Is the tangent-cotangent isomorphism orientation preserving? Consider $(M,g)$ a Riemannian manifold.
Let's define $\varphi : TM\rightarrow T^{\ast}M$ by $\varphi(p,v):=(p,g(v,.))$, for $p\in M$ and $v\in T_{p}M$.
Here, $TM$ stands for the tangent bundle and $T^{\ast}M$ for the cotangent bundle. $\varphi$ is an isomorphism between them.
Now, consider $\eta\in\Omega^{1}(T^{\ast}M)$ given by $\eta_{p,\zeta}(X)=\zeta(\pi_{\ast}(X))$, where $\pi:T^{\ast}M\rightarrow M$ is the canonical projection map such that $\pi(p,\zeta)=p$ and $X\in T_{p,\zeta}(T^{\ast}M)$. One can show that $\omega :=d\eta$ is non-degenerate and thus, $\omega^{n}$ is a volume form in $T^{\ast}M$. 
Considering this orientation for $T^{\ast}M$ and the natural orientation for $TM$, my question is : Is $\varphi$ orientation preserving or reversing ?
Thanks
 A: We can use $\phi$ to pull $\eta$ back to a one-form $\alpha = \phi^* \eta$ on $TM$, which (since $\phi$ is an isomorphism) still has the property that $(d\alpha)^n$ is a volume form on $TM$. It's also not hard to write down a formula for $\alpha$; as a remark before doing so, note that if $X$ is a vector tangent to the manifold $TM$ at $(p, v)$ then 
$$
   \pi_* d\phi(X) = \pi_* X
$$ 
since obviously $d\phi$ doesn't affect the horizontal part of vectors in $TTM$. Then one computes:
$$ 
\alpha_{p,v}(X) = \eta_{\phi(p, v)}(d\phi(X)) = \phi(v) \Big( \pi_* d\phi(X)\Big) = \langle v, \pi_* X \rangle,
$$
which is a useful formula. 
It's also not hard to write down an explicit formula for $d\alpha$. We let $K$ be the natural map $T_{p,v} TM \to T_pM$ taking a vector to its vertical part; then one can show
$$
d\alpha(Y,Z) = \langle KY, \pi_* Z \rangle - \langle \pi_* Y, KZ \rangle.
$$
(Two easy ways to check this: one, check it for coordinate fields using an explicit formula for $d\alpha$ in terms of $\alpha$ and our formula above; two, if you already believe that $d\eta$ has the coordinate representation $\sum d p_i \wedge d x_j$, then it should be clear that $d\alpha$ must have the coordinate representation $\sum d q_i \wedge d x_j$, where $q_i$ is the coordinate associated with $\partial / \partial x_i$; it's easy to see that the above formula follows.)
The point is that this gives a simple way to compute $(d\alpha)^n$ applied to a nice set of coordinate fields for $TM$. In particular, if $x_1, \dots, x_n$ are coordinates on some chart of $M$, and as above $q_i$ are the natural extensions to $TM$, then our formula gives
$$
  (d\alpha)^n\Big(\frac{\partial}{\partial q_1}, \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial q_n}, \frac{\partial}{\partial x_n} \Big) = n!.
$$ 
(This is assuming I didn't screw up a sign somewhere which is unfortunately quite possible; I don't have good references in front of me.)
As Dan Fox points out in his comment, whether you consider this to be the "natural" orientation on $TM$ is entirely up to you.
