5
$\begingroup$

Let $f:M \to M$ be a $C^{1}$ diffeomorphism on a compact Riemannian manifold with a normalized Riemannian volume $\mathrm{Leb}$. Given an $f$-invariant Borel probability $\mu$ in $M$, we call the basin of attraction of $\mu$ the set $B(\mu)$ of the points $x \in M$ such that the averages of Dirac measures along the orbit of $x$ converge to $\mu$ in the weak* sense: $$\lim _{n \rightarrow+\infty} \frac{1}{n} \sum_{j=0}^{n-1} \varphi\left(f^{j}(x)\right)=\int \varphi d \mu$$ for any continuous $\varphi: M \rightarrow \mathbb{R}$. Then we say that $\mu$ is a physical measure for $f$ if the basin of attraction $B(\mu)$ has positive Lebesgue measure in $M$.

A particular type of physical measures are the so-called Sinai–Ruelle–Bowen, or SRB, measures which have the property of having nonzero Lyapunov exponents $\mu$-almost everywhere and admitting a system of conditional measures such that the conditional measures on unstable manifolds are absolutely continuous with respect to the Lebesgue measures $\mathrm{Leb}$ on these manifolds induced by the restriction of the Riemannian structure.

An ergodic SRB measure is physical. The figure-eight attractor has a physical measure with a positive Lyapunov exponent which is not an SRB measure. I want to know whether there is an example of an SRB measure which is not physical.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

You can just take an Anosov map on $T^2$ and multiply by identity on the circle. Then, you will have SRB measures supported on $T^2 \times pt$ which are not physical.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. One silly question; I don't understand why they are hyperbolic; the Lyapunov exponent corresponding to the identity map is not zero? $\endgroup$
    – Adam
    Commented Dec 1, 2021 at 14:38
  • 3
    $\begingroup$ There is a zero exponent. You can define SRB by the property of having conditional measures which are absolutely continuous along unstable manifolds, possibly having some zero exponents. In this case, you can construct SRB which are not physical. Hyperbolic SRB measures (hyperbolic meaning all the Lyapunov exponents are non zero) will always be physical measures. One way to see this is that your measure is absolutely continuous along unstable + absolute continuity of Pesin manifolds. These two things will give you a set of positive volume whose Birkhoff averages converges to the measure. $\endgroup$
    – D. J. Obata
    Commented Dec 1, 2021 at 15:11
  • 2
    $\begingroup$ Just to add to potrie's and Obata's great answers/comments, I think there's a slight misconception by the OP of what's an SRB measure for the general case. In the hyperbolic setting, he is correct, but the general definition (see i.e. Young's What Are SRB Measures, and Which Dynamical Systems Have Them?, def 2.2) only require one positive exponent (probably to exclude ``boring'' unipotent dynamics) and abs. continuity along $W^{u}$. Young explains there (loc. cit def 2.3,2.4 thm 3) the relation to absolute continuity of the Pesin manifolds and the deduction of the theorem Obata referred to. $\endgroup$
    – Asaf
    Commented Dec 1, 2021 at 20:50
  • 1
    $\begingroup$ See also theorem 5 in her article (due to Tsuji) which discuss the figure 8 case $\endgroup$
    – Asaf
    Commented Dec 1, 2021 at 20:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .