Lecture 27:
HW2 session: Thursday, April 2, 5:30 until we are done (with a
break for O3).
Recall defn of P (T , f ) via (n, δ)-separated sets, in particular topological entropy h(T ) : P (T , 0). We also outlined the original defn.
of topological entropy using open covers, but we did not show the
equivalence of the definitions.
Recall defn. of expansiveness
Prop: Expansiveness is a top. conjugacy invariant (but with possibly different expansive constants)
Proof: let φ be a topological conjugacy from T on M to S on
N and T be expansive with expansive constant δ. Since φ−1 is
continuous, there exists > 0 s.t. if x, y ∈ N, d(x, y) < , then
letting u = φ−1(x), v = φ−1(y), d(u, v) < δ.
We claim that S is expansive constant . If not, the for some
x, y ∈ N, x 6= y, and all n ∈ G, d(S nx, S ny) ≤ .
Then for all n ∈ G,
d(T nu, T nv) = d(T nφ−1x, T nφ−1y) = d(φ−1S nx, φ−1S ny) ≤ δ
and u 6= v, a contradiction Note: expansiveness is not necessarily preserved under topological
factor or topological extension.
Recall:
Example: a two-sided shift space is expansive as a Zd-action.
Therefore horseshoe map on Ω is expansive.
Theorem: If T is expansive with expansive constant δ > 0, then
P (T , f ) = Pδ (T , f ).
Proof of Theorem:
We need a defn. and a lemma.
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We first assume that the action is expansive as a Zd+-action.
Defn: Let T be a a cts. Zd+-action. For a positive integer k, δ > 0,
x ∈ M , let
U (k, δ, x) := {y ∈ M : d(T i(y), T i(x)) ≤ δ, ∀i ∈ Dk }
= ∩i∈Dk T −i(B δ (T i(x))
Note: E is (k, δ)-separated iff for all x, y ∈ E, x 6= y, we have
y 6∈ U (k, δ, x).
Lemma (a“finite” version of expansiveness): Let T be an expansive Zd+-action with expansive constant δ > 0. For all 0 < δ 0 < δ,
there exists k = k(δ 0) s.t. for all x ∈ M , we have
U (k, δ, x) ⊆ Bδ0 (x).
Proof: Suppose not. The for some 0 < δ 0 < δ and all positive
integers k, there exist xk , yk ∈ M s.t. d(T i(yk ), T i(xk )) ≤ δ, for all
i ∈ Dk and d(xk , yk ) ≥ δ 0. By compactness, for some x, y ∈ M
and some sequence kj , xkj → x and ykj → y. But then d(x, y) ≥
δ 0 > 0 and for all i ∈ Zd+, d(T i(y), T i(x)) ≤ δ (since Zd+ = ∪k Dk ),
contradicting expansiveness with expansive constant δ. Proof of Theorem: We must show that for all δ 0 < δ, Pδ0 (T , f ) =
Pδ (T , f ).
It follows from the Lemma that for all positive integers n, 0 <
δ 0 < δ and x ∈ M ,
U (n + k(δ 0), δ, x) ⊆ U (n, δ 0, x).
So, any (n, δ 0)-separated set E is (n + k(δ 0), δ)-separated: for x, y ∈
E, x 6= y, we have y 6∈ U (n, δ 0, x) and so y 6∈ U (n + k(δ 0), δ, x).
Thus,
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Pn+k(δ0),δ (T, f ) ≥ Pn,δ0 (T, f )
Thus, Pδ (T , f ) ≥ Pδ0 (T , f ) and the reverse inclusion is obvious.
For an expansive Zd-action, first change the proof by redefining:
U (k, δ, x) := {y ∈ M : d(T i(y), T i(x)) ≤ δ, ∀i ∈ Bk }
(where Bk = [−k, k]d). Then Lemma holds.
And replace “any (n, δ 0)-separated set E is (n+k(δ 0), δ)-separated”
by
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“for any (2n, δ 0)-separated set E, we have that T −k(δ )E is (2n +
2k(δ 0), δ)-separated.”
And observe that while Pδ is defined as a lim sup over Pn,δ , not as
a limit, we get the same value if we take the lim sup over P2n,δ . This result is, in some sense, an analogue of the Sinai Generator
Theorem for ergodic theory and gives us some hope of computing
pressure and topological entropy.
