Topics in Algebraic Geometry- Algebraic Surfaces- Algebraic Surfaces, Lecture 19

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by  Abhinav Kumar

MIT OpenCourseWare
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18.727 Topics in Algebraic Geometry: Algebraic Surfaces
Spring 2008
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ALGEBRAIC SURFACES, LECTURE 19
LECTURES: ABHINAV KUMAR
Corollary 1. If D is an indecomposable curve of canonical type (icoct), then
∼
ω = O , where ω is the dualizing sheaf of D.
D D D
Proof. By Serre duality, h1(ω ) = h0(O ) = 0. We have the short exact sequence
D D
(1) 0 → O (K) → O (K + D) → ω → 0
X X D
so χ(ω ) = χ(O (K +D))−χ(O (K)) = 1 ((K +D)·D) = 0 by Riemann-Roch
D X X 2
(using D2 = 0 and D · K = 0). Thus, h0(ω ) = 1. Since ω has degree 0 along
D D
the E ,
i
(2) deg (O ⊗O (K + D) ⊗O ) = (K + D) · E = 0
Ei D X Ei i
It follows from the proposition last time that ω ∼= O . �
D D

Corollary 2. If D = �n E is an icoct, D� an effective divisor on X s.t.
i i
D� · E = 0 for all i, then D� = nD + D�� where n ≥ 0,D�� an effective divisor
i
disjoint from D.
Proof. Let n be the largest integer s.t. D� − nD ≥ 0, and let D�� = D� − nD,L =
O (D��). ∃ an exact sequence
D
(3) 0 → O (D�� − D) → O (D��) → O (D��) = L → 0
X X D
Let s ∈ H0(X, O (D��)) be s.t. div (s) = D��. Since D�� − D = D − (n +1)D is
X X
not effective, s doesn’t come from H0(O (D�� −D)), so its image in H0(O (D��))
X D
is nonzero. But deg (L| ) = D�� · E = (D� − nD) ·E = 0 =⇒ L ∼= O =⇒
Ei i i D
s(x) =� 0 ∀ x ∈ D, so that the support of D�� must be disjoint from that of D. �
Theorem 1. Let X be a minimal surface with K2 = 0 and K·C ≥ 0 for all curves
on X. If D is an icoct on X, ∃ an elliptic or quasielliptic fibration f : X → B
on X obtained from the Stein factorization of φ : X → P(H0(O (nD))∨) for
|nD| X
some n > 0.
Proof. Idea: use D and K to get an elliptic/quasielliptic fibration. Then show
that the fiber must be a multiple of D.
1
2 LECTURES: ABHINAV KUMAR
Case 1: p = 0. or n ≥ 0, we have the exact sequence
g
(4) 0 → O (nK +(n − 1)D) → O (nK + nD) → O → 0
X X D
obtained by tensoring 0 → O (−D) → O → O → 0 by n(K + D) and
X X D
using O (nK + nD) ⊗O ∼= ω⊗n ∼= O since D is an icoct. We claim
X D D D
that
(5)
H2(O (nK +(n − 1)D)) = H0(−(n − 1)(K + D))0
X
� �
for n ≥ 2. To see this, note that if Δ ∈ �m(K + D)
� for m > 0, then

