Week 11 Tutorial Solutions

1. The only members of $Z$ are $$\begin{array}{rl}a=& \mathtt{0101010101010101}\cdots \\ b=& \mathtt{1010101010101010}\cdots \end{array}$$
2. The only invariant probability measure on $Z$ is $$\nu =\frac{1}{2}({\delta }_a+{\delta }_b)$$
3. The only partitions of $Z$ are $(Z)$ and $(\{a\},\{b\})$. As $T$ fixes the first and permutes the second, we have $$\xi \vee T^{-1}\xi \vee \cdots \vee T^{-(N-1)}\xi =\xi$$ for both and all $N\in \mathbb{N}$. Therefore the entropy of $T$ is zero.
4. The cylinders $[101]$ and $[111]$ are disjoint from $Y$ because $101$ and $111$ are forbidden for sequences in $Y$. The six other cylinders of length 3 have non-empty intersection with $Y$.
5. If the measure has $\mu (Y)=1$ then necessarily $$\begin{array}{rl}0=\mu ([101])& =p(1)q(1,0)q(0,1)\\ 0=\mu ([111])& =p(1)q(1,1)q(1,1)\end{array}$$ both hold.
   
   It may be that $p(1)=0$. In this case the matrix equation gives $$0=p(0)q(0,1)$$ and therefore $q(0,1)=0$ as we would have to have $p(0)=1-p(1)=1$. Forbidding $01$ gives a countable collection of sequences on which the single invariant measure is the point mass at the sequence that is constantly zero.
   
   It may be that $p(1)\ne 0$. In this case we have $q(1,1)=0$ and $q(1,0)q(0,1)=0$. But $q(1,1)=0$ forces $q(1,0)=1$ so we must have $q(0,1)=0$. There are only two sequences satisfying these rules and the only measure is again the point mass at the sequence which is constantly zero.
6. Consider the graph with vertices labeled $0,1,2,3$ and edges as follows.
   • An edge from 0 to 0.
   • An edge from 0 to 1.
   • An edge from 1 to 2.
   • An edge from 1 to 3.
   • An edge from 2 to 0.
   • An edge from 3 to 0.
   Let $\mathrm{\Omega }$ be the corresponding subset of $\{0,1,2,3{\}}^{\mathbb{N}}$ i.e. the one defined by the matrix $$\left[\begin{array}{cccc}1& 1& 0& 0\\ 0& 0& 1& 1\\ 1& 0& 0& 0\\ 1& 0& 0& 0\end{array}\right]$$ and define $\pi$ by $$(\pi \omega )(n)=\{\begin{array}{ll}0& \omega (n)\in \{0,2\}\\ 1& \omega (n)\in \{1,3\}\end{array}$$ for all $\omega \in \mathrm{\Omega }$. Note that $\pi (\mathrm{\Omega })=Y$. Let $\eta$ be any Markov measure compatible with this graph. As a Markov measure it is $T$ invariant. So too is the push-forward $\eta \circ {\pi }^{-1}$ and $$(\eta \circ {\pi }^{-1})(Y)=\eta (\mathrm{\Omega })=1$$ as desired.
   
   Moreover, it is not the point mass at the origin because $$(\eta \circ {\pi }^{-1})([01])=\eta ([01])+\eta ([21])+\eta ([03])+\eta ([23])$$ is positive for many possible values of the parameters defining $\eta$.