# A weak-mixing, zero entropy measure on the 2-shift which gives equal weight to both symbols

I am currently sketching a paper in the general area of symbolic dynamics in which I would like to be able to use the following fact:

Proposition (proposed): there exists a shift-invariant measure $\mu$ on $\{0,1\}^{\mathbb{Z}}$ such that $\mu$ is weak-mixing and has zero entropy with respect to the shift, and such that $$\mu\left(\left\{(x_i)\in \{0,1\}^{\mathbb{Z}} \colon x_0=0\right\}\right)=\mu\left(\left\{(x_i)\in \{0,1\}^{\mathbb{Z}} \colon x_0=1\right\}\right)=\frac{1}{2}.$$ I would like to know if anyone can suggest a reference to an article or textbook which proves that such a measure exists, or failing that if anyone can think of a very quick and crisp proof.

My grounds for believing that the above proposition should be true are that in the set of all shift-invariant probability measures on $\{0,1\}^{\mathbb{Z}}$ equipped with the weak-* topology, there is a dense $G_\delta$ subset all of whose elements are weak-mixing and have zero entropy (see my earlier question for details). It would be somewhat bizarre if this residual set somehow completely missed the one-codimensional affine subspace of measures satisfying the above equation.

(It is not hard to show that there are weak-mixing, zero-entropy measures for which the two quantities in the above equation are arbitrarily close to one half, or to construct weak-mixing measures which satisfy the above equation and have arbitrarily small entropy, but I would like to be able to go the whole distance. I am indifferent to the matter of whether or not $\mu$ is also strong-mixing, but in order to have zero entropy it is well-known that it cannot be Bernoulli or Kolmogorov.)

Edited to add: all of the answers given below have been extremely helpful. After some thought I have decided to accept Tom's answer since in my opinion it most exactly answers the question as specified, but this is not to overlook the fact that Anthony and RW's answers are also very educational and are somewhat broader in their implications.

• Could you build such an example explicitly using substitutions? Take a substitution on symbols {0,1}, let \Sigma be the orbit closure of a periodic point for your substitution, the shift map on \Sigma has zero entropy and is uniquely ergodic. For the Thue Morse substitution, the invariant measure gives measure (1/2,1/2) to each symbol. Lots of people have studied conditions under which these systems are weak mixing (e.g. Dekking and Keane 83, which I don't have access to at home). Maybe you need three symbols to make weak mixing, but if there is an example here it's probably easy and explicit. – Tom Kempton Feb 28 '14 at 9:31
• Dekking and Keane show that the substitution $0 \mapsto 001$, $1 \mapsto 11100$ is weak mixing. The measures of the cylinders $[0]$ and $[1]$ are determined by the limiting ratio of ones to zeros in the substitution, which in turn is given by the ratio of entries in the leading eigenvector of the matrix $(2 2 | 1 3)$. In particular I think that the measures of the cylinder sets are irrational for this substitution. I am a little concerned that irrational eigenvalues might be necessary for weak mixing but this is a good direction of inquiry I think. Thanks! – Ian Morris Feb 28 '14 at 10:49
• Ummm... the leading eigenvector of that matrix is (1|1) with eigenvalue 4: under the substitution, $01\mapsto 00111100$. – Anthony Quas Feb 28 '14 at 13:24
• Anthony: er, so it is! I had somehow miscalculated the determinant as being $2$ and got an eigenvalue of $\frac{5+\sqrt{17}}{2}$. Thank you! – Ian Morris Feb 28 '14 at 14:18
• Take the Chacon map and the stationnary process obtained by coding with a $(1/2,1/2)$-partition. It should work, no ? – Stéphane Laurent Mar 9 '16 at 15:53

So with help from Ian and Anthony it looks like my comment above can be turned into an answer. To summarize:

Let S be the substitution on two symbols given by $0\to001, 1\to11100$. Let $x\in\{0,1\}^{\mathbb N}$ be the limit of $S^n(01)$. Let $\Sigma$ be the orbit closure of $x$ under the shift map.

