arXiv:1604.00866v1 [math.AG] 4 Apr 2016
ACM SHEAVES ON THE DOUBLE PLANE
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Abstract. We show that any aCM vector bundle of arbitrary rank on the
double plane is a direct sum of line bundles. Then we give classification of
aCM sheaves of rank at most 3/2 on the double plane and describe the family
of isomorphism classes of them.
1. Introduction
Ever since Horrocks proved that a vector bundle on the projective space splits as
the sum of line bundles if and only if it has no intermediate cohomology, there have
been two directions of study: one is to find out splitting criterion of coherent sheaves
on a given projective variety X ⊂ Pn , and the other is to classify indecomposable
coherent sheaves that have no intermediate cohomology on X, i.e. H i (X, E(t)) = 0
for any t ∈ Z and i = 1, . . . , dim X − 1; they are called arithmetically CohenMacaulay (for short, aCM). About the former direction, the case of hyperquadrics
was studied in [18]. Madonna (cf. [16]) and Kumar, Rao and Ravindra (cf. [13, 14])
focused on splitting criteria for vector bundles, usually of low rank, on hypersurfaces
of higher degree. About the latter direction, the classification of aCM vector bundles
has been done for several projective varieties such as smooth quadric hypersurfaces
(cf. [11, 12]), cubic surfaces (cf. [4, 8]), prime Fano threefolds (cf. [15]) and
others. In fact, in [7] a complete list of varieties supporting a finite number of aCM
sheaves is provided. Varieties that only support one dimensional families of aCM
bundles, tame varieties, are known by the classical work of Atiyah (elliptic curves,
cf. [1]) and much more recently by work of Faenzi and Malaspina (rational scrolls
of degree four, cf. [9]). Finally, it has been shown that the rest of aCM projective
varieties which are not cones have a wild behaviour, namely they support families
of arbitrary dimension of aCM sheaves (cf. [10]).
In this paper we work on the classification of aCM sheaves on the double plane
X = 2H, i.e. the projective plane H ⊂ P3 with multiplicity two over an algebriacally closed field of characteristic 0. It is known from [6] that any aCM sheaf E
on X admits a locally free resolution of length one, which corresponds to a matrix
factorization [2]. Using the resolution, we observe that the restriction of any aCM
vector bundle on X to H is also aCM and so splits. Our first result is on the
splitness of aCM vector bundle on X of any rank.
2010 Mathematics Subject Classification. Primary: 14F05; Secondary: 13C14, 16G60.
Key words and phrases. arithmetically Cohen-Macaulay, double plane.
The first and third authors are partially supported by MIUR and GNSAGA of INDAM (Italy).
The second author is supported by Basic Science Research Program 2015-037157 through NRF
funded by MEST. The third author is supported by the framework of PRIN 2010/11 ‘Geometria
delle varietà algebriche’, cofinanced by MIUR. The forth author is supported by a FY2015 JSPS
Postdoctoral Fellowship. The third and forth authors thank Sungkyunkwan University for the
warm hospitality.
1
2
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Theorem 1.1. Every aCM vector bundle on X splits.
Indeed we prove a stronger version of splitness of any aCM vector bundle on an ndimensional quadric hypersurface of corank at least three in Pn+1 with n ≥ 2, using
inductive argument. Then we focus our attention to general aCM sheaves on X of
rank at most 3/2. In general the rank need not to be integer (see Definition 2.3).
Our technical ingredient is to twist a given aCM sheaf E by OX (t) with t ∈ Z so
that E is 0-regular, but not (−1)-regular. It guarantees the existence of a non-zero
map u : E → IA (1) for a closed subscheme A that is cut out scheme-theoretically
in X by linear equations. Thus we have candidate for a possible subscheme A and
describe ker(u) in each cases. The structure sheaf OH of H can be shown to be the
unique aCM sheaf of rank 1/2 up to twist. For higher rank we introduce the notion
of a layered sheaf which admits a filtration whose successive quotient is isomorphic
to OH up to twist. It turns out that every aCM sheaf of rank one on X is layered.
Theorem 1.2. Any aCM sheaf of rank one on X is either OX , OH (a) ⊕ OH with
a ∈ Z, or IC for a plane curve C ⊂ H, up to twists.
Following the same idea to the case of aCM sheaves of rank 3/2, we showed
the unique existence of non-layered aCM sheaf of rank 3/2 on X up to twist and
investigate the family of isomorphism classes of layered ones.
Theorem 1.3. Let E be an aCM sheaf of rank 3/2 on X. Then up to a twist we
have the following:
(1) If E is non-layered, then it is unique and it is the only simple one.
(2) If E is layered and indecomposable, then either
(i) E admits a filtration 0 = E0 ⊂ E1 ⊂ E2 ⊂ E3 = E such that Ei /Ei−1 ∼
=
OH for each i = 1, 2, 3,
(ii) 0 −→ OH −→ E −→ IL −→ 0 for a line L ⊂ H, or
(iii) It fits into the following sequence;
0 → IC (a) → E → OH → 0
for a plane curve C ⊂ H and an integer a ≥ deg(C).
The only non-layered sheaf E in (1) admits an extension of Ip (1) by OH (−1), up
to twist, where it is independent on the choice of a point p ∈ X (see the proof
of Proposition 5.9). For the description of type (2-i) we may see Lemma 5.7 with
Remark 5.8. The sheaves of type (2-ii) are parametrized by a P2 -bundle over
P2 = H ∨ from Remark 5.14. By Lemma 5.15 any sheaf fitting into the non-trivial
sequence in (2-iii) is indecomposable. In fact,
such sheaves
2+a−d
d+1with a > deg(C) are
parametrized by a variety of dimension 2+a
−
− 2 −1 with d = deg(C),
2
2
that is a quotient of Ext1X (OH , IC (a)) by the action of Aut(IC (a)), by Corollary
5.17.
Let us summarize here the structure of this paper. In section 2 we introduce the
definition of aCM sheaves and a generalized notion of rank, possibly not an integer.
In section 3 we collect several technical results on the restriction of aCM vector
bundle and show that every aCM vector bundle on X splits. As a generalization
we also show the splitness of any aCM vector bundle on an n-dimensional quadric
hypersurface of corank three. In section 4 we deal with the aCM sheaves of rank
1/2 and 1 to give their complete classification, which induces the wildness of the
double plane. We also show the existence of arbitrarily large dimensional family of
ACM SHEAVES ON THE DOUBLE PLANE
3
indecomposable layered aCM sheaves of any rank at least one, which also implies
the wildness. In section 5 we focus our attention to the case of rank 3/2. We start
from calculating numeric data of extension groups on aCM sheaves of lower ranks.
Our main result in this section is the unique existence of non-layered aCM sheaf of
rank 3/2 up to a twist, which is also semistable and simple. Then we describe the
family of isomorphism classes of non-layered sheaves. Finally in section 6, we collect
several results on the existence of direct factors of decomposable aCM sheaves on
X.
2. Preliminaries
Through out the article our base field k is algebraically closed of characteristic
0. For a coherent sheaf E on a projective scheme X, we denote E ⊗ OX (t) by E(t)
for t ∈ Z. For another coherent sheaf G, we denote by homX (F , G) the dimension
of HomX (F , G).
Definition 2.1. A coherent sheaf E on a projective variety X with an ample
line bundle OX (1) is called arithmetically Cohen-Macaulay (for short, aCM) if the
stalk Ex has positive depth for each x ∈ X and H i (E(t)) = 0 for all t ∈ Z and
i = 1, . . . , dim(X) − 1.
Remark 2.2. In Definition 2.1, the condition that the stalk Ex has positive depth
for each x ∈ X is equivalent to H 0 (E(−n)) = 0 for n ≫ 0 by [19, Théorème 1 in
page 268]. It also implies that it has depth dim(X), namely it is locally CohenMacaulay, again by [19, Théorème 1 in page 268] together with the vanishing of the
intermediate cohomologies of E.
If E is a coherent sheaf on a projective scheme X, then we may consider its
Hilbert polynomial PE (m) ∈ Q[m] with the leading coefficient µ(E)/d!, where d is
the dimension of Supp(E) and µ = µ(E) is called the multiplicity of E.
Definition 2.3. If dim Supp(E) = dim(X), then the rank of E is defined to be
rank(E) =
µ(E)
.
µ(OX )
Otherwise it is defined to be zero.
For an integral scheme X, the rank of E is the dimension of the stalk Ex at the
generic point x ∈ X. But in general rank(E) need not be integer.
Definition 2.4. For an initialized coherent sheaf E on X, i.e. h0 (E(−1)) = 0 but
h0 (E) 6= 0, we say that E is an Ulrich sheaf if it is aCM and h0 (E) = deg(X)rank(E).
Let X ⊂ P3 be the double plane 2H with a projective plane H ⊂ P3 . Then we
have Pic(X) = Z generated by OX (1) associated to the hyperplane class. Since
every line bundle on H is aCM, so is every line bundle on X due to the following
exact sequence
(1)
0 −→ OH (−1) −→ OX −→ OH −→ 0.
In particular, every direct sum of line bundles on X is aCM. Note also that any
extension of an aCM bundle by a line bundle on X splits. By definition 2.3 we get
rank(E) is in (1/2)N. The ideal sheaf IH of H in X is OH (−1) in (1) and so it is
an aCM sheaf of rank 1/2.
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E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Example 2.5. Fix a plane curve C ⊂ H and consider its ideal sheaf in X with the
exact sequence
(2)
0 −→ IC −→ O2H −→ OC −→ 0.
