every k>2. We will prove
H(ll) in the second part of this section.
H(6k+6):
Lemma 4.3 says that H(6k+6), k>l,
holds for s,dj if c +
+ ?,iyi>k + 2, hence if d ~ 2 e > 3 * + 4 . Lemma 8.6 says that H(6* + 5)
holds for s,dj if d-2e>6k
+ 10. Since d>a(6k 4-6) = 6k2 + 17* 4- 12,
H(6* + 5) is always true for * > 2. We will prove H(12) in the second part
of the section.
H(6k + 8): Assume k > 1. Lemma 7.2 shows that H(6k + 8) holds for
s,dj if c 4- 2,-y,- > 4* + 9, c < s - l and, if c = s - l , d j > 3 . If the second condition is not satisfied, then H(6k + 8) follows from lemma 10.6.
The first condition is satisfied if d- 2e > \2k 4- 25. By 8.9 H(6k + 8)
holds if 2 * > 3 0 * + 2 8 . Since d >a(6k + 8) = 6k2 + 2 1 * 4- 18, H(6* + 8)
holds if k > 5. Now look at the proof of lemma 8.3. It gives that if in
E{6k+6)
Z" has w connected components, then for E(6k + 8) Z has
u> — 2* + 1 connected components. The proofs of lemmas 8.1, 8.2 show that
we may take u = v(k>2)
and u> = max (2*, z>). Put x = d — 6k2 — 26* —
- 27, hence 3* + 9 < * < 4* + 11. As in 8.2 we may find (Z", A", B\ C", Q)
satisfying E'(6k + 6), where E'(6k + 6) differ from E(6k+6) in the following way: w = 1, Q has 2k + 8 good secants of type (1,0) to A" U C'.
239
Then as in 8.9 we add x lines of type (1,0) to Q, among them the good
secants to A" U C" and the remaing ones intersecting final lines of B", 0 or
2 of them intersecting each connected component of B". Hence we obtain
H(6k + 8), k>2, if e>9k.
By 7.2 H(6fc + 8) is true for every
k>2.
Alternatively copy the proofs below for H(20),H(26) and H(32).
We will prove H(14) in the second part of the section.
H(6k +4): From 4.2 and 8.5 we obtain H(6k 4-4) for every k > 3.
Now we will prove H(16). We have 57 <d<6\
and w(14) = 48,8. If
c 4- 2 yi > d — 57, we may use the bamboo's trick (we add 9 lines obtaining
a bamboo of degree 57). Hence we may assume e > 12. We apply the bamboo's trick and nilpotents to (Z,A,B,Q) satisfying X(14; 48,8; 0,4,0,0). We
leave to the reader the following chain of implications:
X(10; 28,6,0,0,0,0) => X(12; 37,11; 6,0,0,0) =» X(14; 48,8; 0,6.0,0).
H(6k+7):
If d - s < 4 & + 3 and c < 6 & + 10, then (since
d~s>s~c)
2
we have d < 1 4 & 4- 16. Since d>a{6k+7)
= 6k + 19fc4- 15, this is false
for k > 1. By lemmas 6.3 and 9.9 H(6k 4-7) is true for every k > 4.
H(25); We have 1 2 6 10. We use the bamboo's
trick to obtain (almost) the following chain of implications; to conclude then
use the bamboo's trick:
X(15; 54,6; 0,0,0,0) =* X(17; 66,18;9,0,0,0) =* X(19; 80,20; 0,9,0,0) =»
X(21; 96,8; 0,9,0,0) =* X(23; 112,24,0,9,0,0). The third implication is
not quite true since we obtain a curve with only 4 good secants. But in the
last implication we obtain the good secant we need. Indeed take any two lines
A,B intersecting transversally Q and a line L of type (1,0) on Q\ we may
find 2 good secants of type (0,1) to AUBUL
and to AUBUL',
V suitable modification of L with dim (Z/ n Q) — 0.
Furthermore in the second implication we created some good secants
to the configuration obtained
H(19): We have 7 7 < d < % \ and w(17) = 66,18. If e<6 we use 6.3.
Assume e > 7 . We use the bamboo's trick and the following easy chain of
implications:
X(9; 24,3; 0,0,0,0) => X ( l l ; 32,12; 6,0,0,0) => X(13; 42,14; 5,0,0,0) =>
X(15; 54,6; 0,5,0,0) =• X(17; 66,18; 0,5,0,0) and we conclude with the
bamboo's trick if d>78. If d = 77, we have s>58, hence c>39 and we
use 6.3.
