1
$\begingroup$

Let $(M,\omega)$ be a symplectic manifold of dimension $2n$ with the volume form $\omega^n.$

In this question we associate a Lie algebra $L(M,\omega)$ to $(M,\omega)$. Then we are interested to know:

1) Does the Lie structure of $L(M,\omega)$ depend on symplectic structure $\omega$? At the other extreme can one prove that if two Lie algebras $L(M,\omega)$ and $L(M,\omega')$ are isomorphic Lie algebras, then there is a symplectomorphism $f:(M,\omega) \to (M,\omega')$?

2)In the literature, are there some precise computation of $L(M,\omega)$ for some symplectic manifolds $(M,\omega)$? What can be said about dimension of $L(M,\omega)$?

Here is the Lie algebra we are considering:

$$L(M,\omega)=E(M,\omega)/Z'(M,\omega)$$

where $$E(M,\omega)=\left\{ X\in \chi^{\infty}(M)\mid L_X \omega=(1/n)Div(X)\omega\right \}=\{X\in \chi^{\infty}(M)\mid L_X \omega=f\omega,\;\;\text{ for some }f\in C^{\infty}(M)\}$$

and $Z'(M,\omega)$ is the normalizer of $Z(M,\omega)=\{X\in \chi^{\infty}(M)\mid L_X \omega=0\}$ in $E(M,\omega)$.

$\endgroup$
  • 1
    $\begingroup$ Your question 1 is really two questions: I read "Does the Lie structure depend on the symplectic structure" as the question whether $\forall \omega,\omega' : L(M,\omega) \cong L(M,\omega')$ holds and the question beginning with "in other words" asks whether the invariant $L(M,\omega)$ is enough to classify symplectic structures. $\endgroup$ – Johannes Hahn Jun 22 '18 at 15:45
  • 1
    $\begingroup$ @JohannesHahn Yes they in opposite direction: If the first part has positive answer, that is the Lie structure is sensitive to $\omega$ then it is natural to ask the last part, the classification one. Right? $\endgroup$ – Ali Taghavi Jun 22 '18 at 15:48
  • 1
    $\begingroup$ @AliTaghavi I think he means that an equivalent question must follow "in other words" (In English language), not a completely different question $\endgroup$ – user74900 Jun 22 '18 at 15:58
  • 4
    $\begingroup$ It seems to me that your first space is what is called the "algebra of conformally symplectic vector fields" and the second one the ideal of symplectic vector fields. You can find some results on this in Avez-Lichenrowicz, Sur l'algèbre des automorphismes infinitésimaux d'une variété symplectique, Journal Differential Geometry 40 (1974). $\endgroup$ – Nicola Ciccoli Jun 22 '18 at 20:25
  • 1
    $\begingroup$ @Ali Taghavi Sorry that "ideal" probably shouldn't be there... $\endgroup$ – Nicola Ciccoli Jun 23 '18 at 5:18
4
$\begingroup$

Just to add a little detail to the discussion above and since what I wrote was not clear:

$E(M,\omega)$ is, in Lichnerowicz-Avez-Diaz Miranda (I forgot the third author in the above citation), the Lie algebra of infinitesimal conformal symplectic transformations (Paragraph 5).

$Z(M,\omega)$ is, in Lichnerowicz-Avez, the Lie algebra of symplectic vector fields (equivalently locally hamiltonian vector fields).

$Z^\prime(M,\omega)$ is the normalizer (in Lie algebras you use this term rather than idealizer) of $Z(M,\omega)$ inside $E(M,\omega)$, i.e. the set $\{X\in E(M,\omega)\, : [X,Y]\in Z(M,\omega)\,\forall Y\in Z(M,\omega)\}$.

At page 12 it is shown that $[E(M,\omega),E(M,\omega)]\subseteq Z(M,\omega)$; therefore $Z^\prime(M,\omega)=E(M,\omega)$, if I am not wrong and/or confused by different terminology and notations.

(btw Proposition 2 of the mentioned paper may be of interest to you)

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer and the reference. In dimension $2$(n=1) $E(M,\omega)=\chi^{\infty}(M)$. But I think that $Z$ is not an ideal in $E$. For example $y\partial_x -x\partial_y \in Z,\;\;x^2\partial_x \in E$ but their Lie braket is $2xy\partial_x+x^2\partial_y $ is not in $Z$. $\endgroup$ – Ali Taghavi Jun 23 '18 at 8:17
  • $\begingroup$ Thanks again for sharing such a very interesting paper. I will read it carefuly. But it is strange : It seems that there is a possible contradictory situation. The example in the previous example shows that $Z$ is not an ideal!. $\endgroup$ – Ali Taghavi Jun 23 '18 at 19:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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