The Connectedness of Space Curve Invariants

18 Pages·0.15 MB·English

by  Michele Cook

The Connectedness of Space Curve Invariants
4
9
9 Michele Cook
1

n
February 5, 2008
a
J

8
2

Introduction
1
v
9
Let S = C[x ,...,x ] be the set of polynomials in n+1 variables over
0 1 n+1
0 C (usually corresponding to the homogeneous coordinate ring of Pn). Let
1
≻ be the reverse lexicographical order on the monomials of S. Given a
0
4 homogeneous ideal I in S, gin(I), the generic initial ideal of I, is the
9 Borel-fixed monomial ideal associated to I. (For details see [B], [BM], [BS]
/
m or [Gr].)
o The generic initial ideal reflects much of the structure of the original ideal,
e
for example, it has the same Hilbert function and the same regularity. Also,
g
- many geometric problems can be reduced to combinatorial ones which may
g
l be phrased in terms of generic initial ideals. Thus, we would like to have
a
: some rules governing the kinds of Borel-fixed monomial ideals which can
v
i occur from geometry.
X
The first set of rules are supplied by a result of Gruson and Peskine ([GP]):
r
a
Theorem 1 (Gruson-Peskine) Let Γ ⊂ P2 be a set of points in general
position, with generic initial ideal gin(I ) = (xs, xs−1xλs−1, ..., x xλ1, xλ0).
Γ 1 1 2 1 2 2
Then the invariants, {λ }s−1, satisfy
i i=0
λ +2 ≥ λ ≥ λ +1 for all i < s−1
i+1 i i+1
and we say that the invariants of Γ are connected.
Associated to a space curve there are invariants which generalize the invari-
ants of points in P2 (see below). The main result of this paper is to show
1
that the invariants of a reduced, irreducible, non-degenerate curves in P3
also satisfy a connectedness property:
Let C be a reduced, irreducible, non-degenerate curve in P3, As I is satu-
C
rated, the generators of gin(I) will be of the form xi1xi2xf(i1,i2).
1 2 3
Let
s = min{i | f(i,0) ≤ k},
k
µ (k) = min{j | f(i,j) ≤ k} for 0 ≤ i ≤ s −1.
i k
We call {µ (k)} the invariants of C. For k >> 0, µ (k) = λ , where the
i i i
{λ } are the invariants of a generic hyperplane section, Γ = H∩C, of C. The
i
main result of this paper is to prove:
Theorem 2 (The connectness of curve invariants) Let C be a reduced,
irreducible, non-degenerate curve in P3, then the invariants, {µ (j)}, of C
i
are such that
µ (k)+2 ≥ µ (k) ≥ µ (k)+1 for 0 ≤ i < s −1,
i+1 i i+1 k
and we say that {µ (k)} is connected. Furthermore, if s < s , then
i k 0
µsk−1(k) ≤ 2.
This result greatly restricts the kind of Borel-fixed monomial ideal which can
occur as the generic initial ideal of such curves and thus gives us quite a lot
of control over their Hilbert functions.
The paper is broken into parts as follows: In Section 1, we will prove the
Connectedness Theorem. The proof incorporates the ideas used in the proof
of the Gruson, Peskine result ([GP], [EP]) and a more differential approach
due to Strano ([S]). In Section 2, we will give some further restrictions on
the generic initial ideal of a reduced, irreducible, non-degenerate curve in P3.
We will generalize a result of Strano on the effect of certain generators of the
generic initial ideal of the curve on the syzygies of the hyperplane section
ideal to a larger group of ideals. We will also generalize another result of
Strano on complete intersections to curves whose hyperplane sections have
invariants of which the first few are like those of a complete intersection. I
believe this last result will lead to some interesting results on the gaps of
Halphen ([E]).
Acknowledgement. This paper is taken from my dissertation and I would like
to take this opportunity to thank my advisor Mark Green, for his insight and
patience.
2
1 Proof of the Connectedness Theorem
1.1 A pictorial description of monomial ideals
We will first give an description, due to M. Green ([Gr]), of how an ideal I
whose generators are monomials in three variables may be described pictori-
ally. Hopefully, this desciption will make subsequent explanations clearer.
First, draw a triangle corresponding to the monomials xixj, where i+j = n,
1 2
0 ≤ n ≤ n . Let f(i,j) = min{k | xixjxk ∈ I}. For each i,j, if f(i,j) = ∞
0 1 2 3
(i.e. xixjxk ∈/ I for all k ≥ 0) put a circle in the (i, j) position, if 0 <
1 2 3
f(i,j) < ∞ put the number f(i,j) in the (i, j) position, and if f(i,j) = 0
put an X in the (i, j) position.
