3
$\begingroup$

Reading a paper on hamiltonian mechanics, in a section on classical examples of complete integrability, it is examined the geodesic flow of a triaxial ellipsoid.

Before separating the variables in the Hamilton-Jacobi equation, the original approach followed by Jacobi himself, it is stated that the phase space of this system has none infinitesimal symmetry which is the lift of vector field on the space of the configuration.

So this would be an example of an hamiltonian system of mechanical type with only hidden (or dynamical) symmetries.

My question is: how to prove that there are not Killing vector fields for the triaxial ellipsoid (endowed of the ambient euclidean metric)?

Improve this question
$\endgroup$
1
  • $\begingroup$ I have received two answers that are equally very interesting. My question was not restrictive on the kind of proof. Not being possible multiple choices, I have accepted the first posted answer. $\endgroup$
    – agt
    Commented Jun 9, 2011 at 18:00

2 Answers 2

Reset to default
9
$\begingroup$

This follows from a rigidity theorem for convex surfaces. According to the Wikipedia page on Cauchy rigidity, this was proved by Cohn-Vossen for smooth convex surfaces in $R^3$. So two convex surfaces which are intrinsically isometric are related by an ambient isometry of Euclidean space. So if you have a convex body in Euclidean space which has a killing field, then there would be a 1-parameter family of intrinsic isometries, and therefore a 1-parameter family of rotations. But this is clearly impossible for an ellipsoid with three different axes.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Dear Algol Thanks a lot for the proper reference, and for the explanation. $\endgroup$
    – agt
    Commented Jun 8, 2011 at 20:41
12
$\begingroup$

An alternative, more low-powered proof goes along these lines (and can be found in most older books on differential geometry of curves and surfaces): Compute the Gauss curvature $K$ and the square-norm of the Gauss curvature $|\nabla K|^2_g$ of the induced metric $g$. One then observes that, on a dense open set of the ellipsoid, these are independent functions (i.e., their differentials are linearly independent except on a set of measure $0$). Since any Killing field $Z$ would have its flow preserve both of these functions, it follows that $Z$ must vanish almost everywhere, i.e., it must vanish identically.

Note that this proof is local, whereas the Cohn-Vossen proof is global.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Dear Robert Bryant, I like this argument very much: going along the lines of your sketch, I should have the occasion to give a proof in closed-form. Your and Algol's answers are both very exaustive. Thanks you. $\endgroup$
    – agt
    Commented Jun 8, 2011 at 21:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .