The Frenet Stability Problem - Quantum Calculus

A periodic Frenet curve in $R^n$ is up to isometry uniquely defined by a periodic curve $Q(t)$ in $SO(n)$ where the later describes the motion of the Frenet frame. Differentiation then defines a periodic curve $K(t)$ in the Lie algebra $so(n)$ . We can now reverse the story and start with a periodic curve $K(t)$ in $so(n)$ and reconstruct a curve from it. This always works and the resulting curve does not have to be Frenet but it exists as the first row of $Q(t)$ gives the velocity vector $r'(t)$ of the curve and integration gives then the curve $r(t)$ . The resulting curve does not have to be closed. When is it bounded? Last week, I made thousands of experiments to find out what happens in simplest situations. There is an elegant answer in 2 dimensions telling in which case we have stability. In 3 dimensions already, where we deal with curves in SO(n) or the unit quaternions SU(2) if you like, the non-abelian nature of the Liegroup makes the story already much harder. In that case, I only investigated theoretically cases with constant curvature and periodic torsion. There is also a resonance stability problem there, but the analysis is much harder then. Some pictures of curves given by almost periodic curvature and torsion functions are here. Here is the Frenet stability problem [P.S. I have not seen that in the literature but it almost certainly has been considered already, any references are welcome.

Hacker scene from War games. AI feels like Melvin in that scene!

[If you prompt for “Frenet stability problem”, then (like any Human would), search engines today hallucinate about what it is. The prompt “Frenet stability problem” already tells what it means. If you hear it, you know what the question is. If you ask a trained mathematician about “Frenet stability”, they can figure out in a few seconds what the question is. The only think to know is what a Frenet curve is and what stability is. There is an annoying situation at the moment with search engines. If you ask, they pretend to know and that it is not an original term because they can figure out the problem on the spot and induces that your question was not original. Here is an other trivial example: Last week I wrote about “A new age of Kitsch”. It floats the idea that we not only in art but also in science enter an area of cheapness, the reason being that we have the tools to generate much of it (also non-trivial stuff and even creative stuff) with no effort, like Kitsch art is. I never saw that thought before and consider this meta though as “original”. If you would ask an AI about it, it would smash your creativity because it is an “obvious thought” , similarly as “AI art is kitsch” . If you give high school students the task to write an essay about “AI art is kitsch” , the essays would all look the same. It is too obvious. [An article of John Emmet from October 28, 2024 which could well have been first who formulated such a thought even so it is obvious to everybody who has tried AI generated pictures that it is kitsch. Still, an article like the one of Emmet pointed it out first and should be credited]. AI is annoying when answering a question which has never been asked. It pretends that even the question is well known. Sometimes, the creativity is in the question, but then one should give the person who asks the question also the opportunity to answer it themselves, without pushing them an answer down the throat. That’s what AI search now does. Instead to point you to treasures by giving links to articles of researchers or journalists or artists, it pretend to be the treasure itself. This is annoying. AI feels a bit like Melvin’s behavior in the good old War games movie from 1983 (still one of the best “hacker movies”). Melvin (who only appears in this scene) immediately pretends to know the answer for how to get into the system. He rips the paper away and behaves like a “potato head” (in the word of Jim). It will turn out later in the movie that “Falcon’s Maze” is not the way to get into the system. There was (as Jim, the other computer programmer pointed out in that scene) a backdoor in the system which allowed David to get into the system. What makes “war games” such a good movie is that it is filled with little gems. Like the scene where the father of David butters his maize to find out that it has not been cooked, then exclaims: “it is raw!” whereas his wife tells “Its crisp! Isn’t it wonderful!). Also Hitchcock mastered this technique too. Like in “Frenzy”, where the detective has to eat “Soupe de poisson”.]

The Frenet Stability problem: Given a periodic curve K(t) in so(n), it produces a curve in SO(n) and so a curve $r(t) \in \mathbb{R}^n$ . Under which conditions is the later bounded? Mathematically we can write $Q(t)=\exp( \int_0^s K(s) \; ds)$ and so $r(t) = \int_0^t Q(s)_1 \; ds$ , where $Q_1$ denotes the first column of $Q$ . We want to understand more generally $\int_0^t Q(s) \; ds$ which is now a path in $\mathbb C^n$ . A side problem is to understand the closure of the orbit of Q in the orthogonal group. One can also look more abstractly at the closure of a path $U(t)$ in the unitary group if $U(t)$ is given by a classical or quantum system.

