Euler field for magnectic symplectic structure in $T^*Q$

Question

Let $Q$ be a differentiable manifold, $\pi\colon T^*Q\to Q$ denote its cotangent bundle, and $B \in \Omega^2(Q)$ be a closed form. I'm playing around with some things, and I'm not sure whether it would be more appropriate to post this in MO to ask for references, but anyway, here it goes: consider $\omega$ the standard symplectic structure on $T^*Q$, and let $\omega_B = \omega + \pi^*B$ be the magnectic symplectic form associated to $B$.

We have that if $\alpha \in \Omega^1(T^*Q)$ is the tautological $1$-form, the Euler field associated to $\omega$ is the unique vector field $E \in \mathfrak{X}(T^*Q)$ such that $\iota_E\omega = -\alpha$. Now, if $B = {\rm d}\beta$ is also exact, we also have the Euler field $E_B\in \mathfrak{X}(T^*Q)$ associated to $\omega_B$. I want to write $E_B$ in terms of $\omega$, $E$ and $\beta$ in a coordinate free way, if possible.

The condition $\iota_{E_B}\omega_B = -\alpha+\pi^*\beta$ reads $$\omega(E_B,X) + {\rm d}\beta(\pi_\ast E_B, \pi_*X) = \omega(E,X) + \beta(\pi_*X) \tag{$\ast$}$$for every test field $X \in \mathfrak{X}(T^*Q)$. Using that $\pi_*E=0$, making $X = E$ in $(\ast)$ gives $\omega(E_B,E)=0$. With this, if you now let $X = E_B$ in $(\ast)$, it follows that $\beta(\pi_*E_B)=0$. Another immediate consequence is that $\omega(E_B,X) = \omega(E,X)$ for every vertical field $X \in \ker {\rm d}{\pi}$.

I am unsure how to proceed to get rid of ${\rm d}\beta$. Unless we can actually use the non-degenerability of $\omega$ somehow with only this information. For what's it worth, making $X = (\pi^*\beta)^\sharp$, where $\sharp\colon \Omega^1(T^*Q) \to \mathfrak{X}(T^*Q)$ is induced by $\omega$ actually gives ${\rm d}{\beta}(\pi_*E_B, \pi_*((\pi^*\beta)^\sharp))=0$, but this is just... meh.

Help?

Answer 1

Using your notations, it is explained here that the map $$ \psi : T^*Q \to T^*Q : (q,p) \mapsto (q, p - \beta(q))$$ is a diffeomorphism (clearly) which satisfies $\psi^* \alpha = \alpha - \pi^* \beta =: \alpha_{\beta}$, hence $$\psi^* \omega = \psi^*(-d\alpha) = - d \psi^* \alpha = -d(\alpha - \pi^*\beta) = \omega + \pi^*d\beta = \omega + \pi^*B = \omega_B \, .$$ In other words, $\psi : (T^*Q, \alpha_{\beta}) \to (T^*Q, \alpha)$ is an exact symplectic diffeomorphism. We then compute $$ \iota_{E_B}\omega_B = -\alpha_{\beta} = \psi^*(-\alpha) = \psi^*(\iota_E \omega) = \iota_{\psi^{-1}_* E}\psi^*\omega = \iota_{\psi^{-1}_* E} \omega_B \, . $$ Since $\omega_B$ is nondegenerate, it follows that $E_B = (\psi^{-1})_*E$. This is geometrically 'appealing': $E_B$ is just a fiberwise translation of $E$. In particular $E_B$ is vertical since $E$ is, something you could have deduced from the equality $\omega(E_B, X) = \omega(E,X) = 0$ for $X$ vertical (recalling that the fibers are Lagrangian for $\omega$).