∼ OH (−k) for k = deg(C), so it is aCM and, in particular the map
Since IC,H =
H 0 (OH (t)) → H 0 (OC (t)) is surjective. Thus we get that the map H 0 (O2H (t)) →
H 0 (OC (t)) is surjective, because it factors through H 0 (OH (t)). It implies that IC
is a non-locally free aCM sheaf of rank one, because O2H is aCM. Note that C is
not a Cartier divisor of 2H and so IC is not locally free along C.
3. aCM vector bundles
Let S = k[x0 , . . . , xn ] and f ∈ S be a nonzero homogeneous element in the
irrelevant ideal. Then the maximal Cohen-Macaulay R = S/(f )-modules are the
ones arising from the matrix factorizations of f due to [6, Corollary 6.3]. Via
sheafification we may consider an aCM sheaf E on the hypersurface X = V (f ) as a
cokernel of the map given by a matrix M:
(3)
M
→ E −→ 0,
→ ⊕m
0 −→ ⊕m
i=1 OPn (bi ) −
i=1 OPn (ai ) −
where M = (mij ) is a square matrix of order m with homogeneous entries of degree
bi − aj ∈ Z>0 in S. In particular we get det(M) = f a for some a ∈ Z. If X is
irreducible, then we get a = rank(E) due to [6, Proposition 5.6].
Now assume that n = 3 and set S = k[x, y, z, w]. Let X = 2H be the double
plane given by f = w2 , i.e. X is the double plane whose reduction is the plane H
given by w = 0.
Remark 3.1. An OP3 -sheaf E is an OX -sheaf if and only if w2 E = 0, i.e. wIm(fw ) =
0, where the map fw : E → E(1) is induced by the multiplication by w. For example,
consider an extension E of an OH -sheaf E1 by another OH -sheaf E2 as OP3 -sheaves.
Then Im(fw ) is isomorphic to a subsheaf of E1 and so we get wIm(fw ) = 0. Thus
E is an OX -sheaf.
Proposition 3.2. Any aCM OX -sheaf is an extension of an OH -sheaf by an aCM
OH -sheaf, namely a direct sum of twisted OH sheaves.
Proof. Let E be an aCM OX -sheaf. The morphism fw : E → E(1) induced by the
multiplication by w, produces two exact sequences:
0 −→ ker fw −→ E −→ Imfw −→ 0
0 −→ Imfw −→ E(1) −→ cokerfw −→ 0.
Let A := ker fw and B := Imfw and then they are OH -sheaves. If A is not aCM,
then by the Auslander-Buchsbaum formula, as an OP3 -sheaf we have a locally free
resolution
M
M
M
0 −→
OP3 (a2,i ) −→
OP3 (a1,i ) −→
OP3 (a0,i ) −→ A −→ 0
of length at least 2. Since we can cook up a minimal resolution for E, which has
length 1, from those for A and B
ACM SHEAVES ON THE DOUBLE PLANE
0 −→
A
−→
↑
⊕OP3 (a0,i )
↑
⊕OP3 (a1,i )
↑ψ
⊕OP3 (a2,i )
↑
0
E
−→
5
B
−→ 0
↑
⊕OP3 (b0,i )
↑
⊕OP3 (b1,i )
↑
⊕OP3 (b2,i )
↑
⊕OP3 (b3,i )
↑
0
through the Horseshoe lemma (see [21, 2.2.8]), we are forced to assume that the
length of the resolution of B is 3, otherwise the map ψ could not be split off. Using
the same argument to the second short exact sequence, we would get that length of
the resolution of Cokerw is at least 4, which is impossible, because any OP3 -coherent
sheaf has a resolution of length at most 3 since it is not artinian.
Lemma 3.3. The restriction of aCM bundles on X to H are aCM on H.
Proof. Let E be an aCM bundle with E|H non-aCM. Assume first that E is locally
free. By the Auslander-Buchsbaum formula, we have a locally free resolution of
E|H as OP3 -sheaves
(4)
0 → ⊕OP3 (a2,i ) → ⊕OP3 (a1,i ) → ⊕OP3 (a0,i ) → E|H → 0
of length 2. Applying the horseshoe lemma to the sequence (1) with the locally
free resolutions of E(−1)|H and E|H , we get a OP3 -resolution of E of length 2 from
which is impossible to split out the map ⊕OP3 (a2,i − 1) → ⊕OP3 (a1,i − 1) in the
resolution of E(−1)|H . This contradicts E being aCM.
Lemma 3.4. There are no Ulrich locally free sheaves on X.
Proof. If E is an initialized Ulrich bundle on X, then from (5) we would obtain
that E|H is an (splitting) initialized aCM bundle on H with h0 (E|H ) = 2rank(E|H ),
where rank(E|H ) as an OH -module, contradicting the bound for them.
If E is an aCM sheaf on X, then it is given as the cokernel in the matrix factorization in (3) with det(M) = w2a .
Definition 3.5. Fix r ∈ (1/2)N. A coherent sheaf E of rank r on X is said to be
layered if there exists a filtration 0 = E0 ⊂ E1 ⊂ · · · ⊂ E2r−1 ⊂ E2r = E of E with
Ei /Ei−1 ∼
= OH (ti ) with ti ∈ Z for all i = 1, . . . , 2r.
Remark 3.6. Notice that any layered coherent sheaf on X is aCM, by mimicking
the proof of (3.3). Moreover, any aCM locally free sheaf is layered also from (3.3).
Lemma 3.7. Let E be a layered coherent sheaf on X of rank r ∈ (1/2)N. Then we
have det(M) = w2r .
Proof. By induction on r the result is clear for r = 1/2, i.e. for E ∼
= OH (a), we get
det(M) = w = (w2 )1/2 . In general, we have an exact sequence
(5)
0 −→ E2r−1 −→ E −→ OH (t2r ) −→ 0.
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E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
By inductive hypothesis, if N is the matrix associated to E2r−1 , then we get det(N) =
w2×rank(E2r−1 ) = w2r−1 . Applying the Horseshoe lemma to (5) we obtain an upper
block triangular matrix M associated to E for which det(M) = w2r−1 ×w = w2r . Proposition 3.8. Every aCM vector bundle on X splits.
Proof. Let E be an aCM locally free sheaf of rank r on X. By Lemma 3.3 the
restriction E|H is aCM on H and so it is isomorphic to a direct sum of line bundles,
say E|H ∼
= ⊕ri=1 OH (ai ) with a1 ≥ · · · ≥ ar . Up to a twist we may assume that E is
0-regular but not (−1)-regular, and so we get h2 (E(−2)) = 0 and h2 (E(−3)) > 0.
In particular E is spanned and so is E|H , which implies that ai ≥ 0 for all i. Now
h2 (E(−2)) = 0 implies that h2 (E(−3)|H ) = 0 by (5) and so we get ai ≥ 1. Again
considering (5) with h2 (E(−3)) > 0, we get h2 (E(−4)|H ) > 0 and so ar = 1. In
particular E(−1)|H is spanned.
Note that the restriction map H 0 (E(−1)) → H 0 (E(−1)|H ) is surjective and its
image spans E(−1)|H . Take a nonzero section s in H 0 (E(−1)|H ) induced from the
last factor OH (ar − 1) = OH and then it lifts to a section σ ∈ H 0 (E(−1)) such that
σ|H = s. Since s vanishes at no point of H, so does σ and we get an exact sequence
(6)
σ
0 −→ OX (1) −→ E −→ G −→ 0
with G a vector bundle of rank r − 1 on X. The restriction to H of the injective
map σ in (6) maps OH (1) onto a factor of E|H and so G|H is isomorphic to a direct
sum of line bundles on H, i.e. it is aCM. The exact sequence (5) with G instead
of E gives that G is aCM. By induction on the rank we get that G is isomorphic
to a direct sum of lines bundles and so E is isomorphic to a direct sum of line
bundles.
Corollary 3.9. Any aCM vector bundle E on an n-dimensional quadric hypersurface Y of corank at least 3 in Pn+1 for n ≥ 2, splits.
Proof. By Proposition 3.8 we may assume n ≥ 3. By induction on n, we may also
assume that the assertion holds for all quadric hypersurfaces in Pn with corank
at least 3. Take any hyperplane V ⊂ Pn+1 not included in Y such that V ∩ Y
has corank at least 3. Then the restriction E|Y ∩V is also aCM and so by the
inductive assumption it is isomorphic to a direct sum of line bundles. Now the
results follows word by word from the proof of the classical Horrocks’ theorem (see
[17, Th. 2.3.1]).
4. Wildness of the double plane
Lemma 4.1. Any sheaf of rank 1/2 on X with pure depth 2, is isomorphic to
OH (a) for some a ∈ Z.
Proof. Let K be a sheaf on X with pure depth 2. The natural map K → K∨∨ is
an isomorphism and take K1 ⊂ K with K/K1 of rank 0. Then we have K = K1∨∨ .
Assume now that K has rank 1/2 and let K2 be the subsheaf of K killed by the
multiplication by w. Then K2 is non-zero, because w2 = 0 and so multiplication by
w is not injective; indeed, if it is injective, then we get wK 6= 0 and so wK ⊆ K with
w(wK) = 0. So K2 is a non-zero OH -sheaf of rank at most 1, which is torsion-free
as OX -sheaf and so as OH -sheaf, so K2 is isomorphic to IZ,H (a) for some zerodimensional scheme and of course K/K2 has rank 0. Thus we get K2∨∨ ∼
= OH (a)
and so K ∼
= OH (a).
ACM SHEAVES ON THE DOUBLE PLANE
7
Remark 4.2. For an aCM sheaf E ∼
= OH (a) of rank 1/2 on X, the corresponding
matrix is M = (w) and so E is isomorphic to the cokernel of the map OP3 (−1) → OP3
given by the multiplication by w, up to twist. Recall that H is the plane given by
w = 0.
