It has been shown by Eichhorn, Linz and Hänggi in 2000 that the numerical values of Lyapunov exponents are invariant under any invertible variable transform. This is just a reformulation of the fact that they are metric invariant, because the authors presume the norm $|\cdot|$ to be an arbitrary norm in the given coordinates - just it's basic properties such as linearity are enough.

To get some intuition for this - Lyapunov exponents are linked with the Haussdorf or fractal dimension of the trajectory. Even though the Hausdorff dimension is defined on a metric space, we have an intuition that a fractal dimension is actually more of a property of differential structure rather than of a specific notion of length/surface/volume. The metric is just a handle to get to the fractal dimension, but it's nature is non-metric. We can understand acquiring Lyapunov exponents in a similar way - the metric is just a handle and we choose one arbitrarily.

A second way to get an intuition is through the explicit definition of the exponents $\lambda$ via the linear variation $\delta x(t)$ evolved in time:
$$\lambda = \lim_{t\to \infty} \frac{\log |\delta x(t)|}{t}$$
Let us assume that $\delta x(t) = e^{\mu t}\delta x(0)$. Then out of the linearity of the norm we have $|\delta x(t)| = e^{\mu t} |\delta x(0)|$ and the limit yields
$$\lambda = \lim_{t \to \infty} (\frac{\mu t}{t} + \frac{\log |\delta x(0)|}{t})$$
The second term dies off and we have $\lambda=\mu$ for any positive definite linear $|\cdot|$. I.e. you get the same number with a different norm and thus the Lyapunov exponent gives you something which is connected to a "relative growth rate" independent of the metric. There are some loopholes to this argument and these are covered by the article cited above.

This post imported from StackExchange Physics at 2015-03-23 09:13 (UTC), posted by SE-user Void