Robust Lyapunov Exponents - Quantum Calculus

Abstract: the upper envelope of the Lyapunov exponent functional on SL(2,R) cocycle A over an automorphism T of a probability space can be estimated below by the Herman functional H(A) which is the expectation of log||A||/2 where ||A|| is the Hilbert Schmidt norm of A. We call the upper envelope of the Lyapunov exponent functional on bounded measurable cocycles with uniform topology the robust Lyapunov exponent. This implies for example that if we look at the cocycle dT of the standard map T(x,y) = (2x-y+c sin(x),x) on the two dimensional torus, then there are uniform perturbations smaller than the accuracy of the computer works with, such that the “entropy” computed by this modification is larger or equal than log(c/2). This is what we numerically measure. (see for example page 307 in my thesis, or a javascript online application written in 1999). The proof of that envelope remark is heavily based on Herman’s subharmonic method, which had been used already in my 1992 paper: The fact that the Lyapunov exponent functional is discontinuous for aperiodic systems, was established in this paper published 1991. The envelope remark done here shows that discontinuity can be very common. There are cocycles like $ latex A = Diag(a,1/a)$ for example, where a takes values 10 with probability 1/2 and values 1/10 with probability -1/2 which has has zero Lyapunov exponent by the law of large numbers but which can be driven with arbitrary small rotation wiggles A R(x) with R(x) a rotation matrix with an angle of say 10^-100, to have exponential growth rate larger than 2. The rotation wiggles however need to be constructed carefully.

Let $T$ be an automorphism of a probability space $(X,\mathcal{A},P)$ (an invertible measure preserving map). The number $\lambda(A)$ for a non-commutative random variable $A \in \mathcal{X}=L^{\infty}(X,Sl(q,\mathbb{R}))$ is defined as $\liminf_{n \to \infty} \frac{1}{n} E[ \log(||A^n(x)||)]$ , where $A^n(x)=A(T^{k-1}x) \cdots A(T(x)) A(x)$ . The non-commutativity makes the computation of $\lambda(A)$ difficult in general. This is illustrated in that it is discontinuous in general in the uniform topology given on the Banach manifold $\mathcal{X}$ . If $A=dT$ is the Jacobian matrix of a volume preserving diffeomorphism on a $q$ dimensional compact Riemannian manifold, then $\lambda(A)$ is a lower bound for the Kolmogorov-Sinai entropy = metric entropy of the map. Positive entropy is usually associated with the term chaos. It is a relevant physical quantity because it matters for orbits on sets of positive probability. Positive topological entropy does a priori only matter on a set of zero measure. The maximum over all metric entropies of invariant measures is the topological entropy but in general, the measure of maximal entropy could have zero volume measure.

[For example, we proved in 1995 with Jeng that the billiard in the table $x^p+y^p=1$ has positive topological entropy for all non-integrable cases $p \neq 1,2,\infty$ were the table is a circle or square; the reason was Mather’s theorem for p larger than 2 and Hubacher’s theorem for p smaller than 2 preventing some caustics to work allowing then to apply a theorem of Angenent. Actually as the open AI dump of October 6 also has shown (proving a conjecture of Poritsky from 1950 done here at Harvard), every convex smooth billiard different from an ellipse has positive topological entropy. We do not know of convex billiards with positive entropy but it is possible that the metric entropy is positive for any non-integrable billiard.]

One of the difficulties of the function $\lambda$ is that it is not continuous on $\mathcal{X}$ . It is continuous on a generic set but not in general and that the decision depends on subtle cohomological properties of the dynamical system. It is in particular related to the cohomology of sets. Assume for simplicity that $T$ is ergodic. Take a constant diagonal cocycle $A(x)=Diag(2,-2)$ for example, where $\lambda(A)=\log(2)$ . Now take a measurable subset $Z \in \mathcal{A}$ (an event) and define the rotation $R(x)=1$ for $x \not in Z$ and R(x) a rotation matrix by $\pi/2$ for $x \in Z$ . Then $\lambda A R)$ switches the stable and unstable directions whenever we reach Z. If $Z$ is not a coboundary, meaning if we can not write $Z=T(Y) \Delta Y$ for some $Y$, then $\lambda(A R)=0$ . In my paper, I used Rohlin towers to make this large change distribute on smaller parts so that we can find examples where $\lambda(A)>0$ but $\lambda(A_n) =0$ for a sequence of $A_n \to A$ .

