In this paper we obtain some inequalities for the pressure offXover the transformationT in relation to the pressure offY over the transformationSand offGoverτ

http://jipam.vu.edu.au/

Volume 7, Issue 3, Article 95, 2006

THE PRESSURE OF FUNCTIONS OVER (G, τ)–EXTENSIONS

MOHD. SALMI MD. NOORANI SCHOOL OFMATHEMATICALSCIENCES

UNIVERSITIKEBANGSAANMALAYSIA

43600 BANGI, SELANGOR

MALAYSIA

msn@pkrisc.cc.ukm.my

Received 28 June, 2005; accepted 31 March, 2006 Communicated by S.S. Dragomir

ABSTRACT. Let T: X → X be a (free) (G, τ)–extension of S: Y → Y. Moreover let fX, fY, fG ≥ 0be continuous functions defined onX,Y andGrespectively. In this paper we obtain some inequalities for the pressure offXover the transformationT in relation to the pressure offY over the transformationSand offGoverτ.

Key words and phrases: (G, τ)–extensions, Topological pressure.

2000 Mathematics Subject Classification. 37B40.

1. INTRODUCTION

Let T: X → X be a continuous map of a compact metric space X and τ: G → G be an automorphism of a compact metric group G. Suppose Gacts continuously and freely on the right of T so that the equationT(xg) = T(x)τ(g)holds true ∀x ∈ X, g ∈ G. Moreover let Y be theG–orbit space andS is the natural map onY defined byS(xG) = (T x)G, ∀x ∈ X.

ThenT is called a(G, τ)–extension ofS.

Bowen [1] studied the topological entropy of the aforementioned extension system and, amongst other things, showed that the following formula holds:

h(T) =h(S) +h(τ), whereh(·)is the topological entropy of the appropriate maps.

In this paper, we are interested in the pressure analogue of Bowen’s formula, i.e., we consider the pressure of functions defined on the respective dynamical systems instead of topological entropy. Unfortunately the main result of this paper (i.e., the analogue of Bowen’s formula, see Corollary 4.6) is somewhat short of an equality. Our examples indicate that equality holds but we are unable to prove this in general. The proofs of the results arrived at in this paper are of course modelled along the lines of Bowen.

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

197-05

2. PRESSURE

We shall recall some elementary facts about pressure which are relevant to us in proving our results. The references for this section are [2] and [4].

As before letT: X →Xbe a continuous map of a compact metric space(X, d). Throughout this paper we shall assume thatT have finite topological entropy. LetK be a compact subset ofX. A subsetF ofX is said to be an(n, )-spanning set forK if for givenk ∈K then there exists x ∈ F such that d(Tⁱ(k), Tⁱ(x)) ≤ , ∀0 ≤ i ≤ n −1. Now let f be a continuous real-valued function defined onXand consider the set defined by:

Q_n(T, f, , K) = inf (

X

x∈F

e^Snf(x): F (n, )-spansK )

.

(Here we have used the standard notation: S_nf(x) := f(x) +f(T x) +· · ·+f(Tⁿ⁻¹x).) Then it is easy to see thatQ_n(T, f, , K)≤ ||e^Snf(x)||r_n(T, , K)wherer_n(T, , K)is the cardinality of an(n, )-spanning set forKwith a minimum number of elements. In particular, by virtue of compactness and continuity, we have0< Q_n(T, f, , K)<∞. Now define:

Q(T, f, , K) = lim sup

n→∞

1

nlogQ_n(T, f, , K) Lemma 2.1. Q(T, f, , K)<∞

Proof. We know that

Q_n(T, f, , K)≤ ||e^Snf(x)||r_n(T, , K)<∞.

Hence, since||e^Snf(x)|| ≤e^n||f(x)||, we have

Q_n(T, f, , K)≤e^n||f(x)||r_n(T, , K).

Thus 1

nlogQ_n(T, f, , K)≤ ||f||+ 1

n logr_n(T, , K).

In particular we have

Q(T, f, , K)≤ ||f||+ lim sup

n→∞

1

nlogr_n(T, , K).

It is well known thatlim sup_{n→∞ n}¹logrn(T, , K)<∞. Hence this completes the proof.

