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  Interpretation of the pullback of $P: TM \rightarrow TM^*$

+ 1 like - 0 dislike

Assume that $L = L(q,\dot{q})$ is a time-independent Lagrangian. To go from a Lagrangian description of the dynamics of the system to the Hamiltonian description, the diffeomorphism $P: TM \rightarrow TM^*$ maps the tangent bundle coordinated by $(q,\dot{q})$ to the cotangent bundle coordinated by (q,p) with $p = \frac{\partial L}{\partial \dot{q}}$.

What is the interpretation and physical meaning of the pullback of $P$, defined as

\[P^*dp_i = \left( \frac{\partial^2 L}{\partial\dot{q}^j\partial\dot{q}^i}\right)d\dot{q}^j + \left( \frac{\partial^2L}{\partial q^j\partial \dot{q}^i}\right)dq^j\]

It seems to be some kind of a 1-form living on the cotangent bundle that maps $TM^* \rightarrow \mathbb R$ when acting on a point of the cotangent-bundle. But I have a hard time to see what it actually does and is good for?

asked Nov 30, 2016 in Mathematics by Dilaton (6,240 points) [ revision history ]
edited Dec 1, 2016 by 40227

I have corrected what seemed to me to be a typo in the main formula. I hope that I did not introduce extra mistakes.

I suggest you to give a look at Problem 1 of Chapter 3 of Choquet-Bruhat et al. "Analysis, Manifolds and Physics." Its a solved problem, and there she considered almost the same situation as you have asked.

Observe that the Lagrangian is a mapping \(\mathcal L: TM\longrightarrow\mathbb R\), and from this you managed to build a map \(T(M) \longrightarrow^PT^*(M)\), which happens to be a bundle morphism between the tangent and contangent bundles of the configuration space. Recall that without the Lagrangian, one has no canonical way to identify the space of sections of \(T(M)\), namely the vector fields \(\mathcal X(M)\) , to the space of sections of \(T^*(M)\), namely the one-forms \(\bigwedge^1M\). This is usually performed in an intrinsic way with a metric \(g_{ab}\) on the base space in order to raise or lower indices, speaking in index notation. So the Lagrangian allows you to do this identification like we would do with a metric in GR. However, it may be interesting to note that, if you had a metric on the configuration space and the system evolved like a "freely falling" body (borrowing the term from GR) in \(M\), one would have \(\mathcal L = \frac{1}{2} g_{ab}\dot q^a \dot q^n\), and the pull-back of the momentum coordinate 1-forms on \(T^*(M)\) acting on the velocity coordinate vectors on \(T(M)\) would give \(P^*(dp_a)(\frac{\partial}{\partial \dot q^b})=g_{ab}\). So I think you may imagine that the pull-back induced by your Lagrangian has to do with the effective metric of the system.

To make the argument rigorous, you can think of how the Lagrangian actually gives you the bundle morphism \(T(M) \longrightarrow^PT^*(M)\). To do this intrinsically, you can use the notion of fiber derivative. I believe you can find this in (maybe) chap. 3 of Marsden's book on foundations of mechanics.

But intuitively, what you do you to construct an energy function for the motion of the system on \(T(M)\) from the second-order ODE derived from the Euler-Lagrange equations of \(\mathcal L\). You use the fiber derivative to build from that energy-function a Hamiltonian flow on \(T^*(M)\). The projection of the flow lines onto \(M\) should yield the same trajectories obeyed by the Euler-Lagrange equations, if the Lagrangian is well behaved (or, as stated in the answer below, the "Jacobian" \(\partial_{\dot q_a}\partial_{\dot q_b}\mathcal L\) is nonsingular).

1 Answer

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$dp_i$ is a 1-form on $T^{*}M$ and $P$ is a smooth map $P \colon TM \rightarrow T^{*}M$, so it makes sense to consider the pullback $P^{*}dp_i$ which is a 1-form on $TM$. Concretely, it just means express $p_i$ as a function of $q$ and $\dot{q}$ and then compute $dp_i$ as a function of $dq$ and $d\dot{q}$. An immediate application of the chain-rule gives the formula in the question.

This formula is quite relevant because it tells you that to recover $\dot{q}$ from $p$ (and in particular, to have $P$ diffeomorphism, which is not always the case!), you need to be able to invert the matrix of components

$\frac{\partial^2 L}{\partial\dot{q}^j\partial\dot{q}^i}$.

If you can do it, you obtain an equivalence between the Lagrangian dynamics on $TM$ and the Hamiltonian dynamics on $T^{*}M$. If you can't do it, then the Hamiltonian dynamics on $T^{*}M$ is under-determined and the correct Hamiltonian dynamics equivalent to the original Lagrangian dynamics lives in general on a quotient of a submanifold of $T^{*}M$. It is exactly what happens in gauge theories.

answered Dec 1, 2016 by 40227 (5,140 points) [ revision history ]
edited Dec 5, 2016 by Arnold Neumaier

Thanks for these nice, helpful, and interesting explanations! I am looking forward to getting to the chapters of my differential geometry book where things like gauge theories for example are treated ... :-)

''in a quotient of a subspace of $T^*M$''?? I think it is in a quotient of the space of sections of $T^*M$.

I am ready to agree that this phrase is cryptic but I think it is correct. It refers to the fact that we have first to restrict ourselves to a symplectic submanifold of $T^{*}M$, defined by the second class constraints in the language of Dirac, and then (symplectically) quotient by the ("gauge") transformations generated by the  first class constraints. The resulting object is a symplectic manifold, equipped with an Hamiltonian function, whose dynamics is equivalent to the original Lagrangian dynamics.

ok; so a quotient of a submanifold, not of a subspace - which has a different meaning when $M$ is a Euclidean space, and no meaning in general.

@ArnoldNeumaier Honest question: what did you think 40227 meant by a subspace?

@doetoe: It was not clear. Spaces of sections have subspaces, manifolds usually don't. I care about precision.

@ArnoldNeumaier : you are right, my use of the word "subspace" was sloppy.

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