Computation of the first Chow group of
a Hilbert scheme of space curves

arXiv:1103.0122v2 [math.AG] 20 Jun 2014

Gerd Gotzmann

Abstract
An earlier wrong formula for the dimension of A1 (Hd,g ) ⊗ Q is corrected.

Introduction
The results stated in ([T4], pp. 1) have to be corrected as follows: Let H = Hd,g =
HilbP (P3C ) be the Hilbert scheme , which parametrizes the curves in P3C of degree d and
genus g (i.e., the closed subschemes of P3C with Hilbert polynomial P (T ) = dT − g + 1).
It is always assumed that d ≥ 3 and g is not maximal, i.e. that g < (d − 1)(d − 2)/2.
Theorem 0.1. Let be g(d) := (d − 2)2 /4. Then dimQ A1 (Hd,g ) ⊗ Q = 3 (resp. = 4),
if g ≤ g(d) (resp. if g > g(d)).
Corollary 0.1. NS(H) ≃ Zρ and Pic(H) ≃ Zρ ⊕ Cr , where r := dimC H 1 (H, OH )
and ρ = 3, if g ≤ g(d). If g > g(d), then ρ = 3 or ρ = 4.
Theorem 0.2. Let C ֒→ H×P3 be the universal curve over H. Then dimQ A1 (C) ⊗ Q =
Z

dimQ A1 (H) ⊗ Q + 1.
Z

Corollary 0.2. NS(C) = Zρ+1 and Pic(C) ≃ Zρ+1 ⊕Cs , where s := dimC H 1 (C, OC )
and ρ is defined as in Corollary 1.
That means, the formula (d − 2)(d − 3)/2 for the bound g(d) in ([T4], p.1) is wrong
and has to be replaced by the above formula.

March 2, 2011

Contents
Chapter 1. Summary of earlier results

1

Chapter 2. Subschemes of points in P2 and their Hilbert functions

5

Chapter 3. A rough description of ideals invariant under Γ · T (ρ)

27

Chapter 4. The α-grade.

43

Chapter 5. Estimates of the α-grade in the case ρ1 < 0, ρ2 > 0.

55

Chapter 6. Estimates of the α-grade in the case ρ1 > 0, ρ2 > 0 and r ≥ 1.

65

Chapter 7. Estimates of the α-grade in the case ρ1 > 0, ρ2 > 0 and r = 0.

79

Chapter 8. Borel-invariant surfaces and standard cycles

87

Chapter 9. Relations between B-invariant 1-cycles

95

Chapter 10. Proof of Theorem 1.1

101

Chapter 11. Surfaces in H invariant under Ga · T (4; k)

103

Chapter 12. Surfaces in H invariant under B(4; k)

107

Chapter 13. Relations in B(4; k)-invariant surfaces

115

Chapter 14. Necessary and sufficient conditions

119

Chapter 15. The case d ≥ 5

123

Chapter 16. The cases d = 3 and d = 4

125

Chapter 17. Correction of the results of [T4] and summary

129

Appendix A. Notations

131

Appendix B. Hilbert functions without Uniform Position Property

135

Appendix C. Ideals with many monomials

137

Appendix D. Unipotent groups acting on polynomial rings

139

Appendix E. Standard bases

143

Appendix. Bibliography

145

i

CHAPTER 1

Summary of earlier results
1.1. Description of the starting situation
The result are the same as in [T1]-[T4] and are summed up in Appendix A. The
ground field is C, and H = Hd,g is the Hilbert scheme which parametrizes the curves
C ⊂ P3C of degree d and genus g (i.e. the closed subschemes of P3C with Hilbert polynomial
P (T ) = dT − g + 1). According to F.S.Macaulay, Hd,g is not empty if and only if the

“complementary” Hilbert polynomial Q(T ) = T +r
− P (T ) either has the form Q(T ) =
 T −a+2
 Tr −a+2 T −b+1
T −1+3
T −1+3
+
or the form Q(T ) =
+
+ 1 , where a is an integer ≥
3
2
3
2
1, respectively a and b are integers (Macaulay coefficients), such that 2 ≤ a ≤ b. Between
the degree and genus on the one hand and the Macaulay coefficients on the other hand, one


has the following relations d = a, g = (d − 1)(d − 2)/2, if Q(T ) = T −1+3
+ T −a+2
, and
3
2
 T −a+2 T −b+1
T −1+3
2
d = a−1, g = (a −3a+4)/2−b, if Q(T ) =
+
+
, respectively. One
3
2
1
sees that the first case occurs if and only if one is dealing with plane curves, in which case
the groups A1 (H) and NS(H) both have the rank 2 (cf. [T1], Satz 2a, p. 91). Therefore
in the following we always suppose that d ≥ 3 and g < (d − 1)(d − 2)/2, that means, the



complementary Hilbert polynomial has the form Q(T ) = T −1+3
+ T −a+2
+ T −b+1
,
3
2
1
where 4 ≤ a ≤ b.
We also write HQ instead of Hd,g in order to express that this Hilbert scheme likewise
parametrizes the ideals I ⊂ OP3 with Hilbert polynomial Q(T ), or equivalently, the
saturated graded ideals in C[x, y, z, t] with Hilbert polynomial Q(T ).

In [T1]-[T4] it was tried to describe the first Chow group A1 (H), where we always
take rational coefficients, and we write A1 (H) instead of A1 (H) ⊗ Q. The starting point
Z

is the following consideration: If the Borel group B = B(4; k) operates on H = HQ
in the obvious way, then one can deform each 1-cycle on H in a 1-cycle, whose prime
components are B-invariant, irreducible, reduced and closed curves on H. It follows that
A1 (H) is generated by such B-invariant 1-prime cycles on H. This is a partial statement
of a theorem of Hirschowitz. (Later on we will have to use the general statement, whereas
the partial statement can be proved in a simple way, see [T1], Lemma 1, p. 6.) Now such
a B-invariant 1-prime cycle (i.e. closed, irreducible and reduced curve) C on H can be
formally described as follows: Either each point of C is invariant under ∆ := U(4; k), or
one has C = Gia · η, where η is a closed point of H, which is invariant under T = T (4; k)
and the group Gi , i ∈ {1, 2, 3}. Here Gia is the group Ga , acting by
ψα1 : x 7−→ x, y 7−→ y, z 7−→ z, t 7−→ αz + t
ψα2 : x 7−→ x, y −
7 → y, z 7−→ αy + z, t 7−→ t
ψα3 : x −
7 → x, y −
7 → αx + y, z 7−→ z, t 7−→ t,
1

respectively, on P = k[x, y, z, t], and Gi is
Gia , that means, one defines



1
∗
∗
∗









0
1
∗
∗



,
G
:=
G1 := 
2


0
0
1
0






 0 0 0 1 

the subgroup of ∆, which is complementary to
∗
1
0
0

1
0
0
0

∗
0
1
0


∗ 


∗ 
 ,
∗ 

1 





G3 := 





1
0
0
0

0
1
0
0

∗
∗
1
0


∗ 


∗ 
 .
∗ 

1 

If C has this form, then C is called a curve or a 1-cycle of type i, where i ∈ {1, 2, 3}.
A(H) := Im(A1 (H∆ ) → A1 (H)) is called the “algebraic part” and A1 (H) := A1 (H)/A(H)
is called the “combinatorial part” of the first Chow group of H. Here H∆ denotes the
fixed point scheme which, just as all other fixed point schemes that will occur later on,
is supposed to have the induced reduced scheme structure. (This convention is valid also
for the Hilbert scheme H d := Hilbd (P2C ), see below.)
In order to formulate the results obtained so far, one has to introduce the following
”tautological” 1-cycles on H:

−
C1 = (x, y a , y a−1 z b−a (αz + t)) α ∈ k
C2 = {(x, y a−1 (αy + z), y a−2 z b−a+1 (αy + z))|α ∈ k}−
C3 = {(xa , αx + y, xa−1 z b−a+1 )|α ∈ k}−

D = {(x2 , xy, y a−1, z b−2a+4 (y a−2 + αxz a−3 ))|α ∈ k}−

E = {(x2 , xy, xz, y a, y a−1 z b−a+1 , xtb−2 + αy a−1z b−a )|α ∈ k}−

For the sake of simplicity, we now suppose d ≥ 5 (i.e. a ≥ 6). (The cases d = 3 and d = 4
will be treated separately in Chapter 16.) Then one has the following results:
1. If b < 2a − 4, i.e. if g > γ(d) := (d − 2)(d − 3)/2, then A(H) is generated by E, and
A1 (H) is generated by E, C1 , C2 , C3 .
2. If b ≥ 2a − 4, i.e if g ≤ γ(d), then A(H) is generated by E and D and A1 (H) is
generated by E, D, C2 and C3 ( see [T1], Satz 2, p. 91; [T3], Proposition 4, p. 22; [T4],
Satz 1 and Proposition 2, p. 26).
From reasons of degree it follows that [C2 ] can not lie in the vector space spanned by
[E], [D], [C3 ], so the problem is to decide, if [C3 ] ∈ A(H).

In ([T4], Proposition 3, p. 32) it was erroneously claimed that [C3 ] ∈ A(H), if b ≥
2a − 4. (The error is the wrong computation of the degree in ([T4], p. 28, line 21 to line
30.) Therefore the bound for the genus in ([T4], p. 1) is wrong.
Actually, in ([T2], 3.3.2) it had been proved, that [C3 ] ∈ A(H), if a ≥ 6 is even and
b ≥ a2 /4, i.e. if d ≥ 5 is odd and g ≤ (d − 1)(d − 3)/4. In the case d ≥ 6 even , in
Conclusion 14.3 it will follow that [C3 ] ∈ A(H), if g ≤ (d − 2)2 /4. (This means the bound
of [T2], 3.3.3 is valid if d ≥ 6, already . ). One sees that the condition for g in both cases
can be summed up to g ≤ (d − 2)2 /4.

The major part of the following text serves for the proof that this sufficient condition
is a necessary condition, too (cf. Conclusion 14.1).
2

1.2. Technical tools
The formulas in ([T2], p. 134) and of ([T3], Anhang 2, p. 50) show that it is not
possible to decide by means of the computation of degrees, whether [C3 ] lies in A(H).
Therefore we try to get a grasp of the relations among the B-invariant 1-cycles on H with
the help of the theorem of Hirschowitz ([Hi], Thm. 1, p. 87). We sketch the procedure.
1.2.1. The Theorem of Hirschowitz. There is a closed and reduced subscheme
Z = Z(H) of H, such that Z(k) = {x ∈ H(k)| dim ∆ · x ≤ 1} (cf. [Ho], p. 412 and [T3],
Lemma 1, p. 35). Then one can show, with the help of the theorem of Hirschowitz, that
∼
A1 (Z) → A1 (H) (cf. [T2], Lemma 24, p. 121). As was explained in (1.1), A1 (H) has a
generating system consisting of B-invariant 1-cycles which lie in Z, automatically. As ∆
is normalized by B, B operates on Z and therefore one can form the so called equivariant
Chow group AB
1 (Z), which is isomorphic to A1 (Z) ([Hi], loc. cit.). And the relations
among B-invariant 1-cycles on Z are generated by relations among such cycles, which lie
on B-invariant surfaces V ⊂ Z ( see [Hi], Mode d’emploi, p. 89).
1.2.2. The Restriction morphism. Let Ut ⊂ H be the open subset consisting of
the ideals I ⊂ OP3 with Hilbert polynomial Q, such that t is not a zero divisor of OP3 /I.
Then there is a so called restriction-morphism h
: Ut → 
H d := Hilbd (P2C ), defined by
I 7→ I ′ := I + tOP3 (−1)/tOP3 (−1). E.g., if G :=

1
0
0
0

0
1
0
0

0
0
1
0

∗
∗
∗
1

< ∆, then the fixed point

scheme F := HG is contained in Ut , and the restriction of h to F is denoted by h, again.
In ([G4], Abschnitt 6, p. 672f) the following description of Im(h) is given:
(i) There is a finite set F of Hilbert functions of ideals of colength d on P2 such that
S
Im(h) = {H≥ϕ |ϕ ∈ F }.

(ii) If k = k and if I ⊂ OP2k is an ideal of colength d and Hilbert function ϕ, then
d

P
I ∈ Im(h) ⇐⇒ g ∗ (ϕ) :=
ϕ(n) − d+3
+ d2 + 1 ≥ g.
3
0

(iii) If ϕ ∈ F and if ψ is the Hilbert function of an ideal on P2 of colength d such that
ϕ(n) ≤ ψ(n) for all n ∈ N, then ψ ∈ F .
(iv) Let be ϕ ∈ F and I ⊂ OP2k an ideal with Hilbert function ϕ. Let I ∗ be the ideal
n

in OP3k defined by H 0 (I ∗ (n)) = ⊕ tn−i H 0 (I(i)), then V+ (I ∗ ) ⊂ P3k is a curve of degree d

and genus g ∗ (ϕ).

i=0

Here Hϕ ⊂ H d is the locally closed subscheme (with the reduced induced scheme
structure), which parametrizes the ideals I ⊂ OP2 of colength d with Hilbert function
S
ϕ, and H≥ϕ := {Hψ |ψ ≥ ϕ} is a closed subscheme ( with the induced reduced scheme
structure). The image of C3 under h is the 1-cycle c3 := {(xd , αx + y)|α ∈ k}− . One has
S
Theorem 1.1. Let be d ≥ 5, H := {Hϕ ⊂ H d |g ∗(ϕ) > g(d)} and
A(H) := Im(A1 (HU (3;k) ) → A1 (H)). Then [c3 ] ∈
/ A(H).
3

The proof extends over the chapters 2 to 10 and essentially rests on the apparently
strong condition for an ideal I to have a Hilbert function ϕ such that g ∗(ϕ) > g(d).
1.2.3. Standard cycles on H d . It has been shown, respectively it will be shown
that [C3 ] ∈ A(Hd,g ), if g ≤ g(d) (cf. 1.1). Therefore, we can suppose that g > g(d). If
J ∈ Ut and the restriction ideal I := J ′ has the Hilbert function ϕ, then from (ii) in
(1.2.2) it follows that g ∗ (ϕ) > g(d). It will be shown in Chapter 2 that this implies there
is a linear form ℓ ∈ S1 − (0), an ideal K ⊂ OP2 of colength c and a form f ∈ H 0 (K(m))
such that I = ℓK(−1) + f OP2 (−m), c + m = d and m ≥ c + 2.

Let be C = Ga · η ⊂ H a 1-cycle of type 3 and let be J ↔ η the corresponding
′
′
ideal in OP3 with Hilbert polynomial
Q. Thenthe
 ideal I := J ↔ η := h(η) is
 1 0 ∗ 
invariant under T (3; k) and Γ :=  0 1 ∗  < U(3, k). It follows that either


0 0 1
I = xK(−1) + y m OP2 (−m) or I = yK(−1) + xm OP2 (−m), where K is a monomial
ideal. We say, I has x-standard form or I has y-standard form, respectively, and we call
C ′ := Ga · η ′ a x-standard cycle or y-standard cycle on H d , respectively. With the help of
the theorem of Hirschowitz one can again try to describe the relations between B(3; k)invariant y-standard cycles on H, and one obtains that such relations cannot make the
y-standard cycle c3 disappear modulo A(H) (cf. Proposition 9.1) from which Theorem
0.1 will follow.
1.2.4. 1-cycles of proper type 3. Let be C = Ga · η, η ↔ J , be a 1-cycle of type
3 on H = Hd,g , such that d ≥ 5 and g > g(d). C is called a 1-cycle of proper type 3, if
C ′ = Ga · η ′ is a y-standard cycle on H. Corresponding to Hirschowitz’s theorem one has
to consider B(4; k)-invariant surfaces V ⊂ Z(H), which contain a 1-cycle of proper type
3. It turns out that then V is pointwise invariant under G3 and therefore V is contained
in Ut . Then one can map relations between B-invariant 1-cycles on V by h∗ into relations
between B(3; k)-invariant 1-cycles on h(V ), and one obtains with the aid of Proposition
9.1 the main result of the second part of the paper ( Theorem 14.1), which corresponds
to Theorem 0.1. In Chapter 15 there is complete description of A1 (Hd,g ) if d ≥ 5, and in
Chapter 16 this is done in the cases d = 3 and d = 4 (Theorem 15.1 and Theorem 16.1,
respectively).

4

CHAPTER 2

Subschemes of points in P2 and their Hilbert functions
2.1. General properties
The ground field is C and k denotes an extension field. A closed subscheme Z ⊂ P2k of

− d. If
length d > 0 is defined by an ideal I ⊂ OP2k with Hilbert polynomial Q(n) = n+2
2
0
0
2
the Hilbert function h (I(n)) = dimk H (Pk ; I(n)), n ∈ N, of I is denoted by ϕ(n), then
ϕ′ (n) := ϕ(n) − ϕ(n − 1), n ∈ N, denotes the difference function. If ϕ : N −→ N is any

function, such that ϕ(n) = n+2
− d for n ≫ 0, then the ideals I ⊂ OP2k with Hilbert
2
function ϕ form a locally closed subset Hϕ of the Hilbert scheme H d = Hilbd (P2C ), and we
take Hϕ as a subscheme of H d with the induced reduced scheme structure.
Iarrobino has shown ([I], Lemma 1.3, p.8) that Hϕ 6= ∅ if and only if the difference
function fulfils the following two conditions:
(a) ϕ′ (n) ≤ n + 1, for all n ∈ N and
(b) ϕ′ (n) ≤ max(ϕ′ (n + 1) − 1, 0), for all n ∈ N.
If α = α(ϕ) := min{n ∈ N | ϕ(n) > 0}, then (b) is equivalent to:
(b’) ϕ′ (n) + 1 ≤ ϕ′ (n + 1), for all n ≥ α.
The (Mumford-)regularity e of an ideal I with Hilbert function ϕ as before is characterized by e = reg(ϕ) = min{n ∈ N | ϕ′ (n + 1) = n + 1} (cf. Appendix B, Lemma
2). In principle, the graph of ϕ′ has the shape of Fig. 2.1. If ∅ =
6 Hϕ ⊂ H d , then d is

n+2
determined by the condition ϕ(n) = 2 − d, n ≫ 0, and we call d the colength of ϕ.
It is known that reg(ϕ) ≤ d ([G1], Lemma 2.9, p. 65), and reg(ϕ) = d is equivalent with
I being generated by a linear form ℓ ∈ S1 and a form f ∈ Sd , not divisible by ℓ. Here
S = k[x, y, z] is the graded polynomial ring. Another characterization of reg(ϕ) = d is
that the graph of ϕ′ has the shape of Fig.2.2. One notes that the colength of ϕ is equal to
the number of ”monomials” between the graph of ϕ′ and the line y = x + 1. (For this and
other properties , see [T1]-[T4].) In the following we write P2 instead of P2k and denote by
I an ideal in OP2 , whose finite colength (resp. whose Hilbert function) usually is denoted
by d (resp. by ϕ).
2.2. Numerical and algebraic properties
Lemma 2.1. Let be k = k, I ⊂ OP2 an ideal with colength d, Hilbert function ϕ and
regularity m. We assume that there is a number ε ∈ N, 0 ≤ ε < m−2, such that ϕ′ (n) = n
for all n ∈ N, such that ε + 1 ≤ n ≤ m − 1. Then there is a linear form ℓ = S1 , an ideal
K ⊂ OP2 of colength c and a form f ∈ H 0 (K(m)), such that I = ℓK(−1) + f OP2 (−m). If
5

ℓ1 , ℓ2 are any linear forms in S1 such that ℓ, ℓ1 , ℓ2 is a basis of the k-vector space S1 and
if R := k[ℓ1 , ℓ2 ] is the subring of S, isomorphic to k[x, y], then d = c + m and
(
ℓH 0 (K(n − 1))
if n < m,
H 0 (I(n)) =
0
ℓH (K(n − 1)) ⊕ f Rn−m if n ≥ m.
Proof. By assumption, the graph of ϕ′ has the shape as in Fig.2.3. Then there is a
ℓ ∈ S1 − (0) and an ideal K ⊂ OP2 of regularity ≤ ε such that H 0 (I(n)) = ℓH 0 (K(n − 1))
for all n ≤ m − 1 (cf. [G4] and Appendix B, Lemma 1). If ψ is the Hilbert function of
K, then ϕ′ (n) = ψ ′ (n − 1) for 1 ≤ n ≤ m − 1, and because of the shape of the graphs
of ϕ′ and ψ ′ it follows that ϕ(n) = ψ(n − 1) + (n − m + 1) for all n ≥ m. Therefore
H 0 (I(m)) = ℓH 0 (K(m − 1)) ⊕ f · k, where f ∈ H 0 (I(m)) is a suitable section. Because
of the m-regularity of I it follows that H 0 (I(n)) = ℓH 0 (K(n − 1)) + f Sn−m , n ≥ m. If
n = m + 1, then from ϕ(m + 1) = ψ(m) + 2 it follows that S1 f ∩ ℓH 0 (K(m)) has the
dimension 1. Thus there is a h ∈ S1 − (0), such that hf ∈ ℓH 0 (K(m)). If ℓ would be a
divisor of f , then it would follow that I ⊂ ℓOP2 (−m) and thus I would not have a finite
colength in OP2 . Therefore we may suppose that h = ℓ, and it follows that f ∈ H 0 (K(m)).
We choose ℓ1 , ℓ2 ∈ S1 such that ℓ, ℓ1 , ℓ2 are linear independent and we put R := k[ℓ1 , ℓ2 ].
If there would be a r ∈ Rν − (0) such that rf ∈ ℓH 0 (K(m + ν − 1)), then it would follow
that ℓ is a divisor of f , contradiction. Between the graph of ψ ′ (n − 1) and the line y = x
there are exactly c := colength(ψ) monomials, and therefore d = c + m (cf. Fig.2.3). 
Corollary 2.1. The assumptions and notations are as in Lemma 2.1. Then one
has:
(i) κ := reg(K) ≤ ε, especially κ ≤ m − 3.
(ii) K (respectively the linear form ℓ) is uniquely determined (respectively uniquely up to
a factor out of k different from zero).
(iii) f is uniquely determined up to a factor out of k different from zero, modulo ℓH 0 (K(m−
1)).
(iv) κ and ε are uniquely determined by ϕ.
Proof. (i) ε + 1 = ϕ′ (ε + 1) = ψ ′ (ε). From (Appendix B, Lemma 2) it follows that
κ ≤ ε.
(ii) The regularity only depends on the Hilbert function, and therefore κ = reg(K1 ) =
reg(K2 ) < m − 1. Thus from ℓ1 H 0 (K1 (m − 2)) = ℓ2 H 0 (K2 (m − 2)) it follows that
ℓ1 K1 (−1) = ℓ2 K2 (−1).
If ℓ1 would not be a divisor of ℓ2 then one would have K2 ⊂ ℓ1 OP2 (−1) contradiction.
From this assertion (ii) does follow, and (iii) and (iv) are clear.

Remark 2.1. If ϕ and ψ are two Hilbert functions of colength d, then from ϕ < ψ
(that means ϕ(n) ≤ ψ(n) for all n ∈ N and ϕ(n) < ψ(n) for at least one n ∈ N) it follows
that g ∗ (ϕ) < g ∗ (ψ). This follows immediately from the definition of g ∗(ϕ) in (1.2.2).
6

e−2
P
Remark 2.2. If e := reg(ϕ), d := colength(ϕ) and s(ϕ) :=
ϕ(i), then g ∗(ϕ) =
i=0


n+2
s(ϕ) − e+1
+
d(e
−
2)
+
1.
Because
of
ϕ(n)
=
− d for all n ≥ e − 1 this follows
3
2
from a simple computation with binomial coefficients.

2.2.1. Hilbert functions of colength ≤ 4. We use the formula of Remark 2.2 and
orientate ourselves by the figures 2.4–2.7.
d = 1 There is only one Hilbert function (cf. Fig. 2.4).
 
2
∗
+ 1 · (1 − 2) + 1 = 0.
e = 1, s(ϕ) = 0, g (ϕ) = 0 −
3
d = 2 There is again only one Hilbert function (cf. Fig. 2.5).
 
3
∗
+ 2 · 0 + 1 = 0.
e = 2, s(ϕ) = 0, g (ϕ) = 0 −
3
d = 3 There are two Hilbert functions (Fig. 2.6 a and Fig. 2.6 b).
 
3
∗
+ 3 · 0 + 1 = 0,
e1 = 2, s(ϕ1 ) = 0, g (ϕ1 ) = 0 −
3
 
4
∗
+ 3 · 1 + 1 = 1.
e2 = 3, s(ϕ2 ) = 1, g (ϕ2 ) = 1 −
3
d = 4 There are two Hilbert functions (Fig. 2.7 a and Fig. 2.7 b).
 
4
∗
+ 4 · 1 + 1 = 1,
e1 = 3, s(ϕ1 ) = 0, g (ϕ1 ) = 0 −
3
 
5
∗
+ 4 · 2 + 1 = 3.
e2 = 4, s(ϕ2 ) = 4, g (ϕ2 ) = 4 −
3
2.2.2. Two special ideals. First case: If d ≥ 6 is even, then let be e := d/2 + 1 and
I := (x2 , xy e−2, y e ). The Hilbert function χ can be read from Fig. 2.8. One notes that
n−1

P
colength(I) and reg(I) really are equal to d and e, respectively, and χ(n) =
i = n2 ,
1

if 1 ≤ n ≤ e − 2. Therefore s(χ) =


−

e+1
3

1


i

2

=


e−1
3

and it follows that





e
e
e+1
+ 2(e − 1)(e − 2) + 1 = e−1
−
+
−
+ 2e2 − 6e + 5
3
3
3
3
1
1
= − (e − 1)(e − 2) − e(e − 1) + 2e2 − 6e + 5 = e2 − 4e + 4
2
2
1
= (e − 2)2 = (d − 2)2 .
4

g ∗(χ) =

e−1
3

e−2
P



Second case: If d ≥ 5 is odd, then let be e := (d + 1)/2 and I := (x2 , xy e−1 , y e).
The Hilbert function χ can be read from Fig. 2.9. One notes that colength (I) and reg(I)

are equal to d and e, respectively, and χ(n) = n2 , if 1 ≤ n < e.
7

Therefore s(χ) =

e−2
P
2

g ∗ (χ) =

e−1
3



−

i
2



=

e+1
3



e−1
3



and it follows that

+ (2e − 1)(e − 2) + 1 = −

e−1
2



−

e
2



+ (2e − 1)(e − 2) + 1
3
1
= −(e − 1)2 + 2e2 − 5e + 3 = e2 − 3e + 2 = (d + 1)2 − (d + 1) + 2
4
2
1 2
= (d − 4d + 3).
4

Definition 1. If d ≥ 5, then we set
(
1
(d − 2)2
if d ≥ 6 is even,
g(d) := 41
(d − 1)(d − 3) if d ≥ 5 is odd.
4
g(d) is called the deformation bound for ideals in OP2 of colength d.
The rest of the article is to justify this notation.
2.2.3.
Lemma 2.2. Let be k = k, I ⊂ OP2 an ideal of colength d ≥ 5 and regularity m. Let
be ϕ the Hilbert function of I. If g ∗(ϕ) > g(d), then the assumptions of Lemma 2.1 are
fulfilled by I.
Proof. Let be χ the Hilbert function defined by Fig. 2.8 and Fig. 2.9, respectively.
Let be m = reg(ϕ). If ϕ′ (m) − ϕ′ (m − 1) > 2, then ϕ′ (i) ≤ χ′ (i) and therefore ϕ(i) < χ(i)
for all i, and it would follow that g ∗ (ϕ) ≤ g(χ) (Remark 2.1). If ϕ′ (m)−ϕ′ (m−1) = 1, then
ϕ′ (m−1) = ϕ′ (m)−1 = (m+1)−1 = (m−1)+1, therefore reg(I) ≤ m−1 (cf. Appendix
B, Lemma 2). It follows that ϕ′ (m) − ϕ′ (m − 1) = 2, therefore ϕ′ (m − 1) = m − 1. If
ϕ′ (m−2) = m−2, as well, then the assumptions of Lemma 2.1 are fulfilled with ε := m−3,
for instance. Thus without restriction of generality one can assume ϕ′ (m − 2) ≤ m − 3.

Case 1: ϕ′ (m − 2) < m − 3. Figure 2.10 represents the Hilbert function ϕ as well as the
B(3; k)-invariant ideal M with Hilbert function ϕ. Then one makes the deformation
E(H 0 (M(m))) 7→ E(H 0 (M(m))) − u ∪ v =: E(H 0 (N (m))),

where N is a B(3; k)-invariant ideal with Hilbert function ψ > ϕ. But then it follows
g ∗ (ϕ) < g ∗(ψ) ≤ g ∗ (χ) = g(d), contradiction.

Case 2: ϕ′ (m − 2) = m − 3. If the graph of ϕ′ would have a shape different from that
in Fig. 2.11 a, i.e., if the graph of ϕ′ would have a “jumping place” n < m − 2 (cf. the
terminology in [T1], p. 72), then as in the first case one could make a deformation
E(H 0 (M(m))) 7→ E(H 0 (M(m))) − u ∪ v =: E(N (m)))
(cf. Fig. 2.11b) and would get a contradiction, again. It only remains the possibility
represented in Fig. 2.11a. But then ϕ = χ, which contradicts the assumption g ∗(ϕ) >
g(d).

8

2.3. Numerical conclusions from g ∗ (ϕ) > g(d)
2.3.1. At first we describe the starting situation: In this section we suppose that
g(d) is defined, i.e. d ≥ 5. Moreover let be ϕ a Hilbert function such that Hϕ 6= ∅,
colength(ϕ) = d, reg(ϕ) = m and g ∗ (ϕ) > g(d). Then the assumptions of Lemma 2.2 are
fulfilled for an ideal I, which can be supposed to be monomial. Therefore the assumption
k = k is superfluous. As m and d are uniquely determined by ϕ, c := d − m is uniquely
determined, too.
The aim in this section is to prove the inequality m ≥ c + 2. By Lemma 2.1 and
Lemma 2.2, respectively, one can write I = ℓK(−1) + f OP2 (−m), and c is equal to the
colength of the Hilbert function ψ of K. (As I is monomial, K is monomial, too, and
without restriction one has ℓ ∈ {x, y, z}.)
Lemma 2.3. Let be ψ the Hilbert function of an ideal K of colength c ≥ 5, κ :=
reg(ψ), and m ≥ κ + 2 an integer. If one defines ϕ by ϕ′ (n) := ψ ′ (n − 1), 0 ≤ n ≤
m − 1, ϕ′ (n) := n + 1, n ≥ m, then Hϕ 6= ∅, colength(ϕ) = c + m, reg(ϕ) = m and
g ∗ (ϕ) = g ∗(ψ) + 12 m(m − 3) + c.
Proof. We orientate ourselves by Figure 2.3, but the weaker assumption m ≥ κ + 2
takes the place of the assumption m ≥ κ + 3.

Without restriction one can assume that K is B(3; k)-invariant. Then y m ∈ H 0 (K(m))
and I := xK(−1) + y mOP2 (−m) has the Hilbert function ϕ, the regularity m and the
colength c + m. This follows by considering the figure mentioned above (and has been
shown in a more general situation in [G4], Lemma 4, p. 660). We compute g ∗(ϕ) (cf.
Remark 2.2):
s(ϕ) =

m−2
X

κ−1
X

ϕ(i) =

i=0

i=0

=

κ−2
X

ψ(i) +

i=0

= s(ψ) +

ψ(i − 1) +

m−3
X

m−2
X
i=κ

ψ(i) = s(ψ) +

i=0

i+2
2



m−3
X

i=κ−1

i=κ−1
m−3
X

ψ(i − 1)

−

κ−2
X
i=0

i+2
2





i+2
2



−c



− (m − κ − 1)c.

By Remark 2.2 it follows that:



m+1
g ∗ (ϕ) = s(ψ) + m3 − κ+1
−
(m
−
κ
−
1)c
−
+ (c + m)(m − 2) + 1
3
 3


κ+1
m
m+1
= s(ψ) − 3 + c(κ − 2) + 1 + 3 − 3 − mc + c + (c + m)m − 2m

1
= g ∗ (ψ) − m2 + m2 − 2m + c = g ∗ (ψ) + m(m − 3) + c.
2



2.3.2. The cases c ≤ 4. By Lemma 2.2 (resp. by Corollary 2.1 of Lemma 2.1) one
has ϕ′ (n) = ψ ′ (n − 1), 0 ≤ n ≤ m − 1, ϕ′ (n) = n + 1, n ≥ m, and κ = reg(K) ≤ m − 3.
Then the assumptions of Lemma 2.3 are fulfilled.
9

We use the formula given there and orientate ourselves by the Figures 2.4 - 2.7. The
regularity of the Hilbert function considered each time will now be denoted by κ.
If
If
If
If

c ∈ {0, 1}, then because of d = m + c it follows that m ≥ 4.
c = 2, then κ = 2 and m ≥ κ + 3 = 5.
c = 3 and κ = 2 or κ = 3, then m ≥ κ + 3 = 5.
c = 4, then κ = 3 or κ = 4 and m ≥ κ + 3 ≥ 6.

Thus in the cases 0 ≤ c ≤ 4 one has m ≥ c + 2.
2.3.3. The case g ∗ (ψ) ≤ g(c). This notation implies that c ≥ 5. If κ is the regularity
and c is the colength of any Hilbert function, then because of 1 + 2 + · · · + κ ≥ c, one
√

2c − 1. By Lemma 2.2 the assumptions of
always has κ+1
≥
c,
and
therefore
κ
≥
2
Lemma 2.1 are fulfilled, therefore by Corollary 2.1 it follows that m ≥ κ + 3 > 5.16.
1st case: c and m are even.
By the formulas for g(d) and g ∗ (ϕ) it follows that:
1 2
(c
4

− 4c + 4) + 12 m(m − 3) + c > 41 [(c + m)2 − 4(c + m) + 4]
⇐⇒

1
m(m
2

− 3) + c > 14 [2cm + m2 − 4m]

⇐⇒ m2 − 2(c + 1)m + 4c > 0.

The solutions of the corresponding quadratic equation are 0 and 2c.
Therefore m ≥ 2c + 1 > c + 2.
2nd case: c is even, m is odd.
One obtains the inequality:
1 2
(c
4

− 4c + 4) + 12 m(m − 3) + c > 41 [(c + m)2 − 4(c + m) + 3]

m2 − 2(c + 1)m + 4c + 1 > 0.
√
The solutions of the corresponding quadratic equation are m = c + 1 ± c2 − 2c ≥ 0.
√
√
Because of c + 1 − c2 − 2c < 3, if c ≥ 5, it follows that m ≥ c + 1 + c2 − 2c. Because of
√
c + 1 + c2 − 2c > 2c − 1, if c ≥ 5, it follows that m ≥ 2c, therefore m ≥ 2c + 1 > c + 2.
⇐⇒

3rd case: c is odd, m is even.
One obtains the inequality :
1 2
(c
4

− 4c + 3) + 21 m(m − 3) + c > 14 [(c + m)2 − 4(c + m) + 4]

⇐⇒

m2 − 2(c + 1)m + 4c > 0.

It follows that m ≥ 2c + 1 > c + 2.
4th case: c and m are odd.
One obtains the inequality:
1 2
(c
4

⇐⇒

− 4c + 3) + 21 m(m − 3) + c > 14 [(c + m)2 − 4(c + m) + 4]
m2 − 2(c + 1)m + 4c − 1 > 0.
10

p
The solutions of thepcorresponding quadratic equation are m = c +p1 ± (c − 1)2 + 1.
Because of c + 1 − (c − 1)2 + 1 < 2 it follows that m ≥ c + 1 + (c − 1)2 + 1 > 2c,
therefore m ≥ 2c + 1 ≥ c + 2.
2.3.4. The case g ∗ (ψ) ≥ g(c). As in the proof of Lemma 2.3 one can write I =
xK(−1)+y m OP2 (−m), K a B(3; k)-invariant ideal of colength c, κ = reg(K), d = colength(I) =
c + m, and again m ≥ κ + 3 (Corollary 2.1). We represent the Hilbert function ψ by the
ideal K = xJ (−1) + y κ OP2 (−κ), J a B(3; k)-invariant ideal of colength b and of regularity ε, where κ ≥ ε + 3 (cf. Corollary 2.1). If the Hilbert function of J is denoted by
ϑ, then in principle one has the situation represented by Fig. 2.12. If one assumes that
(m−1) −κ ≤ colength(J ) = b, then one could bring the monomials denoted by 1, 2, 3, . . .
in the positions denoted by 1, 2, 3, . . . (cf. Fig. 2.12). In the course of this the Hilbert
function increases and therefore g ∗ (ϕ) < g ∗ (ϕ1 ) < · · · < g(d), contradiction. Thus one
has (m − 1) − κ > b, i.e., m ≥ κ + b + 2 = c + 2.
2.3.5. Summary.
Lemma 2.4. (Notations as in Lemma 2.1 and Lemma 2.2) If g(d) < g ∗(ϕ), then
m ≥ c + 2.

From the proof of Lemma 2.4 we can conclude one more statement:
Corollary 2.2. Let be g ∗(ψ) ≤ g(c) (which notation implies c ≥ 5). Then m ≥
2c + 1.

2.4. Additional group operations
2.4.1. General auxiliary lemmas. The group Gl(3; k) operates on S = k[x, y, z],
and therefore on H d = Hilbd (P2k ). If ρ = (ρ0 , ρ1 , ρ2 ) ∈ Z3 is a vector such that ρ0 +ρ1 +ρ2 =
0, then
:= { (λ0 , λ1 , λ2 ) ∈ (k ∗ )3 | λρ00 λρ11 λρ22 = 1 } is a subgroup of T = T (3; k), and
nT (ρ) o
1 0 ∗
0 1 ∗
is a subgroup of U(3; k). We let the additive group Ga operate on S by
Γ :=
0 0 1

ψα : x 7→ x, y 7→ αx + y, z 7→ z, α ∈ k,

and σ : Gm → T nearly always denotes the operation σ(λ) : x 7→ x, y 7→ y, z 7→ λz, λ ∈ k ∗ .
Auxiliary Lemma 1. If V ⊂ Sd is a vector space, invariant under G := Γ · T (ρ)
where ρ = (0, −1, 1) or ρ = (−1, 0, 1), then V is monomial, i.e. invariant under T .
Proof. We first consider the case ρ = (0, −1, 1). We take a standard basis of V
consisting of T (ρ)-semi-invariants (see Appendix E). Assuming that the assertion above
is wrong , we conclude that there is a T (ρ)-semi-invariant f ∈ V , such that the monomials
occurring in f are not in V . Then there is such a form with smallest z-degree. From the
invariance of V under T (ρ) it follows that V is invariant under the Gm -action τ (λ) : x 7→
P
λx, y → y, z 7→ z, λ ∈ k ∗ . We write f = Mp, where M = xℓ y m, p = ni=0 ai y n−i z i , ℓ + m +
P
n = d, n ≥ 1 and an 6= 0. It follows that y∂f /∂z = yM n1 iai y n−i z i−1 ∈ V . Now y∂f /∂z
is also a T (ρ)-semi-invariant with smaller z-degree than f . According to the choice of f
11

it follows that g := Myz n−1 ∈ V , therefore y∂g/∂z = (n − 1)My 2 z n−2 ∈ V , etc. One gets
My i z n−i ∈ V, 1 ≤ i ≤ n, therefore My n ∈ V , contradiction.
P
In the case ρ = (−1, 0, 1) we write f = xℓ y m ni=0 ai xn−i z i . Because of x∂f /∂z =
P
xM ni=1 iai xn−i z i−1 we can argue as before.


Auxiliary Lemma 2. Let be I ⊂ OP2 an ideal of colength d, which is invariant under
G := Γ · T (ρ). If ρ0 + ρ1 + ρ2 = 0, ρ0 < 0 and ρ1 < 0, then I is invariant under T .

Proof. Let be n the smallest natural number, such that H 0 (I(n)) is not T -invariant.
Then we have without restriction that n ≥ 1. As H 0 (I(n)) has a standard basis , there
is a proper semi-invariant in H 0 (I(n)), i.e. a form f ∈ H 0 (I(n)) of the shape f =
M(1+a1 X ρ +a2 X 2ρ +· · ·+ar X rρ ), M a monomial, ar 6= 0, r ≥ 1, and no monomial MX iρ is
in H 0 (I(n)), if ai 6= 0. If M would be divisible by z, then g := z −1 f ∈ H 0 (I(n−1)) would
be a proper semi-invariant, too, because from z −1 MX iρ ∈ H 0(I(n − 1)) it follows that
MX iρ ∈ H 0 (I(n)). Therefore, M is not divisible by z. From the proper semi-invariants of
H 0 (I(n)) we chose one, say f , such that the z-degree is minimal. Now from f ∈ H 0 (I(n)),
because of the Γ-invariance, it follows that x∂f /∂z = xM(a1 ρ2 z −1 X ρ +· · ·+rar ρ2 z −1 X rρ)
and y∂f /∂z = yM(a1 ρ2 z −1 X ρ + · · · + rar ρ2 z −1 X rρ ) is in H 0 (I(n)), i.e., g := xMX ρ z −1 p
and h := yMX ρ z −1 p are in H 0 (I(n)), where p(X ρ ) := a1 ρ2 +2a2 ρ2 X ρ +· · ·+rar ρ2 X (r−1)ρ .
As the z-degree of g and of h is smaller than the z-degree of f, g and h are no longer
proper semi-invariants, i.e. the monomials which occur in g or in h, all are in H 0 (I(n)). It
follows that u := z −1 xMX rρ and v := z −1 yMX rρ are in H 0 (I(n)). From the Γ-invariance
|ρ1 |
|ρ |
it follows by applying the operators x∂/∂z and y∂/∂z repeatedly, that xz |ρ00| · yz |ρ1 | · MX rρ ∈
H 0 (I(n)).
Now X ρ = x−|ρ0 | y −|ρ1 | z ρ2 and ρ0 + ρ1 + ρ2 = 0, therefore MX (r−1)ρ ∈ H 0 (I(n)). Applying
the operators mentioned before again gives MX (r−2)ρ ∈ H 0 (I(n)), etc. It follows that
MX iρ ∈ H 0 (I(n)), 0 ≤ i ≤ r − 1, and therefore MX rρ ∈ H 0 (I(n)), contradiction.

