4 H. B. LAUFER
all i. Let 0 = Y0,..., Yk,..., Yt = Z be defined as above, with Ai, the first element
of the computation sequence, satisfying At Z 0. Then for all n 0 and all
Kkl, H\MXQ{L -nZ- Yk_x)) = o! Also, all sections of L - nZ - Yk_x
restricted to Ajk lift to sections of L on M.
III. Tricanonical resolution. In this section, we let V be a purely two-dimen-
sional (reduced) space. V may have nonisolated singularities. Let £: M' - V be a
resolution of V [20]. Let ©(A") be the canonical sheaf on M', i.e. the sheaf of
germs of holomorphic 2-forms (AT is of dimension two). Let &(mK) denote
6(K) 8 ®©(K) with m factors; ® is taken over the structure sheaf of AT. So
&(mK) is the sheaf of m-pluricanonical forms.
Let L be a line bundle on a space M. Recall that x E M is a base point for L if
all elements of T(M, &(L)) vanish at x.
The next theorem is due to Shepherd-Barron [46, Theorem 2, p. 2] in the
important case that L 2K.
THEOREM
3.1. Let V be a Stein two-dimensional space. Let w: M K be the
minimal resolution of V. Suppose that L is a line bundle over M with LAi2KAi
for all irreducible exceptional curves Ai on M. Then L has no base points on M.
PROOF.
The normalization V of V is also Stein; use [13] to see this. V has only
isolated singularities. M is also the minimal resolution of V,TT: M V. 7r^(6(L))
is a coherent sheaf on V [13] and is isomorphic to 0(L) on the regular points of
V. So by Cartan's Theorem A, we need only prove the theorem under the
restrictions that M is strictly pseudoconvex with a connected exceptional set A.
Moreover, we need only show that L has no base points on A.
Let A UAi be the decomposition of A into irreducible components. Recall
that on a minimal resolution, K-AjX) for all /. Moreover, KAt, = 0 if and only
ifg/ = 0and^
/
-^
|
. = -2 .
Recall the Yk used in Proposition 2.1. We shall prove, by induction on k, that L
has no base points on supp(7^). We first consider the case k 1. L-Aj
2K-Aj] 2gjy Let Lx be the restriction to AJx of the line bundle L. By (2.2), Lx
has no base point at any regular point of A-. We may also see, as follows, using
only c(Lx)
2gfJ]
\, that Lx has no base point at any singular point of A-^ We
use (2.1). Let Lx denote the pull-back of Lx to X, a normalization of Ah C. Let
g = gjY
c(Lx-c) = c(Lx)-282g-\.
So H\X, 6(LX - c)) = 0. Then T(C, 6(L,)) - T(C,31) is onto. At a singular
point x of C, 51x ¥= 0 and so T(C, &(LX)) has an element which is nonzero at x.
By Proposition 2.1, elements of T{Ah, &(LX)) lift to T(M, 0(L)).
Now suppose that the induction step is true for k 1, i.e. L has no base
point on supp(7^_j). The computation of (2.4) shows that for K-Aj \,
(L Yk_i)-Aj 2gj . So, as in the previous paragraph, L has no base points on
Aj supp(YA:_1). This completes the induction step for KA: 1.
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