MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What are good ways to think about Lagrangian submanifolds?

Why should one care about them?

More generally: same questions about (co)isotropic ones. Answers from a classical mechanics point of view would be especially welcome.

share|cite|improve this question
Hi Jan, maybe the tag "Poisson geometry" is appropriate as well, at least for the coisotropic submanifolds... – Stefan Waldmann Mar 31 '11 at 18:20
Okay, thanks Stefan I added it. – Jan Weidner Mar 31 '11 at 18:25
up vote 21 down vote accepted

Lagrangian submanifolds arise naturally in Hamiltonian Mechanics, because of the classical Arnold-Liouville theorem. Let me state it here:

Theorem (Arnold-Liouville). Let $(M, \omega, H)$ be an integrable system of dimension $2n$ with integrals of motion $f_1=H$, $f_2, \ldots, f_n$. Let $c \in \mathbb{R}^n$ be a regular value of $f:=(f_1, \ldots, f_n)$. Then the corresponding level $f^{-1}(c)$ is a Lagrangian submanifold of $M$.

Geometrically this means that, locally around the regular value $c$, the map $f \colon M \to \mathbb{R}^n$ collecting the integrals of motion is a Lagrangian fibration, i.e. it is locally trivial and the fibres are Lagrangian submanifolds.

Furthermore, one also shows that the connected components of $f^{-1}(c)$ are of the form $\mathbb{R}^{n-k} \times \mathbb{T}^k$, where $0 \leq k \leq n$ and $\mathbb{T}^k$ is a $k$-dimensional torus. In particular, every compact component must be a lagrangian torus.

For a proof of this result, see for instance the book by Ana Canas Da Silva "Lectures on symplectic geometry".

share|cite|improve this answer
Of course, this shows that most Lagrangian submanifolds aren't like this, since most of them have more interesting topology. – Ben Webster Mar 31 '11 at 17:36
Thanks, this theorem is really beautiful. – Jan Weidner Mar 31 '11 at 17:53

Everything is a lagrangian submanifold (A. Weinstein's lagrangian creed...) Indeed, every manifold is the lagrangian zero section of its cotangent bundle ;)

But more serious: Weinstein's tubular neighbourhood theorem states that every lagrangian submanifold in a symplectic manifold has a neighbourhood symplectomorphic to a neighbourhood of the zero section of the cotangent bundle.

Occurences of lagrangian submanifolds are indeed manifold: they arise as semiclassical support for certain FIO's and can also be thought of as semiclassical version of states in quantum mechanics via the WKB expansion. This point of view is exemplified a lot in the nice booklet of Bates and Weinstein.

Another occurence of coisotropics is in constraint mechanics: In Dirac's theory of constraints a coisotropic submanifold is what he calls a first class constraint. They arise very often in geometric mechanics with degenerate Lagrangians etc.

They are also the natural starting point for reduction: this is perhaps the more modern "coisotropic creed" of Lu (everything is a...)

Finally, to make the deformation quantization aspects a bit more precise: if you are looking for a submanifold $C \subseteq M$ in a Poisson manifold with star product $\star$ which allows for a (say) left module structure on $C^\infty(C)[[\hbar]]$ in such a way that the zeroth order of the module structure is the usual multiplication by the restriction, then you can show quite easily that $C$ has to be coisotropic. Martin Bordemann has a nice point of view how this relates to a theory of quantizing reduciton etc. in his (french!) big preprint :) In particular, the classical vanishing ideal becomes deformed into a left ideal for the star product (this was, I guess, essentially Lu's suggestion)

Note however, that there are other left ideals not of this form, e.g. the Gel'fand ideals of positive functionals, which can be much smaller.

The role of the lagrangian submanifolds $L$ in this context is that the corresponding representation on $C^\infty(L)[[\hbar]]$ becomes "irreducible" in a meaningful way (trivial commutant in the local operators). However, this only makes sense in the symplectic surrounding. In a truly Poisson manifold, only coisotropic makes sense.

