If $M$ is a smooth compact orientable manifold without boundary and $g$ a smooth Riemanniann metric on $M$, then one defines the usual $L^2$-inner product on differential forms by

$$\langle \alpha,\beta \rangle_g = \int_M \ast_g \alpha \wedge \beta,$$

where $\ast_g$ is the Hodge-star operator relative to $g$ and $\alpha$ and $\beta$ are smooth forms of the same degree. The codifferential $\delta_g$ is then defined as the formal adjoint of the exterior differential $d$ (and equals $\pm \ast_g d \ast_g$).

My question is: **Is $\delta_g$ well-defined when $g$ is only continuous?**

Unless I am missing some subtle point, $g$ still defines an $L^2$-inner product on smooth forms, so the formal adjoint of $d$ seems to be well-defined, although it may not be given by the formula $\pm \ast_g d \ast_g$.

The reason this bothers me is the following **example**. Suppose $g$ is only continuous but its Riemannian volume form $\Omega$ is smooth (this can happen). Consider the closed $(n-1)$-form $i_X \Omega$, where $X$ is a smooth divergence-free vector field. Let $\alpha = \ast_g (i_X \Omega)$. Then $\alpha$ is the 1-form dual to $X$, i.e., $\alpha(v) = g(X,v)$, for all $v$, hence only continuous. However, the existence of $\delta_g(i_X \Omega)$ seems to imply that $\alpha$ is weakly differentiable. This can be seen as follows: for any smooth $(n-2)$-form $\omega$, we have:

$$\int_M \alpha \wedge d\omega = \langle i_X \Omega,d\omega \rangle_g = \langle \delta_g(i_X \Omega),\omega \rangle_g = \int_M \beta \wedge \omega,$$

where $\beta = \ast_g \delta_g (i_X \Omega)$. So $-\beta$ is the weak exterior differential of $\alpha$. To me it is far from obvious that differential forms $g(X,\cdot)$ for continuous $g$ should admit a weak differential in general.

Sorry for a long post. Any thoughts would be highly appreciated.