Before we get to the story, let us give still in general a bit more motivation for this. If we look at a finite geometry with n elements, then a quantum wave $\psi(t)$ is a path in $\mathbb{C}^n$ which has constant length. Its motion is described by the Schroedinger equation $i \hbar \psi' = H \psi$ which can also be time periodic $i \hbar \psi' = H'(t) \psi$ , for example if the quantum mechanical system is in a time dependent field. An example of the later would be a system in an external force field. The solution of the Schroedinger equation in the case of constant $H$ is of course $\psi(t) = e^{K t} \psi$ with $K=H/(i \hbar) \in su(n)$ . The unitary evolution $Q(t) = e^{K t}$ is then in $SU(n)$ as $Q(0)=1$ is the identity matrix. In the case of a time dependent Hamiltonian, we have $Q(t) = e^{\int_0^t e^{K(s)} \; ds}$ as one can see by separation of the variables and integrating $d\psi/\psi = K(t) dt$ . [I have a video from 6 years ago answering to a student question why the technique works, told that the technique of separation of variables gives a solution which then can be checked to work. Just verify that $Q' = K Q$ works when plugging in exp(K t). ] Back to the quantum mechanical system, the upshot is that looking at curves in $SU(n)$ is the same as looking at solutions of time periodic Schroedinger equations of a quantum mechanical system with a finite Hilbert space. Quantum mechanics is in principle much easier and much more elegant than classical mechanics. I did this joke before: when the world was created, the architect had to decide about how to implement it. After $10^{-50}$ seconds, it became clear that non-linear dynamics is too hard to control. The architect sighed and decided to use a linear model instead. Quantum mechanics does not have a stability problem when looking at the path Q(t) in the unitary group. What is non trivial is what happens if we look at $\int_0^t Q(s) \; ds$ which is what the Frenet problem is about! It is highly nontrivial even if Q(t) is a given path in the circle. Summing up Q(t) is like producing a random walk, only that the driving force is not random but say almost periodic. Understanding stochastic processes which have correlations is a very nice part of probability theory but can be much harder even in the simplest cases. We wrote once about Dirichlet series or Taylor series where the coefficients are almost periodic. Or see “golden rotations” for a more general overview of such problems.

In the video below I mention that stability problems in general are interesting and very classical. In engineering, one has a system like a flying plane. There are external disturbances of the flight. We of course want that the solution of the differential equations still are reasonable describing a flying plane. We do not want solutions, where the force becomes so big at some points that it breaks the plane. We know in engineering that stability is closely related to “resonances”. The particles circling Saturn in a ring are in positions away from resonances producing its fractal shape. There is a famous paper of Jossi Avron and Barry Simon (the later was my mentor at Caltech) who shed light on why the fractality of the ring structure of Staturn is so close to the Cartesian product of a Cantor set and a circle. The standard map of Chirikov (see my Javascript implementation written July 15, 2015 with a dozen line of code without libraries) or C and Java code written in 1999, also shows that integrability starts to break up in a fractal way. Invariant curves survive longer if they are more Diophantine. This is the content of Kolmogorov Arnold Moser theory. The enemy of stability are resonances which manifests as rational rotation numbers. The golden mean is the real number which is furthest away from all rational numbers. It is the most Diophantine number we know. It is therefore also the most robust. In the Standard map, it is the last stand of stability. After it breaks (it does in a universal manner as my grad school friend Andreas Stirnemann showed once). Andy was like Frank Josellis (and me) working on our PhD’s. All of us in some sense worked on some sort of stability problem. Frank in the context of periodic orbits (which are a source of stability in Hamiltonian dynamics because motion near periodic orbits produce invariant tori nearby), Andy wanted to crack the existence of a unique renormalization fixed point proving universality observations seen in the break-up of invariant tori. I was interested in Lyapunov exponents, which as it describes sensitive dependence on initial conditions is also a stability problem.

Author: oliverknill