We define the so-called conformal Killing operator $K$ mapping (1,0) vectors to (0,2) tensors by $$K(X)_{ab} = \frac{1}{2}\nabla_aX_b+ \nabla_bX_a -\frac{2}{3}(\text{div}X) g_{ab}.$$
Here $g$ is the metric and $\nabla$ is the induced covariant derivative.
I am told that the formal adjoint of $K$ is $-\text{div}.$ This is supposed to be easy, but I am not getting anywhere near the answer and I presume I must be doing something basic incorrectly.
Could anyone point me in the correct direction?
Attempt (WRONG):
Let $h_{ab}$ be a $(0,2)$ tensor. We integrate by parts over a closed manifold:
$$\int -\nabla^a h_{ab} X^b = \int \nabla^a (h_{ab} X^b) + \int h_{ab} \nabla^a X^b= 0 + \int h_{ab}\nabla^aX^b.$$
This would appear to mean that -div is the adjoint of $\nabla,$ which is wrong.
[Edit: see answer below for explanation. The computation is in fact correct, as pointed out by Jack Lee.]
You need to be careful about what space you're taking the adjoint in. In the space of all square-integrable $(0,2)$-tensors, your calculation correctly shows that the adjoint of $-\text{div}$ is indeed $\nabla$.
I think you want to be working with the space of symmetric trace-free $(0,2)$-tensors - $K$ maps into this space (possibly after tweaking the constant $2/3$). When $h$ is symmetric trace-free, we have
$$\begin{align} (K(X) , h) &= \int \nabla_{(a} X_{b)}h^{ab} - \int C \text{div}(X) g_{ab}h^{ab} \\ &=\int \nabla_a X_b h^{ab} - \int C \text{div}(X) \text{tr} (h) \\ &=(\nabla X,h) = (X,-\text{div}(h)). \end{align}$$ The symmetry of $h$ allows you remove the symmetrization on $\nabla X$ and the trace-free property gets rid of the $\text{div}(X)g$ term.