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](https://mathoverflow.net/questions/374816/starting-vector-in-lyapunov-exponents-evaluation), **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)?



Lyapunov vectors along a trajectory
The confusion indeed concerns the order of $Y$ and $Y^\*$ (I prefer to use $\*$ instead of $T$ for transposition) in the definition of the matrix $M$. This is quite common, and the reason is that both orders actually do occur - depending on how the increments are added in the definition of the matrices $Y(t)$. Let me for simplicity assume that the time $t$ is discrete (integer valued).


In your context we are given a group $(T^t)$ of (local) diffeomorphisms (the time $t$ solutions of the differential equation with varying initial points). Your matrices $Y(t)$ are then the derivative maps of these diffeomorphisms, and they satisfy the cocycle condition, which is your formula (5bis) in a somewhat different notation:
$$
Y(x,t) = Y(T^\tau x, t-\tau) Y(x,\tau) \;.
$$
Thus, if we put
$$
X(x) = Y(x,1) \;,
$$
then
$$
Y(x,t) = X(T^{t-1}x)\cdot \ldots \cdot X(Tx) \cdot X(x) \;.
$$
Lyapunov regularity of the sequence $Y(t)=Y(x,t)$ (for a fixed $x$) is equivalent to the existence of a matrix $\Lambda$ such that
$$
Y(t) = \Delta(t) \Lambda^t
$$
with
$$
\tag{\*}
\log \|\Delta(t)\|,\log\|\Delta^{-1}(t)\|=o(t) \;.
$$
If the matrix $\Lambda$ is additionally required to be symmetric, then it is unique and coincides with the limit
$$
M = \lim\_t [Y^\*(t) Y(t)]^{1/2t} \;.
$$
Conversely, if the limit $M$ exists **and** condition (\*) is satisfied, then the sequence is Lyapunov regular. This equivalence is not that hard to verify by taking into account that
$$
\| Y(t) v \|^2 = \langle Y(t) v, Y(t) v \rangle = \langle v, Y^\*(t) Y(t) v \rangle 
$$
for any vector $v$.


In the above situation the increments to the products $Y(t)$ are added on the left. However, quite often one talks about products of random matrices with the increments added on the right, for instance,
$$
Z(t) = A\_1 \cdot A\_2 \cdot \ldots \cdot A\_t \;,
$$
where $(A\_i)$ is a stationary sequence of increment matrices. It is for these products that one has to define the Lyapunov type regularity by considering the limits of $[Z(t)Z^\*(t)]^{1/2t}$.


