Anosov and expanding attractors
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. ARTICLES .
September 2020 Vol. 63 No. 9: 1929–1934 https://doi.org/10.1007/s11425-019-9565-y
Anosov and expanding attractors Dedicated to the Memory of Professor Shantao Liao
Robert F. Williams1,2 1Department
of Mathematics, University of Texas at Austin, Austin, TX 78705, USA; 2Institute for Advanced Study, Princeton, NJ 08540, USA Email: [email protected]
Received March 8, 2019; accepted July 3, 2019; published online August 18, 2020 Abstract
An earlier conjecture is settled with an immersion of a 2-dimensional branched manifold. Possible
obstructions in linear algebra and tiling theory are studied first. Keywords MSC(2010)
Anosov, expanding attractor, Cantor set fiber bundle 37C70
Citation: Williams R F. Anosov and expanding attractors. Sci China Math, 2020, 63: 1929–1934, https://doi.org/ 10.1007/s11425-019-9565-y
Branched manifolds were introduced [5–7] to study hyperbolic attractors, via the following proposition: Proposition 0.1 (See [7]). If a closed neighborhood, N, of an expanding attractor, Λ, of a diffeomorphism f, is foliated by closed disks, then collapsing each leaf of the foliation to a point, yields a map q : N → K, where K is a branched manifold. In addition, f Λ is semi-conjugate to g K and conjugate to the ˆ inverse limit map gˆ K. Here, the “expanding attractor” means the special case of hyperbolic, in which the dimension of the unstable direction is the dimension of the attractor. For the general case of a hyperbolic attractor, a priori the analog to the above, the map g K itself might be hyperbolic (not expanding). I conjectured that branched manifolds would not be necessary in such a theory, i.e., given g K hyperbolic, has dense periodic points, and satisfies the collapsing axiom: for each x ∈ K, there is a neighborhood U, and integer n such that f n (U ) lies in a smooth disk, then Conjecture 0.2. There exists g ′ M , with M being a manifold, with g ′ (shift) equivalent to g, and (thus) having inverse limit conjugate to that of g. It was proved by Wen [4] in a special case. Some time ago I found a putative counter example to Conjecture 0.2, constructed as the union of two tori: g T1 ∪ T2 . However in trying to write this up, I discovered two problems: the first was a statement in linear algebra which is wrong (see Proposition 0.3 below). Let A be an Anosov matrix, that is to say, it has integral entries, determinant greater than or equal to 1, no eigenvalues of absolute value 1, and at least one less than 1. Let X ⊂ Rn be a closed neighborhood of Rn , for example the usual closed unit square. c Science China Press and Springer-Verlag GmbH Germany, part of Springer Nature 2020 ⃝
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1930
Williams R F
Proposition 0.3.
Sci China Math
September 2020
Vol. 63
No. 9
For Anosov A : X → Rn , A(X) contains no translation of X.
Proof. Assume the contrary for the translation f : X → A(X). Let I be one of the stable manifold segments of the maximum diameter in X. Let y ∈ f (I) and x ∈ X with A(x) = y. Let I ′ be the stable segment of X containing x
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