This is probably a very elementary question in symplectic geometry, a subject I've picked up by osmosis rather than ever really learning.

Suppose I have a symplectic manifold $M$. I believe that a Lagrangian fibration of $M$ is a collection of immersed Lagrangian submanifolds so that as a fibered manifold locally $M$ looks like a product. I.e. I can find local coordinates so that the fibers are $\{\text{half the coordinates} = \text{constant}\}$.

Then, at least locally, I can think about the set of fibers as some sort of "space" $N$. My question is: to what extent can I think of $M$ as the cotangent bundle $T^*N$?

Surely the answer is "to no extent whatsoever" globally: the set of fibers is probably not a space in any good way, and certainly not a manifold (see: irrational line in a torus). But what about locally? Then it's really two questions:

Question 1: If I have a lagrangian fibration in $M$, can I find local coordinates $p_i,q^j: M \to \mathbb R$ so that the symplectic form is $\omega = \sum_i dp_i \wedge dq^i$ and the fibers are of the form $\{ \vec q = \text{constant}\}$.

I thought the answer was obviously "yes", and maybe it is, but what I thought worked I can't make go through all the way.

Then the question is about how canonical this is, and that's not really about general Lagrangian fibrations at all:

Question 2: What is a good description of the local symplectomorphisms $\mathbb R^{2n} \to \mathbb R^{2n}$ of the form $\tilde q = \tilde q(q)$ and $\tilde p = \tilde p(p,q)$?

The beginning of the answer is that it is a local symplectomorphism if $\sum_i d\tilde p_i \wedge d\tilde q^i = \sum_i dp_i \wedge dq^j$, but the left-hand-side is $\sum_{i,j,k} \bigl(\frac{\partial \tilde p_i}{\partial q^j}dq^j + \frac{\partial \tilde p_i}{\partial p_j}dp_j \bigr) \wedge \bigl( \frac{\partial \tilde q^i}{\partial q^k}dq^k \bigr)$, so the two conditions are that $\sum_i \frac{\partial \tilde p_i}{\partial q^j} \frac{\partial \tilde q^i}{\partial q^k}dq^k$ is a symmetric matrix, and that $\frac{\partial \tilde p}{\partial p}$ is the (maybe transpose, depending on your convention) inverse matrix to $\frac{\partial \tilde q}{\partial q}$.

Anyway, I guess for completeness I'll also ask the global question:

Question 0: What global conditions on $M$ and the fibration assure that there is a global symplectomorphism with $T^*N$ for some $N$?

  • $\begingroup$ You should change "Lagrangian fibration in a symplectic manifold" to just "Lagrangian fibration" -- I think the former is not the standard way to use the term. I might be wrong but I think people usually just say "M is a Lagrangian fibration [over a base B]". $\endgroup$ Commented Feb 14, 2010 at 21:33
  • $\begingroup$ @Kevin: Well, yeah. I mean, "Lagrangian fibration" doesn't make a lot of sense when the manifold is not symplectic. $\endgroup$ Commented Feb 15, 2010 at 3:26
  • 1
    $\begingroup$ You're right: The only way it makes sense is if the manifold is symplectic. That's why the terminology works :-) $\endgroup$ Commented Feb 15, 2010 at 4:08

7 Answers 7


Your Question 1 is called Darboux theorem for fibrations (see: Arnold, V., Givental, A., Symplectic geometry, Dynamical Systems IV, Symplectic Geometry and its Applications (Arnold, V., Novikov, S., eds.), Encyclopaedia of Math. Sciences 4, Springer-Verlag, Berlin-New York, 1990.)

Here is how to construct suitable Darboux coordinates. Let $q_i$ be local coordinates in the base of the fibration, we identify them with their pullbacks to the symplectic manifold. The functions $q_i$ generate Hamiltonian vector fields $X_{i}$ and these fields are tangent to the fibers (note that $X_{i}$'s commute). Let $\varphi_{i}(t)$ be the flow map generated by $X_{i}$ for time $[0,t]$.

Now we choose (locally) a Lagrangian submanifold $L$ transversal to the fibration. The coordinates $q_i$ give coordinates on $L$, so $(q_1,...,q_n)$ stands for a point on $L$. Here is a construction of a local symplectomorphism $$(p_1,...,p_n,q_1,...,q_n) \mapsto \varphi_{n}(p_n)\circ ...\circ \varphi_{1}(p_1)(q_1,...,q_n).$$ It is easy to check that it is indeed a fibered symplectomorfism sending the symplectic structure to the standard one.