We are now ready to define the pressure: The pressure off with respect to the subsetK of Xover the mapT: X →X is defined by the quantity:

P(T, f, K) = lim

→0Q(T, f, , K).

Remark 2.2.

(1) As is well-known, the metric onX can be arbitrarily chosen as long as it induces the same topology onX.

(2) WhenK =X we obtain the usual definition of the pressure,P(T, f), of the functionf over the mapT:X →X.

(3) Recall thatEis an(n, )-separated set ofK ⊂Xif for any two distinct pointsx, y ∈E there exists some 0 ≤ i < nsuch thatd(Tⁱ(x), Tⁱ(y)) > . It can be checked that the above definition can also be arrived at by using separating sets. In this case we shall be concerned with the quantity

P_n(T, f, , K) = sup (

X

x∈E

e^Snf(x):E (n, )-separatesK )

.

As is well known, the next step is to define P(T, f, K) = lim

→0lim inf

n→∞

1

nlogP_n(T, f, , K).

The following two results are straight forward consequences of the definition of pressure.

Proposition 2.3.

P(T, f, K)≤P(T, f) wheneverK ⊂X

Proof. LetF (n, )-spanX. ThenF also(n, )-spansK ⊂X. Hence inf

( X

x∈F

e^Snf(x): F (n, )-spansK )

≤inf (

X

x∈F

e^Snf(x): F (n, )-spansX )

.

ThusQ_n(T, f, , K)≤Q_n(T, f, ). The result follows by taking the appropriate logarithms and

limits.

Lemma 2.4. Lets_n(T,4, X)denote the cardinality of a(n,4)-separated set ofXwith maxi- mum number of elements. Then

P_n(T, f,8, X)≤e^n||f||s_n(T,4, X).

Proof. For any > 0, it is not difficult to check that P_n(T, f,2, X) ≤ Q_n(T, f, , X) and r_n(T, , X)≤s_n(T, , X). Hence sinceQ_n(T, f, , X)≤e^kSnfkr_n(T, , X)we have

P_n(T, f,8, X)≤Q_n(T, f,4, X)

≤e^kSnfkr_n(T,4, X)

≤e^nkfks_n(T,4, X).

3. GENERALEXTENSIONS

In this section we shall start off with a straightforward modification of a crucial estimate of Bowen and later show how this estimate is used when dealing with pressure. But first recall the following definition:

Let T: X → X and S: Y → Y be continuous maps of compact metric spaces X and Y. Moreover letπbe a continuous surjective map fromX toY such thatπ◦T =S◦π. Then as is well-knownT is called an extension ofS.

With respect to this extension system, we have the following result which is essentially due to Bowen [1].

Lemma 3.1. Let > 0, α > 0 and integer n > 0be arbitrary. Also let f_X be a continuous positive function defined on X. Then there exists some δ > 0 such that if Y_n is an (n, δ)- spanning set for Y with minimum cardinality then for any (n,4)-separating set F of X we have

Card.F ≤Card.Y_n·e^(a+α)(n+M),

wherea= sup_y∈Y P(T, f_X, π⁻¹(y))andM is some finite positive real number.

Proof. Leta = sup_y∈Y P(T, f_X, π⁻¹(y)). For each y ∈ Y, choose an integerm(y) > 0such that

a+α ≥P(T, f_X, π⁻¹y) +α

≥ 1

m(y)logQ_m(y)(T, f_X, , π⁻¹(y)).

(*)

Also for each y ∈ Y let E_y be a (m(y), )-spanning set of π⁻¹(y). Now consider the open neighborhood ofπ⁻¹(y)

U_y = [

z∈Ey

m(y)−1

\

k=0

T^−kB₂(T^kz).

Then it is clear that

(X/U_y)∩\

γ>0

π⁻¹(B_γ(y)) =∅,

where B_γ(y) is the open-ball centered at y with radius γ. Since X is compact, the finite in- tersection property for compact sets then implies there exists some γ = γ(y) > 0such that π⁻¹(B_γ(y)) ⊂ U_y. In particular, sinceX is compact, there existsy₁, y₂, . . . , y_r ∈ Y such that Y is covered by the open ballsB_γ(y_i),i= 1,2, . . . , rand

π⁻¹(B_γ(y_i))⊂U_yi, where

U_yi = [

z∈E_yi

m(yi)−1

\

k=0

T^−kB₂(T^kz) andE_yi is an(m(y_i), )-spanning set ofπ⁻¹(y_i).