2.4.2. Successive construction of Γ-invariant ideals. At first we consider a general situation: Let be K ⊂ OP2 an ideal of colength c and of regularity e; z is supposed
to be a non-zero divisor of OP2 /K; let be R := k[x, y] and ℓ ∈ R1 a linear form. Let be
m > e an integer and f ∈ H 0 (K(m)) a section, whose leading term is not divisible by ℓ,
i.e., if one writes f = f 0 + zf 1 + · · · + z m f m , where f i ∈ Rm−i , then f 0 is not divisible by
ℓ.
Lemma 2.5. The ideal I := ℓK(−1) + f OP2 (−m) has the following properties:
(i) z is not a zero-divisor of OP2 /I.
(ii) H 0 (I(n)) = ℓH 0 (K(n − 1)), if n < m, and
H 0 (I(n)) = ℓH 0 (K(n − 1)) ⊕ f k[x, z]n−m , if n ≥ m and ℓ = αx + y
(respectively H 0 (I(n)) = ℓH 0 (K(n − 1)) ⊕ f k[y, z]n−m , if n ≥ m and ℓ = x).
(iii) colength(I) = c + m, reg(I) = m.
Proof. If ℓ = x, these are the statements of ([G4], Lemma 4, p. 660). If ℓ =
αx + y, α ∈ k ∗ , then let be u the automorphism x 7→ ℓ, y 7→ y, z 7→ z of S. By applying
12

(loc.cit) to
one gets
Now applying u gives

K := u−1 (K), I := u−1(I), f := u−1 (f )
H 0 (I(n)) = xH 0 (K(n − 1))) ⊕ f k[y, z]n−m .
H 0(I(n)) = ℓH 0 (K(n − 1)) ⊕ f k[y, z]n−m .

As k[y, z] = k[ℓ − αx, z] and ℓf ∈ ℓH 0 (K(m)), the statement (ii) follows, if α 6= 0.
If α = 0, we take the automorphism x 7→ y, y 7→ x, z 7→ z and argue as before.



We would like to put the section f in a certain normal form. We first consider the
case ℓ = αx + y. Then we can write f 0 = xm + ℓu, u ∈ Rm−1 , without restriction. As
m − 1 ≥ e, there is v = v 0 + zv 1 + z 2 v 2 + · · · ∈ H 0 (K(m − 1)) such that v 0 = u. As f is
determined only modulo ℓH 0 (K(m − 1)), we can replace f by f˜ := f − ℓv, therefore we
can assume without restriction, that f = f 0 + zf 1 + · · · + z m f m , where f 0 = xm .

We now suppose that K is invariant under Γ, and will formulate conditions that I
is Γ-invariant, too. This is equivalent to the condition that f is Γ-invariant modulo
ℓH 0 (K(m − 1)). By ([T2], Hilfssatz 1, p. 142) this is equivalent to the condition that
hx, yi∂f /∂z ⊂ ℓH 0 (K(m − 1)). It follows that ℓ is a divisor of f i , 1 ≤ i ≤ n, i.e., one has
f = xm + ℓzg, g ∈ Sm−2 .
Write g = g 0 + zg 1 + z 2 g 2 + · · · , where g i ∈ Rm−2−i and choose u = u0 + zu1 + · · · ∈
H 0 (K(m − 2)) such that u0 = g 0 . This is possible, if m − 2 ≥ e. As f is determined only
modulo ℓH 0 (K(m − 1)), one can replace f by f˜ = f − ℓzu. It follows that one can assume
without restriction f = xm + ℓz 2 g, where g ∈ Sm−3 .

Choose u ∈ H 0 (K(m − 3)), where u0 = g 0 ; this is possible, if m − 3 ≥ e. If this is the
case, replace f by f˜ = f − ℓz 2 u. It follows that one can assume without restriction f =
xm +ℓz 3 g, where g ∈ Sm−4 , etc. Finally one obtains f = xm +z m−e ℓg, ℓ = αx+y, g ∈ Se−1 ,
and the Γ-invariance of f modulo ℓH 0 (K(m−1)) is equivalent to hx, yi[(m−e)z m−e−1 ℓg +
z m−e ℓ∂g/∂z] ⊂ ℓH 0 (K(m − 1)), i.e. equivalent to:
(2.1)

hx, yi[(m − e)g + z∂g/∂z] ⊂ H 0 (K(e))

In the case ℓ = x, because of Rm = xRm−1 ⊕ y m · k, one can write f 0 = y m + xu, and
the same argumentation shows that one can write f = y m + z m−e xg, g ∈ Se−1 , and the
Γ-invariance can again be expressed by the inclusion (2.1).
2.4.3. Standard forms. Let I ⊂ OP2 have the colength d and Hilbert function ϕ,
and let I be invariant under G := Γ · T (ρ), where ρ2 > 0. Moreover, we assume that
g ∗ (ϕ) > g(d). By Lemma 2.2 it follows that I = ℓK(−1)+f OP2 (−m), if k = k is supposed.
As H 0 (I(m − 1)) = ℓH 0 (K(m − 2)) is then invariant under G and m − 1 > e = reg(K),
it follows that hℓi and K are G-invariant. Assume that hax + by + zi is Γ-invariant. But
then hax + by + zi = h(a + α)x + (b + β)y + zi, ∀ α, β ∈ k, which is not possible. Thus we
have ℓ = ax + by. From hλ0 ax + λ1 byi = hax + byi, ∀ (λ0 , λ1 , λ2 ) ∈ T (ρ) it follows that
λ0 /λ1 = 1 ∀ (λ0 , λ1 , λ2 ) ∈ T (ρ), if a and b both were different from 0. But then it would
13

follow T (ρ) ⊂ T (1, −1, 0), and therefore ρ2 = 0, contradiction. Therefore we have ℓ = x
or ℓ = y, without restriction.
We consider the case ℓ = x, for example. As it was shown in (2.4.2) we can write
f = y m + z m−e xg, e = reg(K), g ∈ Se−1 .

From Appendix E it follows that xH 0 (K(m − 1)) has a standard basis of T (ρ)-semiinvariants fi = mi pi (X ρ ), i.e., mi is a monomial, pi is a polynomial in one variable with
constant term 1, and such that mi does not occur in fj any longer, if i 6= j. Now each fi is
divisible by x, therefore y m does not occur in fi . If the initial monomial mi of fi appears
in f , then mi has to appear in z m−e xg. By choosing α ∈ k in a suitable way, one can
achieve that mi does not occur in f˜ := f − αfi . As ρ2 > 0 and fi is divisible by x, f˜ still
has the shape y m + z m−e xg̃, g̃ ∈ Se−1 . By repeating this procedure one can achieve that
none of the mi does occur in f = y m + z m−e xg (and f is still invariant under Γ modulo
xH 0 (K(m − 1)). The same argumentation as in the proof of the lemma in Appendix E
then shows that f is automatically a T (ρ)-semi-invariant with initial monomial y m , and
f together with the fi forms a standard basis of H 0 (I(m)). We summarize:
Lemma 2.6. Let be I ⊂ OP2k an ideal of colength d, with Hilbert function ϕ, which
is invariant under G = Γ · T (ρ), where ρ2 > 0. Assume that g(d) < g ∗(ϕ). (It is not
assumed that k = k.) Then I = xK(−1) + f OP2 (−m) or I = yK(−1) + f OP2 (−m),
where K is a G-invariant ideal with colength(K) = c, reg(K) = e, and c + m = d.
Moreover f ∈ H 0 (K(m)) can be written in the form f = y m + z m−e xg respectively in the
form f = xm + z m−e yg, where g ∈ Se−1 . We have hx, yi∂f /∂z ⊂ xH 0 (K(m − 1)) or
hx, yi∂f /∂z ⊂ yH 0(K(m − 1)), respectively, and each of these inclusions are equivalent
to the inclusion (2.1) in section (2.4.2). One has
(
xH 0 (K(n − 1))
if n < m,
0
H (I(n)) =
0
xH (K(n − 1)) ⊕ f k[y, z]n−m if n ≥ m,
respectively
(
yH 0(K(n − 1))
H 0 (I(n)) =
yH 0(K(n − 1)) ⊕ f k[x, z]n−m

if n ≤ m,
if n ≥ m.

If one chooses a standard basis {fi } of xH 0 (K(m−1)) or of yH 0(K(m−1)), respectively,
then one can choose f in such a way that f has the form and the properties mentioned
above and together with the fi ’s forms a standard basis of H 0 (I(m)).
Proof. If k = k this follows from the foregoing discussion. One has, e.g. I ⊗ k =
k

0

yK(−1)+f OP2 (−m), where K ⊂ OP2 ⊗k and f ∈ H (K(m)) have the properties mentioned
above. One sees that K = I ⊗ k : yOP2 ⊗k (−1) and therefore one has the exact sequence:
0 → (OP2 ⊗k /K)(−1) → OP2 ⊗k /I ⊗ k → OP2 ⊗k /I ⊗ k + yOP2⊗k (−1) → 0.
If K := I : yOP2 (−1), then the sequence
0 → (OP2 /K)(−1) → OP2 /I → OP2 /I + yOP2 (−1) → 0
14

is exact, too. Tensoring this sequence with k one obtains a commutative diagram
0
/

(OP2 ⊗k /K ⊗ k)(−1)

·y

/

OP2 ⊗k /I ⊗ k
/

OP2 ⊗k /I ⊗ k + yOP2⊗k (−1)
/

0

OP2 ⊗k /I ⊗ k
/

OP2 ⊗k /I ⊗ k + yOP2⊗k (−1)
/

0



0
/

OP2 ⊗k /K(−1)
/

with exact rows, where the first vertical arrow is obtained from the canonical injection
∼
K ⊗ k ֒→ K by tensoring with OP2 ⊗k (−1). It follows from this, that K ⊗ k → K.
∼
Because of H 0 (I(m)) ⊗ k → H 0 (I(m) ⊗ k) one obtains a standard basis of T (ρ)-semiinvariants of H 0 (I(m) ⊗ k) by tensoring a standard basis of H 0 (I(m)) with ⊗ 1k . As the
k

elements of a standard basis are uniquely determined up to constant factors, it follows
that f = f ⊗k 1k , where f ∈ H 0 (K(m)). Therefore f has the form xm + z m−e yg, g ∈
Se−1 , if f has the form xm + z m−e yg, g ∈ Se−1 ⊗ k. For reasons of dimension it follows
H 0 (I(n)) = yH 0(K(n − 1)), n < m, and H 0 (I(n)) = yH 0(K(n − 1)) + f k[x, z]n−m , n ≥ m.
As the G-invariance of K follows from the G-invariance of I, the remaining statements of

Lemma 2.6 follow by the same argumentation as in the case k = k.
Definition 2. The (uniquely determined) decomposition I = xK(−1)+f OP2 (−m) or
I = yK(−1)+f OP2 (−m) of Lemma 2.6 is called x-standard form or y-standard form of I,
respectively.( N.B. For the Hilbert function ϕ of I this definition implies that g(d) < g ∗ (ϕ)
is fulfilled.)
Corollary 2.3. Let be R = k[x, y]. If I has x-standard form (resp. y-standard
form), then xRm−2 (resp. yRm−2 ) is contained in H 0(I(m − 1)) and thus xRm−1 (resp.
yRm−1 ) is contained in H 0 (I(m)).
Proof. One has m − 2 ≥ c by Lemma 2.4 and thus xRm−2 ⊂ xH 0 (K(m − 2)) (resp.
yRm−2 ⊂ yH 0(K(m − 2))) by Appendix C, Remark 2.

Remark 2.3. If I has x-standard from or y-standard form, respectively, and if Gm acts
on S by σ(λ) : x 7→ x, y 7→ y, z 7→ λz, λ ∈ k ∗ , then I0 := lim σ(λ)I again has x-standard
λ→0

form or y-standard form, respectively. This follows from h0 (I0 (n)) = h0 (I(n)), n ∈ N (cf.
[G2], Lemma 4, p. 542), because then H 0 (I0 (n)) is generated by the initial monomials of
a standard basis of H 0 (I(n)), n ∈ N. The Figures 2.13a and 2.13b show the typical shape
of the pyramid formed by the initial monomials.
Remark 2.4. An ideal cannot have x-standard form and y-standard form at the same
time. For m = reg(I) and c = colength(I) are determined by the Hilbert function ϕ of
I. If I would have x-standard form as well as y-standard form, then E(H 0 (I0 (d))) has
the form shown in Figure 2.14a. As I and I0 have the same Hilbert function, ϕ has the
form shown in Figure 2.14b, and therefore g ∗ (ϕ) ≤ g ∗ (χ) = g(d), where χ is the Hilbert
function of (2.2.2).
Remark 2.5. (Notations and assumption as in Remark 2.3) I∞ := lim σ(λ)I has
λ→∞

again x-standard form or y-standard form, respectively. The reasoning is as follows:
15

The Hilbert function ϑ of I∞ (respectively the regularity µ of I∞ ) is ≥ ϕ (respectively
≥ m). This follows from the semicontinuity theorem. By Lemma 2.4 it follows that
m ≥ c + 2, therefore Rm−2 ⊂ H 0 (K(m − 2)) (cf. Appendix C, Remark 2). If I has xstandard form (respectively y-standard form), then xRm−2 ⊂ H 0 (I(m − 1)) (respectively
yRm−2 ⊂ H 0 (I(m − 1))) follows. Therefore xRm−2 ⊂ H 0 (I∞ (m − 1)) (respectively
yRm−2 ⊂ H 0 (I∞ (m − 1))). It follows that xRµ−2 ⊂ H 0 (I∞ (µ − 1)) (respectively yRµ−2 ⊂
H 0 (I∞ (µ − 1))). Now I∞ has x-standard form or y-standard form, in any case, and the
above inclusions show that I and I∞ have the same kind of standard form.
2.5. The type of a G-invariant ideal
Let I ⊂ OP2k be an ideal of colength d, with Hilbert function ϕ and invariant under
G = Γ · T (ρ), where ρ2 > 0 and k is an extension field of C.
2.5.1.
Definition 3. 1◦ I has the type (−1), if one of the following cases occurs:
1st case: g ∗ (ϕ) ≤ g(d), where d ≥ 5 by convention.
2nd case: 0 ≤ d ≤ 4

2◦ We now assume g ∗ (ϕ) > g(d), which notation implies d ≥ 5. Then one has
I = ℓ0 I1 (−1) + f0 OP2 (−m0 ), where (ℓ0 , I1 , f0 , m0 ) is as in Lemma 2.6. If I1 has type
(−1), then we say I has type 0. If I1 is not of type (−1), then I1 = ℓ1 I2 (−1)+f1 OP2 (−m1 ),
where (ℓ1 , I2 , f1 , m1 ) is as in Lemma 2.6. If I2 has the type (−1), then we say I has the
type 1, etc. As d = colength(I) = colength(I1 ) + m0 , etc., the colengths of the ideals in
question decrease and the procedure will terminate. We have
Lemma 2.7. If I has not the type (−1), then one has a sequence of decompositions
I =: I0 = ℓ0 I1 (−1) + f0 OP2 (−m0 ),
(2.2)

I1 = ℓ1 I2 (−1) + f1 OP2 (−m1 ),

·································

Ir−1 = ℓr−1 Ir (−1) + fr−1 OP2 (−mr−1 ),
Ir = ℓr K(−1) + fr OP2 (−mr ).

For a given ideal Ii , (ℓi , Ii+1 , fi , mi ) is
Ir+1 := K. If di and ϕi is the colength
the inequality g(di ) < g ∗ (ϕi ) is fulfilled.
Hilbert function, the regularity) of K is

defined as in Lemma 2.6, where 0 ≤ i ≤ r and
and the Hilbert function, respectively, of Ii , then
The ideal K has the type (−1). The colength (the
denoted by c (by ψ and κ, respectively).

3◦ We have already noted in Corollary 2.1 that the decompositions of I0 , I1 , · · · , are
uniquely determined, in essence. Therefore the number r is determined uniquely. It is
called the type of I. The types of the ideals occurring in (2.2), their Hilbert functions and
the numbers m0 , · · · , mr are uniquely determined by the Hilbert function ϕ.

16

2.5.2. Properties of an ideal of type r ≥ 0. The assumptions and notations are
as before, and we assume that I has the type r ≥ 0.
Lemma 2.8. In the decompositions (2.2) of I one has:
(a)
(b)
(c)
(d)

colength(Ii ) = colength(Ii+1 ) + mi , i.e., colength(Ii ) = c + mr + · · · + mi , 0 ≤ i ≤ r.
If r = 0, then m0 ≥ c + 2, where c = colength(K).
If r ≥ 1, then m0 ≥ c + 2 + mr + · · · + m1 = colength(I1 ) + 2.
If r ≥ 0, then m0 ≥ 2r (c + 2).

Proof. (a) follows from the decompositions (2.2) and from Lemma 2.6.
(b) follows from Lemma 2.4.
(c) As the statement only depends on the Hilbert functions of I0 , I1 , · · · , Ir+1 = K , one
can assume without restriction ℓi = x, 0 ≤ i ≤ r, and Ii is B(3; k)-invariant, 0 ≤ i ≤ r +1.
Then one has
I = xI1 (−1) + y m0 OP2 (−m0 ) and I1 = xI2 (−1) + y m1 OP2 (−m1 ),
where I2 = K, if r = 1.

We argue as in (2.3.4) and we orientate ourselves by Figure 2.15. If one would have
m0 − 1 − m1 ≤ colength(I2 ), then one could make the deformations
1 7→ 1, . . . , m0 − 1 − m1 7→ m0 − 1 − m1 .

Then we would get g ∗ (ϕ) < · · · ≤ g(d), contradiction, because from type r ≥ 0 it follows
that g ∗ (ϕ) > g(d). Therefore one has m0 − 1 − m1 > colength(I2 ). Now colength(I2 ) = c,
if r = 1 (respectively colength(I2 ) = c + mr + · · · + m2 , if r ≥ 2) as was shown in (a).
(d) If r = 0, this is statement (b). If r = 1, then by (b) and (c) it follows that
m0 ≥ (c + 2) + m1 ≥ 2(c + 2). Now we assume r ≥ 2. We argue by induction and assume
that mr ≥ c + 2, mr−1 ≥ 2(c + 1), . . . , m1 ≥ 2r−1 (c + 2). By (c) it follows that
m0 ≥ (c + 2) + (c + 2) + 2(c + 2) + · · · + 2r−1 (c + 2) = 2r (c + 2).

In the case r = 1, the statement (c) is valid even if K = OP2 , i.e., if c = 0. Because
then one can assume without restriction again
I = xI1 (−1) + y m0 OP2 (−m0 ) and I1 = xOP2 (−1) + y m1 OP2 (−m1 ),
and then, because of colength(I1 ) = m1 , by Lemma 2.4 it follows that m0 ≥ m1 + 2, i.e.,
one gets the statement (c).

Corollary 2.4 (of the proof of Lemma 2.8). If I has type r ≥ 1, then mj + j <
mi + i − 1 for all 0 ≤ i < j ≤ r.
Proof. We refer to Lemma 2.8 and use the same notations. If the sequence of
decompositions (2.2) in (2.5.1) begins with Ii instead of I0 , then from Lemma 2.8c it
follows that mi−1 ≥ c + 2 + mr + · · · + mi . It follows that mi−1 − mi ≥ 2, therefore
mi−1 + (i − 1) > mi + i. One gets
mr + r < mr−1 + (r − 1) < · · · < m1 + 1 < m0 .
17

If mj + j = mi + (i − 1), then it would follow j = i + 1 and therefore mi+1 = mi − 2. We
will show, that this is not possible.
The ideal Ii has the type r −i ≥ 1, the colength c + mr + · · ·+ mi = di , and the Hilbert
function ϕi of Ii fulfils the inequality g(di ) < g ∗(ϕi ) (cf. Lemma 2.7). As to the Hilbert
function ϕi , one obtains the situation of Figure 2.16, which corresponds to the Hilbert
function ϕ. But then it follows that g ∗ (ϕi ) ≤ g ∗ (χ′ ) = g(di ), where the graph of χ′ is
denoted by a dotted line (cf. the first case in (2.2.2), Figure 2.8, and the argumentation
in the proof of Lemma2.2).


18

Fig. 2.1

0

1

2

3

Fig. 2.2

α ... ... ... ... e
Fig. 2.3

0

1

2

3

4 ... ... ... ... d
Fig. 2.4

0

1

2

Fig. 2.5

0

1

2

3

4

ε ε+1 . . . . . . . . . m

Fig. 2.6a

0

1

2 ...

Fig. 2.6b

0

1

2

3

19

0

1

2

Fig. 2.7b
Fig. 2.7a

0

1

2

3 ...
Fig. 2.8

0

1

2

3

4 ...
Fig. 2.9

ye

ye

xy e−2

0

x2
1 2 3 ... ... ... ... ... ... e

20

0

x2
1 2 3

4 ... ... ... ... ... e

Fig. 2.10

u

v

0

1 . . . . . . . . . . . . . . . . . . . . . . . . . . . m−1 m

21

Fig. 2.11a

Fig. 2.11b

u

v

0

2 · · · · · · · · · · · ·m−1 m

1

0

1 ... ... n ... ... ... m

Fig. 2.12

6
5
4
3

ψ ′ (n − 1)

2
1

1
2
3

ϑ′ (n − 2)

4
5
6
0

1

2

3

4

5

6

7

8 ε+2 10 11 12 13 14 15 16 17 18 m
22

Fig. 2.13a

Fig. 2.13b
ym

xm
0

1

2

3

4

5 κ+1 7

8

0

1

2

3

Fig. 2.14a

0

1

2

3

4

5

4

5

6 κ+1 8

Fig. 2.14b

6

7 m

0

23

1

2

3

4

5

6

7 m

Fig. 2.15

m0 −1−m1

2

graph of ϕ′1 (n − 1)

1

1
graph of ϕ′2 (n − 2)

0

1

2

3

4

5

6

7

8

m2 +2 or κ+2

2

10 11 m1 +1 13 14 15 16 17 m0 19

#{monomials between graph of ϕ′2 (n − 1) and line y = x − 1} = colength(I2 )

24

mi

mi+1 +1

Fig. 2.16

25

CHAPTER 3

A rough description of ideals invariant under Γ · T (ρ)
It seems impossible to characterize ideals of colength d in OP2 , which are invariant
under G := Γ · T (ρ). If I is an ideal of type r in the sense of the definition in (2.5.1),
however, then one can give a rough description of the forms f0 , · · · , fr , which occur in the
decomposition (2.2) in (2.5.1). This description will be used later on for estimating the
so called α-grade of I which will be defined in the next chapter.
3.1. Assumptions
Let I ⊂ OP2 be an ideal of colength d, invariant under Γ · T (ρ), where ρ2 > 0, as
usual. We assume that I has the type r ≥ 0. Then according to Lemma 2.7 one has a
decomposition:
I =: I0 = ℓ0 I1 (−1) + f0 OP2 (−m0 ),
I1 = ℓ1 I2 (−1) + f1 OP2 (−m1 ),

(Z)

······························

Ii = ℓi Ii+1 (−1) + fi OP2 (−mi ),

······························

Ir = ℓr Ir+1 (−1) + fr OP2 (−mr ).

Using the notations of (loc.cit.) the ideal Ir+1 := K has the colength c and regularity κ.
If ℓi = x, then
H 0 (Ii (n)) = xH 0 (Ii+1 (n − 1)) + fi k[y, z]n−mi
and if ℓi = y, then

H 0 (Ii (n)) = yH 0(Ii+1 (n − 1)) + fi k[x, z]n−mi
(cf. Lemma 2.6). From (Z) follows
H 0 (I1 (mi + i − 1)) = ℓ1 H 0 (I2 (mi + i − 2)),

H 0 (I2 (mi + i − 2)) = ℓ2 H 0 (I3 (mi + i − 3)),
·································

H 0(Ii−1 (mi + 1)) = ℓi−1 H 0 (Ii (mi )),

H 0 (Ii (mi )) = ℓi H 0 (Ii+1 (mi − 1)) + hfi i,
for by Corollary 2.4 mi + 2 < mi−1 , i = 1, . . . , r. If one starts with H 0 (I1 (mi + i − 2)),
then one obtains a similar system of equations, whose last line is
H 0 (Ii (mi − 1)) = ℓi H 0 (Ii+1 (mi − 2)).
27

Conclusion 3.1. If 2 ≤ i ≤ r (if 1 ≤ i ≤ r, respectively), then H 0 (I1 (mi + i − 1)) =
ℓ1 · · · ℓi−1 H 0 (Ii (mi )) (H 0 (I1 (mi + i − 2)) = ℓ1 · · · ℓi H 0 (Ii+1 (mi − 2)), respectively).

3.2. Notations
We orientate ourselves by Figure 3.1, which shows the initial monomials of H 0 (I(m0 )).
The set of all monomials in Sm0 with z-degree ≥ m0 − (c + r) (with z-degree ≤ m0 − (c +
r + 1), respectively) is called the left domain and is denoted by LB (is called right domain
and is denoted by RB, respectively). The monomials in Sc−1 − in (H 0 (K(c − 1))) form a
basis of Sc−1 /H 0 (K(c − 1)). If we put ℓ = ℓ0 · · · ℓr , then ℓ[Sc−1 /H 0 (K(c − 1))] has a basis
consisting of the c monomials in ℓSc−1 − ℓ· in (H 0 (K(c − 1))).

The initial monomial of ℓ0 · · · ℓi−1 fi · z m0 −mi −i is Miup := xi−ι(i) y mi +ι(i) z m0 −mi −i or
Midown := xmi +i−ι(i) y ι(i) z m0 −mi −i , if ℓi = x or if ℓi = y, respectively. Here we have put, if
1 ≤ i ≤ r + 1, ι(i) := number of indices 0 ≤ j < i such that ℓj = y. We also put ι(0) = 0,
so the formulas give the right result for i = 0, too.
For instance, in Figure 3.1 we have
r = 5, ℓ0 = x, ℓ1 = y, ℓ2 = ℓ3 = x, ℓ4 = ℓ5 = y,
and therefore
ι(1) = 0, ι(2) = ι(3) = ι(4) = 1, ι(5) = 2, ι(6) = 3.
(N.B. ℓ0 · · · ℓi−1 = xi−ι(i) y ι(i) , if 0 < i ≤ r + 1.)

As always we assume ρ2 > 0. If Midown occurs, then Ii = yIi+1(−1) + fi OP2 (−mi ).
If Miup occurs, then Ii = xIi+1 (−1) + fi OP2 (−mi ). By Lemma 2.8 mi ≥ (c + 2) + mr +
· · · + mi+1 , if r ≥ 1 (m0 ≥ c + 2, if r = 0).The colength of Ii+1 equals c + mr + · · · + mi+1 ,
and therefore Rn ⊂ H 0 (Ii+1 (n)) if n ≥ mi − 2 ≥ c + mr + · · · + mi+1 in the case
r ≥ 1(Rn ⊂ H 0(K(n)) if n ≥ c in the case r = 0). This follows from Remark 2 in
Appendix C. The next generating element fi has the initial monomial Miup or Midown .

Conclusion 3.2. Suppose ρ2 > 0 and ρ1 is arbitrary.Then the columns of total degree
mi + i − 2 and mi + i − 1 (in the variables x and y ),which occur in in(H 0(I(m0 ))) , also
occur in H 0 (I(m0 )). Thus xMiup or yMidown , when it exists, is contained in I.
Proof. This follows from , Conclusion 3.1, the decompositions (Z) and the foregoing
discussion.

As always we assume ρ2 > 0. We have to distinguish between two cases (cf. 2.4.1
Auxiliary Lemma 1 and Auxiliary Lemma 2):
Main Case I: ρ0 > 0 and ρ1 < 0
Main Case II: ρ0 < 0 and ρ1 > 0
28

3.3. Description of f0 in Case I
If the initial monomial of f0 equals xm0 , then f0 = xm0 and ℓ0 = y. Therefore
we may assume that f0 is a proper semi-invariant with initial monomial y m0 . Then
I = I0 = xI1 (−1) + f0 OP2 (−m0 ), and we write f0 = y m0 + z µ x·G, µ ∈ N maximal, G a
T (ρ)-semi-invariant (cf. 2.4.3). If G0 is the initial monomial of G, then N : = z µ · x · G0
is called the vice-monomial of f0 , which by assumption is not equal to zero .
The inclusion (2.1) of (2.4.2) reads
(3.1)

hx, yi[µG + z ∂G/∂z] ⊂ H 0 (I1 (m0 − µ))

As for the initial monomial G0 of G, one has
(3.2)

hx, yiG0 ⊂ in(H 0 (I1 (m0 − µ))) = H 0 (in(I1 (m0 − µ)))

where in denotes the subspace or the initial ideal, respectively, generated by the initial
monomials. (Equality follows, for example, from in(I) = lim σ(λ)I and [G2], Lemma 3
λ→0

and 4.) Without restriction we may assume that G0 ∈
/ in(H 0 (I1 (m0 − µ − 1))), because
we can reduce f0 modulo xH 0 (I1 (m0 − 1)).

(N.B.: The same statements are analogously valid in the Case II, that means, if ρ1 > 0,
I = yI1 (−1) + f0 OP2 (−m0 ), f0 = xm0 + z µ yG, etc.)
It follows that one of the following cases occurs, where 1 ≤ i ≤ r:

1◦ N = Nidown := (z/x)Midown = xmi +i−ι(i)−1 y ι(i) z m0 −mi −i+1
2◦ N = Niup := (z/y)Miup = xi−ι(i) y mi +ι(i)−1 z m0 −mi +i+1
3◦ hx, yiN ⊂ ℓin(H 0 (K(m0 − r))), where ℓ := ℓ0 · · · ℓr and N is a monomial in
L := ℓSm0 −(r+1) − ℓin(H 0 (K(m0 − r − 1))).

Notabene: One has L = [ℓSc−1 − ℓ · in(H 0(K(c − 1)))]·z m0 −r−c , because of
m0M
−r−1
ν=c

z m0 −r−1−ν Rν ⊂ H 0 (K(m0 − r − 1)).

As µ is maximal and ρ2 > 0, G0 is not divisible by z. In the cases 1◦ and 2◦ we therefore
have µ = m0 − mi − i + 1. The case 3◦ will be treated in (3.3.6), and we assume the case
1◦ or 2◦ . Then (3.1) can be written as
hx, yi[µG + z ∂G/∂z] ⊂ H 0 (I1 (mi + i − 1)).

If we put h := ℓ1 · · · ℓi−1 , then h = xi−1−ι(i) y ι(i) , because of ℓ0 = x one has ι(i) = number
of indices j such that 0 < j < i and ℓj = y. If i = 1, then h := 1. By Conclusion 1
it follows from (3.1), that G is divisible by h, that means, one has G = hg, g ∈ Smi −1 .
Therefore we can write
f0 = y m0 + z µ xhg
and (3.1) is equivalent to:
(3.3)

hx, yi[µg + z ∂g/∂z] ⊂ H 0 (Ii (mi )).
29

3.3.1. We assume the case 1◦ . As Nidown occurs , Ii = yIi+1 (−1) + xmi OP2 (−mi ). If
g 0 is the initial monomial of the form g (cf. the inclusion (3.3)), then z µ xG0 = hg 0xz µ =
xi−ι(i) y ι(i) z µ g 0 = Nidown . Because of µ = m0 − mi − i + 1 it follows that g 0 = xmi −1 .
Representing the initial monomials of H 0 (Ii (mi )) in the same way as the initial monomials
of H 0 (I0 (m0 )), one sees that southwest of Nidown there is no further monomial which occurs
in f0 .
Conclusion 3.3. Let be ρ1 < 0. If the vice-monomial of f0 equals Nidown , then
f0 = y m0 + αNidown , where α ∈ k. As fi = Midown is a monomial, by Conclusion 3.2
it follows that all monomials which have the same z-degree as Mi and are elements of
in(H 0 (I(m0 ))) also are elements of H 0 (I(m0 )). Therefore we have (x, y)y m0 ⊂ I.

3.3.2. We assume the case 2◦ . Then y m0 X νρ = Niup , where ν > 0 and 1 ≤ i ≤ r.
This statement is equivalent to the equations:
(3.4)

νρ0 = i − ι(i), νρ1 = mi + ι(i) − 1 − m0 , νρ2 = m0 − mi − i + 1.

Lemma 3.1. If one has f0 = y m0 + αNiup + · · · , where 1 ≤ i ≤ r and α ∈ k − (0), then
ρ2 > mi+1 = reg(Ii+1 ), where we put Ir+1 := K and mr+1 := κ = reg(K). Especially Ij is
a monomial ideal for all j > i.
Proof. As we have assumed ρ1 < 0, by (2.4.1 Auxiliary Lemma 2) one has ρ0 > 0.
Therefore ν ≤ i. On the other hand ρ2 = (m0 − mi − i + 1)/ν ≥ (m0 − mi − i + 1)/i; thus
it suffices to show (m0 − mi − i + 1)/i > mi+1 , i.e.
(3.5)

m0 > mi + i · mi+1 + (i − 1).

We start with i = r, in which case one has to show m0 > mr + rκ + (r − 1). This is valid
if r = 0, so we may assume r ≥ 1. By Lemma 2.8 one has m0 ≥ c + 2 + mr + · · · + m1 ,
thus it suffices to show c + 2 + mr + · · · + m1 > mr + rκ + (r − 1). If r = 1, then this
inequality is equivalent to c + 2 > κ, and this is true. Therefore one can assume without
restriction that r > 1, and has to show
c + 2 + mr−1 + · · · + m1 > rκ + (r − 1).
Because of κ ≤ c it suffices to show 2 + mr−1 + · · · + m1 > (r − 1)(κ + 1) which is true
because of κ ≤ c < mr−1 < · · · < m1 (cf. Lemma 2.4 and Corollary 2.4).
Now we assume 1 ≤ i < r, in which case it suffices to show:
c + 2 + mr + · · · + m1 > mi + i · mi+1 + (i − 1)
⇐⇒

(c + 2) + (mr + · · · + mi+2 ) + (mi−1 + · · · + m1 ) > (i − 1)mi+1 + (i − 1).

On the left side of this inequality the second summand (the third summand, respectively)
does not occur, if i + 1 = r (if i = 1, respectively). If i = 1 and r = 2, the inequality
reads c + 2 > 0. If i = 1 and r ≥ 3, the inequality reduces to (c + 2) + mr + · · · + m3 > 0.
Thus the case i ≥ 2 remains, and it suffices to show mi−1 + · · · + m1 > (i − 1)(mi+1 + 1),
which is true because of mi+1 < mi < · · · < m1 (loc.cit.).

30

3.3.3. Description of the ideal Ii . The assumptions are the same as in Lemma
3.1, but we slightly change the notations and write I = Ii , K = Ii+1 , m = mi , e = reg(K).
(Thus e = mi+1 or e = κ.) We have the following situation: I = xK(−1) + f OP2 (−m) is
Γ · T (ρ)-invariant, K is monomial, ρ1 < 0, ρ2 > e = reg(K). From the results of (2.4.2)
and Lemma 2.6 it follows that f = y m + z m−e xg, where g ∈ Se−1 and
(3.6)

hx, yi[(m − e)g + z ∂g/∂z] ⊂ H 0 (K(e))

where f ∈ H 0 (K(m)) is determined only modulo xH 0 (K(m − 1) and we can assume that
f is a T (ρ)-semi-invariant. Then g is a T (ρ)-semi-invariant, too. Thus we can write
g = N(1 + a1 X ρ + · · · ), where N ∈ Se−1 is a monomial. If we assume a1 6= 0 for example,
then NX ρ ∈ Se−1 would have a z-degree ≥ ρ2 . As ρ2 > e (cf. Lemma 3.1 ), this is
impossible.
Conclusion 3.4. Under the assumptions mentioned above, the form g in (3.6) equals
a monomial N ∈ Se−1 − H 0 (K(e − 1)) such that hx, yiN ⊂ H 0 (K(e)). It follows that
(x, y)y m ⊂ I.

3.3.4. We go back to the notations of Lemma 3.1, i.e. one has N = Niup . As in the
case 1◦ one has µ = m0 − mi − i + 1. As y m0 is the initial monomial of f0 , ℓ0 equals x and
h := ℓ1 · · · ℓi−1 equals xi−1−ι(i) y ι(i) as before. If one puts g = µg + z∂g/∂z, then (3.3) can
be written as
(3.7)

hx, yig ⊂ xH 0 (Ii+1 (mi − 1)) + hfi i

where fi has the initial monomial y mi ; thus
(3.8)

g ∈ H 0 (Ii+1 (mi − 1)).

The initial monomial of g equals the initial monomial of g; by construction this is equal
to y mi −1 . (Proof: The initial monomial of xhgz µ = xGz µ is equal to Niup by assumption.
Thus the initial monomial g 0 of g fulfils the equation
xhg 0 z µ = xi−ι(i) y mi +ι(i)−1 z m0 −mi −i+1 .
As xh = xi−ι(i) y ι(i) and µ = m0 − mi − i + 1, it follows that g 0 = y mi −1 .) From (3.7) it
follows that
(3.9)

yg = xF + αfi , F ∈ H 0 (Ii+1 (mi − 1)), α ∈ k ∗ .

Now we can write fi = y mi + z mi −mi+1 xu (cf. the last sentence in (2.4.2)), where u is either
a monomial in Smi+1 −1 − H 0 (Ii+1 (mi+1 − 1)) (cf. Conclusion 2.4), or u = 0.

We consider the first case. Then it follows that z mi −mi+1 xu = βy mi X νρ , where β ∈ k ∗
and ν ≥ 1. As fi can be reduced modulo xH 0 (Ii+1 (mi − 1)), we have z mi −mi+1 xu ∈
/
mi −1
0
,
xH (Ii+1 (mi − 1)) without restriction. From (3.9) it follows that in g, except for y
mi −mi+1
the monomial u := z
xu/y occurs.
Suppose, there is another monomial v in g. Then one would have yv ∈
xH (Ii+1 (mi − 1)), and from (3.8) it would follow that v is an element of the monomial
subspace H 0 (Ii+1 (mi − 1)). Figure 3.2 shows H 0 (Ii (mi )) = xH 0 (Ii+1 (mi − 1)) + hfi i
marked with — in black and H 0 (Ii+1 (mi )) marked with — in blue. Suppose that
0

31

v ∈ H 0 (Ii (mi − 1)) = xH 0 (Ii+1 (mi − 2)). By construction v occurs in g, therefore
xhv = ℓ0 · · · ℓi v occurs in xG and z µ xhv occurs in z µ xG. On the other hand, z µ xhv ∈
z µ xℓ1 · · · ℓi−1 xH 0 (Ii+1 (mi − 2)) = z µ xH 0 (I1 (mi + i − 2)) ⊂ xH 0 (I1 (m0 − 1)), because ℓi = x and so Conclusion 3.1 can be applied. As f0 can be reduced modulo
xH 0 (I1 (m0 − 1)), one can assume without restriction, that v does not occur in g and
therefore does not occur in the inclusions (3.7) or (3.8), which have to be fulfilled. Thus
we can assume without restriction that v ∈ H 0 (Ii+1 (mi − 1)) − H 0 (Ii (mi − 1)). From
yv ∈ xH 0 (Ii+1 (mi − 1)) it follows that v is equal to one of the monomials in Figure 3.2,
which are denoted by ?. Therefore the z-degree of v is ≥ mi − mi+1 .
By construction, the z-degree of u is ≥ mi − mi+1 , too. As u and v occur in the semiinvariant g, both monomials differ by a factor of the form X νρ . As ρ2 > mi+1 (Lemma
3.1), it follows that ν = 0, i.e., u and v differ by a constant factor. Therefore one has
g = βy mi−1 + γz mi −mi+1 xu/y ; β, γ ∈ k ∗ .
We have to describe the position of xu more exactly. From xu ∈
/ xH 0 (Ii+1 (mi+1 − 1)) but
hx, yixu ∈ xH 0 (Ii+1 (mi+1 )) and from the Γ-invariance of Ii+1 it follows that z m0 −mi+1 xu
equals Njdown or Njup , where j is an index r ≥ j > i, or equals a monomial L ∈ L (cf. 3.3).

Suppose z m0 −mi+1 ·xu = Njdown . Then u = z mi −mi+1 xu/y is southwest of Njdown /z m0 −mi
and does not occur in the monomial subspace H 0 (Ii+1 (mi − 1)), which contradicts (3.8).
Finally we note that the monomials of g agree with the monomials of g.
Conclusion 3.5. Assume that f0 has the vice-monomial Niup and that fi is not a
monomial. Then (up to a constant factor) fi = Miup + αNjup , where 1 ≤ i < j ≤ r and
α ∈ k ∗ , or fi = Miup + αL, where L ∈ L is a monomial such (x, y)L ⊂ ℓK(−r − 1), and
α ∈ k∗ .

Then it follows that f0 = y m0 + βNjup + γNjup · (z/y) or f0 = y m0 + βNiup + γL · (z/y),
respectively. Here β and γ are elements of k ∗ . Finally, all monomials, which occur in x f0
or in y 2 f0 also occur in I.
Proof. The shape of f0 and of fi results from the forgoing argumentation. Conclusion
3.4 gives (x, y)L ⊂ ℓK(−r − 1). The statements concerning the monomials in xf0 follow
from Conclusion 3.2. By Lemma 3.1, Ij is monomial, thus yNjup = z Mjup ∈ I and
therefore y Miup ∈ I. Furthermore y 2Njup · (z/y) = y Njup z ∈ I and as well y 2L · (z/y) =
yLz ∈ I. From this the assertion concerning the monomials of y 2 f0 follows.

