Let $M$ be a compact Riemannian manifold, let $0 < \alpha \leq 1$, and let $f: M \to M$ be a $C^{1+\alpha}$ diffeomorphism. Let $\Lambda \subset M$ be a compact uniformly hyperbolic attractor for $f$, let $E^u \to \Lambda$ denote the unstable bundle, and let $\mu$ be an $f$-invariant Borel probability measure supported on $\Lambda$.

Let $\xi$ be a measurable partition of a full $\mu$-measure subset $\Lambda_\xi \subset \Lambda$ such that $\xi$ is subordinate to local unstable manifolds and $f^{-1}\xi$ refines $\xi$. Assume that there is a full $\mu$-measure $\xi$-saturated subset $\Lambda_0 \subset \Lambda_\xi$ such that, for every $x \in \Lambda_0$, the Rokhlin conditional measure $\mu_x^\xi$ on $\xi(x)$ is absolutely continuous with respect to the Riemannian leaf-volume measure $m_x^u$ on the unstable plaque $\xi(x)$. Choose a measurable density version $\rho_x: \xi(x) \to [0,\infty)$ satisfying

\begin{align*} \mu_x^\xi = \rho_x\,m_x^u,\qquad \rho_x(q)>0 \text{ for } m_x^u\text{-a.e. }q\in \xi(x),\qquad \int_{\xi(x)} \rho_x(q)\,d m_x^u(q)=1. \end{align*}

For $p \in \Lambda$, define the unstable Jacobian by

\begin{align*} J^u f(p) := \left|\det\left(df_p\big|_{E^u_p}: E^u_p \to E^u_{f(p)}\right)\right|, \end{align*}

where the determinant is computed with respect to the Riemannian volume forms on $E^u_p$ and $E^u_{f(p)}$.

Assume that the conditional densities satisfy the following compatible Gibbs $u$-transformation rule. For $\mu$-a.e. $x \in \Lambda_0$ with $f^{-1}x \in \Lambda_0$, define

\begin{align*} V_x := f(\xi(f^{-1}x))\cap \xi(x). \end{align*}

There is a constant $c(f^{-1}x)>0$, depending only on the source plaque $\xi(f^{-1}x)$, such that for $m_x^u$-a.e. $q \in V_x$ with $f^{-1}q \in \xi(f^{-1}x)$,

\begin{align*} \rho_x(q)=c(f^{-1}x)\frac{\rho_{f^{-1}x}(f^{-1}q)}{J^u f(f^{-1}q)}. \end{align*}

Finally, assume that for $\mu$-a.e. $x \in \Lambda_0$ and for $m_x^u\otimes m_x^u$-a.e. pair $(y,z)\in \xi(x)\times \xi(x)$ such that $f^{-k}y,f^{-k}z\in \xi(f^{-k}x)$ for every $k\geq 1$, and such that all one-step transformation identities above hold at the corresponding backward iterates, one has

\begin{align*} \lim_{n\to\infty}\frac{\rho_{f^{-n}x}(f^{-n}y)}{\rho_{f^{-n}x}(f^{-n}z)}=1. \end{align*}

Then, for $\mu$-a.e. $x \in \Lambda_0$ and for $m_x^u\otimes m_x^u$-a.e. such backward-compatible pair $(y,z)\in \xi(x)\times \xi(x)$,

\begin{align*} \frac{\rho_x(y)}{\rho_x(z)}=\prod_{k=1}^{\infty}\frac{J^u f(f^{-k}z)}{J^u f(f^{-k}y)}. \end{align*}

The infinite product is understood as the limit of its finite partial products, and this limit is finite and strictly positive.