Lemma 4.3. We have Ext1P3 (OH (a), OH ) ∼
= H 0 (OH (1 − a)).
Proof. Applying the functor HomP3 (OH (a), −) to the standard exact sequence for
H ⊂ P3 , we get
(7)
0 −→ HomP3 (OH (a), OH ) −→ Ext1P3 (OH (a), OP3 (−1)) −→ Ext1P3 (OH (a), OP3 )
−→ Ext1P3 (OH (a), OH ) −→ Ext2P3 (OH (a), OP3 (−1)),
where the last term Ext2P3 (OH (a), OP3 (−1)) ∼
= H 1 (OH (a−3))∨ is trivial. Note that
1
2
ExtP3 (OH (a), OP3 (−1)) is isomorphic to H (OH (a − 3))∨ ∼
= H 0 (OH (−a)) and so
the first map is an isomorphism. Thus we get the assertion.
Remark 4.4. For a fixed plane curve C ⊂ H of degree d, the injection OH (−1) →
O2H in (2) factors through IC . Now the cokernel of the map OH (−1) → IC is
IC,H , which is OH (−d). In particular IC (1) admits an extension in P Ext1P3 (OH (1−
d), OH ). By Lemma 4.3 we get that P Ext1P3 (OH (1 − d), OH ) ∼
= PH 0 (OH (d)) and it
parametrizes the plane curves of degree d. Indeed we get Ext1P3 (OH (1 − d), OH ) ∼
=
Ext1X (OH (1 − d), OH ).
Proposition 4.5. If E is an aCM sheaf of rank 1 on X, then it is isomorphic to
(i) a line bundle on X
(ii) a direct sum of two aCM sheaves of rank 1/2, or
(iii) the ideal sheaf of a plane curve in H, up to twist.
Proof. Up to a twist we may assume that E is 0-regular, but not (−1)-regular.
Since E is aCM and not (−1)-regular, so we have h2 (E(−3)) > 0, i.e. there is a
non-zero map u : E → OX (1). The sheaf Im(u) is isomorphic to a sheaf IA (1) for a
closed subscheme A ( X. Since E is 0-regular, it is spanned and so is IA (1). Thus
A is scheme-theoretically cut out in X by linear equations and so A is one of the
following schemes:
(a) A = ∅;
(b) A ⊇ H;
(c) A is a line in H;
(d) A is zero-dimensional, deg(A) = 2 and A is a complete intersection of X
and two planes 6= H;
(e) A = {p}, scheme-theoretically, for some p ∈ H.
In cases (a), (c), (d) and (e), the surjective map u : E → IA (1) is an isomorphism,
because E has rank 1 with positive depth. In case (a), E is a line bundle. In
case (c), E is a non-trivial extension of OH by OH and so E(−1) is as in (iii)
with a line as the plane curve. Cases (d) and (e) are excluded, because we have
h1 (IA (−1)) = deg(A) from the standard sequence for A ⊂ X, and so A is not aCM.
Finally in case (b), we have A = H, i.e. IA (1) ∼
= OH , because IA (1) is spanned
with h0 (IA (1)) = 1. Note that H is not a Cartier divisor of X. In particular, the
sheaf ker(u) is not trivial. We know that h1 (ker(u)(t)) = 0 for t < 0, because E is
aCM. We also have h2 (ker(u)(−2)) = 0, because h2 (E(−2)) = 0 by 0-regularity of
E and h1 (OH (−2)) = 0. Thus ker(u) is 0-regular and we have the other vanishing,
8
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
implying that ker(u) is aCM. Thus we get ker(u) = OH (a) for some a ∈ Z. Now
we are in one of the cases (i), (ii) or (iii), due to Lemma 4.3 and Remark 3.1 and
the fact that OX is a non-zero extension of OH by OH (−1).
Since IC is uniquely determined by the choice of the curve C, so we get the
following.
Corollary 4.6. The double plane X is of wild type in a very strong sense, that is,
there exist arbitrarily large dimensional families of pairwise non-isomorphic aCM
sheaves of rank one on X
For a fixed r ∈ (1/2)N that is at least one, take two positive integers r1 and r2
→
−
such that r1 + r2 = 2r together with two sequences of integers k = (k1 , . . . , kr1 ) ∈
→
m = (m1 , . . . , mr2 ) ∈ Z⊕r2 . Define two splitting vector bundles on H as
Z⊕r1 and −
follows:
2
1
OH (kj ) , E2 := ⊕rh=1
E1 := ⊕rj=1
OH (mh ).
P
Then Γ := P Ext1P3 (E2 , E1 ) is of dimension −1 + j,h max{0, 3+kj2−mh } by Lemma
4.3 and each element λ ∈ Γ corresponds to a unique aCM sheaf Eλ on X with rank
r, given as an extension of E2 by E1 . Note that all sheaves Eλ are layered.
→
−
→
Proposition 4.7. Assume (r1 , r2 ) = (2r − 1, 1) with k = (k, . . . , k) and −
m = (m)
3+k−m
such that m < k and
≥ 2r − 1. Then for a general λ ∈ Γ, the sheaf Eλ is
2
indecomposable.
Proof. For a general λ ∈ Γ, set E := Eλ . Up to a twist we may assume that
m = 0.
We have λ = (ε1 , . . . , ε2r−1 ) with εi ∈ Ext1P3 (OH , OH (k)). Since 3+k
≥ 2r − 1
2
and λ is general in Γ, so the extensions ε1 , . . . , ε2r−1 are linearly independent.
Assume that E ∼
= F1 ⊕ F2 . Here we consider sheaves such as OH (t), E, F1 and
F2 as OP3 -sheaves, seeing OX as a quotient of OP3 . From this point of view these
sheaves are pure sheaves of depth 2 on P3 and we may apply the notion of (semi)stability for pure sheaves (see [20]). Note that E1 appears in a Harder-Narasimhan
filtration of E, because k is positive and E2 = OH is not an extension of two pure
sheaves of depth 2. Thus one of the two factors of E, say F1 , is a factor of E1 and
so we have E1 ∼
= F1 ⊕ G for some G, while the other one F2 , is isomorphic to either
OH (the case of E1 = F1 ) or an extension of OH by G.
First assume F2 6∼
6 0. Each OH (t) is simple and so there is
= OH , that is, G =
an integer s ∈ {1, . . . , 2r − 2} such that F1 ∼
= OH (k)⊕s and G ∼
= OH (k)⊕(2r−1−s) .
Taking instead of F1 a direct factor of F1 with minimal rank, it is sufficient to
consider only the case s = 1 for contradiction. Since OH (k) is simple, so we
have Aut(E1 ) ∼
= GL(2r − 1, k). Hence, up to an element of GL(2r − 1, k) we may
assume that F1 is the first factor of E1 . With this new basis of OH (k)⊕(2r−1) ,
set λ = (ε1 , . . . , ε2r−1 ). Then ε1 corresponds to the extension of OH by OH (k)
with E/j(G) ∼
= OH (k) ⊕ OH as its middle term, where j : G → F2 → E is the
composition. Thus we get that ε1 is zero, contradicting to the linear independence
of ε1 , . . . , ε2r−1 .
Now assume G = 0, that is, E ∼
= OH ⊕ OH (k)⊕(2r−1) . The extension class λ
induces a surjection E → OH . Since k is positive, the extension class λ is induced
by the projection of OH ⊕ OH (k)⊕(2r−1) onto its first factor. Thus we get λ = 0,
contradicting Lemma 4.3.
ACM SHEAVES ON THE DOUBLE PLANE
9
Lemma 4.8. Assume the same numeric invariants as in Proposition 4.7. For
general λ, λ′ ∈ Γ, we have Eλ ∼
= Eλ′ if and only if there is g ∈ GL(2r − 1, k) such
that g · λ = λ′ .
Proof. Set E := Eλ and E ′ := Eλ′ . Up to shift, E and E ′ are indecomposable extensions of E1 ∼
= GL(2r − 1, k).
= OH by E2 = OH (k)⊕(2r−1) , where we have Aut(E2 ) ∼
Consider all these sheaves as pure sheaves of depth 2 on P3 and use semistability
of pure sheaves with respect to the polarization OP3 (1). Then E2 is semistable and
E2 is the first step of the Harder-Narasimhan filtration of both E and E ′ . Hence
every isomorphism E → E ′ induces an automorphism of E2 . The other implication
is obvious.
Corollary 4.9. For fixed positive integer n and r ∈ (1/2)N with r ≥ 1, there exists
a family ∆ of indecomposable layered aCM sheaves of rank r with dim(∆) ≥ n,
where each isomorphism class of sheaves appears only finitely many times in ∆.
Proof. This is a consequence of Lemmas 4.3, 4.8 and Proposition 4.7.
Proposition 4.10. Let E is an aCM sheaf of rank r ∈ (1/2)N and define t0 to be
the minimal integer such that H 0 (E(t0 )) is non-trivial. If E is simple, then we have
(i) E is t0 -regular,
(ii) E(t0 ) is spanned, and
(iii) there is a non-zero map E(t0 ) → OX (1).
Proof. Up to a twist we may assume that E is 0-regular, but not (−1)-regular.
In particular, E is spanned. Since E is aCM, but not (−1)-regular, we have
h2 (E(−3)) > 0, i.e. there is a non-zero map u : E → OX (1). Then it is sufficient to prove that t0 is zero, i.e. h0 (E(−1)) = 0, since E is spanned. Assume
h0 (E(−1)) > 0 and then we get a non-zero map u1 : OX (1) → E. The map u1 ◦ u
shows that E is not simple.
