Chaos and quasi-periodicity in diffeomorphisms of the solid torus

y [24, Theorem III.6.3] the map Q a ∗ is Collet-Eckmann (see e.g. [24, Section V.4]), that is,
there exist constants κ > 0 and λ > 1 such that
∣ d
∣∣∣
∣dx a ∗(Q a ∗(c)) ≥ κλ n for all n ≥ 0. (22)
Therefore, by combining [38, Theorem 3] with the Collet-Eckmann condition (22) we get
lim
n→∞
d
da Qn a (c) | a=a ∗
d
dx Qn−1 a ∗ (Q a
∗(c))
> 0. (23)
Assume Q k a ∗(c) = p, with p periodic (and repelling) under Q a∗. By p(a) denote the continuation
of p for a close to a ∗ . Then, for all n sufficiently large,
d
da Qn a (c) | a=a ∗ = ∂Qn−k a
∂a
= ∂
∂a Qn−k a
= d da
(Qk a ∗(c)) | a=a ∗ +∂Qn−k a
∂x (Qk a ∗(c)) | d
a=a ∗
(p) | a=a ∗ + ∂
d
(p) | a=a ∗
(
Q
n−k
a (p(a)) ) + ∂
∂x Qn−k a
∂x Qn−k a ∗
da Qk a (c) | a=a ∗=
[
p(a) + Q
k
da
a (c) − p(a) ] | a=a ∗=
(p) d [
Q
k
da a (c) − p(a) ] | a=a ∗ .
(24)
The point Qa
n−k (p(a)) belongs to a hyperbolic periodic orbit, that varies smoothly with the
parameter a. Therefore, its derivative with respect to a (which is the first term in the last
equality) is uniformly bounded in n. On the other hand,
d
dx Qn−1 a ∗
(Q a ∗(c)) = ∂
∂x Qn−k a ∗
(p) d
dx Qk−1 a ∗
(Q a ∗(c)).
Therefore, by (22), (23), and (24) we conclude that
d
da
0 < lim
Qn a (c) | [
d
a=a ∗
n→∞ d
dx Qn−1 a (Q ∗ a ∗(c)) = da Q
k
a (c) − p(a) ] a=a ∗
d
dx Qk−1 a (Q ∗ a ∗(c)) . (25)
This proves that Q a satisfies condition 6 of Definition 4.
We now show that the n-th iterate M a of the quadratic map is a d-family for all n > 1
and for some d ≤ 2 n . For simplicity, we denote Q a ∗ by Q for the rest of this proof. Condition
1 holds for M a ∗ since it holds for Q a ∗. Condition 3 follows from the fact that the composition
of maps with negative Schwarzian derivative also has negative Schwarzian derivative, see
e.g. [24, II.6]. Condition 4 is obviously satisfied.
Condition 2 is now proved by induction on n, where the case n = 1 is obvious. Obviously,
the set Cr M a ∗ of critical points of M a ∗ has cardinality d ≤ 2 n − 1. Moreover,
Cr M a ∗ = Q ( −1 Cr Q n−1) n−1
⋃
∪ Cr Q = (Q −j )(Cr Q). (26)
Suppose that condition 2 holds for a given n ≥ 1. We first show that
j=0
(Q n+1 ) ′′ (x) ≠ 0 for all x ∈ Cr Q n+1 . (27)
By (26), if x ∈ Cr Q n+1 then either x = c, or Q(x) ∈ Cr Q n . If x = c then
(Q n+1 ) ′′ (x) = (Q n ) ′ (Q(c)) · (Q) ′′ (c). (28)
22