Erratum to “Open sets of Axiom A flows with exponentially mixing attractors”
Author(s) -
Vı́tor Araújo,
Oliver Butterley,
Paulo Varandas
Publication year - 2018
Publication title -
proceedings of the american mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.968
H-Index - 84
eISSN - 1088-6826
pISSN - 0002-9939
DOI - 10.1090/proc/14163
Subject(s) - mixing (physics) , axiom , attractor , mathematics , statistical physics , pure mathematics , physics , mathematical analysis , geometry , quantum mechanics
In “Open sets of Axiom A flows with exponentially mixing attractors” there was an oversight concerning the regularity of the Markov partition for higher dimensional flows. Previously [1, p.2978] we claimed that any element of a Markov partition of a non-trivial attractor for an Axiom A vector field, after quotienting out the stable leaves, is a C2 disk, hence a “John domain” [2, Definition 2.1]. The argument for exponential mixing presented in the paper relies on the application of results [2, Theorem 2.7] where the “John domain” condition is required. Bowen [3] showed that, for higher dimensional systems, the boundaries of the Markov partition elements cannot be smooth (here smooth means piecewise C1) , in particular the objects of interest cannot be expected to be C2 disks as previously claimed. In general there is no reason to expect that the elements of the Markov partition are John domains. In general, when the unstable bundle is higher dimensional and the expansion is not isotropic then there seems no hope that the sets are John domains (for evidence of this consult the estimates and comments in [4, §A.2]). Here we show that the originally claimed results remain valid. First observe that, as previously described [1, §4], for any d ≥ 3, it is possible to construct examples of vector fields with Axiom A attractors which have 1D unstable bundle and which satisfy the requirements of the rest of the construction. If the unstable bundle is 1D then the relevant element of the Markov partition is automatically a John domain, simply because it is a connected one-dimensional set. The above observation gives immediately the main results as follows.
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