First I introduce the Lyapunov vectors. Here I follow the notations of a previous answer I got on MO.
We have a dynamic system with discrete time $t$ (integer values). The time evolution is defined by a (local) diffeomorphism $T$: $$ x(t+1) = T[x(t)] \tag{1} $$ where $x \in \mathbb{R}^n$. This gives rise to a one-parameter group $T^t$ of (local) diffeomorphisms.
We define the matrices $Y_{i,j}(x,t)$ as the derivative maps of these diffeomorphisms. I assume that the sequence $Y(x,t)$ is Lyapunov regular. For fixed $x$, it is known that there are a (single) symmetric matrix $\Lambda(x)$ and a matrix $\Delta(x,t)$ such that: $$ Y(x,t) = \Delta(x,t) \cdot \Lambda(x)^t \tag{2} $$ with $$ \log \|\Delta(x,t)\|,\log\|\Delta^{-1}(x,t)\|=o(t) \;. \tag{3} $$
The eigenvectors of $\Lambda(x)$ should be the Lyapunov vectors (please correct me if I'm wrong).
Now my question is: how do such vectors evolve along a trajectory?
I mean this: if $v(x)$ is an eigenvector of $\Lambda(x)$, is there a $v[T(x)]$, eigenvector of $\Lambda[T(x)]$, that can be somehow calculated from $v(x)$?
For what I understand, such vectors are the "covariant Lyapunov vectors". If this is correct, then each of them, say $v$, should evolve according to: $$ v[x(t+1)] = X[x(t)] \cdot v[x(t)] \tag{4} $$ where $X(x)=Y(x,1)$, or equivalently: $$ v[T(x)] = X(x) \cdot v(x) \, . \tag{4bis} $$
Actually, I never saw this claim explicitly written anywhere in this form, so I'm wondering if this is correct.
I found no way to prove it but I tried to do the following. First, I use the property that $Y$ is a cocycle: $$ Y(x,t+1) = Y[T(x),t] \cdot X(x) \tag{5} $$ thus finding: $$ \Delta(x,t+1) \cdot \Lambda(x)^{(t+1)} = \Delta[T(x),t)] \cdot \Lambda[T(x)]^t X(x) \, . \tag{6} $$ This equation contains $X(x)$, as Eq. 4, however I do not know how to go on to prove Eq. 4.
I do not expect that an answer contains the proof, but at least that it confirms that Eq. 4 holds, e.g. by providing a reference, or that it is wrong.
