The variational principle for topological entropy says that if is a compact metric space and is a continuous map, then , where is the topological entropy, and the supremum is taken over all -invariant Borel probability measures. A measure achieving this supremum is called a *measure of maximal entropy* (MME for short), and it is interesting to understand when a system has a unique MME.

Let’s look at this question in the setting of symbolic dynamics. Let be a finite set, which we call an *alphabet*, and let be the set of all infinite sequences of symbols in . This is a compact metric space in a standard way, and the shift map defined by is continuous. We consider a compact -invariant set and ask whether or not has a unique MME.

When is a mixing subshift of finite type (SFT), this was answered in the 1960’s by Parry; there is a unique MME, and it can be obtained by considering the transition matrix for and using some tools from linear algebra. A different proof was given by Bowen in 1974 using the *specification* property; this holds for all mixing SFTs, but also for a more general class of shift spaces.

The purpose of this note is to describe a variant of Bowen’s proof; roughly speaking, we follow Bowen’s proof for the first half of the argument, then give a shorter version of the second half of the argument, which follows comments made in conversation with Dan Thompson and Mike Hochman.

**1. The strategy **

The *language* of the shift space is the set of all finite words that appear in some element of . That is, given we write , where . Then we write . Write for the length of .

In the setting of symbolic dynamics, the topological transitivity property takes the following form: is transitive if and only if for every there is such that . Transitivity by itself gives no control over the length of . We say that shift space has *specification* if there is such that for every there is with such that ; thus specification can be thought of as a uniform transitivity property.

Let be the topological entropy of . Bowen’s proof of uniqueness has the following structure:

- Show that there is such that for every .
- Prove that for every there is such that if is an MME and is a collection of words with , then we have . This can be thought of as a uniform version of the Katok entropy formula.
- Follow the proof of the variational principle to explicitly construct an MME ; then use the specification property and the counting estimates on to show that has the Gibbs property and is ergodic.
- Show that if is another ergodic MME, then the uniform Katok entropy formula for and the Gibbs property for cannot hold simultaneously; this contradiction proves uniqueness.

The proof we give here follows the above structure exactly for steps 1 and 2. Instead of steps 3 and 4, though, we give the following argument; if are two distinct ergodic MMEs, then we can use the uniform Katok entropy formula for together with the specification property to create more entropy in the system, a contradiction. In the next section we make all of this precise.

The advantage of this proof is that it allows us to replace step 3 of Bowen’s proof with a different argument that may be easier and less technical (depending on one’s taste). The disadvantage is that it does not include a proof of the Gibbs property for , which is useful to know about in various settings.

**2. The proof **

** 2.1. Counting estimates **

Write for the set of all words of the form , where and . Then it is clear that , so . In particular, is subadditive, so by Fekete’s lemma , and we get for every .

The upper bound requires the specification property. Define a map by sending to , where are provided by specification. This map is 1-1 so . Taking logs gives

and sending takes the left-hand side to , so ; this gives the counting bounds we claimed earlier, with .

The gluing construction in the previous paragraph will be used later on when we need to derive a contradiction by producing extra entropy.

** 2.2. Uniform Katok formula **

Now suppose is any MME, and given write . Similarly write . Applying Fekete’s lemma to the subadditive sequence , we get .

Given with , we have

Solving for gives

so taking with gives , verifying the uniform Katok formula.

Note that the argument here does not rely directly on the specification property; it only uses the fact that is an MME and that we have the uniform upper bound . In fact, if one is a little more careful with the computations it is possible to show that we can take , where . This has the pleasing property that as , so the lower bound for converges to the lower bound for .

** 2.3. Producing more entropy **

Now suppose are two distinct ergodic MMEs. Then so there are disjoint sets with and . Fix and let and be compact sets such that and . Then the distance between is positive, so there is such that for every , no -cylinder intersects both . In particular, putting , and similarly for with , we have

- for every , and
- and , so by the uniform Katok formula we have , where .

