I just have a quick question about the definition of Lyapunov exponents. My textbook defines them for a smooth map $f:M\to M$, where $M$ is a smooth manifold. For $x\in M$ and $v\in T_xM$: $\lambda(x,v)=$limsup$\frac{1}{n}$log$||d_xf^n(v)||$.
The authors don't explicitly say what they mean by $||\cdot||$, so I just wanted to be sure. I think they mean, $T_{f^n(x)}M$ can have some Riemannian metric $<,>_{f^n(x)}$ defined on it, so since $d_xf^n(v)\in T_{f^n(x)}M$, $||d_xf^n(v)||$ is just $\sqrt{<d_xf^n(v),d_xf^n(v)>_{f^n(x)}}$. Since it isn't explicitly stated, I assume $<,>_{f^n(x)}$ would be the metric induced from embedding $M$ in some high dimensional euclidean space.
In analysis, I've seen $||\cdot||$ defined as the operator norm $||T||=$sup$\{\frac{|Tv|_W}{|v|_V}: v\neq 0\}$ for a linear trasnformation between normed vector spaces $T:V\to W$. But this doesn't seem to make sense here, since $d_xf^n$ is a linear map, not $d_xf^n(v)$.
Here's a copy of the page from my text. Throughout the text, the authors reserve $M$ for smooth manifolds and $X$ for arbitrary sets when defining dynamical systems.
I could not find the book today, so I will have to rely on your question and the page you posted.
Let $M$ be a smooth manifold, and $f \in \mathcal{C}^\infty (M,M)$.
If $M$ is endowed with a Riemannian metric, then we can define the Lyapunov exponent of a pair $(x,v)$ as done in the book.
If $M$ is compact, then we can still work things out. Endow $M$ with an arbitrary Riemannian metric. Define the Lyapunov exponent. Then prove that the Lyapuov exponent does not depend on the Riemannian metric you picked. For the later point, you just have to remark that, on a compact manifold, given two metric $g$ and $g'$, there exists constants $0 < c \leq C$ such that $cg \leq g' \leq Cg$.
If $M$ is not compact, however, the Lyapunov exponent will depend on the metric you choose. For a simple example, take $M = \mathbb{R}$, so that $TM \simeq \mathbb{R} \times \mathbb{R}$, and $f(x) = (x+1)$. Then $d_x f = I$ for all $x$, so that $\| d_x f^n (v) \| = \|v\|_{T_x M}$. Now, if you take $g_x = e^{\lambda x} I$, then the Lyapunov exponent is always $\lambda$ - and as such depends on the metric.
This is not artificial. The translation on $\mathbb{R}$ is smoothly conjugated with any non-trivial homothety on $\mathbb{R}_+^*$, but if you endow both $\mathbb{R}$ and $\mathbb{R}_+^*$ with the euclidean metric, then you get different Lyapunov exponents.
I don't know what happens if you assume that the transformation $f$ is recurrent ; I assume that you could define Lyapunov exponents at the reccurence times in any given compact set which would be independent of any metric. However, between these return times, you would have no control on how they would behave (and thus, the limsup could be very different from the limsup taken at those recurrence times).
Anyway. In the following theorem, the authors work with $\mathbb{S}_1$, which is compact, so any metric works.