240
#(13): We have 4 0 < d < 4 3 and w ( l l ) = 32,12. First assume e > 3. We
may .use the bamboo's trick and the following chain of implications;
B(5) =>X(5; 10,6; 2,0,0,0) => X(7; 16,8; 1,2,0,0) •* X(9; 24,4; 0,3,0,0)
=* X ( l l ; 32,12; 0,3,0,0). Now assume c+e + t>Z. Choose natural numbers c', e\ t' with c'<c, ef <e, t'<:t and c' +e'-\-t' == 3. As above we
prove the existence of something in X ( l l ; 32,12; c',e',t',0)
and win.
Now assume e + c 4 - £ < 2 , hence s < 1 2 . We have d>42 and the same
proof works if £ 4- c 4- £ > 1. If e = c = t = 0, we have 2 zy,« > 17 and we
may use the proofs in § 6 (expecially 6.3) with Wj instead of yit
H(6fe + 3), k>2-. By lemma 9.10 it holds if 2e>13fe + 12. By 6.4 it holds
if c + Xyt> 10k + 6 and d-s> 12k. By 10.5 it holds if c>10k + 16.
Since d-s>s
- c, H(6k + 3) holds if k > 6. We will prove at the end of
the section H(9).
H(33): We have 2 1 0 < d < 2 2 7 and w(31) = 192,32. If * < 1 2 we may
apply 6.4. Assume e > 1 3 . We use the bamboo's trick and the following
chain of implications:
5(27) => X(29; 270,30; £,0,0,0) =* * ( 3 1 ; 192,32; 0,&,0,0) for every k
with 0 < f c < 1 3 .
H(27): We have 1 4 5 < d < 1 5 1 and w(25) = 130,26. Assume c + e>10;
choose e\ c' with e' < £, cr < c, e' 4- c' = 10. We use
£(21) =»X(23; 112,24; 10,0,0,0) =* X(25; 130,26; c', <?',0,0) and win (but
note that if d < 148 we need at most < 4 good secants to curves of degree
< 3; they arise from the construction). From now on we assume c + e < 9,
hence s > 5 3 and d > 148. Assume c4-£ + £ > 7 and choose c' < c, e'<,e,
t'<t
with <;'+*'4-*'= 7. We use 5(21) =• X(23; 112,24; 7,0,0,0) =*
X(25; 130,25; c\ e\ t',0) and win. We assume c + e + t<6. We have Xwt>
>2(d-2e-c-3t)/5
+c + t>2(d-12)/5,
hence Sw,->55. Check that
the proofs in § 6 work with this weaker bound.
H(21): We have 9 2 < d < 9 7 and w(19) = 80,20. If c + e>l and c + * +
+ £ > 7 , choose c' <c, e' <e, tf <t with *' < 6, c'+*'4-*' = 7. We use
5(15) =>X(17; 66,18; 7,0,0,0) => X(19; 80,20; c > ' , t\ 0) (with more
good secants and win. Assume c 4- e < 6 and either c = £ = 0 or c 4- £ + £ <
< 6. Then s < 3 2 , hence d>94. If c + e + t> 5, we win as above. Assume
c 4- e 4-1 < 4. We have 2 zy,- > 40 and we repeat the proofs in § 6.
.H(15): We have 5 1 < < / < 5 5 and w(13) = 42,14. We may use directly
the bamboo's trick only for d = 55. If c + e>4, we use 5(9)=*
241
AT(11; 32,12; 4,0,0,0) => X(13; 42,14; c', e', 0,0) (with more good secants)
with c' +e' = 4, c' < c, e' <e. If c + e < 3, then s < 20, hence d> 53; we
win if c + £ + £ > 2 . Let v be the number of indices i with d , = 5. Assume
c + e + t + k+v>2
and choose c ' < c , <?'<£, t ' < t, k'<k, v'<v with
r
c'+tf' 4- £'+&' 4-z> = 2; to fix the notations we assume k' = 2. We use
B(5) ==>X(7;16,8;2,0,0,0)=*X(9;24,4;0,1,1,0) ==>
X ( l l ; 32,12; 0,1,1,0) ==> X(13; 42,14; 0,0,0,2) and win. If c +e 4-1 + k +
+ v < 1, we have £ zyf- > 26 and we apply carefully the proofs in § 6.
