# Understanding the derivation of this form of the coderivative

Let $M$ be a smooth Riemannian manifold;

Let $E$ be a vector bundle over $M$, equipped with a metric and a compatible connection $\nabla$. Denote by $d:\Omega^k(M,E) \to \Omega^{k+1}(M,E)$ the induced covariant exterior derivative on $E$-valued forms, and let $\delta:\Omega^k(M,E) \to \Omega^{k-1}(M,E)$ be its adjoint.

I am trying to follow the proof of

Claim: Let $E_i$ be a basis of $T_xM$, $X_j \in T_xM$, $g^{st}$ is the inverse matrix of $g(E_s,E_t)$. Then for $\rho \in \Omega^p(M,E)$, $$\delta \rho (X_1,\dots,X_{p-1})=-\sum_{s,t}g^{st} (\nabla_{E_t}\rho)(E_s,X_1,\dots,X_{p-1}).$$

($\, \,$Here $\nabla_{E_t}\rho$ is the induced connection on $\Lambda_p(T^*M) \otimes E \, \,)$

This claim shows as Lemma 1.20 (page 8) in the book "Selected topics in harmonic maps" by Eells and Lemaire.

The proof begins by saying that it suffices to check the formula at the center of a system of normal coordinates, when $X_i$ and $E_s$ are elements of the basis $\frac{\partial }{\partial x_i}$.

The authors then show that $\langle d\sigma , \rho \rangle-\langle \sigma , -\sum_{s,t}g^{st} (\nabla_{\frac{\partial }{\partial s}}\rho)(\frac{\partial }{\partial t},\dots) \rangle$ equals to the divergence of some (compactly supported) vector field.

Then they conclude that we are done, since the integral of such a divergence is zero.

I have two issues with this proof:

1. I don't see why it "suffices to check the formula at the center of a system of normal coordinates". This looks like some locality argument I am missing... (I know $\delta$ turns out to be a local operator-e.g $*d*$- but this is a-posteriori, the definition is a global one).

2. I am not convinced by the divergence argument. It seems to me, that all the authors have shown is that locally, the candidate for $\delta \rho$ makes the difference $\langle d\sigma , \rho \rangle-\langle \sigma , -\sum_{s,t}g^{st} (\nabla_{\frac{\partial }{\partial s}}\rho)(\frac{\partial }{\partial t},\dots) \rangle$ a divergence of a vector field.

(In fact they only showed this equality in one point, the "center" of their coordinate system).

However, in order to use the divergence theorem (Stokes), we need the difference to be a global divergence of a vector field. How do we know we can "glue" all the local vector fields together to a global creature? Or is there another justification for this step?

Any help or clarifications would be appreciated.

References for other proofs are also welcome.

• The (formal $L^2$) adjoint of a differential operator is always a differential operator, no? You should be able to write an arbitrary first-order differential operator $d:\Gamma(E) \to \Gamma(F)$ as something like $(d s)^a = A^{a \mu}_i \nabla_\mu \xi^i$ with $A \in \Gamma(TM \otimes E^* \otimes F)$, then perform integration by parts on $(ds, w)$ to get something like $(\delta w)^i = -\nabla_\mu(A^{a \mu}_i w^a).$ – Anthony Carapetis Oct 30 '17 at 6:29