Suppose is a Riemannian manifold. For each vector field , we can define the conformal killing operator to be the trace free part of Lie derivative , more precisely

Obviously maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation , then what happen to the conformal killing operator ? Notice that

By using the identity for any vector field , here is the volume element, one can get the transfromation of divergence under confromal change

therefore

induces a formal adjoint on trace free 2-tensors. Suppose we have a symmetric 2-tensor , where are coordinates on . If one has

for some vector field and trace free 2-tensor . What does this coorespond to under metric ? To see that, we first need a formula about symmtric 2-tensors,

Now choose any vector field , then

This is equivalent to

Next consider the divergence operator and its adjoints .

It is well know that on 1-forms

where operator turns 1-form to a vector field by using metric . What is the relation of and ? To find that, choose any 1-form ,

using the formula about , one gets

then using is symmetric, continue from (1)

In other words,

In the other way, we can calculate more directly

We get

therefore

One can compare this with (2).