Polarized K3 surfaces of genus 18 and 20 - Research Institute for ...

MUKAI: Polarized K3 surfaces of genus 18 and 20 270
Lemma 3.3 Λ i ≃ Λ 10−i ⊗ O X (i − 5).
Proof. By (1.6), Λ i is isomorphic to (Λ 10−i ) ∨ ⊗ O X (5). Since F is of rank 2, F ∨ is
isomorphic to F ⊗ O X (−1), which shows the lemma.
q.e.d.
We need the decomposition of Λ i into irreducible homogeneous vector bundles. Λ i =
∧ i
(F ⊕5 ), i ≤ 5, has the following vector bundles as its irreducible factors:
(3.4)
Λ 0
Λ 1
O
F
Λ 2
Λ 3
S 2 F, O(1)
S 3 F, F(1)
Λ 4
Λ 5
S 4 F, S 2 F(1), O(2)
S 5 F, S 3 F(1), F(2)
Proposition 3.5 If [N] ∈ Ξ SCI , then S N is a K3 surface.
Proof. By Proposition 2.4 and (2.6), the vector bundle F ⊕5 and the tangent bundle
T X have the same determinant bundle. Hence S N has a trivial canonical bundle. By
Proposition 1.4 and Lemma 1.6, we have the exact sequence 0 −→ K • −→ O SN −→ 0
and the isomorphism K • ≃ Λ • ⊗ O(−5). By (1) of Proposition 2.8 and the Serre duality
H i (K 10 ) ≃ H 12−i (O X ) ∨ , we have
and
H 1 (K 1 ) = H 2 (K 2 ) = · · · = H 10 (K 10 ) = 0
H 1 (K 0 ) = H 2 (K 1 ) = · · · = H 10 (K 9 ) = H 11 (K 10 ) = 0.
Therefore, the restriction map H 0 (O X ) −→ H 0 (O SN ) is surjective and H 1 (O SN ) vanishes.
Hence S N is connected and regular.
q.e.d.
Let F N be the restriction of F to S N . The complex K • ⊗ F gives a resolution of F N .
Since S n F ⊗ F ≃ S n+1 F ⊕ S n−1 F(1), we have the following 4 series of vanishings by (1)
of Proposition 2.8:
(a) H 1 (F ⊗ K 2 ) = H 2 (F ⊗ K 3 ) = · · · = H 9 (F ⊗ K 10 ) = 0,
(b) H 1 (F ⊗ K 1 ) = H 2 (F ⊗ K 2 ) = · · · = H 9 (F ⊗ K 9 ) = H 10 (F ⊗ K 10 ) = 0,
(c) H 1 (F) = H 2 (F ⊗ K 1 ) = · · · = H 10 (F ⊗ K 9 ) = H 11 (F ⊗ K 10 ) = 0
and
(d) H 2 (F) = H 3 (F ⊗ K 1 ) = · · · = H 11 (F ⊗ K 9 ) = H 12 (F ⊗ K 10 ) = 0.
By (c) and (d), both H 1 (F N ) and H 2 (F N ) vanish. By (a) and (b), the sequence
is exact. So we have proved
0 −→ H 0 (F ⊗ K 1 ) −→ H 0 (F) −→ H 0 (F N ) −→ 0
Proposition 3.6 If dim S N = 2, then we have
1) H 1 (S N , F N ) = H 2 (S N , F N ) = 0, and
2) the restriction map H 0 (X, F) −→ H 0 (S N , F N ) is surjective and its kernel coincides
with N.
Let E N be the restriction of E to S N . Arguing similarly for the complexes K • ⊗E, K • ⊗S 2 E
and K • ⊗ sl(E), we have
Proposition 3.7 If dim S N = 2, then we have
(1) all the higher cohomology groups of E N and S 2 E N vanish,
(2) the restriction maps H 0 (X, E) −→ H 0 (S N , E N ) and H 0 (X, S 2 E) −→ H 0 (S N , S 2 E N )
are bijective, and
(3) all the cohomology groups of sl(E N ) vanish.