Darboux's theorem in the symplectic geometry 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.
 A: 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.
