Lecture 33:
Recall general variational principle.
Recall the Theorem that for expansive actions, sup is achieved in
the variational principle.
Review proof of this Theorem, assuming steps 1, 2 and 3. Steps 1
and 3 are not difficult and should be pretty believable.
Idea of Proof of Step 2: for each m and A ∈ αDm ,
∂A ⊆ ∪j∈Dm ∪i∈{1,...,k} ∂T −j (Ai) ⊆ ∪j∈Dm ∪i∈{1,...,k} T −j (∂Ai).
Since α is µ-good and T is measure preserving, we have µ(∂A) = 0.
So, for fixed m,
lim Hµn (αDm ) = Hµ(αDm )
n→∞
So,
lim sup hµn (T , α) ≤ lim sup(1/md)Hµn (αDm ) = (1/md)Hµ(αDm )
n→∞
n→∞
Let m → ∞ and we get:
lim sup hµn (T , α) ≤ lim (1/md)Hµ(αDm ) = hµ(T , α)
n→∞
m→∞
Defn: An equilibrium state for T , f is an element of M(T ) which
achieves the pressure. Let E(T , f ) denote the set of equilibrium
states.
We know that E(T , f ) may be empty, but if T is expansive, then
E(T , f ) is not empty.
There is a notion of Gibbs state for a large set of interactions (call
these admissible interactions. These are much more general than n.n.
1
interactions). They generalize the notion of Gibbs state for the Ising
model introduced ealier in the class.
There is also a corresponding set of continuous functions (call these
admissible functions).
Under a mild assumption on a shift space M , the equilibrium
states for σ|M and admissible function f coincide with the Gibbs
states for admissible interactions on M . This will be shown by Nishant in his talk.
Spencer will show that for 2-dimensional Ising model with sufficiently high β (inverse temperature), there is more than one Gibbs
state and thus by the correspondence above, there can be more than
one equilibrium state. Thus, for very simple functions, coming from
very simple (in fact nearest neighbour) interactions in d = 2, the
equilbrium state can fail to be unique.
However, in d = 1, there is a class of very complicated functions
φ for which the equilibrium state is unique. This is given in Bowen,
Theorem 1.22 below.
Given a 0-1 matrix C, recall the two-sided n.n. Z-shift of finite
type (a.k.a. topological Markov chain)
MC = {x ∈ F Z : Cxi,xi+1 = 1, for all i ∈ Z}
Defn: For a continuous function φ : MC → R and positive integer
k, let
vark (φ) = sup{|φ(x) − φ(y)| : xi = yi, for all |i| ≤ k}
Lets HC denote the set of all exponential continuous functions on
MC i.e., functions for which there exist α ∈ (0, 1), b > 0 s.t. for all
k, vark (φ) ≤ bαk .
2
Clearly each element of HC is continuous.
Theorem (Theorem 1.22, Bowen): Assume MC is topologically
mixing and φ ∈ HC . Then E(σMC , φ) consists of a single point, i.e.,
σMC , φ has a unique equilibrium state.
Notes:
– Recall that MC topologically mixing essentially means that C is
primitive.
– I believe that the result holds for top. transitive MC , i.e. C is
irreducible.
Defn: Given a 0-1 matrix C, let
MC+ = {x ∈ F Z+ : Cxi,xi+1 = 1, for all i ≥ 0}
called a one-sided Z+ shift of finite type (a.k.a. one-sided topological Markov chain).
MC+ has a metric consistent with the product topology, similar to
the metric for MC .
Note that the shift map σ|M + is onto, but typically not 1-1.
C
Let C(M ) denote the set of continuous functions on M .
For a continuous function φ on MC+, define the operator on Lφ :
C(MC+) → C(MC+)
X
Lφf (x) :=
eφ(y)f (y)
y∈σ −1 (x)
Let L∗φ denote the dual operator, which acts on measures, and is
defined: for g ∈ C(MC+),
Z X
L∗φν(g) := ν(Lφ(g)) =
eφ(y)g(y)dν(x)
y∈σ −1 (x)
3
Let HC+ denote the set of exponentially continuous functions on
MC+, i.e., mimic the definition of HC above and replace i ∈ Z with
i ∈ Z+.
Ruelle’s Perron-Frobenius Theorem (1.7 of Bowen): Let MC be
topologically mixing and φ ∈ HC+, there exist
1. λ > 0
2. h ∈ C(MC+), h > 0, s.t. Lφh = λh
3. a Borel probability measure ν on MC+ s.t. L∗φν = λν
R
4. hdν = 1 (normalization)
There is a 5th part which requires topological mixing, but I believe
the above only requires topological transitivity.
In the proof of Theorem 1.22 of Bowen, he first reduces to the case
that φ ∈ HA+, then uses Ruelle’s PF to define a measure µ := hdν
on MC+, and then naturally extends µ to a measure on MC . This
µ := µφ is the unique equilibrium state.
4
Lecture 34:
Recall statement of Bowen, Theorem 1.22 and Ruelle’s PF Theorem (Bowen, 1.7).
We have already proven Bowen, Theorem 1.22 in a very special
case, namely when Φ is a n.n. potential and φ(x) = −Φ(x0x1).
Recall that the unique equilibrium state was a first-order stationary
Markov chain with transition probability matrix:
rj
rj
Pij = CijΦ
= Cij e−Φ(ij)
λri
λri
where λ is the Perron eigenvalue of C Φ and r is a (positive) right
Perron eigenvector for C Φ.
Lemma: the unique stationary vector π for P is
πi = `iri
where ` is a (positive) left Perron eigenvector for C Φ.