share|cite|improve this answer
Thanks Stefan, stupid question: What means FIO? – Jan Weidner Mar 31 '11 at 18:29
sorry: Fourier integral operators. This relation of quite hard lc analysis with symplectic geometry is mainly due to Hörmander and Duistermaat, I think. You find such stuff in Guillemin&Sternberg's Geometric Asymptotics... – Stefan Waldmann Mar 31 '11 at 18:36
@Stefan Waldmann: It is remarkable that first occurrences of Lagrangian submanifolds and even of the Fourier integral operators could be found in the pioneer work of Maslov. – Giuseppe Apr 21 '11 at 7:19
@user40276 Well, in this purely algebraic setting this is not completely the same: What I was able to prove in some cases is that the commutant is trivial. If it would be a Hilbert space and if it would be a reasonably continuous representation then trivial commutant is the same as irreducible (Schur's theorem). But working with formal series, this is of course just an analogy. In some sense, the commutant is more robust than irreducibility (decompositions in invariant subspaces) – Stefan Waldmann May 9 at 8:28
@user40276 For a $C^*$-algebra, the triviality of the commutant is equivalent to irreducibility (both algebraic, which is difficult, and topological, which is easier): this is a standard result in rep-theory of $C^*$-algs. In formal DQ, there are several results scattered in various publications (of other and of mine), you should take a look ;) – Stefan Waldmann May 10 at 7:06

I want to answer the question "why should I care about Lagrangians" by showing how they arise when you try to answer a very natural question which in some sense lies at the heart of classical mechanics.

Suppose we consider some physical system in which the possible configurations correspond to points in a manifold $M$. Given two points $p, q \in M$ a natural question is to ask, is there some initial momentum one can select at $p$ under which after 1 second the configuration will have evolved into the configuration $q$?

In the Hamiltonian formulation of classical mechanics, the position and momentum of a system is described by the points of the cotangent bundle $T^*M$. This is a symplectic manifold in a natural way. The physics of the system are described by a so-called Hamiltonian function $H \colon T^*M \to \mathbb{R}$ and the evolution of the system is given by the flow of the Hamiltonian vector field $X_H$ associated to $H$. We can now rephrase our question as follows: is there a point $x$ in $T^*_pM$ such that the time-1 flow of $X_H$ starting at $x$ lies on the fibre $T^*_qM$? In other words, if I look at the image of $T^*_pM$ under the time-1 flow of $X_H$, does it intersect $T^*_qM$?

What does this have to do with Lagrangians? Well the fibres $T^*_pM$ and $T^*_qM$ are special cases of Lagrangian submanifolds. The image of $T^*_pM$ under the flow of $X_H$ is certainly no longer a fibre but, since the flow of $X_H$ preserves the symplectic form, the image of $T^*_pM$ remains a Lagrangian submanifold.

From this point of view you can think of a Lagrangian submanifold as a generalisation of "the set of possible initial momenta of a given point in configuration space". Moreover, this gerenalisation is forced on you, even if you only care about the very natural question described above. Finally, I hope this answer explains not only why Lagrangians are interesting, but why the possible intersections of different Lagrangians is interesting.

(Credit where credit is due: I guess this picture goes back at least to Arnold. I heard it in a seminar the other week given by Fukaya. The study of intersections of Lagrangians has become a huge industry recently, building essentially on the pioneering ideas of Floer.)

share|cite|improve this answer

Lagrangian submanifolds arise out of Hamiltonian mechanics, where you have position and momentum variables. The beauty of Hamiiltonian mechanics is the symmetry between position and momentum. Inside the combined position-momentum space, you have two natural foliations, where either all the position variables are held fixed or all of the momentum variables are held fixed. The leaves are Lagrangian submanifolds, and any Lagrangian submanifold can be viewed as a leaf of such a foliation.

A Lagrangian submanifold is locally equivalent to the zero section of a cotangent bundle.

share|cite|improve this answer

A more general form of Francisco's answer is that coisotropic submanifolds are those locally defined as the zero set of some Poisson commuting functions, and Lagrangians are those where the number of independent functions is maximal.

Lagrangian submanifolds have a lot of faces though; for example, the graph of a isomorphism from one symplectic manifold $(M_1,\omega_1)$ to another $(M_2,\omega_2)$ is Lagrangian in the symplectic structure $-\omega_1+\omega_2$ if and only if the map is a symplectomorphism. This fact has led many (myself included) to think of Lagrangian submanifolds of the product of two symplectic manifolds as a sort of "generalized map" between them. In this philosophy, a Lagrangian submanifold inside any symplectic manifold would be a "generalized point."

