# 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!

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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 (15,608 points)

@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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