For a Riemannian manifold $(M,g)$ with exterior derivative d, the *codifferential* d$^\ast$ is defined to be the unique map for which
$$
g(\omega,d\omega') = g(d^\* \omega,\omega'), ~~~ \omega,\omega' \in \Omega^{\bullet}.
$$
Now if $\ast$ is the Hodge map for $g$, then it is not too difficult to show that d$= (-1)^k\ast$ d $\ast$, when acting on $\Omega^k(M)$. 


When $M$ is a complex manifold with holomorphic and anti-holomorphic partial derivatives $\partial$, and $\overline{\partial}$, we have a similarly defined $\partial^\ast$, and $\overline{\partial}^\ast$, and a similar relation between these objects and the original derivatives involving the Hodge map (well actually there's a reversal but no matter). For the Lefschetz map something similar also happens.


What I would like to know is whether the adjoint $\nabla^\*$ of the Levi--Civita connection $\nabla$ has some similar re-expression? Or is this too naive?



Adjoint of a Connection Using the Hodge Map?
Yes, even in a more general situation: Let $E\to M$ be a vector bundle with metric and metric connection $\nabla.$ Then there exists $d^\nabla\colon\Omega^k(M,E)\to\Omega^{k+1}(M,E)$ satisfying $d^\nabla(s\otimes\omega)=\nabla s\wedge\omega+s\otimes d\omega,$ and also the Hodge star extends to $\Lambda T^\*M\otimes E.$ With this it is a nice exercise to compute that
the adjoint of $d^\nabla$ is $\delta^\nabla=(-1)^? \* d^\nabla \*,$ where the sign is the same as for the usual codifferential.


