Let $(Q,g)$ be a pseudo-Riemannian manifold, $(q^1,\ldots, q^n)$ be local coordinates in $Q$ and $(q^1,\ldots, q^n, v^1,\ldots, v^n)$ the induced tangent coordinates in $TQ$. I wanted to check that $$\sum_{i,j,k=1}^n v^j \frac{\partial g_{ij}}{\partial q^k}\,{\rm d}q^k \wedge {\rm d}q^i = 0,$$and I'm sure it follows from the standard argument by renaming indexes and exploiting symmetries. I guess I'm just having a bad day and my brain isn't helping. Can someone help me carry this out? Thanks.
(If you want some context, this is what is left to show that if one pulls back the canonical symplectic structure from $T^*Q$ to $TQ$ via the metric, one obtains $\sum_{i,j=1}^n g_{ij}\,{\rm d}q^i \wedge {\rm d}v^j$).
I'll add some work since I'm starting to doubt the result (which makes me very irritated, to be honest). Define ${\rm A} \in \Omega^1(TQ)$ by $${\rm A}_{X_x}(Y_{X_x}) = \langle X_x, {\rm d}\pi_{X_x}(Y_{X_x})\rangle_x$$for all $X_x \in TQ$ and $Y_{X_x}\in T_{X_x}(TQ)$. One can write ${\rm A} = \sum_{i=1}^n a_i\,{\rm d}q^i+b_i\,{\rm d}v^i$. Since the coordinate fields $\partial/\partial v^i$ are vertical, we get $b_i=0$. And $$a_i(X_x) = \langle X_x, \partial/\partial q^i\big|_x\rangle_x = \sum_{j=1}^n v^i(X_x) g_{ij}(x),$$whence ${\rm A} = \sum_{i,j=1}^n (g_{ij}\circ \pi)v^j\,{\rm d}q^i$. Then $$-{\rm dA} = \sum_{i,j=1}^n(g_{ij}\circ \pi){\rm d}q^i\wedge {\rm d}v^j - {\color{blue}{\sum_{i,j,k=1}^nv^j \frac{\partial g_{ij}}{\partial q^k}{\rm d}q^k\wedge {\rm d}q^i}}.$$What do I do with the crap in blue? I have actually done the calculation for the Poincaré half-plane to see what was happening and I got $$-{\rm dA}= \underbrace{\frac{{\rm d}x \wedge {\rm d}v^x + {\rm d}y\wedge {\rm d}v^y}{y^2}}_{\rm expected} - \underbrace{\color{blue}{\frac{2v^x}{y^3}{\rm d}x\wedge {\rm d}y}}_{???} .$$
The term I expected to vanish (five!) years ago does in fact survive. The key to explaining this lies with $$\text{the Levi-Civita connector $K\colon TTQ \to TQ$, defined by $K(\xi) = Z'(0)$,}$$ where $\xi \in T_{(x,v)}(TQ)$ and $Z\colon (-\varepsilon,\varepsilon) \to TQ$ is a curve with $Z(0) = (x,v)$ and $\dot{Z}(0)=\xi$. Here, $\pi\colon TQ\to Q$ is the bundle projection and $Z'(0)$ denotes the covariant derivative at $t=0$ of $Z$ along the projected curve $\pi\circ Z\colon (-\varepsilon,\varepsilon) \to Q$.
Let's show that the bundle isomorphism $TQ\to T^*Q$ provided by $g$ pulls back the canonical symplectic form on $T^*Q$ to $\omega \in \Omega^2(TQ)$ given by $$\omega_{(x,v)}(\xi,\eta) = g_x({\rm d}\pi_{(x,v)}(\xi), K(\eta)) - g_x(K(\xi),{\rm d}\pi_{(x,v)}(\eta)),$$for all $\xi,\eta\in T_{(x,v)}(TQ)$.
Step 1: Understanding $K$.
