E-Arnoldscat

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Lyapunov exponents and measures of Arnold’s cat map
Consider the solid torus T2 = R2/Z2 = S1 × S1. Let
fM
(
x
y
)
def
= f
(
x
y
)
mod 1, f
(
x
y
)
=
(
2x + y
x + y
)
=
(
2 1
1 1
)(
x
y
)
.
The matrix
(
2 1
1 1
)
has eigenvalues
λ1 =
3−
√
5
2
< 1 < λ2 =
3 +
√
5
2
.
with corresponding (orthogonal) eigenvectors
vs =
(
1−
√
5
2
1
)
, vu =
(√
5+1
2
1
)
.
The map fM : T2 → T2 is hyperbolic with splitting
(again we can identify the tangent space T(x,y)M with the Euclidean space)
T(x ,y)M = E
s(x , y)⊕ Eu(x , y),
where E s(x , y) = {tvs : t ∈ R} and Eu(x , y) = {tvu : t ∈ R}.
The hyperbolic splitting is (at every (x, y)) the Oseledets splitting, and
χ1(x , y) = log λ1 < 0 < χ2(x , y) = log λ2.
Theorem
The map fM : (x , y) 7→ (2x + y , x + y) mod 1 is topologically
semi-conjugate to the topological Markov chain σB : ΣB → ΣB
corresponding to the transition matrix
B =

1 0 1 1 0
1 0 1 1 0
1 0 1 1 0
0 1 0 0 1
0 1 0 0 1
 .
(ΣB ,BΣB , σ)
��
ΣB
π
��
σB // ΣB
π
��
ν
π∗
��
(T2,BT2 , fM) T2
fM // T2 µ = π∗ν
Note that ν σB -ergodic implies that π∗ν is fM -ergodic.
measures of maximal entropy of σB : ΣB → ΣB
Theorem
If B = (Bij)Ni ,j=1 is a transition matrix and λ = λ(B) its maximal
eigenvalue with (positive) eigenvectors uB = λu and Bv = λv .
Assume that
∑N
k=1 ukvk = 1. Define p = pB and P = PB by
pk = ukvk , Pij =
vj
λvj
Bij .
Then the (p,P)-Markov measure ν is the Parry measure of
maximal entropy.
The left (right) normalized eigenvector of B for λ2 = 3+
√
5
2 are
1 +
√
5
1 +
√
5
1 +
√
5
2
2
 ,

1 +
√
5
2
1 +
√
5
1 +
√
5
2
⇒ ...⇒ hν(σB) =
5∑
k=1
pk log pk = log
3 +
√
5
2
By Ruelle’s inequality (again this is also true for any other
fM -ergodic measure)
hµ(fM) ≤ χ2(µ) = log
3 +
√
5
2
.
Moreover, by semi-conjugation (extension) hµ(fM) ≤ hν(σB)
Observe that π fails to be invertible only on points which are
contained in the boundary of the elements of the corresponding
Markov partition, ∂P. Note that
S
def
=
⋃
k∈Z
f kM(∂P)
is fM -invariant and has (fM -invariant) T2-Haar measure zero.
Check that this measure is absolutely continuous to the
(fM -ergodic) measure µ = π∗ν, and hence coincides with it.
By the above, the following diagram commutes (that is,
π ◦ σB = fM ◦ π) and the two measure-preserving systems are
isomorph
(ΣB ,BΣB , σ)
��
ΣB
π
��
σB // ΣB
π
��
ν
π∗
��
(T2 \ S ,BT2\S , fM |T2\S) T2 \ S
fM // T2 \ S µ = π∗ν
Hence, their entropies coincide
hµ(fM) = hν(σB) = log
3 +
√
5
2
.