Much of the terminology in symplectic geometry comes from classical mechanics: the symplectic manifold is modeled on a cotangent bundle $T^*N$ of some configuration space $N$ with local position coordinates $q_i$ and momentum coordinates $p_i$ such that $\omega = \sum_i dq_i \wedge dp_i$. Then, a Hamiltonian - a.k.a. a smooth "energy" function - on this phase space induces a flow satisfying Hamilton's equations from classical mechanics: $$\frac{dp_i}{dt} = -\frac{\partial H}{\partial q_i} \hskip{1 in} \frac{dq_i}{dt} = \frac{\partial H}{\partial p_i}$$ In this physical interpretation, what are Lagrangian submanifolds? In particular, why are they named Lagrangian? Is there a relationship between this notion and the Lagrangian formulation of mechanics?

(Note: this question has a lot of great answers providing some physical or geoemtric intuition for Lagrangian submanifolds - in a cotangent bundle, both fibers and the images of closed sections (closed $1$-forms on $N$) are motivating examples of Lagrangian submanifolds - but it does not address the source of the terminology.)

  • 3
    $\begingroup$ You might look at the chapter in da Silva's text on the Legendre transform - this gives you a way of going from a Lagrangian action $L: TM \to \mathbb{R}$ to a description of the Hamiltonian dynamics on $T^*M$, and hence a possible etymology. Authors who might give a more detailed picture are Marsden and possibly Weinstein. $\endgroup$ – dvitek Nov 8 '16 at 21:53
  • $\begingroup$ $T^\ast_xM$ is a Lagrangian submanifold of $T^\ast M$ (as are the images of this cotangent space under the flow generated by the Hamiltonian). $\endgroup$ – Chris Gerig Nov 8 '16 at 22:18
  • 8
    $\begingroup$ Let $(M,\Omega)$ be a symplectic manifold. A submanifold of $N \subset M$ is a Lagrangian submanifold if for each $x \in N$ the tangent space $T_x N$ is a Lagrangian subspace of $T_x M$. Quoting Abraham and Marsden (see page 403 of authors.library.caltech.edu/25029 ), The terminology "Lagrangian subspace" was apparently first used by Maslov [1965], although the ideas were in isolated use before that date. The book by Maslov is in French and can be found maths.ed.ac.uk/~aar/papers/maslovbook.pdf where Lagrangian subspaces are referred to as sous-espace Lagrangiens. $\endgroup$ – Nawaf Bou-Rabee Nov 9 '16 at 0:58
  • 8
    $\begingroup$ The terminology comes from Maslov's pioneering treatise Theory of Perturbations and Asymptotic methods. The reason for the terminology is that Lagrange brackets (a dual version of the Poisson brackets) vanish on Lagrangian submanifolds. $\endgroup$ – alvarezpaiva Jan 3 '17 at 19:51
  • 3
    $\begingroup$ One might add that as noted by Weinstein (1971, p. 339), Lagrangian submanifolds were already studied by Souriau (1953, 1954; pdfs 16, 17) under the name V.I.S. (variété isotrope saturée, pronounced “vis”). Of course, as @alvarezpaiva said, opticians had long been calling (some of) them normal congruences. $\endgroup$ – Francois Ziegler Jul 28 '18 at 16:18

This echos the 2017 comments, but since the question has now been bumped to the front page it might be helpful to give the actual source in Maslov's book [1].

[1] V.P. Maslov, Perturbation Theory and Asymptotic Methods (in Russian, Moscow, 1965); Théorie des pertubations et méthodes asymptotique (French translation, Paris, 1972).

The Lagrangian manifold (variété lagrangienne) is introduced on page 114-115:

114 115 338

| cite | improve this answer | |
  • $\begingroup$ Consider a variety, $k$-dimensional, $x_0 = x_0 (\alpha)$, $\alpha = \alpha_1, \dots, \alpha_k (k \le n))$, diving in $\mathbb{R}^n$. A generalization of the transverse condition can be written in the form (2.1) It is more convenient to consider in the phase space $q = x p = \mu \dot{x}$ the non-sigular differentiable variety of dimension $n$ $q = q(\alpha), p = \mu \dot{q}(\alpha), \alpha = \alpha_1, \dots \alpha_n$. The Condition (2.1) can be written (2.2) that is to say that in any local coordinate system of the variety the Lagrange brackets are null. Such a variety is called Lagrangian $\endgroup$ – ChoMedit Nov 16 '19 at 6:53
  • $\begingroup$ and the form (2.2) is Lagrange brackets.. $\endgroup$ – ChoMedit Nov 16 '19 at 6:57

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.