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.

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