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I'm reading one of the classical theorems presented in Bowen's lecture notes, "Equilibrium States and the Ergodic Theory of Anosov Diffeomorphisms." I'm trying to figure out a very short line of reasoning that's eluding me in the proof of the following:

Let $f$ be a transitive $C^2$ Anosov diffeomorphism. If $\mbox{det}~ (Df^n: T_xM \to T_xM) = 1$ for each $n$-periodic point, then $f$ admits an invariant measure of the form $d\mu = h dm$, with $h$ positive and Holder and $m$ the natural volume form.

The sketch of the proof is to use the periodicity condition to construct a function $h(x) = e^{u(x)}$, where the Holder function $u : M\to \mathbb{R}$ is related to the log of the Jacobian $\phi(x) = \mbox{log}~\mbox{Jac} (Df: T_xM \to T_xM)$ via the cohomology relation $\phi(x) = u(fx) - u(x)$. One easily derives the fact $h(fx) = h(x) e^{\phi(x)}$ from this relation. I understand the method of proof up to here. It remains to show that the measure $hdm$ is $f$-invariant.

Bowen proceeds as follows:

Thinking of $\mu = h dm$ as the absolute value of a form,

$\begin{eqnarray*}f^* (hdm)(fx) &=& h(x) e^{\phi(x)} \\ &=& dm(fx) \\ &=& h(fx) dm(fx) \\ &=& (hdm)(fx),\end{eqnarray*}$

hence $\mu$ is $f$-invariant.

The first equality is what's giving me trouble (the second and third follow from our aforementioned fact and the definition of the form hdm). If $f^*$ is the usual pullback of forms, things seem not to transform as I am expecting (maybe I forgot all my calculus...) can someone with more experience here please enlighten me as to what he's doing in this string?

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  • $\begingroup$ I think maybe $\phi(x)$ should be $-\mbox{log}~\mbox{Jac} (Df: T_xM \to T_{f(x)}M)$ and after that modification the first equality seems to be OK. $\endgroup$
    – Siming Tu
    Apr 10, 2015 at 13:23

2 Answers 2

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There is indeed a typo. The correct normalisation should be $h= e^{-u}$ and thus $Jac(f)=h/h\circ f$. Then we get for all $g$

$$ \int g\ h\ dm = \int g\circ f\ h \circ f\ Jac(f)\ dm = \int g\circ f\ h\ dm $$

See the book of Katok-Hasselblatt, introduction to the modern theory of dynamical systems, theorem 19.2.7 (same theorem but without the error sign).

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  • $\begingroup$ Thanks very much for the correction and reference! This does clear it up. $\endgroup$
    – IL123
    May 3, 2015 at 19:39
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So it looks to me like there's a typo in the LHS, which I think should be $f^*(h \ dm )(x)$. Let's see what happens if we evaluate at $x$ instead of $fx$. And maybe it will help to put the arguments in square brackets and write evaluation points as subscripts, viz. $(f^*(h \ dm))_x[X] = (h \ dm)_{fx}[df_x(X)] = (h(fx) \ dm_{fx})[df_x(X)] = (h(x) e^{\phi(x)} dm_{fx})[df_x(X)]$. Now look what happens when you ignore what's in square brackets.

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  • $\begingroup$ Your comment is helpful, but I'm still a little puzzled. For instance, given your equalities it seems hard to salvage the desired result that the evaluations agree for each $x \in M$ (resp. $fx \in M$). Also, Bowen's notation seems to say that the forms are equal when the evaluations agree for each $X$, but given your comment that doesn't seem true even for the first equality with the possible typo corrected (to put this another way, why are we allowed to 'ignore what's in the square brackets' and say that $(f^*(hdm))_x = h(x) e^{\phi(x)}dm_{f(x)}$?) $\endgroup$
    – IL123
    Apr 9, 2015 at 1:27

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