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11 votes
0 answers
212 views

Factor map between subshifts preserving topological pressure (or measure-theoretic entropy)

Let $G$ be a countable amenable group and let $X,Y$ be subshifts with finite alphabet over $G$. Suppose that $h(X) = h(Y)$ (equal topological entropy). I am interested in continuous factor maps $\pi: ...
Sophie M's user avatar
  • 695
8 votes
1 answer
280 views

Connection between entropy and the set of factors of a sequence

Let $a = (a_n)_{n=0}^\infty$ be a bounded real-valued sequence. By a factor of $a$ I mean a finite block $w \in \mathbb R^l$ that appears in $a$, that is, there exists $n \geq 0$ such that $a_n a_{n+1}...
Jakub Konieczny's user avatar
5 votes
1 answer
432 views

Is there a one-dimensional subshift of positive entropy s, all of whose sub-subshifts also have entropy s?

A subshift is a subset $X$ of $A^\mathbb{N}$ or $A^\mathbb{Z}$ (with $A$ finite), such that $X$ is topologically closed and closed under the shift operation. The shift operation is defined by $\sigma(...
Linda Brown Westrick's user avatar
4 votes
1 answer
176 views

Measures maximizing entropy in a set of measures with fixed average for some observable

Let $\Omega$ be the set of all infinite binary sequences $(x_i)_{i\ge 0}$ endowed with the product topology coming from discrete topology on $\{0,1\}$. Consider $0<\alpha<1$ and let $$K_\alpha=\{...
Dominik Kwietniak's user avatar
4 votes
2 answers
445 views

Entropy equals zero?

Imagine you have a shift invariant ($\sigma$-invariant) probability measure $\eta$ in the Bernoulli space $\{0,1\}^{\mathbb{N}}$. Define $\mathcal{P} = \{[0],[1]\}$; $\mathcal{P}^{n} = \mathcal{P}\...
Bruno Brogni Uggioni's user avatar
2 votes
1 answer
131 views

Morse-Hedlund\Coven-Hedlund theorem for non-Abelian groups

There is a well know theorem by Coven and Hedlund, in Sequences with minimal block growth, stating that the complexity function of an aperiodic sequence\configuration $\omega\in \mathcal{A}^{\mathbb{Z}...
Keen-ameteur's user avatar
2 votes
1 answer
267 views

entropy and d-bar: how do we estimate continuity?

Let $G = \{0,1\}^{\mathbb{N}} = \mathbb{Z}_{2}^{\mathbb{N}}$ be the Bernoulli space of two symbols, let $\sigma$ be the shift map and $M(G)$ the set of $\sigma$-invariant probabilities. Let $\bar{d}$ ...
Bruno Brogni Uggioni's user avatar
2 votes
2 answers
269 views

probability measures with entropy equal to nonnegative number

Is it true that for a given nonnegative number, there exists a measure-theoretical entropy value (supremum of entropies of all partitions under a measure-preserving transformation) that equals this ...
Ivy's user avatar
  • 63
2 votes
1 answer
261 views

Embeddings of subshifts

Consider $(X,\sigma_X)$ and $(Y, \sigma_{Y})$ be subshifts of the one sided shift in two symbols. Assume that $(X,\sigma_X)$ is a transitive subshift of finite type and $(Y, \sigma_{Y})$ is a ...
Rafael Alcaraz Barrera's user avatar
1 vote
1 answer
410 views

joining or coupling

given two shift invariant measures in the Bernoulli space $\{0,1\}^{\mathbb{N}}$, is there a way to construct joinings of them? It's very diffcult, in general, to find exactly the minimal joining i.e, ...
Bruno Brogni Uggioni's user avatar
1 vote
0 answers
139 views

weak-* versus entropy growth

General question. Let $\eta_{n}$ be a sequence of invariant measures on $\{0,1,2,...,p-1\}^{\mathbb{N}}$ and $B$ the Bernoulli uniform measure. Knowing that $\eta_{n} \rightarrow B$ in the weak-* ...
Bruno Brogni Uggioni's user avatar
1 vote
0 answers
190 views

Entropy of factors of Bernoulli schemes

Let $X$ be a Bernoulli scheme. A factor $\psi: X \to Y$ is finitary if for almost every $x \in X$ there exist integers $m \leq n$ such that the zero coordinates of $\psi(x)$ and $\psi(x')$ agree for ...
Stephen Shea's user avatar
0 votes
1 answer
172 views

entropy growth of invariant measures - General question

In general, given a sequence of shift-invariant measures $\eta_{n}$ on $\{0,1\}^{\mathbb{N}}$ what to do to guarantee this convergence of entropies: $$h(\eta_{n}) \rightarrow \log2?$$ Because I'm ...
Bruno Brogni Uggioni's user avatar