# Lyapunov vectors along a trajectory

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)?

• 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?
– R W
Commented Oct 29, 2020 at 15:24
• 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. Commented Oct 29, 2020 at 15:52
• Yes - I think there is a bit of confusion here - I will write a detailed answer shortly
– R W
Commented Oct 29, 2020 at 20:17

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 $$[Z(t)Z^*(t)]^{1/2t}$$.
• 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? Commented Nov 4, 2020 at 10:48