To solve the geodesic equations for a specific metric in General Relativity I can find conserved quantities $F = \xi_{\mu}\frac{\mathrm{d}x^{\mu}}{\mathrm{d}\lambda}$ along geodesics by using Killing Vectors $\xi_{\mu}$. I was told that to I can use the Liouvilles/Arnold Theorem of Integrability to check if the geodesic equations are solvable by quadratures. It states (Arnold, Mathematical Methods of Classical Mechanics):

If a system with $n$ degrees of freedom (i.e., with a 2$n$-dimensional phase space), $n$ independent first integrals in involution ($(F_i,F_j) \equiv 0 \qquad i,j = 1,2,\dots n$) are known, then the system is integrable by quadratures.

Here the bracket denotes the Poisson-Bracket. More exactly Arnold then states for a part of the theorem :

Suppose that we are given $n$ functions in involution on a symplectic 2$n$-dimensional manifold \begin{equation} F_1,\dots,F_n \qquad (F_i,F_j) \equiv 0 \qquad i,j = 1,2, \dots n \,. \end{equation} Then the canonical equations with hamiltonian function $H$ can be integrated by quadratures.

My questions are: Why is one allowed to state this theorem in this less exact form without stating the need of a symplectic manifold, when one would suppose that you need a symplectic structure to define Poisson-Brackets? Where is the symplectic manifold, i.e. the symplectic form coming from when I am working on a Pseudo-Riemannian manifold? Can I just somehow take the cotangent bundle to do this?

I apologize if the question is trivial or for eventual mistakes in my question.

  • $\begingroup$ Yes, the canonical symplectic structure comes from the cotangent bundle if that's what you're asking. $\endgroup$ – Qmechanic Apr 24 at 5:30
  • $\begingroup$ If the canonical symplectic $2$-form $\omega$ on $T^{*}X$ is closed, then it can also be exact. If the symplectic $2$-form $\omega$ on an even dimensional space $R^{2n}$ is closed, it can't be exact. Also, since $\omega$ is closed, i.e., $d\omega=0$, the space may have to be flat - but that might only apply to symplectic forms on Riemannian spaces. $\endgroup$ – Cinaed Simson Apr 25 at 2:32
  • $\begingroup$ @Cinaed Can you elaborate on those points? Things that are not clear to me: How does the dimension of a manifold ensure that a closed form $\omega$ is not also exact? Why does closedness of a form imply the flatness of my manifold? And why especially on Riemannian manifolds? I think I am getting several things wrong here. $\endgroup$ – horropie Apr 25 at 7:29
  • $\begingroup$ The dimension of the space needs to be even and ω is born closed. When ω comes from the cotangent bundle, it can also be exact. It it's not from cotangent bundle, then it's still closed but it can't be exact. Since ω is closed, then dω=0 which is like saying it's Ricci flat and things may only work on a flat manifold. In general relativity it could be a problem if the manifold needs to be flat - I don't know. It's not a problem with Minkowski space-time because it's flat. $\endgroup$ – Cinaed Simson Apr 25 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.