Now we prove H{n) for 8 < « < 12 by a case by case checking.
H(S): We have 1 8 < i < 2 1 and w(6)=13,6. Thus the bamboo's trick
works if d < 20; but if d = 19 we have to check that we may apply 3.5 (iii)
i.e. that we add at most 8 nilpotents; indeed Z in B(6) has only 7 singular
points. If d - 21 we are in the injective range b) of H(9). Since 21*8 +
+ s+a = 165 with — 7 < # < 7 , we have s < 4 and | f l | < s —1. Hence we
have dx > 3 and we may use the proofs in § 6 and remark 3.6.
H(9): We have 22 < d < 25. and w(7) = 16,8. The bamboos's trick works
if d > 23;. indeed for d = 25 we have 25 ' 9 H-s - 220 > s - 1 and we are
in range b), hence we may apply remark 3.6. If d = 22, we have s> 13,
hence ds = 1; we find (Z,A,B,Q) satisfying X(7; 16,8; 1,0,0,0) and we
conclude as in § 10 using the bamboo's trick.
H(10): We have 2 6 < d < 2 9 and zu(8) = 20,5. The bamboo's trick works
only if d = 26. For d = 27, it is sufficient to have c + 2 j / j > l ; this is
true because d is odd. Assume d = 2S. If c + X y , - > 3 we may use the
bamboo's trick, 3.5(iii) and the following remark: if we put / > 10 nilpotents, we have s > ; 4- 1, we are in the range b) and we may loose ;'— 5
conditions. If c 4- 2 yi < 2, we have e > 11 and we use £(8) and the same
remark; if A U B satisfies £(8), we add aa even number of lines to each
connected component of B.
H ( l l ) : We have 3 0 < d < 3 3 (since #(11) = 0) and w (9) = 24,4. If
c + 2 yi > (d — 29), we may apply the proofs of § 6. If this inequality does
not hold, we find e>16~d.
We use 5(3) =» X(5; 10,6; 5,0,0,0)=*
X(7; 17,8; 1,5,0,0) => X(9; 23,13; 0,6,0,0) which gives the thesis if
d = 30; a similar chain of implications works even if d > 30.
H(12): We have 3 5 < d < 3 8 and w(10) = 28,6. The bamboo's trick
works if c 4- Sy,' > 35; indeed note that if d = 38, we may apply 3.5 (iii)
because for d = 38, we have s < 10, hence we add less than 11 nilpotents.
242
If the inequality does not hold, we have e > 42 - d. As in § 8 we use E{6) =•
X(8; 19,33; 5,1,0,0) -• X(10; 27,16; 0,6,0,0) and conclude.
12. H(n),
n<7.
In this section we prove that H(n), w < 7 , holds for all data, except
the 5 cases listed in theorem 1. A priori for n < 4 it is not sufficient to consider only the maximal data since H(n), n < 4, is not true. However it is
easy to check that for n < 4, every not excluded not maximal datum for
H(n) can be deduced from a maximal not excluded datum as in the proof
of lemma 2.1 and remark 2.2. Thus from now on we will consider only
maximal data. We fix a maximal datum s,d{ for H(n) with dj>dj for
z > / . Put d = dx 4- ... + ds. We assume d>s and s > l since in the remaining cases the result is known ([6] and the proofs of [1]).
H(l) is obviously true.
H{2) is true for s = 1, dx = 4, s — 2, dx = 3, d2 — 1 or s = d = 3,
false for s — dx—d2—2
(see the introduction and the proof of B(2) in
§5).
H(3): H(3) is false for s = 2, dx = 4, d2 = 2 and s = 3, dx = d2 =
= d3 — 2. Start from the union T of 2 disjoint lines, hence satisfying H(1).
Using a quadric Q we obtain every maximal case with dx < 3, s < 4 and
at most 2 indices i with d^2.
Starting from a rational normal curve we
obtain every case with dx > 4, d2 < 1, without adding any nilpotent.
Recall that w(n) = x,y means that for Bin) Z has degree x and
y connected components.
H(4). We have «(4) = 7, r(4) = 8, hence 7 < d < 9 , and w(2).= 4,2.