Example: Let M = XF be a (two-sided) shift space (with shift
action σ = σ|M ), which is expansive as a cts Zd-action with 1 as
expansive constant.
Let GAn = GAn(M ) be the set of configurations on Dn =
[0, n−1]d which extend to an element of M (GA stands for “globally
admissible”).
Recall that E ⊂ M is (n, 1)-separated iff for all x, y ∈ M, x 6= y,
we have x|Dn 6= y|Dn , i.e., x and y disagree on at least one site of
Dn.
Thus, a maximal (n, 1)-separated set has size |GAn|. Thus,
h(σ) = P1(σ, 0) = lim (1/nd) log |GAn|
n→∞
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Lecture 28:
Recall: For any (two-sided) shift space M = XF , the shift action
T = σ|M ) is expansive as a cts Zd-action with expansive constant =
1.
And we showed:
Prop:
h(T ) = h(σ|M ) = P1(σ|M , 0) = lim (1/nd) log |GAn|
n→∞
where GAn = GAn(M ) be the set of configurations on Dn = [0, n −
1]d which extend to an element of M .
Idea: For each n, let En be a maximal (n, 1)-separated set En
that achieves Pn,1(T , 0). The map En → x|Dn is a bijection from
En onto GAn.
Note that the limit exists because of subadditivity.
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Example: For full (two-sided) shift on two symbols, |GAn| = 2n
and so h(σ) = log 2.
Example: For Ledrappier 3-dot, we have d = 2 and |GAn)| ≤ 22n
and so h(σ|M ) = 0.
Expansiveness is not always needed to compute topological entropy or pressure:
Example: a rotation of the circle is not expansive: for all δ if
d(x, y) < δ, then for all n, d(T nx, T ny) < δ.
Claim: h(T ) = 0.
Proof: Any (n, δ) separated set is (1, δ)-separated because T is
an isometry and so if d(T i(x), T i(y)) > δ, then d(x, y) > δ. So,
Pn,δ (T, 0) is constant in n. So for all δ > 0, Pδ (T, 0) = 0. So,
h(T ) = 0. Back to shift spaces:
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The problem of deciding if a given w ∈ F Dn belongs to GAn is
undecidable.
Defn: For a positive integer n, let LAn = LAn(M ) denote the set
of configurations on Dn which do not contain an element of a forbidden list F that defines M = XF (LA stands for “locally admissible,”)
Think of n.n. SFT defined by a list F of forbidden adjacencies:
w ∈ LAn iff for all nearest neighbor configs 6∈ F.
Clearly, GAn ≤ LAn. There are often LA configs that are not
GA, even for n.n. SFT’s. But there is a decision procedure to decide
membership in LAn(M ).
Friedland’s Theorem: For a (two-sided) shift space M :
h(σ|M ) = lim (1/nd) log |LAn(M )|
n→∞
Limit exists again by subadditivity.
Proof: Clearly, LHS ≤ RHS.
For simplicity, consider d = 2.
Let GAn, LAn be the same as GAn, LAn except that we replace
Dn by Bn = [−n, n]d. Limits on both side of eqn. above are the same
if you replace GAn, LAn by GAn, LAn and replace nd by (2n + 1)d.
Claim: ∀k ∃a(k) such that for all u ∈ LAk ,
u ∈ GAk iff u extends to LAa(k).
Proof: By compactness, for u ∈ LAk \ GAk , there exists a(u)
s.t. u does not extend to an element of LAa(k). Set a(k) :=
maxu∈LAk \GAk a(u).
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By defn of a(k), if u ∈ LAk extends to LAa(k), then u ∈ GAk .
Clearly, if u ∈ GAk , then extends to LAa(k). Fix k. Choose n large and partition Bn into translates of Bk ,
reserving a border of width a(k).
|LAn| ≤ |GAk |(
2n+1−2a(k) 2
)
2k+1
|F |(8n+4)a(k)
log |LAn|
2n + 1 − 2a(k) 2 log |GAk |
≤
(
)(
)
(2n + 1)2
2k + 1
(2n + 1)2
(8n + 4)a(k) log |F |
)
+
(2n + 1)2
log |LAn| log |GAk |
≤
.
n→∞ (2n + 1)2
(2k + 1)2
lim
(1)
Let k → ∞. For nn Zd SFT M . Let Φ be a nn interaction on M , i.e., a
real-valued function defined on all allowed (i.e., non-forbidden) configurations on edges and vertices.
