In my symplectic geometry homework I have this exercise: Let $Q$ be a manifold, $\pi:T^*Q\to Q$ be the projection map, and $\kappa:T(T^*Q)\to T^*Q$ be the connection map. Prove that the standard symplectic form $\omega_0$ on $T^*Q$ for $\xi,\eta\in T(T^*Q)$ has the form $\omega_0(\xi,\omega)=\kappa(\xi)[d\pi(\eta)]-\kappa(\eta)[d\pi(\xi)]$.
Due to my lacklustre differential geometric background I do not know what the connection map is, despite having tried searching online. Is someone able to explain or define it in a self-contained manner, or provide a link where this is done? My background is a course on smooth manifolds, and riemannian geometry 'as taught in physics'.
Here is an attempt to reverse engineer what $\kappa$ should be, by using Darboux coordinates $(x^i, p_i)$ for $T^*Q$. Write $$\xi = \xi^i \partial_{x^i}+\xi_i\partial_{p_i} \quad\mbox{and}\quad \eta = \eta^i\partial_{x^i} + \eta_i \partial_{p_i}, $$and here I'm using Einstein's convention. We know that $$ \omega_0 = {\rm d}x^i\wedge {\rm d}p_i $$implies that $$\omega_0(\xi,\eta) = \xi^i\eta_i -\eta^i\xi_i. $$Noting that $${\rm d}\pi(\xi) = \xi^i\partial_{x^i} \quad \mbox{and}\quad {\rm d}\pi(\eta) = \eta^i\partial_{x^i},$$one guess would be that $\kappa(\xi)({\rm d}\pi(\eta))$ is $\xi^i\eta_i$, but the index balance required by Einstein's convention hints that this is not the case, as we'll need $\kappa(\xi)\colon TQ \to \Bbb R$ and here we have $\eta_i$ instead of $\eta^i$. So the only other choice is to say that $\kappa(\eta)({\rm d}\pi(\xi)) = -\eta_i\xi^i$. This would lead us to $\kappa(\eta)(v) = -\eta_iv^i = -\eta_i\,{\rm d}x^i(v)$.
That said, the challenge is writing this in a coordinate-free way. But $TT^*Q$ has natural horizontal and vertical distributions (namely, spanned by the $\partial_{x^i}$ and the $\partial_{p_i}$ -- and this does not depend on the choice of coordinates), and $\eta_i\,{\rm d}x^i$ is the covector corresponding to the vertical projection of $\eta$ under the isomorphism $\eta_i\,{\rm d}x^i \cong \eta_i\partial_{p_i}$: any vertical tangent vector in $T_{(x,p)}(T^*Q)$ is identified with an element of $T^*_xQ$, since the tangent space to a vector space at any point is itself (note that the index balance required by Einstein's convention is still respected, as the $i$ in $\partial_{p_i}$ counts as an upper index).
This means we get our conclusion: $\kappa(\xi)$ should be minus the covector corresponding to the vertical projection of $\xi$.