Up to now all of the arguments that we have made appear in Bowen’s proof; the approximation argument just given appears in step 4 of his proof, where one uses the Gibbs property of the constructed MME to derive a contradiction with the uniform Katok formula. It is at this point that our argument diverges from Bowen’s, since we have not proved a Gibbs property. Instead, we apply the specification property to the collections of words to get a lower bound on that grows more quickly than , a contradiction.

Before giving the correct argument, let me describe three incorrect arguments that illustrate the main ideas and also show why certain technical points arise in the final argument; in particular this will describe the process of arriving at the proof, which I think is worthwhile.

**First wrong argument**

Here is a natural way to proceed. With as above, consider for each and each the set of all words

where if and if , and are the gluing words provided by specification. Note that each choice of produces at least words in . Moreover, the collections of words produced by different choices of are disjoint, because and are disjoint. We conclude that

so

If then this would be enough to show that , a contradiction. Unfortunately since this argument does not work if , so we must try again…

**Second wrong argument**

Looking at the above attempted proof, one may describe the problem as follows: each of the words makes us lose a factor of from our estimate on . If , then instead of letting and range over and then gluing them, we could replace the words with the words . In particular, this would replace the estimate with the estimate .

This suggests that given , we should only keep track of the set for which , since if we can avoid losing the factor of by avoiding the gluing process.

So, let’s try it. Given , let (with and ), and consider the map

given by specification (whether the product ends with or depends on the parity of ).

Let be the image of , and note that

If the collections , were disjoint for different choices of , then we could fix and sum (1) over all with to get

where we use Sitrling’s approximation for the last inequality. Taking logs and dividing by gives

For sufficiently small this is , which gives , a contradiction.

The problem with this argument is that the collections need not be disjoint for different choices of ; this is because may have for some value of , so that we cannot necessarily recover uniquely from knowing .

**Third wrong argument**

Let’s try to address the issue just raised, that we cannot recover uniquely from because may have subwords in . We address this by only using words where there are `not many’ such subwords. More precisely, given for , let

be the set of `bad’ times, and similarly with the roles of and reversed. Let , and observe that since , invariance of gives

for every . We conclude that

Let , and note that

so

We conclude that

so taking , the uniform Katok estimate gives . A similar definition and argument gives .

Now we repeat the argument from the previous section (the second wrong argument) using in place of . Given , let be the image of the map with the corresponding restricted domain, and note that the estimate (1) still holds (with the new value of ).

The final piece of the puzzle is to take and estimate how many other collections can contain it; that is, how many possibilities there are for once we know . We would like to do the following: write and then argue that must be contained in the union of the sets of times corresponding to the . The problem is that this only works when there is a disagreement between which of the sets the maps are trying to use, and so I cannot quite make this give us the bounds we want.

**The correct argument**

To fix this final issue we change the construction slightly; instead of letting mark the times where we transition between , we do a construction where at each we transition from to and then immediately back to . Then will impose strong conditions on the set .

Let’s make everything precise and give the complete argument. As in the very beginning of this section we fix and let be disjoint compact sets with , where are distinct ergodic MMEs. Let be such that for every , no -cylinder intersects both .

Let , and similarly for . We repeat verbatim part of the argument from the last section; given for , let

be the set of `bad’ times. Let , and observe that since , invariance of gives

for every . We conclude that

Let , and note that

so

We conclude that

so taking , the uniform Katok estimate gives .

Now given and , let for (putting and ) and define a map

by the specification property, so that for and we have

where have length and are provided by specification. Let be the image of and note that

Given and , let denote the set of such that . Writing , note that implies that for each , either or there are consecutive elements of such that , and in this latter case we have that , so . We conclude that

By our choice of , for each the set on the right has at most elements. In particular, we have

This bounds the number of that can contain a given , and since there are distinct choices of with , the bound in (2) gives

Taking logs gives

Dividing by and sending gives

Putting this gives

Thus it suffices to make the appropriate choice of at the beginning of the proof. More precisely, let be as in the uniform Katok lemma, and let be small enough that . Then and so the estimate above gives

which contradicts our original assumption that was the topological entropy. This contradiction shows that there is a unique measure of maximal entropy.