Tag Archives: equilibrium state

Generic Anosov does not admit fat horseshoe

To start let’s describe an interesting proposition in QIU Hao’s paper (Commun. Math. Phys. 302 (2011), 345–357.)

Assume f\in\mathrm{Diff}^1(M) and \Lambda be a basic (isolated and transitive, or mixing) hyperbolic set of f. It is well known (Anosov) that there exists an open neighborhood \mathcal{U}\ni f and an open set U\supset\Lambda such that for each g\in\mathcal{U},

1). \Lambda_g=\bigcup_{n\in\mathbb{Z}}g^nU is an isolated hyperbolic set of g. Moreover \Lambda_g\to\Lambda as g\to f.
2). there exists a (Holder) homeomorphism h_g:\Lambda\to\Lambda_g such that h_g\circ f(x)=g\circ h_g(x) for every x\in\Lambda. Moreover h_g\to \mathrm{Id} as g\to f.

Now let’s consider the unstable log-Jacobian \phi_g\in C(\Lambda_g,\mathbb{R}) as
\phi_g(x)=-\log\det(D_xg:E^u_g(x)\to E^u_g(gx)), x\in\Lambda_g.

By classical hyperbolic theory (Sinai, Ruelle and Bowen), we know that for each g\in\mathcal{U}\cap \mathrm{Diff}^2(M), the topological pressure P(\phi_g;g,\Lambda_g)=0 and there exists a unique equilibrium state of \phi_g with respect to (g,\Lambda_g).

Define a map \Phi:\mathcal{U}\to C(\Lambda,\mathbb{R}) by \Phi(g)(x)=\phi_g(h_g(x)). It takes a few seconds to see that \Phi is continuous.

Proposition 3.1 (Qiu) For each g\in\mathcal{U}, P(\phi_g;g,\Lambda_g)=0.
Proof. Since topological pressure is invariant under topological conjugation, we have
P(\phi_g;g,\Lambda_g) = P(\Phi(g);\Lambda,f).
Now we pick g_k\in\mathcal{U}\cap \mathrm{Diff}^2(M) with g_k\to f. Note this also implies h_{g_k}\to\mathrm{Id}, \Phi(g_k)\to\phi_f. So
This finishes the proof.

Remark: In particular for all Anosov diffeomorphisms and all Axiom A diffeomorphisms with no cycle condition, we have P(\phi_f;f)=0.

Proposition 3.1 (continued). for generic g\in\mathcal{U}, there exists a unique equilibrium state for \phi_g with respect to (g,\Lambda_g).
Proof. Since (f,\Lambda) is expansive, the entropy map \mathcal{M}(f,\Lambda)\to\mathbb{R},\mu\mapsto h(f,\mu) is upper semicontinuous and there is a residual subset \mathcal{R}\subset C(\Lambda,\mathbb{R}) such that each \phi\in\mathcal{R} has a unique equilibrium state with respect to (f,\Lambda). Since \Phi is continuous, the pre-image \Phi^{-1}(\mathcal{R}) is

1. a G_\delta set in \mathcal{U} since the pre-images of open sets are open;

2. a dense set in \mathcal{U} since \Phi(g)\in\mathcal{R} for all g\in\mathcal{U}\cap\mathrm{Diff}^2(M).

In particular \Phi^{-1}(\mathcal{R}) is residual in \mathcal{U}. The proof is complete.

We focus on a special case of QIU’s main result. Let \mathcal{A}^r be the set of C^r Anosov diffeomorphisms on M (might be empty). For an invariant measure \mu, we let B(\mu) be the set of points with \frac{1}{n}\sum_{0\le k<n}\delta_{f^kx}\to \mu.

Theorem A (Qiu). Generic f\in\mathcal{A}^1 has a unique SRB measure \mu_f: m(M\backslash B(\mu_f))=0.
Indeed, \mu_f is the unique equilibrium state of \phi_f (hence ergodic).

Robinson and Young constructed an Anosov diffeomorphism with nonabsolutely continuous foliations, by embedding an Bowen horseshoe \Lambda_B to some f\in\mathcal{A}(\mathbb{T}^2). Although f is transitive, every point in \Lambda_B can not be a transitive point. Theorem A implies that this phenomenon fails generically:

Observation: generic f\in\mathcal{A}^1(M) does not admit Bowen’s fat horseshoe.
Proof. A priori, we donot know if every Anosov is transitive. So we divide \mathcal{A}^1(M)=\mathcal{A}_t(M)\sqcup \mathcal{A}_e(M) into transitive ones and exotic ones. But we know they are always structurally stable. Therefore both parts are open.

By Theorem A, we know that, for generic f\in\mathcal{A}_t(M), \mu_f is ergodic and fully supported. Hence every point in B(\mu_f) is a transitive point. In particular every closed invariant set of f has trivial volume: 0 or 1.

For maps in the exotic ones f\in\mathcal{A}_e(M), at least they can be viewed as Axiom A system, and Smale’s Spectra Decomposition Theorem applies: \Omega(f)=\Lambda_1\sqcup\cdots\sqcup\Lambda_n, where n=n(f) is locally constant. Some of them are attractors, say A_1,\cdots, A_k, some are repellers, say R_1,\cdots, R_l. Let \mathcal{R}_0 be the residual subset given by QIU for all repellers. Clearly \mathcal{R}=\mathcal{R}_0\cap\mathcal{R}^{-1}_0 is also generic. For each f\in \mathcal{R}, there is an SRB \mu_{u,i} relative to R_i(f) with m(\bigcup_i B(\mu_{u,i}))=1 and an SRB \mu_{s,j} relative to A_{j} for f^{-1} with m((\bigcup_j B(\mu_{s,j}))=1. Incorrect conclusion. The following are void.

To derive a contradiction, suppose that there was a fat horseshoe H of f, then

1. either H\cap B(\mu_{u,i})=\emptyset, (contradicts m(\bigcup_i B(\mu_{u,i}))=1);

2. or H\cap B(\mu_{s,j})=\emptyset, (contradicts m(\bigcup_j B(\mu_{s,j}))=1);

3. or there exist x\in H\cap B(\mu_{u,i}) and y\in H\cap B(\mu_{s,j}). In particular H has nontrivial intersections with the attarctor A_j and the repeller R_i simultaneously, which contradict the transitivity of Horseshoe. Q.E.D.