n
either Δ = 0 =⇒ mK ∼ −mD =⇒
K ·
H = −D · H < 0 for an ample
divisor H, giving a contradiction, or Δ > 0 with a similar contradiction.
Also, H2(O ) = 2 since D has support of dimension 1, implying that
D
H2(O (nK + nD)) = 0, and H1(O ) = H0(ω ) = H0(O ) �= 0 gives
X D D D
H1(O (nK + nD)) =� 0. We know from Riemann-Roch that
X
1
χ(O (nK + nD)) = χ(O )+ (nK + nD)(nK + nD − K)
X X
(6) 2
= χ(O ) = 1 − q
X
(since p = 0). Noether’s formula states that
g
(7) 12 − 12q = 12 − 12q − 12p = K2 +2 − 2b + b
g 1 2
with b = 2q since the irregularity Δ = 0 because p = 0. So
1 g
(8) 10 − 8q = b ≥ 1 =⇒ q ≤ 1 =⇒ χ(O ) = 0, 1
2 X
and χ(O (nK + nD)) = 0 or 1 for n ≥ 2. Since H1(O (nK + nD)) �= 0
X X
and H2(O (nK + nD)) = 0, we must have H0(O (nK + nD)) �= 0 for
X X
n ≥ 2. So ∃ D ∈ |nK + nD|.

As before, we see that D =� 0.
n n
�
We claim that D is of canonical type. Letting D
= n E , we find
n i i
that
(9) D ·E = n(K · E )+ n(D· E ) = 0
n i i i

�
This implies that D = aD + k F for some a ≥ 0,k > 0 integers, F
n j j j j
distinct irreducible curves that don’t intersect D. Now K ·
F ≥ 0, and
j
�
by our hypothesis ( k F ) ·K ≥ 0. But it equals K · nK + nD − D = 0,
j j
so K · F = 0 for all j. Finally,
j
(10) D · F = n(K ·F )+ n(D· F ) = 0
n j j j
so D is of canonical type.
n
Now, D can’t be a multiple of D for all n, For then D = mD =⇒
n n
nK ∼ λ D for some integer λ for each n ≥ 2 =⇒ K = 3K · 2K is a
n n
multiple of D, say λ · D = K. If λ < 0, this contradicts K · H ≥ 0. If
λ ≥ 0, then |K| = |λD| = ∅ which contradicts p = 0. So ∃ a curve of
g
ALGEBRAIC SURFACES, LECTURE 19 3
canonical type D� on X s.t. removing the multiple of D and decomposing
to get an icoct, we get something disjoint from D. So let D� be an icoct,
disjoint from D. Then
(11) 0 → O (2K + D + D�) → O (2K +2D +2D�) → O ⊕O → 0
X X D D�
(using ω ∼= O ,ω ∼= O ). As before, we can show that H2(O (2K +
D D D� D� X
D+D�)) = 0, and therefore H2(O (2K +2D+2D�)) = 0. So χ(O (2K +
X X
2D+2D�)) = χ(O ) = 0 or 1, while h1(O (2K +2D+ D�)) ≥ 2 (because
X X 2
h1(O ),h1(O ) ≥ 1) implies that h0(O (2K + 2D + 2D�)) ≥ 0. Now,
D D� X
take
(12) Δ ∈ |2K +2D +2D�| , Δ > 0, Δ2 = 0, dim |Δ| ≥ 1
Since D,D� are of canonical type, so is Δ (easy exercise).
We now claim that |Δ| is composed with a pencil (i.e. it gives a map
to a curve). To see this, let C be the fixed part of |Δ|, then since Δ is
of canonical type, we get (Δ − C)2 ≤ 0 (the self-intersection of a divisor
supported on a curve of canonical type is ≤ 0). So the rational map
(13) φ : X → φ (X) = B ⊂ |Δ|
|Δ| |Δ|
is defined everywhere (else would have (Δ−C)2 > 0. Use C ·C = C˜ ·C˜ +
1 2 1 2
m m for a single blowup at p if C ,C pass through p with multiplicity
1 2 1 2
�
m ,m and apply to two elements of Δ − C with zero intersection after
1 2
the blowup). Since dim |Δ| ≥ 1,B can’t be a point. And it can’t be
a surface, else we would have Δ � C = φ∗(H) =⇒ ((Δ − C)2) > 0.
So Δ is composed with a pencil and φ is a morphism. Now Δ · D =
|Δ|
D · (2K +2D +2D�) = 0 and D ·(Δ − C) ≥ 0 and D· C ≥ 0 (write C as

