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
$$frac{d}{dt}omega(X(t),Y(t)) = -frac{d}{dt}omega(JJX(t),Y(t)) = -omega(Jnabla_t JX,Y(t)) + omega(X(t),nabla_t Y).$$
But then I am stucked.

Kajelad · Answer

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,
$$
nabla_V(omega(X,Y))=omega(nabla_VX,Y)+omega(X,nabla_VY)iff nablaomega=0
$$
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.

Derivative of a symplectic form

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