Let $X$ be a compact metric space and let $f\colon X\to X$ be a continuous expansive map. Let $\mathcal{V}$ denote the space of Hölder continuous potential functions $\phi\colon X\to \mathbb{R}$, and let $\mathcal{W}$ denote the set of potential functions (not necessarily Hölder continuous) that have a unique equilibrium state.

It is well known that if $f$ satisfies the specification property, then every $\phi\in \mathcal{V}$ has a unique equilibrium state, and hence $\mathcal{V} \subset \mathcal{W}$ (Rufus Bowen, Some systems with unique equilibrium states, Math. Systems Theory 8 (1974/75), no. 3, 193–202).

There are many systems without the specification property for which something is known about the set of potentials $\mathcal{W}$. For example, intrinsic ergodicity (existence of a unique measure of maximal entropy), which is equivalent to the statement that $\mathcal{W}$ contains the constant functions, has been studied in a number of recent works (Buzzi–Fisher, Bufetov–Gurevich, Climenhaga–Thompson), and there are stronger results for particular examples, such as $\beta$-shifts, for which $\mathcal{W}$ is known to contain the space of Lipschitz functions (Peter Walters, Equilibrium states for $\beta$-transformations and related transformations, Math. Z. 159 (1978), no. 1, 65–88).

However, I do not know of any examples of expansive maps without specification for which the inclusion $\mathcal{V} \subset \mathcal{W}$ is known. Does anybody know of such an example?

(Note that it should be not too difficult to adapt the answers to this question to obtain a system for which $\mathcal{W}$ contains the constant functions, but does not contain all of $\mathcal{V}$.)

Edit: Of course if $f$ is uniquely ergodic then $\mathcal{W}$ contains all potential functions. The most obvious examples of uniquely ergodic systems, irrational circle rotations (or rather, their symbolic counterparts, which are expansive) possess a weak version of the specification property, but I don't know if this weak specification holds for every uniquely ergodic system.

What I'd really like to know is if there is an expansive map that is not uniquely ergodic and does not have the specification property for which $\mathcal{V} \subset \mathcal{W}$. I'd also be interested in knowing whether unique ergodicity implies weak specification.

(By "weak specification" I mean that orbits can be consecutively shadowed with uniformly bounded gaps, as in the usual specification property, but that we do not require the shadowing orbit to be periodic, and we allow the length of the gaps to vary.)


2 Answers 2


In the time since I asked this question, Dan Thompson and I have posted a preprint showing, among other things, that every Hölder continuous potential function on a β-shift has a unique equilibrium state. Since β-shifts do not have specification for most β>1, this gives an example of the sort I was after.


Let $T_1 \colon X_1 \to X_1$ be an Anosov diffeomorphism and let $T_2 \colon X_2 \to X_2$ be a uniquely ergodic expansive homeomorphism which is not a periodic orbit. Let $T \colon X_1 \times X_2 \to X_1 \times X_2$ be given by the direct product of the two maps. Clearly $T$ is expansive, and $T$ does not have specification because it has no periodic orbits. The invariant measures of $T$ are precisely the products of the invariant measures of $T_1$ with the unique invariant measure of $T_2$ (right...?). So, calculating the equilibrium state(s) of a Hölder function defined on $X$ is the same problem as calculating the equilibrium state(s) of the function on $X_1$ defined by integrating $f$ along fibers against the unique invariant measure of $T_2$. The fiberwise integral has to be Hölder because $f$ is Hölder, and it follows that $f$ has a unique equilibrium state.

Edit: the sentence beginning "The invariant measures of $T$ are precisely..." is probably wrong - see comments below.

  • $\begingroup$ That might work, but I'm not entirely convinced yet. One potential issue is that some uniquely ergodic maps, such as irrational rotations, have a weak version of the specification property (for every $\epsilon>0$ there exists $\tau$ such that any collection of orbit segments can be $\epsilon$-shadowed by a single orbit that takes at most $\tau$ iterates to pass from one segment to the next, and is not required to be periodic). This version does not imply the existence of any periodic orbits. I'm not sure whether it still holds for the symbolic coding of such a rotation, which is expansive. $\endgroup$ Dec 3, 2010 at 14:50
  • $\begingroup$ Another potential issue is that even if two maps $T_1$ and $T_2$ are uniquely ergodic with measures $\mu_1, \mu_2$, the direct product $T_1 \times T_2$ can have other invariant measures besides $\mu_1\times \mu_2$. For example, if $R_\alpha$ is an irrational circle rotation, then $R_\alpha\times R_\alpha$ has lots of invariant measures, despite the fact that $R_\alpha$ is uniquely ergodic. So you need the measure-preserving systems $(T_1,\mu_1)$ and $(T_2,\mu_2)$ to be "disjoint", which may well be true in the case you mention, but I don't immediately know how to prove. $\endgroup$ Dec 3, 2010 at 14:53
  • $\begingroup$ Ah, good point. It seems very unlikely to me that an irrational rotation will be disjoint from all of the invariant measures simultaneously. Introducing skewing might solve the problem, but more likely is just going to make things complicated. $\endgroup$
    – Ian Morris
    Dec 3, 2010 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.