# The natural metric of a phase space and the Lyapunov exponent

For me, it seems that there is no apparent metric on a phase space of a dynamical system. Of course one can naively define an Euclidean metric on it, but it seems that this metric has not much to do with the peculiar features of a phase space.

But in many textbooks on dynamical systems, it is the metric employed in the definition of the Lyapunov exponent.

Is this really a nice approach?

• I think this is a great question but I am not good enough to answer it right now. My direct answer would have been "it's not a mertic space, it's a symplectic space", but it doesn't quite answer anything. By searching the web I did find these notes which I cannot fully comprehend yet but might hold some answers for you. – gatsu Sep 8 '14 at 16:22

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.

• Minor comment to the post (v1): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Sep 8 '14 at 17:58
• A closely related question if I may: Neighboring trajectories in a chaotic system diverge exponentially (rate given by Lyapunov's exp), does this imply that the manifold of such systems has negative Riemannian curvature? (for which neighboring geodesics must diverge exponentially.) Just trying to understand why it has to be exponentially diverging. – Ellie Dec 30 '14 at 22:08
• @Phonon Geodesic flows on Riemannian manifolds with negative sectional curvature and bound geodesics are indeed chaotic. Especially in relativity, it is often possible to transform the pseudo-Riemannian geodesic flow into a Riemannian one to prove chaoticity. The exponential definition is somewhat "numerically-empiric" - it just seems to be exclusive to non-quasiperiodic behaviour whereas integrable systems always display linear to polynomial error growth. – Void Dec 31 '14 at 11:18
• @Void thank you for getting back to me. Makes a lot of sense. I was always strongly under the impression that only for an exponential divergence the ergodic hypothesis would hold, i.e. even though trajectories are spread apart, linear divergence is not enough to guarantee the system will cover every measurable subset of phase space. Last question, in classical mechanics, for either regular or chaotic dynamic systems the phase space has to be compact, but is it correct to say for a chaotic one the topology cannot be a flat(nor torus-like) one? (obviously not symplectic anymore anyway) – Ellie Dec 31 '14 at 12:21
• @Phonon Not really, the best known Lorenz model has a $\mathbf{R}^3$ phase space. The famous kicked oscillators have a formal $\mathbf{T}^2 \times \mathbf{R}^2$ phase space which is also the case of the planar double pendulum. There are some topological considerations in chaos but the phase space only has to have dimension $\geq 3$ to allow for continuous-time chaos. – Void Dec 31 '14 at 13:51