If d = 7, we can use the bamboo's trick and win; indeed we apply lemma
3.5 for a = 1, Z? = 4. Now assume d = 8. We start from A = B U C (disjoint union) with d e g £ = 3 , d e g C = l , hence satisfying H{2). Using T
and Q we obtain the solutions to H(4) for every datum with 1 < d 2 < 3,
d 3 < 1 and dj > 3. If 2 = dx = ... = d s , then s = 4 (exceptional case) or
s < 3 (not maximal case); 5 = 4, dx — d2 = d3 = 2, d 4 = 1 follows (using
Q) from the union of 3 disjoint lines plus a general point of F 3 . If s = 4,
î = 3, d2 = d3 = 2, d4 = 1 (hence we want to prove H(4), b)) we start
from B 6 B ( 3 ; 3,2,1), B not contained in any cubic surface, and we add a
reducible conic in a plane; the same construction works if s = 3, dx = d2 =
= 3, d3 = 2 , starting from B G 5 ( 2 ; 3,3).
243
If d = 9, then s < 2 . If d2<3, we start from BGB(2-} 3,1). If
d\ — 5, d2 = 4 we may start from A E B ( 2 ; 3 , 2 ) and add 4 lines in a quadric
Q (indeed we have to prove H(4), b)).
H(5). We have a(5) = 9, r(5) = 11, $(5) = 0, hence 9 l , we may apply the bamboo's trick
because we have to use lemma 3.5 for a = 1 and b = 5. It remains only the
case s = 5, d1 = ... = ds = 2. We start from 5 disjoint lines Y not contained
in any cubic surface ([6]). We add in a smooth quadric 5 lines, each of them
intersecting one of the lines in F, and a point. Now assume d — 11. We start
from the union Y of 5 lines not contained in any cubic surface. Any 3 disjoint lines are contained in a smooth quadric. Hence we may find a quadric
Q containing t < 3 suitable good secants to Y. We may obtain any configuration we want without using any nilpotent because, d - 5 = 5 + 1.
H(6). We have a(6) = 12, r(6) = 13, hence 12<d<U,
and w(4) =
= 8,3. If d = 12 we apply the bamboo's trick and 3.5(iii) since we add at
most 5 nilpotents.
If d = 13 the bamboo's trick works if c + 2y,- > 1; this is true because
13 is odd. Now assume d = 14, hence s < 5 . If we have c + 2 y , > 2 , we
may apply the bamboo's trick because we are in H(6), b), 14 • 6 - 84 + s = s
and we may apply remark 3.6. The inequality is always satisfied because
s<5.
H(7). We have a(7) = 15, r(7) = 17, q{l) = 0, hence 1 5 < d < 1 7 ,
and 1^(5) =10,6. The bamboo's trick works if J < 1 6 . If J = 17 we
cannot apply lemma 3.5 but since 17*7-1-5 — 120 = s — 1, we may add at
most s — 1 nilpotents because we can "loose" s — 1 conditions to prove
this case of H(7), b), hence we may apply remark 3.6.
APPENDIX. Not every curve has maximal rank.
In this appendix we will prove the existence of a few curves which have
not maximal rank. These examples will not be counterexamples to any
"maximal rank conjecture" since they are not general in the irreducible component of Hilb (IP^) containing them. However they will show how (hopeless?) difficult should be to understand the postulation of every curve in a
projective space.
This appendix was stimulated by [5].
244
Define integers a(n ,N), b (n ,N), r(n ,N), q (n ,N) by the relations
(n + l)a(n,N) + b(n,N)=(
/N + n\
J ,
0<b(n,N)<n
fN + n\
nr(n,N) + q(n,N) = [
1 , 0 <g(«,AT)< n - 1 .
By [6] a(n,N) is the maximal number of general lines whose disjoint
union has rY1B (n) surjective.
PROPOSITION A.l. Take positive integers kynyN with N>3,
n-Kk<
<:a(n,N). Then there exists a union Y of k disjoint lines with rYW (n)
surjective but rYW (n — 1) not surjective, Y spanning a linear space of
dimension min (N, 2d — n).
Note that if k <#(ra — 1, N), such a Y has not maximal rank and that
if k > a(n — 1, N) the proposition follows from [6].
It is sufficient to find a union Y of k disjoint lines with rY1P (n)
surjective and a line D with card ( F f l D ) > « + l, Indeed any hypersurface
of degree n containing Y contains D, hence the homogeneous ideal of Y
is not generated by forms of degree < n. By Castelnuovo's lemma [9] p. 99,
hl{WN,JtYvN(n-l))*0.
We will show that proposition A.l follows from the proof of [6].
