SANDOR J. KOVACS
A line bundle
on X is called big if X is proper and the global sections of
define a birational map for some m 0.
is called nef if deg(Cic)
0 for every
In particular every ample line bundle is nef and big.
In this paper a log variety (X, D) consists of a proper variety X and an effective
divisor D, called the boundary. A log morphism takes boundary to boundary, in
particular the image of the boundary is a divisor. (X, D) is of log general type if
Kx +Dis big. It is log canonically polarized if Kx + D is ample. Note that in this
is necessarily projective.
For a log variety,
D) (resp. Bir(X, D)) denotes the set of au-
tomorphisms (resp. proper birational automorphisms) of
lP'N, then Lin(
denotes the set of linear automor-
phisms that leave both
fixed. A smooth log variety will mean simply that
X is smooth. Note that this is very different from the notion of log smooth.
Let G be a finite group and
G. Then IGI (resp.
denotes the order of
G (resp. the order
§1. Log varieties
§§1.1 Log canonical embeddings.
Let (X, D) be a log canonically polarized
smooth log variety of dimension n. Let m
N be such that the complete linear
system of (m + 1)Kx + mD gives a birational morphism ¢ : X
lP'N for some
N 0 such that
= ¢(D) is still a divisor in
= ¢(X), i.e.,
is a log
variety and ¢ is a log morphism. This can be achieved for instance by requiring
that ¢ separate points.
1.1.1 FACT. In characteristic 0, l(m + 1)Kx + mDI separates the points of X
as soon as m ~ (n!
by [Angehrn-Siu95] ( cf. [Kollar97, 5.8]).
Observe that any linear automorphism of lP'N leaving the pair
induces a proper birational automorphism of (X, D). In fact ¢* gives an isomor-
phism between Lin(
and Bir(X, D). Now Bir(X, D) is finite by [Iitaka82,
11.12], hence so is Lin(
and in order to estimate I Aut(
I it is enough to
1.1.2 LEMMA. Let (X, D) be a log canonically polarized smooth log variety of
dimension n and
lP'N a birational log morphism given by the complete
linear system l(m + 1)Kx + mDI. Let
(a) If D is nef, then
((m + 1)(Kx +
((m + 1)(Kx +
(b) If K x is nef, then
degD:::; degX = ((m + 1)Kx + mD)n.
(a) In this case we actually prove that
degX + degD:::; ((m + 1)(Kx +
(m + 1)(Kx +D), D, and (m + 1)Kx + mD are nef, so for all i = 1, ... , n- 1,
((m + 1)Kx + mD)i · D · ((m + 1)Kx + mD)n-i-l