Thus the triangle
(cid:20)T
(cid:20) kT
(cid:20) k kT
(cid:20) T
kk k
(cid:20) T
1k1k3kk
(cid:20) T
(cid:20)X X X X 1kT
corresponds to the ideal (x3x ,x2x x ,x x2x3,x4,x3x ,x2x2,x x3,x4x ,x5).
1 3 1 2 3 1 2 3 1 1 2 1 2 1 2 2 3 2
Note: For (i, j) not in the picture, it is assumed that xixj ∈ I.
1 2
Remark: It is the Borel-fixedness that ensures the step like look.
In this example the invariants are as follows:
µ (0) = 5 µ (0) = 3 µ (0) = 2 µ (0) = 1
0 1 2 3
µ (1) = 4 µ (1) = 3 µ (1) = 1
0 1 2
µ (2) = 4 µ (2) = 3 µ (2) = 1
0 1 2
µ (3) = 4 µ (3) = 2 µ (3) = 1
0 1 2
and µ (k) = µ (3) for k ≥ 3.
i i
Note: The invariants of this monomial ideal are connected.
To prove Theorem 1, we will first show in section 1.2, that if there is an
i and j, with 0 ≤ i < s − 1, such that µ (j) + 2 < µ (j), then for
j i+1 i
generic hyperplanes H,H′, the ideal J = (I| : (H′ ∩ H)j) can be “split”
H
so that there exists a homogeneous polynomial X of degree i+1 such that
gin(X) ∩ gin(J) = gin(X ∩ J). We will then show in section 1.3, that if I
arises as the ideal of a reduced, irreducible, non-degenerate curve C ∈ P3,
such an X cannot exist.
3
1.2 Splitting a non-connected ideal
Definition. An elementary move e for 1 ≤ k ≤ n is defined by:
k
e (xJ) = xJˆ,
k
where Jˆ = (j1,...jk−1,jk + 1,jk+1 − 1,jk+2,...,jn+1) and where we adopt
the convention that xJ = 0 if j < 0 for some m.
m
One can show that a monomial ideal I, is Borel-fixed if and only if for all
xJ ∈ I and for every elementary move e , e (xJ) ∈ I.
k k
Theorem 3 (syzygy configuration) Let I be a Borel-fixed monomial ideal
with generators xJ1, ...,xJN. Then the first syzygies of I is generated by
{xi⊗xJj−xLij⊗xJlij | 1 ≤ j ≤ N, 0 < i < max(Jj), min(Lij) ≥ max(Jlij)},
where max(J) = max{i | j > 0} and min(J) = min{i|j > 0}.
i i
The proof of this is due to Green and may be found in [Gr].
Theorem 4 Let I be a homogeneous ideal generated in degree ≤ n, with
generators xJ1, ...,xJN of gin(I) in degree ≤ n. Then any generator P of
gin(I) in degree n+1 is such that P ≺ x xJj, for some J such that |J | = n
i j j
and i < max(J ).
j
Proof.
We will first use the syzygies of (xJ1, ...,xJN) to obtain some new generators
of gin(I) which satisfy the conditions stated above.
n+1
(xJ1, ...,xJN) is a Borel fixed monomial ideal so by Theorem 3, the syzygies
among the xJj are generated by syzygies of the form
{ xk ⊗xJj −xLkj ⊗xJlkj | 1 ≤ j ≤ N, 1 ≤ k < max(Jj)}.
By Galligo’sTheorem ([Ga]) we may assume, after a possible change of basis,
thatgin(I) = in(I). Letg ∈ I besuchthatin(g ) = xJi andassumetheg are
i i i
monic. Given asyzygy asabove, leth = x g −xLkjg , thenin(h ) ≺ x xJj.
1 k j lkj 1 k
Given hi, if in(hi) = xKi+1xJji+1, let hi+1 = hi −ai+1xKi+1gji+1, where ai+1 is
the leading coefficient of h , so that in(h ) ≺ in(h ) ≺ x xJj.
i i+1 i k
This process must terminate, so for i sufficiently large we get either h = 0,
i
4
(in particular this will occur if degh ≤ n) or xJj does not divide in(h ) for
i i
all 1 ≤ j ≤ N, in which case in(h ) is a new generator of gin(I) satisfying
i
the conditions stated in the theorem.
Now let P = in(h) be a generator of gin(I) , h = Pf g .
n+1 i i
Let i be such that, in(f g ) is maximal, and of the maximal in(f g ), g
0 i0 i0 i0 i0 i0
has maximal degree, and among those g of the same maximal degree in(g )
i0 i0
is minimal.