To compute $K(\xi)$ relative to tangent coordinates $(q^1,\ldots, q^n, v^1,\ldots, v^n)$ for $TQ$, write $$\xi = \xi^{q^i}\frac{\partial}{\partial q^i}\bigg|_{(x,v)} + \xi^{v^i}\frac{\partial}{\partial v^i}\bigg|_{(x,v)},\qquad \xi^{q^i},\xi^{v^i}\in\Bbb R,$$and $Z(t) = (q^i(t),v^i(t))$, with $\dot{q}^i(0) = \xi^{q^i}$ and $\dot{v}^i(0) = \xi^{v^i}$ for all $i$. Hence $$K(\xi) = Z'(0) = \left(\dot{v}^k(0) + \varGamma_{ij}^k(q(0))\dot{q}^i(0)v^j(0)\right)\frac{\partial}{\partial q^k}\bigg|_{Z(0)} = \left(\xi^{v^k} + \varGamma_{ij}^k(x)\xi^{q^i}v^j\right)\frac{\partial}{\partial q^k}\bigg|_{(x,v)},$$where we also write $v = v^j \partial/\partial q^j |_x$ in $T_xQ$. That is, we have that $$K = ({\rm d}v^k + \varGamma_{ij}^k\, v^j\,{\rm d}q^i)\otimes \frac{\partial}{\partial q^k}$$
Step 2: pulling back the canonical symplectic form on $T^*Q$.
We just need to use the relation $p_k = g_{k\ell}v^\ell$. Namely:$${\rm d}p_k = v^\ell\frac{\partial g_{k\ell}}{\partial q^r}{\rm d}q^r + g_{k\ell}{\rm d}v^\ell \implies {\rm d}q^k\wedge {\rm d}p_k = v^\ell\frac{\partial g_{k\ell}}{\partial q^r}\,{\rm d}q^k\wedge {\rm d}q^r + g_{k\ell}{\rm d}q^k\wedge{\rm d}v^\ell.$$
Step 3: checking that this equals the proposed $\omega\in\Omega^2(TQ)$ defined in terms of $K$.
Clearly, ${\rm d}\pi = {\rm d}q^k\otimes \partial/\partial q^k$, so distributing the terms in $$\omega = g_{k\ell} {\rm d}q^k \otimes ({\rm d}v^\ell +\varGamma_{ij}^\ell\,v^j\,{\rm d}q^i) - g_{k\ell} ({\rm d}v^k +\varGamma_{ij}^k v^j\,{\rm d}q^i)\otimes {\rm d}q^\ell$$ leads to $$\omega = g_{k\ell}{\rm d}q^k\wedge {\rm d}v^\ell + \underbrace{\left(g_{k\ell}v^j\varGamma_{ij}^\ell{\rm d}q^k\otimes {\rm d}q^i - g_{k\ell}v^j\varGamma_{ij}^k{\rm d}q^i \otimes {\rm d}q^\ell\right)}_{(\ast)},$$meaning that the only thing left to do is to simplify $(\ast)$. By symmetry of $g$, we may switch the roles of $k$ and $\ell$ in the second term in $(\ast)$ and rename $i\mapsto r$ to obtain $$(\ast) = g_{k\ell}v^j\varGamma_{rj}^\ell{\rm d}q^k\wedge {\rm d}q^r = \frac{v^j}{2}\underbrace{\left(\frac{\partial g_{kj}}{\partial q^r} + \frac{\partial g_{rk}}{\partial q^j} - \frac{\partial g_{rj}}{\partial q^k}\right)}_{(\dagger)}{\rm d}q^k\wedge {\rm d}q^r = v^j\frac{\partial g_{kj}}{\partial q^r}{\rm d}q^k\wedge{\rm d}q^r,$$as required. We use that the second term in $(\dagger)$ is symmetric in $r$ and $k$ (so it vanishes when contracted against ${\rm d}q^k\wedge {\rm d}q^r$), while the first and third are opposites (eliminating the factor of $1/2$).