For w ∈ LAn(M ), define the energy function
X
Φ
U (w) :=
Φ(w(η)),
edges, vertices η contained in Dn
Define the partition function:
X
Φ
Φ
Zn :=
e−U (w).
w∈LAn
Example: Ising model: M is full Zd shift action, with F = {1, −1}
and Φ(wiwj ) = −βwiwj , for all adjacent vertices i, j in Zd.
Define
d
X
f Φ(x) := −Φ(x0) −
Φ(x0xei )
i=1
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Observe that Snf Φ(x) depends only on x|Dn+1 .
Prop: (a special case of Ruelle, Theorem 3.4) For nn Zd SFT and
f = f Φ for nn potential Φ,
P (T , f ) = lim (1/nd) log ZnΦ
n→∞
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Lecture 29:
Recall statement of:
Prop: (a special case of Ruelle, Theorem 3.4) For n.n. Zd SFT M
and a n.n. potential Φ,
P (σ|M , f Φ) = lim (1/nd) log ZnΦ
n→∞
Note: The RHS is often taken to be the defn of the pressure, P (Φ),
of a n.n. interaction on a n.n. SFT.
Note: Special case Φ = 0 reduces to computation of topological
entropy as in Friedland’s theorem.
Proof: For simplicity, consider d = 2. Write T = σ|M , f =
f Φ, U = U Φ, Zn = ZnΦ.
Let En be a maximal (n, 1)-separated set En that achieves Pn,1(T , f ).
The map En → x|Dn is a bijection from En onto GAn.
Let M1 = max Φ, M0 = min Φ.
Then
X
−U (w) −2nM1
e
e
w∈GAn
≤
X
Sn f (x)
e
X
≤
x∈En
e−U (w)e−2nM0
w∈GAn
So
2
P (T , f ) = P1(T , f ) = lim sup(1/n ) log(
n→∞
= lim (1/n2) log(
n→∞
X
eSnf (x))
x∈En
X
e−U (w))
w∈GAn
and by a modification of proof of Friedland’s Theorem,
X
2
= lim (1/n ) log(
e−U (w)) = lim (1/n2) log Zn
n→∞
n→∞
w∈LAn
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Modification of Friedland:
X
e−U
Φ (w)
w∈LAn
X
≤ ((
e−U (w))(
2n+1−2a(k) 2
)
2k+1
((e−M0 )8k+4)(
2n+1−2a(k) 2
)
2k+1
(|F |e−3M0 )a(k)(8n+4)
w∈GAk
Take log, divide by (2n + 1)2 and let n → ∞:
X
X
Φ
2
−U Φ (w)
2
lim (1/(2n+1) ) log
e
≤ (1/(2k+1) ) log
e−U (w)
n→∞
w∈LAn
w∈GAk
−(4M0)/(2k + 1)
Let k → ∞. Next, we compute this explicitly for d = 1.
Will use: Perron-Frobenius Theorem:
Let A be an irreducible matrix with spectral radius
λA := max eigenvalues of A |λ|. Then
1. λA > 0 and is an eigenvalue of A.
2. λA has a (strictly) positive eigenvector and is the only eigenvalue
with a nonnegative eigenvector.
Terminology: λA is called the Perron eigenvalue of A and its
corresponding eigenvectors are called Perron eigenvectors.
Brief discussion of proof of P-F:
Recall the geometric argument that for a primitive stochastic matrix P , with strictly positive eigenvector π, we have Pijk → πj :
Let W be the unit simplex in Rn. Define
g : W → W,
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x 7→
xA
xA1
In the primitive case, g contracts W into a single point, giving a
unique eigenvalue λ with positive eigenvector.
And λ has maximum modulus; for otherwise we would have leakage out of W .
Modify to take care of the general irreducible case: if A is irreducible, then for > 0, A + I is primitive.
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