Ngô Quốc Anh

January 17, 2013

Conformal Killing Operator

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:14

In this notes, I want to summarize some properties of the so-called Conformal Killing Operator relative to the metric $g$, say $\mathbb L_g$.

1. First, we start with its definition. Roughly speaking, the conformal killing operator is a generalization of the Killing operator relative to the metric $g$. It maps any vector field on $M$ to some tensor of type $(2,0)$. More precisely, in components, we have $\displaystyle (\mathbb L_g v)^{ij} = \nabla^iv^j + \nabla^jv^i - \frac{2}{n}\nabla_kv^k g^{ij},$

where $v$ is a vector field on $M$.

Immediately, one can check that $\mathbb L_gv$ is traceless as can be seen in the following $\displaystyle {g_{ij}}{({\mathbb{L}_g}v)^{ij}} = {g_{ij}}{\nabla ^i}{v^j} + {g_{ij}}{\nabla ^j}{v^i} - 2{\nabla _k}{v^k} = 0.$

2. We now define the so-called Conformal Vector Laplacian associated to the metric $g$. Basically, it is given by $\displaystyle (\Delta_{g,\text{conf}}v)^i= \nabla_j(\mathbb L_g v)^{ij}.$

In components, we have $\displaystyle\begin{array}{lcl} {({\Delta _{g,{\text{conf}}}}v)^i} &=&\displaystyle {\nabla _j}{\nabla ^i}{v^j} + {\nabla _j}{\nabla ^j}{v^i} - \frac{2}{n}{\nabla ^i}{\nabla _k}{v^k} \hfill \\ &=&\displaystyle {\nabla ^i}{\nabla _j}{v^j} + R_j^i{v^j} + {\nabla _j}{\nabla ^j}{v^i} - \frac{2}{n}{\nabla ^i}{\nabla _k}{v^k} \hfill \\ &=&\displaystyle \frac{{n - 2}}{n}{\nabla ^i}{\nabla _j}{v^j} + R_j^i{v^j} + {\nabla _j}{\nabla ^j}{v^i}. \hfill \\ \end{array}$

3. Let us now determine the kernel of $\Delta_{g,\text{conf}}$. By definition, one can easily see that $\ker \mathbb L_g \subset\ker \Delta_{g,\text{conf}}$. However, we shall prove that they are actually the same. For any vector field $v$, we first have $\begin{array}{lcl}\displaystyle\int_M {{v_j}{{({\Delta _{g,{\text{conf}}}}v)}^j}}& =& \displaystyle\int_M {{v_j}{\nabla _l}{{({\mathbb{L}_g}v)}^{jl}}} \hfill \\ &=&\displaystyle \int_M {{\nabla _l}[{v_j}{{({\mathbb{L}_g}v)}^{jl}}] - {{({\mathbb{L}_g}v)}^{jl}}} {\nabla _l}{v_j} \hfill \\ &=&\displaystyle - \int_M {{{({\mathbb{L}_g}v)}^{jl}}} {\nabla _l}{v_j}, \hfill \\ \end{array}$

where the Gauss-Ostrogradsky theorem has been used to get the last line. In view of the right-hand side integrand, we see that $\begin{array}{lcl}{g_{ij}}{g_{kl}}{({\mathbb{L}_g}v)^{ik}}{({\mathbb{L}_g}v)^{jl}} &=&\displaystyle {g_{ij}}{g_{kl}}[{\nabla ^i}{v^k} + {\nabla ^k}{v^i}]{({\mathbb{L}_g}v)^{jl}} - \frac{2}{n}{\nabla _m}{v^m}\underbrace {{g^{ik}}{g_{ij}}{g_{kl}}}_{{g_{ij}}\delta _l^i}{({\mathbb{L}_g}v)^{jl}} \hfill \\ &=&\displaystyle [{g_{kl}}{\nabla _j}{v^k} + {g_{ij}}{\nabla _l}{v^i}]{({\mathbb{L}_g}v)^{jl}} - \frac{2}{n}{\nabla _m}{v^m}\underbrace {{g_{jl}}{{({\mathbb{L}_g}v)}^{jl}}}_0 \hfill \\ &=& 2{g_{ij}}{\nabla _l}{v^i}{({\mathbb{L}_g}v)^{jl}}, \hfill \\ \end{array}$

where we have used the symmetry and the traceless property of $(\mathbb L_g v)^{ij}$. Therefore, we can write $\displaystyle\int_M {{v_j}{{({\Delta _{g,{\text{conf}}}}v)}^j}} = - \frac{1}{2}\int_M {{g_{ij}}{g_{kl}}{{({\mathbb{L}_g}v)}^{ik}}{{({\mathbb{L}_g}v)}^{jl}}} = - \frac{1}{2}\int_M {{{\left\| {{\mathbb{L}_g}v} \right\|}^2}} .$

Assume that $(\Delta_{g,\text{conf}}v)^i=0$, we then see that $\mathbb L_g v =0$ since the metric $g$ is positive definite.

By integration by parts, we have the following $\begin{array}{lcl}\displaystyle\int_M {{u_j}{{({\Delta _{g,{\text{conf}}}}v)}^j}}& =& \displaystyle\int_M {{u_j}{\nabla _l}{{({\mathbb{L}_g}v)}^{jl}}} \hfill \\ &=&\displaystyle \int_M {{\nabla _l}[{u_j}{{({\mathbb{L}_g}v)}^{jl}}] - {{({\mathbb{L}_g}v)}^{jl}}} {\nabla _l}{u_j} \hfill \\ &=&\displaystyle \int_{\partial M}(\mathbb L_gv)^{jl}\nu_j u_l - \int_M {{{({\mathbb{L}_g}v)}^{jl}}} {\nabla _l}{u_j}, \hfill \\ \end{array}$

for any vector fields $u$ and $v$ and $\nu$ is the outward normal vector field on $\partial M$. As above, we can write $\displaystyle \int_M {{u_j}{{({\Delta _{g,{\text{conf}}}}v)}^j}}=\int_{\partial M}(\mathbb L_gv)^{jl}\nu_j u_l - \frac{1}{2}\int_M (\mathbb L_gv)^{jl} (\mathbb L_gu)_{jl}.$

4. Solutions to the Conformal Vector Poisson Equation. Let now discuss the existence and uniqueness of solutions to the conformal vector Poisson equation $\displaystyle (\Delta_{g,\text{conf}}v)^i=S^i$

where the vector field $S$ is already given. We shall prove that a necessary condition for the solution to exist is that the source must be orthogonal to any vector field in the kernel, in the sense that $\displaystyle\int_M {{C_j}{S^j}} = 0 \quad \forall C \in \ker \Delta_{g,\text{conf}}.$

This is clear as the following. First we know that $\displaystyle\int_M {{C_j}{S^j}} = \int_M {{C_j}{{({\Delta _{g,{\text{conf}}}}v)}^j}} = - \frac{1}{2}\int_M {{g_{ij}}{g_{kl}}{{({\mathbb{L}_g}C)}^{ik}}{{({\mathbb{L}_g}v)}^{jl}}} .$

Because ${{{({\mathbb{L}_g}C)}^{ik}}}=0$, we obtain the desired identity.

Clearly if $(M,g)$ admits no conformal Killing vector, the identity is trivial. To see why this condition also gives us a sufficient condition, we use the Fredholm theory.