Now we come to the case that fi = y mi . The above reasoning had shown that y mi −1
occurs in g. Suppose that another monomial v occurs in g. The same argumentation as
before shows that v equals one of the monomials in Figure 3.2 denoted by ?. Thus v is
equal to Njup or Njdown for an index i < j ≤ r, or is equal to a monomial L ∈ L, such that
(x, y)L ⊂ ℓK(−r − 1). Furthermore, there can be only one such monomial v.
Conclusion 3.6. If f0 has the vice-monomial Niup and if fi = y mi , then f0 = y m0 +
αNiup +βNjup or f0 = y m0 +αNiup +βNjdown , where 1 ≤ i < j ≤ r, or f0 = y m0 +αNiup +βL,
32

where L ∈ L is a monomial such that (x, y)L ⊂ ℓK(−r − 1). All monomials occurring in
xf0 or yf0 also occur in I.
Proof. The statements concerning the shape of f0 follow from the foregoing discussion. As Ii+1 is monomial by Lemma 3.1 and as fi is a monomial, Ii is monomial and
(x, y)L, (x, y)Niup and (x, y)Njdown are contained in I.

3.3.5. Suppose in the forms f0 and fj , where 1 ≤ j ≤ r , as in Conclusion 3.5
or Conclusion 3.6 there are three monomials with coefficients different from zero. We
call f0 (and fj , respectively) a trinomial and we then have f0 = y m0 + αNiup + βE0
and fj = Mjup + γNkup + δEj , where the “final monomials” E0 and Ej have the shape
described in Conclusion 3.5 and Conclusion 3.6, and where α, β, γ, δ ∈ k ∗ . If i > k,
then mi ≤ mk+1 < ρ2 (Lemma 3.1). As we are in the case ρ0 > 0, ρ2 > 0, it follows that
|ρ1 | > mi , and as Niup and E0 both occur in the semi-invariant f0 , one has E0 = Niup · X νρ ,
where ν ≥ 1. Looking at Figure 3.2, one sees that then E0 cannot be an element of Sm0 ,
contradiction. In the same way it follows that i < k is not possible, and thus we have
i = k. As f0 and fj are T (ρ)-semi-invariants, it follows that y m0 · X νρ = Mjup , where
ν ≥ 1. We conclude from this that
νρ0 = j − ι(j), νρ1 = mj + ι(j) − m0 , νρ2 = m0 − mj − j,
where 1 ≤ j ≤ r. We want to show that this implies ρ2 > mj+1 . Similarly as in the proof
of Lemma 3.1 it suffices to show m0 > mj + jmj+1 + j. We start with the case j = r.
Then again mr+1 := κ and one has to show m0 > mr + rκ + r. The case r = 0 cannot
occur. If r = 1, the inequality reads m0 > m1 + κ + 1, and because of m0 ≥ c + 2 + m1
(Lemma 2.8) and κ ≤ c this is right. Therefore we can assume r ≥ 2.

Because of (loc.cit.) it suffices to show c + 2 + mr + · · · + m1 > mr + rκ + r. As κ ≤ c it
suffices to show 2+mr−1 +· · ·+m1 > (r−1)κ+r ⇐⇒ 1+mr−1 +· · ·+m1 > (r−1)(κ+1).
Because of κ ≤ c < mr < · · · < m1 this is true (cf.Lemma 2.4 and Corollary 2.4).
We now assume 1 ≤ j < r, in which case it suffices to show:

c + 2 + mr + · · · + m1 > mj + jmj+1 + j
If j = 1 this inequality reads c + 2 + mr + · · · + m1 > m1 + m2 + 1, and this is true,
because r ≥ 2. Thus we can assume 2 ≤ j < r and the inequality which has to be shown
is equivalent to
(c + 1) + (mr + · · · + mj+2 ) + (mj−1 + · · · + m1 ) > (j − 1)(mj+1 + 1)
Because of mj+1 < mj < · · · < m1 (loc.cit.), this is true.
We thus have proved that from y m0 X νρ = Mjup

(3.10)

ρ2 > mj+1

follows. As k > j by definition, we have mk ≤ mj+1 < ρ2 and the same argumentation as
before carried out with Niup and E0 shows that Nkup and Ej cannot simultaneously occur
in fj .
33

Conclusion 3.7. Among the forms fi , 0 ≤ i ≤ r, which have the shape described in
Conclusion 3.5 and Conclusion 3.6, there can only occur at most one trinomial.
3.3.6. We now consider the case 3◦ . The following argumentation is first of all
independent of the sign of ρ1 . We write f0 = f 0 + z u g, where f 0 = y m0 if ℓ0 = x and
f 0 = xm0 , if ℓ0 = y. If we choose µ maximal then we have µ ≥ m0 − (κ + r). For as
Rκ ⊂ in(H 0 (K(κ))), the initial monomial of g has a z-degree > m0 − (κ + r + 1), i.e., the
initial monomial occurs in a column of total degree in x and y smaller or equal κ + r.
From the Γ-invariance of f0 modulo ℓ0 H 0 (I1 (m0 − 1)) it follows that hx, yi∂f /∂z =
hx, yi [µz µ−1 g + z µ ∂g/∂z] ⊂ ℓ0 H 0 (I1 (m0 − 1)). From the decompositions in (Z) (cf. 3.1)
we conclude that H 0 (Ii (n)) = ℓi H 0 (Ii+1 (n − 1)) if n < mi . Now
m0 − µ < κ + r + 1 < mr + r < · · · < m1 + 1 < m0
(cf. Corollary 2.4) and thus
H 0 (I1 (m0 − µ)) = ℓ1 H 0 (I2 (m0 − µ − 1)),

H 0 (I2 (m0 − µ − 1)) = ℓ2 H 0 (I3 (m0 − µ − 2)),

0

················································

H (Ir−1 (m0 − µ − r + 2)) = ℓr−1 H 0 (Ir (m0 − µ − r + 1)),
H 0 (Ir (m0 − µ − r + 1)) = ℓr H 0 (K(m0 − µ − r)).

It follows that H 0 (I1 (m0 − µ)) = ℓ1 · · · ℓr H 0 (K(m0 − µ − r)) and therefore
hx, yi [µg + z ∂g/∂z] ⊂ ℓH 0 (K(m0 − µ − r)) where ℓ := ℓ0 · · · ℓr = xa y b
and a (respectively b) is the number of ℓi = x (respectively the number of ℓi = y). This
implies that g is divisible by ℓ, and changing the notation we can write f0 = f 0 + ℓz µ g,
where ℓ = ℓ0 · · · ℓr , µ ≥ m0 − (κ + r) is maximal, g ∈ Sm0 −µ−r−1 and
hx, yi [µg + z ∂g/∂z] ⊂ H 0 (K(m0 − µ − r)).
Now f0 ∈ H 0 (K(m0 )) or f0 ∈ H 0 (I1 (m0 )) and m0 ≥ colength(K) + 2, if r = 0 or
m0 ≥ colength(I1 ) + 2, if r ≥ 1, respectively (cf. Lemma 2.6 and Lemma 2.8). Therefore
Rm0 ⊂ H 0 (K(m0 )) or Rm0 ⊂ H 0 (I1 (m0 )), respectively (cf. Appendix C, Remark 2). It
follows ℓz µ g ∈ H 0 (K(m0 )) (or ℓz µ g ∈ H 0 (I1 (m0 )) and thus ℓg ∈ H 0 (K(m0 − µ)) (or
ℓg ∈ H 0 (I1 (m0 − µ)) = ℓ1 · · · ℓr H 0 (K(m0 − µ − r)), respectively).

From this we conclude that ℓ0 g ∈ H 0 (K(m0 − µ − r)), in any case. We also note that
g∈
/ H 0 (K(m0 − µ − r − 1)) without restriction, because otherwise
ℓg ∈ ℓH 0 (K(m0 − µ − r − 1)) = ℓ0 H 0 (I1 (m0 − µ − 1)),
and thus z µ ℓg ∈ ℓ0 H 0 (I1 (m0 − 1)) would follow. But then ℓg could be deleted.
34

Conclusion 3.8. If the vice-monomial of f0 is different from all monomials Niup and
Njdown , then we can write
f0 = f 0 + z µ ℓg, where µ ≥ m0 − (κ + r)is maximal, ℓ := ℓ0 · · · ℓr ,
f 0 = y m0 , if ℓ0 = x (or f = xm0 , if ℓ0 = y, respectively),

g ∈ Sm0 −µ−r−1 − H 0 (K(m0 − µ − r − 1)), ℓ0 g ∈ H 0 (K(m0 − µ − r))

and hx, yi [µg + z ∂g/∂z] ⊂ H 0 (K(m0 − µ − r)). 

3.4. Order of f0 in the case 3◦
We keep the notations of Conclusion 3.8, but now we write h := ℓ0 · · · ℓr . In order to
simplify the notations , the cases ρ1 > 0 and ρ1 < 0 will be treated separately. As always
we assume ρ2 > 0.
3.4.1. Let be ρ1 > 0 (and thus ρ0 < 0). Conclusion 3.8 is still valid, if x and y are
interchanged. Thus one can write f0 = xm0 + z µ hg, ℓ0 = y and
H 0(I(d)) = yH 0(I1 (d − 1)) ⊕ h{f0 xn z d−m0 −n |0 ≤ n ≤ d − m0 }i,
where Ii := K if r = 0 (cf. Lemma 2.6).

Suppose there is a 0 ≤ ν ≤ d−m0 such that xν+1 f0 ≡ xm0 +ν+1 modulo hH 0 (K(m0 +ν−r)).
Because of
hH 0 (K(m0 + ν − r)) = yℓ1 · · · ℓr H 0 (K(m0 + ν − r)) ⊂ yH 0(I1 (m0 + ν))
then follows that
H 0 (I(d)) = yH 0(I1 (d − 1))

⊕ h{f0 xn z d−m0 −n |0 ≤ n ≤ ν}i ⊕ h{xm0 +n z d−m0 −n |ν + 1 ≤ n ≤ d − m0 }i.

˙ α (H 0 (I(d)) (cf. Chapter 4), then
If one wants to determine the so called α-grade of ∧ψ
it is easier to estimate the contribution coming from the monomials in H 0 (I(d)). In the
following definition we assume ρ2 > 0 and the sign of ρ1 is arbitrary.
Definition 4. The order of f0 is the smallest natural number ν such that xν+1 f0 ≡
xm0 +ν+1 modulo hH 0 (K(m0 + ν − r)) (such that y ν+1 f0 ≡ y m0 +ν+1 modulo hH 0 (K(m0 +
ν − r)), respectively) if ρ1 > 0 (if ρ1 < 0, respectively).
Here we have put h := ℓ0 · · · ℓr (cf. (Z) in 3.1).
Remark 3.1. If ρ1 > 0, then ρ0 < 0, and from Lemma 2.6 it follows that f0 has
the initial monomial M0down = xm0 . Then Conclusion 3.2 gives yxm0 ∈ I. On the other
hand, if ρ1 < 0 the same argumentation shows that f0 has the initial monomial y m0 and

xy m0 ∈ I.
We now take up again the assumption ρ1 > 0 from the beginning of (3.4.1).
Subcase 1: ℓ0 = x. As ρ0 < 0 one has f0 = y m0 .
Subcase 2: ℓ0 = y. Putting h := xa y b one can write
f0 = xm0 + z µ hg = xm0 (1 + xa−m0 y b z µ g).
35

As f0 is a T (ρ)-semi-invariant, one has
xa−m0 y bz µ g = X γρ p(X ρ ), where γ ∈ N−(0), p(X ρ ) = a0 +a1 X ρ +. . .+as X sρ and ai ∈ k.
We now assume f0 is not a monomial. If both µ and γ are chosen maximal, then a0 6= 0,
and we can write g = Mp(X ρ ), where M := xm0 −a+γρ0 y γρ1 −b and µ = γρ2 . From the
Γ-invariance of f0 modulo ℓ0 H 0 (I1 (m0 − 1)) it follows that
hx, yihz µ−1 [µg + z∂g/∂z] ⊂ ℓ0 H 0 (I1 (m0 − 1)),
thus
hx, yiℓ1 · · · ℓr [µg + z∂g/∂z] ⊂ H 0 (I1 (m0 − µ)).
We had already obtained H 0 (I1 (m0 − µ)) = ℓ1 · · · ℓr H 0 (K(m0 − µ − r)) in (3.3.6).
We conclude that
hx, yi[µg + z∂g/∂z] ⊂ H 0 (K(m0 − µ − r))
and therefore:
xg ≡ − µ1 xM[ρ2 a1 X ρ + 2ρ2 a2 X 2ρ + · · · + sρ2 as X sρ ] modulo H 0 (K(m0 − µ − r))
Assume that s = degree (p) > 0. Then as 6= 0 and j := min {i > 0|ai 6= 0} is an integer
between 1 and s inclusive. Then one can write :
xg ≡ − µ1 xMX jρ [jρ2 aj + (j + 1)ρ2 aj+1 X ρ + · · · + sρ2 as X (s−j)ρ ] modulo H 0 (K(m0 − µ − r))
From this it follows that
xf0 = xm0 +1 + h z µ xg ≡ f˜ modulo hH 0 (K(m0 − r)),
where f˜ := xm0 +1 + h z µ̃ M̃ p̃(X ρ ) and µ̃ := µ + jρ2 > µ, M̃ := xMxjρ0 y jρ1 ,
p̃(X ρ ) := ã0 + ã1 X ρ + · · · + ãs−j X (s−j)ρ
and finally ã0 := − µ1 jρ2 aj 6= 0, · · · , ãs−j := − µ1 sρ2 as 6= 0.
Remark 3.2. f˜ is again a T (ρ)-semi-invariant, because
x−(m0 +1) hz µ̃ M̃ = x−m0 −1 xa y b z µ+jρ2 x · xm0 −a+γρ0 · y γρ1 −b · xjρ0 · y jρ1
= x(γ+j)ρ0 y (γ+j)ρ1 z (γ+j)ρ2 . 

Remark 3.3. f˜ is only determined modulo ℓ0 H 0 (I1 (m0 )), as hH 0 (K(m0 − r)) ⊂
ℓ0 H 0 (I1 (m0 )).

We continue the above discussion and get at once xf0 ≡ xm0 +1 modulo hH 0 (K(m0 −r)),
if s = 0. In any case we have degree (p̃) < degree (p), and continuing in this way, we get
finally p̃·(s+1) = 0. This means, there is an integer 0 ≤ ν ≤ s such that
(3.11)

xν+1 f0 ≡ xm0 +ν+1 modulo hH 0 (K(m0 − r + ν)).
36

3.4.2. We now assume ρ1 < 0. If f0 is a monomial, then the order of f0 is zero
by definition. Thus we may assume without restriction that ρ0 > 0 (c.f. 2.4.1 Auxiliary
Lemma 2).
Subcase 1: ℓ0 = y. Then f0 = xm0 (Lemma 2.6).
Subcase 2: ℓ0 = x. With notations analogous to that of (3.4.1) one has: f0 = y m0 (1 +
xa y b−m0 z µ g) is a semi-invariant, therefore xa y b−m0 z µ g = X γρ p(X ρ ), γ ∈ N − (0). If f0 is
not a monomial and µ and γ are chosen maximal, then one can write g = Mp(X ρ ), where
µ = γρ2 , M = xγρ0 −a y m0 −b+γρ1 and p(X ρ ) = a0 +a1 X ρ +· · ·+as X sρ . Furthermore one gets
yf0 ≡ f˜ modulo hH 0 (K(m0 − r)), where f˜ := y m0 +1 + hz µ̃ M̃ p̃(X ρ ), µ̃ := µ + jρ2 , M̃ :=
yMxjρ0 y jρ1 and p̃ is defined as in (3.4.1). It is clear that one gets the corresponding
statements as in (3.4.1), if x and y as well as a and b are interchanged. This means, there
is an integer 0 ≤ ν ≤ s such that
(3.12)

y ν+1f0 ≡ y m0 +ν+1 modulo hH 0 (K(m0 − r + ν)).

3.4.3. Estimate of the order of f0 . The order has been defined as the smallest
natural number ν such that xν+1 f0 ≡ xm0 +ν+1 (respectively y ν+1f0 ≡ y m0 +ν+1 ) modulo
hH 0 (K(m0 − r + ν)). We keep the notations of (3.4.1) and (3.4.2). From γρ2 = µ ≥
m0 − (κ + r) (cf. 3.3.6) it follows that γ ≥ [m0 − (κ + r)]/ρ2 . On the other hand,
xm0 −a+γρ0 · xsρ0 has to be a polynomial in case that ρ1 > 0 ( y m0 −b+γρ1 · y sρ1 has to be
a monomial in case that ρ1 < 0, respectively). That means m0 − a + (γ + s)ρ0 ≥ 0
(m0 − b + (γ + s)ρ1 ≥ 0, respectively). Now one has ρ0 < 0, if ρ1 > 0, and ρ0 > 0, if
ρ1 < 0 (cf. 2.4.1 Auxiliary Lemma 2). It follows that
γ + s ≤ (m0 − a)/|ρ0 | if ρ1 > 0,

γ + s ≤ (m0 − b)/|ρ1 | if ρ1 < 0.

Here a (respectively b) is the number of ℓi , 0 ≤ i ≤ r, such that ℓi = x (respectively
ℓi = y).
In case that ρ1 > 0 we obtain , because of |ρ0 | = ρ1 + ρ2 :
s≤

m0 − a m0 − (κ + r)
−
ρ1 + ρ2
ρ2

And in case that ρ1 < 0, because of |ρ1 | = ρ0 + ρ2 , we obtain
s≤

m0 − b m0 − (κ + r)
−
.
ρ0 + ρ2
ρ2

From the congruences (3.11) and (3.12) we get
Conclusion 3.9. The order s of f0 fulfils the inequality:



a
 κ − m0 1 − 1
+ ρr2 − ρ1 +ρ
if ρ1 > 0,
ρ2
ρ2
ρ1 +ρ2
2


s≤
b
 κ − m0 1 − 1
+ ρr2 − ρ0 +ρ
if ρ1 < 0.
ρ2
ρ2
ρ0 +ρ2
2
37

3.5. Summary in the Case I
We recall that this means ρ0 > 0, ρ2 > 0 and ρ1 < 0.
Proposition 3.1. Let be ρ2 > 0 and ρ1 < 0.
(a) The following cases can a priori occur:
1st case: One of the fi , say fi0 has as vice-monomial an “upper empty corner” Njup
. Then
0
ρ2 > κ, thus K is monomial, and the fi have one of the following forms:
(1)
(2)
(3)
(4)
(5)

fi = Midown
fi = Miup + αNjdown
fi = Miup + αNjup + βNkup
fi = Miup + αNjup + βNkdown
fi = Miup + αNjup + βL, L ∈ L a monomial
such that
(x, y) · L ⊂ ℓK(−r − 1), ℓ := ℓ0 · · · ℓr .

(6) fi = Miup + αNjup + βNkup · (z/y), α, β ∈ k ∗
(7) fi = Miup + αNjup + βL · (z/y), L ∈ L a monomial
such that

(x, y) · L ⊂ ℓK(−r − 1), α, β ∈ k ∗ .
Here α and β are (a priori) arbitrary elements of k and 0 ≤ i < j < k ≤ r.

2nd case: Not any of the fi has as vice-monomial an “upper empty corner” Njup . Then
each of the fi has one of the following forms:
(1) fi = Miup + αNjdown
(2) fi = Midown
P
(3) fi = Miup + F, F = αj Lj , Lj ∈ L monomial, αj ∈ k ∗ , such that xF ∈ ℓK(−r −
P
1), and for suitable βj ∈ k ∗ and G := βj Lj one has (x, y) · G ⊂ ℓK(−r − 1).

(b) In the 1st case there is only one trinomial at most, i.e. there is only one fi at most,
which has the shape as in one of the cases 1.3 - 1.7, with α and β different from zero.
S
(c) Let be M :=
{monomials in H 0 (I(n))} and let be hMi the subspace of S generated
n≥0

by these monomials. In the 1st case one has:

• xfi ∈ hMi, for all 0 ≤ i ≤ r;
• yfi ∈ hMi, if fi has the shape as in one of the cases 1.1–1.5
• y 2 fi ∈ hMi, if fi has the shape as in one of the cases 1.6 and 1.7.
(d) In the 2nd case one has hx, yifi ⊂ hMi, if fi has the shape as in case 2.1.
If fi has the shape as in case 2.3, then xMiup ∈ M , and the order s of fi fulfils the
inequality


1
1
r
κ
− mi
−
+ ,
s≤
ρ2
ρ2 ρ0 + ρ2
ρ2

where 0 ≤ i ≤ r.

38

Proof. (1.1) and (1.2) result from Conclusion 3.2 and Conclusion 3.3; (1.3)–(1.5)
result from Conclusion 3.6; and (1.6), and (1.7) results from Conclusion 3.5. (2.1) results
again from Conclusion 3.2 and Conclusion 3.3; (2.2) is clear, and (2.3) results from Conclusion 3.8, for if Miup occurs, then ℓi = x, and this corresponds to the linear form ℓ0 in
Conclusion 3.8.
(b) results from Conclusion 3.7.
(c) results from the Conclusions 3.3, 3.5 and 3.6.
(d) results in the case (2.1) from the fact that fj = Mjdown is a monomial and therefore all
monomials with the same z-degree as Mjdown , which occur in in(H 0(I(m0 ))), also occur in
H 0 (I(m0 )) (Conclusion 3.2). In the case (2.3) the first part of the statement results from
Conclusion 3.2, and the statement concerning the order results from Conclusion 3.9. 
Remark 3.4. If r = 0, then only the cases (2.2) and (2.3) can occur, i.e., one has
either f0 = xm0 or f0 = y m0 + F , where F has the properties mentioned above.
3.6. Summary in the Case II
We recall that this means ρ0 < 0, ρ1 > 0, ρ2 > 0.
We translate formally the results of (3.5): As always ρ2 > 0 is assumed, one has ρ0 < 0.
By definition ι(i) = #{ℓj = y|0 ≤ j < i} and therefore i − ι(i) = #{ℓj = x|0 ≤ j < i}.
We conclude that if one interchanges x and y in the monomial Miup (respectively Midown )
and if one simultaneously replaces ι(i) by i − ι(i) and vice versa i − ι(i) by ι(i), then one
obtains Midown (respectively Miup ). Therefore the statements of Proposition 3.1 can be
translated to the Case II by interchanging x and y and “up” and “down”:
Proposition 3.2. Let be ρ2 > 0 and ρ1 > 0.
(a) The following cases can a priori occur:
1st case: One of the fi , say fi0 has as vice-monomial a “lower empty corner” Njdown .
Then ρ2 > κ, thus K is monomial and the fi have one of the following forms:
fi = Miup
fi = Midown + αNjup
fi = Midown + αNjdown + βNkdown
fi = Midown + αNjdown + βNkup
fi = Midown + αNjdown + βL, L ∈ L a monomial
such that (x, y) · L ⊂ ℓK(−r − 1), ℓ := ℓ0 · · · ℓr
(6) fi = Midown + αNjdown + βNkdown · (z/x), α, β ∈ k ∗
(7) fi = Midown + αNjdown + βL · (z/x), L ∈ L a monomial
such that (x, y) · L ⊂ ℓK(−r − 1), α, β ∈ k ∗ .

(1)
(2)
(3)
(4)
(5)

2nd case: Not any of the fi has as vice-monomial a “ lower empty corner” Njdown . Then
each of the fi has one of the following forms:
(1) fi = Midown + αNjup
(2) fi = Miup
39

P
(3) fi = Miup + F, F =
αj Lj , Lj ∈ L monomial,αj ∈ k ∗ , such that yF ∈ ℓK(−r −
P
1), and for suitable βj ∈ k ∗ and G := βj Lj one has (x, y) · G ⊂ ℓK(−r − 1).
(b) The same statement as in Proposition 3.1.
(c) The same statement as in Proposition 3.1, if x and y are interchanged.
(d) The same statement as in Proposition 3.1, if one replaces xMiup by yMidown .
And the order s of fi fulfils the inequality


r
1
1
κ
+ ,
− mi
−
s≤
ρ2
ρ2 ρ1 + ρ2
ρ2
where 0 ≤ i ≤ r.
Remark 3.5. If r = 0, then only the cases (2.2) and (2.3) can occur.

40



Fig. 3.1
M0

M2
N2

M3
N3

N5 M 5
N4 M 4
N1 M 1
7

8

9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

Explanation: reg(K) ≤ c ⇒ Rn ⊂ H 0 (K(n)), n ≥ c
Sc−1 /H 0 (K(c − 1)) has a basis of c monomials, namely the monomials
of Sc−1 − inH 0 (K(c − 1)) ⇒ ℓ[Sc−1 /H 0(K(c − 1))] has a basis
of c monomials, namely the monomials of ℓSc−1 − ℓ · inH 0 (K(c − 1))
They generate a subspace L. ℓ := ℓ0 · · · ℓr = xr+1−ι(r+1) y ι(r+1)

41

m0

6

m1 +1

5

m2 +2

4

m3 +3

3

m4 +4

2

m5 +5

1

κ+6

0

Fig. 3.2
y mi
Niup
Ii+1

?
?
Ii = xIi+1 (−1) + fi OP2 (−mi )
?
?

?
?

42

mi

mi+1 +1

mi+1

?

CHAPTER 4

The α-grade.
4.1. Notations.
We let Gm (respectively Ga ) operate on S = k[x, y, z] by σ(λ) : x → x, y 7→ y, z 7→ λz
(respectively by ψα : x 7→ x, y 7→ αx + y, z 7→ z).
m
m
V
V
Let be G = Grassm (Sn ). If V ∈ G(k), then V has the dimension 1 and V 7→ V
m
m
V
V
p
defines a closed immersion G → P( Sn ) = PN , N := dim Sn − 1, the so called Plücker
embedding.
If one numbers the monomials in Sn , then one gets a basis {e1 , e2 , · · · } of Sn , and
m
V
therefore e(i) = ei1 ∧ · · · ∧ eim is a basis of Sn , where (i) = (i1 , · · · , im ) runs through all

. If one puts
sequences of natural numbers, such that 1 ≤ i1 < · · · < im ≤ n+2
2

ψα (e(i) ) := ψα (ei1 ) ∧ · · · ∧ ψα (eim ), then Ga operates in an equivariant manner on G
and PN , and the same is valid for the operation σ of Gm . If ξ ∈ G(k) corresponds to
m
V
the vector space V ⊂ Sn , then Cξ := { ψα (V )|α ∈ k}− is a point or a curve in PN ,
and there are polynomials f0 , · · · , fN in one variable with coefficients in k, such that
Cξ = {(f0 (α) : · · · : fN (α))|α ∈ k}− . At least one of the fi is equal to 1, and if ξ is
not invariant under Ga , then Cξ is a Ga -invariant closed curve in PN of degree equal to
max{deg(fi )|0 ≤ i ≤ N} (cf. [T1], Bemerkungen 2 und 3, p. 11). This number is denoted
by α-grade (V ).

Let now be ei , 1 ≤ i ≤ ℓ := n+2
, the monomials in Sn , ordered in the inverse
2
ℓ
P
lexicographic manner. If fi =
aji ej , 1 ≤ i ≤ m, is a basis of V , then f1 ∧ · · · ∧ fm =
j=1


a
·
·
·
a
i
1
i
m
1
1
P
P(i) e(i) , where P(i) = det  · · · · · · · · · · · ·  is the Plücker coordinate for the index
(i)
aim 1 · · · aim m
m
V
P
(i) = (i1 , · · · , im ). It follows that ψα ( V ) = h P(i) ψα (e(i) )i and we conclude from this:
(i)

α − grade (V ) ≤ max{α − grade ψα (e(i) ) | P(i) 6= 0}

(4.1)

(i)

where we define the α-grade of ψα (e(i) ) to be the α-grade of the monomial subspace
ℓ
P
pjν (α)ej , 1 ≤
hei1 , · · · , eim i of Sn . This can be computed as follows: Write ψα (eiν ) =
j=1

ν ≤ m, where the pνj are polynomials in one variable with coefficients in Z. Then
P
ψα (e(i) ) = P(j) (α)e(j) , where the P(j) (α) are the Plücker coordinates of the vector space
(j)

43

hψα (ei1 ), · · · , ψα (eim )i. The P(j) are polynomials in one variable with coefficients in Z. As
P(i) (α) = 1, the α-grade of hei1 , · · · , eim i is equal to max{deg(P(j) )}.
(j)

Whereas it seems practically impossible to find a formula for the α-grade of an arbitrary vector space V ⊂ Sn , the α-grade of a monomial subspace V ⊂ Sn can be computed
n
L
as follows: Write V =
z n−i Vi where Vi ⊂ Ri is a monomial subspace and R = k[x, y].
i=0

If we put m(i) := dim Vi then Vi has a basis of the form {xi−aij y aij |1 ≤ j ≤ m(i)}, where
n
P
0 ≤ ai1 < · · · < aim(i) ≤ i is a sequence of integers. As α-grade (V ) =
α-grade (Vi ), we
i=0

can consider V as a graded vector space in R, which has a basis of monomials of different
degrees. In ([T1], 1.3, p. 12f.) it was shown:
If 0 ≤ c1 < · · · < cr ≤ i are integers, and
W := hxi−c1 y c1 · · · , xi−cr y cr i ⊂ Ri ,
then α-grade (W ) = (c1 + · · · + cr ) − (1 + · · · + r − 1).

(4.2)

Later on we will need an estimate of the α-grade of an ideal I ⊂ OP2 of colength d,
which is invariant under G = Γ · T (ρ). This will be done by estimating the α-grade of
the vector space V = H 0 (I(n)), if n is sufficiently large. By means of (4.1) the estimate
will be reduced to the computation of the α-grade of monomial subspaces, and because
of (4.2) this can be regarded as a combinatorial problem, the formulation of which needs
some more notations.

4.2. The weight of a pyramid.
Definition 5. A pyramid with frame c and colength d, 1 ≤ d ≤ c, is a set P of
monomials in R = k[x, y] with the following properties: The i-th “column” Si of P
consists of monomials xi−aij y aij , 0 ≤ ai1 < . . . < aim(i) ≤ i, for all 0 ≤ i ≤ c − 1, such that
the following conditions are fulfilled:
(1)

c−1
S

Si = P

i=0

(2) #({xi−j y j |0 ≤ j ≤ i ≤ c − 1} \ P ) = d
Then w(Si ) := (ai1 + · · · + aim(i) ) − (1 + · · · + m(i) − 1) is called the weight of the i-th
c−1
P
column Si , and w(P ) :=
w(Si ) is called the weight of the pyramid P .
i=0

Example. Let be I ⊂ OP2 an ideal of colength d which is invariant under T (3; k).

Then reg(I) ≤ d, therefore h0 (I(d − 1)) = d+1
− d, and we can write H 0 (I(d − 1)) =
2
d−1
L d−1−i
z
Vi , where Vi ⊂ Ri are monomial subspaces. If Si is the set of monomials in
i=0

Vi , then P :=

d−1
S

Si is a pyramid with frame and colength equal to d. From (4.2) one

i=0

concludes that w(P ) = α-grade (H 0 (I(d − 1)). (N.B. One has Rn ⊂ H 0 (I(n)) if n ≥ d.)
44

Remark 4.1. Let 0 ≤ c1 < · · · < cr ≤ i be a sequence of integers. w(c1, · · · , cr ) :=
(c1 + · · · + cr ) − (1 + · · · + r − 1) is maximal if and only if cν = i − r + ν, 1 ≤ ν ≤ r, i.e.
if (c1 , · · · , cr ) = (i − r + 1, · · · , i).

The aim is to determine those pyramid P of type (c, d), i.e., with frame c and colength
d, for which w(P ) is maximal. Because of Remark 4.1 we will consider without restriction
only pyramids with Si = {xi−a(i) y a(i) , · · · , xy i−1 , y i }, where a(i) := i+1−m(i) is a number
between 0 and i inclusive. We call xi−a(i) y a(i) the initial monomial and a(i) initial degree of
Si . For simplicity we write Si = (a(i), a(i) + 1, · · · , i) and P = hxi−a(i) y a(i) |0 ≤ i ≤ c − 1i.
Remark 4.2. w(Si ) = ia(i) + a(i) − a(i)2 .
Proof. w(Si ) = (a(i) + · · · + i) − (1 + · · · + i − a(i)) = (1 + · · · + i) − (1 + · · · + a(i) −



a(i)
i−a(i)+1
1) − (1 + · · · + i − a(i)) = i+1
−
−
, and a direct computation gives the
2
2
2
assertion.

Taking away xi−a(i) y a(i) from Si and adding xj−a(j)+1 y a(j)−1 to Sj , if Si 6= ∅, a(j) > 0
and j 6= i, then gives a pyramid
P ′ = P − {xi−a(i) y a(i) } ∪ {xj−a(j)+1 y a(j)−1 }.

We express this as Si (P ) = (a(i), · · · , i), Sj (P ) = (a(j), · · · , j), Si (P ′ ) = (a(i) + 1, · · · , i),
Sj (P ′ ) = (a(j) − 1, · · · , j) and we get w(Si (P )) = ia(i) + a(i) − a(i)2 ; w(Si (P ′ )) =
i(a(i) + 1) + (a(i) + 1) − (a(i) + 1)2 ; w(Sj (P )) = ja(j) + a(j) − a(j)2 , w(Sj (P ′ )) = j(a(j) −
1) + a(j) − 1 − (a(j) − 1)2 . It follows that w(P ′) − w(P ) = [w(Si (P ′)) − w(Si (P ))] +
[w(Sj (P ′ )) − w(Sj (P ))] = [i + 1 − 2a(i) − 1] + [−j − 1 + 2a(j) − 1]. We get the following
formula:
(4.3)

w(P ′) − w(P ) = 2(a(j) − a(i)) − (j − i) − 2,

where we have made the assumption that i 6= j, Si (P ) 6= ∅, and a(j) > 0.

Now let P be a pyramid of type (c, d), such that w(P ) is maximal. Then for all
deformations P 7→ P ′ as above, the right side of (4.3) is ≤ 0, i.e. a(j) − a(i) ≤ 12 (j − i) + 1.
If i < j (if i > j, respectively) this is equivalent to
1
1
a(j) − a(i)
1
1
a(j) − a(i)
≤ +
and
≥ +
, respectively.
j−i
2 j−i
j−i
2 j−i
Putting in j = i + 1 (respectively j = i − 1) gives a(i + 1) − a(i) ≤ 1.5 (a(i) − a(i − 1) ≥
−0.5, respectively). As the left side of these inequalities are integers, it follows that
a(i + 1) − a(i) ≤ 1 (a(i) − a(i − 1) ≥ 0, respectively) for all 0 ≤ i < c − 1 (for all
0 < i ≤ c − 1, respectively). Note that we can apply (4.3) only under the assumption
a(i + 1) > 0 (respectively a(i − 1) > 0). But if a(i + 1) = 0 (respectively a(i − 1) = 0),
then the two last inequalities are true, too. So we get
Remark 4.3. If the pyramid P of type (c, d) has maximal weight, then one has
a(i) ≤ a(i + 1) ≤ a(i) + 1, for all 0 ≤ i ≤ c − 2.


Remark 4.4. If P = {xi−aij y aij |i, j} is a pyramid of type (c, d), then P ′ = yP :=
{xi−aij y aij +1 |i, j} is called shifted pyramid, and w(P ′) = w(P ) + #P .
45

Proof. The column Si = (ai1 , · · · , aim(i) ) goes over to Si′ = (ai1 + 1, · · · , aim(i) + 1),
and w(Si′ ) = (ai1 + 1) + · · · + (aim(i) + 1) − (1 + · · · + m(i) − 1) = w(Si ) + m(i). Thus
c−1
P
w(P ′) = (w(Si ) + m(i)) = w(P ) + #P .

i=0

Remark 4.5. A pyramid with maximal weight does not contain a step of breadth
≥ 4, i.e. it is not possible that a(j) = · · · = a(i) > 0, if i − j ≥ 3.

Proof. If one would have the situation shown in Figure 4.1, then one could make the
deformation xi−a(i) y a(i) 7→ xj−a(j)+1 y a(j)−1 and then from (4.3) one would get: w(P ′) −
w(P ) = 2 · 0 − (j − i) − 2 = i − j − 2 ≥ 1, contradiction.

Remark 4.6. In any pyramid two consecutive “normal” steps can be replaced by a
step of breadth 3, without a change of weight. In the course of this, the sum over the
x-degrees of all monomials in the pyramid increases by 2, however.
Proof. (cf. Fig. 4.2). By assumption one has a(i) + 1 = a(i + 1) and a(i + 1) + 1 =
a(i + 2). If one makes the deformation xi−a(i) y a(i) 7→ xj−a(j)+1 y a(j)−1 , where j = i + 2,
then one gets w(P ′) − w(P ) = 2 · 2 − 2 − 2 = 0.

N.B. The possibility a(i) = 0 is not excluded.
Remark 4.7. There is a pyramid of maximal weight, which does not contain two
consecutive normal steps, that means, there is no index i, such that a(i + 1) = a(i) +
1, a(i + 2) = a(i + 1) + 1. This is called a “prepared” pyramid. (N.B. a(i) may equal zero.)
Proof. Apply Remark 4.6 for several times.



Remark 4.8. A prepared pyramid of maximal weight does not contain two steps of
breadth 3.
Proof. We consider two cases:
1st case: The steps of breadth 3 are situated side by side. Then one makes the deformation
described in Fig. 4.3 and one gets: w(P ′) − w(P ) = 2[a(i − 5) − a(i)] + 5 − 2 = 1,
contradiction.
2nd case: Between the steps of breadth 3 there are ν ≥ 1 double steps. One then makes
the deformation described in Fig. 4.4. Putting j = i − 2(ν + 1) one gets: w(P ′) − w(P ) =
2(a(j) − a(i)) − (j − i) − 2 = 2(−ν) + 2(ν + 1) − 2 = 0. Then P ′ would have maximal
weight, too. But P ′ contains a step of breadth 4, contradicting Remark 4.5.

Remark 4.9. Each positive natural number d can uniquely represented either in the
form d = n(n + 1) − r, 0 ≤ r < n (1st case) or in the form d = n2 − r, 0 ≤ r < n (2nd
case). Both cases exclude each other.
Proof. Choosing n sufficiently large, then the sequence n(n+1), n(n+1)−1, · · · , n(n+
1) − (n − 1), n(n + 1) − n = n2 , n2 − 1, · · · , n2 − (n − 1), n2 − n = (n − 1)n, · · · contains
any given set {1, · · · , m} ⊂ N.

46

We now assume that P is a prepared pyramid of type (c, d), with d ≥ 3 and maximal
weight. If P does not contain a step of breadth 3, then P has the form described either
in Fig. 4.5a or in Fig. 4.5b, according to if either d = n(n + 1) or d = n2 . (One has
n
n
P
P
2 · ν = n(n + 1) and (2ν − 1) = n2 .) If there is 1 step of breadth 3, then from the
1

1

Remarks 3,5,6,7 and 8 it follows that P has the form described either in Fig. 4.6a or in
Fig. 4.6b. Here the step of breadth 3 may lie quite left or right in Fig. 4.6a or Fig 4.6b,
respectively. One sees that Fig. 4.6a and Fig. 4.6b result from Fig. 4.5a and Fig. 4.5b,
respectively, by removing the monomials marked by −.

We first compute the weights of the pyramids shown in Fig. 4.5a and Fig. 4.5b, and
then the weights of the “reduced” pyramids P in Fig. 4.6a and Fig. 4.6b.
1st case: (Fig. 4.5a)
i
a(i)
c−1
n
c−2
n
.......... .....
c − 2ν − 1 n − ν
c − 2ν − 2 n − ν
.......... .....
c − 2n + 1
1
c − 2n
1

w(Si ) = ia(i) − a(i)(a(i) − 1)
(c − 1)n − n(n − 1)
(c − 2)n − n(n − 1)
........................................
(c − 2ν − 1)(n − ν) − (n − ν)(n − ν − 1)
(c − 2ν − 2)(n − ν) − (n − ν)(n − ν − 1)
........................................
(c − 2n + 1) · 1 − 1 · 0
(c − 2n) · 1 − 1 · 0

2nd case: (Fig 4.5b)
i
c−1
c−2
c−3
..........
c − 2ν
c − 2ν − 1
..........
c − 2n + 2
c − 2n + 1

a(i)
n
n−1
n−1
.....
n−ν
n−ν
.....
1
1

w(Si ) = ia(i) − a(i)(a(i) − 1)
(c − 1)n − n(n − 1)
(c − 2)(n − 1) − (n − 1)(n − 2)
(c − 3)(n − 1) − (n − 1)(n − 2)
........................................
(c − 2ν)(n − ν) − (n − ν)(n − ν − 1)
(c − 2ν − 1)(n − ν) − (n − ν)(n − ν − 1)
........................................
(c − 2n + 2) · 1 − 1 · 0
(c − 2n + 1) · 1 − 1 · 0

1st case: We sum up w(Si ) and w(Si−1 ) if i = c − 2ν − 1 and we get:
n−1
X
ν=0

(n − ν)(2c − 2n − 2ν − 1) =

n
X
ν=1

ν(2c − 4n + 2ν − 1)

= (2c − 4n − 1) · 12 n(n + 1) + 13 n(n + 1)(2n + 1)
= n(n + 1)(c − 2n − 0.5 + 32 n + 13 )

= n(n + 1)(c − 34 n − 16 ).

In the reduced pyramid the initial terms a(i) of the column S i , if i = c−2, c−4, · · · , c−
2r, then are equal to n − 1, n − 2, · · · , n − r. This means, a(i) = n − ν, if i = c − 2ν,
and w(S i ) = (c − 2ν)(n − ν) − (n − ν)(n − ν − 1), 1 ≤ ν ≤ r. If i = c − 2ν we get:
47

w(S i )−w(Si ) = (c−2ν)(n−ν)−(n−ν)(n−ν −1)−[(c−2ν)(n−ν +1)−(n−ν +1)(n−ν)] =
−(c − 2ν) + (n − ν) · 2 = 2n − c.
It follows that w(P ) − w(P ) = r(2n − c).

