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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)?

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  • $\begingroup$ There is an inconsistency in how you write the cocycle condiditon between eqs (5) and (6) and the definition of the limit matrix $M$ as a result of which your increment $Y(x_A,\tau)$ appears 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? $\endgroup$
    – R W
    Commented Oct 29, 2020 at 15:24
  • $\begingroup$ To get (6), I just plug $Y$ from (5bis) into (4), so the $Y(x_A,\tau)$ appears inside the product (between the $Y(x_B,t)$. Of course this is strange also for me: I was wondering if $M$ should be rather defined with $Y^T Y$, so that $Y(x_A,\tau)$ appears outside. Can you confirm? I also added the definition of Lyapunov exponent in the question as requested. $\endgroup$ Commented Oct 29, 2020 at 15:52
  • $\begingroup$ Yes - I think there is a bit of confusion here - I will write a detailed answer shortly $\endgroup$
    – R W
    Commented Oct 29, 2020 at 20:17

1 Answer 1

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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 $[Z(t)Z^*(t)]^{1/2t}$.

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  • $\begingroup$ My question addressed an additional aspect: is it true that the eigenvectors of $\Lambda$ change, point by point along a trajectory, according to eq. (2)? Can we see it from your definitions? $\endgroup$ Commented Nov 4, 2020 at 10:48
  • $\begingroup$ I posted it as a new question: mathoverflow.net/questions/375718/… $\endgroup$ Commented Nov 5, 2020 at 13:40
  • $\begingroup$ I even bountied it. $\endgroup$ Commented Nov 7, 2020 at 20:31

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