Derivative of a symplectic form

Question

Let $(M,\omega)$ be a symplectic manifold and $\gamma \in C^\infty(I,M)$ a path in $M$. For $X,Y \in \Gamma_\gamma(TM)$ I would like to compute the derivative $$\frac{d}{dt}\omega(X(t),Y(t)).$$ My intuition somehow tells me that this is equal to $$\omega(\dot{X}(t),Y(t)) + \omega(X(t),\dot{Y}(t)),$$ but working in local coordinates this does not look right. I think that one has to interpret the dot as a covariant derivative. Thus I tried to use an $\omega$-compatible almost complex structure $J$ to derive $$\frac{d}{dt}\omega(X(t),Y(t)) = -\frac{d}{dt}\omega(JJX(t),Y(t)) = -\omega(J\nabla_t JX,Y(t)) + \omega(X(t),\nabla_t Y).$$ But then I am stucked.

Answer 1

The expression you write down implicitly requires a connection (since it involves derivatives of vector fields), so doesn't mean anything on an arbitrary symplectic manifold until you choose a connection. Unlike Riemannian manifolds, there is no canonical choice of connection on a symplectic manifold, unless it is endowed with some extra structure.

Having chosen a connection $\nabla$, the statement you give, $$ \nabla_V(\omega(X,Y))=\omega(\nabla_VX,Y)+\omega(X,\nabla_VY)\iff \nabla\omega=0 $$ may or may not hold. When it does, we say $\nabla$ is compatible with $\omega$. Such connections always exist, but they are not unique.

A symplectic connection on $(M,\omega)$ is a torsion free connection compatible with $\omega$. These always exist as well, but are again not unique.

Your choice of connection seems to assume that $(M,\omega)$ has additional structure. If we instead have a Kähler manifold $(M,\omega,g,J)$, there is a canonical connection in the form of the Levy-Civita connection $\nabla_g$ associated with $g$. The compatibility conditions on the Kähler structure imply that $\nabla_g$ is compatible with $\omega$. Not every symplectc manifold admits a Kähler structure.