2nd case: We sum up w(Si ) and w(Si−1 ) if i = c − 2ν and we get:
n−1
X
ν=1

(n − ν)(2c − 2n − 2ν + 1) =

n−1
X
ν=1

ν(2c − 4n + 2ν + 1)

1
1
= (2c − 4n + 1) · (n − 1)n + (n − 1)n(2n − 1)
2
3
2
1
= (n − 1)n(c − 2n + 0.5 + n − )
3
3
4
1
= n(n − 1)(c − n + ).
3
6

Beside this, we have to add the weight (c − 1)n − n(n − 1) = n(c − n), if i = c − 1,
and we get w(P ) = n[(c + 0.5)n − 43 n2 − 61 ].
In the reduced pyramid the initial terms a(i) of the column S i , if i = c−1, c−3, · · · , c−
2r + 1, then are equal to n − 1, n − 2, · · · , n − r. This means a(i) = n − ν, if i = c − 2ν + 1,
and w(S i ) = (c − 2ν + 1)(n − ν) − (n − ν)(n − ν − 1), 1 ≤ ν ≤ r. If i = c − 2ν + 1 we get:
w(S i ) − w(Si ) = (c − 2ν + 1)(n − ν) − (n − ν)(n − ν − 1) − [(c − 2ν + 1)(n − ν + 1) − (n −
ν + 1)(n − ν)] = 2n − c − 1, 1 ≤ ν ≤ r. We get w(P ) − w(P ) = r(2n − c − 1).
Remark 4.10. The maximal weight of a pyramid P of type (c, d) is equal to
n[(c − 1, 5)n + (c + 2r − 1/6) − 4/3 · n2 ] − rc

(4.4)

if d = n(n + 1) − r and 0 ≤ r < n and it is equal to

n[(c + 0.5)n + (2r − 1/6) − 4/3 · n2 ] − r(c + 1)

(4.5)

if d = n2 − r and 0 ≤ r < n.
Proof. If d ≥ 3, this follows from the foregoing computation. If d = 2 or if d = 1,
then n = 1 and r = 0. The formula (4.4) and the formula (4.5) give the weights 2c − 3
and c − 1, which is confirmed by Fig. 4.7a and Fig. 4.7b, respectively.

Remark 4.11. The formulas (4.4) and (4.5) agree in the ends of the intervals. This
is shown by putting in r = n in (4.4) and r = 0 in (4.5) and by putting in n − 1 instead
of n and r = 0 in (4.4) and r = n in (4.5), respectively, and then checking equality.

We denote the maximal weight of a pyramid of type (c, d) by w(Pc,d).
Remark 4.12.
(
− 43 n3 − 1, 5n2 + (2r − 16 )n + dc,
w(Pc,d) =
− 34 n3 + 0.5n2 + (2r − 16 )n − r + dc,

if d = n(n + 1) − r,
if d = n2 − r

Thus w(Pc,d) is a strictly increasing function of c ≥ d, if d is fixed.
48



Remark 4.13. Fixing the integer c ≥ 5, then w(Pc,d) is a strictly increasing function
of 1 ≤ d ≤ c.
Proof. w(Pc,d) = − 43 n3 − 1, 5n2 − 16 n + cn(n + 1) + r(2n − c), if d = n(n + 1) − r, 0 ≤
r ≤ n, and w(Pc,d) = − 34 n3 + 0.5n2 − 16 n + cn2 + r(2n − c − 1), if d = n2 − r, 0 ≤ r ≤ n.
√
From n(n + 1) − r = d ≤ c and from n2 − r = d ≤ c it follows that n ≤ c. From c ≥ 5
we conclude 2n − c < 0. From d ≥ 1 it follows that n ≥ 1. As a function of r both terms
for w(Pc,d) are strictly decreasing in the intervals 0 ≤ r ≤ n, and the assertion follows
from Remark 4.11 and Remark 4.12.

Remark 4.14. Fixing the integer 1 ≤ c ≤ 4, w(Pc,d) is an increasing function of
1 ≤ d ≤ c.
Proof. By drawing the possible patterns one finds:
d
1 2
w(P2,d ) 1 1

d
1 2 3
w(P3,d) 2 3 3

d
1 2 3 4
w(P4,d ) 3 5 6 7


We define Pc := Pc,c and w(∅) = 0.
Proposition 4.1. w(Pc ) ≤ (c − 1)2 for all c ∈ N.
Proof. 1st case: c = d = n(n + 1) − r, 0 ≤ r < n. Because of Remark 4.12 one has
to show:
− 34 n3 − 1, 5n2 + (2r − 16 )n + c2 ≤ (c − 1)2 ⇔
4 3
n + 1, 5n2 − (2r − 61 )n − 2[n(n + 1) − r] + 1 ≥ 0 ⇔
3
4 3
n − 0.5n2 − (2r + 11
)n + 2r + 1 ≥ 0 ⇐
3
6
4 3
11
2
n − 0.5n − (2n + 6 )n + 2n ≥ 0 ⇐
3
4 3
n − 2, 5n2 + 61 n ≥ 0. This is true if n ≥ 2. If n = 1, then r = 0, and by substituting
3
one can convince oneself that the inequality is true in this case, too.
2nd case: c = d = n2 − r, 0 ≤ r < n. One has to show:

− 43 n3 + 0.5n2 + (2r − 16 )n − r + c2 ≤ (c − 1)2

(4.6)

⇔ 43 n3 − 0.5n2 − (2r − 61 )n + r − 2[n2 − r] + 1 ≥ 0

⇔ 43 n3 − 2, 5n2 − (2r − 61 )n + 3r + 1 ≥ 0.

Assuming n ≥ 2, this inequality follows from
4 3
n
3

− 2, 5n2 − (2n − 61 )n + 3n + 1 ≥ 0
⇐

⇔

4 3
n −
3
4 2
n
3

4, 5n2 + 3 16 n ≥ 0

− 4, 5n + 3 61 ≥ 0.

This is true if n ≥ 3. Putting n = 1 and n = 2 in (4.6) gives the inequalities r ≥ 0 and
2 − r > 0, respectively, which are true by assumption. As P1,1 = ∅, the assertion is true
if c = 0 or c = 1, too.

49

4.3. Preview
Let be V ⊂ Sn a m-dimensional subspace and V ↔ ξ ∈ G(k) the corresponding point
in G = Grassm (Sn ). We let Gm and Ga operate on S as described in (4.1). We assume
V not to be invariant under Gm or Ga . Then C := {ψα (ξ)|α ∈ k}− is a closed irreducible
curve, which is to have the induced reduced scheme structure and which we imagine as a
curve in PN by means of the Plücker embedding p. Let be h the Hilbert polynomial of
p(C) ⊂ PN , X = Hilbh (G) ֒→ Hilbh (PN ) and σ : Gm → X the morphism λ 7→ σ(λ)C. It
has an extension σ : P1 −→ X , which induces a family of curves
C ֒→ G × P1
ց

f

↓ p2
P1

such that f is flat and Cλ := f −1 (λ) = σ(λ)C for all λ ∈ P1 − {0, ∞}. As f is dominant,
we get (cf. [Fu], p.15): If C0/∞ := p1 (C0/∞ ), then [C0 ] = [C∞ ] in A1 (G).
Let be ξ0/∞ = lim σ(λ)ξ and C0/∞ := {ψα (ξ0/∞ )|α ∈ k}− . The central theme of
λ→0/∞

[T1]–[T4] is the question, what is the connection of [C0 ] and [C∞ ]. The essential tool,
which was already used in [T1] is the α-grade of V , which is nothing else than the degree
of C, imbedded in PN by means of p (cf. [T1], 1.3).
We paraphrase the recipe for estimating the α-grade of V : Let M1 < · · · < Mℓ , ℓ =
ℓ

P
n+2
,
be
the
inverse-lexicographically
ordered
monomials
of
S
,
and
let
be
f
=
aij Mj ,
n
i
2
j=1

1 ≤ i ≤ m, a basis of V . If Mj1 < · · · < Mjm is a sequence of monomials in Sn , then the
Plücker coordinate PV (Mj1 , · · · , Mjm ) is defined to be the determinant of (aijν )1≤i,ν≤m .
In the following , V is a T (ρ)-invariant subspace of Sn , and fi = Mi (1 + a1i X ρ + a2i X 2ρ +
· · · + aiν(i) X ν(i)ρ ), 1 ≤ i ≤ m, is a basis of T (ρ)-semi-invariants. From formula (4.1) in
(4.1) it follows that α − grade(V ) ≤ max{α − grade hM1 X j(1)ρ , · · · , Mm X j(m)ρ i}
(j)

where (j) = (j(1), · · · , j(m)) ∈ [1, ℓ]m ∩ Nm runs through all sequence such that
PV (M1 X j(1)ρ , · · · , Mm X j(m)ρ ) 6= 0.

It is possible to choose the semi-invariants fi so that the initial monomials Mi (the
final monomials Mi X ν(i)ρ =: Ni , respectively) are linearly independent, i.e. different from
each other (cf. Appendix E or the proof of Hilfssatz 6 in [T2], Anhang 1, p. 140). As
the Plücker coordinates of a subvector space, up to a factor different from zero, do not
depend on the basis which one has chosen, one has
(4.7)

PV (M1 , · · · , Mm ) 6= 0 and PV (N1 , · · · , NM ) 6= 0.

Define V0 = hM1 , · · · , Mm i ↔ ξ0 and V∞ := hN1 , · · · , Nm i ↔ ξ∞ . As the function deg
is constant on flat families of curves, we get
α-grade(V ) = deg(C) = deg(C0 ) = deg(C∞ )
As C0/∞ ⊂ C0/∞ we conclude that
(4.8)

α-grade(V ) ≥ max(α-grade(V0 ), α-grade(V∞ )).
50

Now it is scarcely possible to see if PV (M1 X j(1)ρ , · · · , Mm X j(m)ρ ) is different from zero.
Therefore we introduce the number
j(1)

j(m)ρ
max −α-grade(V ) := max{α − grade ha1 M1 X j(1)ρ , . . . , aj(m)
i}
m Mm X
(j)

where (j) runs through all sequences (j(1), · · · , j(m)) ∈ Nm , such that 0 ≤ j(i) ≤ ν(i)
for all 1 ≤ i ≤ m and a0i := 1.
Remark 4.15. (a) Clearly α-grade (V ) ≤ max −α-grade (V ).
(b) In the definition, the monomials need not be ordered.
j(i)
(c) If one coefficient ai is equal to zero or if two of the monomials Mi X j(i)ρ are equal
for different indices i, then the m-times exterior product of the monomial space and its
α-grade are zero (cf. 4.1).
To say it differently, take from each semi-invariant fi a monomial Mi X j(i)ρ , whose
m
j(i)
coefficient ai is different from zero, form ∧ ψα (Mi X j(i)ρ ), and determine the highest
i=1

power of α occurring in such an exterior product. Finally, define max −α-grade (V ) to
be the maximum of these degrees, if (j) runs through all such sequences.
Accordingly, one defines
j(1)

j(m)ρ
min −α − grade(V ) := min{α-gradeha1 M1 X j(1)ρ , · · · , aj(m)
i}
m Mm X
(j)

where (j) runs through all sequences of the kind described above and which give an
α-grade different from zero.
As α-grade (V0/∞ ) = deg(C0/∞ ), from (4.7) we conclude that
(4.9)

min −α-grade(V ) ≤ min(deg C0 , deg C∞ ).

Later on, the vector space V will always be equal to H 0 (I(n))), where I ⊂ OP2 is a
G = Γ · T (ρ)-invariant ideal of y-standard form (cf. 2.4.3 Definition 2). We will see that
max −α-grade (I) := max −α-grade (H 0 (I(n))) and min −α-grade (I) := min −α-grade
(H 0 (I(n))) not only are independent of n ≥ colength (I), but also can be computed with
the help of smaller numbers n. The real aim of the following estimates is to prove the
following inequality: If I ⊂ OP2 is an ideal of y-standard form and if reg(I) = m, then
(!)

Q(m − 1) + min −α-grade(I) > max −α-grade(I).

From this it will follow that C0 and C∞ do not contain any y-standard cycle besides
C0 and C∞ , respectively (cf. Lemma 9.2).

51

Fig. 4.1

Fig. 4.2

Fig. 4.3

a(i+2)

a(j)

Fig. 4.4

a(i)

a(i)

a(i)

a(i−5)
a(i)

a(j)

Fig. 4.5a
n
n−1 n−1

n−ν n−ν

52

−

c−1

c−2

n

c−2ν−1

c−2ν−2

c−2n+1

c−2n

−

−

−

−

−

n

e−2n
c−2n+1

e−2n+1

c−2n+2

−

n−ν n−ν

Fig. 4.5b

−

Fig. 4.6a

53

c−2ν−1
c−2ν

−
−

c−2
c−1

−

n

e−1

−

n−1 n−1

e−2

n

n

Fig. 4.6b

Fig. 4.7a

e−1

e−2

e−2n+2

e−2n+1

n

Fig. 4.7b

c−1

c−1

54

CHAPTER 5

Estimates of the α-grade in the case ρ1 < 0, ρ2 > 0.
5.1. Preliminary remarks.
We refer to Proposition 3.1 in section 3.5 and we treat case (I.1) at first. If in fi the
vice-monomial Njup occurs, then Ik is monomial for all k ≥ j + 1. Especially, fj+1 , · · · fr
are monomials, which do not cause a deformation of the pyramid.
We show that the z-degree of such a monomial Njup cannot be equal to the z-degree
of an initial monomial Mk . For then it would follow mj + j − 1 = mk + k for another
index k, which is not possible by Corollary 2.4. For the same reason it is not possible
that the z-degree of Njup · (z/y) is equal to the z-degree of Nkdown or of Mk . The corresponding statements are true, if “up” and “down” are exchanged, as it follows from the
corresponding definition in (3.2) and (3.3).
Finally, if there occurs a deformation of the form (1.6) of Proposition 3.1, then it can
be that the final monomial Nkup (z/y) of fi has the same z-degree as the initial monomial
Mℓ of fℓ . But then ℓ > k > j > i, and therefore Iℓ is a monomial ideal by Lemma 3.1
. But then fℓ does not define a deformation, at all. It follows from this remarks that
one can separately consider the deformations defined by the different fi , if one wants to
determine the changes of α-grade caused by these deformations. (N.B. This statement is
analogously valid in the situation described by Proposition 3.2, too).
At first we determine the change of the α-grade, if in one fi the initial monomial Miup
is replaces by another monomial occurring in fi :
1◦ Miup 7−→ Njdown , if 0 ≤ i < j ≤ r;
2◦ Miup 7−→ Njup , if 0 ≤ i < j ≤ r;
3◦ Miup 7−→ L, L ∈ L monomial such that (x, y)L ⊂ ℓK(−r − 1), 0 ≤ i ≤ r;
4◦ Miup 7−→ Nkup · (z/y) = Mkup · (z/y)2 , 0 ≤ i < k ≤ r;
5◦ Miup 7−→ L · (z/y), L ∈ L monomial such that (x, y)L ⊂ ℓK(−r − 1), 0 ≤ i ≤ r.

The deformation 4◦ (resp. 5◦ ) comes from the case 1.6 (resp. 1.7) of Proposition 3.1,
and therefore there is at most one such a deformation, whereas in the deformations 1◦ and
2◦ (resp. 3◦ ) the index i may a priori run through all integers 0 ≤ i < r (resp. 0 ≤ i ≤ r).
Then for the index j in the cases 1◦ and 2◦ (resp. for the monomial L in the case 3◦ )
there are several possibilities. But if one has chosen i, then one has to decide for an index
j (resp. for a monomial L), and we will give an uniform estimate of the corresponding
changes of α-grades.
We identify the set L, which was introduced in Section (3.3) with the vector space
generated by the monomials in this set.
55

We denote by LB (left domain) the vector space generated by all monomials in Sm0
with z-degree ≥ m0 − (c + r), i.e., generated by all monomials xa y bz m0 −(a+b) , where
a + b ≤ c + r.

As to the deformation 4◦ (resp. 5◦ ), there is still the possibility yMiup 7→ yMkup (z/y)2
(resp. yMiup 7→ yL · (z/y)). This is because in the case 1.6 (resp. 1.7) of Proposition 3.1,
fi has the order 1, whereas in the remaining cases fi has the order 0.
Remember that (cf. Figure 3.1 and 5.1):
Miup = xi−ι(i) y mi +ι(i) z m0 −mi −i
Niup = Miup · (z/y) = xi−ι(i) y mi +ι(i)−1 z m0 −mi −i+1
Midown = xmi +i−ι(i) y ι(i) z m0 −mi −i
Nidown = Midown · (z/x) = xmi +i−ι(i)−1 y ι(i) z m0 −mi −i+1
Ekup := Mkup · (z/y)2 = xk−ι(k) y mk +ι(k)−2 z m0 −mk −k+2
Ekdown := Mkdown · (z/x)2 = xmk +k−ι(k)−2 y ι(k) z m0 −mk −k+2 .
5.2. Estimates in the case I.
We determine one after the other the changes of the α-grade in the deformations:
1 Miup → Njdown .
First we note ϕ′ (mi + i) = mi + 1 and ϕ′ (mj + j − 1) = mj − 1 (see Fig.5.2 ). The α-grade
of the column, in which Miup occurs changes by −(mi + ι(i)) + ϕ′ (mi + i) − 1 = −ι(i)
(cf. the formula (4.2) in 4.1). The α-grade of the column, to which Njdown is added,
changes by ι(j) − ϕ′ (mj + j − 1) = ι(j) − mj + 1 ( loc . cit.). Therefore the α-grade
changes by −mi − ι(i) + ϕ′ (mi + i) − 1 + ι(j) − ϕ′ (mj + j − 1) = ι(j) − ι(i) − mj + 1.
As 0 ≤ ι(i) ≤ ι(j) ≤ j , the absolute value of this difference is ≤ max(j, mj − 1), where
0 ≤ i < j ≤ r.
2◦ Miup → Njup .
The α-grade of the column, to which Miup belongs changes by −(mi +ι(i))+ϕ′ (mi +i)−1 =
−ι(i); the α-grade of the column, to which Njup is added, changes by mj +ι(j)−1−ϕ′ (mj +
j − 1) = ι(j). Therefore the change of α-grade is equal to 0 ≤ ι(j) − ι(i) ≤ j, where
0 ≤ i < j ≤ r.
3◦ Miup 7−→ L ∈ L.
The α-grade of the column, to which Miup belongs, changes by −ι(i). From the domain L
the monomial L is removed, such that by Proposition 4.1 one gets the following estimate
of the α-grade: The α-grade after the deformation 3◦ of that part of the pyramid, which

belongs to the left domain is ≤ (c − 1)2 + ι(r + 1)[ c+1
− (c − 1)]. For one has #L = c, and
2

c+1
0
there are 2 − c initial monomials of ℓH (K(c − 1)) in the left domain LB. Therefore
the expression in the bracket gives the number of monomials in the corresponding part
of the pyramid after the deformation. Besides this one has to take into account that the
pyramid is pushed upwards by ι(r + 1) units (cf. Remark 4.4). We recall that LB (resp.
RB) is the vector space generated by the monomials of total degree ≤ c + r (resp. of
total degree ≥ c + r + 1) in x and y (cf. Fig. 5.3). The change of α-grade caused by
the deformation 3◦ can be expressed as follows: The change in the domain RB is −ι(i);
◦

56

the α-grade of the left domain of the pyramid after the deformation is estimated as given
above.
4◦ Miup 7−→ Ekup .
At first we consider the case that Ekup belongs to the right domain, i.e. mk +k−2 ≥ c+r+1,
and we orientate ourselves by Figure 5.1. The deformation 4◦ changes the α-grade of the
column of Miup by −ι(i) (cf. 3◦ ), and the α-grade of the column to which Ekup is added
changes by mk +ι(k)−2−ϕ′ (mk +k −2). As we have remarked above, ϕ′ (mk +k) = mk +1
and therefore ϕ′ (mk +k−2) = mk −2. Therefore the α-grade of the column of Ekup changes
by ι(k). Altogether the deformation 4◦ causes a change of α-grade by 0 ≤ ι(k) − ι(i) ≤ k,
where 0 ≤ i < k ≤ r.
This deformation occurs only once, yet one has to take into account the deformation
4◦ bis (y/z)Miup 7→ Nkup ( Proposition 3.1c). In the column of yMiup this gives a change of
the α-grade by −(mi + ι(i) + 1) + ϕ′ (mi + i + 1) − 1 = −mi − ι(i) − 1 + mi + 2 − 1 = −ι(i).
In the column of Nkup the α-grade changes by mk + ι(k) − 1 − ϕ′ (mk + k − 1) = mk +
ι(k) − 1 − (mk − 1) = ι(k). Altogether the deformation 4◦ bis gives a change of α-grade
by 0 ≤ ι(k) − ι(i) ≤ k.
Now to the case mk + k − 2 ≤ c + r. Due to the deformation 4◦ (resp. 4◦ bis) the
α−grade in the right domain of the pyramid changes by −ι(i). In any case the deformation
4◦ (resp. 4◦ bis) gives a change of α-grade in the right domain of absolute value ≤ r.
5◦ Miup 7−→ L · (z/y).
Removing Miup (resp. (y/z)Miup ) gives a change of α-grade by −ι(i) in the corresponding
column (cf. case 4◦ ). The changes in the left domain will be estimated later on.

The deformations 1◦ − 5◦ exclude each other, i.e., there are at most r + 1 such deformations plus two deformations 4◦ bis and 5◦ bis. The changes in the right domain can be
estimated in the cases 1◦ and 2◦ by max(j, mj − 1) ≤ r + mi+1 , where i runs through the
numbers 0, · · · , r − 1. The absolute value of the change in the case 3◦ can be estimated
by r, and the same is true for the deformations 4◦ , 4◦ bis, 5◦ and 5◦ bis.
We now consider possible trinomials.
6 We assume there is a trinomial of the form 1.3. We want to determine the change of
α-grade,if Njup is replaced by Nkup , where we start from a pyramid containing Njup instead
of Miup . The changes of α-grade in the following diagram follow from the computation in
2◦ .
Miup
ι(j) − ι(i) ւ
ց ι(k) − ι(i)
δ
up
Nj
−→ Nkup
◦

The change of α-degree is therefore 0 ≤ δ := ι(k) − ι(j) ≤ r.
7◦ The trinomial has the form 1.4.
Miup
ι(j) − ι(i) ւ
ց ι(k) − ι(i) − mk + 1
δ
up
Nj
−→ Nkdown

(cf. 1◦ and 2◦ ) Therefore δ = ι(k) − ι(j) − mk + 1, and as in 1◦ we obtain the estimate
0 ≤ |δ| ≤ max(k, mk − 1) ≤ mk + k < mi+1 + r.
57

8◦ The trinomial has the form (1.5).
Miup

ι(j) − ι(i) ւ
ց −ι(i)
δ
up
Nj
−→ L
(cf. 3◦ ) Therefore δ = −ι(j) and 0 ≤ |δ| ≤ r.
9◦ The trinomial has the form (1.6).
Miup
ι(j) − ι(i) ւ
ց ι(k) − ι(i)(resp. − ι(i))
δ
up
Nj
−→ Nkup · (z/y)

(cf. 4◦ ) It follows δ = ι(k) − ι(j) (resp. δ = −ι(j)) and therefore |δ| ≤ r.
10◦ The trinomial has the form (1.7).
Miup
ι(j) − ι(i) ւ
ց −ι(i)
δ
up
Nj
−→ L · (z/y)
(cf. 5◦ ) It follows that δ = −ι(j) and |δ| ≤ r.
N.B. Because of Njup · (y/z) = Mjup the cases 9◦ bis and 10◦ bis do not occur.

Summarizing the cases 1◦ − 10◦ one sees that the total change of α-grade in the right
r
P
domain has an absolute value ≤ (r + 1)r + 2r +
mi . In order to formulate this result
i=1

in a suitable manner, we have to introduce some notations.

We take up the decomposition (Z) of Section (3.1) and choose a standard basis of
H (K(c)). Then we multiply the elements in this basis as well as the forms fi by monomials
in the variables x, y, z to obtain a basis of T (ρ)- semi-invariants of H 0 (I(n)) with different
initial monomials. By linearily combining one can achive that the initial monomial of
each semi-invariant does not appear in any other of these semi-invariant, i.e., one gets a
standard basis. (Fig. 3.1 is to show these initial monomials.) The set of all monomials
which occur in this basis form a pyramid, which is denoted by P. Here n ≫ 0, e.g. n ≥ d.
0

From each element of the basis we take a monomial the coefficient of which is different
from zero and such that the monomials are different from each other .Then we compute
the α-grade of the vector space generated by these monomials. The maximum and the
minimum of the α-grades which one obtains in this way had been denoted by
max −α − grade(V ) and by

min −α − grade(V ),

respectively. One chooses from such a sequence of monomials, which gives the maximal αgrade (which gives the minimal α- grade , respectively) those monomials the total degree
in x and y of which is ≥ c + (r + 1). Then one forms the α-grade of the subspaces
generated by these monomials and denotes it by max −α-grade (P ∩ RB) ( by min −αgrade (P ∩ RB), respectively ). If one chooses from such sequences of monomials those
58

monomials the total degree in x and y of which is ≤ c + r, then max −α-grade (P ∩ LB)
and min −α-grade (P ∩ LB) are defined analogously.

Of course this is valid in the case ρ1 > 0, too, but the assumption ρ2 > 0 is essential .
We make the
Definition 6. A := max −α-grade (P ∩ RB) − min −α-grade (P ∩ RB).
Then we can formulate the result obtained above as
Conclusion 5.1. In the case I.1 one has
r
X
mi .
A ≤ r(r + 3) +



i=1

N.B.. If r = 0 one has actually A = 0.
Now to the case I.2 (cf. Proposition 3.1, 2nd case).
1 Miup 7−→ Njdown gives a change of α-grade of absolute value ≤ max(r, mi+1 ), where
0 ≤ i ≤ r − 1 (cf. the case I.1).
2◦ Miup 7−→ L ∈ L gives a change of α-grade in the right domain by −ι(i) (see above).
Further possible deformations are yMiup 7→∈ L, y 2Miup 7→∈ L, · · · , y ν Miup 7→∈ L, so long
as mi + i + ν < mi−1 + (i − 1) − 1 (cf. Conclusion 3.2 ). This gives in the column of yMiup
(of y 2 Miup , · · · , y ν Miup , respectively) a change of α-grade by
◦

−(mi + ι(i) + 1) + [ϕ′ (mi + i + 1) − 1] = −(mi + ι(i) + 1) + [mi + 1] = −ι(i)

(by − (mi + ι(i) + 2) + [ϕ′ (mi + i + 2) − 1] = −(mi + ι(i) + 2) + [mi + 2] = −ι(i),

··· ,

−(mi + ι(i) + ν) + [ϕ′ (mi + i + ν) − 1] = −(mi + ι(i) + ν) − [mi + ν] = −ι(i), respectively)
as long as mi + i + ν < mi−1 + (i − 2), see above.) This procedure can be repeated at
most c times, until L is full. As ι(r) ≤ r, the total change of α-degree in the right domain
caused by deformations 2◦ has an absolute value ≤ cr. If A is defined as before one gets
Conclusion 5.2. In the case I.2 one has
r
X
A ≤ r(r + c) +
mi .



i=1

N.B. If r = 0, then one really has A = 0. For removing M0up = y m0 does not change
the α-grade of the column of z-degree 0 (this α-grade is equal to 0), or one has f0 = xm0 .
5.3. The case r ≥ 1.
We start with an ideal I of type r ≥ 0 such that ℓ0 = y, and we refer to Proposition
3.1 again. The aim is to prove the inequality (!) in (4.3), where now V = H 0 (I(n)). In
the course of the following computations it will turn out that α-grade (V ) is independent
of n, if n is sufficient large, e.g. if n ≥ d.
59

If one simply writes α-grade (I) instead of α-grade (H 0 (I(n))), where n is sufficient
large, then one has to show:
Q(m0 − 1) + min −α − grade (I) > max −α − grade (I)

(!)

We orientate ourselves by Fig. 5.4. From the Remarks 4.4 , 4.13 , 4.14 and Proposition
4.1 it follows that


c+1
− c]
min −α-grade(LB ∩ P) ≥ ι(r + 1)[
2


c+1
2
− (c − s − 1)]
max −α-grade(LB ∩ P) ≤ (c − 1) + ι(r + 1)[
2

where s + 1 = total number of all deformations Miup 7→∈ L, · · · , y ν Miup 7→∈ L, even
with different indices i (s = −1 means, there are no such deformations). For proving (!)
it is sufficient to show
Q(m0 − 1) > (c − 1)2 + (s + 1) · ι(r + 1) + A

(*)

where A = max −α-grad (P ∩ RB) − min −α-grade (P ∩ RB) by definition (cf. Section
5.2 for the notations).
N.B. If c = 0 there are no deformations into the left domain of the pyramid. Therefore
the inequality (*) reduces to
Q(m0 − 1) > A.

(*bis)

These statements are independent of ρ1 < 0 (Case I) or ρ1 > 0 (Case II).
Unfortunately one has to distinguish these two cases in the following estimates and we
start with Case I.1 (cf. Proposition 3.1).
Because of 0 ≤ s + 1 ≤ c, 1 ≤ ι(r + 1) ≤ r + 1 and the estimates of

as well as Lemma 2.8) it is sufficient to show:
r
r
X
X

m0 +1
2
− (c +
mi ) > (c − 1) + c(r + 1) + r(r + 3) +
mi
2
i=0

i=1

1
m0 (m0 − 1) > c2 + cr + r(r + 3) + 1
2
The case r = 0 has to be treated separately (see below Section 5.4), such that we can
assume r ≥ 1. If c = 0, then m0 ≥ 2r+1 ( Lemma 2.8), and (5.1) follows from 2r (2r+1 −1) >
r(r + 3) + 1, which inequality is true if r ≥ 1. Therefore we can assume c > 0. Because of
m0 ≥ 2r (c+2) the inequality follows from 2r−1 (c+2)·2r (c+1) > c2 +cr +r(r +3)+1 ⇐⇒
22r−1 (c2 + 3c + 2) > c2 + cr + r(r + 3) + 1, which is true if r ≥ 1 and c > 0.
(5.1)

⇐⇒

Conclusion 5.3. In the case I.1 the inequality (*) is valid, if r ≥ 1.



Now to the case I.2. Because of Conclusion 5.2 one has to replace in the inequality
above r(r + 3) by r(r + c), i.e. one has to show that 22r−1 (c2 + 3c + 2) > c2 + 2rc + r 2 + 1,
if r ≥ 1. This is true, if c ≥ 0.
Conclusion 5.4. In the case I.2 the inequality (*) is valid, if r ≥ 1.
60



5.4. The case r = 0.
As always we suppose g ∗(ϕ) > g(d). As ℓ0 = y and ρ1 < 0 by assumption, one has
f0 = xm0 (cf. Fig. 5.5). Therefore there are no deformations into L, i.e. A = 0 and (*)
reads Q(m0 − 1) > (c − 1)2 . If for simplicity one writes m instead of m0 , then one has to
show
(5.2)

m(m − 1) > 2c2 − 2c + 2

Now c = colength (K), K an ideal of type (−1) and the following cases can occur:
1. If ψ is the Hilbert function of K, then g ∗ (ψ) ≤ g(c). Especially one has c ≥ 5 and by
Corollary 2.2 m ≥ 2c + 1, from which (5.2) follows.
2. One has c ≤ 4. If c = 0, then I is monomial and there are no deformations at all. It
follows that (∗bis) is fulfilled.
A little consideration shows that because of m ≥ c + 2 the inequality (5.2) is fulfilled
in the cases 1 ≤ c ≤ 4, too. Thus (*) and (∗bis) are proved in the case r = 0, also. Using
Conclusion 5.4 gives
Proposition 5.1. Assume that I has the type r ≥ 0 and has y-standard form; assume
ρ1 < 0 and ρ2 > 0. Then (*) and (∗bis) are fulfilled, respectively.

N.B. Hence the inequality (!) follows (see the corresponding argumentation in 5.3).

61

Fig. 5.1
Mkup
Nkup

mk +k

mk +k−2

Ekup

Fig. 5.2

ϕ′ (mi + i) = mi + 1
ϕ′

0

1

2

3

4

5

6

m2 +2

8

m1 +1

10 m0

62

Fig. 5.3

c+r+1

c+r

column with c + 1 monomials
is complete

mr +r

right domain

#{ monomials in the left domain } = h0 (K(c − 1)) = c+1
−c
2

n+2
as K has the Hilbert polynomial 2 − c.

m0

left domain

#{ monomials in the right domain } = Q(m0 − 1) − #{ monomials in the left domain }




= m02+1 − d c+1
−c
2

63

Fig. 5.4a

Fig. 5.4b

· · · · · · · · · · · · · · · · · · m1 +1 · · · · · · m0

· · · · · · · · · · · · · · · · · · m1 +1 · · · · · · m0

Fig. 5.5 (r = 0)

column filled
with monomials

#L=c
c c+1
LB

RB

64

CHAPTER 6

Estimates of the α-grade in the case ρ1 > 0, ρ2 > 0 and r ≥ 1.
We refer to Proposition 3.2 in (3.6) and take over the notations from there. As has
been remarked in (5.1) one can compute the changes of α-grade by the single deformations
fi separately.

6.1. Estimates in the case II
At first we compute the changes of the α-grade , if we replace the initial monomial
Mi of fi by another monomial occurring in fi (cf. 5.1):
1◦ Midown 7−→ Njup , if 0 ≤ i < j ≤ r.
In the column of Midown there is a change of the α-grade by −ι(i)+ϕ′ (mi +i)−1 = mi −ι(i).
In the column of Njup there is a change of the α-grade by mj + ι(j) − 1 − ϕ′ (mj + j − 1) =
mj +ι(j)−1−(mj −1) = ι(j). Therefore the deformation 1◦ gives a change of the α-grade
by mi + ι(j) − ι(i), 0 ≤ i < j ≤ r.
2◦ Midown 7−→ Njdown , 0 ≤ i < j ≤ r.
In the column of Midown there is a change of the α-grade by mi − ι(i); in the column of
Njdown is a change of α-grade by ι(j) − ϕ′ (mj + j − 1) = ι(j) − mj + 1. Therefore the
deformation 2◦ gives a change of α-grade, whose absolute value is |ι(j)−ι(i)+mi −mj +1| ≤
max(|mi + ι(j)|, |mj + ι(i) − 1|) ≤ mi + j, where 0 ≤ i < j ≤ r.
3◦ Midown 7−→ L ∈ L.
In the column of Midown is a change of α-grade by 0 < mi − ι(i) ≤ mi + i, 0 ≤ i ≤ r. The
change of α-grade in the left domain can be estimated as in Case I.
4◦ Midown 7−→ Ekdown = Mkdown · (z/x)2 , 0 ≤ i < k ≤ r.
At first we consider the case that Ekdown belongs to the right domain, i.e. mk + k − 2 ≥
c+r +1. The change of α-grade in the column of Midown is mi −ι(i); the change of α-grade
in the column of Ekdown is ι(k) − ϕ′ (mk + k − 2) = ι(k) − (mk − 2). Therefore the absolute
value of the change of α-grade by the deformation 4◦ is |mi − ι(i) + ι(k) − mk + 2| ≤
max(|mi + ι(k)|, |mk + ι(i) − 2|) = mi + ι(k) ≤ mi + k, if 0 ≤ i < k ≤ r. This deformation
can occur only once, yet one has to take into account the deformation
4◦ bis (x/z)Midown 7→ Nkdown (cf. Proposition 3.2c).
In the column of xMidown this gives a change of the α-grade by −ι(i) + ϕ′ (mi + i + 1) − 1 =
mi − ι(i) + 1. In the column of Nkdown the α-grade changes by ι(k) − ϕ′ (mk + k − 1) =
ι(k) −mk + 1. Thus the absolute value of the change of α-grade in the right domain due to
4◦ bis is |mi −ι(i)+1+ι(k)−mk +1) ≤ max(|mi +ι(k)|, |mk +ι(i)−2|) = mi +ι(k) ≤ mi +k,
where 0 ≤ i < k < r. This deformation occurs only once.
Now to the case that mk + k − 2 ≤ c + r. Removing Midown ( resp. (x/z)Midown ) gives a
65

change of α-grade by mi − ι(i) ( by mi − ι(i) + 1, respectively), whose absolute value is
bounded by mi + r.
5◦ Midown 7−→ L · (z/x).
Removing Midown (resp. (x/z)Midown ) causes a change of α-grade of the column of Midown
(resp. (x/z)Midown ) by mi − ι(i) (resp. by mi − ι(i) + 1), which are estimated by mi + i
(resp. mi + i + 1), where 0 ≤ i ≤ r. The deformation 5◦ (resp. 5◦ bis) can occur only
once. The changes in the left domain will be estimated later on.
The deformation 1◦ −5◦ exclude each other, i.e. there are at most r+1 such deformation
plus two deformations 4◦ bis and 5◦ bis. The changes of α-grade in the right domain in the
cases 1◦ − 3◦ have an absolute value ≤ mi + r, 0 ≤ i ≤ r. The same estimate is valid for
the deformations 4◦ and 4◦ bis, even if Ek belongs to the left domain, as we have assumed
r ≥ 1. As for the deformations 5◦ (resp. 5◦ bis) we estimate the change of the α-grade by
mi + r (resp. mi + r + 1).
We now consider possible trinomials.
6 We assume there is a trinomial of the form 1.3. Similarly as in the Case I in Chapter
5, we have a diagram
Midown
γ(j) ւ
ց γ(k)
δ
down
Nj
−→ Nkdown
where γ(j) := mi − mj − ι(i) + ι(j) + 1 and γ(k) := mi − mk − ι(i) + ι(k) + 1 (cf. 2◦ ).
Therefore δ = mj − mk − ι(j). It follows that |δ| ≤ max(|mj + ι(k)|, |mk + ι(j)|) ≤ mi + r.
◦

7◦ The trinomial has the form 1.4.
Midown
γ(j) ւ
ցβ
δ
down
−→ Nkup
Nj

where β := mi − ι(i) + ι(k) (cf. 1◦ and 2◦ ). It follows that δ = mj − ι(j) + ι(k) − 1 and
|δ| ≤ mi + r.
8◦ The trinomial has the form 1.5.
Midown
γ(j) ւ
ցβ
δ
down
Nj
−→ L

where β := mi − ι(i) (cf. 3◦ ). It follows that δ = mj − ι(j) − 1 and |δ| ≤ mi + r.
9◦ The trinomial has the form 1.6.
Midown
γ(j) ւ
ցβ
δ
down
Nj
−→ Nk · (z/x)

where β := mi − mk − ι(i) + ι(k) + 2, respectively β := mi − ι(i) (cf. 4◦ ). It follows that
δ = mj − mk − ι(j) + ι(k) + 1 (resp. δ = mj − ι(j) − 1) and |δ| ≤ max(|mj − ι(k)|, |mk +
66

ι(j) − 1|) ≤ mj + r.
10◦ The trinomial has the form 1.7.
Midown
γ(j) ւ
ց mi − ι(i)
δ
down
Nj
−→ L · (z/x)

(cf. 5◦ ). It follows that δ = mj − ι(j) − 1 and |δ| < mi + r.
Notabene. Because of Njdown · (x/z) = Mjdown the case 9◦ bis or 10◦ bis does not occur.

Summarizing the cases 1◦ − 10◦ one sees that the total change of α-grade in the
r
P
mi . If one estimates the changes
right domain has an absolute value ≤ (r + 1)r +
i=0

in the cases 4◦ bis and 5◦ bis by mi + r and mi + r + 1, respectively, one obtains A ≤
r
P
r(r+3)+1+3m0 + mi . As we have assumed that r ≥ 1, we have m0 ≥ c+2+mr +· · ·+m1
i=1

( Lemma 2.8) and we obtain

Conclusion 6.1. If r ≥ 1 is assumed,in the case II.1 one has A ≤ 4m0 + r(r + 3) −
c − 1.

Now we come to the case II.2 (cf. Proposition 3.2, 2nd case).
1◦ Midown 7−→ Njup again gives a change of the α-grade in the right domain, whose
absolute value is ≤ mi + r, 0 ≤ i ≤ r − 1 (see above).
2◦ Midown 7−→ L ∈ L gives a change of α-grade in the right domain by mi − ι(i), 0 ≤
i ≤ r (see above). Further possible deformations are xMidown 7→∈ L, x2 Midown 7→∈
L, · · · , xν Midown 7→∈ L, as long as mi + i + ν < mi−1 + (i − 2) (cf. Conclusion 3.2).
This gives in the column of xMidown (of x2 Midown , · · · , xν Midown , respectively) a change
of α-grade by −ι(i) + ϕ′ (mi + i + 1) − 1 = −ι(i) + (mi + 2) − 1 = mi − ι(i) + 1 (by
−ι(i) + ϕ′ (mi + i + 2) − 1 = mi − ι(i) + 2, · · · , −ι(i) + ϕ′ (mi + i + ν) − 1 = mi − ι(i) + ν,
respectively), as long as mi + i + ν < mi−1 + (i − 2) and ν ≤ c − 1.
Remark 6.1. One has |mi − ι(i)| ≤ m0 for all 0 ≤ i ≤ r.
Proof. The inequality −ι(i) + mi ≤ mi ≤ m0 is true, and ι(i) − mi ≤ i − mi ≤
r − mi ≤ r ≤ m0 is true if r = 0. If r ≥ 1 one has m0 ≥ c + 2 + mr + · · · + m1 > r (Lemma
2.8).

