In his Applied Differential Geometry book, William Burke says the following after telling that the action should be the integral of a function $L$:

A line integral makes geometric sense only if it's integrand is a 1-form. Is $Ldt$ a 1-form? Well, that's the wrong question. The correct question is: On what space is $Ldt$ a 1-form? It is not a 1-form on configuration space, the space of positions, because it can have a non-linear dependence on velocities. A 1-form must be a linear operator on the tangent vectors. The correct space for $L dt$ is the line-element contact bundle of the configuration space.

Now, why intuitively the correct setting for lagrangian mechanics is on the contact bundle? I understand the contact bundle as pairs $(p,[v])$ where $p$ is a point in configuration space and $[v]$ is an equivalence class of vectors, explicitly $v\sim kv$.

Thinking not on all that arguments for selecting the space on which $Ldt$ is a $1$-form, physically, how can we intuit that the contact bundle is useful for that? I mean, is there some observation in classical mechanics that guides us in building the theory on that space?


The special feature of contact geometry is the contact 1-form $\lambda$, which satisfies $\lambda\wedge d\lambda\ne0$ (let's restrict to 3-dimensions). In our Lagrangian mechanics example, $\lambda = dq-vdt$. You want this to pull-back to zero on the ``permissible'' curves in phase space -- these curves represent the motions of your system.

For a more detailed but tangential explanation, see:


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.