5. aCM sheaves of rank 3/2
Let us consider the case when the rank of E is 3/2.
Lemma 5.1. We have HomX (OH (a), OX ) = 0 for all a ≥ 0.
Proof. Applying HomX (OH (a), −) to the exact sequence
0 −→ OH (−1) −→ OX −→ OH −→ 0,
we get HomX (OH (a), OX ) = 0 if a ≥ 1. We also get the assertion for a = 0,
because this exact sequence does not split, since OX is a simple sheaf.
Lemma 5.2. We have Ext1P3 (OH (a), OX ) ∼
= H 0 (OH (1 − a)) for all a ∈ Z
Proof. Applying the functor HomP3 (OH (a), −) to the standard exact sequence for
X ⊂ P3 , we get
(8)
0 −→ HomP3 (OH (a), OX ) −→ Ext1P3 (OH (a), OP3 (−2)) −→ Ext1P3 (OH (a), OP3 )
−→ Ext1P3 (OH (a), OX ) −→ Ext2P3 (OH (a), OP3 (−2)),
10
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
where the last term Ext2P3 (OH (a), OP3 (−2)) ∼
= H 1 (OH (a − 2))∨ is trivial. The first
map is an isomorphism, because we have by sequence (1) and Serre’s duality
HomP3 (OH (a), OX ) ⊇ HomP3 (OH (a), OH (−1)) ∼
= H 0 (OH (−1 − a))
Ext1P3 (OH (a), OP3 (−2)) ∼
= H 0 (OH (−1 − a)).
= H 2 (OH (a − 2))∨ ∼
Now the assertion follows from the isomorphism Ext1P3 (OH (a), OP3 ) ∼
= H 0 (OH (1 −
a)).
Lemma 5.3. We have Ext1P3 (OX , OH (a)) ∼
= H 0 (OH (a + 2)) for a ≥ 0.
Proof. Applying [5, Lemma 13 in §4] to H ⊂ P3 with a pair (F , G) = (OX , OH (a)),
we get
3
Ext1P3 (OX , OH (a)) ∼
= HomH (T or1P (OX , OH ), OH (a)),
because we have Ext1H (OH , OH (a)) = 0 and Ext2H (OH , OH (a)) ∼
= H 0 (OH (−a −
3
∨
P3
∼
3)) = 0. By tensoring (1) with OH , we get T or1 (OX , OH ) = T or1P (OH (−1), OH ) ∼
=
OH (−2). Thus we get Ext1P3 (OX , OH (a)) ∼
= H 0 (OH (a + 2)).
Lemma 5.4. We have Ext1X (OX , OH (a)) = 0 for all a ∈ Z.
Proof. Let E be an extension of OX by OH (a). Since OH (a) is aCM, so the restriction map H 0 (E) → H 0 (OX ) is bijective for a < 0. Thus if a < 0, then H 0 (E) is
non-trivial and the image of any nonzero map OX → E induces a splitting of the
extension.
Assume a ≥ 0 and then apply HomX (OX (−a), −) to the exact sequence (1). By
Serre’s duality, we get Ext1X (OX (−a), OX ) ∼
= H 1 (OX (−a − 2))∨ , which is trivial.
2
Similarly ExtX (OX (−a), OX ) is also trivial and thus we get
(9)
∼
=
Ext1X (OX (−a), OH ) −→ Ext2X (OX (−a), OH (−1)).
Again apply HomX (OX (−a + 1), −) to the sequence (1) twisted by OX (−1), we
get a surjection Ext2X (OX (−a + 1), OX ) → Ext2X (OX (−a + 1), OH ). But since
Ext2X (OX (−a+1), OX ) ∼
= H 0 (OX (−a−1))∨ is trivial, so is Ext2X (OX (−a+1), OH ).
Thus the groups in (9) is trivial and we get the assertion.
Proposition 5.5. If E is a simple layered aCM sheaf on X, then it is either OH
or OX up to twist.
Proof. Let r ∈ (1/2)N be the rank of E and then the result is trivial for r = 1/2.
By Proposition 4.5 it is sufficient to prove that r ≤ 1. Assume r > 1 and fix a
filtration with 0 = E0 ⊂ E1 ⊂ · · · ⊂ E2r−1 ⊂ E2r = E of E with Ei /Ei−1 ∼
= OH (ai )
with ai ∈ Z for i = 1, . . . , 2r. Up to a twist we may assume that E is 0-regular, but
not (−1)-regular. Since E is aCM, so we get h2 (E(−3)) > 0 and by duality there is
a non-zero map u : E → OX (1). Set IA (1) := Im(u). If a1 > 0, then composing the
inclusion IA (1) ⊆ OX (1) with the surjection OX (1) → OH (1) and a non-zero map
OH (1) → OH (a1 ) ⊂ E, would imply that E is not simple. Thus we get a1 ≤ 0.
(a) Assume for the moment aj > 0 for some j ≥ 2 and define s to be the
minimal such an integer.
Claim: There is another layering filtration 0 = F0 ⊂ F1 ⊂ F2 ⊂ · · · F2r = E
with either (i) F1 ∼
= OH (as ) or (ii) as = 1, aj = 0 for some j < s and F2 ∼
= OX (1).
Proof of Claim: We use induction on s. By assumption Es /Es−2 is an extension
of OH (as ) by OH (as−1 ). If this extension splits, then we may find another layering
filtration 0 = G0 ⊂ G1 ⊂ · · · ⊂ G2r−1 ⊂ G2r = E of E such that
ACM SHEAVES ON THE DOUBLE PLANE
11
• Gi = Ei if i ∈
/ {s − 1, s},
• Gs−1 /Gs−2 ∼
= OH (as ), and
• Gs /Gs−1 ∼
= OH (as−1 ).
If s = 2, then we may take Fi = Gi for all i. If s > 2, then we use the inductive
hypothesis. Now assume that Es /Es−2 is a non-trivial extension of OH (as ) by
OH (as−1 ). Since as > 0 and as−1 ≤ 0, Lemma 5.2 and Remark 4.4 give as = 1,
as−1 = 0 and Es /Es−2 ∼
= OX (1). If s = 2, then Claim is proved. Now assume s > 0.
Since aj ≤ 0 for all j < 0 we may apply s − 2 times the twist by −1 of Lemma
5.4 to get a new filtration Fi such that F2 ∼
= OX (1), Fj = Ej for all j ≥ s, and
Fi /Fi−1 ∼
= OH (ai−2 ) for i = 2, . . . , s.
By Claim we get either a non-zero map IA (1) → F1 or a non-zero map IA (1) →
F2 . So by composing with u we get that E is not simple, a contradiction.
(b) Assume aj ≤ 0 for all j. Since E is 0-regular, so it is spanned. In particular,
E/E2r−1 is spanned, i.e. a2r ≥ 0. Our assumption gives a2r = 0. If a1 ≥ 0, then
we get a non-zero map E/E2r−1 → E1 and so E is not simple. If a1 < 0 and
so h2 (OH (a1 − 2)) > 0, then we get h2 (E(−2)) > 0 since E/E1 is aCM. But it
contradicts the 0-regularity of E.
Lemma 5.6. For a plane curve C of degree d in X, we have
• dim Ext1P3 (OH (a), IC ) = 2−a
2 + max{2− a − d, 0};
• dim Ext1X (OH (a), IC ) = 2−a
− 2−a−d
,
2
2
n
where 2 is zero for n ≤ 1.
Proof. Recall that IC is an extension of OH (−d) by OH (−1) with d = deg(C).
Applying the functor HomP3 (OH (a), −) to the extension for IC , we get
(10)
0 −→ HomP3 (OH (a), OH (−d)) −→ Ext1P3 (OH (a), OH (−1)) −→ Ext1P3 (OH (a), IC )
−→ Ext1P3 (OH (a), OH (−d)) −→ Ext2P3 (OH (a), OH (−1)),
because we have an isomorphism HomP3 (OH (a), OH (−1)) ∼
= HomP3 (OH (a), IC ),
i.e. each morphism OH (a) → IC factors through OH (−1). If a ≥ 1 and so a+d ≥ 2,
we get that Ext1P3 (OH (a), OH (−1)) ∼
= H 0 (OH (−a)) and Ext1P3 (OH (a), OH (−d)) ∼
=
H 0 (OH (1 − a − d)) are trivial by Lemma 4.3. Thus Ext1P3 (OH (a), IC ) is trivial.
Now assume a ≤ 0. By Serre’s duality and Lemma 4.3, we get
Ext2 3 (OH (a), OH (−1)) ∼
= H 0 (OH (a − 2))∨ ,
= Ext1 3 (OH (−1), OH (a − 4))∨ ∼
P
P
which is trivial. Thus the sequence (10) becomes
(11)
0 −→ H 0 (OH (−a − d)) −→ H 0 (OH (−a))
−→ Ext1P3 (OH (a), IC ) −→ H 0 (OH (1 − a − d)) −→ 0.
1
0
∼
If a +
whose
dimension is
d ≥ 2, then we get H (OH (−a)) = ExtP3 (OH (a), IC ),2−a
2−a
.
If
a
+
d
=
1,
then
similarly
we
get
the
dimension
+
1.
Finally if
2
2
a + d ≤ 0 and so each term in (11) is non-zero, then we get
dim Ext1P3 (OH (a), IC ) = h0 (OH (−a)) + h0 (OH (1 − a − d)) − h0 (OH (−a − d))
2−a
2
= (a − 5a − 2d + 6)/2 =
+ (2 − a − d)
2
and so we get the assertion for Ext1P3 (OH (a), IC ).