From this we conclude: Replacing Midown , xMidown , · · · , xν(i) Midown by monomials in L,
even with different indices i as long as ν(i) ≤ c − 1, gives a change of α-grade in the right
ν(i)
P
(m0 + j), because |mi − ι(i) + j| ≤ m0 + j by the
domain whose absolute value is ≤
j=0

remark above . One gets A ≤

r−1
P

(mi + r) +

i=0

r ν(i)
P
P

(m0 + j). As ν(0) + 1 + · · ·+ ν(r) + 1 ≤ c,

i=0 j=0

67

it follows that A ≤
we obtain

r−1
P
i=1

r−1
P

(mi +r)+

i=0

c−1
P

(m0 +j). As m0 ≥ c+2+mr +· · ·+m1 and mr ≥ c+2,

j=0

mi ≤ m0 − 2(c + 2). This estimate is valid if r ≥ 1. In the case r = 0 one

only has the deformations M0down 7→∈ L, · · · , xs M0down 7→∈ L, and s can be estimated as
in ( Proposition 3.2).
If M0down occurs, the α-grade in the column of M0down , · · · , xs M0down increases by
m0 , · · · , m0 + s, respectively.

Conclusion 6.2. In the case II.2 one has A ≤ (c + 1)m0 + r 2 − 2(c + 2) + 2c , if r ≥ 1.

If r = 0, if I has y-standard form and κ := reg(K), then A ≤ (s + 1)m0 + s+1
, where
2
s ≤ κ/ρ2 − m0 (1/ρ2 − 1/(ρ1 + ρ2 )).

6.2. The case r ≥ 2.
We recall that we started from an ideal I of type r ≥ 0 with y-standard form, and
the aim was to show the inequalities
Q(m0 − 1) > (c − 1)2 + (s + 1)ι(r + 1) + A

(*)

and
respectively (cf. Section 5.3).

Q(m0 − 1) > A (*bis)

At first we treat the case II.1, where A ≤ 4m0 + r(r + 3) − c − 1, if r ≥ 1.

Auxiliary Lemma 1. If I has the type r = 2 (the type r = 1, respectively), then
m0 ≥ 14 ( m0 ≥ 7, respectively).
Proof. We use the results of Lemma 2.8 and write in the case r = 2 : I = I0 =
yI1 (−1) + f0 OP2 (−m0 ), I1 = ℓ1 I2 (−1) + f1 OP2 (−m1 ), I2 = ℓ2 K(−1) + f2 OP2 (−m2 ). I2
has the type 0, therefore colength (I2 ) = c + m2 ≥ 5 and it follows that m2 ≥ 5 − c. As
I1 has the type 1, we get m1 ≥ m2 + c + 2 ≥ 7. Because of m0 ≥ c + 2 + m2 + m1 it
follows that m0 ≥ c + 2 + 5 − c + 7.

To begin with, let c = 0. Then (∗bis) reads 21 m0 (m0 + 1) − (mr + · · · + m0 ) >
r
P
4m0 + r(r + 3) − 1. Because of m0 − (c + 2) ≥
mi it is sufficient to prove
1

(6.1)

m0 (m0 − 11) > 2r(r + 3) − 6

If r = 2 (respectively r = 3) the right side of (6.1) is equal to 14 (equal to 30, respectively).
As m0 ≥ 14 by the Auxiliary Lemma 1 (m0 ≥ 23 (c + 2) = 16 by Lemma 2.8, respectively),
(6.1) is true in these cases.
Let be r ≥ 4. Then m0 ≥ 2r+1 and (6.1) follows from 2r (2r+1 − 11) > r(r + 3). Now
2r > r if r ≥ 2, and 2r+1 − 11 > r + 3 is equivalent to 2r+1 > r + 14, which is true if r ≥ 4.
68

Thus we can assume c ≥ 1 in the following. Because of 1 ≤ ι(r + 1) ≤ r + 1, 0 ≤
r
P
s + 1 ≤ c, mi ≤ m0 − (c + 2) the inequality (∗) will follow from:
i=1

(6.2)

1
m0 (m0 + 1) − (2m0 − 2) > (c − 1)2 + c(r + 1) + 4m0 + r(r + 3) − c − 1
2
⇐⇒ m0 (m0 − 11) > 2c2 + 2(r − 2)c + 2r(r + 3) − 4

Because of m0 ≥ 2r (c + 2) it suffices to show:
(6.3)

2r (c + 2)[2r c + 2r+1 − 11] > 2c2 + 2(r − 2)c + 2r(r + 3) − 4

If r = 2 this inequality reads 4(c + 2)(4c − 3) > 2c2 + 16 ⇔ 7c2 + 10c − 20 > 0 and
this is true if c ≥ 2.

In the case r = 2, c = 1, the inequality (6.2) reads m0 (m0 − 11) > 18, which is
true because m0 ≥ 14 (cf. Auxiliary Lemma 1). Therefore we can now suppose without
restriction r ≥ 3, c ≥ 1. But 2r+1 > 11 and thus (6.3) follows from 2r (c + 2) · 2r c >
2c2 + 2cr + 2r(r + 3) which is equivalent to :
(22r − 2)c2 + (22r+1 − 2r)c > 2r(r + 3)

(6.4)

The left side of (6.4) is a monotone function of c, and if c = 1, then (6.4) reads
2 − 2 + 22r+1 − 2r > 2r(r + 3) ⇔ 22r + 22r+1 > 2r 2 + 8r + 2 ⇔ 22r−1 + 2r > r 2 + 4r + 1.
This is true if r ≥ 3. Summarizing all subcases we obtain
2r

Conclusion 6.3. In the case II.1 the inequality (∗) is fulfilled for all r ≥ 2.
We now consider the case II.2 and assume r ≥ 2. With the help of Conclusion 6.2 and
r
P
the estimates
mi ≤ m0 − (c + 2), s + 1 ≤ c, ι(r + 1) ≤ r + 1 one sees that (∗) follows
i=1

from
 
1
c
2
2
.
m0 (m0 + 1) − (2m0 − 2) ≥ (c − 1) + c(r + 1) + (c + 1)m0 + r − 2(c + 2) +
2
2
A simple computation shows that this is equivalent to
(6.5)

m0 (m0 − 2c − 5) > 3c2 + 2cr − 7c − 10 + 2r 2 .

Now we have m0 ≥ 2r (c + 2) ≥ 4(c + 2). If c = 0, then (6.5) follows from 2r+1 >
−10 + r 2 ,which is true for all r ≥ 2. Therefore we can assume c > 0. Then (6.5) follows
from 2r (c + 2)(2c + 3) > 3c2 + 2cr + 2r 2 ⇔ 2r (2c2 + 7c + 6) > 3c2 + 2cr + 2r 2 which is
true for all c ≥ 1 and r ≥ 2. We get
Conclusion 6.4. In the case II.2 the inequality (*) is fulfilled for all r ≥ 2.
6.3. The case r = 1.
Then I = yI1 (−1) + f0 OP2 (−m0 ), I1 = ℓ1 K(−1) + f1 OP2 (−m1 ), where I1 has the type
0 and K has the type −1.
69

6.3.1.

We start with the case II.1 of Proposition 3.2 .

Subcase 1: ℓ1 = y: Then one has the situation shown in Figure 6.1 and there are the
following possibilities (case II 1.5 and II 1.7, respectively):
1◦ f0 = xm0 + αN1down + βL, L ∈ L monomial such that (x, y)L ⊂ ℓK(−2).
2◦ f0 = xm0 + αN1down + βL · (z/x), L ∈ L monomial such that (x, y)L ⊂ ℓK(−2).

We treat the case 1◦ . At first, one has the possibility xm0 7→ N1down . The α-grade
of the column of xm0 changes by m0 ; the α-grade of the column of N1down changes by
ι(1) − ϕ′ (m1 ) = 1 − (m1 − 1) = 2 − m1 . Therefore the change of α-grade in the right
domain is m0 − m1 + 2. The deformation xm0 7→ L gives a change of α-grade by m0 in
the right domain. As the order of f0 is equal to 0 in the case II 1.5 (cf. Proposition 3.2c),
there are no other changes of α-grade caused by f0 .

By Proposition 3.2 again, it follows that f1 has the form of case II.1.5, where α = 0,
and the order of f1 is equal to 0. The deformation M1down 7→∈ L gives a change of αgrade by −ι(1) + ϕ′ (m1 + 1) − 1 = −ι(1) + m1 = m1 − 1. Thus in the case 1◦ one has
A ≤ max(m0 − m1 + 2, m0 ) + m1 − 1 = m0 + m1 − 1, because m1 ≥ c + 2 (Lemma 2.4).
2◦ At first, f0 defines a deformation as in the case 1◦ and gives a change of α-grade
≤ max(m0 , m0 − m1 + 2) = m0 in the right domain. But as f0 has the order ≤ 1 by
Proposition 3.2c, there is still the possibility xm0 +1 7→∈ L, which gives a change of αgrade by m0 + 1 in the right domain. As f1 again has the same form as in the case 1◦ , it
follows that A ≤ 2m0 + m1 .
Because of s + 1 ≤ c the inequality (*) follows from

1
m0 (m0 + 1) − (c + m0 + m1 ) > (c − 1)2 + 2c + 2m0 + m1 .
2
As m1 ≤ m0 − (c + 2), this inequality follows from m0 (m0 − 9) > 2c2 − 2c − 6. Because of
m0 ≥ 2c + 4 it suffices to show (2c + 4)(2c − 5) > 2c2 − 2c − 6 ⇔ 2c2 > 14. This inequality
is fulfilled, if c ≥ 3. The cases c = 0, 1, 2 will be treated later on (see below).
Subcase 2: ℓ1 = x: Then from Figure 6.2 we conclude that only the second case of
Proposition 3.2 can occur, and we have
Conclusion 6.5. In the case II.1 the inequality (*) is fulfilled except if ℓ0 = y, ℓ1 = y
and c = {0, 1, 2}.
6.3.2.

We now treat the case II.2 of Proposition 3.2).

Subcase 1: ℓ1 = y: Figure 6.1 shows that only the case II.2.3 is possible. Then
there are s + 1 deformations M0 7→∈ L, · · · , xs M0 7→∈ L ( and t + 1 deformations
M1 7→∈ L, · · · , xt M1 7→∈ L, respectively). The changes of α-grade in the columns of
M0 , · · · , xs M0 (of M1 , · · · , xt M1 , respectively) is m0 , · · · , m0 +s ( and m1 −1, m1 , · · · , m1 +
t − 1, respectively). Here s and t fulfil the inequalities of ( Proposition 3.2d). Thus the

total change of α-grade in the right domain fulfils the inequality: A ≤ (s + 1)m0 + s+1
+
2
70

(t + 1)m1 +

t
2



− 1, where

κ
1
1
1
− m0 ( −
)+
and
ρ2
ρ2 ρ1 + ρ2
ρ2
κ
1
1
t≤
− m1 ( −
).
ρ2
ρ2 ρ1 + ρ2
Estimate of s: Because of κ ≤ c and m0 ≥ 2(c + 2) one obtains :
1
1
1
c
− 2(c + 2)( −
)+
s≤
ρ2
ρ2 ρ1 + ρ2
ρ2
c+1
1
1
1
1
=
− 2(c + 1)( −
) − 2( −
)
ρ2
ρ2 ρ1 + ρ2
ρ2 ρ1 + ρ2
1
1
1
2
− ) − 2( −
)
= (c + 1)(
ρ1 + ρ2 ρ2
ρ2 ρ1 + ρ2
s≤

We first consider the possibility s ≥ 0. This implies

2
ρ1 +ρ2

−

1
ρ2

> 0, i.e. ρ2 > ρ1 .

2
− x1 ; i.e. x corresponds to ρ2 and a corresponds to ρ1 , therefore 1 ≤
Let be fa (x) = x+a
√
√
a < x. fa′ (x) = −2/(x+a)2 +1/x2 < 0 ⇔ 2x2 > (x+a)2 ⇔ 2x > x+a ⇔ x > a(1+ 2).
√
√
It follows that fa (x) has the maximum for x = a(1 + 2), and fa (a(1 + 2)) = 0.171···
.
a
Therefore s ≤ 0.172(c + 1).

Estimate of t: Because of m1 ≥ c + 2 (cf. Lemma 2.4) one has


c
1
1
t≤
− (c + 2)
−
ρ2
ρ2 ρ1 + ρ2




c
1
1
1
c
−c
−
−2
−
=
ρ2
ρ2 ρ1 + ρ2
ρ2 ρ1 + ρ2


c
1
1
=
−2
−
ρ1 + ρ2
ρ2 ρ1 + ρ2

Therefore t ≤ c/2, if ρ2 ≤ ρ1 and t ≤ c/3, if ρ2 > ρ1 .

First possibility: ρ2 ≤ ρ1 : Then s < 0, i.e. there are no deformations defined by f0 ,

and A ≤ (c/2 + 1)m1 + c/2
− 1.
2

Second possibility: ρ1 < ρ2 . Then s ≤ 0.172(c + 1), t ≤ c/3 and A ≤ (0.172c +


c/3
1, 172)m0 + 0.172c+1,172
+
(c/3
+
1)m
+
− 1.
1
2
2
As m1 ≤ m0 − (c + 2) (Lemma 2.8), one obtains the following estimates:

First possibility: A ≤ (c/2+1)[m0 −(c+2)]+c2/8−1 ⇒ A ≤ (0.5c+1)m0 −3/8c2 −2c−3

Second possibility: A ≤ (0.172c + 1, 172)m0 + 0.5(0.172c + 1, 172)2 + (c/3 + 1)[m0 −
(c + 2)] + c2 /18 − 1 ⇒ A ≤ (0.51c + 2, 172)m0 − 0.25c2 − 1, 46c − 2, 3

As we have assumed that ℓ0 = y, ℓ1 = y it follows that ι(r + 1) = ι(2) = 2. As
mentioned above, only the case II.2.3 of Proposition 3.2 can occur. If c = 0, there are no
deformations at all, so that one can assume c ≥ 1. Replacing s + 1 by c, one sees that it
suffices to show:
(6.6)

Q(m0 − 1) > c2 + 1 + A
71


First possibility: A ≤ (0.5c + 1)m1 + 0.5c
− 1 (see above). One sees that it suffices
2
to show:
1
m (m0 + 1) − (c + m0 + m1 ) > c2 + 1 + (0.5c + 1)m1 + 81 c2 − 1 ⇐⇒ 12 m0 (m0 + 1) −
2 0
m0 − (0.5c + 2)m1 > 89 c2 + c.
Because of m1 ≤ m0 −(c+2) this follows from 21 m0 (m0 +1)−(0.5c+3)m0 +(0.5c+2)(c+2) >
9 2
c + c ⇐⇒ m0 (m0 − c − 5) > 1, 25c2 − 4c − 8.
8
As m0 ≥ 2c+4 this follows from (2c+4)(c−1) > 1, 25c2 −4c−8 ⇐⇒ 0.75c2 +6c+4 > 0.
This inequality is fulfilled for all c.
Second possibility: A ≤ (0.51c + 2, 172)m0 − 0.25c2 − 1, 46c − 2, 3. Putting this into
(6.6), one has to show m0 (m0 +1)−2(c+m0 +m1 ) > 1, 5c2 −2, 92c−2, 6+(1.02c+4, 344)m0.
Because of m1 ≤ m0 −(c+2) this follows from m0 (m0 +1)−2(2m0 −2)−(1, 02c+4, 344)m0 >
1, 5c2 − 2, 92c − 2, 6 ⇐⇒ m0 (m0 − 1, 02c − 7, 344) > 1, 5c2 − 2, 92c − 6, 6.
As m0 ≥ 2c + 4 this follows from (2c + 4)(0.98c − 3, 344) > 1, 5c2 − 2, 92c − 6, 6 ⇐⇒
0.46c2 + 0.152c − 6, 776 > 0.
This inequality is fulfilled if c > 3, 676 and the cases c = 0, 1, 2, 3 will be treated later on
in (6.3.3).
Subcase 2: ℓ1 = x. Figure 6.2 shows that in Proposition 3.2 f1 = M1up has to be a
monomial and that for f0 one of the two cases 2.1 or 2.3 can occur. We first treat the case
2.1, i.e., one has the deformation xm0 7→ N1up . The α-grade of the column of xm0 changes
by m0 and the α-grade of the column of N1up changes by 1. Therefore A ≤ m0 + 1. There
are no further deformations in the right domain. Now in the inequality (*) one has s = 0
and ι(2) = 1 and therefore has to show:
1
m (m0 + 1) − (c + m0 + m1 ) > (c − 1)2 + 1 + m0 + 1. Because of m1 ≤ m0 − (c + 2)
2 0
this follows from m0 (m0 − 5) > 2(c − 1)2 . As m0 ≥ 2c + 4 this follows from (2c +
4)(2c − 1) > 2(c − 1)2 ⇐⇒ 2c2 + 10c − 6 > 0. This inequality is fulfilled, if c ≥ 1.
In the case c = 0 one has (∗bis), i.e., one has to prove Q(m0 − 1) > A, i.e. to prove
1
m (m0 + 1) − (c + m1 + m0 ) > m0 + 1.
2 0
One sees that this follows from m0 (m0 − 5) > −2. As one has m0 ≥ 7 by (6.2 Auxiliary
Lemma 1), this inequality is fulfilled.
Now we treat the case 2.3, that means f0 = M0 + F as in Proposition 3.2. The only
possible deformations are M0 7→∈ L, xM0 7→∈ L, · · · , xs M0 7→∈ L. As c = 0 implies that
I is monomial, we can assume c > 0.
We again distinguish two cases:

First possibility: ρ2 ≤ ρ1 . Then s + 1 = 0, i.e. there is no deformation at all, and
therefore one has A = 0. Then (*) reads 21 m0 (m0 + 1) − (c + m0 + m1 ) > (c − 1)2. Because
of m1 ≤ m0 − (c + 2) this follows from m0 (m0 − 3) > 2c2 − 4c − 2. Because of m0 ≥ 2c + 4
it suffices to show (2c + 4)(2c + 1) > 2c2 − 4c − 2 ⇐⇒ 2c2 + 14c + 6 > 0 which is true
for all c.
Second possibility: ρ1 < ρ2 . Then s ≤ 0, 172(c + 1), as was shown above. We have
already remarked at the beginning of (6.3.2) that the s + 1 deformations noted above give

a change of α-grade A ≤ (s + 1)m0 + s+1
. It follows that A ≤ (0.172c + 1, 172)m0 +
2
72

0.5(0.172c + 1, 172)2.
In order to prove (*) it suffices to show Q(m0 − 1) > (c − 1)2 + (0.172c + 1, 172)[1 + m0 +
0.5·(0.172c+1, 172)]. Because of m1 ≤ m0 −(c+2) this follows from m0 (m0 +1)−2(2m0 −
2) > 2(c − 1)2 + (0.344c + 2, 344)(m0 + 0.086c + 1, 586) ⇐⇒ m0 (m0 − 0.344c − 5, 344) >
2, 029584c2 − 3, 252832c + 1, 717584.
Now c ≥ 1 and m0 ≥ 2c + 4 so that it suffices to show (2c + 4)(1, 656c − 1, 344) >
2, 029584c2 − 3, 252832c + 1, 717584 ⇐⇒ 1, 282416c2 + 7, 188832c − 7, 093584 > 0.
This is true if c ≥ 1. Therefore the inequality (*)is fulfilled in Subcase 2 .
Conclusion 6.6. In the case II, if r = 1, then (*) is fulfilled except in the case II.1
if ℓ0 = y, ℓ1 = y and c ∈ {0, 1, 2} or in the case II.2 if ℓ0 = y, ℓ1 = y and c ∈ {0, 1, 2, 3}.
6.3.3. The cases 0 ≤ c ≤ 4. c = 0 From Figure 6.3 it follows that M0 7→ N1
is the only possible deformation, which is the case II.1. The change of α-grade is
A = m0 + m1 − 2(m1 − 1) = m0 − m1 + 2 and the inequality (∗bis) reads Q(m0 − 1) >
m0 −m1 +2 ⇐⇒ m0 (m0 −3) > 4, and this is true, as m0 ≥ 7 by (6.2 Auxiliary Lemma 1).
c = 1 At first, we note that K is monomial. From Figure 6.4 it follows that there are
three deformations, which can occur, with the “total” change of α-grade B:
1. M0 7→ L, B = m0 + 2
2. M0 7→ N1 and M1 7→ L, B = (m0 − m1 + 2) + (2m1 − m1 − 1) + 2 = m0 + 3
3. M1 7→ L, B = m1 + 1.
Then (∗bis) follows from Q(m0 −1) > m0 +3 ⇐⇒ 12 m0 (m0 +1)−(1+m0 +m1 ) > m0 +3.
As m1 ≤ m0 − 3 (Lemma 2.8), one sees that it suffices to show m0 (m0 − 5) > 2. This is
true, because of m0 ≥ 2(c + 2) = 6 (loc.cit.).
c = 2 Then K is monomial, and the possible deformations are shown in Figure 6.5
a, b. If the case II.1 occurs, then f0 = m0 + αN1 + βF and f1 = M1 + γL. One sees that
the change of α-grade in the right domain becomes maximal, if the deformations M0 7→ F
and M1 7→ L occur. Therefore A ≤ m0 + 2m1 − m1 − 1 = m0 + m1 − 1. The inequality (*)
follows from Q(m0 − 1) > 12 + 2 · 2 + (m0 + m1 − 1) ⇐⇒ 21 m0 (m0 + 1) − (2 + m0 + m1 ) >
4 + m0 + m1 ⇐⇒ 21 m0 (m0 + 1) − 2m0 − 2m1 > 6. Because of m1 ≤ m0 − (c + 2) = m0 − 4
it suffices to show 12 m0 (m0 + 1) − 4m0 > −2 ⇐⇒ m0 (m0 − 7) > −4. This is true as
m0 ≥ 2c + 4 = 8.
If the Figure 6.5a occurs, then only the case 1◦ of (6.3.1) is possible and the change
of α-grade in the right domain is A ≤ m0 + m1 − 1. One gets the same estimate of A if
Figure 6.5a occurs in the case II.2, because xF and xL are elements of K and thus the
order of f0 and of f1 is equal to 0. Then (*) reads Q(m0 − 1) > 12 + 2 · 2 + (m0 + m1 − 1),
and this is true as was shown just before.
If Figure 6.5b occurs, then the order of f0 can be equal to 1, but it is not possible
that f1 = M1 + αF , where α 6= 0, because K is monomial and F ∈
/ K, whereas f1 ∈ K by
Lemma 2.6.
73

We want to sharpen the estimate of (6.3). M0 7→ F and xM0 7→ L yield A =
2m0 + 1; M0 7→ F, M1 7→ L yield A = m0 + m1 − 1 and M0 7→ N1 , M1 7→ L yield
A = (m0 − m1 + 2) + m1 − 1. Therefore A ≤ 2m0 + 1, and (*) reads 12 m0 (m0 + 1) − (2 +
m0 +m1 ) > 12 +2·2+2m0 +1. Because of m1 ≤ m0 −4 it suffices to show m0 (m0 −7) > 8.
As m0 ≥ 2c + 4 = 8, the case c = 2, m1 = 4, m0 = 8 remains (cf. Fig. 6.5c). One sees
that the deformations M0 7→ F, xM0 7→ L give the maximal change of “total” α-grade

B = (8 + 9) + 1 + 2 = 20. As Q(7) = 92 − 14 = 22 the inequality (!) in (5.3) is fulfilled.
c = 3 Because of Conclusion 6.6 one has only to consider deformations of the form
f0 = M0 + F, f1 = M1 + G, where F, G ∈ L (cf. Proposition 3.2). From the computations
in (6.3.2) it follows that the order of f0 is ≤ 0.172(3 + 1) < 1 and the order of f1 is ≤ 1.
Therefore the change of α-grade in the right domain is A ≤ m0 + (m1 − 1) + m1 . If one
replaces m1 by m0 − 5 ≥ m1 in the inequality (*) of (5.3), then one has to show:
1
m (m0 + 1) − (2m0 − 2) > 22 + 3 · 2 + m0 + 2(m0 − 5) − 1 ⇐⇒ m0 (m0 − 9) > −6. This
2 0
is true because of m0 ≥ 2 · (3 + 2) = 10 (cf. Lemma 2.8).
Conclusion 6.7. Even in the cases c ∈ {0, 1, 2, 3} the inequalities (*) and (!) respectively, are fulfilled.
Summarizing the Conclusions 6.1–6.7, we obtain:
Proposition 6.1. If ρ1 > 0, ρ2 > 0, I has the type r ≥ 1 and has y-standard form,
then the inequality (!) of Section (4.3) is fulfilled.

74

Fig. 6.1

monomials

N1 M1
0

1

2

3

4

5

6

7

m1 m1 +1 10

11

12

13

M0
m0

c c+1
LB

RB

Fig. 6.2
M1
N1

monomials

0

1 . . . . . . . . . c+1c+2 . . .
LB

m1 m1 +1 . . .

RB

75

. . . . . . . . . m0

Fig. 6.3

N1 M1
0

1

2

3

M0
4 m1 m1 +1 . . . m0

Fig. 6.4

L
N1 M1
0

1

2 . . . . . . . . .m1 +1. . . m0
Fig. 6.5a

L
F
N1 M1
0

1

2

3

4

5 m1 +1. . . m0

76

Fig. 6.5b

F

Fig. 6.5c

L

F

L

N1 M1
0

1

2

3

4

5

6

N1 M1
M0
7 8

0

77

1

2

3

4

5

6

M0
7 8

CHAPTER 7

Estimates of the α-grade in the case ρ1 > 0, ρ2 > 0 and r = 0.
Let I be an ideal of type r = 0 with y-standard form. Then one has by definition:
I = yK(−1) + f OP2 (−m), colength (K) = c, reg(K) = κ, colength (I) = d = c + m, I
and K invariant under G = Γ · T (ρ), and if the Hilbert functions of K and I are ψ and
ϕ, respectively, then g ∗ (ϕ) > g(d). By definition, K has the type (−1), that means, the
following cases can occur:
1st case: g ∗ (ψ) ≤ g(c), where c ≥ 5 by convention.
2nd case: 0 ≤ c ≤ 4.
As usual the aim is to prove (*) in (5.3), and we write m and f instead of m0 and f0 .

7.1. The case g ∗ (ψ) ≤ g(c).
7.1.1. We first assume that there is no deformation into L, at all. That means f =
x (cf. Fig. 7.1), A = 0 and one has to prove the inequality (*), i.e. Q(m − 1) > (c − 1)2 .
Because of Q(m−1) = 21 m(m+1)−(c+m), this is equivalent to m(m−1) > 2c2 −2c+2. As
m ≥ 2c+1 by Corollary 2.2, it suffices to show (2c+1)·2c > 2c2 −2c+2 ⇐⇒ c2 +2c−1 > 0,
and this is true for c ≥ 5.
m

m
s
7.1.2. We now assume that there are
 M0 = x 7→∈ L, · · · , x M0 7→∈
 deformations
1
L. Then by Proposition 3.2d s ≤ ρκ2 − m ρ12 − ρ1 +ρ
.
2

Auxiliary lemma 1. If g ∗ (ψ) ≤ g(c) and if there is the deformation M0 7→∈ L, then
ρ1 < ρ2 .

Proof. Because M0 X νρ = L is a monomial in L, the slope of the line connecting M0
and L is ≤ the slope of the line connecting M0 and L0 (see Figure 7.1, take into account
the inclusion (3.1) in 3.3 and the following remark concerning the vice-monomial of f .)
It follows that ρ1 /ρ2 ≤ κ/(m − κ). Now κ/(m − κ) < 1 is equivalent to 2κ < m, and
because of κ ≤ c and Corollary 2.2 this is true.

Auxiliary lemma 2. If g ∗ (ψ) ≤ g(c), then κ = reg(K) ≤ c − 2.
Proof. One has κ ≤ c and from κ = c it follows that g ∗ (ψ) = (c − 1)(c − 2)/2 > g(c)
(cf. [T1], p. 92 and Anhang 2e, p. 96). By considering the figures 7.2a and 7.2b one can
convince oneself that κ = c − 1 implies g ∗ (ψ) > g(c).

79

It is clear that the change of α-grade in the right domain caused by the deformations
mentioned above is equal to


s+1
.
(7.1)
A = m + (m + 1) + · · · + (m + s) = (s + 1)m +
2

Because of κ ≤ c − 2 and m ≥ 2c + 1 (Corollary 2.2) one has


1
1
c−2
.
− (2c + 1)
−
s≤
ρ2
ρ2 ρ1 + ρ2
It follows that

c−2
s≤
− 2(c − 2)
ρ2
and therefore
(7.2)



1
1
−
ρ2 ρ2 + ρ2




−5

2
1
s < (c − 2)
−
ρ1 + ρ2 ρ2



1
1
−
ρ2 ρ2 + ρ2





Auxiliary lemma 3. If g ∗ (ψ) ≤ g(c) and if there is a deformation M0 7→∈ L, then
s < 61 (c − 2).
Proof. As 1 ≤ ρ1 < ρ2 (cf. Auxiliary lemma 1), one has to find an upper bound for
2
the function f : [N − {0}] × [N − {0}] → R, f (x, y) := x+y
− y1 , on the domain 1 ≤ x < y.
One can convince oneself that f (1, 2) is the maximal value.

In the case r = 0 the inequality (*) reads 12 m(m + 1) − (c + m) > (c − 1)2 + (s + 1) + A.
Putting in the expression (7.1), one gets 12 m(m + 1) − m > c2 − c + s + 2 + (s + 1)m +

s+1
⇐⇒ m(m + 1) − 2m > 2c2 − 2c + 2(s + 2) + 2(s + 1)m + s(s + 1) ⇐⇒ m(m − 1) >
2
2c2 − 2c + 2(s + 2) + 2(s + 1)m + s(s + 1). Because of the Auxiliary lemma 3 it suffices to
 73 2 29
c − 18 c + 28
. As m ≥ 2c + 1 (Corollary 2.2) this follows from:
show: m m − 13 c − 37 > 36
9

5
4
73 2
29
28
(2c + 1) 3 c − 3 > 36 c − 18 c + 9 ⇐⇒ 47
c2 + 11
c − 40
> 0. As this is true if c ≥ 2, we
36
18
9
have proven (*) in the 1st case.
7.2. The cases 0 ≤ c ≤ 4.
If c = 0, then I = (y, xm ) is monomial and (*) is true, obviously. Unfortunately,
one has to consider the cases c = 1, · · · , 4 separately. The ideal K can have the Hilbert
functions noted in (2.2.2).

7.2.1. c = 1 . From Figure 7.3 we see that deg C0 = 2 + · · · + m − 1 = m2 − 1

and deg C∞ = 1 + · · · + m = m+1
. The deformation of C0 into C∞ is defined by
2
m
m−1
f0 = x + yz
, and this is a simple deformation in the sense of ([T1], 1.3).
One has α-grade (I) = deg(C) = max(deg C0 , deg C∞ ) (loc.cit. Hilfssatz 2, p. 12).
In order to prove (*) one has to show: Q(m − 1) > deg C∞ − deg C0 = m + 1 ⇐⇒
1
m(m + 1) − (1 + m) > m + 1 ⇐⇒ m2 − 3m − 4 > 0. This is true if m ≥ 5.
2
Conclusion 7.1. If c = 1, then (*) is fulfilled except in the case c = 1, m = 4, which
will be treated in (9.3).

80

7.2.2. c = 2 . There are two subcases, which are shown in Fig. 7.4a and Fig. 7.4b.
Here only simple deformations occur, again.

1st subcase (Fig. 7.4a). One gets deg C0 = 1 + 3 + 4 + · · · + m − 1 = m2 − 2; deg C∞ =

2 + 3 + · · · + m = m+1
− 1. The same argumentation as in (7.2.1) shows that one has
2
to show Q(m − 1) > m + 1, i.e. m2 − 3m − 6 > 0. This is fulfilled if m ≥ 5.
In the case m = 4, by the formula of Remark 2.2, it follows that g ∗ (ϕ) = 4. As g(6) = 4
and g ∗(ϕ) > g(d) by assumption, the case m = 4 cannot occur.
2nd subcase (Fig. 7.4b). deg C0 = 2 + · · · + m − 1, deg C∞ = 2 + · · · + m; Q(m − 1) >
deg C∞ − deg C0 = m ⇐⇒ m2 − 3m − 4 > 0. This is true if m ≥ 5, and the case m = 4
cannot occur.
Conclusion 7.2. If c = 2, then (*) is fulfilled.



7.2.3. c = 3 . Here 5 deformations are possible, which are all simple (Fig. 7.5a-7.5e).

1st subcase (Fig. 7.5a). deg C0 = 3 + · · · + m − 1 = m2 − 3; deg C∞ = 2 + 4 + 4 + · · · +

m − 1 = m2 . Q(m − 1) > 3 ⇐⇒ m2 − m > 12. This is true because of m ≥ c + 2 = 5.
2nd subcase (Fig. 7.5b). deg C0 = 3 + · · · + m − 1; deg C∞ = 1 + 3 + 4 + · · · + m.
Q(m − 1) > deg C∞ − deg C0 = m + 1 ⇐⇒ m2 − 3m − 8 > 0. This is true because of
m ≥ 5.

3rd subcase (Fig. 7.5c). deg C0 = 3 + · · · + m − 1; deg C∞ = 2 + · · · + m. Q(m − 1) >
deg C∞ − deg C0 = m + 2 ⇐⇒ m2 − 3m − 10 > 0. This is fulfilled if m ≥ 6. As
m ≥ c + 2 = 5, the case m = 5 remains. But if m = 5, then from Remark 2.2 it follows
that g ∗ (ϕ) = 8 < g(8) = 9, which contradicts the assumption.
4th subcase (Fig. 7.5d). deg C0 = 1 + 2 + 4 + · · · + m − 1; deg C∞ = 1 + 3 + · · · +
m; Q(m − 1) > deg C∞ − deg C0 = m + 1 ⇐⇒ m2 − 3m − 8 > 0. This is fulfilled if
m ≥ 5.

5th subcase (Fig. 7.5e). deg C0 = 2+4+4+· · ·+m−1; deg C∞ = 2+· · ·+m; Q(m−1) >
deg C∞ − deg C0 = m − 1 ⇐⇒ m2 − 3m > 4. This is fulfilled if m ≥ 5.
Conclusion 7.3. If c = 3, then (*) is fulfilled.

7.2.4. c = 4 . There are 8 subcases which are shown in Fig. 7.6a1 –7.6e. At first we
make the additional assumption that m ≥ 7. The slope ρ1 /ρ2 of the line connecting the
monomials denoted by + and - is equal to 1/2 in Fig. 7.6b1 , is ≤ 1/4 in Fig. 7.6b2 and
is ≤ 2/5 in Fig. 7.6b3 . It follows that the deformations of Fig. 7.6b1 and of Fig. 7.6b2 (
respectively the deformations of Fig. 7.6b1 and of Fig. 7.6b3 ) cannot occur simultaneously,
that means, we have only simple deformations.
1st subcase (Fig. 7.6a1 ). deg C0 = 2 + 4 + · · · + m − 1; deg C∞ = 1 + 2 + 4 + · · · +
m; Q(m − 1) > deg C∞ − deg C0 = m + 1 ⇐⇒ m2 − 3m − 10 > 0. This is fulfilled if
m ≥ c + 2 = 6.

2nd subcase (Fig. 7.6a2 ). deg C0 = 2 + 4 + · · · + m − 1; deg C∞ = 3 + 4 + · · · +
m; deg C∞ − deg C0 = m + 1, etc, as in the first subcase.
81


3rd subcase (Fig 7.6b1 ). deg C0 = 4 + 4 + 5 + · · · + m − 1 = m2 − 2; deg C∞ =
2 + 4 + 6 + 5 + · · · + m − 1; Q(m − 1) > deg C∞ − deg C0 = 4 ⇐⇒ m(m − 1) > 16. This
is fulfilled if m ≥ 5.
4th subcase (Fig. 7.6b2 ). deg C0 = 4 + 4 + 5 + · · · + (m − 1); deg C∞ = 3 + · · · +
m; Q(m − 1) > deg C∞ − deg C0 = m − 1 ⇐⇒ m2 − 3m − 6 > 0. This is fulfilled if
m ≥ 5.

5th subcase (Fig. 7.6b3 ). deg C0 = 4 + 4 + 5 + · · · + m − 1; deg C∞ = 2 + 4 + 4 + · · · +
m; Q(m − 1) > m + 2 ⇐⇒ m2 − 3m − 12 > 0. This is fulfilled if m ≥ 6.

6th subcase (Fig 7.6c). deg C0 = 3 + · · · + m − 1; deg C∞ = 3 + · · · + m; Q(m − 1) >
m ⇐⇒ m2 − 3m − 8 > 0. This is fulfilled if m ≥ 5.

7th subcase (Fig. 7.6d). deg C0 = 1 + 2 + 3 + 5 + . . . + m − 1; deg C∞ = 1 + 2 + 4 +
· · · + m; Q(m − 1) > m + 1 ⇐⇒ m2 − 3m − 10 > 0. This is fulfilled if m ≥ 6.

8th subcase (Fig. 7.6e). deg C0 = 2 + 4 + 6 + 5 · · · + m − 1 = m2 + 2; deg C∞ =

; Q(m − 1) > m − 2 ⇐⇒ m2 − 3m − 4 > 0. This is fulfilled,
2 + 4 + 4 + · · · + m = m+1
2
if m ≥ 5.

As m ≥ c + 2 = 6, the case m = 6 remains. All inequalities are fulfilled if m ≥ 6;
therefore the possibility remains that in Fig. 7.6b1 and Fig. 7.6b3 , if m = 6 and ρ =
(−3, 1, 2), the deformations f = (x3 y + ay 2 z 2 ) · z 2 and g = x6 + by 2 z 4 simultaneously
occur. Now f ∧ g = x3 yz 2 ∧ x6 + bx3 yz 2 ∧ y 2 z 4 + ay 2 z 4 ∧ x6 and therefore
max −α-grade (I) ≤ max(deg C0 in Fig. 7.6b1 , deg C∞ in Fig. 7.6b3 , deg C∞ in Fig. 7.6b1 )

 m


 m

m
m+1
= max m2 − 2, m+1
,
+
2
and
min
−
α
−
grade(I)
=min
−
2,
,
+
2
.
2
2
2

 2
2
m+1
m
m+1
Now 2 > 2 + 2, if m ≥ 3, and therefore max −α − grade(I) = 2 , and



m
min − α − grade(I) = m2 − 2. Thus (*) follows from Q(m − 1) > m+1
−
+ 2 ⇐⇒
2
2
2
m − 3m − 12 > 0, and this is true if m ≥ 6.
Conclusion 7.4. If c = 4, then (*) is fulfilled.
7.2.5.



Summarizing the Conclusions 7.1–7.4 we obtain

Proposition 7.1. If ρ1 > 0, ρ2 > 0 and r = 0, then (*) is fulfilled except in the case
c = 1, m = 4, I = (y(x, y), x4 + yz 3 ).

Notabene. Hence the inequality (!)in Section 4.3 is fulfilled with one exception .

82

Fig. 7.1

first column filled
with monomials
of H 0 (K(c))
L0

monomial domain

0

1

2

3

4

5

6

7

8

κ κ+1

M0
m0

c c+1

LB

RB

Fig. 7.2b
Fig. 7.2a

0

1 2 3 4 5 6 7 8 9
colength(K) = 10, reg(K) = 9

0

83

1 2 3 4 5 6 7 8 9 10
colength(K) = 11, reg(K) = 10

Fig. 7.3

Fig. 7.4a

Fig. 7.4b

−

0

1

2

M
3 ... ... m

0

+
2 ... ... m

1

−
0

1

2

+
3 ... ... m

Fig. 7.5c
Fig. 7.5a

−

0

1

2

Fig. 7.5b

−

+
3 ... ... m

0

1

2

−
+
3 ... ... m

0

1

Fig. 7.5d

2

+
3 ... ... m

Fig. 7.5e

−

0

1

2

+
3 ... ... m

84

−
0

1

2

+
3 ... ... m

Fig. 7.6a1

Fig. 7.6a2

−
−
0

1

2

3

+
4 ... ... m

0

1

2

Fig. 7.6b1

−

0

1

2

Fig. 7.6b2

−

+
3

3

+
4 ... ... m

4 ... ... m

0

85

1

2

3

+
4 ... ... m

Fig. 7.6b3
Fig. 7.6c

−

0

1

2

−

3

+
4 ... ... m

0

1

2

Fig. 7.6d

3

+
4 ... ... m

Fig. 7.6e

−

0

1

2

3

−

+
4 ... ... m

0

86

1

2

3

+
4 ... ... m

CHAPTER 8

Borel-invariant surfaces and standard cycles
8.1. Preliminary remarks.
d
The group GL(3; k) operates on S by
matrices, thus

 GL(3; k) operates on H =
 1 0 ∗ 
d
2
Hilb (Pk ). We recall the subgroups Γ =  0 1 ∗  < U(3; k) and T (ρ) < T :=


0 0 1
T (3; k), already introduced in (2.4.1). As each subspace U ⊂ Sn is invariant under
D := {(λ, λ, λ)|λ ∈ k ∗ }, each ideal I ⊂ OP2 and each point of H d is invariant under D.
Instead of the operation of T on H d one can consider the operation of the group T /D,
which is isomorphic to T (2; k).

According to the theory of Hirschowitz, the 1-cycles in H d which are invariant under
B = B(3; k), form a generating system of the first Chow group of H d , and relations among
them are realized in B-invariant surfaces V ⊂ H d ([Hi], Mode d’emploi, p. 89).

We distinguish between the cases, whether V is pointwise invariant under the Gm operation σ(λ) : x 7→ x, y 7→ y, z 7→ λz, or not. This we call the homogeneous case and
the inhomogeneous case, respectively.
8.2. The inhomogeneous case.

