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Bracket on $\mathcal C^\infty(TM)$ in classical mechanics ?

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For a configuration manifold $M$ we expect there to be a bracket on $\mathcal C^\infty (T^*M)$ endowing the space with a Lie algebra structure. Here we assume that $T^*M$ is also equipped with a non-degenerate symplectic 2-form $\omega_H$ and that the bracket can be written $\omega_H(X_f,X_g)$ for two Hamiltonian vector fields $X_f,X_g\in \chi(T^*M)$. We note that there exists a 1-form $\theta_H$ on $T^*M$ such that $\omega_H=-d\theta$ and that $\theta_H$ can be pulled back to $TM$ to give $\theta_L$. In charts we may write, 

\begin{equation}\theta_L=\frac{\partial L}{\partial \dot q^i}dq^i\end{equation} 

Hence we expect the exterior derivative to give an induced 2-form on $TM$. 

\begin{equation}\omega_L=\frac{\partial ^2}{\partial q\partial v}dq\wedge dq +\frac{\partial ^2L}{\partial v\partial v}dq\wedge dv \end{equation}

At this point I am interested in evaluating $\omega _L(X_f,X_g)$ in order to see what expression is generated. My attempt is to assume that $X_f$ and $X_g$ are now Lagrangian vector fields on $TM$ and hence can be written,

\begin{equation}X_f=\dot q\frac{\partial f}{\partial q}+\dot v\frac{\partial f}{\partial v}\end{equation}

This approach however does not successfully lead anywhere (could be my shoddy maths skills to blame however, so I would be extremely interested to see if it did in fact lead somewhere?). If we cannot develop a bracket on $TM$ does this mean that $f,g\in\mathcal C^\infty (TM)$ do not have a Lie algebra structure, and at what point would they inherit such a structure during their transition to the cotangent bundle?

Thank you for your time and I hope this question is appropriate for this forum (?) apologies if not! 

asked Dec 30, 2015 in Theoretical Physics by AngusTheMan (15 points) [ revision history ]

1 Answer

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Your bracket on $T^*M$ is correct. If you do all computations in $T^*M$ then things will be consistent.

But something is wrong with the way you try to go from $T^*M$ to $TM$. This cannot always be done.

If you start with a Lagrangian 1-form on $TM$ you need a nondegeneracy condition in order to get a symplectic form and hence a Poisson bracket on its sections. (This excludes, e.g., gauge theories; there the bracket is defined only on the quotient space of gauge equivalence classes..) Moreover, the definition of a Hamiltonian vector field is now different.

For more details see Section 18.4 of my online book Classical and Quantum Mechanics via Lie algebras.

answered Dec 31, 2015 by Arnold Neumaier (12,570 points) [ no revision ]

@ArnoldNeumaier Thank you for you answer and link to your work, it really helped a lot and is an amazing resource! I obtained the same expression for the bracket after a bit more computation. I would love to know a little more about equations 18.29, where do they come from for instance, and their significance? With regard to the above bracket, does this satisfy the conditions to make $\mathcal C^\infty (TM)$ into a Lie algebra? Thank you again for your time. @ArnoldNeumaier 

@AngusTheMan: The equations come from (18.1) and the definition of a locally Hamiltonian vector field, directly above Proposition 18.1.2. Their significance is that, in the singular case, they single out the gauge invariant  functions. [See Example 18.3.1, which one can also cast in a lagrangian form - exercise!] Indeed, these equations are responsible for being able to derive the Jacobi identity: Theorem 18.1.3 proves that one gets a Poisson algebra, in particular a Lie algebra. But this Poisson algebra is $C^\infty(TM)$ only in the regular case!

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