Codifferential relative to a continuous metric

Question

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.

Answer 1

I think I may have the answer to my question. The main issue is that the exterior differential $d$ is an unbounded linear operator relative to the $L^2$-norm so its adjoint $\delta_g$ with respect to $\langle \cdot, \cdot \rangle_g$ isn't necessarily defined everywhere. To be more precise, denote by $L^2(\Lambda^k,M)$ the space of all $k$-forms on $M$ with coefficients (relative to any smooth local coordinate system) in $L^2$; this space is Hilbert relative to $\langle \cdot, \cdot \rangle_g$ and $d$ is defined on the dense subspace of smooth $k$-forms. By standard theory of unbounded operators on Hilbert spaces, $\xi \in L^2(\Lambda^k,M)$ is in the domain of the adjoint $\delta_g$ of $d$ if the linear functional $f_\xi : \eta \mapsto \langle \xi,d\eta \rangle_g$ is continuous. Since

$$\eta \mapsto \langle \xi,d\eta \rangle_g = \int_M \ast_g \xi \wedge d\eta,$$

$f_\xi$ is continuous if and only if $\ast_g \xi$ has a weak exterior differential with coefficients in $L^2$, i.e., iff $\ast_g \xi$ is in the space $\Omega^{n-k}_{2,2}(M)$ (in the notation of Vladimir Gol’dshtein and Marc Troyanov, Sobolev inequalities for differential forms and $L_{q,p}$-cohomology, Journal of Geometric Analysis 16 (2006), no. 4, 597–631). If $g$ and $\xi$ are smooth, this is of course always the case. So it appears that for a continuous metric $g$, $\xi \in L^2(\Lambda^k,M)$ is in the domain of $\delta_g$ iff $\ast_g \xi$ has a weak exterior differential with coefficients in $L^2$.

Comments or corrections would be greatly appreciated.