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?