We let T = T (3; k) operate by diagonal matrices and let Ga operate by ψα : x 7→
x, y 7→ αx + y, z 7→ z on S. Then the group G = Ga · T operates on H d . Let V ⊂ H d be
a G-invariant, closed, two-dimensional variety, which is not pointwise invariant under Ga
and is not pointwise invariant under the Gm -operation σ(λ) : x 7→ x, y 7→ y, z 7→ λz. (For
abbreviation we speak of an inhomogeneous surface.)
We suppose k = k. If ξ ∈ H d (k) is a closed point, then Tξ and Gξ denote the inertia
group of ξ in T and G, respectively.
8.2.1. Auxiliary lemma 1. There is a point ξ ∈ V (k) such that V = T · ξ.
Proof. If dim T · ξ < 2 for all ξ ∈ V (k), then ξ is T -invariant or T · ξ is a T -invariant,
irreducible curve, ∀ ξ ∈ V (k). The first case can occur for only finitely many points; in the
second case one has Tξ = T (ρ), ρ ∈ Z3 − (0), such that ρ0 + ρ1 + ρ2 = 0 ([T1], Bemerkung
1, p.2).
There are only finitely many ρ, such that there exists an ideal I ⊂ OP2 of fixed
colength d, which is invariant under T (ρ), but not invariant under T (see [T2], Hilfssatz
6, p. 140). In other words, there are only finitely many ρ, such that (H d )T (ρ) 6⊇ (H d )T .
87

S
From the assumption follows V = {V T (ρi ) |i ∈ I}, where I is a finite set of indices. As
V is irreducible, it follows that V = V T (ρ) for a certain ρ. Now one has
Gg(ξ) = gGξ g −1 ∀ ξ ∈ V (k), ∀g ∈ G(k),
and therefore
Gg(ξ) ⊃ T (ρ) ∪ gT (ρ)g −1, ∀ ξ ∈ V (k),

∀ g ∈ G(k).

We show there are λ0 6= λ1 in k ∗ such that τ = (λ0 , λ1 , 1) ∈ T (ρ). We assume that
(λ0 , λ1 , λ2 ) ∈ T (ρ) implies λ0 /λ2 = λ1 /λ2 , i.e. λ0 = λ1 . Then each element in T (ρ) has
the form (λ, λ, λ2), thus T (ρ) ⊂ Gm · D, where D = {(λ, λ, λ)|λ ∈ k ∗ }. But then ρ2 = 0
and V is pointwise Gm -invariant, contradiction.
We thus take τ = (λ0 , λ1 , 1) ∈ T (ρ) such that λ0 =
6 λ1 and take


1 α 0

0 1 0  ∈ Ga < G, where α 6= 0.
g=
0 0 1

Then

gτ g −1
is an element of Gg(ξ) and thus
τ −1 gτ g −1


λ0 (λ1 − λ0 )α 0
λ1
0 
= 0
0
0
1



1 β 0
=  0 1 0
0 0 1


β := λ−1
0 (λ1 − λ0 )α 6= 0

is an element of Gg(ξ) , too. It follows that g(ξ) is invariant under ψβ , thus invariant under
Ga and therefore ξ is invariant under Ga . This is valid for all ξ ∈ V (k), contradicting the
assumption.

8.2.2. If one lets T (2; k) operate on S by x 7→ λ0 x, y 7→ λ1 y, z 7→ z then one sees
that G = T · Ga operates on H d in the same way as the group T (2; k) · Ga which for
simplicity is again denoted by G (and is equal to B(2; k)). If ξ is as in the Auxiliary
lemma 1, then from V = G · ξ it follows that Gξ < G is 1-dimensional. There is the
·
S
decomposition Gξ = gi H, H := G0ξ , in finitely many cosets. As H is a 1-dimensional
connected group, two cases can occur:
1st case: H ≃ Ga . Now the unipotent elements in B(2; k) are just the matrices
ψα = ( 10 α1 ). It follows that there is α 6= 0 such that ψα ∈ Gξ . But then ξ is invariant
under Ga . As Ga is normalized by T , T ·ξ is pointwise Ga -invariant. Because of V = T · ξ,
V is pointwise Ga -invariant, contradiction.
2nd case: H ≃ Gm . As each element of Gm is semi-simple, so is each element of
the isomorphic image H. Thus the commutative group H < GL(2; k) consists of semisimple elements. Then there is an g ∈ GL(2; k), such that g −1Hg consists of diago∼
nal matrices ([K], Lemma 1, p. 150). Because of Gm ≃ H → g −1 Hg < T (2; k) one
has a 1-psg f : Gm → T (2; k). Let p : T (2; k) → Gm be the projection onto the
88

n
first component. Then
p ◦ af : Gm → Gm has the form λ → λ , n a suitable inteλ 0
λ ∈ k ∗ =: T (a, b), a and b suitable integers. It
ger. Thus g −1 Hg =
0 λb


a11 a12
−1
follows H = gT (a, b)g ⊂ G = Ga · T (2; k) = B(2; k). Writing g =
gives
a21 a22


a22 −a12
−1
−1
g =D
, D := det(g). We compute:
−a21
a11



  a
a11 a12
λa 0
λ a11 λb a12
and
=
a21 a22
0 λb
λa a21 λb a22
 a




λ a11 λb a12
a22 −a12
λa 0
−1
−1
=
g =D
g
−a21
a11
0 λb
λa a21 λb a22

 a
λ a11 a22 − λb a12 a21 −λa a11 a12 + λb a11 a12
=
D −1
λa a a − λb a21 a22 −λa a12 a21 + λb a11 a22
 a 21 22

λ a11 a22 − λb a12 a21 (λb − λa )a11 a12
.
D −1
(λa − λb )a21 a22
λb a11 a22 − λa a12 a21

This matrix is an upper triangular matrix if and only if (λa − λb )a21 a22 = 0, ∀λ ∈ k ∗ .

Subcase 1. a = b ⇒ H = T (a, a).

Subcase 2. a 6= b and a21 = 0. Write






a11 a12
a11 0
1 c
g=
= uτ, where τ =
,u =
and c := a12 /a22 .
0 a22
0 a22
0 1

Then H = gT (a, b)g −1 = uT (a, b)u−1 .



λa 0
0 λb



Subcase 3. a 6= b and a21 6= 0. Then a22 = 0 and g
g −1 =


−λb a12 a21 (λb − λa )a11 a12
−1
D
. As D = −a12 a21 , this matrix equals
0
−λa a12 a21

 b
λ (λa − λb )c
, where c = a11 a12 /a12 a21 = a11 /a21 .
0
λa


 b
λ (λa − λb )c
−1
∗
Thus H = gT (a, b)g =
λ ∈ k , where c := a11 /a21 . If u :=
0
λa


 b



1 −c
λ (λa − λb )c
λb 0
−1
, then u
u =
and thus H = u−1 T (b, a)u.
0 1
0
λa
0 λa
We have proved
Auxiliary lemma 2. There is an element u ∈ Ga such that H = uT (a, b)u−1, where a
and b are suitable integers.

8.2.3. Let ξ and u be as in Auxiliary lemma 1 and 2, respectively. Set ζ := u−1 (ξ).
Then Gζ = u−1 Gξ u ⊃ u−1 Hu = T (a, b) < T (2; k) and thus dim T (2; k) · ζ ≤ 1. If this
dimension would be equal to 0, then Gζ and Gξ would have the dimension 2. Because
of V = G · ξ the dimension of V would be 1, contradiction. Thus dim T · ζ = 1 and
(Appendix E, Corollary) gives
89

Auxiliary lemma 3. The inertia group Tζ of ζ in T (3; k) has the form T (ρ), where
ρ = (ρ0 , ρ1 , ρ2 ) and ρ2 6= 0.

8.2.4. Summary.
Proposition 8.1. Let V ⊂ H d be a closed 2-dimensional subvariety, invariant under
G = Ga · T (3; k), where Ga operates by ψα : x 7→ x, y 7→ αx + y, z 7→ z and T (3; k)
operates by diagonal matrices on S. We suppose that V is not pointwise invariant under
this Ga -operation and is not pointwise invariant under the Gm -operation σ(λ) : x 7→
x, y 7→ y, z 7→ λz. Then there is a point ξ ∈ V (k) and u ∈ Ga such that :
(i) V = T (3; k) · ξ (ii) The inertia group Tζ of ζ := u(ξ) in T (3; k) has the form T (ρ),
where ρ = (ρ0 , ρ1 , ρ2 ) and ρ2 6= 0. (iii) V = Gm · Ga · ζ.
Proof. The statements (i) and (ii) follow from (8.2.1) – (8.2.3). Put G := Gm · Ga =
Gm × Ga . If the statement (iii) were wrong, then the inertia group Gζ would have
a dimension ≥ 1 and thus would contain a subgroup H isomorphic to Gm or Ga . If
H ≃ Gm , then ζ would be invariant under p1 (H) = {(λn , 1)|λ ∈ k ∗ }, n ∈ Z − (0). Then
ζ and ξ would be invariant under Gm and thus V would be pointwise Gm - invariant,
contradiction.
If H ≃ Ga then ζ and ξ would be invariant under Ga . As Ga is normalized by T (3; k),
T (3; k) · ξ would be pointwise Ga -invariant and the same would follow for V .

8.3. The homogeneous case.
We now assume V ⊂ H d is a 2-dimensional subvariety, invariant under G :=
Ga · T (3; k), not pointwise invariant under Ga , but now pointwise invariant under the
Gm -operation σ as in (8.1). As there are only finitely many T (3; k)-fixed points in H d , it
follows that V is not pointwise fixed by the Gm -operation τ (λ) : x 7→ λx, y 7→ y, z 7→ z.

Let ξ ∈ V (k) be not Ga -invariant and not Gm -invariant. We define a morphism f by
f : Ga × Gm → V , (α, λ) 7→ ψα τ (λ)ξ.

Assume that f has an infinite fibre. Then there is an element (β, µ) ∈ Ga × Gm such
that ψα τ (λ)ξ = ψβ τ (µ)ξ, i.e.
(8.1)

ψ(α−β)λ−1 (ξ) = τ (λ−1 µ)ξ

for infinitely many (α, λ) in Gm × Ga .

By assumption, C := {ψα (ξ)|α ∈ k}− and D := {τ (λ)(ξ)|λ ∈ k ∗ }− are curves in V . If
one assumes that only finitely many different λ can occur in (8.1), then on the left side
of (8.1) also only finitely many α can occur. For ξ is not a fixed point of Ga , so that from
ψα (ξ) = ψβ (ξ) it follows that α = β. This contradicts the last assumption. Thus C and
D have in common infinitely many points and hence are equal (as subschemes with the
induced reduced scheme structure).
The fixed-point set of C under Gm consists of the two points ξ0/∞ := lim τ (λ)ξ. As
λ→0/∞

C has an unique Ga -fixed-point, and as Ga is normalized by Gm , one of the two points, say
90

ξ∞ , is fixed by Ga . Thus C = {ψα (ξ0)|α ∈ k}− and ξ0 corresponds to a monomial ideal.
n
S
There are only finitely many T (3; k)-fixed points ξi , 1 ≤ i ≤ n, in H d . Set M := Ga · ξi .
1

Then C \ M is open and non empty, and choosing ξ ∈ C \ M it follows that f as defined
above has no infinite fiber.

Proposition 8.2. Let V ⊂ H d be a closed 2-dimensional subvariety, invariant under
G = Ga · T (3; k), not pointwise Ga -invariant, but pointwise invariant under the Gm operation σ(λ) : x 7→ x, y 7→ y, z 7→ λz. Then V is not pointwise invariant under the
Gm -operation τ (λ) : x 7→ λx, y 7→ y, z 7→ z. And for all closed points ζ in an open non
empty subset of V one has V = Ga · Gm · ζ.

8.4. Standard cycles.
In the following we suppose d ≥ 5 and we recall to memory the closed subscheme
[
H = {Hϕ ⊂ H d | g ∗ (ϕ) > g(d)}

of H d (cf. 1.2.2). As usual, we let Ga operate by ψα : x 7→ x, y 7→ αx + y, z 7→ z and we
let Gm operate by σ(λ) : x 7→ x, y 7→ y, z 7→ λz (by σ(λ) : x 7→ λx, y 7→ y, z 7→ z) on S in
the inhomogeneous case (in the homogeneous case, respectively).
8.4.1.
Definition 7. Let C ⊂ H be a B = B(3; k) -invariant curve, which is not pointwise
−
Ga -invariant. Then C={ψα (ξ)|α ∈k}
 , where ξ ↔ I is an ideal, invariant under
 1 0 ∗ 
T = T (3; k) and Γ =  0 1 ∗  < ∆ := U(3; k), which is uniquely determined by


0 0 1
C (cf. [T1], Proposition 0, p. 3). We call C a x-standard cycle respectively a y-standard
cycle, if I has x-standard form respectively y-standard form (see 2.4.3 Definition 2).
8.4.2. Let V ⊂ Z = Z(H) be a 2-dimensional, B-invariant subvariety, where Z is
defined as in (1.2.1). We suppose, that V contains a y-standard cycle. Then V is not
pointwise Ga -invariant, so that we can write V = Gm · Ga · ζ, where ζ ∈ V (k) is as in
Proposition 8.1 or Proposition 8.2 , respectively. By the definition of Z, the orbit ∆ · ζ
has a dimension
ζ is not ∆-invariant,the inertia group ∆ζ of ζ in ∆ has the form
 ≤ 1. As 

 1 α ∗
G(a : b) =  0 1 β  ∈ ∆|aα + bβ = 0 , where a, b ∈ k and not both elements are


0 0 1
zero (cf. Appendix D, Lemma 1). Let I be the ideal corresponding to ζ. If ϕ is the Hilbert
function of I, then g ∗ (ϕ) > g(d) by the definition of H and thus I = ℓK(−1)+f OP2 (−m),
where K ⊂ OP2 has the colength c, f ∈ Sm , c+m = d and m = reg(I) ≥ c+2 ( see Lemma
2.1 – Lemma 2.4). If ν := min{n|H 0(I(n)) 6= (0)}, then ν < m . This follows from Lemma
2.2 and Corollary 2.1 . As G(a : b) is unipotent, there is an eigenvector f ∈ H 0(I(ν)).
From x ∂f /∂z ∈ hf i and bx ∂f /∂y − ay∂f /∂z ∈ hf i we conclude that f = xν , if we
assume b 6= 0, which we do now. Let be η ∈ V (k) any point. If L is the corresponding
91

ideal, then xν ∈ H 0 (L(ν)). ( Proof : This is first of all true for all η ∈ W := Gm · Ga · ζ.
By means of the mapping J 7→ H 0 (J (d)) we can regard H d as a closed subscheme of
G = GrassQ(d) (Sd ). If J ∈ H d is any ideal, the condition xν ∈ H 0 (J (ν)) is equivalent to
the condition xν Sd−ν ⊂ H 0 (J (d)). An element of G(Spec A) is a subbundle L ⊂ Sd ⊗ A
of rank Q(d), and the condition xν Sd−ν ⊂ L defines a closed subscheme of G. Thus the
condition xν ∈ H 0 (L(ν)) defines a closed subset of V . As V = W , this condition is fulfilled
for all points of V .) Assume that L has y-standard form. Then L = y·M(−1)+gOP2 (−n),
where e := colength (M), n := reg(L) ≥ m, and e + n = d. As ν < m ≤ n we get
xν ∈ H 0 (L(ν)) = yH 0(M(ν − 1)), contradiction.
Lemma 8.1. If V ⊂ Z(H) is a B-invariant surface which contains a y-standard cycle,
then V is point-wise invariant under Γ.
Proof. From the foregoing reasoning it follows that b = 0, i.e., ζ is invariant under
Γ = G(1 : 0). As Γ is normalized by Ga and T , it follows that Gm · Ga · ζ is point-wise
invariant under Γ, and the same is true for V .

8.4.3. We suppose that V ⊂ Z(H) is a B-invariant surface, which contains a ystandard cycle. We represent V in the form V = Gm · Ga · ζ, according to Proposition
8.1 or Proposition 8.2, respectively.
Lemma 8.2. (a) One can assume without restriction that J ↔ ζ has y-standard form.
(b) I0/∞ are monomial ideals and have y-standard form.
Proof. Let be ζ ↔ I as in Proposition 8.1 (resp. in Proposition 8.2 ).
(a) By Lemma 2.6), I has x- or y-standard form. First we consider the inhomogeneous
case . From I = xK(−1) + f OP (−m) and the Γ-invariance of I (Lemma 8.1), the Γinvariance of K follows. By (Appendix C, Remark 2) we have Rc ⊂ H 0 (K(c)), and
because of m ≥ c + 2 we get Rm−2 ⊂ H 0 (K(m − 2)), thus xRm−2 ⊂ xH 0 (K(m − 2)) =
H 0 (I(m − 1)). We conclude that xRm−2 ⊂ H 0 (J (m − 1)), hence xRm−2 · Sd−m+1 ⊂
H 0 (J (d)), if J ↔ η is any point of Gm · Ga · ζ. The same reasoning as in (8.4.2) shows
that xRm−2 · Sd−m+1 ⊂ H 0 (J (d)) and hence xRm−2 ⊂ H 0 (J (m − 1)) is true for all
J ↔ η ∈ V . As J = ℓM(−1) + gOP2 (−n), e = colength (M), n = reg(J ) ≥ m, e + n = d
and ℓ ∈ R1 − (0), it follows that H 0 (J (m − 1)) = ℓH 0 (M(m − 2)) ⊂ xRm−2 , hence ℓ = x.
It follows, that no ideal in V has y-standard form, contradiction. Thus the statement (a)
is proved in the inhomogeneous case. If the homogeneous occurs and if we assume that
ζ ↔ I would have x-standard form, the same argumentation as in the inhomogeneous case
gives a contradiction. By Lemma 2.2 we have a representation I = ℓK(−1) + f OP2 (−m),
and if the homogeneous case occurs, because of the Γ-invariance of I, we conclude that
ℓ = −βx + y, where β ∈ k. Replacing I by ψβ (I) = yψβ (K)(−1) + ψβ (f )OP2 (−m), we
can assume without restriction that I has y-standard form.
(b) If I = yK(−1) + f OP2 (−m), colength (K) = c, reg(I) = m, c + m = d, m ≥ c + 2,
then Rm−2 ⊂ H 0 (K(m − 2)) (Appendix C, Remark 2), thus yRm−2 ⊂ H 0 (I(m − 1)),
hence yRm−2 ⊂ H 0 (σ(λ)I(m − 1)), λ ∈ k ∗ . As yRm−2 ⊂ H 0 (J (m − 1)) is equivalent
to yRm−2 · Sd−m+1 ⊂ H 0 (J (d)), the condition yRm−2 ⊂ H 0 (J (m − 1)) defines a closed
92

subscheme of V , which is invariant under the Gm -operation σ, as one sees by the same
reasoning as in the proof of Lemma 8.1 . It follows that yRm−2 ⊂ H 0(I0/∞ (m − 1)). By
Lemma 2.2 we can write I0 = ℓM(−1) + gOP2 (−n), where n = reg(I0 ) ≥ m. But then
yRm−2 ⊂ H 0 (I0 (m − 1)) = ℓH 0 (M(m − 2)), showing that ℓ = y. The same argument
shows that I∞ has y-standard form, too.

In the inhomogeneous case ζ is fixed by T (ρ), where ρ2 6= 0 ( Proposition 8.1) , hence
ζ0/∞ are fixed by T (ρ) · Gm = T (3; k)
In the homogeneous case ζ0/∞ are invariant under the two Gm -operations σ and τ , hence
are invariant under T (3; k).


93

CHAPTER 9

Relations between B-invariant 1-cycles
9.1. Generalities on relations between 1-cycles
We describe the method of Hirschowitz in our special situation. B = B(3; k) operates
on H d and we take a closed subscheme X ⊂ H d , which is B-invariant. Z1B (X) is the
free group generated by B-invariant curves in X. It is easy to see that the canonical
homomorphism Z1B (X) → A1 (X) is surjective (cf. [T1], Lemma 1, p. 6). By ([Hi],
Theorem 1, p.87) the kernel RatB
1 (X) can be described as follows. One has to consider all
1
operations of B on P and all two dimensional subvarieties B ⊂ X × P1 with the following
k

properties:

(i) p2 : B → P1 is dominant, hence surjective and flat.
(ii) The operation of B on P1 fixes 0 and ∞.
(iii) B is invariant under the induced operation (ξ, µ) 7→ (gξ, gµ), ξ ∈ X, µ ∈ P1 , g ∈ B,
of B on X ×k P1 .
N.B. According to a general theorem of Fogarty, the fixed-point scheme (P1 )U (3;k) is
connected; hence from (iii) it follows that U(3; k) operates trivially on P1 .
If Bµ := p−1
2 (µ) is the fibre and Bµ := p1 (Bµ ) is the isomorphic image in X, then one
has:
g(Bµ ) = { (gξ, gµ) | ξ ∈ X, such that (ξ, µ) ∈ B }

= { (ξ, gµ) | ξ ∈ X, such that (ξ, gµ) ∈ B }

= Bgµ , for all g ∈ B, µ ∈ P1
We conclude that
(9.1)

gBµ = Bgµ , for all g ∈ B, µ ∈ P1 .

From property (ii) it follows that B0 and B∞ are B-invariant 1-cycles in X. Then
B
RatB
1 (X) is defined to be the subgroup of Z1 (X) generated by all elements of the form B0 −
B
B
B∞ , and the theorem of Hirschowitz says that AB
1 (X) := Z1 (X)/ Rat1 (X) is canonically
isomorphic to A1 (X) (loc.cit.). We consider V := p1 (B) as a closed subscheme of X with
the induced reduced scheme structure. As B ⊂ V ×k P1 , dim V ≥ 1. If dim V = 1, then
B = V × P1 and B0 − B∞ is equal to zero in Z1B (X). Thus we can assume without
restriction that V is a B-invariant, 2-dimensional subvariety of X.
95

9.2. The standard construction
Let X ⊂ H d be a B(3; k)-invariant subvariety. Start with a subvariety B ⊂ X ×k P1 as
in (9.1), such that V := p1 (B) ⊂ H d is a 2-dimensional subvariety, which is automatically
B(3; k)-invariant. Assume V is not pointwise invariant under the Ga -operation as in 8.2.
Write
(9.2)

V = Ga · Gm · ξ

where ξ ∈ V (k) is a suitable point and Gm operates by σ(λ) : x 7→ x, y 7→ y, z 7→ λz or by
σ(λ) : x 7→ λx, y 7→ y, z 7→ z, respectively (cf. Proposition 8.1 and Proposition 8.2). Then
Gm is a subgroup of T and fixes 0 = (0 : 1) and ∞ = (1 : 0) by assumption. Assume
that σ operates trivially on Gm . From (9.1) it follows that σ(λ)Bµ = Bµ , ∀λ ∈ k ∗ .
If ξ is the point of (9.2), then one chooses µ in such a way that (ξ, µ) ∈ Bµ . Then
ψα (ξ, µ) = (ψα (ξ), µ) ∈ Bµ , ∀α ∈ k, hence ψα (ξ) ∈ Bµ , ∀α ∈ k. Because of (9.1) and (9.2)
we would get V ⊂ Bµ . Then the closed subscheme Bµ ⊂ B has the dimension two and
it follows Bµ = B, contradiction, as p2 is dominant by assumption. This argumentation
also shows ξ ∈
/ B0 ∪ B∞ , i.e. there is µ ∈ P1 − {0, ∞} such that (ξ, µ) ∈ Bµ . Thus one
can find λ ∈ k ∗ such that σ(λ)µ = (1 : 1). Replacing ξ and µ by ξ ′ = σ(λ)ξ and σ(λ)µ,
respectively, then (9.2) is fulfilled with ξ ′ instead of ξ. Thus one can assume without
restriction that µ = (1 : 1). As A1 = P1 − {∞} is fixed by σ, Gm operates by σ on A1 and
fixes (0 : 1). Then there is m ∈ Z such that σ(λ)(a : 1) = (λm a : 1) for all a ∈ k, λ ∈ k ∗ .
As the action of Gm on P1 is not trivial, m 6= 0.
C := {ψα (ξ)|α ∈ k}− ⊂ B(1:1) is a curve in V ; let h be its Hilbert polynomial
with respect to the usual embedding of H d in a projective space (cf. [T2], 4.1.1). The
association λ 7→ σ(λ)C defines a morphism Gm → X := Hilbh (V ). It has an extension to
a morphism P1 → X , which defines a flat family of curves
C ֒→ V ×k P1
↓ p2
P1
∗
If Cλ := p−1
2 (λ), then Cλ := p1 (Cλ ) = σ(λ)C, ∀λ ∈ k and

(9.3)

[p1 (C0 )] = [C] = [p1 (C∞ )] ∈ A1 (V ).

The finite morphism A1 − (0) → A1 − (0) defined by λ 7→ λm has an extension
f : P1 → P1 such that (λ : 1) 7→ (λm : 1), ∀λ ∈ k, and (1 : 0) 7→ (1 : 0). For simplicity, the
morphism 1V × f : V × P1 → V × P1 is denoted by f , again. By construction C ⊂ B(1:1)
and hence σ(λ)C ⊂ σ(λ)B(1:1) = Bσ(λ)(1:1) . The fibre Cλ = σ(λ)C ×{(λ : 1)} is mapped by
f into σ(λ)C × {(λm : 1)} = σ(λ)(C × {(1 : 1)}) ⊂ σ(λ)(B(1:1) × {(1 : 1)}) = σ(λ)B(1:1) =
Bσ(λ)(1:1) , ∀λ ∈ k ∗ .

This construction of the family C is called standard construction and the proof in
([T1], Lemma 1, p.6) shows that C ⊂ V ×k P1 is a subvariety. As C0 and C∞ are closed in
o

C, C := C − (C0 ∪ C∞ ) is open in C. Because C and B are irreducible and f is projective,
96

o

from f (C) ⊂ B we conclude that f (C) = B. As V × {0} and V × {∞} are mapped by f
into itselve, C0 ⊂ V × {0} and C∞ ⊂ V × {∞} are mapped by f into B ∩ V × {0} = B0
and B ∩ V × {∞} = B∞ , respectively. As f (C) = B, we get f (C0 ) = B0 and f (C∞ ) = B∞ ,
i.e. C0 = B0 and C∞ = B∞ . The curves p1 (C0 ) and p1 (C∞ ) are called the limit curves (of
the standard construction) and are denoted by C0 = lim σ(λ)C and C∞ = lim σ(λ)C,
λ→0

respectively, and one has
B0 − B∞ = C0 − C∞

(9.4)

λ→∞

in Z1B (X).

We note that the relation (9.4) was derived under the assumption that V is not pointwise invariant under Ga . If V is pointwise invariant under Ga , the standard construction
cannot be carried out. We get
P
Lemma 9.1. Let X be as before. Elements of RatB
qi Ci ,
1 (X) are either of the form
Ga
where qi ∈ Q and Ci ⊂ X is a B(3; k)-invariant 1-prime cycle in X or they occur in
the following way: One considers all B(3; k)-invariant 2-dimensional subvarieties V ⊂ X,
which are not pointwise invariant under Ga . Then there are points ξ ∈ V (k) such that
V = Gm · Ga · ξ, where Gm operates by σ(λ) : x 7→ x, y 7→ y, z 7→ λz on S (inhomogeneous
case) or by σ(λ) : x 7→ λx, y 7→ y, z 7→ z (homogeneous case). C := {ψα (ξ)|α ∈ k}− is
a curve in V (with the reduced induced scheme structure). By means of the standard
construction it defines a family of curves C ⊂ V ×k P1 , which is flat over P1 and with the
Ga
fibres Cλ = σ(λ)C for all λ ∈ P1 − {0, ∞}. RatB
1 (X) is generated by the relations in X
noted above as well as by the relations C0 − C∞ , where C0/∞ = lim σ(λ)C are the limit
λ→0/∞

curves of C in V and these are Ga -invariant 1-cycles.
Proof. As σ(λ)ψα = ψαλ σ(λ) in the homogeneous case, σ(λ)C = {ψα σ(λ)ξ|α ∈ k}−
is Ga -invariant, ∀λ ∈ k ∗ . And the same is true in the inhomogeneous case .

9.3. Relations containing a y-standard cycle
Suppose d ≥ 5. The closed subscheme H ⊂ H d (cf. Section 8.4) is clearly B-invariant.
As U(3; k) is normalized by T (3; k), the closed subscheme X = Z(H) is B-invariant,
too (cf.Section 1.2). From Lemma 8.1 it follows that relations in Z1B (X) containing a
y-standard cycle are defined by 2-dimensional B-invariant surfaces V ⊂ X, which are
pointwise Γ-invariant but not pointwise Ga -invariant.
We can write V = Gm · Ga · ζ, where ζ ∈ V (k) is a point as in Proposition 8.1
or Proposition 8.2. By Lemma 8.2 we can assume that I ↔ ζ has y-standard form.
Let be ζ0/∞ = lim σ(λ)ζ , where σ(λ) denotes one of the two Gm - operations menλ→0/∞

tioned in Lemma 9.1 .Then carry out the standard construction by using the curve
C := { ψα (ζ) | α ∈ k }− . By Lemma 8.2 C0/∞ are y-standard cycles. We conclude from
Proposition 8.1 and Proposition 8.2 that W := Gm · Ga · ζ is B-invariant and therefore
V − W is a union of B-invariant curves and points. As C0/∞ is invariant under Gm and
Ga , from W ∩ C0/∞ 6= ∅ it would follow that V ⊂ C0/∞ . Hence C0/∞ ⊂ V − W , and
C0/∞ is a union of B-invariant curves, too. As ζ0/∞ ∈ C0/∞ , one has C0/∞ ⊂ C0/∞ . Now
97

from the Propositions 5.1, 6.1 and 7.1 it follows that the inequality (!) in ( 4.3) is fulfilled
with one exception (cf. Proposition 7.1 ). Let be m = reg(I). Because of max −α-grade
(I) ≥ α-grade (I) = deg C = deg C0 = deg C∞ and min(deg C0 , deg C∞ ) ≥ min −αgrade (I) (cf. the definitions and the inequality (4.9) in Section 4.3), from (!) it follows
that
Q(m − 1) + min(deg C0 , deg C∞ ) > deg C0 = deg C∞ .

(**)

As V = W , each ideal J in V has a regularity n ≥ m. If J is monomial and has ystandard form , then from Remark 4.4 it follows that the y-standard cycle generated by
J has a degree ≥ Q(n − 1) ≥ Q(m − 1).
Lemma 9.2. C0 and C∞ are the only y-standard cycles contained in the B-invariant
cycles C0 and C∞ , respectively, and they both occur with the multiplicity 1.
Proof. Except when c = 1, m = 4 the statements follow from (**) and the foregoing
discussion. In the remaining exceptional case (cf. Proposition 7.1 ) we carry out the
standard construction with I = (y(x, y), x4 + yz 3 ) ↔ ζ. Then ζ0 ↔ (y(x, y), x4), ζ∞ ↔
(y, x5 ) and C0 and C∞ are y-standard cycles of degrees 5 and 10, respectively, as one can
see from Figure 9.1.
If η := lim ψα (ζ), the corresponding ideal L is invariant under Ga , hence invariant
λ→∞

under U(3; k). One sees that xz 3 ∈ L, and hence x ∈ L follows. One concludes from this
that L = (x, y 5 ).

As σ and ψα commute, it follows that η is the only point in σ(λ)C for all λ, which is
S
S
U(3; k)-invariant. From V = p1 (C) = {p1 (Cλ )|λ ∈ P1 } = {σ(λ)C|λ ∈ k ∗ } ∪ C0 ∪ C∞
it follows that
V = W ∪ {η} ∪ C0 ∪ C∞
where W := Gm · Ga · ζ. Now η0 := lim ψα (ζ0 ) ↔ (x(x, y), y 4) and η∞ := lim ψα (ζ∞ ) ↔
α→∞

α→∞

(x, y 5 ) are the only B(3; k)-invariant points in C0 and C∞ , respectively. By the theorem
of Fogarty V U (3;k) is connected, hence is a union of pointwise U(3; k)-invariant, closed
and irreducible curves. If E is such a curve, then E ⊂ C0 ∪ C∞ follows. Now deg C0 =
deg C∞ = deg C = 10. Hence C∞ = C∞ and E ⊂ C0 follows. As each y-standard cycle
in V has a degree ≥ Q(3) = 5, C0 is the only y-standard cycle contained in C0 , and C0
occurs with multiplicity 1.

Proposition 9.1. Let be X = Z(H). Relations in RatB
1 (X) , which are defined by a
B-invariant surface in X and which contain a y-standard cycle, are generated by relations
of the form C1 − C2 + Z where C1 and C2 are y-standard cycles and Z is a 1-cycle in X,
whose prime components are B-invariant but are not y-standard cycles.
Proof. This follows from Lemma 9.1 and Lemma 9.2 .
98



Fig. 9.1

Fig. 9.2

C∞

C0
−
0

1

2

3

+
4

0

1

2

3

99

4

5

CHAPTER 10

Proof of Theorem 1.1
This had been formulated in Chapter 1 and we repeat the notations and assumptions
S
introduced there: d ≥ 5, g(d) = (d − 2)2 /4, H = {H≥ϕ ⊂ H d |g ∗(ϕ) > g(d)}, A(H) =
Im(A1 (HU (3;k) ) → A1 (H)), c3 = {(αx + y, xd )|α ∈ k}− .
P
We make the assumption [c3 ] ∈ A(H). Then [c3 ] =
qi [Ui ] in A1 (H), where qi ∈ Q,
and Ui ⊂ H are T -invariant curves, which are pointwise U(3; k)-invariant. (A1 (HU (3;k) )
is generated by such curves, as follows from the theorem of Hirschowitz or more directly
∼
from an application of Lemma 1 in [T1], p. 6.) If X := Z(H), then A1 (X) → A1 (H)
∼
([T2], Lemma 24, p.121) and AB
1 (X) → A1 (X) ([Hi], Theorem 1, p.87).
P
Hence we can regard α := c3 − qi Ui as a 1-cycle in Z1B (X) ,whose canonical imB
B
B
age in AB
1 (X) := Z1 (X)/ Rat1 (X) vanishes. (We recall that Z1 (X) is the free group
generated by all closed, reduced and irreducible curves in X.) Define B(X) to be the subgroup of Z1B (X), which is generated by all B(3; k)-invariant curves in X (closed, reduced,
irreducible), which are not y-standard cycles. Define:
F1 (X) = Z1B (X)/B(X), R1 (X) = RatB
1 (X) + B(X)/B(X)
F1 (X) is the free group generated by the y-standard cycles, and by Proposition 9.1 R1 (X)
is generated by elements of F1 (X) of the form C1 − C2 , where C1 and C2 are (different)
y-standard cycles. It follows that the canonical image of α in F1 (X)/R1 (X) on the one
hand vanishes and on the other hand is equal to the canonical image of c3 in this Qvector space. Hence c3 ∈ R1 (X). We will show that this is not possible. To each of the
finitely many y-standard cycles we associate a canonical basis element ei ∈ Qn , 1 ≤ i ≤ n.
Especially, we associate c3 to the element en = (0, . . . , 1). From c3 ∈ R1 (X) it follows
that en ∈ h{ei − ej |1 ≤ i < j ≤ n}i, which is not possible. All in all we have
Theorem 10.1. [c3 ] ∈
/ A(H).



Corollary 10.1. dimQ A1 (H) ≥ 3.
Proof. E := {(x2 , xy, y d−1 + αxz d−2 )|α ∈ k}− and F := {(x, y d−1 (αy + z))|α ∈ k}−
are 1-cycles in H, which are pointwise invariant under U(3; k) and G(0, 1), respectively
(see the notation in Appendix D). If [c3 ] = q1 [E] + q2 [F ], q1 , q2 ∈ Q, we compute the
intersection numbers with the tautological line bundles and get
 
d
= q1 + q2 (n − d + 1)
2
for all n ≥ d − 1. Hence q2 = 0. If q1 6= 0, then [c3 ] ∈ A(H) would follow.
101



CHAPTER 11

Surfaces in H invariant under Ga · T (4; k)
We let Ga operate by ψα : x 7→ x, y 7→ αx + y, z 7→ z, t 7→ z and let T (4; k) operate by
diagonal matrices on P = k[x, y, z, t], hence on H = HQ . Let V ⊂ H be a 2-dimensional
subvariety, which is invariant under Ga · T (4; k) but not pointwise invariant under Ga .
11.1. The inhomogeneous case
We suppose that V is not pointwise invariant under the Gm -operation σ(λ) : x 7→
x, y 7→ y, z 7→ z, t 7→ λt.
11.1.1. Auxiliary Lemma 1. There is a ξ ∈ V (k) such that V = T (4; k) · ξ.
Proof. If dim T (4; k) · ξ ≤ 1, then the inertia group Tξ ⊂ T (4; k) has the dimension
≥ 3. If the dimension is 4, then ξ corresponds to a monomial ideal with Hilbert polynomial
Q, and there are only finitely many such ideals. If dim Tξ = 3, then Tξ = T (ρ) ([T2],
Hilfssatz 7, p.141). If ξ corresponds to the ideal I with Hilbert polynomial Q, then
H 0 (I(b)) has a basis of the form fi = mi pi (X ρ ), where m is a monomial and pi is a
polynomial in 1 variable. At least for one index i the polynomial pi (X ρ ) contains a
positive power of X ρ . Then mi X ρ ∈ Pb . As mi ∈ Pb and ρ = (ρ0 , . . . , ρ3 ) ∈ Z4 − (0) is a
vector such that ρ0 + · · ·+ ρ3 = 0, this implies |ρi | ≤ b. Hence there are only finitely many
such vectors ρ such that the fixed-point scheme HT (ρ) does not consist of only finitely
many T (4; k)-fixed points.
n
S
Suppose that dim T (4; k) · ξ ≤ 1 for all ξ ∈ V (k). Then V = V T (ρi ) , ρi ∈ Z4 − (0),
1

hence V = V T (ρ) , ρ ∈ Z4 − (0) suitable vector. We show there is τ = (λ0 , . . . , λ3 ) ∈ T (ρ)
such that λ0 6= λ1 . If not, it follows that T (ρ) ⊂ T (1, −1, 0, 0), hence νρ = (1, −1, 0, 0), ν ∈
Z. But then ρ3 = 0 and V would be pointwise invariant under Gm , contradiction.


1 α(λ−1
−1
−1
0 λ1 − 1)
= ψβ ,
Take this τ and an arbitrary α 6= 0. Then τ ψα τ ψα =
0
1
where β := α(λ−1
0 λ1 −1) 6= 0. The same argumentation as in the proof of ( 8.2.1 Auxiliary
Lemma 1) gives a contradiction.


11.1.2. If 0 ≤ i < j ≤ 3 define Tij = {(λ0 , λ1 , λ2 , λ3 ) ∈ T (4; k)|λℓ = 1 if ℓ 6= i and
ℓ 6= j}. We let G := Ga · T23 operate von V . If ξ ∈ V (k) is as in Auxiliary Lemma 1,
let Gξ be the inertia group of ξ in G. From 1 ≤ dim Gξ ≤ 2 it follows that there is a
1-dimensional connected subgroup H < G0ξ . Then H ≃ Ga or H ≃ Gm . In the first case it
103

follows that H = Ga and hence T (4; k) · ξ is pointwise Ga -invariant, contradiction. Hence
H ≃ Gm . In the diagram
i

Gm ≃ H ֒→ G = Ga × T23
p1 ւ
ց p2
Ga
T23
p1 ◦ i is the trivial map, hence H = {(1, 1, λr , λs )|λ ∈ k ∗ } =: T23 (r, s), where r, s ∈ Z are
not both equal to zero.



λ α
∗
11.1.3. Let be G :=
λ, µ ∈ k , α ∈ k and let ξ be as in Auxiliary
0 µ
Lemma 1. Then there is again a subgroup H < G0ξ isomorphic to Gm . It has the form
 a


λ (λb − λa )c
∗
H=
λ ∈ k , where a, b ∈ Z are not both equal to zero and c ∈ k
0
λb


1 −c 0 0
0 1 0 0

is a fixed element (cf. 8.2.2). Let be u = 
0 0 1 0.
0 0 0 1



a
λ






b

λ
∗
−1


=: T01 (a, b). If ζ := u(ξ) one gets
λ
∈
k
Then uHu =

1 






1
Tζ ⊃ T01 (a, b). From (11.1.2) it follows Tζ ⊃ T23 (r, s), too. As Tζ contains the diagonal
group, dim Tζ ≥ 3 follows. If ζ were fixed by T (4; k), then ξ = u−1 (ζ) would be fixed by
T23 , contradiction. It follows that Tζ = T (ρ). Here ρ3 6= 0, because ξ is not invariant
under σ, for otherwise V would be pointwise Gm -invariant. If c = 0, then Tξ = T (ρ),
which contradicts the choice of ξ.
Conclusion 11.1. The element u is different from 1, and putting ζ := u(ξ) one has
V = Ga · Gm · ζ.
Proof. Put G := Ga · Gm = Ga × Gm . If dim Gζ ≥ 1, then G0ζ would contain a
subgroup H isomorphic to Ga or Gm . But then ζ would be invariant under Gm or Ga ,
and then the same would be true for ξ, which gives a contradiction as above.

Conclusion 11.2. V \ Ga · Gm · ζ is a union of points and curves which are invariant
under Ga · T (4; k).
Proof. If τ ∈ T (ρ), then τ · Ga · Gm · ζ = Ga · τ · Gm · ζ = Ga · Gm · τ · ζ = Ga · Gm · ζ,
and if τ ∈ Gm , the same is true.

11.2. The homogeneous case
11.2.1. We first suppose that V is pointwise invariant under the Gm -operation σ(λ) :
x 7→ x, y 7→ y, z 7→ z, t 7→ λt, but not pointwise invariant under the Gm -operation τ (λ) :
x 7→ x, y 7→ y, z 7→ λz, t 7→ t. Then V is invariant under T (3; k) ≃ {(λ0 , λ1 , λ2 , 1)|λi ∈ k ∗ }
and we have the same situation as in Chapter 8. We use the same notations introduced
104

there
ξ (8.2.1 Auxiliary Lemma 1). We let G := Ga · T (2; k) =

 and get V = T (2; k) · 
λ α
λ, µ ∈ k ∗ , α ∈ k operate on V . Then there is a subgroup H of Gξ , which
0 µ
is isomorphic to Ga or Gm . In the first case ξ would
 bea Ga -invariant,
 hence Vwould
λ (λb − λa )c
be pointwise Ga -invariant. In the second case H =
λ ∈ k ∗ . The
0
λb
same argumentation as in (11.1.3) shows that the inertia group of ζ = u(ξ) in T (3; k) has
a dimension ≥ 2. If Tζ = T (3; k), then ζ would be invariant under the Gm -operation τ ,
hence ξ invariant under τ , too, and V would be pointwise invariant under τ . It follows
that Tζ = T (ρ), where ρ = (ρ0 , ρ1 , ρ2 , 0) and ρ2 6= 0. We note that u 6= 1, for otherwise
the inertia group of ξ in T (4; k) would have a dimension ≥ 3.
Conclusion 11.3. The element u is different from 1 and putting ζ = u(ξ) one has
V = Ga · Gm · ζ.
The same argumentation as in (11.1.3) gives
Conclusion 11.4. V \ Ga · Gm · ζ is a union of points and curves which are invariant
under Ga · T (4; k).

