Coordinate-Free Computation of Covariant Derivative Suppose $S$ is an oriented 2-dimensional manifold equipped with a Riemannian metric $\langle \cdot,\cdot \rangle$. Let $\theta$ be a 1-form on $S$. By non-degeneracy, there is an identification of the tangent and cotangent bundle via the metric. In particular, there is a bundle map $Y \colon TS \rightarrow TS$ such that $$\langle Y_x(v),w \rangle = d\theta_x(v,w). $$ Likewise, there exists a vector field $Z$ on $S$ with $$\langle Z(x),v \rangle = \theta_x(v).$$
I am interested in computing the covariant derivative of $Z$.
Of course, this is possible in local coordinates, but the computation becomes super messy since the Riemannian metric comes into play several times. Hence, I would like to know if anybody can come up with a coordinate-free computation.
(edited:) Even if no explicit solution can be written down, I hope to verify the following formula: $$\langle v+Z,\nabla_wZ \rangle = \langle \nabla_{(v+Z)}Z,w \rangle - \langle Y(v+Z),w \rangle.$$ (For the interested: this is an intermediate step to verify that the symplectic gradient flow of the energy Hamiltonian for a magnetic form is conjuagted to the symplectic gradient flow of the Hamiltonian $E \circ \mathcal{L}^{-1}$ ($E$ energy, $\mathcal{L}$ Legendre transform) for the standard symplectic form via the Legendre transform.)
In my context, $Y$ maps a vector $v \in T_xS$ to $s(x)Jv$, where $s$ is a smooth function on $S$ and $J$ is the canonical almost complex structure induced by the metric. However, I feel like the computation can be done without explicitly knowing $Y$.
 A: I sat down once more and was able to verify the identity in question, which is what I needed. Nonetheless, I still hope someone come up with a formula for $\nabla Z$ itself, because it still poses an interesting question, I believe.
The identity should hold for all $v$, so we may simplify the equation by setting $u = v + Z$ and show it for all $u$. Hence, we need to show
$$\langle u, \nabla_wZ \rangle = \langle \nabla_uZ, w \rangle - \langle Y(u),w \rangle.$$
Let $\gamma$ be a path with $\gamma(0) = x$ and $\dot{\gamma}(0) = u$.
Let $W$ be a vector field with $W(x) = w$.
Then
$$(d\iota_W\theta)_x(u) = \frac{d}{ds}\Big|_{s=0} \langle Z \circ \gamma, W \circ \gamma \rangle = \langle \nabla_uZ,w \rangle + \langle Z,\nabla_uW \rangle.$$
Further,
$$(\iota_Wd\theta)_x(u) = \langle Y(w),u \rangle.$$
Let $\phi_t$ denote the flow of $W$ and observe that
$$\nabla_t((d\phi_t)_x(u))(0) = \nabla_t \partial_s (\phi_t \circ \gamma)(0)(0) = \nabla_s \partial_t (\phi_t \circ \gamma)(0)(0) = \nabla_uW(x).$$
Combining all these with Cartan's formula for the Lie derivative we obtain
$$
\begin{align*}
\langle \nabla_uZ,w \rangle + \langle Z,\nabla_uW \rangle + \langle Y(w),u \rangle &= (d\iota_W\theta)_x(u) + (\iota_Wd\theta)_x(u) \\
&= (\mathcal{L}_W\theta)_x(u) \\
&= \frac{d}{dt}\Big|_{t=0} \theta_{\phi_t(x)}((d\phi_t)_x(u)) \\
&= \frac{d}{dt}\Big|_{t=0} \langle Z \circ \phi_t, d\phi_t(u) \rangle \\
&= \langle \nabla_wZ,u \rangle + \langle Z,\nabla_uW \rangle.
\end{align*}
$$
We are finished since
$$-\langle Y(u),w \rangle = -d\theta(u,w) = d\theta(w,u) = \langle Y(w),u \rangle.$$
Remark: This does not need $S$ to be an oriented 2-manifold. Any Riemannian manifold works.
