Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

It is well-known that the total space of the cotangent bundle $T^*X$ of a given smooth manifold $X$ admits a symplectic form $\omega$. It is actually exact: $\omega=d\lambda$. The $1$-form $\lambda$ is called the Liouville form and can be defined in a quite tautological way: given an element $p=(x,\xi)\in T^*X$ (i.e. $x\in X$ and $\xi\in T^*_xX$) and a vector $v\in T_p(T^*X)$, we define $\lambda_p(v):=\xi(\pi_*(v))$, where $\pi:T^*X\to X$ (and thus $\pi_*(v)\in T_xX$).

Does the total space of the second order cotangent bundle $T^*_2X$ also admits a "natural" geometric structure ? It seems that there is similarly a tautological "second order form" on $T^*_2X$ (in place of the Liouville tautological $1$-form $\lambda$). But then I don't see any analog for $\omega$...

Recall that the second order cotangent bundle can be defined as the dual of the second order tangent bundle $T^2X$, which is the bundle of order $2$ differential operators on $X$ that vanish on constants.

In case you would be an algebraic geometer, the fiber of $T^*_2X$ at a point $x$ ($X$ is now a smooth algebraic variety) is just $I_x/I_x^3$ if $I_x$ is the maximal ideal corresponding to $x$ (w.r.t. to an affine open neighbourhood of it).

share|improve this question

1 Answer 1

up vote 13 down vote accepted

It's not completely clear to me what form of answer you would accept. In one sense, the answer is 'the structure on $T^\ast_2M$ is the pseudogroup structure that is induced by prolongation of the pseudogroup $\mathrm{Diff}(M)$, but this kind of tautological answer is not useful.

If you are asking how one would characterize this kind of pseudogroup structure, it is worth going back and thinking about the cotangent bundle $T^\ast M$ for a moment. Yes, there is a $\mathrm{Diff}(M)$-invariant symplectic structure on $T^\ast M$, but that's not the whole structure since, as you have remarked, the Liouville form $\lambda$ is also $\mathrm{Diff}(M)$-invariant and could legitimately be regarded as part of the structure on $T^\ast M$. Note, by the way that the pseudogroup preserving $\lambda$ does not act transitively on $T^\ast M$, as it preserves the zero section (which is where $\lambda$ vanishes), so the 'natural' structure on $T^\ast M$ is not homogeneous. There are various ways that one can 'forget' some of the natural structure and get something homogeneous. One way is to just throw away everything except $\omega = d\lambda$ and consider only the symplectic structure, but you don't have to go that far. For example, you could retain $\omega$ and the Lagrangian foliation whose leaves are the fibers of $T^\ast M\to M$, and this would be a homogeneous structure.

I suspect that you might be interested in something similar for $T^\ast_2M$, that is, you would like to know whether there is some kind of weakening of the natural pseudogroup structure on $T^\ast_2M$ that is homogeneous and more 'familiar' than the full structure (which, by the way, also isn't homogeneous). One way to do this is to recognize that, because there is a canonical submersion $\phi: T^\ast_2 M\to T^\ast M$, one can simply pull back the symplectic form $\omega$ and see that it is well-defined on $T^\ast_2M$. (Of course, it's degenerate there, so it's no longer symplectic.) You can then see that there is a way to interpret an element $e\in T^\ast_2M$ as an $\omega$-Lagrangian plane in the tangent space $T_{\phi(e)}(T^\ast M)$, so you can regard $T^\ast_2 M$ as a certain open subset of the $\omega$-Lagrangian Grassmannian of the symplectic manifold $(T^\ast M,\omega)$, namely the subset that is transverse to the fibers of the map $T^\ast M\to M$. This weaker structure is homogeneous under the pseudogroup of symplectomorphisms and so might be thought of as a natural weakening that is more familiar that the full structure.

This kind of thinking might not be exactly where you want to go, but maybe it will be helpful as you try to think about what sort of weaker structure you really want to understand on $T^\ast_2 M$.

share|improve this answer
2  
When I asked the question, it's not completely clear to me what form of answer I would accept. I must say that the kind of answer you gave was not what I had in mind. But it is a GREAT answer (btw, it took me half an hour to understand why a point in $T^*_2M$ would determine a lagragian subspace in the tangent space of $T^*M$), which might turn to be what I was looking for (without knowing it). Do you happen to know if this structure you describe has been used by people working on ODE and PDE. PS: I'll wait (in case other answers would show up) a bit before validating your answer. –  DamienC Aug 29 '12 at 13:47
    
@Prof. Bryant: I was wondering how far your "tautological" answer goes. Suppose I remember only $T^*M$ as a smooth manifold together with its action of Diff(M) and forget any other structure it carries. Can one recover the Liouville form from this information alone? –  Michael Bächtold Aug 29 '12 at 20:21
    
@DamienC: Certainly the induced structures on prolonged spaces of $\mathrm{Diff}(M)$ are used to study the geometry of differential equations (both ODE and PDE) under changes of variables or other transformations. There is quite a lot of classical literature about this subject by Lie, Bäcklund, Darboux, Goursat, Engels, and Cartan. Most of this has been subsumed in to the theory of exterior differential systems, contact geometry, Gromov's h-principle, the theory of jet spaces, etc., but the focus isn't so much on developing the theory of the 'naked' structures, but their interaction with DEs. –  Robert Bryant Aug 29 '12 at 23:35
1  
@Michael: You can 'recover' the Liouville form $\lambda$ up to a constant multiple, since, up to a constant multiple, it is the only $1$-form on $T^\ast M$ that is invariant under the action of $\mathrm{Diff}(M)$. You can't 'recover' the Liouville form exactly because there is an diffeomorphism of $T^\ast M$, namely scaling by a constant in the fibers, that commutes with the action of $\mathrm{Diff}(M)$. –  Robert Bryant Aug 29 '12 at 23:47
1  
@DamienC: Yes, there are such special submanifolds, but they are more somewhat more subtle than in the symplectic case. The defining feature of coisotropic submanifolds in the Lagrangian case is that they have lots of submanifolds that are Lagrangian in the ambient symplectic manifold. Similarly, there are special submanifolds $\Sigma\subset T^\ast_2X$, the involutive submanifolds, with the property that they contain 'many' submanifolds that are the $1$-graphs in $T^\ast_2X$ of closed $1$-forms on $X$. Cartan wrote a very nice paper studying these submanifolds when $X$ has dimension $3$. –  Robert Bryant Sep 1 '12 at 15:40

Your Answer

 
discard

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.