Topics in Algebraic Geometry- Algebraic Surfaces- Algebraic Surfaces, Lecture 21

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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 21
LECTURES: ABHINAV KUMAR
From last time: f : X → B is an elliptic/quasi-elliptic fibration, F = m P
bi i i
multiple fibers, R1f O = L ⊕ T , for L invertible on B and T torsion. b ∈
∗ X
Supp(T ) ⇔ h0(O ) ≥ 0 ⇔ h1(O ) ≥ 2 ⇔ F is an exceptional/wild fiber.
Fb Fb b

Theorem 1. With the above notation, ω = f∗(L−1 ⊗ ω ) ⊗O (� a P ), where
X B X i i
0 ≤ a < m,a = m − 1 unless F is exceptional, and deg(L−1 ⊗ ω ) =
i i i bi B
2p (B) − 2+ χ(O )+ �(T ), where �(T ) is its length as an O -module.
a X B
Proof. We have proved most of this: specifically, we have that ω = f∗(L−1 ⊗
X
ω ) ⊗ O (� a P ) for 0 ≤ a < m. We have a Leray spectral sequence Epq =
B X i i i 2
Hp(B,Rqf O ) =⇒ Hp+q(X, O ). The smaller order terms give us a short
∗ X X
exact sequence
0 → H0(O ) → H1(O ) → H0(R1f O ) → H2(O ) = 0
B X ∗ X B
(1)
0 → H2(O ) → H1(R1f O ) → 0
X ∗ X
Using this, we see that
χ(O ) = h0(O ) − h1(O )+ h2(O )
X X X X
= h0(O ) − h1(O ) − h0(L ⊕ T )+ h1(L ⊕ T )
B B
(2)
= χ(O ) − χ(L) − h0(T )
B
= −deg L − �(T )
by Riemann-Roch, so deg L = −χ(O ) − �(T ). Since deg ω = 2p (B) − 2, we
X B a
have deg (L−1 ⊗ ω ) = 2p (B) − 2 + χ(O ) + �(T ). It remains to show that
B a X
a = m − 1 if F is not exceptional. If fact, we can prove something stronger:
i i bi
let α be the order of O (P ) ⊗O in Pic(P ). Then we claim that
i X i Pi i
(1) α divides m and a + 1,
i i i
(2) h0(P , O ) ≥ 2 and h0(P , O ) = 1, and
i (αi+1)Pi i αiPi
(3) h0(P ,nP ) is a nondecreasing function of n.
i i
Assuming this, if a < m − 1, then α < m , so m P is exceptional by (b) and
i i i i i i
(c), since then h0(O ) ≥ 2.
miPi
We now prove the claim. If m > n ≥ 1, then O → O → 0 gives
mP nP
H1(P, O ) → H1(P, O ) → 0, implying that n �→ h1(P, O ) is nondecreas­
mP nP nP
ing. But by Riemann-Roch and the definition of canonical type, χ(O ) = 0, so
nP
1
2 LECTURES: ABHINAV KUMAR
h0 = h1 is also nondecreasing. Now, by the definition of α , O (α P ) ⊗O ∼=
∼ i X i i Pi
O , implying that O (−n P ) ⊗ O = O as well. We thus obtain an exact
Pi X i i Pi Pi
sequence 0 → O (−α P ) ⊗ O = O → O → O → 0, inducing a
X i i Pi Pi (αi+1)Pi αiPi
long exact sequence
(3) 0 → k ∼= H0(O ) → H0(O ) → H0(O ) → ···
Pi (αi+1)Pi αiPi
and h0(O ) ≥ 2. But for 1 ≤ j < α ,L = O (−jP ) ⊗ O is an
(αi+1)Pi i j X i Pi
invertible O -module whose degree in each component of P equals 0. Since
L �∼= O ,HP0(i L ) = 0, and 0 → L → O → O → 0 givie s H0(O ) ∼=
Hj0 (O P)i . Sincej H0(O ) ∼= k forj P ico(cj+t,1 )HPi 0(O j)P i ∼= ··· ∼= H0(O ()j +∼=1) Pki as
jPi P 2P αP
well.
Finally,
(4) (O (P ) ⊗O )mi ∼= O (F ) ⊗O ∼= O
X i Pi X bi Pi Pi
This is proved as follows, Since the fiber is cut out by a rational function f,
H0(O (F ) ⊗O ) �= 0. Via the exact sequence
X bi Pi
(5) 0 → O → O (F ) 1/f→ �→1/f O (F ) ⊗O → 0
X X bi X bi Fbi
we get a global section of O (F ) ⊗O . But this also has degree 0 along the
X bi Fbi
components. So it must be trivial, but what we proved for icoct. We also have
∼
O ((a + 1)P ) ⊗O = ω ⊗O (P ) ⊗O
(6) X i i Pi X X i Pi
∼ ∼
= ω = O
Pi Pi
implying that α |a + 1 as desired. �
i i
Corollary 1. K2 = 0.
Corollary 2. If h1(O ) ≤ 1, then either a +1 = m or a + α +1 = m .
X i i i i i
Proof. Exercise. �
Remark. Raynaud showed that m /α is a power of p = char(k) (or is 1 if
i i
char(k) = 0). Therefore, there are no exceptional fibers in characteristic 0.
1. Classification (contd.)
If f : X → B is an elliptic/quasi-elliptic fibration, then

�
(7) ω = f∗(L−1 ⊗ ω ) ⊗O ( a P ), 0 ≤ a < m
X B X i i i i
If n ≥ 1 is a multiple of m ,...,m , then
1 r

