Clarification on the definition of a symplectic integrator According to the notes that I am reading, a numerical one-stop method $y_{n+1}=\Phi_h(y_n)$
is said to be symplectic if, when applied to a Hamiltonian system, the discrete
flow $y\mapsto \Phi_h(y)$ is a symplectic transformation for all sufficiently
small steps.
I am not sure how to understand this. Is this the same as saying that when we ignore the $h^2$ and higher order powers of $h$ in $y\mapsto \Phi_h(y)$, we get a symplectic transformation (defined as a mapping those differential is a symplectic linear mapping everywhere)?
Thanks in advance!
 A: If a method is symplectic, then that means that the exact method conserves a symplectic scalar product. Where the step size might come in is in implicit methods. In the presence of a non-separable Hamiltonian, all (or almost all?) symplectic methods are implicit. If the number of solution steps (fixed-point iteration or Newton-like) for the implicit step equations is fixed, then indeed for large step sizes the actual method may cease to be (nearly) symplectic.
As a consequence of the symplectic condition, linear first integrals, if they exist, are conserved exactly. Other first integrals are conserved to a higher order with some low-order perturbation, that is, for example for the Hamiltonian you get $H_h(p,q)=H_0(p,q)+h^dH_1(p,q)+O(h^{d+1})$, where $H_0$ is the original Hamiltonian, $H_1$ some perturbation term formed from the partial derivatives of $H_0$, $d$ is the order of the symplectic method and $H_h$ the conserved energy for fixed step size $h$. In many cases the last term is actually $O(h^{d+2})$.
This formula is only valid far away from singularities of the Hamiltonian. It also only gives the practical advantage of staying on or close to the energy surface and other manifolds of conserved quantities for fixed-step integrations. If the step size is varied, then the conserved energy function changes according to the step size, for each different step size a different function is kept (almost) constant. In total this gives that no function is constant (that is, better than the order of the method predicts) over a variable-step integration.