12
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Now consider Ext1X (OH (a), IC ). If a ≥ 1, then the same argument works verbatim and so we may assume a ≤ 0. Note that Ext1P3 (OH (a), OH ) ∼
= Ext1X (OH (a), OH )
for all a ∈ Z by Remark 4.4. Applying HomX (−, OH (−1)) to the sequence (1)
twisted by OX (a), we get
0 = Ext1X (OX (a), OH (−1)) −→ Ext1X (OH (a − 1), OH (−1))
(12)
ϕ
−→ Ext2X (OH (a), OH (−1)) −→ Ext2X (OX (a), OH (−1))
by Lemma 5.4. Here we have Ext2X (OX (a), OH (−1)) = 0 from the proof of Lemma
5.4. Thus the map ϕ is an isomorphism and so
Ext2 (OH (a), OH (−1)) ∼
= Ext1 (OH (a − 1), OH (−1)) ∼
= H 0 (OH (1 − a)).
X
X
Thus the last map in (10) is the map H 0 (OH (1 − a − d)) → H 0 (OH (1 − a)) given
by multiplication by the defining equation of C, and so it is injective. Thus we get
dim Ext1X (OH (a), IC ) = h0 (OH (−a)) − h0 (OH (−a − d))
and so the assertion for Ext1X (OH (a), IC ).
From the proof of Lemma 5.6, we get that hom(OH , IL (1)) = hom(IL (1), OH ) =
⊕2
1 for any line L ⊂ H, which is essentially equivalent to IL (1) ∼
6 OH
by the proof
=
of Lemma 5.7).
Lemma 5.7. Let E be a sheaf of rank 3/2 with the filtration
0 = E0 ⊂ E1 ⊂ E2 ⊂ E3 = E
∼
such that Ei /Ei−1 = OH for all i ∈ {1, 2, 3}. Setting eL := dim Hom(OH , E) and
eR := dim(E, OH ), we have the following.
(i) 1 ≤ eL , eR ≤ 3.
⊕3
(ii) eL = 3 ⇔ eR = 3 ⇔ E ∼
.
= OH
⊕2
(resp.
(iii) eL = 2 (resp. eR = 2) if and only if E is an extension of OH by OH
⊕2
OH by OH ).
(iv) eL = eR = 1 if and only if (13) is the unique filtration of E with Ei /Ei−1 ∼
=
OH for all i. In this case E2 ∼
= IL (1) for a line L ⊂ H uniquely determined
by E.
(v) eL = eR = 2 if and only if E ∼
= IL (1) ⊕ OH for a line L ⊂ H.
(13)
Proof. Certainly we have eL , eR ≥ 1. Let εi ∈ Ext1X (E3 /Ei , Ei /Ei−1 ) corresponding
to E3 /Ei−1 for i ∈ {1, 2}. Note that λ(ε1 , ε2 ) is associated to a sheaf isomorphic to
⊕3
E for each λ 6= 0. Since OH
is a flat limit of a family of sheaves isomorphic to E
as λ goes to 0, so by semicontinuity we get eL , eR ≤ 3 and so the assertion (i).
Note that E is a semistable sheaf on P3 of pure depth with the same normalized
⊕k
Hilbert polynomial as OH . Since OH
is polystable for any positive integer k, so
⊕k
⊕c
any nonzero map u : OH → E has the image isomorphic to OH
, where c is the
linear span of the subspace of Hom(OH , E) induced by the components of u. Using
⊕k
Hom(E, OH
) we get in the same way all statements for eL , except the last one.
∼
If E = IL (1) ⊕ OH for some line L ⊂ H, then we have eL = eR = 2, because
homX (OH , IL (1)) = homX (IL (1), OH ) = 1. Conversely assume eL = eR = 2 and
⊕2
⊕2
then there exist an inclusion j : OH
֒→ E and a surjection u : E ։ OH
. Due
⊕2
to the rank counting, we have u ◦ j 6= 0 and so we get OH ⊂ j(OH ) mapped
⊕2
isomorphically by u onto some OH ⊂ OH
. Hence OH is a factor of E and so we
∼
get E = IL (1) ⊕ OH for some line L ⊂ H.
ACM SHEAVES ON THE DOUBLE PLANE
13
Remark 5.8. Consider the case (iii) in Lemma 5.7 with (eL , eR ) = (2, 1). Let
W be the set of all (e1 , e2 ) ∈ Ext1X (OH , OH )⊕2 such that e1 and e2 are linearly
independent, and then W is an irreducible variety of dimension 6. Consider a
sheaf E of this type, corresponding to (e1 , e2 ), and then E has a unique subsheaf
⊕2
. Take another sheaf E ′ corresponding to (e′1 , e′2 ), which is isomorphic
E2 ∼
= OH
⊕2
to E. Then E ′ also has a subsheaf E2′ ∼
and we have f (E2 ) = E2′ for any
= OH
′
isomorphism f : E → E . We get that two sheaves E and E ′ are isomorphic if
and only if there is M ∈ GL(2, k) with M (e1 , e2 ) = (e′1 , e′2 ). So the isomorphism
classes are parametrized by the orbits of this action of GL(2, k) on W , i.e. each
isomorphism class corresponds to a plane in Ext1X (OH , OH ) ∼
= k3 and so the family
2
of the sheaves of this type is parametrized by P . A similar description may be
applied to the case with (eL , eR ) = (1, 2).
Proposition 5.9. There exists a unique non-layered aCM sheaf of rank 3/2 on X
up to twist.
Proof. Up to a twist we may assume that E is 0-regular, but not (−1)-regular.
Since E is aCM, but not (−1)-regular, then we have h2 (E(−3)) > 0, i.e. there is a
non-zero map u : E → OX (1). The sheaf Im(u) is isomorphic to a sheaf IA (1) for
a closed subscheme A ( X. Since E is 0-regular, it is spanned and so is IA (1). In
particular, A is scheme-theoretically cut out in X by linear equations and so A is
one of the schemes described in the proof of Proposition 4.5. Recall that in case
(b) we have A = H and IA (1) ∼
= OH . Since E has rank greater than 1, so ker(u) is
not trivial in each cases.
Note that the sheaf IA (1) has positive depth and E has depth 2. So ker(u) has
depth 2. In each cases except (b), ker(u) has rank 1/2 and so ker(u) ∼
= OH (a) for
some a ∈ Z by Lemma 4.1. Thus in cases (a) and (c), E is layered.
Take A as in case (b) and so IA (1) ∼
= OH . The sheaf ker(u) has rank 1 and depth
2. Since h0 (OH (t)) = 0 for all t < 0 and E is aCM, so we get h1 (ker(u)(t)) = 0 for
all t < 0. Assume for the moment h2 (ker(u)(−2)) = 0 and then ker(u) is 0-regular.
In particular we have h1 (ker(u)(t)) = 0 for all t ≥ 0 and so ker(u) is aCM ; in this
case ker(u) is in one of the cases described in Proposition 4.5 and so E is layered.
Now assume h2 (ker(u)(−2)) > 0 and so by duality there exists a non-zero map
v : ker(u) → OX . Set IB := Im(v) with B ( X a closed subscheme. Assume first
B + H and then IB has rank 1 and so v is injective. It implies h0 (ker(u)) ≤ 1,
where the equality holds if and only if B = ∅, and in this case we get that E
is an extension of OH by OX . If B 6= ∅, then we get h0 (E) ≤ 1, contradicting
the spannedness of E. Now assume B ⊇ H. The sheaf ker(v) has rank 1/2 and
depth 2, because it is the kernel of a map from ker(u) with depth 2 to OX with
positive depth. Thus we get ker(v) ∼
= OH (a) for some a ∈ Z by Lemma 4.1. From
H 0 (ker(v)) ∼
= H 0 (ker(u)), the natural map H 0 (ker(u)) → H 0 (E/ker(v)) is a zero
map. Since E is an extension of OH by ker(u), so E/ker(v) is spanned by a unique
section and it has rank 1. Thus we get E/ker(v) ∼
= OX , which is impossible since
h2 (E(−2)) = 0.
Assume case (d) and then h1 (ker(u)(t)) is trivial for all t < 0, since E is aCM.
For an integer t ∈ Z, let
ρt : H 0 (E(t)) → H 0 (IA (t + 1))
be the restriction map. Note that h1 (E) = 0, IA (1) is spanned by the image of
H 0 (E) and IA (1) needs two sections to be spanned. It implies h1 (ker(u)) = 0.
14
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Since A is the complete intersection of X and two planes, its homogeneous ideal in
P3 is generated by the equation w2 of X and two linear equations, say u and v of the
two planes. Thus for each n > 1, H 0 (IA (n)) is generated by the monomials um v n−m
with 0 ≤ m ≤ n. Take u1 , v1 ∈ H 0 (E) such that ρ0 (u1 ) = u and ρ0 (v1 ) = v. Take
n = t + 1. If m > 0, we have ρt (um−1 v t+1−m u1 ) = um−1 v t+1−m u = um v t+1−m . If
m = 0, we have ρt (v t v1 ) = v t+1 . Thus ρt is surjective and so h1 (ker(u)(t)) = 0 for
all t > 0, concluding that ker(u) is aCM and so it is isomorphic to OH (a) for some
a ∈ Z. From an exact sequence
0 = H 1 (E(−1)) −→ H 1 (IA ) ∼
= k −→ H 2 (OH (a − 1)) −→ H 2 (E(−1)) = 0,
we get a = −2. Since E is aCM with h2 (E(−2)) = 0, so we get h1 (IA (−1)) =
h2 (OH (−4)) = 3. But since A is a complete intersection with an exact sequence
0 −→ OX (−2) −→ OX (−1)⊕2 −→ IA −→ 0
and h2 (IA (−1)) = 0, so we get h1 (IA (−1)) = h2 (OX (−3)) − h2 (OX (−2)⊕2 ) = 2,
a contradiction.
