To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N? - MathOverflow

To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2010-02-14T20:03:43Z

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 $N$?

Answer by Kevin Lin for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2010-02-14T21:13:38Z

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.

Answer by Santiago Canez for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2010-02-15T01:10:20Z

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.

Answer by Petya for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2010-03-18T19:36:44Z

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.

Answer by unknown (google) for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2010-05-18T05:06:47Z

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.

Answer by Giuseppe Tortorella for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2011-04-29T14:22:49Z

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.

Answer by Jay for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2012-02-04T03:27:23Z

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

Answer by Nicola Ciccoli for To what extent can I think of a Lagrangian fibration in a symplectic manifold as T*N?

2012-02-04T07:57:28Z

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.