24.04.2014 Views

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

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

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

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

MUKAI: <strong>Polarized</strong> <strong>K3</strong> <strong>surfaces</strong> <strong>of</strong> <strong>genus</strong> <strong>18</strong> <strong>and</strong> <strong>20</strong> 267<br />

Proposition 1.8 A vector bundle E is generated by its global section if <strong>and</strong> only if the<br />

tautological line bundle O P (1) is so.<br />

Proposition 1.9 Let σ be the global section <strong>of</strong> the tautological line bundle O P (1) corresponding<br />

to s ∈ H 0 (U, E) via (1.7). If U is smooth, then the following are equivalent:<br />

i) s is nondegenerate <strong>and</strong> (s) 0 is smooth, <strong>and</strong><br />

ii) the divisor (σ) 0 ⊂ P(E) is smooth.<br />

Pro<strong>of</strong>. Since the assertion is local, we may assume E is trivial, i.e., E ≃ OU ⊕r . Let<br />

f 1 , · · · , f r be a set <strong>of</strong> generators <strong>of</strong> the ideal I defining (s) 0 . A point x ∈ (s) 0 is singular<br />

if <strong>and</strong> only if df 1 , · · · , df r are linearly dependent at x, that is, there is a set <strong>of</strong> constants<br />

(a 1 , · · · , a r ) ≠ (0, · · · , 0) such that a 1 df 1 + · · · + a r df r = 0 at x. This condition is equivalent<br />

to the condition that the divisor<br />

(σ) 0 : f 1 X 1 + · · · + f r X r = 0<br />

in P(E) ≃ U × P r−1 is singular at x × (a 1 : · · · : a r ). There<strong>for</strong>e, ii) implies i). Since (σ) 0<br />

is smooth <strong>of</strong>f (s) 0 × P r−1 , i) implies ii).<br />

q.e.d.<br />

By these two propositions, Bertini’s theorem (see [7, p. 137]) is generalized <strong>for</strong> vector<br />

bundles:<br />

Theorem 1.10 Let E be a vector bundle on a smooth variety. If E is generated by its<br />

global sections, then every general global section is nondegenerate <strong>and</strong> its scheme <strong>of</strong> zeroes<br />

is smooth.<br />

2 A homogeneous space <strong>of</strong> SO(9, C)<br />

Let X be the subvariety <strong>of</strong> Grass(P 2 ⊂ P 7 ) consisting <strong>of</strong> 2-planes in the smooth 7-<br />

dimensional hyperquadric<br />

(2.1)<br />

Q 7 : q(X) = X 1 X 5 + X 2 X 6 + X 3 X 7 + X 4 X 8 + X 2 9 = 0<br />

in P 8 . The special orthogonal group G = SO(9, C) acts transitively on X. Let P be the<br />

stabilizer group at the 2-plane X 4 = X 5 = · · · = X 9 = 0. X is isomorphic to G/P <strong>and</strong> the<br />

reductive part L <strong>of</strong> P consists <strong>of</strong> matrices<br />

⎛<br />

⎜<br />

⎝<br />

A 0 0<br />

0 t A −1 0<br />

0 0 B<br />

with A ∈ GL(3, C) <strong>and</strong> B ∈ SO(3, C). We denote the diagonal matrix<br />

⎞<br />

⎟<br />

⎠<br />

[x 1 , x 2 , x 3 , x 4 , x −1<br />

1 , x −1<br />

2 , x −1<br />

3 , x −1<br />

4 , 1]<br />

by < x 1 , x 2 , x 3 , x 4 >. All the invertible diagonal matrices < x 1 , x 2 , x 3 , x 4 > <strong>for</strong>m a maximal<br />

torus H <strong>of</strong> G contained in L. Let X(H) ≃ Z ⊕4 be the character group <strong>and</strong> {e 1 , e 2 , e 3 , e 4 }<br />

its st<strong>and</strong>ard basis. For a character α = a 1 e 1 +a 2 e 2 +a 3 e 3 +a 4 e 4 , let g α be the α-eigenspace<br />

{Z : Ad < x 1 , x 2 , x 3 , x 4 > ·Z = x a 1<br />

1 x a 2<br />

2 x a 3<br />

3 x a 4<br />

4 Z} <strong>of</strong> the adjoint action Ad on the Lie algebra<br />

g <strong>of</strong> G. Then we have the well-known decomposition<br />

g = h ⊕<br />

⊕<br />

0≠α∈X(H)<br />

g α .

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!