Discrete Holomorphic Local Dynamical Systems

88 Eric Bedford
with multiplicity 1” means, in particular, that C is regular at p1. In a similar manner,
we see that if C is a curve with the cohomology class H − P1 − P2, thenC must be
the line p1p2.
If X is a surface obtained by repeated blowups, the total cohomology is given by
H ∗ (X;C)=H 0 (X;C)⊕H 1,1 (X;C)⊕H 4 (X;C).If f is a holomorphic map, then the
total map f ∗ acts on each of these factors. We have H 0 (X;C) ∼ = C, and f ∗ | H 0 = 1.
Similarly, the dimension of H 4 is the number of connected components (which is
equal to 1), and f ∗ | H 4 is multiplication the mapping degree of f , which is 1 in the
case of an automorphism. Thus the Lefschetz Fixed Point Formula takes the form:
Theorem 2.7. If X is obtained from P 2 by iterated blowups and if each f n has
isolated fixed points, then
Pern = 2 + trace( f ∗n )
where f ∗ denotes the restriction of f ∗ to H 1,1 , and Pern denotes the number of
solutions of {p ∈ X : f n (p)=p}, counted with multiplicity.
2.3 Invariant Currents and Measures
Let f be an automorphism of a compact Kähler surface X.Since f ∗ ∈ GL(H 1,1 ;Z),
the determinant of f ∗ must be ±1. The pullback and push-forward preserve the
intersection product: (ω,η)=(f ∗ ω, f ∗ η)=(f∗ω, f∗η). In fact, the push-forward
and pullback are adjoint: f ∗ α · β = α · f∗β .Further,( f −1 ) ∗ =(f ∗ ) −1 . And since
limn→∞ || f n ∗ || 1/n = limn→∞ || f ∗n || 1/n ,wehaveλ ( f )=λ ( f −1 ).
Theorem 2.8. Let f ∈ Aut(X) be an automorphism of a Kähler manifold with
λ ( f ) > 1. Thenλ is an eigenvalue of f ∗ with multiplicity 1, and it is the unique
eigenvalue with modulus > 1.
Proof. Let ω1,...,ωk denote the eigenvectors for f ∗ for which the associated eigenvalues
μ j has modulus > 1. For 1 ≤ j ≤ k we have
(ω j,ωk)=(f ∗ ω j, f ∗ ωk)=μ j ¯μk(ω j,ωk),
so (ω j,ωk) =0. Letting L denote the linear span of ω1,...,ωk, we see that each
element ω = ∑cjω j ∈ L satisfies (ω,ω)=0. By the Signature Theorem, it follows
that the dimension of L is ≤ 1. On the other hand, since λ ( f ) > 1, L is spanned by
a unique nontrivial eigenvector. If ω has eigenvalue μ, then ¯ω has eigenvalue ¯μ, so
we must have μ = ¯μ = λ .
Now we claim that λ has multiplicity one. Otherwise there exists θ such that
f ∗ θ = λθ+ cω. In this case,
(θ,ω)=(f ∗ θ, f ∗ ω)=(λθ + cω,λω)=λ 2 (θ,ω),
so (θ,ω)=0. Similarly, we have (θ,θ)=0, so by the Signature Theorem again, the
space spanned by θ and ω must have dimension 1, so λ is a simple eigenvalue. ⊓⊔