Symplectic integrators and divergence free vector fields Imagine a simple, divergence free vector field (such as a 2D series of concentric circles).
If one seeds a triangle (three different particles) and tracks their evolution with an integrator, can one infer that if the integrator is symplectic, the triangle's area will be preserved?
I know this works in the phase space, but what about position/geometry related projections of the trajectories?
 A: Divergence vector fields are area-preserving, in the sense that if your particles evolve according to the equation $y' = f(y)$ and they fill out an area $A$ at $t=0$, then the area filled out by the particles at any later time is also $A$. However, the shape of the area is not preserved. In particular, your triangle at $t=0$ may deform into another shape; the edges are no longer straight but become curved. Thus, even for the exact flow it is not true that the area of the triangle defined by three particles is preserved. However, if you are looking at an infinitesimal triangle then the edges will remain straight an area is preserved.
With that proviso, the answer to your question is yes, symplectic integrators do preserve the area (of infinitesimal triangles). I don't understand the last question. If your system is defined by a first-order differential equation on $n$ variables, then the phase space is $R^n$ is the position space.
Added: For second-order equations, the easiest setting is when your system has a Hamiltonian $H(q,p) = \frac12 |p|^2 + V(q)$, where $q$ is (generalized) position, $p$ is (generalized) momentum, and $V$ is the potential energy. The system of equations is now $q' = p$ and $p' = -\nabla V(q)$ where $\nabla V$ is the gradient of $V$. A symplectic integrator applied to this system will preserve area, assuming $q$ and $p$ are both scalars. If $q$ and $p$ are vectors, so that the phase space is four-dimensional, then it preserves something called "symplectic area", and as a consequence it also preserves four-dimensional volume.
The two equations $q' = p$ and $p' = -\nabla V(q)$ can be combined in the single equation $q'' = -\nabla V(q)$, which may be more familiar. In particular, $-\nabla V(q)$ is the acceleration. The system is symplectic if the acceleration can be written in this form, as the gradient of some potential energy. This is the case if the acceleration is caused by a conservative force (see Wikipedia).