Now letδ >0be the Lebesgue number of this cover ofY and letY_nbe a(n, δ)-spanning set forY with minimum cardinality. Hence for eachy∈ Y_nwe can definec_j(y)as the element in {y₁, y₂, . . . , y_r}satisfyingB_δ(S^j(y))⊂B_γ(c_i(y))for eachj = 0,1, . . . , n−1.

Next define recursively the positive integers t₀(y) = 0, t_s(y) =

s−1

X

r=0

m c_tr(y)(y)

for each s = 1,2, . . . , q, where q = q(y) having the property t_q+1(y) ≥ n. Now for each q+ 1-ple

(x0, x1, . . . , xq)∈E_ct

0(y)(y)×E_ct

1(y)(y)× · · · ×E_ctq(y)(y) define the set

V(y; (x₀, x₁, . . . , x_q)) =

x∈X:d T^t+ts(y)(x), T^t(x_s)

<2

∀0≤t < m c_ts(y)(y)

& 0≤s≤q}. Then it is easy to check that

[

(y;(x0,x1,...,xq))

V(y; (x₀, x₁, . . . , x_q)) =X and ifF is an(n,4)-separated subset ofX, then

(3.1) Card.(F ∩V(y; (x₀, x₁, . . . , x_q))) = 0or1

for eachq+ 2-ple(y; (x₀, x₁, . . . , x_q)),wherey∈Y_nandx_s∈E_cts(y)(y),s= 0,1, . . . , q.

To complete the proof of this lemma, we shall now obtain an estimate for the numberN_y of theq+ 1-ple’s(x₀, x₁, . . . , x_q(y)). Since

N_y =

q(y)

Y

s=0

r_m(^cts(y)(y)) T, , π⁻¹ c_ts(y)(y) we have by virtue of(∗)andf_X ≥0

logN_y ≤

q(y)

X

s=0

logr_m(^cts(y)(y)) T, , π⁻¹ c_ts(y)(y)

≤

q(y)

X

s=0

logQ_m(^cts(y))(T, f_X, , π⁻¹(c_ts(y)))

≤

q(y)

X

s=0

m c_ts(y)(y)

(a+α).

Recall that

t_q+1(y) =

q(y)

X

r=0

m c_ts(y)(y) . Alsot_q+1(y) =t_q(y) +m c_tq(y)(y)

. Therefore

t_q+1(y)≤n−1 +m c_tq(y)(y)

≤n+M, whereM = max{m(y₁), m(y₂), . . . , m(y_r)}.

Hence

logN_y ≤(n+M)(a+α) so thatN_y ≤e^(n+M)(a+α). In particular, (3.1) now implies

Card.F ≤Card.Y_n·N_y

≤Card.Y_n·e^(n+M)(a+α)

and this completes the proof of this lemma.

Some remarks are in order:

(1) Apart from the choices of integersm(y), the rest of the proof of the above lemma is an exact copy of Bowen’s theorem [1, Theorem 17].

(2) WhenF is a(n,4)-separating set ofXwith maximum cardinality, then s_n(T,4, X)≤r_n(S, δ, Y)e^(n+M)(a+α)

where as befores_n(T,4, X)is the cardinality of suchF andr_n(S, δ, Y)is the cardinal- ity ofY_n.

The following theorem is the pressure analogue of Bowen’s Theorem [1, Theorem 17] (see also [3]).

Theorem 3.2. Letf_X andf_Y be continuous real-valued functions defined onX andY respec- tively such thatf_X ≥0andf_Y ≥0. Then

P(T, f_X)≤ ||f_X||+P(S, f_Y) + sup

y∈Y

P(T, f_X, π⁻¹y).

Proof. Let > 0, α >0and integern >0be arbitrary anda = sup_yP(T, f_X, π⁻¹y). Also let s_n(T,4, X)denote the cardinality of a(n,4)-separated set of X with maximum cardinality.