Finally assume case (e). As in case (d) we have h1 (ker(u)(t)) is trivial for all
t ≤ 0. On the other hand, we have h2 (ker(u)(−1)) = 0, because h1 (Ip ) = 0. Thus
ker(u) is 1-regular and so we get h1 (ker(u)(t)) > 0 for t > 0, concluding that ker(u)
is aCM and so ker(u) ∼
= OH (a) for some a ∈ Z. If a ≤ −2, we get
H 1 (Ip ) = 0 −→ H 2 (OH (a − 1)) −→ H 2 (E(−1)) = 0
with h2 (OH (a−1)) > 0, a contradiction. If a ≥ 0, we get h1 (E(−2)) ≥ h1 (Ip (−1))−
h2 (OH (a − 2)) = 1, a contradiction. Now assume a = −1 and then we get the exact
sequence
(14)
π
0 −→ OH (−1) −→ E −→ Ip (1) −→ 0.
From the exact sequence 0 → OH → Ip (1) → Ip,H (1) → 0, we may set G :=
π −1 (OH ). Since G is an extension of OH by OH (−1), so G is isomorphic to either
OX or OH ⊕ OH (−1) by Lemma 4.3 and Remark 4.4.
(i) Assume G ∼
= OH ⊕ OH (−1). We see that E/OH is an extension of Ip,H (1)
by OH (−1). By [5, Lemma 13 in §4], we get
Ext1P3 (Ip,H (1), OH (−1)) ∼
= Ext1H (Ip,H (1), OH (−1)),
= Ext1X (Ip,H (1), OH (−1)) ∼
because we have HomH (Ip,H , OH (−1)) = 0. Thus E/OH is isomorphic to either
⊕2
OH
or OH (−1) ⊕ Ip,H (1). In the former case, E is as in case (b) with A ⊇ H,
while the latter case is impossible, because E/OH must be globally generated.
(ii) Assume G ∼
= OX . We have Ext1X (Ip,H (1), OX ) ∼
= k and
= H 1 (Ip,H (−1))∨ ∼
so for each point p ∈ H there is a unique sheaf E up to isomorphism; if we need to
specify p we write Ep for the sheaf in (14). Note that we have homX (E, OX (1)) =
h2 (E(−3)) = 3 and so a 2-dimensional projective space P := P HomX (E, OX (1))
of morphisms E → OX (1). None of these maps may be surjective, because with
kernel isomorphic to OH (l) for some l ∈ Z it would give a sheaf E with different
Hilbert polynomial. At least one of these maps is not surjective only at a point
and so this is true for a non-empty subset of P. Since dim Ext1X (Ip,H (1), OX ) = 1,
for each p ∈ H there is an open neighborhood Up of p such that for every q ∈ Up
there is a surjection Ep → Iq (1). Thus we get Eq ∼
= Ep for all q ∈ Up . Since any
two non-empty open subsets of H meet, we get Eq ∼
= Ep for all q ∈ H, i.e. there is
ACM SHEAVES ON THE DOUBLE PLANE
15
a unique E in case (e). We also see that for every v ∈ P there is q ∈ H such that
Im(v) = Iq (1), and so we get an identification P ∼
= H.
Claim 1 : E is aCM.
Proof of Claim 1 : By (14) we have h1 (E(t)) < 0 for all t < 0. On the
other hand, we have h1 (Ip (−1)) = h2 (OH (−3)) = 1 and the boundary map δ :
H 1 (Ip (−1)) → H 2 (OH (−3)) corresponds to the extension class [E]. Thus δ is an
isomorphism and so we get h2 (E(−2)) = 0. Since h1 (E(−1)), we get that E is
0-regular and so h1 (E(t)) = 0 for all t ≥ 0. Hence E is aCM.
Claim 2 : If E is layered, then there is a surjection E ։ OH .
Proof of Claim 2 : Take a filtration 0 = E0 ⊂ E1 ⊂ E2 ⊂ E3 = E with
Ei /Ei−1 ∼
= OH (ti ) for i = 1, 2, 3. Since h0 (E(−1)) = 0, we have ti ≤ 0 for all i.
0
Since h (E) = 3, then ti = 0 for all i and in particular E/E2 ∼
= OH .
Assume the existence of a surjection u′ : E → OH . Composing with the inclusion
OH → OX (1), we get a morphism v ∈ P such that Im(v) ∼
6 Iq (1) for any q ∈ H, a
=
contradiction. Thus E is not layered.
Notation 5.10. Let us denote the unique non-layered aCM sheaf of rank 3/2 on
X in Proposition 5.9 by E, i.e. E is 0-regular, but not (−1)-regular.
Corollary 5.11. For each r ∈ (1/2)N at least 3/2, there exists a non-layered aCM
sheaf of rank r on X.
Proof. In Proposition 5.9 we showed the existence of a unique non-layered aCM
sheaf E of rank 3/2, which is 0-regular with h0 (E) = h2 (E(−3)) = 3 and h0 (E(−1)) =
⊕(2r−3)
0. Define a sheaf G := E ⊕ OH
and then it is not layered. Indeed, assume
the existence of a filtration 0 = G0 ⊂ G1 ⊂ · · · ⊂ G2r−1 ⊂ G2r = G of G with
Gi /Gi−1 ∼
= OH (ti ) for some ti ∈ Z. From h0 (G(−1)) = 0 and h0 (G) = 2r, we get
ti = 0 for all i. Since there is no non-zero map E → OH , so we have E ⊂ G2r−1 . By
descending induction on i we get E ⊂ Gi for all i, a contradiction.
Recall that the non-layered and 0-regular aCM sheaf E of rank 3/2 in Proposition
⊕3
5.9 has the same Hilbert function of OH (−1) ⊕ Ip (1), or of OH
. The sheaf E is
0
0
spanned with h (E) = 3 and h (E(−1)) = 0.
Proposition 5.12. E is simple and semistable as a sheaf on P3 with pure depth 2.
Proof. Assume that E is not semistable and then we get an exact sequence of OX sheaves
0 −→ A −→ E −→ B −→ 0
such that B is torsion-free,
the normalized Hilbert polynomial of A is strictly bigger
than the one, t+2
,
of
O
and rank(A) + rank(B) = 3/2. Since B has positive
H
2
depth, A has depth 2. If A has rank 1/2, then Lemma 4.1 gives A ∼
= OH (a) for
some a ∈ Z, because A has depth 2. Since A destabilizes E and h0 (E) = 3, we get
a = 1. But the sheaf B of rank 1 is spanned by a linear subspace of H 0 (E) not
contained in H 0 (A), which is impossible. Now assume rank(B) = 1/2 and then
we get B ∨∨ ∼
= OH (b) for some b ∈ Z by Lemma 4.1. Since the map B → B ∨∨
is injective, we get B ∼
= IZ,H (b) for some some zero-dimensional scheme Z ⊂ H.
Since the Hilbert function of IZ,H (b) is strictly less that the one of OH , we get that
either b < 0 or b = 0 and Z 6= ∅. In both cases we get h0 (B) = 0, contradicting the
spannedness of E. Thus E is semistable.
16
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Assume that E is not simple. Note that the image of any non-zero map E → E
is simultaneously a quotient of E and a subsheaf of E. Thus it has the normalized
Hilbert polynomial of OH . Since k is algebraically closed, there is a non-zero map
f : E → E, which is not an isomorphism. Since f is not surjective, we get that
• ker(f ) 6= 0,
• f (E) is saturated in E and so with depth 2, and
• both ker(f ) and f (E) have the normalized Hilbert polynomial of OH .
First assume that ker(f ) has rank 1/2 and so, since it has depth 2 and with the
Hilbert polynomial of OH , we get ker(f ) ∼
= OH . Since h0 (E) = 3 and h1 (OH ) = 0,
0
f (E) is spanned with h (f (E)) = 2. Since f (E) is saturated in E, the quotient
E/f (E) is a spanned sheaf with positive depth and rank 1/2. Thus we get E/f (E) ∼
=
IZ,H (d) for some d ∈ Z and some 0-dimensional scheme Z ⊂ H. Since IZ,H (d) has
the Hilbert function of OH , we get d = 0 and Z = ∅. But it would give a filtration
0 ⊂ ker(f ) ⊂ f (E) ⊂ E, a contradiction.
Now assume that f (E) has rank 1/2 and so, having depth 2 and the Hilbert
polynomial of OH , we have f (E) ∼
= OH . Since E is aCM and h0 (OH (t)) = 0 for all
1
t < 0, we get h (ker(f )(t)) = 0 for all t < 0. Since h2 (E(−2)) = h1 (OH (−2)) = 0,
we get h2 (ker(f )(−2)) = 0 and so ker(f ) is 0-regular. Therefore h1 (ker(f )(t)) = 0
for all t ≥ 0, and so ker(f ) is aCM. By Proposition 4.5, ker(f ) is layered and so E
is layered, a contradiction.
Our next job is to describe the family of isomorphism classes of aCM sheaves,
not extension classes, of rank 3/2 on X, i.e. to check when the middle terms of two
extensions in the proof of Proposition 5.9 are isomorphic. First of all, the theorem of
Krull-Remak-Schmidt says that for any coherent sheaf F on X its indecomposable
factors are uniquely determined, up to a permutation of the factors and isomorphism
as abstract OX -sheaves, since X is projective. So we only look to indecomposable
ones. Up to a shift we may assume that E is 0-regular, but not (−1)-regular. In
the proof of Proposition 5.9 we have 5 cases (a)∼(e), where cases (a) and (c) only
give decomposable sheaves and case (d) does not occur. In case (e), if E is layered,
then it is as in Lemma 5.7. Thus there remains only case (b).
