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Hamiltonian systems, symplectic flows, classical integrable systems

7 votes
Accepted

neighborhood of symplectic surfaces

Here is the version of the Darboux-Weinstein theorem that you want to use: Let $(M_1,\omega_1)$ and $(M_2,\omega_2)$ be symplectic manifolds of dimension $2n$ and let $\iota_i:P\to M_i$ be smooth emb …
Robert Bryant's user avatar
15 votes

what prevents a manifold to be symplectic?

Well, already the $4$-sphere, which is even dimensional and orientable, does not carry a symplectic structure. It does not even support a $2$-form that is nowhere degenerate, much less a closed $2$-f …
Robert Bryant's user avatar
8 votes

Lagrangian surgery

Actually, one can always do a (local) surgery to 'resolve' transverse intersection points of Lagrangian submanifolds, and this works in all dimensions, not just dimension $2$. Here is a sketch of an …
Robert Bryant's user avatar
10 votes
Accepted

More pseudoholomorphic curves than complex valued functions

I think that, instead of complex-valued functions $f:M\to\mathbb{C}$ you mean $J$-holomorphic functions, i.e., complex-valued functions $f:M\to\mathbb{C}$ that satisfy $f'(x)(Jv) = i\,f'(x)(v)$ for al …
Robert Bryant's user avatar
7 votes

coisotropic submanifolds

I think that it's possible that the OP was interested in knowing something about the 'generality' of coisotropic submanifolds, even though, as José and Nicola point out, it is hard to give an explicit …
Robert Bryant's user avatar
20 votes

Is there a physical intuition for Darboux's theorem?

Another way to interpret this question is: Is there a 'heuristic' reason that all closed nondegenerate $2$-forms in $2n$ dimensions are locally equivalent? (It becomes more reasonable to ask this wh …
Robert Bryant's user avatar
14 votes
Accepted

What structure on the second order cotangent bundle ?

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 pse …
Robert Bryant's user avatar
16 votes
Accepted

A Question on Exterior Forms

No, such an $a$ does not always exist. The lowest dimension in which an example without an $a$ could possibly exist is dimension $5$, and, low and behold, there is such an example. I will give one n …
Robert Bryant's user avatar
3 votes

Is the space of half-dimensional symplectic linear subspaces of $\mathbb{R}^{4n}$ which triv...

The answer is 'no', but seeing this depends on whether your 'standard half-dimensional subspace' $\mathbb{R}^{2n}\times \lbrace0\rbrace$ is symplectic or not as a subspace of $\mathbb{R}^{4n}$. (You …
Robert Bryant's user avatar
7 votes

example of special lagrangian submanifold

If the question is "Are there examples of compact real-analytic Riemannian manifolds that cannot be isometrically embedded as a special Lagrangian submanifold of a compact Calabi-Yau manifold?", then …
Robert Bryant's user avatar
4 votes
Accepted

Existence of solution to a system of linear PDEs with boundary conditions

The general solution of your equations in a simply connected domain on which $r_2\not=0$ and $r_1\not=\pm1$ is $$ \beta = \frac12 + \frac1{{(r_1}^2{-}1)}\, \left(\frac{\partial a}{\partial\theta_1}+b( …
Robert Bryant's user avatar
9 votes
Accepted

First Chern class vanishes on a Lagrangian submanifold

Yes, if you ignore 2-torsion: After all, if you give $TM$ a complex structure compatible with $\omega$, say $J$, then you'll have that $TM$ pulls back to $L$ to become isomorphic to $TL\oplus T^*L\si …
Robert Bryant's user avatar
5 votes

Symplectic blow-up

I believe that you will find a careful discussion of symplectic blow-up of symplectic submanifolds in D. McDuff, Examples of simply-connected symplectic non-Kählerian manifolds, J. Differential Geom. …
Robert Bryant's user avatar
11 votes
Accepted

Grassmannians of planes isotropic with respect to general tensors

What you are asking about is very classical in the theory of exterior differential systems. The subspaces of $D$ that you are calling `isotropic' are what Élie Cartan called the integral elements of …
Robert Bryant's user avatar
5 votes
Accepted

$\omega$ is a symplectic form then $L^k_{\omega}:\Omega^q(M)\to \Omega^{q+2k}(M)$ is injecti...

The injectivity case is well-known and follows quite easily from the statement (usually attributed to Lefschetz) that $L^k_\omega:\Omega^{n-k}(M)\to \Omega^{n+k}(M)$ is a isomorphism for $0\le k\le n$ …
Robert Bryant's user avatar

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