Consider first the case N= 3. We define Hn,H^k-x
as in [6] but
adding the condition that for the union Y of a(n) disjoint lines (resp.
(k - 1)(3& - 2)/2 disjoint lines and 2k reducible conies) there exists a line
D with d i m ( y n D ) = 0 and card (Y f) D) > n + 1 (resp. > 3k). H2 is
obvious since 3 disjoint lines are in a smooth quadric. H3 follows from Hx;
take a smooth quadric Q, L 4 ,L 5 not in Q and 3 disjoint lines in Q. H 4
follows from 5(2); take the union Y of a bamboo T, degT— 3, and a
disjoint line; call PtP' the singular points of T\ take a smooth quadric Q
containing P,P' and a secant to T; in Q add 3 lines Lj, L2, L3, x(P) anc*
X(P'); let Z be the union of Y, Lit x(P), x(^'); we may deform (2, T) to
(Z', T'), Z' D I,- for i < 3, in such a way that there is a line D in Q intersecting the Lj and 2 lines of T'. Then the proof of Hn-2 => Hn (n =
= 0,2 mod (3)) and H 3 ^_ 3 ^ H ^ - j =»// 3 ^ + 1 holds verbatim: at each step
D intersects transversally Q and we add in Q many disjoint lines, 2 of them
intersecting D. We can control the postulation of the intersection of such a
configuration with Q either using [1] lemma 6.2 or using the proofs in § 3.
245
For N>4 we use Hn N, H'n N with a similar modification, while we
use the same H% N _ t as in [6]. Note that Hl,H'1 are true because any two
points are collinear. At each step of the induction (made using a hyperplane
H) we take H intersecting transversally D and add in H a line passing
through DCiH. This is possible because a(n,N)> r(n - 1, N) 4- 1 (hence
with the notations of [6] rn>sn^1)
for JV>4, n<2. Indeed from the
definition of a(n,N),r(n — 1,N) we obtain:
(n + l)(a(n,N)-a(n-l,N))+a(n-l,N)
+ b(n,N)-b(n-l,N)
={
Assume a(n,N)<:r(n - 1 , A/). Since (n ~ \)r(n - 1 , N) <*na(n - 1 , JV).+
, .
/n+N-l\
/iV4-«-l\
+ n, we obtain 2r(n - 1, N) 4- 2n > (
1, hence 2 (
+
/n+N-l\
\
N _ 1
/
\
.N
J
+ 2n(n~ 1)> (n - 1) f
_
1, which is false for w = 2, AT> 6 or w> 3,
N>5 or JV=4, w > 5 . For low n we use a(2,N)>N+ 1, a(3,4) = 8,
r(2,4) = 7, a(4,4) = 14, r(3,4) = l l .
The postulation of the intersection of the configurations we obtain with
a hyperplane H cause no trouble. We take d suitable lines in the configuration constructed for Hn N.
In the same way it is possible to prove the existence of many reducible
rational curves which are not of maximal rank. But we are interested in
finding smooth rational curves which are not of maximal rank. If we find in
FN a rational, connected curve Y and a line D with dim (D C\Y) = 0 and
c a r d ( F n D ) > n + 1 , we want to show that Y is a limit of a family Yt,
t E T, of smooth rational curves with card(F,'. C\ D ) > n 4-1 for every t.
We have to generalize the construction of [2] prop. 2.1. A similar results holds
even for higher genera if the hyperplane section is not special. We need a few
notations.
Fix a complete curve X and a very ample line bundle L on X; for
simplicity we assume X irreducible. Put N = h°(X,L)— 1, ^ = degL. Fix
n<N. Let Pr^(L,P w ) be the closure in Pw of the set of CCHilb(P M )
such that there is an isomorphism h : X-* C with h*(Oc(l)) — L. Note that
if <pL is the embedding given by the complete linear system H°(X,L), we
may see such a C as a projection of yL (X) from P ^ into IP„ (up to an
element of PGL(N+1)). In [2] it was showed that if 3 <n <N, P^(L,P„)
contains many reducible curves. We want to prove that some reducible curve
is a limit of smooth curves C with card (C O D)>s for some fixed s.
246
PROPOSITION A.2. Let CGP„, n>3, be a smooth, connected curve. Put
M : = 0 C ( 1 ) and assume M not special. Fix s distinct points Plt...,Ps
in
C. Let D be a general line of P„ and Llt...,Ls
be general lines intersecting D and with Pj G L, for every i. Then C U Lx U ... U Ls is a limit of
a family {Ct}, tST, of smooth curves in PiîMiPi + ... + PS), P„), d =
= deg C 4- s, with card (Ct C\D)>s.