$(\Sigma,\sigma)$ is uniquely ergodic and weak mixing but not strong mixing (Dekking and Keane '83). Let $m$ be the invariant measure. We have that $m[0]=m[1]=\frac{1}{2}$, just by looking at the leading eigenvalue of the matrix (22|13).

Furthermore, primitive substitution systems have zero entropy. I don't know the original reference for this, but much stronger results are known, for example that the complexity function of any primitive substitution has sublinear growth. Apparently this is proved in J.-J. PANSIOT: Complexite des facteurs des mots innis enegendres par morphismes iteres, 1984, or in Queffelec 'substitution dynamical systems'. I read it as a comment on page 5 of Ferenczi 'complexity in sequences and dynamical systems'.

Thus the system $(\{0,1\}^{\mathbb N},\sigma,m)$ satisfies the requirements of your proposition.

To prevent any further beating around the bush let me explain Anthony's answer - as it is in fact much easier than what he actually wrote.

Take any weakly mixing zero entropy transformation $T$ on a probability space $(X,m)$, and take a subset $A\subset X$ with $m(A)=1/2$. Then the symbolic coding of $T$ determined by the partition $(A,X\setminus A)$ satisfies OP's Proposition.

• In fact, the same is true if we replace "weak mixing" with "mixing", since there are mixing, zero-entropy mpt's. – EvaristoCarriego Apr 23 '14 at 3:11
• Thanks for correctly figuring out what I should have said! This is a nice answer - sorry I didn't see it before. – Anthony Quas May 30 '14 at 19:54

Take a non-atomic weak-mixing measure $\mu_0$ with 0 entropy on two symbols. Equip $X^+=\{0,1\}^{\mathbb N}$ with the lexicographic order and for $z\in X^+$ define $[0,z]=\{x\in X^+\colon x\le z\}$ and $\Phi(z)=\mu_0([0,z])$. Since $\mu_0$ is non-atomic, it's easy to check that $\Phi(z_0z_1\ldots z_n01111\ldots)=\Phi(z_0z_1\ldots z_n10000\ldots)$. In particular, we can define a map $\phi$ from $[0,1]$ to $[0,1]$ by $\phi(t)=\Phi(z(t))$ where $z(t)$ is one of the binary expansions of $t$.

Now notice that $\phi$ is a continuous increasing function with $\phi(0)=0$ and $\phi(1)=1$. Hence there exists $\alpha$ such that $\phi(\alpha)=\frac 12$.

Finally define a map $\Psi$ from $\{0,1\}^{\mathbb Z}$ to itself by $$\Psi(x)_n=\begin{cases} 0&\text{if \sum_{k=0}^\infty 2^{-k-1}x_{n+k}<\alpha;}\\ 1&\text{otherwise.} \end{cases}$$

Now defining $\mu=\mu_0\circ\Psi^{-1}$ gives a weak-mixing measure of 0 entropy with frequency of 1's exactly $\frac 12$.

• Why is $\mu$ shift invariant? – R W Feb 28 '14 at 22:18
• @R W: It wasn't due to a typo. Now I've fixed the definition of $\Psi$ so that it's shift-commuting. From this it's easy to check that the push-forward of an invariant measure is invariant. – Anthony Quas Feb 28 '14 at 22:48
• @ Anthony Quas: OK - so in plain language you just take a subset $A$ in $X^+$ of measure $\frac12$, and then take the symbolic realization determined by the partition $\xi=(A,X^+\setminus A)$. That's fine - but how do you know that $\xi$ is a generator? (as otherwise you may lose weak mixing when passing to the quotient measure). – R W Mar 1 '14 at 14:04
• @R W: Weak mixing is preserved under taking factors. Weak mixing means there are no eigenfunctions (other than 1). If the factor has an eigenfunction, then so does the original transformation. – Anthony Quas Mar 1 '14 at 15:30
• @ Anthony Quas: Oups - indeed – R W Mar 1 '14 at 15:43