Let us define a deformation operator $\operatorname{Def}$ on a Riemannian manifold $(M,g)$, acting on divergence-free vector fields as $$ \operatorname{Def} X = \frac{1}{2} \mathcal L_X g, $$ where $\mathcal L_X g$ is a Lie derivative of metric tensor $g$. In coordinates $$ \left(L_X g\right)_{lk} = g_{kl} {X^l}_{;\,j} + g_{jl} {X^l}_{;\,k}, \qquad {X^l}_{;\,j} = {(\nabla X)^l}_{j}. $$ Above by $\nabla$ is denoted Levi-Civita connection and the whole relation is taken from Taylor's "Partial Differential Equations I" (3.33).
Then we may took deformation operator's formal adjoint $\operatorname{Def}^*$: $$ \int_M \left< \operatorname{Def} Y, \operatorname{Def} X \right> \, \mu = \int_M \left< Y, \operatorname{Def} ^* \operatorname{Def} X \right> \, \mu. $$
What is an explicit expression for $\operatorname{Def}^* \operatorname{Def} X$ if $\operatorname{div} X = 0$? It should somehow be related to divergence of $\operatorname{Def} X$ or Laplacian plus Ricci tensor of $X$.
Even better though is to get $\left(\operatorname{Def}^* \operatorname{Def} X\right)^\flat$ --- a corresponding covector field.
If we restrict $\operatorname{Def}^*$ to act only on covariant symmetric tensors, like $\operatorname{Def} X$ itself, then it is minus divergence $\operatorname{Def}^* = -\operatorname{div}$: $$ \left(\operatorname{Def}^* A \right)_i = -\nabla_j A_{ij}, \quad A_{ij} = A_{ji} $$ The adjoint property is due to $$ \langle \operatorname{Def} Y, A \rangle = \frac{1}{2} \left(\nabla_j Y_i + \nabla_i Y_j \right) A_{ij} = \ldots - \frac{1}{2} \left(\nabla_j A_{ij} \right) Y_i - \frac{1}{2} \left(\nabla_i A_{ij} \right) Y_j = \ldots - \left(\nabla_j A_{ij} \right) Y_i, $$ where $\ldots$ denotes terms of the full divergence form (boundary terms). In the above consistently was used the fact that Levi-Civita connection is compatible with metric ($\nabla g = 0$).