Then Lemmas 2.4 and 3.1 give us

P_n(T, f_X,8, X)≤e^nkfXks_n(T,4, X)

≤e^nkfXkr_n(S, δ, Y)e^(n+M)(a+α)

≤e^nkfXkQ_n(S, f_Y, δ, Y)e^(n+M)(a+α), where the last line follows from the positivity off_Y.

Hence

logP_n(T, f_X,8, X)≤nkf_Xk+ (n+M)(a+α) + logQ_n(S, f_Y, δ, Y).

In particular, on dividing bynand taking limit superior, we have as→0 P(T, f_X)≤ kf_Xk+P(S, f_Y) +a+α.

The result then follows sinceαis arbitrary.

Corollary 3.3. Let X and Y be compact metric spaces and T: X → X and π: X → Y be continuous such thatπ◦T =π. Moreover letf ≥0be continuous onX.Then

sup

y∈Y

P(T, f, π⁻¹y)≤P(T, f)≤ ||f||+ sup

y∈Y

P(T, f, π⁻¹y).

Proof. The first inequality follows from Prop. 2.3. Then in the above theorem take S = Id,

f_Y = 0andf_X =f.

By takingf = 0in the above corollary, we have

Corollary 3.4. Let X and Y be compact metric spaces and T: X → X and π: X → Y be continuous such thatπ◦T =π. Then

h(T) = sup

y∈Y

h(T, π⁻¹y).

The last corollary is contained in [1, Corollary 18].

4. (G, τ)−EXTENSIONS

As in the introduction let T: X → X be a(G, τ) - extension ofS: Y → Y. For the rest of this paper letf_Y andf_G be positive real-valued functions defined onY andGrespectively.

Now define the functionf:X → Rasf(xg) =f_Y(xG) +f_G(g)so thatf is also positive and continuous.

Lemma 4.1. The functionf is well-defined.

Proof. Letxg = zg⁰ wherex, z ∈X andg, g⁰ ∈ G. Consider the projection mapπ: X → Y. Then π(xg) = π(zg⁰) implies xG = zG so that x = zg for some g ∈ G. Hence f(xg) = f(zgg) = f_Y(zG) +f_G(gg). Alsof(zg⁰) = f_Y(zG) +f_G(g⁰). And this impliesf(xg) = f(zg⁰) iffG(gg) = fG(g⁰). But this is true sincex=zg implieszgg =zg⁰and in turn by virtue of free acting this impliesgg =g⁰. In other wordsfG(gg) = fG(g⁰)and this completes the proof.

Lety ∈Y. Then recall thatQn(T, f, , π⁻¹y)is defined as Q_n(T, f, , π⁻¹y) = inf

( X

x∈F

e^Snf(x):F (n, )−spansπ⁻¹y )

. We have

Proposition 4.2. Giveny∈Y, integern ≥1and >0,

Q_n(T, f, , π⁻¹y)≤e^n||fY||Q_n(τ, f_G, δ) for someδ >0.

Proof. Let d andd⁰ be the metrics associated withX and G respectively. Then since Gacts continuously onX, we have by uniform continuity, there existsδ > 0such thatd(xg, xg⁰)≤ wheneverd⁰(g, g⁰) ≤ δ. Now letx ∈ π⁻¹yand E_n be a(n, δ)-spanning set for G. Then it is easy to check thatxEn is a (n, )-spanning set forπ⁻¹y. Observe that by commutativity ofT andS (viaπ) and the relationT(xg) =T(x)τ(g)we have

e^Snf(xg)=e^SnfY(xG)e^SnfG(g) with respect to the appropriate mapsT,Sandτ, so that

X

g∈En

e^Snf(xg) =e^SnfY(xG) X

g∈En

e^SnfG(g) or

X

xg∈zEn

e^Snf(xg) =e^SnfY(zG) X

g∈En

e^SnfG(g). Therefore

Q_n(T, f, , π⁻¹y)≤e^SnfY(xG)Q_n(τ, f_G, δ).

Note that the above manipulation is independent ofx ∈ π⁻¹ysince ifx⁰ ∈ π⁻¹ythenx⁰ =xg for someg ∈Gso thatx⁰G =xGwhich in turn impliesSnfY(x⁰G) = SnfY(xG). The result

follows sincee^SnfY(xG) ≤e^nkfYk.

