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)


\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)


\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).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: