Picture bottom is from Topping's Lectures on the Ricci flow (author website, zbMath link).
I wanto to show the formal adjoint of $\delta$ is covariant derivative. The definition of formal adjoint is in picture below (in fact, I am not sure which definition is suitable).
Definition 2.2 The (formal) adjoint of a linear operator $D: \Gamma(E) \rightarrow \Gamma(F)$ is the operator $D: \Gamma(F) \rightarrow \Gamma(E)$ defined by $$ \int_{M}\langle D \xi, \eta\rangle d v_{g}=\int_{M}\left\langle\xi, D^{*} \eta\right\rangle d v_{g} \tag{2.14} $$ for all $\xi \in \Gamma(E), \eta \in \Gamma(F)$ with compact support. (screenshot)
I want to prove the formal adjoint of $\delta$ is covariant derivative. For simple, I choice $T= T_{ij} dx^i\otimes dx^j$ and $G= dx^b$. What I calculate: $$ \delta (T)(\partial_k) = -g^{ij} (\nabla T)(\partial_i, \partial_j, \partial_k ) = -g^{ij} (\nabla_{\partial_i} T)(\partial_j, \partial_k ) \\ =-g^{ij}[\partial_i(T(\partial_j, \partial _k))- T(\nabla _{\partial_i}\partial_j, \partial_k) - T(\partial_j, \nabla _{\partial_i}\partial_k) ] \\ = -g^{ij}\frac{\partial T_{jk}}{\partial x_i} +g^{ij}\Gamma_{ij}^aT_{ak} + g^{ij}\Gamma_{ik}^aT_{ja} $$ where $\partial_i$ is the basis vector. Therefore, I have $$ \delta (T) = [-g^{ij}\frac{\partial T_{jk}}{\partial x_i} +g^{ij}\Gamma_{ij}^aT_{ak} + g^{ij}\Gamma_{ik}^aT_{ja}] dx^k $$ so $$ \langle \delta (T), G\rangle _g = -g^{ij}g^{kb}\frac{\partial T_{jk}}{\partial x_i} + g^{ij}g^{kb}\Gamma_{ij}^aT_{ak} + g^{ij}g^{kb}\Gamma_{ik}^aT_{ja} \tag{2} $$ On the other hand, there is $\nabla dx^b = -\Gamma_{ij}^b dx^i \otimes dx^j$. So, I have $$ \langle T, \nabla G \rangle_g = T_{ij}\Gamma_{kl}^b g^{ik} g^{jl} \tag{3} $$ Next, I want to show $$ \int_M \langle \delta(T), G\rangle dV_g = \int_M\langle T, \nabla G \rangle dV_g \tag{4} $$ About (4), I have $$ \text{LHS} = \int_M (-g^{ij}g^{kb}\frac{\partial T_{jk}}{\partial x_i} + g^{ij}g^{kb}\Gamma_{ij}^aT_{ak} + g^{ij}g^{kb}\Gamma_{ik}^aT_{ja} ) dV_g \\ \text{RHS} = \int_M (T_{ij}\Gamma_{kl}^b g^{ik} g^{jl}) dV_g $$ According to Deane's hint, I guess I should integral by parts to get $$ \int_M (-g^{ij}g^{kb}\frac{\partial T_{jk}}{\partial x_i}) dV_g = \int_M (\frac{\partial g^{ij}}{\partial x_i}g^{kb} T_{jk} ) dV_g +\int_M ( g^{ij}\frac{\partial g^{kb}}{\partial x_i} T_{jk} ) dV_g $$ since $\partial M = \varnothing$. Next, I think I should use the $$ \Gamma_{ij}^k = \frac{1}{2}g^{kl}( \frac{\partial g_{jl}}{\partial x_i} + \frac{\partial g_{li}}{\partial x_j} - \frac{\partial g_{ij}}{\partial x_l} ) $$ and integral by parts. But its really complex. So before do it, I want to know whether the way is right ?
Besides, I am not familiar with the integral by parts on manifold.
We will be regularly using the divergence operator $\delta: \Gamma\left(\otimes^{k} T^{*} \mathcal{M}\right) \rightarrow$ $\Gamma\left(\otimes^{k-1} T^{*} \mathcal{M}\right)$ defined by $\delta(T)=-\operatorname{tr}_{12} \nabla T$. Again, $\operatorname{tr}_{12}$ means to trace over the first and second entries (here of $\nabla T$ ).
Remark 2.1.1. $\color{red}{\textbf(}$ The formal adjoint of $\delta$ acting on this space of sections is the covariant derivative $\nabla: \Gamma\left(\otimes^{k-1} T^{*} \mathcal{M}\right) \rightarrow \Gamma\left(\otimes^{k} T^{*} \mathcal{M}\right) .$$\color{red}{\textbf)}$ However, if one restricts $\delta$ to a map from $\Gamma\left(\wedge^{k} T^{*} \mathcal{M}\right)$ to $\Gamma\left(\wedge^{k-1} T^{*} \mathcal{M}\right)$ then its formal adjoint is the exterior derivative (up to a constant depending on one's choice of inner product). Moreover, we shall see later that if $k=2$ and one restricts $\delta$ to a map from $\Gamma\left(\operatorname{Sym}^{2} T^{*} \mathcal{M}\right)$ to $\Gamma\left(T^{*} \mathcal{M}\right)$, then its formal adjoint is $\omega \mapsto \frac{1}{2} \mathcal{L}_{\omega^{\#}} g$ where $\#$ represents the musical isomorphism $\Gamma\left(T^{*} \mathcal{M}\right) \mapsto \Gamma(T \mathcal{M})$ (see $[14$, $(2.66)]$, for example). (screenshot)
