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Hamiltonian systems, symplectic flows, classical integrable systems
37
votes
Accepted
decompositions for exceptional Lie algebras $E_6$, $E_7$ and $E_8$
The $E_6$ and $E_7$ decompositions you list are explained in Cartan's 1894 thesis (see pages 89–92 for these formulae). For $E_8$, Cartan instead gives a decomposition (a $\mathbb{Z}_3$-grading) of t …
30
votes
Accepted
The Lefschetz operator
There is an elementary proof in our 2003 book Exterior Differential Systems and Euler-Lagrange Partial Differential Equations (Bryant, et al, University of Chicago Press). It does not use any represen …
29
votes
Accepted
Pseudo-holomorphic curves in the six-sphere
It might be immodest of me to mention my own work, but there is quite a lot known about the pseudo-holomorphic curves in $S^6$. For example, it is known that the pseudoholomorphic rational curves are …
22
votes
Accepted
Can a Lagrangian submanifold of ${\mathbb R}^{2n}$ be dense ($n>1$)?
Your question already has the answer in it for $n=2$. Take a connected complex curve $L\subset\mathbb{C}^2$ that is dense in $\mathbb{C}^2$. Then $L$ is Lagrangian for the real part of the holomorph …
21
votes
Asking whether there is a compact Lie group containing affine symplectic group
The answer is 'no', the affine symplectic group cannot appear as a Lie subgroup of any compact Lie group. The reason is that the affine symplectic group contains $\mathrm{SL}(2,\mathbb{R})$ as a Lie …
21
votes
Accepted
Darboux-like theorems
While Francois' answer is fine, as far as it goes, I think that it is important to bear in mind the history of this problem.
The original problem of 'flatness' or integrability (and more generally, e …
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 …
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 …
16
votes
Accepted
Questions on R. Bryant's paper "Calibrated embeddings in the special Lagrangian and coassoci...
I'm afraid that that article does not do a lot of details in the introductory Section 0, just because more complete explanations were already available in earlier articles of mine. Here are some brief …
15
votes
Accepted
Recognizing the stablizer of a degenerate three forms in six dimension
First of all, it is not true that there are only two types of nondegenerate $3$-forms in $\Lambda^3(V^\ast)$ when $\dim_\mathbb{R}V=6$. There are actually $3$ such nondegenerate orbits. Here is the …
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 …
15
votes
Accepted
Non-isomorphic compact Kähler manifolds that are biholomorphic, symplectomorphic and isometric
The answer is 'no, not necessarily'.
Consider the following example: Let $M=N=\mathbb{CP}^2$, let $(\omega_0,J_0)$ be the standard Fubini-Study Kähler structure on $M$. Now let $f$ be an arbitrary, …
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 …
13
votes
Accepted
Darboux like theorem for non-degenerate 3-forms in 6-manifolds
This depends on what you mean by 'Darboux-like'. It is certainly not true that a closed nondegenerate 3-form on a 6-manifold is necessarily locally equivalent to one of the 'flat' models, so there is …
13
votes
Accepted
Which curves are boundary of pseudoholomorphic curves?
The 'moment conditions' that Ben McKay mentions are simply this: A closed curve $C$ in $\mathbb{C}^n$ bounds a compact Riemann surface (which might be singular) if and only if the integral around $C$ …