1864
S. Zakeri
isometrically to the unique Poincare´ geodesic between c(f (U)) and f (p) ∈ ∂f (U). (Note
that by our assumption f (U) = U unless U is a fixed Siegel disk for which the α fixed
point is the center. So in any case α is still a fixed point of F.) Apply this construction to
every bounded Fatou component and let F = f everywhere else. The map F will be the
required modification of f which satisfies the following properties.
(F1) F(c(U)) = c(F(U)) for every bounded Fatou component U. In particular, by (C2),
whether or not the critical point 0 belongs to the Fatou set, F(0) = f (0) = c is
always the critical value of f .
(F2) F = f on the closure of the basin of attraction of infinity.
(F3) F(z) = F(z0) ⇔ z = ꢁz0.
(F4) α and β are the only fixed points of F.
Also, since the support of the Brolin measure is the Julia set where f and F agree, it
follows that properties (iii) and (v) in §1.1 also hold for F. In other words,
(F5) µ(F−1(A)) = µ(A) for any measurable set A ⊂ C, and
(F6) µ(F(A)) = 2µ(A) for any measurable set A ⊂ C for which F|A is one-to-one.
LEMMA 2. Let x, y, . . . , z ∈ K. Suppose that the critical point 0 is not an interior
point of the tree [x, y, . . . , z]. Then F maps [x, y, . . . , z] homeomorphically to
[F(x), F(y), . . . , F(z)].
In this case, we simply write
'
F : [x, y, . . . , z] −→ [ F(x), F(y), . . . , F(z)].
Proof. First let us show that F restricted to [x, y, . . . , z] is injective. If not, it follows
from (F3) that [x, y, . . . , z] contains a pair ꢁa of symmetric points. By (1.3), we see that
[a, −a] = −[a, −a]. Hence the 180◦ rotation from the arc [a, −a] to itself must have
a fixed point, namely the critical point 0. But this implies that 0 is an interior point of
[x, y, . . . , z], contrary to our assumption.
Therefore, F restricted to [x, y, . . . , z] is injective. The image tree F([x, y, . . . , z]) is
evidently connected and contains all of the image points F(x), F(y), . . . , F(z). Since all
the ends of F([x, y, . . . , z]) are among F(x), F(y), . . . , F(z), we conclude that it is also
minimal. To finish the proof, it is enough to show that the image of every regulated arc in
[x, y, . . . , z] is a regulated arc. But this follows from (F1) since F preserves the centers,
hence the radial arcs, in bounded Fatou components of f .
Definition 1. By the spine of the filled Julia set K we mean the unique regulated arc
[−β, β] between the β-fixed point and its preimage −β, which are the landing points
of the unique external rays R0 and R1/2, respectively. By (1.3), the spine is invariant under
the 180◦ rotation z → −z. In particular, the critical point 0 always belongs to the spine.
Let z ∈ J be a biaccessible point, with a ray pair (Rt , Rs) landing at z and 0 <
t < s < 1. If z ∈/ [−β, β], it follows that both t and s satisfy 0 < t < s < 1/2 or
1/2 < t < s < 1. Consider the orbit of the ray pair (Rt , Rs) under f . Since there exists
an integer n > 0 such that 1/2 ≤ 2ns − 2nt < 1, the corresponding rays f ◦n(Rt) and
f
◦n(Rs) must belong to different sides of the curve R1/2 ∪ [−β, β] ∪ R0 (see Figure 3).