# what is the domain of the Lagrangian of a surface embedding?

If we view our Lagrangian particle mechanics geometrically, then we describe a particle trajectory as a map from R to a manifold, and the Lagrangian $L(x,\dot{x})$ as a function on the tangent bundle of the manifold.

I am curious how we would describe geometrically the Lagrangian function $L(u,\partial_x u,\partial_y u)$ for an embedded surface, $R^2$ to the manifold. Like for solving for a minimal surface. What is the domain of this Lagrangian function, is it $\Lambda^2 TM$? So it eats bivectors? No, that doesn't seem right. We don't expect $L(u,\partial_x u,\partial_y u) = -L(u,\partial_y u,\partial_x u)$ or any (anti)symmetry in the arguments. And not a tensor product either. What is its domain?

-
I like the question. May I suggest that you increase the chance of it being answered by accepting some answers to your previous questions? I see that you personally thanked the authors of some of them, so I'm guessing you found them acceptable :-) – joriki Dec 22 '11 at 7:20
Thank you for the suggestion. I had not forgotten those threads and I do intend to close them with credit given where it's due. However I was waiting for some pending items first. In the mean time, I gave upvotes to the responders. Will dealing with those older threads will improve my chances of getting responses to newer threads? – ziggurism Dec 22 '11 at 8:20
I would say that the domain is the $1$-jet of maps from $\mathbb{R}^2 \to M$. Note that this works also for the Lagrangian mechanics example too, since the $1$-jet bundle of maps from $\mathbb{R} \to M$ is the tangent bundle to $M$. – Sam Lisi Dec 22 '11 at 13:34
Thank you, Sam. I don't know much about jet bundles, though I've heard they're the natural way to handle higher derivatives. Now I'm even more intrigued. But since there isn't a more specific question, and you've pointed me in the right direction, I guess I will close this question. Thanks again! – ziggurism Feb 6 '12 at 4:36