Thurston equivalence of topological polynomials
By Laurent Bartholdi, Volodymyr V. Nekrashevych. 2005
115 citations recorded by the source corpus.
We answer Hubbard's question on determining the Thurston equivalence class of “twisted rabbits”, i.e. composita of the “rabbit” polynomial with nth powers of the Dehn twists about its ears. The answer is expressed in terms of the 4-adic expansion of n. We also answer the equivalent question for the other two families of degree-2 topological polynomials with three post-critical points. In the process, we rephrase the questions in group-theoretical language, in terms of wreath recursions.
Subfields: Computational Theory and Mathematics, Geometry and Topology, Mathematical Physics
View Paper: https://www.semanticscholar.org/paper/35bd598621ac318d085b950463c9a089f4eabb20
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In Computational Theory and Mathematics
- Rostislav I. Grigorchuk
- Volodymyr V. Nekrashevych
- Zoran Sunic
- Ievgen V. Bondarenko
- Yevgen Muntyan
- Dmytro Savchuk
- Rostyslav V. Kravchenko
- Miklós Abért
In Geometry and Topology
- Rostislav I. Grigorchuk
- Volodymyr V. Nekrashevych
- Zoran Sunic
- Ievgen V. Bondarenko
- Yevgen Muntyan
- Dmytro Savchuk
- Rostyslav V. Kravchenko
- Enrique Pardo
In Mathematical Physics
- Rostislav I. Grigorchuk
- Zoran Sunic
- Volodymyr V. Nekrashevych
- Ievgen V. Bondarenko
- Yaroslav Vorobets
- Enrique Pardo
- R. Exel
- Laurent Bartholdi
Citations in the corpus
- Classification of groups generated by 3-state automata over a 2-letter alphabet (2008, methodology): “s vertex is equal to a−1c. Hence, a−1chas infinite order. We have ac−1 = σ(ba−1,1), ba−1 = σ(1,cb−1), cb−1 = (ac−1,1), Thus the subgroup generated by these elements is isomorphic to IMG(1 − 1 z2) (see [BN06]). We have c −1b= (a c,1), ac−1ba = (1,ca−1). Thus, by Lemma 4 the group is not free. 748 ∼= G”
- Groups St Andrews 2005: Self-similarity and branching in group theory (2007, background): “This led to the solution of the longstanding ‘Twisted Rabbit Problem’ by Bartholdi and Nekrashevych [2].”