From the Darboux's theorem in the symplectic geometry, we know that symplectic manifolds with the same dimension is locally "equivalence". I have a little puzzle with the meaning of "equivalence".

I do not think the "equivalence" means isometric because we can change the symplectic form with number factor. From wiki here, the local "equivalence" means locally symplectomorphic which is a diffeomorphism march with symplectic structure. If we forgot the symplectic structure, that is only the diffeomorphism, but why do we call it totally different with the Riemann manifold, specitally no curvature? Curvature is not a topology concept, I guess that the symplectic structure contains a "weak isometric" which make the curvature meanless. Am I right? And I need more details about the weak isometric.

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    $\begingroup$ "Equivalence" means locally symplectomorphic. Why would you forget the symplectic structure? $\endgroup$ Mar 5 '13 at 6:44
  • $\begingroup$ We say it's "totally different from Riemannian" because for a Riemannian manifold you have local structure (i.e. curvature) which distinguishes such manifolds, whereas Darboux implies there is no local information on symplectic manifolds, they're all (locally) the same, and hence we have to pay attention to the global structure instead. $\endgroup$ Mar 5 '13 at 7:30
  • $\begingroup$ You can only talk about "isometric" when a Riemannian metric is present. Symplectic structures can be defined on smooth manifolds that aren't equipped with Riemannian metrics; in such a case, "isometric" doesn't make sense. $\endgroup$ Mar 5 '13 at 7:31

Darboux's theorem says that any symplectic manifold $(M^{2n}, \omega)$ is locally symplectomorphic to the "trivial" symplectic manifold $( \Bbb R^{2n}, \omega_0)$, where $$\omega_0 = \sum_{j = 1}^n dx^j \wedge dy^j$$ is the standard symplectic form on $\Bbb R^{2n}$. What this means is the following. Given any point $p \in M$, there is a neighborhood $U \subset M$ of $p$ and a diffeomorphism $$\psi: U \longrightarrow \Bbb R^{2n}$$ such that $\psi^\ast \omega_0 = \omega$ on $U$, i.e. $\psi$ is a symplectomorphism. Note that we do not throw away the symplectic structure in this notion of equivalence; without the symplectic structure there is no symplectic geometry to speak of in the first place. "Equivalence" between two symplectic manifolds means that they are symplectomorphic.

A symplectic manifold $(M, \omega)$ is different from a Riemannian manifold $(N, g)$. A symplectic form $\omega$ is a nondegenerate $2$-form, i.e. a nondegenerate antisymmetric rank $2$ tensor. A Riemannian metric $g$, on the other hand, is a nondegenerate symmetric rank $2$ tensor. So right away we see that symplectic manifolds and Riemannian manifolds are fairly different objects. But this is not the reason we say that Riemannian geometry and symplectic geometry are totally different subjects. Darboux's theorem implies that there are no local invariants of symplectic manifolds; any such invariant would also be a symplectomorphism invariant of $(\Bbb R^{2n}, \omega_0)$, and hence we wouldn't be able to discern anything special about a symplectic manifold $(M, \omega)$ by inspecting it locally. This is in stark contrast to Riemannian geometry, where the curvature of a Riemannian manifold is a local object.

Therefore the reason we say symplectic geometry is a totally different subject than Riemannian geometry is that symplectic manifolds have no local symplectomorphism invariants, while local isometry invariants of Riemannian manifolds such as curvature are quite central to the field of Riemannian geometry.

Added: The proof of Darboux's theorem relies on the fact that a symplectic form $\omega$ is closed, i.e. $d\omega = 0$. If $\omega$ is not closed (i.e. we have an almost symplectic manifold) then we do not necessarily have Darboux's theorem. So while Riemannian geometry is in some sense about nondegenerate symmetric bilinear forms on tangent spaces, symplectic geometry is about nondegenerate antisymmetric bilinear forms on tangent spaces with an extra condition described by the differential equation $d\omega = 0$. The extra condition is what makes the theory qualitatively different.

We can impose a differential equation for the metric on a Riemannian manifold to get a "version of Darboux's theorem" for Riemannian geometry. The analogous condition in Riemannian geometry is having a metric $g$ whose Riemann curvature tensor is everywhere zero (i.e. a flat metric); such a manifold is locally isometric to $\Bbb R^n$ with the standard metric.

  • $\begingroup$ Thanks, but I want to know how does the symplectic structure work, how does it make the "Equivalence" much stronger and for which step the difference between Riemannian manifold and symplectic manifold appears. $\endgroup$
    – Strongart
    Mar 8 '13 at 15:18
  • $\begingroup$ @Strongart: I added something to answer your comment at the end of my answer. $\endgroup$ Mar 8 '13 at 17:51
  • $\begingroup$ Thanks, maybe I should check the proof of Darboux's theorem carefully. $\endgroup$
    – Strongart
    Mar 10 '13 at 14:44

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