Coisotropic manifolds are also shadows of the representation theory of a deformation quantization; any such representation must have coisotropic limit (this is essentially Gabber's theorem) [perhaps it's better to say should have coisotropic limit, for some definition of "should"].

share|cite|improve this answer
Ben, What do you mean by "any such representation much have coisotropic limit." Does this mean that a module over deformation quantization becomes a module supported on a coisotropic subvariety when $\epsilon \to 0$? – Jeremy Pecharich Mar 31 '11 at 17:57
The last remark I don't understand: there are nice representations in deformation quantization where no coisotropic submanifold is involved at all. Take e.g. the Wick star product on $\mathbb{C}^n$ and the delta-functional at $0$. This is positive and the corresponding GNS representation yields the Bargmann-Fock like representation (all OK in the formal power series setting) But the point $0$ is certainly not coisotropic... – Stefan Waldmann Mar 31 '11 at 17:58
Thanks Ben. Do you refer to the theorem of Gabber, that says that characteristic varieties of $D$-modules are coisotropic or is there a more general theorem? @Stefan I have to admit, that I don't know much about deformation quantization. However in the D-module case the characteristic variety of the D-module generated by the deltafunction at a point is the cotangentspace at that point, so indeed coisotropic. – Jan Weidner Mar 31 '11 at 18:35
@Jan: hmm... So this perhaps means that things really start to differ here. You have quite a bunch of interesting representations in DQ where the classical limit is (in a specific way) a point but then they probably do not correspond that easily to $D$-modules. As far as I understand these relations, $D$-module theory corresponds somehow to DQ on cotangent bundles. The representations I have in mind correspond more to a Kähler situation and, in a more analytic framework, to Berezin-Toeplitz quantization schemes. Lot of questions... – Stefan Waldmann Mar 31 '11 at 18:43
In order to have an appropriate notion of characteristic variety for a module over a DQ algebra, one needs an analog of a good filtration for a D-module. The appropriate analog is a lattice, as it is explained in the book of kashiwara-schapira on DQ-modules. – DamienC Jul 15 '12 at 11:45

More on the dynamical relevance of Lagrangian submanifolds for autonomous Hamiltonian systems:

Any Lagrangian submanifold which is contained in a regular energy level is automatically invariant for the dynamics.

When applied to Lagrangian graphs inside cotangent bundles, this observation leads to the stationary Hamilton-Jacobi equation.

share|cite|improve this answer

Another (classical) way to think of a Lagrangian submanifold: the differential $df$ of a function on a manifold can be thought of as an $n$-dimensional submanifold in the cotangent space $T^*M$. How to characterize all n-dim submanifolds in $T^*M$ which are of this form? They are precisely the Lagrangian submanifolds (which are additionally transversal to the projection $T^*M\to M$). This geometric point of view comes in handy when trying to solve Hamilton-Jacobi equations $H(x,p)=0$.

share|cite|improve this answer
Well, a <em>closed</em> one-form $A$ will also have a Lagrangian submanifild as image. Those will give you all projectable Lagrangian submaniflds of $T^*M$. Interestingly enough, this is what "minimal coupling" of a classical mechanical system to a magnetic field $B = dA$ is all about. – Stefan Waldmann Apr 1 '11 at 9:04
Right, I was just thinking locally, thanks for pointing it out – Michael Bächtold Apr 1 '11 at 9:41
...there could even be transversal Lagrangian submanifolds which don't correspond to any closed one forms but "multivalued forms" – Michael Bächtold Apr 1 '11 at 10:10
OK, you got me. I didn't read carefully. I was thinking of projectible ones, which are the closed oneforms... – Stefan Waldmann Apr 1 '11 at 13:03
So (transversal) Lagrangian submanifolds are, in this case, an implementation of multivalued forms? Does this interpretation actually come up in practice? – Mariano Suárez-Alvarez Apr 4 '11 at 2:00

Since you've already gotten lots of classical mechanical answers, I'll give my favorite source of Lagrangians. Let $\sigma$ be a holomorphic section of a Hermitian line bundle $\mathcal L$ with curvature $\omega$, over a K\"ahler manifold $M$. An algebraic geometer would tell you to look at where $\sigma$ is $0$. But I'll tell you to look at where $|\sigma|$ is maximized.

This is always isotropic, and usually Lagrangian. One example to think about: $\mathcal L = O(n)$ over $\mathbb{CP}^1$. Then the section $x^k y^{n-k}$ is maximized on a circle for $0 < k < n$, and at a point for $k =0,n$.

share|cite|improve this answer
It seems strange that this set will be usually Lagrangian, unless you impose some additional symmetry, for example that you consider a toric varitety and the situation is symmetric (or equivariant) with respect to $T^n$ action. In general the maximum is achieved just at a collection of points. Is not this true? – Dmitri Apr 4 '11 at 13:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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