Why is the determinant of the Jacobian of symplectic integrators always 1?

Question

My numerics books says that a symplectic integrator has the property that the determinant of $det \frac{\partial F}{\partial \xi}=1$ for the state vector $\xi = (X,V)$ for $F_\epsilon: \xi _t \xrightarrow{} \xi _{t+\epsilon}$.

Definition 4.8 - Symplectic integrator. A time-integrator is a map advancing the state vector $( \xi:=(X, V) )$ of any pair of a coordinate ( X ) and its canonically conjugate momentum ( V ) from time ( t ) to time $( t+\epsilon )$, i.e. $\boldsymbol{F}_{\epsilon}: \boldsymbol{\xi}_{t} \mapsto \boldsymbol{\xi}_{t+\epsilon} . $

A symplectic time-integrator is the sub-class of integrators for which $ \operatorname{det} \frac{\partial F}{\partial \xi}=1 $ which guarantees conservation of $ d X \wedge d V $.

I wonder how one comes up with this property and if this is the Jacobian, e.g how to differentiate a state vector?

Answer 1

Because by definition symplectic methods have steps that are symplectomorphisms, i.e., preserve the symplectic form and with that also the volume form of the state space (position+impulse). The latter is equivalent to the determinant of the Jacobian being one, as that determinant is the rate of (infinitesimal) volume change.