11.2.2. We now suppose V is pointwise invariant under T23 = { (1, 1, λ2, λ3 ) | λi ∈ k ∗ }.
Then V is not pointwise invariant under the Gm -operation σ(λ) : x 7→ λx, y 7→ y, z 7→
Gm
z, t 7→ t. Let ξ ∈ V \ (V
∪ V Ga ) be a closed point, and put G := Ga ⋉ Gm =


λ α
λ ∈ k∗ , α ∈ k .
0 1

Assume that dim Gξ ≥ 1. Then H := G0ξ is 1-dimensional and connected. As ξ is not

 a

λ (1 − λa )c
∗
, a ∈ Z − (0). Putting
Ga -invariant, H ≃ Gm hence H =
λ∈k
0
1


1 −c 0 0
0 1 0 0
−1
a
∗

u = 
0 0 1 0 and ζ = u(ξ) we get uHu = { (λ , 1, 1, 1) | λ ∈ k }. It follows
0 0 0 1
that ζ is Gm -invariant, hence T (4; k)-invariant. Thus ξ corresponds to an ideal I such
that H 0 (I(b)) has a generating system of the form Mi (y − cx)ni , where Mi is a monomial
without y and ni ∈ N. There are only finitely many points ζi ∈ V (k) which are T (4; k)
S
invariant and it follows that ξ ∈ Ga · ζi .
S
Conclusion 11.5. If ξ ∈ V \ [ Ga · ζi ∪ V Ga ∪ V Gm ] is a closed point, then V =
Ga · Gm · ξ.

11.3. Summary
Let V ⊂ HQ be a 2-dimensional subvariety, invariant under G := Ga · T (4; k) but not
pointwise invariant under Ga . We distinguish between three cases:
1. V is not pointwise invariant under the Gm -operation σ(λ) : x 7→ x, y 7→ y, z 7→ z, t 7→
λt.
105

2. V is pointwise invariant under the Gm -operation σ as in the first case, but not pointwise
invariant under the Gm -operation τ (λ) : x 7→ x, y 7→ y, z 7→ λz, t 7→ t.
3. V is pointwise invariant under the Gm -operations σ and τ as in the 1th and 2nd case.
Then V is not pointwise invariant under the Gm -operation ω(λ) : x 7→ λx, y 7→ y, z 7→
z, t 7→ t.
Lemma 11.1. (a) In the 1st case (respectively in the 2nd case) there is ζ ∈ V (k) with
the following properties:
(i) The inertia group Tζ of ζ in T (4; k) has the form T (ρ) where ρ3 > 0 (respectively
ρ2 > 0 and ρ3 = 0).
(ii) V = Ga · Gm · ζ and V \ Ga · Gm · ζ is a union of points and curves invariant under
Ga · T (4; k).
(iii) There is u ∈ Ga , different from 1, such that V = T (4; k) · ξ, where ξ := u(ζ).
(b) In the 3rd case, if one chooses ζ ∈ V (k) such that ζ is neither Ga – nor Gm -invariant

and does not lie in the set ξ ∈ V (k) ∃u ∈ Ga and ∃µ ∈ V T (4;k) (k) such that ξ = u(µ) ,
then V = Ga · Gm · ζ.


106

CHAPTER 12

Surfaces in H invariant under B(4; k)
12.1. The operation of the unipotent group on H
12.1.1.

Let be p = (a : b : c) ∈ P2 (k)

1 α ∗



0 1 β
G(p) := 

0
0 1


 0 0 0

and


∗




∗
aα + bβ + cγ = 0 .
γ



1

This is a 5-dimensional subgroup of ∆ := U(4 : k) and each 5-dimensional subgroup of ∆
has this form, where p ∈ P2 (k) is uniquely determined (Appendix D, Lemma 1).

Especially one has the groups 
G
: 0
i = G(pi ), where

 p1 = (0
1
1
∗
∗
∗










0
0 1 ∗ ∗
, G2 = 
0), p3 = (1 : 0 : 0) and G1 = 



0
0
0
1
0






 0
 0 0 0 1 


1
0
∗
∗







0
1
∗
∗
 .

0 0 1 ∗



 0 0 0 1 


1 α 0 0
 0 1 0 0
−1

Remark 12.1. If ψα = 
0 0 1 0, then ψα G(p)ψα = G(p).
0 0 0 1

: 1), p2 
=(0 : 1 :
∗ ∗ ∗ 


1 0 ∗
 , G3 =
0 1 ∗

0 0 1 

Remark 12.2. If τ = (λ0 , λ1 , λ2 , λ3 ) ∈ T (4; k), then τ G(p)τ −1 = G(τ p), where τ p :=
−1
−1
(aλ−1
0 λ1 : bλ1 λ2 : cλ2 λ3 ).
12.1.2. In ([T2], 3.2.1) and ([T3], 10.2) we had introduced a closed, reduced subscheme Z = Z(HQ ) of HQ such that Z(k) = {x ∈ HQ (k)| dim ∆ · x ≤ 1}. From the
∼
theorem of Hirschowitz it follows that A1 (Z) −→ A1 (HQ ) ([T2], Lemma 24, p. 121). In
the following we consider a surface V ⊂ Z (i.e. a closed 2-dimensional subvariety) which
is B(4; k)-invariant, but not pointwise invariant under ∆ = U(4; k).
12.1.3. Auxiliary Lemma 1. Let be ξ ∈ Z(k) but ξ not invariant under ∆ ,hence
∆ξ = G(p) . If τ ∈ Tξ , then τ p = p .
Proof. If τ ∈ Tξ , then τ G(p)τ −1 τ ξ = τ ξ and thus G(τ p)ξ = ξ. If τ p 6= p , then
G(τ p) 6= G(p) and ξ would be fixed by the subgroup of ∆ , which is generated by G(p)
and G(τ p) , i.e. fixed by ∆ .
107


12.1.4. Auxiliary Lemma 2. Let ξ be as in Auxiliary Lemma 1 . If p = (a : b : 0)
and a, b 6= 0 , then Tξ ⊂ T (1, −2, 1, 0) .
Proof. This follows from Remark 12.2 and Auxiliary Lemma 1 .



12.2. The case p = (a : b : c) where a, b, c 6= 0.
Let be V ⊂ Z a B-invariant surface and ξ ∈ V (k) a point such that ∆ξ = G(a : b : c),
−2 3
2
∗
where a, b, c 6= 0. Then Tξ ⊂ Λ := {(λ0 , λ1 , λ−1
0 λ1 , λ0 λ1 )|λ0 , λ1 ∈ k }. As dim T (4; k)·ξ ≤
2, one has dim Tξ ≥ 2 and thus Tξ = Λ.



λ
α
0
0


0





0
λ
0
0
1
∗


operate on V .
α ∈ k, λ0 , λ1 ∈ k
We let G := Ga · T01 = 
0 0 1 0





 0 0 0 1
Remark 12.3. T (4; k)ξ ∩ G = (1).

−2 3
−1 2
−2 3
2
Proof. If (λ0 , λ1 , λ−1
0 λ1 , λ0 λ1 ) ∈ G, then λ0 λ1 = λ0 λ1 = 1 which implies λ0 /λ1 =
1, and then λ0 = λ1 = 1 follows.


Remark 12.4. V = T01 · ξ.
Proof. Because of Remark (12.3) dim T01 · ξ < 2 is not possible.



From Remark (12.4) it follows that Gξ < G is 1-dimensional. If H := G0ξ would be
isomorphic to Ga
, then
be invariant
under
 Ga , hence invariant under ∆, contradic mξ would

λ
(λn − λm )c
tion. Thus H =
λ ∈ k ∗ , m, n ∈ Z not both equal to zero, c ∈ k
0
λn


 m


1 −c
λ
0
−1
∗
(see 8.2.2 Auxiliary Lemma 2). If u :=
, then uHu =
λ ∈ k =:
0 1
0 λn
T01 (m, n). Putting ζ := u(ξ) one gets Tζ ⊃ T01 (m, n). Because of Remark (12.1) from
∆ξ = G(p) it follows that ∆ζ = G(p), too. Hence Tζ = Λ ⊃ T01 (m, n), which is not
possible. We have proved
Lemma 12.1. If V ⊂ Z is a B(4; k)-invariant surface, which is not pointwise∆ invariant , then there is no point ξ ∈ V (k), whose inertia group ∆ξ in ∆ has the form
G(a : b : c), where a, b, c 6= 0.

12.3. 1-cycles of proper type 3.
12.3.1. Recalling the restriction morphism h. The ideals I ⊂ OP3 with Hilbert
polynomial Q such that t is not a zero divisor of OP3 /I form an open non empty subset
108

Ut ⊂ HQ , and I 7→ I ′ := I + tOP3 (−1)/tO
P3 (−1)

1 0



0 1
phism h : Ut → H d = Hilbd (P2 ). If Γ := 

0 0


 0 0
in Ut .

defines
the so called restriction mor0 ∗ 


0 ∗
 < ∆, then HΓ is contained
Q
1 ∗


0 1

12.3.2. 1-cycles of proper type 3. We recall, respectively introduce, the notations:
An ideal J ⊂ OP3k with Hilbert polynomial Q corresponds to a point ξ ∈ HQ (k). J has
the type 3, if J is invariant under T (4; k) and G3 = G(1 : 0 : 0), but not invariant under
∆. The curve C = Ga · ξ = {ψα (ξ)|α ∈ k}− in HQ is called the 1-cycle of type 3 defined
by ξ. We say C is a 1-cycle of proper type 3, if I := J ′ = h(J ) has y-standard form (cf.
2.4.3 Definition 2). If ϕ is the Hilbert function of I ↔ ξ ′ ∈ H d (k), then g ∗(ϕ) > g(d)
m
by definition, and one has I = yK(−1) + x
O
 K ⊂ OP2 has the colength
P2 (−m),where
 1 0 ∗ 
′
c and is invariant under T (3; k) and G3 := 0 1 ∗ < U(3; k). Moreover one has


0 0 1
d = m + c and m ≥ c + 2.



T −a+2
T −b+1
If Q(T ) = T −1+3
+
+
is the Hilbert polynomial of J , then the
3
2
1
3
closed subscheme V+ (J ) ⊂ P defined by J has the “complementary” Hilbert polynomial
P (n) = dn − g + 1, where d = a − 1, g = g(J ) = (a2 − 3a + 4)/2 − b (cf. [T1], p. 92).
Lemma 12.2. Let J be of proper type 3 and put ν := min{n|H 0 (J (n)) 6= (0)}. If
xν ∈ H 0 (J (ν)), then g(J ) < 0.
Proof. We start with an ideal J fulfilling these conditions. There are subspaces
Ui ⊂ Si , invariant under T (3; k)·G′3 such that S1 Ui ⊂ Ui+1 , i = 0, 1, 2, . . . and H 0 (J (n)) =
n
L
tn−i Ui , for all n ∈ N. Besides this, Un = H 0 (I(n)), at least if n ≥ b, where I = J ′

i=0

is the restriction ideal of J (cf. [G78], Lemma 2.9, p. 65). We replace Un by H 0 (I(n))
for all ν ≤ n ≤ b − 1 and we get an ideal J ∗ ⊃ J such that H 0(J ∗ (n)) = (0), if
n
L
n < ν, and H 0 (J ∗ (n)) =
tn−i H 0 (I(i)), if n ≥ ν. (N.B.: xν ∈ H 0 (J (ν)) ⇒ xν ∈
0

−→

i=ν

0

Im(H (J (ν)) res H (I(ν)))). Then (J ∗ )′ is equal to I, J ∗ is of proper type 3, the



n−a+2
n−b∗ +1
Hilbert polynomial of J ∗ is equal to Q∗ (n) = n−1+3
+
+
, the length of
3
2
1
∗
∗
∗
∗
2
J /J is equal to Q (n) − Q(n) = b − b ≥ 0, and because of g(J ) = (a − 3a + 4)/2 − b∗
one has g(J ∗ ) ≥ g(J ). Thus it suffices to show g(J ∗ ) < 0 We thus can assume without
n
L
restriction J = J ∗ , i.e. H 0 (J (n)) =
tn−i H ( I(i)) for all n ≥ ν, and xν ∈ H 0(I(ν)).
i=ν

Using the terminology of [T1]–[T4], one can say the pyramid E(J ) is complete up to the
level ν over the platform H 0 (I(n)), where n ≥ b, for instance (cf. Fig. 12.1). Further
note that
(12.1)

H 0 (I(n)) = yH 0(K(n − 1)) ⊕ xm k[x, z]n−m

for all n ∈ N (cf. Lemma 2.6). We associate to I the ideal Ĩ represented in Figure
12.2; that means Ĩ arises from I by shifting yH 0(K(n − 1)) into yH 0(K̃(n − 1)), where
109

the Hilbert functions of K and K̃ agree and hence I and Ĩ have the same Hilbert function
n
L
ϕ. Then J˜ is defined by H 0 (J˜(n)) =
tn−i H 0 (Ĩ(i)) and H 0 (J˜(n)) = (0), if n < ν.
i=ν

Then Ĩ and J˜ fulfil the same assumptions as I and J do, and g(J ) = g(J˜). Thus we
can assume without restriction that I has the shape as represented by Fig. 12.2. Then
•
•
one makes the graded deformations • 7→ 1 (or • → 2 , etc.) in E(J ). Each of the
orbits which are to be exchanged, have the same length. One gets an ideal J˜ with the
same Hilbert polynomial, which is again of proper type 3. If ϕ̃ is the Hilbert function of
Ĩ := (J˜)′ , then ϕ̃ > ϕ, hence g ∗(ϕ̃) > g ∗ (ϕ) > g(d) (cf. Remark 2.1). As J and J˜ have
the same Hilbert polynomial, one has g(J˜) = g(J ). The colength of Ĩ in OP2 is the same
as the colenght d of I in OP2 , hence the coefficient a, remains unchanged. Now one can
again pass to (J˜)∗ and has again xν ∈ H 0((J˜)∗ (ν)), ν = min{n|H 0 ((J˜)∗ (n)) 6= (0)}, (J˜)∗
of proper type 3. Thus it suffices to show g((J˜)∗ ) < 0. Continuing in this way one sees
that one can assume without restriction: J is of proper type 3, I = J ′ has the shape of
n
L
Figure 12.4, H 0 (J (n)) =
tn−i H 0 (I(i)), for all n ≥ ν, H 0 (J (n)) = (0), if n < ν and
i=ν

xν ∈ H 0 (J (ν)), i.e., xν ∈ H 0 (I(ν)). From (12.1) it follows that ν ≥ m. As m ≥ c + 2, we

have h0 (K(n)) = n+2
− c, n ≥ m − 1, hence h0 (I(n)) = h0 (K(n − 1)) + (n − m + 1) =
2




n−1+2
n−1+2
n−1+1
n+2
−
c
+
(n
−
m
+
1)
=
+
+
1
−
(c
+
m)
=
− d, n ≥ m. But then
2
2
1
2



n
n
X
X
i+2
−d
h0 (J (n)) =
h0 (I(i)) =
2
i=ν
i=ν

 X
ν−1 
n 
X
i+2
i+2
− (n − ν + 1)d
−
=
2
2
i=0
i=0

 

ν+2
n+3
− (n − ν + 1)d.
−
=
3
3

From this we get P (n) = (n − ν + 1)d + ν+2
, thus:
3

(12.2)
g(J ) = (ν − 1)d − ν+2
+1.
3
We regard g(J ) as a function of ν ≥ m, and we
1
1
g(x) := (x − 1)d − x3 − x2 −
6
2

have to determine the maximum of
1
x + 1, x ≥ m.
3
q
1 2
1
′
We have g (x) = − 2 x − x + (d − 3 ) = 0 ⇔ x = −1 ± 2d + 31 , and we show
q
m ≥ −1 + 2d + 31 . This last inequality is equivalent to (m + 1)2 > 2d + 31 ⇔ m2 + 23 ≥ 2c
(because of d = c + m). As m ≥ c + 2, this is true.
It follows that g ′(x) ≤ 0, if x ≥ m, hence g(x) is monotone decreasing if x ≥ m. Now
1
1
1
g(m) = (m − 1)d − m3 − m2 − m + 1
6
2
3
1 3 1 2 1
= (m − 1)(c + m) − m − m − m + 1
6
2
3
1 3 1 2 1
≤ (m − 1)(2m − 2) − m − m − m + 1.
6
2
3
110

The right side of this inequality is smaller than 0 if and only if 18 < m3 − 9m2 + 26m,
which is true as m ≥ c + 2 ≥ 2.

12.4. B(4; k)-invariant surfaces containing a 1-cycle of proper type 3.
Let be V ⊂ Z = Z(HQ ) a B(4; k)-invariant surface containing a 1-cycle D of proper
type 3. Then V is not pointwise invariant under the Ga -operation ψα : x 7→ x, y 7→
αx+y, z 7→ z, t 7→ t. According to Lemma 11.1 one can write V = Ga · Gm · ζ. The inertia
group ∆ζ has the form G(p) and by Lemma 12.1 it follows that p = (a : b : c), a, b, c 6= 0,
is not possible.
12.4.1. The case p = (a : 0 : c), a, c 6= 0. Then V is not pointwise invariant under
the Gm -operation σ (notations as in Lemma 11.1), as the following argumentation will
show: Let J ↔ ζ be the corresponding ideal, let P be an associated prime ideal, which is
G(p)-invariant. If t ∈ P, then from (Appendix D, Lemma 2) it follows that P = (x, y, z, t),
contradiction. Thus t is a non zero divisor of OP3 /J . If J would be invariant under σ,
then J would be generated by elements of the form f · tn , where f ∈ Sm , m, n ∈ N. It
follows that f ∈ J and thus J is invariant under Γ (cf. 12.3.1). But by assumption
∆ζ = G(a : 0 : c) 6⊃ Γ.
The inertia group Tζ ⊂ T (4; k) has the form T (ρ), where ρ3 > 0 ( Lemma 11.1 a). By
Appendix E H 0 (J (b)) has a standard basis fi = Mi pi (X ρ ).

Let be W := Ga · Gm · ζ The morphism his defined on V ∩ Ut ⊃ W . As W = V , it
follows that h(W ) = {ψα (ζ ′)|α ∈ k} is dense in h(V ∩ Ut ), where ζ ′ = h(ζ) ↔ J ′ . As
ζ ∈ Ut , it follows that ζ0 = lim σ(λ)ζ ∈ Ut too (see [G88], Lemma 4, p. 542, and [G89],
λ→0

1.6, p.14). As ρ3 > 0 it follows that ζ0′ = ζ ′. Now write D = {ψα (η)|α ∈ k}− , where
η ∈ V (k) is invariant under T (4; k) and G3 , hence C ⊂ V ∩Ut and h(η) ∈ {ψα (ζ ′)|α ∈ k}− .
′
As η is not Ga -invariant , h(η) is not Ga -invariant, hence 
h(η)
∈ k}. Then
 ∈ {ψα (ζ )|α
1 0 ∗ ∗ 





0 1 ∗ ∗
′
′


< G(p) is
(Appendix D, Lemma 3) shows h(η) = ζ = ζ0 . H := 
0 0 1 0



 0 0 0 1 
normalized by σ, hence ζ0 is H-invariant. As ζ0 ∈ Ut is Gm -invariant , it follows that ζ0
is Γ-invariant. But then ζ0 is invariant under G3 , and as ζ0′ = η ′ corresponds to an ideal
in y-standard form, J0 ↔ ζ0 is of proper type 3.

As G(p) is unipotent there is an eigenvector f 6= 0 in H 0 (J (ν)), ν := min{n|H 0(J (n)) 6=
(0)}. From x∂f /∂t ∈ hf i it follows that ∂f /∂t = 0. From y∂f /∂z ∈ hf i it follows
∂f /∂z = 0 and from cx∂f /∂y − az∂f /∂t ∈ hf i we deduce f = xν (cf. Appendix D,
Lemma 2). But then xν ∈ J0 , too. Now h0 (J0 (n)) = h0 (J (n)), n ∈ Z (cf. [G88], Lemma
4, p.542 and [G89], 1.6, p.14). But then from Lemma 12.2 it follows that g < 0.
Conclusion 12.1. In the case p = (a : 0 : c), a, c 6= 0, V does not contain a 1-cycle
of proper type 3, if g > g(d) is supposed.
111

12.4.2. The case p = (a : b : 0), a, b 6= 0. As Tζ ⊂ T (1, −2, 1, 0) (cf. Auxiliary
Lemma 2 ), it follows from Lemma 11.1 that V is pointwise invariant under σ(λ) : x 7→
x, y 7→ y, z 7→ z, t 7→ λt. As V is pointwise invariant under Γ, one has V ⊂ GΦ , where Φ
is the Hilbert function of J ↔ ζ and GΦ is the corresponding “graded Hilbert scheme”.
This had been defined in ([G4], Abschnitt 2) as follows: Gm operates on HQ by σ. Then
it is shown in (loc. cit.) that G := (HQ )Gm ∩ Ut is a closed subscheme of HQ , and G is a
disjunct union of closed subschemes GΦ , where GΦ parametrizes the corresponding ideals
with Hilbert function Φ.
Now suppose D = {ψα (η)|α ∈ k}− ⊂ V is a 1-cycle of proper type 3, where η
corresponds to an ideal L of proper type 3. Then L and J ↔ ζ have the same Hilbert
function.
Let ν and f ∈ H 0 (J (ν)) be defined as in (12.4.1). From x∂f /∂z ∈ hf i and y∂f /∂t ∈
hf i it follows that ∂f /∂z = ∂f /∂t = 0. But then from bx∂f /∂y − ay∂f /∂z ∈ hf i it
follows that ∂f /∂y = 0, hence f = xν . (cf. Appendix D, Lemma 2). If I corresponds to
any point ξ ∈ Ga · Gm · ζ, then xν ∈ H 0 (I(ν)), and the same argumentation as in (8.4.2)
shows this is true for all points in V . But from xν ∈ H 0 (L(ν)) it follows that g < 0.
Conclusion 12.2. In the case p = (a : b : 0), a, b 6= 0, V does not contain a 1-cycle
of proper type 3, if g > g(d) is supposed.

12.4.3. If V contains a 1-cycle of proper type 3, then V cannot be pointwise Ga invariant, so the case p = (0 : b : c) cannot occur, at all.
Lemma 12.3. Suppose g > g(d). If V ⊂ Z(HQ ) is a B(4; k)-invariant surface containing a 1-cycle of proper type 3, then V is pointwise invariant under G3 = G(1 : 0 : 0). 
Fig.: 12.1
c monomials are missing

xm z n−m
112

Fig.: 12.2

xm z n−m
Fig.: 12.3

2
1

xm z n−m

113

Fig.: 12.4

c monomials are missing

xm z n−m

114

CHAPTER 13

Relations in B(4; k)-invariant surfaces
We suppose g > g(d) and let V ⊂ X = Z(HQ ) be a B(4; k)-invariant surface, which
contains a 1-cycle of proper type 3. From it follows that V is pointwise G3 -invariant. Then
from ([T1], Proposition 0, p.3) we conclude that any B-invariant 1-prime cycle D ⊂ V is
either pointwise ∆-invariant or a 1-cycle of type 3. The aim is to describe the relations in
Z1B (X) defined by the standard construction of Section 9.2 carried out with regard to V .
13.1. First case : V is not pointwise invariant under the Gm -operation σ
We use the notation of (11.1). According to Lemma 11.1 V = Gm · Ga · ζ, where
Tζ = T (ρ) and ρ3 > 0, for otherwise ζ would be Gm -invariant and hence V would be
pointwise Gm -invariant. Let J ↔ ζ and C := {ψα (ζ)|α ∈ k}− . If one chooses a standard
basis of H 0 (J (b)) consisting of T (ρ)-semi-invariants, then one sees that h(V ) = C ′ :=
{ψα (ζ ′)|α ∈ k}− , where ζ ′ = h(ζ). As V contains a 1-cycle of proper type 3, C ′ is a
y-standard cycle, generated by ζ ′ ↔ J ′ .

Now if D ⊂ V is a 1-cycle of proper type 3, then h(D) = C ′ . If D is of type 3, but
not of proper type 3, then h(D) is not a y-standard cycle. Write D = Ga · η, η ∈ V (k)
invariant under T (4; k). It follows that η ′ = h(η) is one of the two T (3; k)-fixed points of
C ′ (cf. Appendix D, Lemma 3). If η ′ = ζ ′, then D would be of proper type 3. Hence η ′
is the unique point of C ′ , invariant under B(3; k) and h(D) equals the point η ′ .
If D is pointwise ∆-invariant, then h(D) is pointwise invariant under U(3; k) and hence
equals the point η ′ , too .
Let C0/∞ = lim σ(λ)C be the B-invariant limit curves of C coming from the stanλ→0/∞

dard construction. We write in Z1B (V ):
X
C0 =
mi Ai + Z0

and C∞ =

X

nj Bj + Z∞

where Ai and Bj are the 1-cycles of proper type 3, occurring in C0 and C∞ , respectively,
and mi , nj ∈ N − (0). All the other prime cycles occurring in C0 and C∞ are summed up
in Z0 and Z∞ , respectively.

Let Mn be a tautological line bundle on HQ . Then (Mn · Ai ) = δn + ai , (Ln · Bj ) =
δn + bj . Here δ ∈ N − (0) is independent of i and j, as h(Ai ) = h(Bj ) = C ′ for all i, j,
whereas the constants ai and bj still depend on Ai and Bj , respectively. From [C0 ] = [C∞ ]
P
P
it follows that
mi (δn + ai ) =
nj (δn + bj ) + c for all n ≫ 0, where c depends only
P
P
on Z0 and Z∞ (see Appendix D, Lemma 4) . It follows that
mi = nj . Therefore we
115

can write
(13.1)

C0 − C∞ =

X
k

(Ek − Fk ) +

X

Gk

k

where (E1 , E2 , . . . ) := (A1 , . . . , A1 , A2 , . . . , A2 , . . . ) and (F1 , F2 , . . . ) := (B1 , . . . , B1 , B2 ,
. . . , B2 , . . . ). Here A1 (respectively B1 ) are to occur m1 -times (respectively n1 -times) etc.
P
P
By the way, the arbitrary association Ek 7→ Fk is possible because of
mi =
nj . If
Ek = Fk , the summand Ek − Fk is left out. Gk is composed of either pointwise U(4; k)invariant curves or 1-cycles of type 3, which are not proper.
13.2. Second case: V is pointwise invariant under the Gm -operation σ, but
not pointwise invariant under the Gm -operation τ
We use the same notations as in (11.3).
1st subcase: h(V ) is not 2-dimensional. By assumption V contains a 1-cycle D of proper
type 3, hence h(V ) = h(D) =: D ′ is a y-standard cycle and all the other 1-cycles in V ,
which are of proper type 3, are mapped by h onto D ′ . Then one carries out the standard
construction by means of the operation τ and one gets formally the same relations as
(13.1).
′
2nd subcase: h(V
H d is a B(3; k)-invariant surface, which is pointwise invariant
) = V ⊂

 1 0 ∗ 
under G′3 = 0 1 ∗ < U(3; k). As V ′ contains a y-standard cycle, V ′ is not
 0 0 1 
pointwise Ga -invariant.

At first, the standard construction is carried out in V = Gm · Ga · ζ (cf. Lemma
11.1): C := {ψα (ζ)|α ∈ k}− is a closed curve in V with Hilbert polynomial p and
λ 7→ τ (λ)C defines a morphism Gm → Hilbp (V )Ga , whose extension to P1 defines a
∗
flat family C ֒→ V × P1 over P1 such that Cλ := p−1
2 (λ) = τ (λ)C × {λ}, if λ ∈ k .
C0/∞ =: p1 (C0/∞ ) are called the limit curves of C and [C0 ] − [C∞ ] is the “standard
relation” defined by V .
Put U := {τ (λ)ψα (ζ)|α ∈ k, λ ∈ k ∗ }.

Put U ′ := {τ (λ)ψα (ζ ′ )|α ∈ k, λ ∈ k ∗ }.

Then U = V and U ′ = V ′ . Carrying out the standard construction by means of C ′ :=
p2
{ψα (ζ ′)|α ∈ k}− one gets a flat family C ′ ֒→ V ′ × P1 −→ P1 . One has a morphism
h×id
C ֒→ V × P1 −→ V ′ × P1 , which is denoted by f .
Put U := {(τ (λ)ψα (ζ), λ)|α ∈ k, λ ∈ k ∗ } ⊂ C.

Put U′ := {(τ (λ)ψα (ζ ′), λ)|α ∈ k, λ ∈ k ∗ } ⊂ C ′ .

C and C ′ are reduced and irreducible (see [T1], proof of Lemma 1, p.6). Hence U = C and
U′ = C ′ . As f (U) = U′ and f is projective, f (C) = C ′ follows. As the diagram
C

f

p2 ց

−→
P1
116

ւ p2

C′

′
is commutative, f (C0 ) = C0′ and f (C∞ ) = C∞
follows. As the diagram
h×id

V × P1 −→ V ′ × P1

p1 y
V

h

−→


yp1
V′

′
is commutative, it follows that C′0/∞ := p1 (C0/∞
) = p1 f (C0/∞ ) = h(C0/∞ ). Let be
′
′
ζ0/∞ := lim τ (λ)ζ and ζ0/∞ := lim τ (λ)ζ .
λ→0/∞

λ→0/∞

′
′
Now from Lemma 9.2, it follows that C0/∞
:= {ψα (ζ0/∞
)|α ∈ k}− are the only ystandard cycles in C′0/∞ and they both occur with multiplicity 1. We want to show that
C0/∞ := {ψα (ζ0/∞ )|α ∈ k}− are the only 1-cycles of proper type 3 in C0/∞ , and they both
occur with multiplicity 1. In order to show this, we consider the extension of τ : λ 7→ τ (λ)ζ
to a morphism τ : A1 → V . Then ζ0 = τ (0). As there is a commutative diagram

Gm

τ

−→
τ′ ց

V

yh
V′

where τ ′ (λ) := τ (λ)ζ ′ and as the extensions are determined uniquely, it follows that
τ ′ = h ◦ τ , hence h(ζ0 ) = ζ0′ . By constructing the corresponding extensions to P1 , it
′
′
follows in the same way that h(ζ∞ ) = ζ∞
. Hence we get h(C0/∞ ) = C0/∞
. Therefore C0
and C∞ are 1-cycles of proper type 3, and from ζ0/∞ ∈ C0/∞ we conclude C0/∞ ⊂ C0/∞ .

Assume there is another 1-cycle D of proper type 3 contained in C0 , or assume that
C0 occurs with multiplicity ≥ 2 in C0 . Then there is an equation [C0 ] = [C0 ] + [D] + · · ·
in Z1B (V ), where D = C0 , if C0 occurs with multiplicity ≥ 2. From Lemma 9.2 it follows
that h(D) = C0′ . It follows that
h∗ ([C0 ]) = h∗ ([C0 ]) + h∗ ([D]) + · · ·

= deg(h|C0 )[h(C0 )] + deg(h|D)[h(D)] + · · ·
= h∗ ([C]) = deg(h|C)[C ′ ].

As C = {ψα (ζ)|α ∈ k}− and α 7→ ψα (ζ) is injective and the same is true for C ′ and
α 7→ ψα (ζ ′ ), h|C is an isomorphism. The same argumentation shows that h|C0 and h|C∞
are isomorphisms. Hence we get [C0′ ] + [C0′ ] + · · · = [C ′ ] = [C′0 ], which means that C0′
occurs with multiplicity ≥ 2 in C′0 , contradiction.
The same argumentation shows that C∞ is the only 1-cycle of proper type 3 in C∞ , and
it occurs with multiplicity 1. This gives a relation of the form
(13.2)

C0 − C∞ = C 0 − C ∞ − Z

where C0 and C∞ are 1-cycles of proper type 3 and Z is a 1-cycle whose components are
either pointwise U(4; k)-invariant curves or 1-cycles of type 3, which are not proper.
117

13.3. Third case:V is pointwise invariant under σ and τ
Then we write V = Gm · Ga · ζ as in the 3rd case of Lemma 11.1.

1st subcase : h(V ) is not 2-dimensional. The same argumentation as in the first subcase
of (13.2), using the Gm -operation ω as in the third case of Lemma 11.1 instead of the
Gm -operation τ , gives relations of the form (13.1).
2nd subcase: If h(V ) is 2-dimensional, the same argumentation as in the second subcase
of 13.2 , with ω instead of τ , gives relations of the form (13.2).
Proposition 13.1. Assume g > g(d) and let V ⊂ X := Z(HQ ) be a B(4; k)-invariant
surface containing a 1-cycle of proper type 3. Then each relation in Z1B (X) defined by V ,
which contains a 1-cycle of proper type 3 is a sum of relations of the form C1 − C2 + Z.
Here C1 and C2 are 1-cycles of proper type 3, and Z is a 1-cycle whose prime components
either are pointwise ∆-invariant or 1-cycles of type 3 , which are not proper .
Proof. This follows from the foregoing discussion.

118



CHAPTER 14

Necessary and sufficient conditions
We take up the notations introduced in Chapter 1. Let be d ≥ 5, g(d) = (d −
2) /4, H = Hd,g , A(H) = Im(A1 (HU (4;k) ) −→ A1 (H)). Obviously, the cycle C3 =
{(xa , αx + y, xa−1 z b−a+1 )|α ∈ k}− , where d = a − 1, g = (a2 − 3a + 4)/2 − b, is a 1cycle of proper type 3.
2

14.1. The necessary condition.
In the proof of Theorem 1.1 in Chapter 10 we replace H, U(3; k), B(3; k) and “ystandard cycle” by H, U(4; k), B(4; k) and “1-cycle of proper type 3”, respectively. Then
using Proposition 13.1 instead of Proposition 9.1 , the same reasoning as in the proof of
Theorem 1.1 gives the
Conclusion 14.1. If d ≥ 5 and g > g(d), then [C3 ] ∈
/ A(H).



14.2. The sufficient condition.
14.2.1. The case d ≥ 5 and d odd. In ([T2], 3.3.2, Folgerung, p.129) it had been
shown [C3 ] ∈ A(H), if a ≥ 6 and b ≥ a2 /4. We express this condition by means of the
formulas in (1.1):
1
1
1
b ≥ a2 /4 ⇔ g ≤ [(d + 1)2 − 3(d + 1) + 4] − (d + 1)2 = (d2 − 4d + 3).
2
4
4
1
2
As d is odd, this is equivalent to g ≤ g(d) = 4 (d − 2) .
Conclusion 14.2. If d ≥ 5, d is odd and g ≤ g(d), then [C3 ] ∈ A(Hd,g ).



14.2.2. The case d ≥ 6 and d even. We will show that the bound given in
([T2], 3.3.3 Folgerung, p.129) is already valid, if a ≥ 7.
1◦ We consider the Hilbert function ϕ of the monomial ideal (x2 , xy e−2 , y e), e := d/2 + 1 ≥
4, which is represented in Fig. 14.1. In Section (2.2.3) this Hilbert function had been
denoted by χ and it had been shown that g ∗ (ϕ) = 41 (d−2)2 =: g(d). The figures 14.1 - 14.6
show all possible monomial ideals with Hilbert function ϕ. Besides this, we consider the
ideal I := (xy, xe , y e, xe−1 + y e−1) which corresponds to a closed point ξ in the Iarrobino
variety Iϕ . This variety parametrizes all sequences (U0 , U1 , . . . ) of subspaces Ui ⊂ Ri with
dimension ϕ′ (i) = ϕ(i) − ϕ(i − 1), such that R1 Ui ⊂ Ui+1 , i ∈ N . Here R is the polyno mial ring in two variables .
2◦ We show that V = Ga · Gm · ξ is 2-dimensional, where Gm operates by τ (λ) : x 7→
f

λx, y 7→ y, z 7→ z. If this would not be the case, then Ga × Gm −→ Iϕ defined by
119

(α, λ) 7→ ψα τ (λ)ξ would have a fibre with infinitely many points. The argumentation in
(8.3) then showed that one of the points ξ0/∞ = lim τ (λ)ξ is invariant under Ga . As
λ→0/∞

ξ0 ↔ I0 = (xy, xe , y e−1) and ξ∞ ↔ I∞ = (xy, xe−1, y e ), this is wrong as e ≥ 4. Thus we
can carry out the standard construction with C = {ψα (ξ)|α ∈ k}− , if e ≥ 4. As we have
already noted in the proof of Lemma 9.1 , the curves τ (λ)C are Ga -invariant, hence the
construction of the family C takes place in (Hilbp (Iϕ ))Ga . Therefore the limit curves C0/∞
are invariant under B(3; k).
As usual, we put C0/∞ = {ψα (ξ0/∞ )|α ∈ k}− .
3◦ Let I ↔ (U0 , U1 , · · · ), where Ui = xyRi−2 , if 0 ≤ i ≤ e − 2, Ue−1 = xyRe−3 + hxe−1 +
y e−1i, Ui = Ri , i ≥ e. We get
ψα (Ue−1 ) = x(αx + y)Re−3 + hxe−1 + (αx + y)(αx + y)e−2i

⇒

= x(αx + y)Re−3 + hxe−1 + (αx + y)y e−2i

˙ α (Ue−1 ) = αxe−1 ∧ αxe−2 y ∧ · · · ∧ αx2 y e−3 ∧ αxy e−2
∧ψ
+ terms with smaller powers of α.

Hence α-grade (Ue−1 ) = e − 1. On the other hand, by formula (4.2) in 4.1 we get α-grade
(hxe−2 y, . . . , y e−1i) = e − 1, too. From this it follows that C0 = C0 .
4◦ Besides C∞ , the limit cycle C∞ contains other prime components D1 , D2 , . . . which
are B(3; k)-invariant curves in Iϕ .
As char(k) = 0 is supposed, if m ≤ n, G := Grassm (Rn ) has only one Ga -fixed point,
namely the subspace hxn , · · · , xn−m+1 y m−1 i. Then ([T1], Proposition 0, p.3) shows that
each B(2; k)-invariant curve in G has the form Ga · η, where η ∈ G(k) corresponds to a
monomial subspace of Rn . It follows that each B(3; k)-invariant curve in Iϕ also has the
form Ga · η, where η ∈ Iϕ (k) corresponds to a monomial ideal.
5◦ Figure 14.1 - Figure 14.6 show all possible monomial ideals with Hilbert function ϕ,
and using formula (4.2) we compute the degrees of the corresponding 1-cycles:




e−2
e−2
e−2
deg c2 = e−2
,
deg
c
=
+
(e
−
1),
deg
c
=
2
+
(e
−
1),
deg
c
=
2
+ (e −
3
4
5
2
2
2
2
2), deg c6 = 1.
We see that c2 (respectively c3 ) is equal to C∞ (respectively C0 ). If C0 (respectively
Di ) occurs in C∞ with the multiplicity n (respectively ni ), then from deg C∞ = deg C0 =


e−2
deg C0 it follows that e−2
+
(e
−
1)
=
n
+ n1 deg D1 + · · · where n > 0, ni ≥ 0.
2
2

e−2
1st case: e ≥ 6. Then 2 > e − 1, hence n = 1. From D1 ∈ {c4 , c5 , c6 } and e − 1 ≥
n1 deg D1 we conclude that D1 = c6 and
[C0 ] = [C∞ ] + (e − 1)[c6 ]
2nd case: e = 5. Then deg C0 = 7, deg C∞ = 3, deg c4 = 10, deg c5 = 9, and we get
[C0 ] = 2[C∞ ] + [c6 ] or [C0 ] = [C∞ ] + 4[c6 ].
3rd case: e = 4. Then deg C0 = 4, deg C∞ = 1, deg c4 = 5, deg c5 = 4. Therefore one gets
[C0 ] = n[C∞ ] + m[c6 ]
where n, m ∈ N, n ≥ 1, n + m = 4.
6◦ One has a closed immersion Iϕ → Hd,g(d) defined by I 7→ I ∗ , where I ∗ is the ideal
120

generated by I in OP3 . That means H 0 (P3 ; I ∗ (n)) =

case, one gets an equation

∗
[C0∗ ] = n[C∞
] + m[c∗6 ],

n
L

i=0

tn−i H 0 (P2 ; I(i)), n ∈ N. In any

m, n ∈ N, n ≥ 1.

∗
Now from ([T1], Lemma 5, p.45) it follows that one can deform C0∗ (respectively C∞
) by
a sequence of graded deformations of type 3 (cf. [T1], p.44) modulo A(H) in the cycle C3
(respectively in the zero cycle). The cycle c∗6 is equal to the cycle Dα , α = (d + 2)/2 = e,
which had been introduced in ([T4], p.20f). By ([T4], Lemma 5, p.25) one gets De ≡ D2
modulo A(H), and in ([T4], Abschnitt 3.3, p.25) it had been noted that [D2 ] = [D], where
D is the cycle introduced in (1.1). From ([T4], Satz 1, p.26) it follows [C3 ] ∈ A(Hd,g(d) ).
Applying the shifting morphism (cf. [T3], Folgerung, p.55 and Proposition, p.56) we get:

Conclusion 14.3. If d ≥ 6 is even and g ≤ g(d), then [C3 ] ∈ A(Hd,g ).



14.3. Summary.
Theorem 14.1. Let be d ≥ 5. Then [C3 ] ∈ A(Hd,g ) if and only if g ≤ g(d).



Proof. This follows from Conclusion 14.1- 14.3.