Proof. First assume C linearly normal. Consider Pw as a hyperplane of
P w + 1 = P(H°(C,Af(Pj))). Take W CWn+u j : C-+W an isomorphism with
j*(Oc(l))=M(P1).
Let p be the projection of P w + 1 into Pw from/(Pj).
We may assume C—p{W). Take a general line D' in P„ + 1 and put D =
= p ( D ) . Take a general line i?! with ; ( ? , ) G Rlt Rx h D' =#0. Moving D',
i?! we may assume that the projection p , of W into PM from a general
point ^4^ of i?j is an embedding. The family {pt(W)} has as special fiber
when At tends to Pj the union of C and the line in Pw through RxC\Wn
and the intersection with Wn of the tangent line to W at Px (see [2] prop.
2.1 and the picture in [2]). This solve the case 5 = 1 . If s > 1, by induction
for general lines Rj, i = 2,... ,s, with ;(P,) Gi?,- and D C\ Rj=£0 for every
i such that t y U i ? 2 U . , . U # $ is a limit of smooth curves in Prd~1(M(P2 +
+ . . . + P S , P w +i) each of them intersecting D' at s~ 1 points. We may
repeat the proof for the case s = 1 if we are sure that for general At G R:
pt embedds W U R2 U ... U i?s into P w . We have to check that a general
At is not on a line intersecting 2 of the R^s with i > 2 or one R{, i>2,
and W. The first assertion follows from the fact that /(Pi) lies in no 3-secant line to W because M is very ample. Assume that every point of Rt
lies on the scheme V closure of the union of the lines intersecting Rs\{j(Ps)}
and W. We may move D' and the Rk, k<s, keeping fixed Rs, hence V,
and obtain a contradiction. If C is not linearly normal, we embedd C with
the complete linear system H°(C,M), we repeat the construction just made
in P(H°(C,M)) and then project the configuration obtained into P Mi *•
PROPOSITION A.3. Fix integers n,d with » + 2 «fe'J<r(»), n>4. Then
there exists a smooth rational curve Y of degree d with rY(n) surjective
and rY(n — \) not surjective.
Proof. First we assume d> In -I- 3 and n > 5. We steal the proofs in [4].
We may repeat the proofs in [1] using always reducible curves, union of a
smooth rational curve C and a finite number of disjoint trees, each tree intersecting C exactly at a point and quasi-transversally; it is sufficient to
247
control the intersection of such a curve with a quadric Q\ we use [1] lemma
6.2 and the construction at the beginning of [1], 6.6, or our § 3. Indeed a
preliminary version of [1] was written in this more general setting, smoothing
the reducible curve only at the end of the paper, with deg C = 3. The next
assertion A(n) is similar to the assertion H^\^x)
of [1]:
A(n), n>4:
There exists (T,Z,R,S,D)
such that:
1) T is the union of a smooth rational curve Z of degree ri— 1 and r(n) —
— n + 1 disjoint lines L,, each L,- intersecting Z quasi-transversally
and only at a point.
2) R is a line intersecting Z, SCR,
card(S) = #(«) and h°(J>zus(ri))
= 0.
3) D is a line intersecting n 4- 1 of the lines L, in T.
A(4) is true since it follows from EF2 (Pi) of [1] adding 4 lines in quadric.
The proof of A(5) is similar. Now we can prove that A{n — 2) implies
A{n) for n > 6 using [1] and the proof of proposition A.l. For example we
show how to prove that A(6k + 3) implies A(6k 4- 5). We take (T,Z,K,S,D)
given by A(6k + 3). We take a smooth quadric Q containing R; deforming
T,D we may assume that Q intersects transversally T,D. We take in Q
4&+4 lines Dj, i = 1,... ,4& 4-4, each D,- intersecting /?, with D,- intersecting D for i = 4&4-.3, 4 & + 4 and D,- intersecting Z if and only if
2 < z ' < 2 & + 2. Since 2 ( « - 3 ) > w - l w*e. may assume 5 = { D , n i ? } ,
2<i<2k
+2.