�
(8) H0(X,ω⊗n) = H0(B,L−n ⊗ ωn ⊗O ( a (n/m )b ))
X B B i i i
Now we recall the 4 classes of surfaces:
(a) ∃ an integral curve C on X s.t. K · C < 0.
ALGEBRAIC SURFACES, LECTURE 21 3
(b) K ≡ 0.
(c) K2 = 0, K ·C ≥ 0 for all integral curves C, and ∃C� s.t. K · C� > 0.
(d) K2 > 0, and K · C ≥ 0 for all integral curves C.
Lemma 1. If X is in (a), then κ(X) = −∞, i.e. p = 0 for all n ≥ 1. If X is
n
in (b), then κ(X) ≤ 0. If X has an elliptic or quasielliptic fibration f : X → B,

and if we let λ(f) = 2p (B) − 2+ χ(O )+ �(T )+ � ai , then X is not in class
a X mi
(d) and
• X is in (a) iff λ(f) < 0, in which case κ(X) = −∞,
• X is in (b) iff λ(f) = 0, in which case κ(X) = 0,
• X is in (c) iff λ(f) > 0, in which case κ(X) = 1.
Proof. If K · C < 0, then X is ruled, and κ(X) = ∞. We did this before,
and there is an easy way to see that p = 0 for all n ≥ 1. For every divisor
n
D ∈ Div(X), ∃n s.t. |D + nK| = ∅ for n > n . (Since (D + nK) · C =
D D
D·C + n(K· C) becomes negative eventually. Now C is effective. We claim that
C2 ≥ 0, so by our useful lemma, 6 |D + nK| can’t have an effective divisor. If
C2 < 0, then C · K < 0 would imply that C was an exceptional curve of the first
kind, contradicting the minimality of X. Thus, C2 ≥ 0.) In particular, D = K
gives |nK| = ∅ for large enough n, implying that |nK| = ∅ for all n (since
p < p ).
n mn
Next, assume K ≡ 0 (case (b)). If p ≥ 2, then dim |nK| ≥ 1 =⇒ ∃ a
n
strictly positive divisor Δ > 0 in |nK|. Then Δ · H > 0 for a hypersurface
section, contradicting nK · H = 0 since K ≡ 0. So p ≤ 1 for all n, implying
n
that κ(X) ≤ 0.
Now assume X has an elliptic/quasielliptic fibration, and let M = f (ω ) =
∗ X
L−1 ⊗ ω from last time. Then M has degree λ(f). Let H be a very ample
B
divisor on X. Then π = f| : H → B is some finite map of degree = H · F > 0.
H
Now n(K · H) = deg(ωn | ) = deg (π∗M) = (deg π)(deg M) = (H · F )λ . So
X H H B f
if λ < 0, then K · H < 0 and X is in (a).
f
Similarly, λ(f) = 0 =⇒ K · H = 0 for every irreducible hyperplane section
H, and any curve C can be written, up to ∼, as the difference of 2 such. This
implies that K · C = 0 ∀ C =⇒ K ≡ 0. Lastly, λ(f) > 0 =⇒ K · C > 0 for
all horizontal irreducible C. For vertical C, K · C = 0 by the formula for K,
implying that K · C ≥ 0 for all C integral, (K2) = 0 by the formula, implying
that we are in class (c). �
Let X be a minimal surface with K2 = 0,p ≤ 1 (in particular, every surface in
g
class (b) is of this form. Then Noether’s formula gives 10 − 8q +12p = b +2Δ.
g 2
Since p ≤ 1, 0 ≤ Δ ≤ 2p ≤ 2, also Δ = 2(q − s) is even, we obtain the following
g g
possibilities.
(1) b = 22,b = 0,χ(O ) = 2,q = 0,p = 1, Δ = 0.
2 1 X g
(2) b = 14,b = 2,χ(O ) = 1,q = 1,p = 1, Δ = 0.
2 1 X g
4 LECTURES: ABHINAV KUMAR
(3) b = 10,b = 0,χ(O ) = 1, and either q = 0,p = 0, Δ = 0 or q = 1,p =
2 1 X g g
1, Δ = 2.
(4) b = 6,b = 4,χ(O ) = 0,q = 2,p = 1, Δ = 0.
2 1 X g
(5) b = 2,b = 2,χ(O ) = 0, and either q = 1,p = 1, Δ = 0 or q = 2,p =
2 1 X g g
0, Δ = 2.
Note. If X is in class (b) and p = 1, then K ∼ 0 (because K = 0,H0(K) �= 0
g
imply that K ∼ 0.).
Let’s deal with case 4 of class (b).
Proposition 1. Let X be minimal in class (b), and b = 2,b = 2. Then s = 1,
2 1
Alb(X) is an elliptic curve, and X → Alb(X) gives an elliptic/quasielliptic
fibration.
Proof. Let’s see that the fibers of f are irreducible. If not, we would have ρ >
2(F,H, component of F ) and b ≥ ρ > 2, contradicting b = 2. Now, to see that
2 2
the fibers are not multiple, note that χ(O ) = 0 from the list.
X
(9) deg (L−1 ⊗ ω ) = 2p (B) − 2+ χ(O )+ �(T ) = �(T ) ≥ 0
B a X

Since ω = f∗(L−1 ⊗ω )⊗O ( �a P ) ≡ 0, we see that �(T )·f−1(y)+ �a P ≡
X B X i i i i
0. But it is an effective divisor, implying that all the a = 0,�(T ) = 0 and thus
i
a = m − 1 ∀ i (there are no wild fibers since T = 0). So m = 1 ∀ i. Thus, we
i i i
have integral fibers, which is the case of a bielliptic surface. �