## Conformal killing operator and divergence transformation under conformal change

Suppose ${(M,g)}$ is a Riemannian manifold. For each vector field ${V}$, we can define the conformal killing operator ${\mathcal{D}}$ to be the trace free part of Lie derivative ${\mathcal{L}_Vg}$, more precisely

$\displaystyle \mathcal{D}V=\mathcal{L}_Vg-\frac{2}{n}(div_g V)g$

Obviously ${\mathcal{D}}$ maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation ${\tilde{g}=e^{2f}g}$, then what happen to the conformal killing operator ${\tilde{\mathcal{D}}}$? Notice that

$\displaystyle \mathcal{L}_V\tilde g=\mathcal{L}_V(e^{2f}g)=2e^{2f}V(f)g+e^{2f}\mathcal{L}_v g$

By using the identity ${\mathcal{L}_V d\mu_g=(div_g V)d\mu_g}$ for any vector field ${V}$, here ${d\mu_g}$ is the volume element, one can get the transfromation of divergence under confromal change

$\displaystyle div_{\tilde g}V=div_g V+nV(f)$

therefore

$\displaystyle \tilde{\mathcal{D}}V=\mathcal{L}_V\tilde g-\frac{2}{n}(div_{\tilde g} V)\tilde g=e^{2f}\mathcal{D}V.$

${\mathcal{D}}$ induces a formal adjoint ${\mathcal{D}^*}$ on trace free 2-tensors. Suppose we have a symmetric 2-tensor ${h=h_{ij}dx^i\otimes dx^j}$, where ${x^i}$ are coordinates on ${M}$. If one has

$\displaystyle \tilde{\mathcal{D}}^*(h-\tilde{\mathcal{D}}V)=0$

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

$\displaystyle \langle h,w\rangle_{\tilde g}=\int_M h_{ij}w_{kl}\tilde g^{ik}\tilde g^{jl}d\mu_{\tilde {g}}=\langle e^{(n-4)f}h,w\rangle_g$

Now choose any vector field ${W}$, then

$\displaystyle 0=\langle h-\tilde{\mathcal{D}}V,\tilde{\mathcal{D}}W\rangle_{\tilde g}=\langle h-e^{2f}\mathcal{D}V, e^{2f}\mathcal{D}w \rangle_{\tilde g}=\langle e^{nf}(e^{-2f}h-\mathcal{D} V),\mathcal{D} W\rangle_{g}$

This is equivalent to

$\displaystyle \mathcal D^*(e^{nf}(e^{-2f}h-\mathcal DV))=0$

Next consider the divergence operator ${\delta:\mathscr{S}^{p+1}M\rightarrow \mathscr{S}^p M}$ and its adjoints ${\delta^*}$.

$\displaystyle \delta T=-g^{ik}\nabla_iT_{k....}$

It is well know that on 1-forms

$\displaystyle \delta^*\alpha(X,Y)=\frac{1}{2}{\nabla_X\alpha(Y)+\nabla_Y\alpha(X)}=\frac 12 (L_{\alpha^\sharp}g)(X,Y).$

where ${\sharp}$ operator turns 1-form to a vector field by using metric ${g}$. What is the relation of ${\delta h}$ and ${\tilde \delta h}$? To find that, choose any 1-form ${\alpha}$,

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle \tilde \delta h,e^{-(n-2)f}\alpha\rangle_{\tilde g}=\langle h,\tilde\delta^*(e^{-(n-2)f}\alpha)\rangle_{\tilde g} \ \ \ \ \ (1)$

using the formula about ${\delta^*}$, one gets

$\displaystyle \tilde \delta^*(e^{-(n-2)f}\alpha)=e^{-(n-2)f}\tilde \delta^*\alpha-(n-2)e^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

$\displaystyle =e^{-(n-2)f}\delta^*\alpha+e^{-(n-2)f}\alpha(f)g-ne^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

then using ${h}$ is symmetric, continue from (1)

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle e^{(n-4)f},\tilde \delta^*(e^{-(n-2)f}\alpha)\rangle_{g}=\langle e^{-2f}h,\delta^*\alpha+\alpha(f)g-ndf\otimes\alpha\rangle_{g}$

$\displaystyle =\langle\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f,\alpha\rangle_g$

In other words,

$\displaystyle \tilde \delta h=\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f$

$\displaystyle =\delta h-(n-2)e^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f \ \ \ \ \ (2)$

In the other way, we can calculate more directly

$\displaystyle \nabla_k h_{ij}=\frac{\partial}{\partial x^k}h_{ij}-\Gamma^p_{ki}h_{pj}-\Gamma^p_{kj}h_{ip}$

$\displaystyle \tilde \Gamma^k_{ij} = \Gamma^k_{ij}+ \delta^k_i\partial_j f + \delta^k_j\partial_i f -g_{ij}\nabla^k f$

We get

$\displaystyle \tilde \delta h=-\tilde g^{ki}\tilde \nabla_kh_{ij}=-e^{-2f}g^{ki}\nabla_kh_{ij}+e^{-2f}g^{ki}h_{pj}(\delta_k^p\partial_if+\delta^p_i\partial_kf-g_{ki}\nabla^pf)$

$\displaystyle +e^{-2f}g^{ki}h_{ip}(\delta^p_k\partial_j f+\delta^p_j\partial_kf-g_{kj}\nabla^pf)$

$\displaystyle \tilde \delta h=e^{-2f}\delta h+e^{-2f}(g^{ki}h_{kj}\partial_if+g^{ki}h_{ij}\partial_kf-nh_{pj}\nabla^p f+g^{ki}h_{ik}\partial_j f+g^{ki}h_{ij}\partial_k f- h_{jp}\nabla^p f)$

therefore

$\displaystyle \tilde \delta h=e^{-2f}\delta h-(n-2)e^{-2f}g^{ki}h_{kj}\partial_if+e^{-2f}(tr_gh)\nabla f$

One can compare this with (2).