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 twodimen
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 2forms (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 mpluricanonical 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 ShepherdBarron [46, Theorem 2, p. 2] in the
important case that L — 2K.
THEOREM
3.1. Let V be a Stein twodimensional 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, KAjX) 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. LAj
2KAj] 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 pullback of Lx to X, a normalization of Ah — C. Let
g = gjY
c(Lxc) = 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 KAj \,
(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.