# Starting vector in Lyapunov exponents evaluation

Let us consider the equation: $$\dot{x}_i = F_i(x)$$ with $$x\in \mathbb{R}^n$$ and $$i=1\dots n$$, and the equation for small displacements: $$\dot{\delta x} = \sum_j \frac{\partial}{\partial x_j} F_i(x) \delta x_j$$ I often read (and checked in practice) the following: starting from a random initial $$\delta x$$, for increasing time it approaches a given "regime", such that its growth is given by the largest Lyapunov exponent $$\lambda_1$$.

However, in principle, it should be possible to find a particular $$\delta x$$ such that it will evolve with a different (lower) Lyapunov exponent $$\lambda_2<\lambda_1$$. Such initial vector $$\delta x(t=0)$$ could depend on the initial $$x(t=0)$$.

My first question is the following: is such a vector really dependent on $$x(t=0)$$, or it is constant for every $$x$$?

The question could appear strange, but it is related to the definition of "Lyapunov vector" that I saw often. This definition is based on the following procedure. 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: $$M = \lim_{t\to +\infty} \frac{\log Y Y^T}{t}$$ According to this definition, the Lyapunov exponents and vectors are the eigenvalues and eigenvectors of $$M$$. For this reason, I would say that the Lyapunov vector is a single vector for the whole trajectory, not depending on $$x$$. Hence my question about its relation with the above-mentioned vector field $$\delta x(x)$$.

Since I think that the vector field $$\delta x(x)$$ is not the Lyapunov exponent, I'm asking about literature about it, where the properties of such a field are studied and discussed.

To begin with, there is no reason whatsoever for the dynamical system determined by an arbitrary vector field on $$\mathbb R^n$$ to be Lyapunov regular.

If the system is Lyapunov regular, then the associated filtrations of the tangent space start from the bottom of the Lyapunov spectrum. If you are interested just in the top Lyapunov exponents (assuming for simplicity that they are simple - I use plural because they may very well depend on the starting point), then you will have an $$(n-1)$$-dimensional distribution in the tangent bundle (generated by the directions with the lower exponents), and you are asking about (the existence of?) a vector field consisting of tangent vectors outside of this distribution. Probably, you would also want this field to be invariant with respect to the dynamics.

If you think about your questions in the coordinate-free form, then you will see that there is no reason for the answer to question (1) to be positive. Indeed, the very notion of a "constant" (or parallel, in a more rigorous terminology) vector field depends on the linear structure on $$\mathbb R^n$$ (how do you identify tangent vectors at different points? what happens if one changes coordinates by applying a diffeomorphism of $$\mathbb R^n$$?).

(2) What do you mean by "the Lyapunov vector"?

(3) The distributions determined by the Lyapunov filtration have been considered in numerous publications - see, for instance Smooth ergodic theory and nonuniformly hyperbolic dynamics by Barreira & Pesin and the references therein.

EDIT (after OP's clarifications). The vector field you are talking about is then the field orthogonal to the distribution I mentioned (the difference is that you are implicitly using the standard Euclidean structure; it seems to be a very natural thing until you think about what happens if one changes coordinates). The argument I had outlined shows that there is no reason for this vector field to be constant (in your terminology).

By the way, your definition of Lyapunov regularity in terms of the matrix $$M$$ is incomplete. The missing condition is that the increments $$Y_t^{-1}Y_{t+1}$$ have to be subexponential (although it is usually satisfied automatically - for instance, in your situation it would follow from the boundedness of $$F_i$$ - it does not have to hold in general). The simplest example is $$Y_t=\begin{pmatrix}1 & 0 \\ e^t & 1\end{pmatrix} \;.$$ Here the limit $$M$$ exists, but this family is not Lyapunov regular.

• Actually, my question is "what is a Lyapunov vector?" I added the definition of Lyapunov vector that I saw often, although not in reliable references. Maybe this helps to clarify the question. Oct 25 '20 at 9:08
• I see the point, there are two things (the distributions in the tangent bundle and the eigenvectors of $M$) which are connected and which (both) depend on the point $x$ (and there are no reasons to wonder about this fact). However, my concern is that, it seems to me, that $M$ is the same for two different initial points $x_1$ and $x_2$ on the same trajectory. This is puzzling for me, exactly for the reasons highlighted by R W. Even worse, imagine what happens if the trajectory densely covers a volume. Oct 25 '20 at 17:54
• @DorianoBrogioli Of course, for any consistent definition of the vector field in question it has to be invariant with respect to the flow. However, the invariance with respect to dynamics is not at all "constancy" in your sense.
– R W
Oct 25 '20 at 21:57
• So we have $M(x)$, and $M(x_A)$ is not always equal to $M(x_B)$ (comparing component by component), not even if $x_A$ and $x_B$ are on the same trajectory. However, you say that $M(x)$ is invariant under the flow. For me (a poor physicist) this means that all the components of $M$ are constant... Do you mean: "taken the flow as a coordinate transformation, then $M(x)$ is covariant with respect to that transformation"? Sorry if the question looks trivial: multidisciplinarity requires to cope with such language differences. Oct 26 '20 at 10:47
• Now I have difficulty understanding the meaning of "covariant". I wanted to say the following. There are two completely different ways ways of identifying the tangent spaces at two points: the "coordinate" and the "flow" ones. For the first one we introduce global coordinates, and then just compare these coordinates. For the second one we notice that if $A$ is mapped to $B$ by the flow, then the tangent space at $A$ is also mapped onto the tangent space at $B$. In the case of Lyapunov vectors we deal with the invariance in the second sense - which by no means implies coordinate invariance.
– R W
Oct 26 '20 at 13:04