In Remark 5.8 we considered a parameter space W of a finite-dimensional vector
space with a group action by GL(2, k) to show that the family of isomorphism
classes of the sheaves is isomorphic to P2 . In general, we follow the same strategy,
even in which W has not a so simple description and there is a group G of linear
transformations of W . In all cases we first prove that E has a unique E2 ⊂ E and
a unique map E → E/E2 . So if f : E → E ′ is an isomorphism, then E ′ also has
a unique E2′ with f (E2 ) = f (E2′ ) and G = Aut(E2 ) so that E ′ corresponds to a
point of W in the same G-orbit as the point corresponding to E. The question is
the converse. Indeed, if o, o′ ∈ W correspond respectively to E, E ′ such that there
exists g ∈ Aut(E2 ) with go = o′ , then g induces a map of extensions and so a map of
their middle terms. This map is an isomorphism, because g −1 induces the inverse
map for their middle terms.
Consider case (b) with an exact sequence
(15)
u
0 −→ ker(u) −→ E −→ OH −→ 0,
where ker(u) is aCM and so it is one of the sheaves in Proposition 4.5 with
h2 (ker(u)(−2)) = 0.
ACM SHEAVES ON THE DOUBLE PLANE
17
Lemma 5.13. If E is an indecomposable sheaf fitting into (15), then either
(1) ker(u) is as in (iii) of Proposition 4.5,
(2) E is as in Lemma 5.7, or
(3) 0 → OH → E(−1) → IL → 0 for a line L ⊂ H.
Proof. If ker(u) is isomorphic to OX (t) for some t ∈ Z, then (15) splits since
Ext1X (OH , OX (t)) ∼
= H 1 (OH (−t − 2))∨ is trivial.
Assume that ker(u) ∼
= OH (a) ⊕ OH (b) with a ≥ b. Indeed we get b ≥ 0 from
2
h (E(−2)) = 0. If b ≥ 2, then we get E ∼
= OH (a) ⊕ OH (b) ⊕ OH by Lemma
4.3. We also describe the case (a, b) = (0, 0) in Lemma 5.7 and Remark 5.8. If
a ≥ 2 and b = 1, then from (15) we get that E/OH (1) is an extension of OH by
OH (a) and so E/OH (1) ∼
= OH ⊕ OH (a). Thus we get a surjection E → OH (a).
Since a > b, (15) gives that OH (a) is a factor of E. Assume a = b = 1 and then
we have homX (E, OH ) = 1 and so there is a unique subsheaf E2 of E, isomorphic to OH (1)⊕2 . By (15) E is induced by (e1 , e2 ) ∈ Ext1X (OH (1)⊕2 , OH ). Since
dim Ext1X (OH (1), OH )) = 1 by Lemma 4.3, e1 and e2 are linearly dependent and
so OH (1) is a factor of E.
Lastly let us consider the case (a, b) = (1, 0). Then the quotient sheaf E/OH (1) is
⊕2
an extension of OH by OH , and so it is isomorphic to either OH
or IL (1) for some
line L ⊂ H. In the former case E is given by two elements of Ext1 (OH (1), OH ),
which must be linearly dependent by Lemma 4.3) and so OH is a factor of E. Thus
we may assume E/OH (1) ∼
= IL (1) with the exact sequence
(16)
0 −→ OH (1) −→ E → IL (1) −→ 0.
Now it remains to check that if (16) is a non-trivial extension, then E is indecomposable. Certainly OH (1) is not a factor of E, because (16) is assumed to be a nontrivial extension and homX (OH (1), E) = 1 by (15). Assume E ∼
= OH ⊕ G for some G
and then G is an aCM sheaf with the same Hilbert polynomial of OH (1) ⊕ OH , and
so G is isomorphic to either OH (1) ⊕ OH or OX (1). The former case is excluded
since OH (1) is not a factor of E, and the latter case is also excluded by (16), because
h0 (IL ) = 0.
Remark 5.14. Consider the sheaves in (3) of Lemma 5.13. Since HomX (OH (1), IL (1))
is trivial, we have homX (OH (1), E) = 1 by (16). Thus L is uniquely determined by
E and different lines in H give non-isomorphic sheaves. Applying HomX (−, OH (1))
to the extension IL (1) of OH by OH , we get
0 −→ HomX (OH , OH (1)) −→ Ext1X (OH , OH (1)) −→ Ext1X (IL (1), OH (1))
η
−→ Ext1X (OH , OH (1)) −→ Ext2X (OH , OH (1)),
because the map HomX (OH , OH (1)) → HomX (IL (1), OH (1)) is an isomorphism,
i.e. every map IL (1) → OH (1) factors through OH . Note that the map η is the
map H 0 (OH (2)) → H 0 (OH (3)) by Remark 4.4 and the proof of Lemma 5.6. Since
η is given by the multiplication by the defining equation of L ⊂ H, so it is injective.
Thus we get dim Ext1X (IL (1), OH (1)) = 3, i.e. there is a 2-dimensional projective
space of isomorphism classes of such sheaves.
Now assume ker(u) ∼
= IC (a) for some degree d curve C and a unique integer a,
as in (iii) of Proposition 4.5. Since h0 (OH ) = 1 and E is spanned,
have
we
a ≥ d.
2+a−d
Conversely, if a ≥ d, then we get dim Ext1X (OH , IC (a)) = 2+a
−
, which
2
2
is positive.
18
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
Lemma 5.15. Any non-trivial extension of OH by IC (a) with a ≥ d is indecomposable.
Proof. Let ε be the non-trivial extension class of OH by IC (a) corresponding to E
and assume that E is decomposable with E ∼
= OH (b) ⊕ G for some aCM sheaf G of
rank 1, where G is described in Proposition 4.5. Since E is spanned, G is spanned
and b ≥ 0.
(i) First assume b > 0 and then we have HomX (OH (b), OH ) = 0 with a
surjective map G → OH . Since G is spanned, by Proposition 4.5, G is isomorphic
⊕2
to either OX , OH
or IL (1) for some line L ⊂ H. If G ∼
= OX , then we get
a+1
a−d+2
b+2
0
(17)
1+
+
= h (E) = 1 +
.
2
2
2
Since HomX (OH (b), OH ) = 0, we get a non-zero map j : OH (b) → IC (d) with
j(OH (b)) saturated in E and so saturated in IC (d). Since IC (a) is an extension of
OC (a−d) by OC (a−1), so we get either b = a−1 or b ≤ a−d, which are impossible
⊕2
by (17). Thus we get either G ∼
OH
or G ∼
= IL (1) for some line L ⊂ H, with the
= b+2
a+1
a−d+2
equality 1 + 2 +
=
+
2.
The non-zero map j : OH (b) → IC (d)
2
2
gives that either b = a − 1 or b ≤ a − d, and so we get a = 1 and so d = 1, from
which we get b = 0, a contradiction.
(ii) Now assume b = 0 and then the map E → OH induces the zero-map
OH → OH , because ε 6= 0 and any non-zero map OH → OH is an isomorphism.
Thus we get a surjective map G → OH , and so as before G is isomorphic to either
⊕2
or IL (1) for some line L ⊂ H. If G ∼
OX , OH
= OX , then h0 (E) = 2 and so
⊕2
h0 (IC (a)) = 1, a contradiction. If either G ∼
or G ∼
= OH
= IL (1), then we get
h0 (E) = 3 and h0 (IC (a)) = 2, i.e. a = d = 1. Thus we are in the set-up of Lemma
5.7, and it is indecomposable.
The sheaf IC (a) is an extension of OH (a − d) by OH (a − 1). First assume a ≥ d + 1.
Since ωC ∼
= OC (d − 3), the exact sequence
0 −→ IC −→ OX −→ OC −→ 0
gives h2 (IC (a−3)) = 0. Therefore h2 (E(−3)) = 1 and so there is a unique surjection
E → OH , up to a scalar, i.e. a unique subsheaf isomorphic to IC (a) for some integer
a and some curve C. Hence if f : E → E ′ is an isomorphism, then E ′ contains IC (a)
and f (IC (a)) = IC (a). We get that the isomorphism classes of E are parametrized
by the non-zero extensions of OH by IC (a) by the action of Aut(IC (a)). By Lemma
5.16 we have dim(Aut(IC (a)) = 1 + d+1
2 .
Lemma 5.16. For a curve C ⊂ H of degree d, we get dim End(IC ) = 1 + d+1
2 .
Proof. Since Aut(IC ) is a non-empty
open subset of End(IC ), it is sufficient to
prove that dim Aut(IC ) = 1 + d+1
2 . If d ≥ 2, then there is a unique subsheaf A
of IC isomorphic to OH (−1), because OH (−1) is stable with a normalized Hilbert
polynomial as a sheaf with pure depth 2 on P3 and this Hilbert polynomial is strictly
bigger than the one of OH (−d). In the case d = 1 we use that homX (OH (−1), IC ) =
⊕2
(see the proof of Lemma 5.7 for more details) to get the
1, because IC (1) 6∼
= OH
same assertion. Hence every automorphism of IC preserves the exact sequence
0 −→ OH (−1) −→ IC −→ OH (−d) −→ 0.
ACM SHEAVES ON THE DOUBLE PLANE
19
Use that homX (OH (−d), OH (−1)) = d+1
and that if an isomorphism f : IC →
2
IC induces the multiplication by c on the subsheaf OH (−1), then f − c · IdIC is
uniquely determined by the associated map OH (−d) → OH (−1).
Corollary 5.17. The aCM sheaves of rank 3/2 given as an extension of OH by
IC (a) with a > deg(C), are parametrized by a quotient variety of dimension 2+a
2 −
2+a−d
d+1
−
−
1.