It seems Question 2 is a special case of: What are the fiber-preserving symplectomorphisms of $T^*M$? This has a nice answer.

First, any diffeomorphism $f$ of $M$ defines a fiber-preserving symplectomorphism of $T^*M$, its cotangent lift, by $(q,\xi) \mapsto (f(q),((df_q)^*)^{-1}\xi)$. These are exactly the fiber-preserving symplectomorphisms of $T^*M$ which preserve the canonical 1-form. Second, any closed 1-form $\beta$ on $M$ defines a fiber-preserving symplectomorphism of $T^*M$ by $(q,\xi) \mapsto (q,\xi+\beta_q)$. These are exactly the ones which preserve the fibers of the projection $T^*M \to M$.

Then it is not hard to show that any fiber-preserving symplectomorphism of $T^*M$ is a composition of a cotangent lift with fiber translation by a closed 1-form.

  • $\begingroup$ Oof, this is where I wish I'd split this as two questions, because I should accept your answer, but then I might not get an answer to question 1. Anyway, +1! Thanks. $\endgroup$ Commented Feb 15, 2010 at 3:31
  • $\begingroup$ A clean and succint proof of this can be found at pages 30-31 (theorem 3.29 and corollary 3.31) of "Lectures on the Geometry of Quantization" by Sean Bates and Alan Weinstein. $\endgroup$
    – Alex M.
    Commented Jul 17, 2016 at 7:27
  • $\begingroup$ This also shows that the leaves of a lagrangian foliation have a natural affine structure. Another way would be to use the so-called "Bott connection" of the (co)normal bundle along the leaves, which is flat, and "transport" it via the symplectic form to the tangent bundle of the leaves, where one checks that it is torsion free. Being flat, this puts an affine structure on the leaves. $\endgroup$
    – BS.
    Commented Jul 28, 2018 at 22:09

Hey Theo --- I don't think it is reasonable to expect Lagrangian fibrations to be cotangent bundles globally. Easy example: take a 2d torus, give it a symplectic form (equivalently a volume form in this case); every 1d submanifold is automatically Lagrangian; the torus is a circle bundle over a circle; realizing it this way, it is a fibration over the circle with fibers being Lagrangian circles. Certainly this is not a cotangent bundle.

Another example, integrable systems yield Lagrangian fibrations over R^n: these are usually not cotangent bundles. See the section on integrable systems in Cannas da Silva's book.

  • $\begingroup$ So, the theorem I know that neighborhoods of Lagrangians look like cotangent bundles says this: If I have an individual Lagrangian submanifold, then I can find coordinates near it so that the manifold its its tangent bundle. But: it's not at all obvious to me that the parallel sections of this cotangent bundle (that depend on local coords anyway) give my original fibration. But I'm really asking the opposite question: given a Lagrangian fibration, can I find a Lagrangian local cross-section? $\endgroup$ Commented Feb 15, 2010 at 3:29
  • $\begingroup$ No, not in general. There are Lagrangian fibrations with multiple fibers. E.g. take $D\times E$ where $D$ is a 1-dim complex disk with a coordinate $u$ and $E$ is an elliptic curve with a standard one form $\lambda$. Then $du\wedge \lambda$ is a symplectic form on $D\times E$ and the projection $D\times E \to D$ is a Lagrangian fibration. Now take the quotent $X$ of $D \to E$ by the involution which acts as $u \to -u$ on $D$ and as a translation by a point of order $2$ on $E$. The symplectic form descends to $X$ and the fibration descends to a fibration with a double fiber over $0$. $\endgroup$ Commented Feb 15, 2010 at 3:59
  • $\begingroup$ Hi Tony, I am not sure what "multiple fiber" or "double fiber" mean? $\endgroup$ Commented Feb 15, 2010 at 5:29
  • $\begingroup$ Hi Kevin! It means that the homology class of the fiber is divisible. In the example let $\Delta = D/(\pm 1)$ and let $p : X \to \Delta$ be the projection induced from the projection $D\times E \to D$. Then this is a Lagrangian fibration for which the map $p$ is submersive everywhere outside of $0 \in \Delta$. The fiber over $t \neq 0\in \Delta$ is a smooth copy of $E$. The fiber over $0$ consists of critical points for $p$ but is a submanifold isomorphic to $F = E/(\text{tanslation})$. In $H_{2}(X)$ we have $[E] = 2[F]$ so $p$ can not have local sections in a neighborhood of $0 \in \Delta$. $\endgroup$ Commented Feb 15, 2010 at 13:37
  • $\begingroup$ Hmm. I don't think this is a Serre fibration. It doesn't have the path lifting property. It is a fibration in Theo's sense as far as I can tell. If Theo meant locally trivial fibrations, then the existence of a local Lagrangian section is automatic so I don't understand the question then. $\endgroup$ Commented Feb 15, 2010 at 18:32

