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For a $n$ dimensional smooth manifold $M$, I consider the cotangent bundle $T^*M$ with the canonical symplectic form $\omega$. A symplectic map $\phi : T^*M \to T^* M$ is a map which leaves the symplectic form invariant, i.e. $\phi^* \omega = \omega$.

Question: Is there a notion of symplectic maps between the spaces of volume forms over the phase space $T^*M$?

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In more detail: Let me denote the space of all volume forms over $T^*M$ as $\Omega^{2n}(T^*M) := \Gamma( \Lambda^{2n} (T^*M) )$ (the notation is from John M. Lee's book 'Intro to smooth manifolds'.) Then for a map $\psi$ which transforms volume forms, i.e. $$\psi : \mathrm{\Omega}^{2n}(T^*M) \to \mathrm{\Omega}^{2n}(T^*M),$$ I am looking for a condition which ensures that $\psi$ is compatible with the symplectic structure of $T^*M$?

Example: A symplectic map $\phi : T^* M \to T^*M$ implies a map $$\psi : \mathrm{\Omega}^{2n}(T^*M) \to \mathrm{\Omega}^{2n}(T^*M) : \eta \mapsto \phi^* \eta.$$ Such a map should be compatible for example.

But I would expect that there are more maps which are compatible and that not all of them are derived from symplectic maps like in the example.

Background: For numerical simulation of Hamiltonian equations, it is good to use symplectic integrators, such as symplectic Euler. However, I am interested in solving Liouville's equations and this raised the question what a corresponding symplectic integrator would be in that case?

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    $\begingroup$ I suppose you mean by $\Omega^n(T^*M)$ the $n$-th antisymmetric tensor power of $T^*M$. This is a vector bundle of fiber dimension one over $M$. So the total space has dimension $n+1$. This makes it very unlikely to be symplectic (in ca half of all cases) $\endgroup$
    – Stefan Waldmann
    Commented Jul 10, 2020 at 10:57
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    $\begingroup$ But $\Omega^{2n}(T^*M)$ is the space of volume forms on $T^*M$. $\endgroup$
    – Ben McKay
    Commented Jul 10, 2020 at 11:31
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    $\begingroup$ Is $\psi$ a vector bundle automorphism of the volume form bundle on $T^*M$? Or is it a map on sections of that bundle? $\endgroup$
    – Ben McKay
    Commented Jul 10, 2020 at 11:34
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    $\begingroup$ A couple of questions for clarification: First, is it important that the underlying symplectic manifold is the total space of a cotangent bundle? I can't see why you question requires this. Second, since a symplectomorphism preserves the volume form, its induced action on $\Omega^{2n}$ is the identity. So what does it mean for a self-map on $\Omega^{2n}$ to be compatible with the symplectic structure? Perhaps I've missed something. $\endgroup$
    – Paul Reynolds
    Commented Jul 10, 2020 at 12:32
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    $\begingroup$ You're correct though, my assertion that a symplectomorphism preserves all volume forms is not right. Perhaps you are interested in the diffeomorphisms that preserve $\omega^n$, with no further conditions, of which there are many. $\endgroup$
    – Paul Reynolds
    Commented Jul 10, 2020 at 14:24

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