Fig.: 14.2

Fig.: 14.1

c2

0

1

2

3

4 ... ... e

0
121

1

2

3

4 ... ... e

Fig.: 14.3

Fig.: 14.4

c3

0

1

2

3

c4

4 ... ... e

0

1

2

3

4 ... ... e

Fig.: 14.6
Fig.: 14.5

c5
c6

0

1

2

3

4 ... ... e

0

122

1

2

3

4 ... ... e

CHAPTER 15

The case d ≥ 5
We suppose d ≥ 5, i.e. a ≥ 6, and g <

d−1
2



, i.e. g not maximal.

From ([T1], Satz 2, p.91) and ([T4], Proposition 2, p.26) it follows that A1 (Hd,g )/A(Hd,g )
is generated by the so-called combinatorial cycles C1 , C2 , C3 . Using the formulas in (1.1),

one shows that g ≤ γ(d) := d−2
is equivalent to b ≥ 2a − 4.
2
1st case: g > γ(d). This is equivalent to b < 2a − 4. By ([T4], Satz 1, p.26) one has
A(Hd,g ) = hEi. Assume there is a relation q0 [E] + q1 [C1 ] + q2 [C2 ] + q3 [C3 ] = 0, qi ∈ Q.
Computing the intersection numbers with the tautological line bundles Ln by means of
the formulas (1)-(5) in ([T2], p.134f) and the formula in ([T3], Hilfssatz 1, p.50), one
sees that q2 = 0. Put r := 2a − 4 − b and denote the r-times applied shifting morphism by f (see [T3], p.52 ). The images of E, C1 , C3 under f in Hd,γ(d) are denoted
by e, c1 , c3 . By ([T2], 3.2.2 Folgerung, p.124) and ([T3], Anhang 2, p.54f) it follows that
[f (E)] ≡ [e], hf (C1)i ≡ hc1 i and hf (C3 )i ≡ hc3 i modulo A(Hd,γ(d) ). By ([T3], Proposition
4, p. 22) [c1 ] ∈ A(Hd,γ(d) . As γ(d) > g(d) if d ≥ 5, from Theorem 14.1 it follows that
[c3 ] ∈
/ A(Hd,γ(d) ). Hence f (C3 ) is not a single point, hence deg(f |C3 ) 6= 0. Applying f∗
to the relation q0 [E] + q1 [C1 ] + q3 [C3 ] = 0 then gives q3 deg(f |C3 ) · [c3 ] ∈ A(Hd,γ(d) ). If
q3 6= 0 it would follow that [c3 ] ∈ A(Hd,γ(d) ), contradiction. But from q0 [E] + q1 [C1 ] = 0
it follows q0 = q1 = 0 (cf. [T2], 4.1.3).
2nd case: g(d) < g ≤ γ(d). Then b ≥ 2a − 4 and in (1.1) it was explained that in this case
A1 (Hd,g ) is generated by E, D, C2 and C3 . If q0 [E]+q1 [D]+q2 [C2 ]+q3 [C3 ] = 0, then q2 = 0
by ([T2], loc.cit.). If one assumes that q3 6= 0, it follows that [C3 ] ∈ hE, Di ⊂ A(Hd,g ),
contradicting Theorem 14.1. As in the first case we get q0 = q1 = 0.
3rd case: g ≤ g(d). Then A(Hd,g ) = hE, Di, [C1] ∈ A(Hd,g ) and [C3 ] ∈ A(Hd,g )
(see.[T3], Proposition 4, p.22 ; [T4], Satz 1, p.26 ; and finally Theorem 14.1 ). Thus
A1 (Hd,g ) = hE, D, C2 i.

All in all we get: Suppose that d ≥ 5 and g < d−1
, i.e. g is not maximal. Put
2

d−2
2
g(d) := (d − 2) /4 and γ(d) := 2 .

Theorem 15.1. (i) If g > γ(d), then A1 (Hd,g ) is freely generated by E, C1 , C2 , C3.
(ii) If g(d) < g ≤ γ(d), then A1 (Hd,g ) is freely generated by E, D, C2 , C3 .
(iii) If g ≤ g(d), then A1 (Hd,g ) is freely generated by E, D, C2.


N.B.: If g = d−1
, then A1 (Hd,g ) is freely generated by C1 := {(x, y d (αy + z))|α ∈ k}−
2
and C2 := {(αx + y, xd+1 )|α ∈ k}− ([T1], Satz 2, p.91).

123

CHAPTER 16

The cases d = 3 and d = 4
16.1. The case d = 3.



T −4+2
T −b+1
If g is not maximal, i.e., if g ≤ 0, then Q(T ) = T −1+3
+
+
, where
3
2
1
b ≥ 4. If b = 4, then in ([T2], p.137) it had been shown that [C3 ] ∈ A(H). Applying the
shifting morphism , (see [T3] ,p .54 ) then shows that this statement is true for all g < 0
(cf. [T3], Anhang 2, Folgerung, p.55 and Proposition, p.56). By ([T1], Satz, p.91, [T3],
Prop. 4, p.22 and Satz 1, p.26) it follows that A1 (H3,g ) is generated by [E], [D], [C2], if
g ≤ 0. The three cycles are linear independent, as already was noted in Chapter 15.
16.2. The case d = 4



If g is not maximal, i.e., if g < 3, then Q(T ) = T −1+3
+ T −5+2
+ T −b+1
and b ≥ 5.
3
2
1
4
If b = 5, then g = 2, and then A1 (H) = Q , as had been shown in ([T3], Anhang 1).
We now treat the case b = 6, i.e., g = 1. As the condition b ≥ 2a − 4 is fulfilled,
A(H) = hE, Di by ([T4], Satz 1, p.26). The quotient A1 (H)/A(H) is generated by C1 , C2
and further cycles C3 , C4, C5 of type 3 (cf. [T1], Satz 2c, p.92). By ([T3], Proposition 4,
p.22) [C1 ] ∈ A(H), and we are going to simplify the reductions of C3 , C4 , C5 described in
([T3], Abschnitt 8).
16.2.1. The cycle C3 . By definition a cycle of type 3 has the form C = Ga · ξ, where
ξ corresponds to a monomial ideal J with Hilbert polynomial Q, which is invariant under
the subgroup G3 < U(4; k) (cf. Chapter 1). Here Ga operates as usual by ψα : x 7→
x, y 7→ αx + y, z 7→ z, t 7→ t. Let I := J ′ ∈ H 4 (k) be the image of J under the restriction
morphism. If I is Ga -invariant, then I is B(3; k)-invariant, hence deg(Ln |C) = constant
and [C] ∈ A(H) (cf. [T3], Anhang 2, Hilfssatz 2, p.50). Therefore we can assume without
restriction that I is not Ga -invariant. As the colength of I in OP2 is equal to 4, the Hilbert
function of I is equal to one of the functions represented in Figure 2.7a and Figure 2.7b.
In (2.2.1) we had obtained g ∗ (ϕ1 ) = 1 and g ∗ (ϕ2 ) = 3. The Hilbert function ϕ1 leads to
two possible 1-cycles of proper type 3, namely to
F1 := {ψα (ξ1 )|α ∈ k}− ,

F2 := {ψα (ξ2 )|α ∈ k}− ,

ξ1 ↔ (xy, y 2, x3 ),

and

ξ2 ↔ (x2 , y 2).

The Hilbert function ϕ2 leads to different 1-cycles of proper type 3, which however by
means of admissible G3 -invariant deformations all can be deformed modulo A(H) in
C3 = {ψα (ξ3 )|α ∈ k}− , ξ3 ↔ (x5 , y, x4z 2 ).
125

With such admissible G3 -invariant deformations we can deform C3 in the cycle generated
by (x4 , xy, y 2, yz 2 ), and afterwards this cycle can be deformed by the graded deformation
(yz 2 , yz 3 , . . . ) 7→ (x3 , x3 z, . . . ) in the cycle F1 .

We deform F1 into F2 in the following way: Put K := (x3 , x2 y, xy 2, y 3). If L ⊂
R2 = k[x, y]2 is a 2-dimensional subspace, then hx, yiL ⊂ H 0 (K(3)) = R3 and z n L ∩
H 0 (K(n + 2)) = (0) for all n ≥ 0. It follows that the ideal (L, K) ⊂ OP3 has the Hilbert
polynomial Q, and L 7→ (L, K) defines a closed immersion P2 ≃ Grass2 (R2 ) → HQ . Let
hxy, y 2i ↔ η1 ∈ P1 , hx2 , y 2i ↔ η2 ∈ P2 , ℓi := {ψα (ηi )|α ∈ k}− , i = 1, 2. Because of
reasons of degree one has [ℓ1 ] = 2[ℓ2 ] in A1 (P2 ), hence [F1 ] = 2[F2 ] in A1 (HQ ). Now
F2 = {(x2 , xy 2, y 3 , (αx + y)2 |α ∈ k}− = {(x2 , xy 2 , y 3, λxy + µy 2 )|(λ : µ) ∈ P1 } is equal to
the cycle D3 which had been introduced in ([T4], Abschnitt 3.2, p.20). By Lemma 5 in
(loc.cit.) D3 ≡ D2 modulo A(H), where D2 := {(x2 , xy, y 4, λxz 2 + µy 3)|(λ : µ) ∈ P1 } is
the cycle , which had been denoted by D in Chapter 1. It follows that [C3 ] ∈ A(H), and
applying the shifting morphism gives
Conclusion 16.1. [C3 ] ∈ A(H4,g ) for all g ≤ 1.



Remark 16.1. If d = 3 and g > g(3) = 1/4, C3 does not occur at all. If d = 4 and
g > g(4) = 1, then A(H4,2 ) = hEi ([T4], Satz 1, p.26) and hence [C3 ] ∈
/ A(H4,2 ).
16.2.2.

[C4 ] ∈ A(H) is true for arbitrary d and g ([T4], Proposition 2, p.26).

16.2.3. The cycle C5 . In (loc.cit.) [C5 ] ∈ A(H) was shown, too, but the proof
required tedious computations, which can be avoided by the following argumentation.
One has only to treat the cases g = 0 and g = −2, that means, the cases b = 7 and b = 9
([T1], Satz 2c(iii), p.92).
We start with b = 7. Then C5 = Ga · ξ0 , where ξ0 ↔ J0 := (y 2, K), K := (x3 , x2 y, xy 2,
y 3, x2 z, y 2 z). Put J := (x2 + y 2, K) ↔ ξ. As J0 and J have the same Hilbert function, ξ0
and ξ both lie in the same graded Hilbert scheme Hϕ (cf. Appendix A). If Gm operates
by τ (λ) : x 7→ λx, y 7→ y, z 7→ z, t 7→ t, then one sees that indeed ξ0 = lim τ (λ)ξ and
λ→0

ξ∞ := lim τ (λ)ξ ↔ J∞ := (x2 , K). Put C := {ψα (ξ)|α ∈ k}− , C0/∞ := {ψα (ξ0/∞ )|α ∈
λ→∞

k}− . One sees that α-grade H 0 (J (n)) = α-grade H 0 (J0 (n)) = α-grade H 0 (J∞ (n)) + 2,
for all n ≥ 2, hence deg C = deg C0 = deg C∞ + 2, relative to an appropriate imbedding
of Hϕ into PN .
Put V := Gm · Ga · ξ. As ξ is invariant under G := G3 · T23 and as G3 and T23 =
{(1, 1, λ2, λ3 )|λ2 , λ3 ∈ k ∗ } are normalized by Gm · Ga , V is pointwise G-invariant. Let p
be the Hilbert polynomial of C ⊂ PN . Then the standard construction by means of V
takes place in X := Hilbp (V )Ga , as C is Ga -invariant. Thus the limit curves C0/∞ are
pointwise G-invariant and invariant under Ga and Gm , hence they are B(4; k)-invariant.
Now C0/∞ ⊂ C0/∞ and the above computation of the degrees shows that C0 = C0 ,
whereas C∞ except from C∞ has to contain a further 1-cycle Z of degree 2, i.e., C∞
has to contain an irreducible curve F of degree ≥ 1, which is B(4; k)-invariant. As V is
pointwise G-invariant, F is either pointwise U(4; k)-invariant or an 1-cycle of type 3. As
126

deg(Ln |F ) is constant it follows that [F ] ∈ A(H) (see [T3], Anhang 2, Hilfssatz 2, p.50).
It follows that Z ∈ A(H).
From [C5 ] = [C0 ] = [C∞ ] + Z and C∞ = C4 (cf. [T1], Satz 2, p. 91), because of
(16.2.2) it follows that [C5 ] ∈ A(H).

We now treat the case b = 9. Here C5 = Ga · η, where η ↔ (x3 , x2 y, y 2, x2 z 3 ). By
the G3 -admissible deformation y 2 7→ x2 z 2 C5 is deformed in the cycle C5′ := Ga · ξ0 ,
where ξ0 ↔ J0 := (x3 , x2 y, xy 2, y 3 , y 2z, x2 z 2 ). By ([T1], 1.4.4) [C5 ] = q[C5′ ] modulo
A(H). Hence we can argue with C5′ and J0 in the same way as in the case b = 7: Let
K := (x3 , x2 y, xy 2, y 3, x2 z 2 , y 2z 2 ) and J := (x2 z + y 2 z, K) ↔ ξ. If ξ0/∞ := lim τ (λ)ξ ,
λ→0/∞

then ξ0 is as above and ξ∞ ↔ J∞ = (x3 , x2 y, xy 2, y 3, x2 z, y 2 z 2 ).

Obviously, one has the same computation of degree as in the case b = 7 and the
same argumentation shows that [C5′ ] = [C∞ ] modulo A(H). C∞ is deformed by the G3 admissible deformation y 2 z 2 7→ x2 in the cycle Ga · ζ, where ζ ↔ (x2 , xy 2 , y 3, y 2z 3 ) As
this is the cycle C4 , by 16.2.2 we get [C5 ] ∈ A(H).
Conclusion 16.2. [C5 ] ∈ A(H4,g ) if g = 0 or g = −2.



16.2.4. Summary. The results of 16.1 and 16.2 give
Theorem 16.1. (i) If g ≤ 0, then A1 (H3,g ) is freely generated by [E], [D], [C2 ].
(ii) A1 (H4,2 ) ≃ Q4 and if g ≤ 1, then A1 (H4,g ) is freely generated by [E], [D], [C2 ].

127



CHAPTER 17

Correction of the results of [T4] and summary
In [T4] the computation of the degree on page 28 is wrong, and hence Conclusions 1-3
and Proposition 3 on the pages 29-32 are wrong. As well ([T4], Lemma 7, p.33) is wrong.
The statement of ([T4], Proposition 4, p.33) is right, but with regard to Theorem 15.1(i) it
is irrelevant. The statement in ([T4], Satz 2, p.35) is wrong and has to be replaced by the
statements of Theorem 15.1 and Theorem 16.1 , respectively. The results of the sections 8
and 9 in [T4] remain valid, if one replaces the old bound by the bound g(d) = (d − 2)2 /4.
Furthermore, ([T4], Satz 3, p.35) has to be replaced by:
Theorem 17.1. Let be d ≥ 3 and g not maximal, let C be the universal curve with
degree d and genus g over Hd,g .

(i) If g > γ(d) := d−2
, then A1 (C) is freely generated by E ∗ , C1∗ , C2∗ , C3∗ and L∗ .
2
(ii) If g(d) < g ≤ γ(d), then A1 (C) is freely generated by E ∗ , D ∗ , C2∗ , C3∗ and L∗ .
(iii) If g ≤ g(d), then A1 (C) is freely generated by E ∗ , D ∗ , C2∗ and L∗ .
The statements of ([T4], Satz 4 and Satz 5, p. 36) are correct, if the bound mention
there is replaced by g(d) = (d − 2)2 /4. The reason is that the arguments used in (loc.cit.)
formally do not depend on the bound.
All in all, one gets the results which had been stated in the introduction.
Concluding Remark: Having arrived at this point, it is not so difficult any more to
explicitly determine the cone of curves and the ample cone of Hd,g (and of C).

129

APPENDIX A

Notations
The ground field is C; all schemes are of finite type over C; k denotes an extension
field of C. P = k[x, y, z, t], S = k[x, y, z], R = k[x, y] are the graded polynomial rings.
T = T (4; k) group of diagonal matrices
∆ = U(4; k) unitriangular group
B = B(4; k) Borel group
T (ρ)
subgroup
of T
(3;k) or of T (4; k) (cf. 2.4.1 and [T1], p.2).

1
0
0
∗







0
1
0
∗


< U(4; k)
Γ= 

 0 0 1 ∗

 0 0 0 1 
G1 , G2 , G3 subgroups of U(4; k) (cf. 1.1)
H = Hd,g Hilbert scheme of curves in P3 with degree d ≥ 1 and genus g, i.e. H =
Hilbp (P3k ), where P (T ) = dT − g + 1.

Q(T ) = T +3
− P (T ) complementary Hilbert polynomial
3
HQ = Hilbert scheme of ideals I ⊂ OP3 with Hilbert polynomial Q(T ), i.e. H = Hd,g =
HQ .





HQ 6= ∅ if and only if Q(T ) = T −1+3
+ T −a+2
or Q(T ) = T −1+3
+ T −a+2
+ T −b+1
,
3
2
3
2
1
where a and b are natural numbers 1 ≤ a ≤ b. The first case is equivalent with d = a
and g = (d − 1)(d − 2/2, i.e., equivalent with the case of plane curves. We consider
only the case g < (d − 1)(d − 2)/2. In this case we have the relations d = a − 1 and
g = (a2 − 3a + 4)/2 − b.
It was not possible to reserve the letter d for denoting the degree of a curve. If
necessary d denotes a number large enough, e.g. d ≥ b = bound of regularity of all ideals
in OP3 with Hilbert polynomial Q (cf. [G1], Lemma 2.9, p.65).
G = Grassm (Pd ) Grassmann scheme of m-dimensional subspaces of Pd .

Let ϕ : N → N be a function with the following properties: There is an ideal I ⊂ OP2
of finite colength with Hilbert function h(n) = h0 (I(n)), such that 0 ≤ ϕ(n) ≤ h(n) for
all n ∈ N and ϕ(n) = h(n) for n ≥ d, where n is large enough, e.g. n ≥ d := colength(I).
On the category of k-schemes a functor is defined by
Hϕ (Spec A) = {(U0 , · · · , Ud )|Un ⊂ Sn ⊗ A subbundle of rank ϕ(n) such that S1 Un−1 ⊂
Un , 1 ≤ n ≤ d}
Hϕ is a closed subscheme of a suitable product of Grassmann schemes; it is called
graded Hilbert scheme.

131

To each ideal J ⊂ OP3k with Hilbert polynomial Q corresponds a point ξ ∈ H(k),
which we denote by ξ ↔ J .
h(J ) denotes the Hilbert function of J , that means h(J )(n) = dimk H 0 (J (n)), n ∈ N.

If ϕ is the Hilbert function of an ideal in OP2k of colength d, then
Hϕ := {I ⊂ OP2k |h0 (I(n)) = ϕ(n), n ∈ N}

is a locally closed subset of Hilbd (P2 ), which we regard to have the induced reduced scheme
structure.
If G is a subgroup of GL(4; k), then HG denotes the fixed-point scheme, which is to
have the induced reduced scheme structure. The same convention is to be valid for all
fixed-point subschemes of H d = Hilbd (P2 ) .
If C ֒→ H is a curve, then by means of the Grothendieck-Plücker embedding H −→ PN
we can regard C as a curve in a projective space, whose Hilbert polynomial has the form
deg(C) · T + c. Here deg(C) is defined as follows : If I is the universal sheaf of ideals
on X = H × P3k , then F := OX /I is the structure sheaf of the universal curve C over
H, and the direct image π∗ (F (n)) is locally free on H of rank P (n) for all n ≥ b. The
˙ ∗ (F (n)) are called the tautological line bundles on H,which are
line bundles Mn := ∧π
very ample and thus define the Grothendieck - Plücker embeddings in suitable projective
spaces. Here ∧˙ is to denote the highest exterior power. Then deg(C) is the intersection
number deg(Mn |C) := (Mn · C). (If C is a so called tautological or basis cycle one can
compute this intersection number directly, see [T2], Section 4.1.)
After these more or less conventional notations we introduce some notations concerning
monomial ideals. If J ⊂ OP3 is T -invariant, then H 0 (P3k ; J (d)) ⊂ Pd is generated by
monomials . To each monomial xd−(a+b+c) y a z b tc in H 0 (J (d)) we associate the cube [a, a +
1] × [b, b + 1] × [c, c + 1] in an y − z − t - coordinate system, and the union of these
cubes gives a so called pyramid, which is denoted by E(J (d)). Usually we assume that
d
L
J is invariant under ∆ or Γ. Then we can write H 0 (J (d)) =
td−n Un , where Un ⊂ Sn
n=0

are subspaces such that S1 · Un ⊂ Un+1 , 0 ≤ n ≤ d − 1, which we call the layers of the
pyramid. (In [T1]–[T4] we made extensive use of this concept, but here it occurs only
once in 12.3.3.)
A1 (−) denotes the group of rational equivalence classes with coefficients in Q.

132

Fig.: A.1

1
0
0

1

2 ... ... ... ... ... ...

133

APPENDIX B

Hilbert functions without Uniform Position Property
Lemma B.1. Let be k be an algebraically closed field, I ⊂ OP2k an ideal of finite
colength with Hilbert function ϕ and difference function ϕ′ (n) = ϕ(n) − ϕ(n − 1). Let m
be a natural number such that ϕ′ (m + 1) = ϕ′ (m) + 1. The ideal J ⊂ OP2k generated by
H 0 (I(m)) has the following properties:
(i) J is m-regular;
(ii) H 0 (J (n)) = H 0 (I(n)) for all n ≦ m + 1;
(iii) If δ := m + 1 − ϕ′ (m) > 0, then there is a form f ∈ Sδ and an ideal L ∈ OP2 of finite
colength such that J = f · L(−δ).
Proof. Let be In := H 0 (I(n)), I :=

∞
L

In . The ideal I is called Borel-normed, if

n=0

in(I) is invariant under B(3; k), where in(I) is the ideal generated by the initial monomials
of all forms in I. According to a theorem of Galligo, there is a g ∈ GL(3; k) such
that g(I) is Borel-normed. (In [G4], Anhang IV, in the special case of three variables,
there is an ad-hoc-proof.) Therefore we can assume without restriction that I is Borelnormed . Then Fig.A.1 shows not only the graph of ϕ′ , but also the monomials in
H 0 (I0 (n)), where I0 := [in(I)]∼ (cf. [G4], Anhang V, Hilfssatz 1, p.116). One sees that
S1 in(Im ) = in(Im+1 ) and this implies the statements (i) and (ii) (cf. loc.cit., Lemma, p.
116 or [Gre], Proposition 2.28, p. 41).
Let ψ be the Hilbert function of J . Then ψ is also the Hilbert function of J0 = in(J )
(cf. [G4], Hilfssatz 1, p.114), and the further development of the graph of ψ ′ is marked
by · · · in Fig. A.1. The line l : y = x − δ + 1 is marked by - - - - . If c is the number

of monomials between the graphs of ψ ′ and l, then ψ(n) = n−δ+2
− c, n ≥ m. Then the
2

n+2
Hilbert polynomial of OP2 /J is equal to p(n) = 2 − ψ(n) = δn + 1.5δ − 0.5δ 2 + c.
Hence V+ (J ) ⊂ P2k is 1- codimensional and there is an irreducible component which is
equal to a hypersurface V (f ), f ∈ Sν an irreducible form (Hauptidealsatz of Krull). From
J ⊂ Rad(J ) ⊂ (f ) it follows J = f · K(−ν), where K ⊂ OP2 is an ideal with Hilbert

function χ(n) := ψ(n + ν) = n−(δ−ν)+2
− c. If δ − ν > 0, one argues with K in the same
2
way as with J , and one finally gets the statement (iii) .

Lemma B.2. Let the assumptions and the notations be as before. Then reg(I) =
min { n ∈ Z | ϕ′ (n) = n + 1 }.
Proof. As in the proof of Lemma 1, we can assume that I is Borel-normed. We let
2
Gm operate on S by σ(λ) : x 7→ x, y 7→ λg y, z 7→ λg z, where g is a high enough natural
number. Then lim σ(λ)I is equal to the ideal I0 as in the proof of Lemma 1, I and I0
λ→0

have the same Hilbert function, and reg(I) = reg(I0 ) ( cf. [Gre], Theorem 2.27). Hence
135

one can assume without restriction that I is monomial. But then the statement follows
from ([T1], Anhang 2), for instance.


136

APPENDIX C

Ideals with many monomials
If k is a field, let be S = k[x1 , . . . , xr , t] and R = k[x1 , . . . , xr ]. Gm operates on S by
σ(λ) : xi 7→ xi , 1 ≤ i ≤ r, and t 7→ λt, λ ∈ k ∗ . Let H be the Hilbert scheme of ideals

− Q(n) is
I ⊂ OPr with Hilbert polynomial Q, i.e., H = HilbP (Prk ), where P (n) = n+r
r
r
the complementary Hilbert polynomial of the subscheme V+ (I) ⊂ P . We suppose that
H is not empty. Then the ideals I ⊂ OPr with Hilbert polynomial Q, for which t is a
non-zero divisor of OP r /I, form an open, non-empty subset Ut ⊂ H.
If K/k is an extension field and if I ∈ H(K), then the limit ideals I0/∞ := lim σ(λ)I
λ→0/∞

are in H(K) again, and if I ∈ Ut , then I0 ∈ Ut , too (cf. [G2], Lemma 4). We say that I
fulfils the limit condition, if I∞ ∈ Ut .
Remark C.1. If I is fixed by the subgroup Γ : xi 7→ xi , t 7→ α1 x1 + · · · + αr xr + t of
U(r + 1; k), then I does fulfil the limit condition (cf. [G2], proof of Lemma 3, p. 541).
If I ∈ Ut , then I ′ := I + tOPr (−1)/tOPr (−1) can be regarded as an ideal in OPr−1
with Hilbert polynomial Q′ (T ) = Q(T ) − Q(T − 1).
C.0.5. Lemma. Let I ∈ H(k) ∩ Ut be an ideal which fulfils the limit condition.
(i) If d ≥ max(reg(I0 ), reg(I∞ )), then H 0 (Prk , I(d)) ∩ Rd has the dimension Q′ (d).
(ii) If d ≥ reg(I ′ ) and H 0 (I(d))∩Rd has a dimension ≥ Q′ (d), then d ≥ max(reg(I0 ), reg(I∞ )).
Proof. (i) There is a basis of M := H 0 (I(d)) of the form gi = tei gi0 + tei −1 gi1 + · · · ,
such that 0 ≤ e1 ≤ e2 ≤ · · · ≤ em , m := Q(d), gij ∈ R and gi0 ∈ Rd−ei , 1 ≤ i ≤ m, linear
independent. Then M∞ := lim σ(λ)M = h{tei gi0 |1 ≤ i ≤ m}i (limit in Grassm (Sd )) has
λ→∞

the dimension m. As d ≥ reg(I∞ ) by assumption, it follows that Q(d) = h0 (I∞ (d)), and
L 0
hence M∞ = H 0 (I∞ (d)). Now t is a non-zero divisor of S/
H (I∞ (n) by assumption,
n≥0

Thus it follows that H 0 (I∞ (n)) = h{tei −(d−n) gi0 |ei ≥ d − n}i for all 0 ≤ n ≤ d. If n = d−1
one gets H 0 (I∞ (d −1)) = h{tei −1 gi0 |ei ≥ 1}i, hence Q(d −1) = |{i|ei ≥ 1}|. It follows that
Q′ (d) = |{i|ei = 0}|. Thus M ∩ Rd ⊃ h{gi0|ei = 0}i has a dimension ≥ Q′ (d). Because
of reg(I ′ ) ≤ reg(I) one has h0 (I ′ (d)) = Q′ (d) and the canonical restriction mapping
ρd : M = H 0 (I(d)) −→ H 0 (I ′ (d)) is injective on M ∩ Rd . It follows that the dimension
of M ∩ Rd can not be greater than Q′ (d).
(ii) From the exact sequence
(1)

·t

0 −→ I(−1) −→ I −→ I ′ −→ 0
137

it follows that H i (I(n − i)) = (0) if i ≥ 2 and n ≥ e := reg(I ′ ) (see [M], p.102). The
sequence
ρ

d
0 −→ H 0 (I(d − 1)) −→ H 0 (I(d)) −→
H 0 (I ′ (d)) −→ H 1 (I(d − 1)) −→ H 1 (I(d)) −→ 0

is exact as d ≥ e, where ρ is induced by the canonical restriction mapping S −→
S/tS(−1) = R. As ρd is injective on H 0 (I(d)) ∩ Rd and h0 (I ′ (d)) = Q′ (d), it follows
from the assumption that ρd is surjective. From the e - regularity of I ′ it follows that
R1 H 0 (I ′ (n)) = H 0 (I ′ (n + 1)), for all n ≥ e. Hence ρn is surjective for all n ≥ d. Hence
0 −→ H 1 (I(n − 1)) −→ H 1 (I(n)) −→ 0 is exact for all n ≥ d, thus H 1 (I(n − 1)) = (0)
for all n ≥ d. It follows that reg(I) ≤ d. One again has the exact sequences:
(2)

·t

′
0 −→ I0/∞ (−1) −→ I0/∞ −→ I0/∞
−→ 0

As (I ′ )0/∞ = (I0/∞ )′ ⊃ I ′ and all these ideals have the Hilbert polynomial Q′ , it follows
that (I ′ )0/∞ = I ′ . As H 0 (I(d)) ∩ Rd is fixed by σ(λ), it follows that H 0 (I(d)) ∩ Rd ⊂
H 0 (I0/∞ (d)). Then one argues as before, using (2) instead of (1).

Remark C.2. Let I ⊂ OP2 be an ideal of colength d, let be S = k[x, y, z], R = k[x, y],
and let Gm operate by σ(λ) : x 7→ x, y 7→ y, z 7→ λz . We assume I to be invariant under
Γ (see above). As d ≥ reg(I) for all ideals I ⊂ OP2 of colength d, the assumption of part

(i) of the lemma is fulfilled, hence H 0 (I(n)) ∩ Rn has the dimension Q′ (n) = n+1
for all
1
n ≥ d and therefore:
H 0 (I(n)) ⊃ Rn

(3)

for all n ≥ d.

This inclusion has been used in the text for several times, e.g. in Section (2.2).
Remark C.3. Let I ⊂ OP2 be an ideal of finite colength, with Hilbert function ϕ,
which is invariant under Γ · T (ρ). Let Gm operate on S by σ(λ) : x 7→ x, y 7→ y, z 7→ λz. If
in(I) is the initial ideal with regard to the inverse lexicographical order, then in(I) is equal
to the limit ideal I0 = lim σ(λ)I. As h0 (I0 (n)) = ϕ(n), it follows that in(H 0 (I(n))) =
λ→0

H 0 (I0 (n)), for all n ∈ N (cf. Appendix E and [G2], Lemma 3 and Lemma 4, pp. 541).
Thus the number of the initial monomials and of the monomials in H 0 (I0 (n)), which are
represented in our figures, can be determined by means of the Hilbert function, alone.

138

APPENDIX D

Unipotent groups acting on polynomial rings
Lemma D.1. The 5-dimensional subgroups of ∆ = U(4; k) have the form



1 α ∗ ∗







0
1
β
∗


G(p) := 
aα + bβ + cγ = 0
0 0 1 γ




 0 0 0 1


where p = (a : b : c) ∈ P2 (k) is uniquely determined.


∗ 


∗
 ⊂ ∆ is a normal subgroup. Let G be a 5-dimensional
0

1 
−→ ∆/N 
is 
an injective homomorphism
and ∆/N ≃ G3a .

1 0 x y







0
1
0
z
 x, y, z ∈ k, ax + by + cz = 0
First case: dim G∩N = 2. Then G∩N = 
0 0 1 0





 0 0 0 1


1 α x y 





0 1 β z
2


, where
where (a : b : c) ∈ P (k) is a suitable point. It follows that G = 
0 0 1 γ 



 0 0 0 1 
α,
β, γ are any element

 of k, and x, y, z ∈ k have to fulfil the conditions noted above. If
′
′
′
1
α
x
y





′
′ 
0
1
β
z


0 0 1 γ ′  is any other element of G, then



 0 0 0 1 

1



0
Proof. N := 
0


 0

0 ∗
1 0
0 1
0 0
subgroup of ∆. Then G/G ∩ N

a(x′ + αβ ′ + x) + b(y ′ + αz ′ + xγ ′ + y) + c(z ′ + βγ ′ + z) = 0.

As α, β, γ, α′, β ′ , γ ′ are any elements of k, we conclude that a = b = c = 0, contradiction.
Second case: N ⊂ G. Then G/N ֒→ G3a is 2-dimensional, and one concludes from this
that G has the form noted above and that p ∈ P2 (k) is uniquely determined. Furthermore
it is easy to see that G(p) is a subgroup.

Lemma D.2. Let be P = k[x, y, z, t], let V ⊂ P be a subspace which is invariant under
G(p). If f ∈ P is invariant under G(p) modulo V , then the polynomials
x∂f /∂z, y∂f /∂t, x∂f /∂t, bx∂f /∂y − ay∂f /∂z, cx∂f /∂y − az∂f /∂t, cy∂f /∂z − bz∂f /∂t all
lie in V .
139



1 α 0 0
0 1 β 0 
r s m n

Proof. If g = 
0 0 1 γ  ∈ G(p) and M = x y z t , then g(M) − M =
0 0 0 1
r
s
m
x (αx+y) (βy +z) (γz +t)n −M = αsxr+1 y s−1z m tn +βmxr y s+1z m−1 tn +γnxr y s z m+1 tn−1
+ terms containing αi , β i , γ i , where i ≥ 2.
If g(f )−f ∈ V , then it follows that αx∂f /∂y +βy∂f /∂z +γz∂f /∂t+ terms containing
α , β i , γ i , where i ≥ 2, lies in V .
First case: c 6= 0. Then γ = −(aα/c + bβ/c), hence αx∂f /∂y + βy∂f /∂z − (aα/c +
bβ/c)z∂f /∂t + terms containing α2 , αβ, β 2 · · · ∈ V . It follows that α(cx∂f /∂y−az∂f /∂t)+
β(cy∂f /∂z − bz∂f /∂t)+ terms containing α2 , αβ, β 2, · · · ∈ V . Put α = 0 and β = 0,
respectively. It follows that cy∂f /∂z − bz∂f /∂t ∈ V and cx∂f /∂y − az∂f /∂t ∈ V . Multiplication by a and b, respectively, and then subtracting the two polynomials from each
other gives c(bx∂f /∂y − ay∂f /∂z) ∈ V . As c 6= 0, the last three statements of the assertion follow.
Second case : c = 0, b 6= 0. Then we can choose γ ∈ k arbitrarily and −aα/b = β.
It follows that αx∂f /∂y − (aα/b)y∂f /∂z + γz∂f /∂t + terms containing α2 , γ 2 , · · · ∈ V .
Putting α = 0 gives z∂f /∂t ∈ V . Putting γ = 0 gives bx∂f /∂y − ay∂f /∂z ∈ V .
Third case: b = 0, c = 0. Then a = 1 and β and
γare any elements
 of k, whereas α = 0.


 1 0 ∗ ∗ 

0 1 0 ∗


⊂ G(p), the same reaThis gives y∂f /∂z and z∂f /∂t ∈ V . As N = 
0 0 1 0



 0 0 0 1 
i

soning as in the proof of ([T2], Hilfssatz 1, p. 142) shows that x∂f /∂z, x∂f /∂t, y∂f /∂t ∈
V.


Lemma D.3. Let I ⊂ OP2 be a monomial ideal of colength d > 0 and let ξ be the
corresponding point in H d (k). If ξ is not invariant under the Ga -operation ψα : x 7→
x, y 7→ αx + y, z 7→ z, then C := Ga · ξ contains exactly two fixed points under T (3; k),
namely the point ξ and the Ga - fixed point ψ∞ (ξ) := lim ψα (ξ).
α−→∞


Proof. Embedding H d in Grassq (Sd ), where q := d+2
−d, one sees that it is sufficient
2
to prove the corresponding statement for a T (3; k)-invariant q - dimensional subspace
d
L
U ⊂ Sd and the corresponding point in Grassq (Sd ). As one can write U =
z d−i Ui ,
i=0

where Ui ⊂ Ri , is a subspace, it suffices to prove the corresponding statement for an
r-dimensional subspace U ⊂ Rn , which is invariant under T (2; k), but not invariant under
Ga . As lim ψα (U) is a Ga - invariant subspace and as char(k) = 0, it follows that this
α−→∞

subspace is equal to hxn , xn−1 y, . . . , xn−r+1 y r−1i. It follows that C has two fixed-points
under T (2; k). In order to prove there are no more fixed-points, it suffices to show the
following: If there is an element α 6= 0 in k such that ψα (U) is T (2; k)-invariant, then
ψα (U) is T (2; k)-invariant for all α 6= 0. If one takes a monomial M = xn−r y r ∈ U, then
ψα (M) = xn−r (αx + y)r ∈ ψα (U). As this is a monomial subspace by assumption, it
140

follows that xn−r xi y r−i ∈ ψα (U), 0 ≤ i ≤ r. Thus one has M ∈ ψα (U) and it follows
that U = ψα (U), But this implies ψnα (U) = U for all n ∈ N and thus ψα (U) = U for all
α ∈ k.

Lemma D.4. Suppose C ⊂ HQ is a B-invariant, irreducible, reduced curve, which is
pointwise invariant under Γ such that the image of C under the restriction morphism h
is a single point. Then (Mn · C) is constant for n ≫ 0 .
Proof. From ([T1], Proposition 0, p.3) it follows that C is either pointwise ∆invariant or a 1-cycle of type 3. We consider the first case. Then C = {σ(λ)ξ|λ ∈ k ∗ }− ,
where σ is a suitable Gm -operation and ξ ∈ H(k) corresponds to a ∆-invariant ideal J .
Let be I = h(J ) and I ∗ the ideal on P3 ,which is generated by I (cf. 1.2.2). Then
n

H 0 (J (n)) ⊂ ⊕ tn−i H 0 (I(i)) for all n (cf. [G5], Hilfssatz 3.2, p.295). Now the assertion
i=0

follows, as I is fixed by σ.
In the second case on has C = {ψα (ξ)|α ∈ k}− , where ψα is the usual Ga -operation
and ξ ∈ H(k) corresponds to a monomial ideal J . Then one argues as in the first case,
with ψα instead of σ.


141

APPENDIX E

Standard bases
Let k be an extension field of C. If ρ = (ρ0 , . . . , ρr ) ∈ Zr+1 − (0), then T (ρ) :=
{ λ = (λ0 , · · · , λr ) | λi ∈ k ∗ and λρ := λρ00 · · · λρr r = 1 } is a subgroup of dimension r of
Gr+1
m .
Auxiliary lemma: If σ, τ ∈ Zr+1 − (0) such that T (σ) ⊂ T (τ ), then there is an integer
n such that τ = n · σ.
Proof. Write σ = (a0 , · · · , ar ), τ = (b0 , · · · , br ) . As the dimension of T (σ) is equal
to r, there is an index i such that ai 6= 0 and bi 6= 0.Without restriction one can assume
that a0 6= 0 and b0 6= 0 . Choose p, q ∈ Z such that pa0 = qb0 and (p, q) = 1. Then
1 −qb1
r −qbr
λpa
· · · λpa
= 1 for all λ ∈ T (σ) follows. Because of dim T (σ) = r one gets
1
r
pai − qbi = 0, 0 ≤ i ≤ r, and thus σ = qρ, τ = pρ, where ρ ∈ Zr+1 − (0) is a suitable
vector. If ǫ is any q-th root of unity in C, one can choose λ ∈ (C∗ )r+1 such that λρ = ǫ.
From λqρ = λσ = 1 it follows that ǫp = λpρ = λτ = 1, too, and q = 1 follows.

We let Gr+1
operate on S = k[X0 , · · · , Xr ] by Xi 7→ λi Xi . If ρ = (ρ0 , . . . , ρr ), then
m
ρ0
ρ
ρr
X := X0 · · · Xr .
Lemma E.1. Let V ⊂ Sd be a T (ρ)-invariant subspace. Then V has a basis of the
form fi = mi · pi (X ρ ), where the mi are different monomials, pi is a polynomial in one
variable with constant term 1 and mi does not appear in mj · pj (X ρ ) if i 6= j.
Proof. By linearly combining any basis of V one obtains a basis fi = mi + gi , where
the mi are different monomials, each gi is a sum of monomials, each of which is greater
than mi in the inverse lexicographic order, and mi does not appear in gj . If g ∈ T (ρ),
then g(fi ) contains the same monomials as fi and from g(fi ) ∈ h{fi }i we conclude that
each fi is a semi-invariant, i.e, hg(fi )i = hfi i for all g ∈ T (ρ) .
P
P
ai λαi X αi =
Now let be f =
ai X αi any T (ρ)-semi-invariant. Let be λ ∈ T (ρ). Then
P
α (0)
α (r)
c(λ) · ai X αi , where λαi := λ0 i . . . λr i . It follows that λαi = c(λ), if ai 6= 0, and
therefore λαi −αj = 1 for all i, j, such that ai 6= 0 and aj 6= 0. Thus T (ρ) ⊂ T (αi − αj ), and
the Auxiliary lemma gives αi − αj = nij ρ, nij ∈ Z, if ai 6= 0 and aj 6= 0. One sees that
P
there is an exponent α0 ∈ {αi } and natural numbers ni , such that f = X α0 · ai X ni ρ . 
Corollary E.1. Let V ⊂ Sd be a m-dimensional subspace, let x ∈ Grassm (Sd ) be
the closed point of Grassm (Sd ) defined by V . If the orbit T · x has the dimension 1, then
the inertia group Tx of x has the form T (ρ), where ρ ∈ Zr+1 − (0).
143

Proof. This follows by similar argumentations as before (see[T2], Hilfssatz 7,p.141).


144

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