Here's another example. Take a cotangent bundle T*M and add to the canonical symplectic structure the pullback from M of a closed but nonexact 2-form. In the resulting symplectic manifold, the cotangent fibres still form a lagrangian fibration, but there is no local lagrangian cross section.


Dear Theo Johnson-Freyd, I hope to have at least partially understood the content of your question, and that my answer could be useful.

0.Setting and specification of the terminology.
In a symplectic $2n$-dimensional manifold $(M,\omega)$, let be given a lagrangian foliation $\mathcal{F}$, i.e. a foliation of $M$ whose leaves are lagrangian w.r.t. $\omega$. (Instead, I mean a lagrangian fibration of $(M,\omega)$ as a surjective summersion $f:M\to B$ whose fibers are lagrangian w.r.t. $\omega$. Any fibration determines a foliation but the converse is not true. The difference will be immaterial in my point(1), but not so in my point(2).)

1.Local Existence of lagrangian submanifolds transversal to $\mathcal{F}$. For any $p\in M$, there exists a lagrangian submanifold of $(M,\omega)$ which passes through $p$ and is transversal to $\mathcal{F}$.

Infact, for any $p\in M$, there exists a chart $(U,\phi)$ for $M$ centered at $p$, such that:
$\omega= \sum_{i=1}^{n}{d\phi_i \wedge d\phi_{n+i}}$,
the restriction of $\mathcal{F}$ on $U$ is generated by $\frac{\partial}{\partial\phi_{n+1}},\ldots,\frac{\partial}{\partial\phi_{2n}}$,
and consequently $\phi_{n+1}=\ldots=\phi_{2n}=0$ is a local lagrangian submanifold of $(M,\omega)$ passing through $p$ and transverval to $\mathcal{F}$.

This is just the Caratheodory-Jacobi-Lie theorem, applied starting with a system $d\phi_1,\ldots,d\phi_n$ of $1$-forms which locally generates the distribution corresponding to the lagrangian foliation $\mathcal{F}$.

2. A relative globalization.
If $L$, a lagrangian submanifold of $(M,\omega)$, is transversal to $\mathcal{F}$, then there exists a diffeomorphism $f$ from an open neigborhood of $L$ in $M$ onto an open set in $T^*L$ such that:
$f|_L$ is the zero section of $\tau_L^{\ast}:T^{\ast}L\to L$,
$f_{\ast}\omega$ is the canonical symplectic on $T^{\ast}L$,
and $f$ takes the leaves of $\mathcal{F}$ in the fibers of $\tau^{\ast}_L$.

This is just Theorem 7.1 in "Symplectic Manifolds and their Lagrangian submanifolds" of A.Weinstein.


An example of global result answering Question 0 is the following (From Woodhouse, Geometric Quantization, Proposition 4.7.1) Let $P$ be a real polarization of a symplectic manifold $M$ with leaves that are simply connected and geodesically complete and let $Q$ be a Lagrangian submanifold of $M$ that intersects each leaf transversally in exactly one point. Then $M$ is symplectomorphic to $T^*Q$, $P$ coincide with the vertical foliation and $Q$ with the zero section.

More generally if $P$ is a real polarization of $M$ such that the space of leaves $M/P$ is a manifold, leaves are complete and simply connected and $H^2(M/P;\mathbb R)=0$ then $M$ is symplectomorphic to a cotangent bundle. To each real polarization one can associate a cohomology class which is an obstruction to the existence of a global Lagrangian section.


good reference: Duistermaat, J. J. (1980), On global action-angle coordinates. Communications on Pure and Applied Mathematics, 33: 687–706


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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