As P 6= in(f g ), there exists i such that in(f g ) = in(f g ). We have
i0 i0 1 i1 i1 i0 i0
picked i in such a way that either degg > degg , or degg = degg and
0 i0 i1 i0 i1
in(g ) ≺ in(g ).
i0 i1
We now want to show that x |in(f ) for some i < max(J ), where in(g ) =
i i0 i0 i0
xJi0.
Claim.
Let xA,xB be generating monomials of a Borel-fixed monomial ideal, I, such
that |A| > |B| and xMxA = xNxB for some monomials xM and xN. Then
there exists x |xM such that i < max(A).
i
Proof of Claim.
Let A = (a ,...,a ,0,...,0), and B = (b ,...,b ,b ,...,b ), assume
1 s 1 s s+1 n+1
that s = max(A). Suppose b ≤ a for all i ≤ s, then we may apply elemen-
i i
tary moves to B to get Bˆ such that xBˆ ∈ I with bˆ ≤ a for all i, and as
i i
|Bˆ| = |B| < |A|, this would imply xA cannot be a generator. Therefore there
exists b > a for some i ≤ s, and if i < s we are done. If b ≤ a for all i < s
i i i i
but b > a we may again apply elementary moves to B and get the same
s s
conclusion. Therefore there exists b > a for i < s.
i i
If degg = degg and in(g ) ≺ in(g ), let in(g ) = xA, in(g ) = xB where
i0 i1 i0 i1 i0 i1
A = (a ,a ,...a ) and B = (b ,b ,...b ). Then there exists s such that
1 2 n+1 1 2 n+1
a = b for all k > s and a > b . As the degrees are the same, there must
k k s s
exist a < b for some i < s and hence x |in(f ) for some i < s ≤ max(J ).
i i i i0 i0
Thus, in either case we have x |in(f ) with i < max(J ). Consider the
i i0 i0
syzygy
xi ⊗xJi0 −xLii0 ⊗xJlii0
∗
and let h be the element of I obtained from this syzygy, as above.
h∗ = xigi0 −xLii0glii0 − PaixKigi where in(xKigi) ≺ in(xigi0), and∗by The-
orem 3 either |J | > |J | or in(g ) ≻ in(g ). Let h = h−e h , where
i0 lii0 lii0 i0 1 i0
e = a in(f )/x and a is the leading coefficient of f .
i0 i0 i0 i i0 i0
∗ ∗
If dege = 0, in(h ) (cid:22) max{in(h),in(h )}. If in(h) = in(h ), we are done
i0 1
5
as inh∗ ≺ xixJi0. Therefore we may assume in(h) 6= in(h∗), and so either
∗
in(h ) = in(h ) or in(h ) = in(h). In the first case we get P = in(h) ≺
1 1
in(h∗) ≺ xixJi0, and we are done.
∗
If dege ≥ 1, h = 0 and in(h ) = in(h) = P.
i0 1
Thereforeweareleftwiththecasein(h ) = P,however inaddingthemultiple
1
∗
of h , we only added terms which either have initial term less than the term
in(f g ), or if the term has thesame initialterm as the g term, correspond-
i0 i0 i0
ingtothexLii0glii0 term, theneither glii0 oflesser degree, orin(glii0) ≻ in(gi0).
Therefore we may proceed by induction and get the required result. ⋄
Now suppose that gin(I) is such that there exists i and j such that
µ (j)+2 < µ (j), where 0 ≤ i < s −1.
i+1 i j
Consider the ideal J = (I| : (H′ ∩ H)j) for generic hyperplanes H,H′.
H
Then gin(J) = (gin(I) : xj) and J has invariants ν (k) = µ (k + j), and in
3 i i
particular
ν (0)+2 < ν (0).
i+1 i
For example, pictorially we could be in the following situation:
gin(I) = (cid:20)T gin(I|H : (H ∩H′)2) = (cid:20)T
(cid:20) kT (cid:20) kT
(cid:20) kkT (cid:20) k kT
(cid:20)4kk kT (cid:20)2kk kT
(cid:20) T (cid:20) T
1k1k2k k X X X k
(cid:20) T (cid:20) T
(cid:20)X X X X 3kT (cid:20)X X X X 1kT
Let K be the ideal generated by elements of degree ≤ i + ν (0) + 2 in
i+1
J. We want to show that there exists an ideal K ⊆ Kˆ ⊆ J such that
gin(Kˆ) = (xi+1)∩gin(J).
1
Corollary 5 All elements of gin(K) are divisible by xi+1.
1
Proof.