Proof:
(πP )j =
X
πiPij =
i
=
X
`iCijΦ
i
X
ri`iCijΦ
i
rj
λri
rj λ`j rj
=
= `j rj . = πj
λ
λ
Let’s see how the ordinary PF theorem gives Ruelle’s PF Theorem
in this case, i.e., φ(x) = −Φ(x0x1).
Define
λ := λC Φ , h(x) := `x0
Then
Lφh(x) =
X
φ(y)
e
h(y) =
y∈σ −1 (x)
X
i: Cix0 =1
5
e−Φ(ix0)`i
= (`C Φ)x0 = λ`x0 = λh(x)
(since ` is a left Perron eigenvector).
Define Ea0...am−1 = {x ∈ MC+ : x0 = a0, . . . , xm−1 = am−1} (an
initial cylinder set) and
Qm−2 Φ
i=0 Cai ai+1 ram−1
ν(Ea0...am−1 ) =
λm−1
ν is a probability measure because r is a right Perron eigenvector,
e.g.
P Φ
X
b Ca0 b rb
= ra0 = ν(Ea0 )
ν(Ea0b) =
λ
b:Ca0 b =1
And it suffices to check L∗φν = λν on initial cylinder sets.
Z X
L∗φν(Ea0...am−1 ) =
eφ(y)χEa0...am−1 (y)dν(x)
y∈σ −1 (x)
= e−Φ(a0a1)ν(Ea1...am−1 )
Qm−2 Φ
i=1 Cai ai+1 ram−1
= λν(Ea0...am−1 )
= e−Φ(a0a1)
λm−2
Now we verify that this in this special case, the measure µ = hdν
is the Markov chain given by P and π above.
Z
µ(Ea0...am−1 ) =
h(x)dν(x) = `a0 ν(Ea0...am−1 )
Ea0 ...am−1
Qm−2
= `a0
i=0
CaΦiai+1 ram−1
λm−1
= `a0 ra0
m−2
Y
i=0
m−2
Y
r
a
i+1
Φ
Caiai+1
= π a0
Paiai+1
λrai
i=0
Being proof of 1.22, assuming Ruelle PF Theorem.
Bowen first reduces to φ ∈ C(MC+) (by simple lemmas 1.5 and
1.6).
6
Construct µ on MC from µ := hdν on MC+.
Lemma 0: for all f, g ∈ C(M+
C ), we have
P
a. (Lmf )(x) = y∈σ−m(x) eSmφ(y)f (y)
b. (Lmf ) · g = L(f · (g ◦ σ)).
Proof:
a. m = 1: defn; check by induction, e.g.
X
2
(L f )(x) = (L(Lf ))(x) = L(
eφ(y)(Lf )(y))
y∈σ −1 (x)
=
X
X
e
φ(y) φ(z)
e
f (z) =
y∈σ −1 (x) z∈σ −1 (y)
X
eS2φ(z)f (z)
z∈σ −2 (x)
b.
m
((L f )(x))·g(x) =
X
Sm φ(y)
e
f (y)g(x) =
y∈σ −m (x)
X
eSmφ(y)f (y)g(σ m(y))
y∈σ −m (x)
= L(f · g(σ m))
Lemma 1: µ is σ-invariant.
Proof: By Lemma 0,
Z
Z
Z
µ(g) = hgdν = (1/λ) (Lh) · gdν = (1/λ) L(h · (g ◦ σ)dν
= (1/λ)L∗ν(h · (g ◦ σ)) = ν(h · (g ◦ σ)) = µ(g ◦ σ)).
Lemma 2: σ is mixing (in particular, ergodic) w.r.t. µ.
Note: mixing is the same for one-sided and two-sided shifts.
Proof of Lemma 2:
First, we mention Part 5 of Ruelle’s PF Theorem:
7
For all g ∈ C(MC+),
lim ||λ−mLm
φg−(
m→∞
Z
gdν)h|| = 0.
Idea: Think of the iterating of induced mapping on the simplex, in
the finite dimensional case, that squashes the simplex to a single
point.
This requires top. mixing.
Suffices to check mixing for cylinder sets E and F . We have
µ(E ∩ σ −n(F )) = µ(χE · (χF ◦ σ n)) = ν(h · χE · (χF ◦ σ n))
= (1/λn)((L∗)nν)(h·χE ·(χF ◦σ n)) = (1/λn)ν(Ln(h·χE ·(χF ◦σ n)))
= ν((1/λn)(Ln(h · χE )) · χF )
by Lemma 0 (part b).
By part 5 of Ruelle PF Theorem,
(1/λn)(Ln(h · χE )) → ν((h · χE )h
uniformly. Thus,
(1/λn)(Ln(h · χE )) · χF → ν((h · χE )h · χF
uniformly. Now, integrate w.r.t. ν and we get:
µ(E∩σ −n(F )) = ν((1/λn)(Ln(h·χE ))·χF ) → ν(ν((h·χE )h·χF ) = µ(E)µ(F ).
Defn: µ ∈ M(σ|MC ) is called Gibbs-Bowen measure for φ if there
exist constants c1, c2 > 0 and P ∈ R s.t.
c1 ≤
µ(Ex0...xm−1 )
≤ c2
exp(−P m + Smφ(x)
8
for all x ∈ MC and all m ∈ Z+.
Lemma 3: P must be P (σ|MC , φ).
Proof: For fixed m, let Tm be a collection of exactly one element
of each initial cylinder set of length m that achieves max exp(Smφ(x)
on Ex0...xm−1 . Then
Pm
e
=e
Pm
(
X
X
µ(Ex0...xm−1 )) ≤ c2
x∈Tm
x∈Tm
= c2Pm,1(σ|MC , φ)
Take log, divide by m and let m → ∞, to get
P ≤ P1(σ|MC , φ) = P (σ|MC , φ)
Similarly, P ≥ P (σ|MC , φ). 9
e(Smφ(x)