�
k E + F , where F doesn’t have any of the E as components). This
i i i i
forces D · (Δ − C) = 0. Since D is connected, it is contained in one of
the fibers and D2 = 0. We see that D is a rational multiple of one of the
fibers of the Stein factorization f : X → B� → B. Since the gcd of the
coefficients of D is 1, the fiber must be a positive integral multiple of D.
It is easy to see that the genus of the fiber is 1, implying that it is an
elliptic/quasielliptic fibration.
Case 2: p > 0. As before, it is enough to show that dim H0(O (Δ)) ≥ 2
g X
for some divisor Δ of canonical type. We’ll show that ∃n > 0. s.t.
dim H0(O (nD)) ≥ 2. Let F = O (nD)/O . So we have
X n X X
(14)
0 → O → O (nD) → F → 0 =⇒ H0(O (nD)) → H0(F ) → H1(O )
X X n X n X
It is enough to show that H0(F ) → ∞ as n → ∞ since the dimension
n
of H1(O ) is fixed. Let L = F = O (D)/O (note that F = 0). Then
X 1 X X 0
4 LECTURES: ABHINAV KUMAR
L is an invertible sheaf on D, and
(15) 0 → F → F → O ((n + 1)D)/O (nD) ∼= Ln → 0
n−1 n X X
implies that n �→ h0(F ) is nondecreasing. By Riemann-Roch,
n
(16) χ(O (nD)) = χ(O ) =⇒ χ(F ) = 0
X X n
for all n. One finds that H2(O (nD)) = 0 for n >> 0 since K − nD has
X
h0 = 0 (D is effective). Thus, H1(F ) =� 0 for n >> 0 since h2(O ) =
n X
p > 0 and we have the exact sequence
g
(17) H1(F ) → H2(O ) → H2(O (nD))
n X X
This implies that h0(F ) = h1(F ) > 0 for n >> 0. If the sequence of
n n
integers {h0(F )} is bounded above, let n be the largest s.t. h0(F ) <
n n−1
h0(F ). (There exists such an n because h0(F ) = 0,h0(F ) > 0 for
n 0 n
n >> 0.) We must have h0(F ) = h0(F ) = ··· , and we obtain a
n n+1
nonzero global section of Ln coming from s ∈ H0(F ) not in the image of
n
H0(F ). D is an icoct and Ln has degree 0 on every component of D,
n−1
so s| does not vanish anywhere on D. Supp (F ) = D =⇒ s generates
D n
F as an O -module at all points of X, and thus defines a surjection
n X
O → F = O (nD)/O with kernel O (−nD) and an isomorphism
X n X X X
∼
O /O (−nD) = O (nD)/O . The tensor power gives an isomorphism
X X X X
∼
O /O (−nD) → O (mnD)/O ((m − 1)nD)
= F /F for all
X X X X mn (m−1)n
m > 1. Now, we have

0 0
� � � �
0 �� O �� O ((m − 1)nD) �� F �� 0
X X (m−1)n
=
�� � � � �
(18) 0 �� O �� O (mnD) �� F �� 0
X X mn
� � � �
F F
n n
� � � �
0 0
ALGEBRAIC SURFACES, LECTURE 19 5
implying
H1(F ) α �� H1(F ) �� H1(F ) �� 0
(m−1)n mn n
�� ��
(19) H2(O ) = �� H2(O )
X X
�� ��
0 = H2(O ((m − 1)nD) H2(O (mnD)) = 0
X X
for m >> 0. So α is nonzero (because H2(O ) is nonzero), h1(F ) >
X mn
h1(F ) for m >> 0, implying that h0(F ) > h0(F ), a contradiction.
n mn n
�
Theorem 2. Let X be a minimal surface with K2 = 0,K · C ≥ 0 ∀ curves C on
X. Then either 2K ∼ 0 or X has an icoct.

Proof. Next time. �