Then we take as Z' for A(n) a deformation of Z U R U Dx and we
add 2& + 1 nilpotents x(^)» P^S. It is easy to control the postulation of
Z C\Q. For T O Q the only new problem is given by the » — 1 lines intersecting both Z and D. Take the smooth quadric Q' spanned by D and 2
lines A, A' in Q intersecting i?. We rriay assume that Q' intersects Z at
2(n— 1) points. Hence we may.find ra — 1 lines in Q' intersecting Z, in
particular intersecting Q at »— 1 points on A and w — 1 points on A'.
Now we may repeat the construction outlined at the beginning of the proof
of [1] prop. 6.6 (or of lemma 3.3). The theorem follows from A(n) and
proposition A.2. Now assume d<*2n + 2. We consider the following assertion I(n):
I(n), n>6:
There exists (T,C,D) such that:
1) C is a rational normal curve in JP3, T is the union of C and 2n — 1 disjoint lines, each of them intersecting quasi-transversally and only at a point
C; rT(n) is surjective.
248
2) D is a line 2 secant to C and intersecting n - 1 of the lines in T.
We prove by induction that I(n~2) implies I(n) adding in a smooth
quadric 4 lines intersecting C, 2 of them intersecting D (hence we have to
choose carefully Q). If d < In 4- 1 we take a suitable F C T, YD C, deg F =
= d. But one that we cannot apply the statement of A.2 because we impose
that 2 of the points of T H D are in C. In the proof of A.2 take as D' a
general secant line to W.
For n = 4,5 use a statement similar to /(w) but with r(n)- 3 instead
of In- 1. •
We note the irreducibility of the set of triples (T,Z,D) where Z is a
smooth rational curve of degree d — n — 1, D is a line, T is the union of Z
and n 4- 1 lines L,- intersecting D, each L,- intersecting Z quasi-transversally and only at a point. Hence it is possible to show that for d<^r(n — 1)
we may take Y in the statement of A.3 with b1 (^Y(n - 1)) = 1. A similar
remark holds for proposition A.l.
Consider the union of something not contained in a surface of degree t
and n + 1 suitable lines; by irreducibility it shows that we may assume in A.3
that h°( J>Y(t)) = 0 if r(t) 4- n + 2 < d.
REFERENCES
[1]
Ballico (E.) and Ellia (Ph.): Generic curves of small genus in P 3 are of maximal
rank, Math. Ann., Vol. 264, 1983, pp. 211-225.
[2]
Ballico (E.) and Ellia (Ph.); On degeneration of projective curves, Algebraic geometry-open problems, Proc. Conf., Ravello/Italy 1982, Springer Lect. Notes Math.
997,1983, pp. 1-15.
[3]
Ballico (E.) and Ellia (Ph.): On the postulation of many disjoint rational curves in
P N , N > 4, Boll. U.M.I., (to appear).
[4]
Ballico (E.) and Ellia (Ph.): Sur la postulation des courbes de P w et de leurs projec1
tions, preprint.
[5]
Gruson (L.), Lazarsfeld (R.) and Peskine (C): On a theorem of Castelnuovo, and
the equations defining space curves, Invent. Math., Vol. 72, 1983, pp. 491-506.
[6]
Hartshorne (R.) and Hirschowitz (A.): Droites en position generale dans Vespace
projectif, Algebraic Geometry, Proc. La Rabida 1981, Springer Lect. Notes Math.
961, 1982, pp. 169-189.
249
[7]
Hartshorne (R.) and Hirschowitz (A.): Cohomology of a general instanton bundle,
Ann. Scient. E.N.S. Vol. 15, 1982, pp. 365-390.
[8]
Hirschowitz (A.): Sur la postulation generique des courbes rationelles, Acta Math.,
Vol. 146, 1981, pp. 209-230.
[9]
Mumford (D.): Lectures on curves over an algebraic surface, Ann. of Math. Studies
59, Princeton Univ. Press N.J., 200 0., 1966.
[10] Rao (P.): A note on cohomology modules of rank two bundles, J. of Algebra, Vol.
86, 1984, pp. 23-34.
[11] Tannenbaum (A.): Deformations of space curves, Arch. Math., Vol. 34, 1980,
pp. 37-42.
EDOARDO BALLICO - Scuola normale superiore - 56100 Pisa - Italy.
Lavoro pervenuto in redazione il 21/XI/1984 e in forma definitiva il 21/Vl/I985h

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Edoardo Ballico ON THE POSTULATION OF DISJOINT RATIONAL

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