Defining Jacobian of differential map on Riemannian manifold I'm reading Rufus Bowen's notes on ergodic properties of Anosov diffeomorphisms, and I've come across an important-seeming function but I'm confused about how it's defined. 
Let $M$ be a Riemannian manifold, and let $f : M \to M$ be a $\mathcal{C}^2$ Axiom A diffeomorphism. (For our purposes, this means there is a closed invariant $\Omega_s \subseteq M$ for which there is a $Df$-invariant splitting of the tangent bundle $T\Omega_s = E^s \oplus E^u$.) Bowen writes the following (first paragraph of Section 4B): 

For $x \in \Omega_s$, let $\phi^{(u)}(x) = -\log \lambda(x)$, where $\lambda(x)$ is the Jacobian of the linear map $$Df : E^u_x \to E^u_{fx}$$ using inner products derived from the Riemannian metric.

My question is, how exactly is $\lambda(x)$ defined? My first thought was it's the determinant of the Jacobian matrix of partial derivatives, but (a) this requires writing $Df$ in coordinates and I'm not sure the determinant is coordinate-idnependent, and (b) this determinant has seemingly nothing to do with the Riemannian structure on $M$, but the dependence of $\phi^{(u)}(x)$ on the Riemannian metric plays a key role in subsequent arguments of his lecture notes. 
So how do we typically define the Jacobian of the differential of a diffeomorphism? Any help is appreciated! 
 A: levap has already given a very nice answer. I will give an alternative discussion based on the determinant as a ratio of volume elements. Hopefully this will be useful for those reading Bowen- the interpretation of determinant as a volume ratio is the reason for its importance in smooth ergodic theory.
Let $T : V \to V'$, where $V, V'$ are inner product spaces. A choice of inner product on $V$ gives rise to a choice of scaling of Lebesgue measure $m$, i.e., the one assigning mass $1$ to an orthogonal $\dim(V)$-cube of side length $1$. Let $m'$ be the corresponding Lebesgue volume for $V'$.
With $m, m'$ as above, the determinant $\det(T)$ is (or, more accurately, may be characterized as) a volume ratio: we have
$$
\det(T) = \frac{m'(T U)}{m(U)}
$$
where $U \subset V$ is any nonempty open subset. 
In this way, we see directly how the inner product structures on each of $V, V'$ affect the corresponding notion of determinant. Indeed, taking a different view, we need not regard $V, V'$ as inner product spaces- for the purposes of having a well-defined determinant, it suffices to specify scalings of Lebesgue measure on each of $V, V'$. 
Returning to Bowen's context: The determinant of $df$ between tangent spaces depends on the volume elements on each of $T_x M, T_{fx}M$- these are precisely the scalings of Lebesgue measure specified by the inner products (a.k.a. Riemannian metrics) on each of $T_x M, T_{f x} M$.
A: Given a linear map $T \colon V \rightarrow W$ between two $n$-dimensional vector spaces, there is no meaningful notion of $\det(T)$ without some additional choices of structure. If you try and define $\det(T)$ to be the determinant of the square matrix $A$ representing $T$ with respect to some choice of bases for $V$ and $W$ then with respect to a difference choice of bases the map $T$ will be represented by the matrix $Q^{-1}AP$ but in general $\det(Q^{-1}AP) = \det(Q^{-1})\det(A)\det(P) \neq \det(A)$ so this is ill-defined. 
However, if $V,W$ are real inner product spaces then there is an invariant notion of the absolute value of the determinant of $T$. Namely, define $|\det(T)|$ to be $|\det(A)|$ where $A$ is any square matrix which represents $T$ with respect to some choice of orthonormal bases for $V$ and $W$. If $T$ is represented by $A$ with respect to some choice of orthonormal bases then it will be represented by $Q^{-1}AP$ with respect to a different choice of orthonormal bases but now $P,Q$ are orthogonal matrices and since $\det(P),\det(Q) = \pm 1$ we have
$$ |\det(Q^{-1}AP)| = |\det(Q^{-1})| |\det(A)| |\det(P)| = |\det(A)| $$
so this notion is independent of the orthonormal bases used to represent $T$. Geometrically, on an inner product space one can talk about the volume of $n$-dimensional parallelotopes and if $P$ is such a parallelotope in $V$ then $\operatorname{Vol}(T(P)) = |\det(T)| \operatorname{Vol}(P)$ so $\det(T)$ measures the effect of $T$ on volumes of parallelotopes. This definitions depends on the inner products on $V$ and $W$ and if we change the inner products, $|\det(T)|$ might change. If in addition you endow $V,W$ with an orientation, you can even get rid of the sign ambiguity and define an invariant notion of $\det(T)$ as a real (not neccesarily non-negative) number.
In your case, the Riemannian structure on $M$ gives you an inner product on each tangent space so $Df_{x} \colon T_x M \rightarrow T_{f(x)}M$ is a map between inner product spaces and $|\det(Df_x)|$ is defined as above. Similarly, if you have a subspace $E \subseteq T_x M$ and $Df_{x}$ is an isomorphism then $Df_x|_{E} \colon E \rightarrow Df_x(E)$ is a map between equidimensional inner product spaces (where the inner products are simply the restrictions on $E,Df_x(E)$ of the inner products on $T_xM, T_{f(x)}M$) so $|\det(Df_x|_{E})|$ makes sense. Since you are taking the logarithm of the determinant, I assume that in any case you are interested only in the absolute value of the determinant so this is enough. 