Theorem 4.3. Giveny∈Y, we have

P(T, f, π⁻¹y)≤ kf_Yk+P(τ, f_G).

Proof. From Proposition 4.2 we have

Q_n(T, f, , π⁻¹y)≤e^n||fY||Q_n(τ, f_G, δ).

Therefore

lim sup 1

nlogQ_n(T, f, , π⁻¹y)≤ kf_Yk+ lim sup 1

nlogQ_n(τ, f_G, δ), that is

Q(T, f, , π⁻¹y)≤ ||fY||+Q(τ, fG, δ).

The result follows by taking→0andδ→0.

Combining Theorem 3.2 and Theorem 4.3 and takingf =f_X , we have Proposition 4.4.

P(T, f)≤ kfk+kf_Yk+P(S, f_Y) +P(τ, f_G) We also have:

Proposition 4.5.

P(T, f)≥P(S, f_Y) +P(τ, f_G)

Proof. Let > 0and let d⁰⁰ be the metric onY. Then by the uniform continuity ofπ and the fact thatG-acts freely onX there exists someδ >0such that

a. d⁰⁰(π(x), π(z))≤whend(x, z)≤δ

b.d(xg, xg⁰)> δ whenx∈Xandd(g, g⁰)> .

Now letG_n ⊂ Gbe(n, )-separated andY_n ⊂ Y be(n, )-separated and chooseX_n ⊂ X so thatπ|_Xn: X_n →Y_nis a bijection. ThenX_nG_nis(n, δ)-separated. Thus

P_n(T, f, δ)≥ X

xg∈XnGn

e^(Snf)(xg)=P_n(S, f_Y, )·P(τ, f_G, ).

In particular

P(T, f, δ)≥P(S, f_Y, ) +P(τ, f_G, ).

The result follows by taking→ ∞andδ→ ∞.

Corollary 4.6.

P(S, f_Y) +P(τ, f_G)≤P(T, f)≤ kfk+kf_Yk+P(S, f_Y) +P(τ, f_G).

And by takingf_Y ≡0≡f_G, we recover Bowen’s formula Corollary 4.7.

h(T) =h(S) +h(τ).

5. FINALREMARKS

Whenf_Y andf_Gare both constant, by using the variational formula for pressure and Bowen’s formula, it is easy to deduce thatP(T, f) = P(S, fY) +P(τ, fG), i.e., equality holds in this trivial case.

Perhaps, a non-trivial example supporting the equality is as follows:

Example 5.1. Let Y = {−1,1}^Z and σ: Y → Y be the full two-shift. Consider the group extension given by

ˆ

σ: Y ×Z3 →Y ×Z3

(y, g)7→(σy,(g+ 2y₀)mod3).

Of course, in this caseτ =Id. Also, letf(y, g) = f_Y(y) +f_G(g),wheref_Y(y) = 0ify₀ =−1, fY(y) = 1ify0 = 1andfG = 2, constant. Then one can easily check thatP(σ, fY) = log(1+e) andP(Id, fG) = 2. Moreover it is not difficult to see thatP(ˆσ, f) = log(e²+e³). In particular, we haveP(ˆσ, f) =P(σ, f_Y) +P(Id, f_G).

We end with the following conjecture:

Conjecture 5.1. LetT: X → X be a (free)(G, τ)–extension ofS: Y → Y such thatT has finite topological entropy. Also let f: G → Rbe defined asf(xg) = f_Y(xG) +f_G(g)where fY,fGare positive real-valued functions onY andGrespectively. Then

P(T, f) =P(S, f_Y) +P(τ, f_G).

REFERENCES

[1] R. BOWEN, Entropy for group endormorphisms and homogeneous spaces, Trans. Amer. Math. Soc., 153 (1971), 401–414.

[2] A. KATOK AND B. HASSELBLATT, Introduction to the Modern Theory of Dynamical Systems, Cambridge University Press, Cambridge, 1995.

[3] H.B. KEYNES, Lifting topological entropy, Proc. Amer. Math. Soc., 24 (1970), 440–445.

[4] P. WALTERS, An Introduction To Ergodic Theory, GTM 79, Springer-Verlag, Berlin, 1982.