Existence of adjoint operators on manifolds

Question

Let $(M,g)$ be an oriented Riemannian manifold and $V$ a finite-rank vector bundle equipped with a non-degenerate bundle metric $\langle\cdot,\cdot\rangle_{V}$. This bundle metric, in turn, gives rise to a non-degenerate bilinear pairing on $\Gamma(V)$ defined by $$\langle\cdot,\cdot\rangle_{\Gamma(V)}=\int\langle\cdot,\cdot\rangle_{V}\,\mathrm{vol}_{g}$$ for every two elements in $\Gamma(V)$ with compactly overlapping support. Now, by non-degeneracy, there is a well-defined injection

$$\Gamma(V)\to\Gamma_{c}^{\prime}(V)$$ $$s\mapsto \mathcal{D}_{s}(\cdot):=\langle s,\cdot\rangle_{\Gamma(V)}$$

I have seen in several occasions that this should imply that for every linear continuous operator $A:\Gamma_{c}(V)\to\Gamma(V)$ there exists a formal adjoint $A^{\dagger}:\Gamma_{c}(V)\to\Gamma(V)$, i.e. for all $s,t\in\Gamma_{c}(V)$:

$$\langle As,t\rangle_{\Gamma(V)}=\langle s,A^{\dagger}t\rangle_{\Gamma(V)}$$

However, I don't see how this is the case. The proof of existence of proper adjoints in the Hilbert space setting is done using the Riesz representation theorem, which requires completeness. This is not the case here, since we are not talking about Hilbert spaces and only consider formal adjoints. Of course, if $A$ is a differential operator, the claim is clear, since we can locally write $A=\sum a_{\alpha}\partial^{\alpha}$ and define the adjoint by Stoke's theorem, but in the general case, I don't see how to show existence.

Answer 1

You prove this as follows: As a first step you show the existence of the formally adjoint differential operator on a small open set $U$. This set $U$ is chosen, such that it carries a chart and such that you can trivialize the bundle $V$ on $U$. Make such a trivialization orthonormal and then write down the equation defining the formally adjoint operator in a chart. The volume form then gives a contribution $\sqrt{\det(g_{ij})}$ as usual. Then standard partial integration in $\mathbb{R}^n$ of the operator $A$ (in coordinates) yields the formally adjoint operator $A^*$ (in coordinates). You have to take into account that the partial integration will also give you contributions from the derivatives of $\sqrt{\det(g_{ij})}$. The formula obtained this way defines a differential operator only on $U$, and a priori the formula might also depend on the coordinates and the trivialization. We now show that the defining equation, i.e. your last displayed equation holds for all $t$ and $s$. This shows the existence of $A^*$ on $U$, denoted in the following as $A_U^*$.

Now as second step take two such open sets $U_1$ and $U_2$ with choices of charts and trivializations which we suppress in notation. Then it is easy to see from the defining equation that the formal adjoints $A_{U_i}^*$ defined via $U_i$, $i=1,2$ agree on $U_1\cap U_2$, more precisely: On $U_1\cap U_2$ we have $A_{U_1}^*=A_{U_2}^*$ as differential operators.

The property in the second step does not only provide uniqueness on small open sets, but it also allows to glue the locally defined operators $A_U^*$ to a globally defined differential operator, which is the requested formally adjoint operator.

Note that this operator is only formally adjoint, it is a different question, whether and how it extends to an adjoint operator, and this extension usually also depends on the type of Sobolev space you use and on possible boundary conditions.