Let xaxbxc ∈ gin(K) ford ≤ i+ν (0)+2. Ifa ≤ i, thenbyBorel-fixedness
1 2 3 d i+1
xixνi+1(0)+2 ∈ gin(K) ⊂ gin(J), but xixνi(0) is a generator of gin(J) and so
1 2 1 2
ν (0)+2 ≥ ν (0). But ν (0) > ν (0)+2.
i+1 i i i+1
Suppose by induction, all elements of gin(K) are divisible by xi+1 for d ≥
d 1
i+ν (0)+2.
i+1
6
Claim.
If d ≥ i+ν (0)+2, then any generator of gin(K) has an x term.
i+1 d 3
Proof of claim.
If xaxbxc a generator of gin(K) , then a ≥ i+1. Let a = i+1+j, we have
1 2 3 d
xi+1+jxνi+1+j(0) ∈ gin(J), and i+1+j+ν (0) ≤ i+1+ν (0) < d, hence
1 2 i+1+j i+1
b < ν (0) and c > 0.
i+1+j
LetP = xaxbxc beageneratorofgin(K) . Ifa ≤ i, thenbyBorel-fixedness
1 2 3 d+1
xixd+1−i ∈ gin(K) ,andasxi+1 dividesallelementsofdegree≤ d,xixd+1−i
1 2 d+1 1 1 2
is a generator of gin(K) , and hence by Theorem 4 xixd+1−i ≺ x xJ for xJ
d+1 1 2 k
some generator of gin(K) and k < max(J). However, by the claim xJ has
d
an x term and thus xixd+1−i ≻ x xJ, which is a contradiction. ⋄
3 1 2 k
Lemma 6 Let K be an ideal in C[x ,x ,x ] such that gin(K) ⊆ (xk), with
1 2 3 1
k ≥ 1 and k maximal. Then, after a possible change of basis, there exists a
homogeneous polynomial X such that in(X) = xk and K = XK′.
1
We will need to use the following Proposition to prove Lemma 6 and later,
to prove Theorem 2
Proposition 7 Let {X } be a family of homogeneous polynomials in the
H
polynomial ring S = C[x ,...,x ], (with n ≥ 2) parametrized by generic
1 n+1
hyperplane sections H = Pt x in such a way that the coefficients of X
i i H
are differentiable functions in the t . Suppose that, if we differentiate with
i
respect to H = Pt x we get
i i
∂X ∂X
H H
x −x ∈ (X ) mod(H).
j i H
∂t ∂t
i j
Then X is a constant up to a multiple of H, for generic H.
H
Proof (Green).
LetY = X | ,as(x ∂XH−x ∂XH)| ∈ (Y),wemaylet(x ∂XH−x ∂XH)| =
H H j ∂ti i ∂tj H j ∂ti i ∂tj H
l Y.
ij
∂X ∂X ∂X ∂X ∂X ∂X
H H H H H H
x (x −x )−x (x −x )+x (x −x ) = 0,
k j i j k i i k j
∂t ∂t ∂t ∂t ∂t ∂t
i j i k j k
7
thus
(x l −x l +x l )Y = 0.
k ij j ik i jk
Assuming Y 6= 0 generically (otherwise we are done), we get
x l −x l +x l = 0,
k ij j ik i jk
and hence
l = αx +βx , l = αx +γx , l = −βx +γx ,
ij j i ik k i jk k j
for some α, β and γ.
Therefore
∂X ∂X
H H
(x ( −αX )−x ( +βX ))| = 0,
j H i H H
∂t ∂t
i j
and we have
∂X ∂X
H H
−αX = x U mod(H), +βX = x U mod(H)
H i H j
∂t ∂t
i j
and similarly
∂X
H
+γX = x U mod(H).
H k
∂t
k
Letting α = α , β = −α , γ = −α we get ∂XH − x U = α X mod(H).
i j k ∂ti i i H
We may vary X continuously by any multiple of H without changing the
H
′
hypothesis or claims of the Proposition, so letting X = X −HU we get
H H
′
∂X ∂X ∂U ∂U
H = H −x U −H = α X −H mod(H) = α X′ mod(H)
∂t ∂t i ∂t i H ∂t i H
i i i i
Therefore we may assume that X is such that
H
∂X
H
= α X +HU .
i H i
∂t
i
Differentiating again we get
∂2XH = ∂αiX +α ∂XH +x U +H∂Ui
∂tj∂ti ∂tj H i ∂tj j i ∂tj
= ∂αiX +α (α X +HU )+x U +H∂Ui
∂tj H i j H j j i ∂tj
8
and
∂2XH = ∂αjX +α ∂XH +x U +H∂Uj
∂ti∂tj ∂ti H j ∂ti i j ∂ti
= ∂αjX +α (α X +HU )+x U +H∂Uj.
