Dual Riemannian metric and the Dual Metric Form Let $M$ be a Rieamnnian manifold with metric $g: X(M) \times X(M) \to C^{\infty}(X)$, where $X(M)$ are the vector fields of $X$.
As is well known, we can induce a bilinear pairing 
$$
\langle \cdot , \cdot \rangle_g: \Omega^1(M) \times \Omega^1(M) \to C^{\infty}(M)
$$
by setting 
$$
\langle \omega, \omega' \rangle_g = g(\omega^{\sharp}, (\omega')^{\sharp}),
$$
where, as usual, $\sharp$ is defined by $g(\omega^\sharp, X) = \omega(X)$, for $X \in X(M)$.
On the other hand, as is also well known, there exists a unique element $\omega_g \in \Omega(M) \otimes \Omega(M)$ such that, for $(X,Y) \in X(M) \times X(M)$, 
$$
\omega_g (X,Y) = g(X,Y),
$$ 
where $\omega_g$ is applied to $(X,Y)$ in the obvious way.
Thus, we have a pairing on $T^\ast(M)$ and an element of $\Omega(M) \times \Omega(M)$ both coming from $g$. I would like to know if there exists a simple relationship between these two objects. (By simple, I suppose I mean something global and algebraic, free from messy local expressions.)
Moreover, what does metric compatibility for a connection look like for either of these?
 A: It is worth noting that this is really a question in linear algebra. An inner product on a vector space $V$ defines an element in $ \omega \in V^*\otimes V^ * $. The same inner product also induces an inner product on $V^* $.  How is the inner product on $V^* $ related to $\omega$? It is, I suppose, a reasonable question, but you should be able to figure it all out using a basis of $V$, the dual basis of $V^*$, and the matrices representing the inner products as well as the tensor $\omega$.
[ADDED] As for the relationship between $\omega_g$ and the inner product on the cotangent bundle, isn't it the same as the relationship you give between $g^*$ and the inner product on the tangent bundle? In other words
$$
\omega_g(\theta^\sharp, \phi^\sharp) = g^*(\theta, \phi)
$$
A: What comes to my mind is a co-algebra structure, however, I'm not sure if it is of big importance. The coproduct $\Delta: T^\ast (M)\to T^\ast (M)\otimes T^\ast (M)$ is a homeomorhpism which preserves the inner product.
A: In some cases, the notion of a Riemann manifold is overly general.  If should happen that the problem(s) one has in mind admit a Kählerian complex structure, then the vanishing of the Nijenhuis tensor—which expresses an integrability condition—can provide the "non-messy" global structure that is requested.  Associated to the vanishing Nijenhuis tensor is a closed symplectic form that in dynamical models usefully specifies Hamiltonian vector fields.
