Constant slope maps on the extended real line

Abstract

Abstract. For a transitive countably piecewise monotone Markov interval map we consider the question whether there exists a conjugate map of constant slope. The answer varies depending on whether the map is continuous or only piecewise continuous, whether it is mixing or not, what slope we consider, and whether the conjugate map is defined on a bounded interval, half-line or the whole real line (with the infinities included).

Constant Slope Maps and the Vere-Jones Classification

“…This implies that the map T a(0) is strongly recurrent hence by Theorems 5 and 8 also linearizable for any ℓ. [15] observed that the map T a(1) is not conjugate to any map of constant slope. It can be shown that for each choice of a sequence a = (a n ) n≥1 such that the corresponding T has finite topological entropy the following dichotomy is true: either T is recurrent and then equation (5) has no λ-solution for λ > e h(T ) , or T is transient and then equation (5) does not have any λ-solution.…”

Constant slope, entropy, and horseshoes for a map on a tame graph

“…It is enough to construct a countably affine graph G of infinite total length and a constant slope map f ∈ CMM(G), whose transition matrix M has no summable eigenvectors. This was done in [15] on the extended real line [−∞, ∞].…”

“…It is enough to construct a metric tame graph G of infinite total length and a constant-slope map f ∈ CMM(G) whose transition matrix M has no summable eigenvectors. This was done in [13] on the extended real line [−∞, ∞]. Note that the constant slope is with respect to the standard metric on R, which cannot be extended to [−∞, ∞], so to fit the example to our setting we formally realize [−∞, ∞] as a double-infinite countable zig-zag in R 2 of infinite length.…”

“…Finally, we construct an example to show that condition (i) in Theorem 3.3 is not sufficient. It is a modification of a piecewise-continuous interval map from [13], made continuous by working on a dendrite. r n = 0, lim n θ n = 0.…”

“…Thus the metric is compatible with the original topology τ on G. Let x, y ∈ Q ∩ G. Let α ∈ A xy satisfy (x, y) = A(α) and β ∈ A f (x)f (y) . Since each vertex of G belongs to P 0 and f (P n ) ⊂ P n−1 ⊂ P n , every element of the set P(β, n − 1) of arcs [i 1 • • • i n ] ∈ P n−1 contained in the arc β has its preimages in the set P(α, n) of arcs [i 0 • • • i n ] ∈ P n contained in the arc α; using (15) and the definition of we can write for each n…”