I have the equation: $$ \dot{x}_i = F_i(x) \tag{1} $$ with $x\in \mathbb{R}^n$. To deal with the Lyapunov exponents, we write the equation for small displacements $\delta x_i$: $$ \dot{\delta x}_i = \sum_j \frac{\partial}{\partial x_j} F_i(x) \delta x_j \tag{2} $$ The rate of increase of the vectors is related to the Lyapunov exponent $\lambda$: $$ | \delta x (t) | \approx e^{\lambda t} | \delta x (t=0) | $$ Here I assume that the system is Lyapunov regular.

The definition of "Lyapunov vector" that I saw is the following. First, a matrix $Y_{i,j}(t)$ is considered, with equation: $$ \dot{Y_{i,j}}= \sum_k \frac{\partial}{\partial x_k} F_i Y_{k,j} $$ Then a matrix $M$ is defined as: $$ M = \lim_{t\to +\infty} \frac{\log Y Y^T}{2t} \tag{3} $$ According to this definition, the Lyapunov exponents and vectors are the eigenvalues and eigenvectors of $M$.

I tried to investigate how the Lyapunov vectors depend on the starting point $x$, taking two points $x_A$ and $x_B$ along a trajectory: $x_A=x(t=0)$ and $x_B=x(t=\tau)$.

I calculate $M$ in the two points: $$ M(x_A) = \lim_{t\to +\infty} \frac{\log Y(x_A,t) Y^T(x_A,t)}{2t} \tag{4} $$ and: $$ M(x_B) = \lim_{t\to +\infty} \frac{\log Y(x_B,t) Y^T(x_B,t)}{2t} \tag{5} $$ Since $Y$ is a cocycle: $$ Y(x_A,t) = Y(x_B, t-\tau) Y(x_A, \tau) \tag{5bis} $$ Then: $$ M(x_A) = \lim_{t\to +\infty} \frac{\log Y(x_B, t-\tau) Y(x_A, \tau) Y^T(x_A, \tau) Y^T(x_B, t-\tau)}{2t} \tag{6} $$ If the $Y$s commuted, we would write the logarithm of the products as the sum of logarithms of the factors, and thus get $M(x_A)=M(x_B)$ (Eq. 6 would give the same limit as Eq. 5, since $\tau$ is constant), i.e. $M$ would be constant along a trajectory. However, they do not commute, so maybe $M$ changes along the trajectory.

**My question is:** Is this correct? Actually, according to a previous answer I got on MO, **it is believed that $M$ changes if we evaluate it starting from $x_A$ or $x_B$ along the same trajectory.** Moreover, it seems that the "covariant Lyapunov vectors" evolve along a trajectory according to Eq. (2). If they correspond to the eigenvectors of $M$ (altough it is not stated clearly anywhere), then it means that $M$ does not only change along the trajectory, but that its eigenvectors $M$ evolve according to Eq. (2). Is this correct? If so, how can we see it from Eq. (6)?

inside(rather than outside) in formula (6). To set things straight - what is in your notation the standard definition (not the matrix one) of the Lyapunov exponents as growth rates? $\endgroup$