2
2
There remains the case a = d. If h2 (E(−3)) = 1, then again E fits in a unique
extension of OH by IC (d) and so we have the same description as in Corollary 5.17.
Now assume h2 (E(−3)) > 1. If a = d = 1, then we are in the set up of Lemma
5.7 with eR ≥ 2. Now assume a = d > 1. Note that E contains a unique copy of
⊕2
OH (d − 1) with F := E/OH (d − 1), where F is isomorphic to OH
or IL (1) for some
′
line L ⊂ H. Thus any isomorphism f : E → E sends OH (d−1) to the corresponding
copy of OH (d − 1) of E ′ , and so f induces an isomorphism of extensions of F by
OH (d − 1). We get that the isomorphism classes of E are parametrized by the
quotient of the family of non-zero extensions of F by OH (d − 1), by the action of
Aut(F ). See Lemma 5.16 for the computation of Aut(F ) in the case F ∼
= IL (1).
In all cases E is decomposable only if either
• E∼
= OH (d−1)⊕F , which is possible only if the extension of F by OH (d−1)
is the trivial one, or
• E∼
+ 1.
= OH ⊕ G for some aCM sheaf G that is 0-regular with h0 (G) = d+1
2
If the inclusion j : OH → E as a factor, does not split the extension E of OH
by IC (d), then the surjection E → OH maps j(OH ) onto zero and so j(OH ) is a
saturated subsheaf of IC (d). Its existence would imply that IC (d) ∼
= OH ⊕ OH (d −
1), contradicting the fact that IC (d) is not an OH -sheaf.
6. Decomposability
In this section we collect several results on the existence of direct factors of
decomposable aCM sheaves on X.
Lemma 6.1. For any zero-dimensional subscheme Z ⊂ H and a ≤ −2, we have
Ext1X (OH (a), IZ,H ) = 0.
Proof. We may use the exact sequence
0 −→ IZ,H −→ OH −→ OZ −→ 0
together with HomX (OH (a), OZ ) = 0 and Ext1X (OH (a), OH ) = 0 for all a ≤ −2 by
Lemma 4.3.
Lemma 6.2. Let F be an OX -sheaf fitting in an exact sequence
(18)
⊕α
0 −→ OH (−1)⊕β −→ F −→ OH
−→ 0
for positive integers α, β ∈ Z. If γ is the largest non-negative integer such that F
⊕γ
has OH
as a factor, then we get
⊕(α−γ)
⊕γ
F∼
⊕ OH (−1)⊕(β−α+γ) ⊕ OH
.
= OX
Proof. By induction on the integer α + β, we reduce to the case γ = 0. Note that
F is aCM with h0 (F ) = α > 0. Take a non-zero map j : OX → F .
First assume that j is not injective. Since ker(j) has rank 1/2 and depth 2, we
have ker(j) ∼
= OH (t) for some t ∈ Z by Lemma 4.1. Since Im(j) is a spanned and
20
E. BALLICO, S. HUH, F. MALASPINA AND J. PONS-LLOPIS
torsion-free sheaf of rank 1/2 on X, we have Im(j) ∼
= IZ,H (b) with b ≥ 0, and
we get b = 0 only if Z = ∅. Indeed we get b = 0 and Z = ∅, because we have
HomX (OH (b), OH ) = 0 for b > 0. Thus j induces an inclusion g : OH ֒→ F . Since
⊕α
⊕α
any non-zero map OH → OH
maps OH onto a factor of OH
, (18) shows that OH
is a factor of (18).
Now assume that any non-zero map j is injective. From γ = 0 together with
dim Ext1X (OH (−1), OH ) = 1 by Lemma 4.3, row/column operations on the (β × α)⊕α
matrix B representing the extension class [F ] ∈ Ext1X (OH
, OH (−1)⊕β ) show that
β ≥ α. Similarly, if β > α, then F has OH (−1) as a factor. Thus we may assume
⊕α
that α = β by induction on α + β. For a fixed factor R ∼
, we may find
= OH of OH
⊕α
∼
a factor L = OH (−1) of OH (−1)
up to a change of basis so that the extension
(18) has a non-zero entry for L and zero for the other α − 1 entries. Now we change
⊕α
the matrix B by operations on OH
so that the column associated to L has 0 except
for the entry corresponding to R. With this choice the extension of R by L has the
property that its middle term, isomorphic to OX , is a direct factor of F . Then we
may write off the factors R and L to reduce to the case with α′ = α − 1, and we
may apply induction.
Lemma 6.3. For any zero-dimensional subscheme Z ⊂ H and a < 0, we have
Ext1X (OX (a), IZ,H ) = 0.
Proof. If a ≤ −2, then we may use Lemma 6.1 and the exact sequence
0 −→ OH (a − 1) −→ OX (a) −→ OH (a) −→ 0.
Now assume a = −1. As in the proof of Lemma 6.2 we reduce to the case Z = ∅.
Take any sheaf F fitting in
(19)
u
v
0 −→ OX (−1) −→ F −→ OH −→ 0.
and then we get h0 (F ) = 1. Let j : OX → F be a non-zero map. If j is not
injective, we get the existence of a subsheaf of F isomorphic to OH and this subsheaf
induces a splitting of (19). Now assume that j is injective. Since the maps j∗ :
H 0 (OX ) → H 0 (F ) and v∗ : H 0 (F ) → H 0 (OH ) are isomorphisms and no proper
subsheaf of OH has a non-zero section, the map v|j(OX ) is surjective. Thus we get
j(OX ) ∩ u(OX (−1)) ∼
= OH (−1). But it is impossible, because u is injective and
there is no non-zero map OH (−1) → OX (−1) by Lemma 5.1.
Let E be an aCM sheaf of rank r ∈ (1/2)N and let A := ⊕si=1 OH (ti ) with
t1 ≥ · · · ≥ ts be the kernel of the map fw : E → E(1) induced by the multiplication
by w as in Proposition 3.2. The image B := Im(fw ) ⊂ E(1) is a torsion-free sheaf on
H with rank 2r − s. Since w2 = 0, we have 2s ≥ r and B ⊆ A(1) with A := ker fw .
We obviously have s ≤ 2r, and s = 2r if and only if E ∼
= ⊕si=1 OH (ti ).
Proposition 6.4. Let E be an aCM sheaf as above and σ be the minimal integer
such that E(σ) is spanned. Let α > 0 (resp. β ≥ 0) be the number of indices j with
tj = ts (resp. tj = ts + 1). If ts ≤ σ − 2, then either
(i) E has OH (ts ) as a direct factor, or
(ii) E has OX (ts − 1)⊕α as a factor and α ≤ β.
Proof. Up to a twist we may assume σ = 0 and so B is spanned. The inclusion
of A ⊂ E induces an inclusion j : OH (ts )⊕α → E and set G := coker(j). Since
ACM SHEAVES ON THE DOUBLE PLANE
21
OH (ts )⊕α is saturated in A and A is saturated in E, G has positive depth. G fits
in an exact sequence
s−α
0 −→ ⊕i=1
OH (ti ) −→ G −→ B −→ 0.
Since B ∨∨ is a generically spanned vector bundle on H of rank 2r−s, there is a zerodimensional scheme Z1 ⊂ H and an inclusion j : IZ1 ,H ֒→ B such that B/IZ1 ,H is
torsion-free of rank 2r − s− 1 as an OH -sheaf. Since the latter is obviously spanned,
we get a filtration 0 = B0 ⊂ B1 ⊂ · · · ⊂ B2r−s = B with zero-dimensional schemes
Zi ⊂ H such that
• Bi /Bi−1 ∼
= IZi ,H for i = 1, . . . , 2r − s − 1, and
• B/B2r−s−1 ∼
= IZ2r−s ,H (b) for some b ∈ Z.
∼ ⊕s−α OH (ti ) ⊕ B and consider the exact sequence
Lemma 6.1 gives G =
i=1
u
s−α
OH (ti ) ⊕ B −→ 0.
0 −→ OH (ts )⊕α −→ E −→ G ∼
= ⊕i=1
If β = 0, then the filtration of B shows that E has OH (ts )α as a factor, because
ts ≤ −2 and ts ≤ ts−α−1 − 2 if s 6= α. Now assume β > 0 and set F := u−1 (A′ ),
s−α
where A′ is the factor of ⊕i=1
OH (ti ) ⊕ B, isomorphic to OH (ts − 1)⊕β . Consider
the exact sequence
s−α−β
OH (ti ) −→ 0.
0 −→ F −→ E −→ B ⊕ ⊕i=1
Lemma 6.2 gives the existence of an integer γ with 0 ≤ γ ≤ α, β ≥ α − γ and
F∼
= OX (ts − 1)⊕(α−γ) ⊕ OH (ts − 1)⊕(β−α+γ) ⊕ OH (ts )⊕γ . If γ > 0, then OH (ts )⊕γ
is a factor of E by Lemma 4.3 and the filtration of B together with the assumption
ts−α−β+1 ≥ ts + 2 for s > α + β. If γ = 0 and so β ≥ α, then E has OX (ts − 1)⊕α
as a factor by Lemma 6.3.
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Università di Trento, 38123 Povo (TN), Italy
E-mail address: edoardo.ballico@unitn.it
Sungkyunkwan University, Suwon 440-746, Korea
E-mail address: sukmoonh@skku.edu
Politecnico di Torino, Corso Duca degli Abruzzi 24, 10129 Torino, Italy
E-mail address: francesco.malaspina@polito.it
Department of Mathematics, Kyoto University, Kyoto, Japan
E-mail address: ponsllopis@gmail.com