Discrete Holomorphic Local Dynamical Systems

Dynamics of Rational Surface Automorphisms 95
Inspection shows that the exceptional curves are given by
Σ0 : = {x0 = 0} ↦→ p1 :=[0:1:0]
Σ β : = {x + b = 0} = {x1 + bx0 = 0} ↦→ p2 :=[0:0:1]
Σγ : = {y + a = 0} = {ax0 + x2 = 0} ↦→ q :=[1:−a :0].
This means that Σ0 minus the indeterminacy locus maps to p1, etc.Inorderto
find the exceptional curves in a more systematic way, we may take the jacobian of
the homogeneous form of f , which gives 2x0(bx0 + x1)(ax0 + x2) and which shows
us the exceptional curves directly.
If we set p∗ =(−b,−a)=[1:−b : −a], then the points of indeterminacy are
given by
p2 = Σ0 ∩ Σ β ↦→ p1p2 = Σ0
p1 = Σ0 ∩ Σγ ↦→ p1q =: ΣB
p∗ = Σ β ∩ Σγ ↦→ p2q =: ΣC
We note that the map f is not actually defined at the points of indeterminacy, and
we interpret these formulas to mean f −1 : Σ0 ↦→ p2,etc.
Let π : Y → P 2 be the space obtained by blowing up p1 and p2. Letusshow
that Σ0 is not exceptional for the induced map fY . We may use local coordinates
(t,x) ↦→ [t : x :1] at Σ0 = {t = 0}. Local coordinate chart U ′ at p1 =[0:1:0] is
given by (s1,η1) ↦→ [s1 :1:s1η1]. In these coordinates the map becomes:
(t,x) ↦→ [t : x :1] ↦→ f [t : x :1]=[f0 : f1 : f2]
=[f0[t : x :1]/ f1[t : x :1] :1: f2[t : x :1]/ f1[t : x :1]]
=[t/x :1:(1 + at)/(x(x + bt))] = [s1 :1:s1η1]
So we have
(t,x) ↦→ (s1,η1)=(t/x,(1 + at)(bt + x)),
which means that we have Σ0 = {t = 0}∋[0 :x :1] ↦→ (0,1/x) ∈ P1, soΣ0 is not
exceptional. Similarly, we have
Σ β → P2 → Σ0 → P1 → ΣB = {y = 0}. (10)
Exercise: Carry out the details of the remark concerning (10). In particular, show
that the only exceptional curve for the rational map fY : Y ��� Y is Σγ, and the only
point of indeterminacy is p∗.
Exercise: Compare Figures 7 and 10, and show that a map f corresponding to
Figure 10 must be linearly conjugate to a map of the form L ◦ J, whereL is linear,
and J is the involution from the previous section.
Now we can look at H 2 (Y )=〈H,P1,P2〉. We observe that ΣB is a line which
contains exactly one center of blowup, namely p1. Thus we have ΣB = H − P1 ∈
H 2 (Y). Similarly, since Σ0 contains both p1 and p2,wehaveΣ0 = H − P1 − P2.