∂ti H j i H i i j ∂ti
Thus
∂α ∂α ∂U ∂U
j i j i
x U −x U = ( − )X +H(α U −α U + − )
j i i j H j i i j
∂t ∂t ∂t ∂t
i j i j
and so
∂α ∂α ∂α ∂α ∂α ∂α
j i k i k j
(x ( − )−x ( − )+x ( − ))X | = 0.
k j i H H
∂t ∂t ∂t ∂t ∂t ∂t
i j i k j k
Therefore ∂αj = ∂αi for all i,j, and hence there exists α such that α = ∂α
∂ti ∂tj i ∂ti
for all i.
Thus we may assume ∂XH = ∂αX +HU . Let X′ = e−αX , then
∂ti ∂ti H i H H
′
∂X ∂X ∂α
H = e−α( H − X ) = 0 mod(H)
H
∂t ∂t ∂t
i i i
Therefore we may assume ∂XH = HU .
∂ti i
Let ∂XH = HkU , with k ≥ 1. Let I be such that |I| = k, then
∂ti i
∂k+1X
H = k!xIU mod(H).
∂tI∂t i
i
If I + v = I′ + v , then ∂k+1XH = ∂k+1XH and so (xI′U − xIU )| = 0.
i j ∂tI∂ti ∂tI′∂tj j i H
Now, xI′ = x xJ and xI = x xJ and so (x U − x U )| = 0, therefore
i j i j j i H
U = x V + HV and ∂XH = Hk(x V + HV ). Let X′ = X − 1 Hk+1V,
i i i ∂ti i i H H k+1
then
∂XH′ = ∂XH −Hkx V − 1 Hk+1∂V
∂ti ∂ti i k+1 ∂ti
= Hk+1V − 1 Hk+1∂V
i k+1 ∂ti
= Hk+1W .
i
Therefore we may assume ∂XH = 0 and hence that X is a constant up to a
∂ti H
multiple of H. ⋄
Proof of Lemma 6
K ⊆ Ksat, therefore gin(K) ⊆ gin(Ksat) and gin(Ksat) is generated by
{xaxb|xaxbxc ∈ gin(K) for some c ≥ 0}.
1 2 1 2 3
9
Therefore gin(Ksat) ⊆ (xk) so we may assume K is saturated.
1
Let H be a generic hyperplane section, then gin((K| )sat) = (xk), and hence
H 1
(K| )sat = (X ).
H H
Let l be such that (K| : ml) = (K| )sat. Then for all monomials xI ∈ ml,
H H
thereexists apolynomialA , such thatxIX +HA ∈ K. Differentiating
H,I H H,I
with respect to H = Pt x we get
i i
∂X
xI H +x A ∈ K| ,
i H,I H
∂t
i
and thus
∂X ∂X
x H −x H ∈ (K| : ml) = (K| )sat = (X ).
j i H H H
∂t ∂t
i j
Thus considering the X as elements of C[x ,x ,x ], the X satisfy the
H 1 2 3 H
hypothesis of Proposition 7 and so X = X | is independent of H.
H H
Let Y ∈ K, Then Y| ∈ K| ⊆ (K| )sat = (X) and Y = A X+HB ∈ K.
H H H H H
Differentiatingwithrespect toH weget0 = (∂AHX+x B )| ∈ K| ⊆ (X).
∂tj i H H H
′
Therefore (x B )| ∈ (X) and thus B | ∈ (X). Let B = XC +HB ,
i H H H H H H H
then Y = XA′ +H2B′ ∈ (X).
H H
Let Y = A X + HkB , with k ≥ 2. Differentiating with respect to H, k
H H
times, we get
∂kA
0 = ( HX +k!xIB )| ∈ K| ⊆ (X).
∂tI H H H
Therefore (xIB )| ∈ (X) and thus B | ∈ (X). Letting B = XC +
H H H H H H
HB′ , we get Y = XA′ + Hk+1B′ ∈ (X). Therefore we may assume
H H H
B = O and hence Y = XA ∈ (X). ⋄
H H
Let K be an ideal contained in J maximal with respect to the properties
that K = XK′ with in(X) = xi+1, and (K) = (J) for d ≤ i+ν (0)+2.
1 d d i+1
K = X ∩J.
Lemma 8 gin(X)∩gin(J) ⊆ gin(X ∩J)
Proof.
By Galligo’s Theorem ([Ga]), we may assume gin(I) = in(I) for all ideals
I = (X),J or K.
10