When looking over my old paper from 1992 on the density of positive Lyapunov exponents during the last week, it appeared to me that the Herman method of subharmonicity actually can turn quite a large class of cocycles to appear having $\lambda(A)>0$ and actually with a lower bound $H(A)$ which is defined as $H(A)=E[ \log(||A||_2/2)]$ where $||A||$ is the HIlbert Schmidt norm of $A$ . This works even for cocycles that clearly have zero exponents like $A(x) = diag(a, a^{-1})$ , where $a$ are IID random variables taking values in say $2,1/2$ with probability 1/2 each. Actually, by Birkhoffs ergodic theorem (or even the strong law of large numbers), such a cocycle has $\lambda(A)=0$ . But the robust $\Lambda(A)>0$ . Lets just make a bold definition:

Definition. Let $\Lambda(A)$ be the maximal limit $\lambda(A_n)$ you can get for sequences $A_n \to A$ . We call this upper envelope function the robust Lyapunov exponent. For points $A$ where $\lambda$ is continuous (a generic set by a theorem of Baire), we have $\Lambda(A)=\lambda(A)$ . Define also the Herman norm $H(A) = E[ \log(||A||_2/2)]$ of $A \in \mathcal{A}$ . It is a modification of a quantity introduce by Herman as $\frac{1}{2\pi} \int_0^{2\pi} \lambda(A R(\beta)) \; d\beta \geq H(A)$ . (See Proposition 2.4 in my paper.) A robust modification can be defined for any function $f$ on a metric space. Take for example $f_{robust}(x) = \lim_{r \to 0} \sup \{ f(y), y \in B_r(x) \}$ if $B_r(x)$ is the ball of radius $r$ . It is know as an upper envelope or the smallest upper semi continuous function that is bounded below by f. For continuous functions, the robust function of course agrees with the function.

The claim is now $\Lambda(A) \geq H(A)$ . As always, we can assume without loss of generality that $T$ is ergodic (otherwise build the ergodic decomposition and do it on each ergodic fiber). The idea is simple: By Herman, we can find $B=A R(\beta)$ with $\lambda(B) \geq H(A)$ . Now partition $X = \bigcup_{k=1}^n X_k$ of $X$ with $P[X_k] =1/n$ . The Abramov formula shows $\lambda(B_{X_k}) \geq H(A);/P[Z_k] = n H(A)$ (See Lemma 2.2 in my paper.) But now we can use Corollary 2.5 to see that there is an angle $\beta_k -\beta$ where $\beta_k \leq 1/(H(A) n)$ such that $\lambda B_{X_k} R(\beta_k-\beta) \geq n H((A)$ which by Lemma 2.3 in my paper shows that $\lambda B R(1_{Z_k} (\beta_k-\beta)) \geq n H(A)$. Now start turning one after the other “wheel” to make the rotation everywhere small. Start with $k=1$ to get $\lambda B R(1_{Z_1} (\beta_1-\beta)) \geq n H(A). Call this new cocycle again $B$ . It still has $\lambda(B) \geq H(A)$ but now it is on the set $Z_1$ close to $A$ . Now induce on $Z_2$ . We can turn the second wheel such that $\lambda B R(1_{Z_2} (\beta_2-\beta)) \geq n H(A)$ and call this cocycle again $B$ . Now this cocycle is close to $A$ on $Z_1 \cup Z_2$ and still $\lambda(B) \geq H(A)$ . Continue like this until $B$ is everywhere close to $A$ but still $\lambda(B) \geq H(A)$ .

This is constructive in the following way. Assume we had an oracle that could evaluate $\lambda(A)$ for $A \in \mathcal{X}$ . Given a $A \in \mathcal{X}$ with $H(A)>0$ and given $\epsilon>0$ , we can find constructively, using finite many evaluations of the function $\lambda$ a perturbation $B$ of $A$ that is $\epsilon$ close and has $\lambda(B) \geq H(A)$ . This allows using a sequence of evaluations of the function $\lambda$ to get a sequence $A_n$ that converts to $A$ and for which $\lambda(A_n) \geq H(A)$ . This establishes especially the existence of many examples (diagonal for example), where $\lambda(A)=0$ but $\Lambda(A) \geq H(A)$ can be large. Discontinuity of $\lambda$ on the Banach manifold $\mathcal{X}$ is therefore very common. It might even be “the rule” away from the uniform hyperbolicity case, even so we know that the set of continuity points is Baire generic by Baire’s theorem. By “the rule” could mean that if we take a probability space of cocycles not in the uniform hyperbolic part, then almost all elements produce discontinuity.

Philosophically, this has consequences: if we are in a commutative classical world, then we can compute expectations of random variables exactly using simple integration by the laws of large numbers or using ergodic theorems. But now, if we allow even the smallest “noncommutative leak” (provided for example by quantum mechanical effects), the non-commutative predictions become discontinuous and stuff which has been zero can jump to values that can differ from the commutative case. The take-away is that “quantization” is a severe step that does not allow “perturbation theory”. Discontinuity is fundamentally diametral to any perturbation approach. Assume we have a commutative model for a classical theory and we introduce a “Planck constant” $h$ which for $h=0$ is the classical commutative theory and which for $h>0$ becomes non-commutative. The computation just shown indicates that expectation quantities do in general not respond continuously to turning on the value $h$ to a non-zero value. Perturbation theory can fail on a fundamental level even if we look at time averages of bounded quantities.

Author: oliverknill