TO1 CHAOTIC TIME SERIES
LYAPUNOV EXPONENTS AND SMOOTH
CONJUGACY
Let f,g : M → M (where M is a compact manifold, or a bounded domain in Rm) be discrete dynamical
systems; recall that f and g are (smoothly) conjugate if there is a smooth 1-1 map Φ : M → M with a
smooth inverse such that:
g
Φ °f°Φ-1
=
First note that if µf is an invariant measure for f then we can define an induced measure µg by
µg(B)
µf(Φ-1 (B))
=
for all measureable sets B ⊂ M. Then
µg(g-1 (B))
=
µf(Φ-1 (g-1 (B)))
=
µf(f-1(Φ-1 (B)))
=
µf(Φ-1 (B))
=
µg(B)
and hence µg is invariant for g.
Observe that since M is bounded (or compact) and DΦ, DΦ -1 are continuous, there exists a constant
C > 0 such that
C -1 v
≤
D xΦ.v
≤
C v
(*)
for all x∈M and all v. We will use this to show that the Lyapunov exponents of g with respect to µg
are the same as those of f with respect to µf .
Suppose that
[
] [D f ]
1
2n
[
][
1
2n

lim  D x f n
n→∞ 
†
n
x



=
Λf
=
Λg
and

lim  D y g n
n→∞ 
†
]

Dyg 

n
where y = Φ(x). We aim to show that Λf and Λg have the same eigenvalues.
Observe that α is an eigenvalue of Λf if and only if there exists a vector v ≠ 0 such that
 D f n .v
x
lim 
n→∞ 
v





1
n
=
α
i.e. so that the length of v grows as αn. Similarly α is an eigenvalue of Λg if and only if there exists a
vector v ≠ 0 such that
 D g n .v
y
lim 
n→∞ 
v





1
n
α
=
Now
D ygn.D xΦ
D x Φ. Dxfn
=
n
where xn = fn(x). Thus if we write u = DxΦ.v and apply (*) we have
C -1 v
≤
≤
u
C v
and
≤
C -1 D xfn.v
D ygn.u
≤
C D xfn.v
Thus
C −1 D x f n .v
D y g n .u
≤
Cv
u
C D x f n .v
≤
C −1 v
and hence
 D f n .v
x
lim 
n→∞ 
v





1
n
=
 D g n .u
y
lim 
n→∞ 
u





1
n
We thus see that α is an eigenvalue of Λf if and only if α is an eigenvalue of Λg, and therefore the
Lyapunov exponents of g with respect to µg are the same as those of f with respect to µf . A slightly
more careful analysis can also be used to show that the multiplicity of each exponent is identical for
both systems. This is left as an exercise.