# Ngô Quốc Anh

## April 24, 2010

### Conformal invariant operators: Laplacian operators

Filed under: Riemannian geometry — Ngô Quốc Anh @ 9:23

Let $M$ be an $n$-dimensional differentiable connected Riemannian manifold with metric tensor $g$. In order to distinguish $g$ from other metrics on $M$, we shall denote the manifold $M$ with $g$ by $(M, g)$.

Definition. If $g'$ is another metric on $M$, and there is a function $\omega$ on $M$ such that $g' = e^{2\omega}g$, then $g$ and $g'$ are said to be conformally related or conformal to each other, and such a change of metric $g \to g'$ is called a conformal change of Riemannian metric.

We are now in a position to define the so-called “conformal invariant operators“.

Definition. We call a metrically de fined operator $A$ conformally covariant of bi-degree $(a,b)$ if under the conformal change of metric $g_\omega = e^{2\omega}g$, the pair of  corresponding operators $A_{g_\omega}$ and $A$ are related by

$\displaystyle {A_{{g_\omega }}}(\varphi ) = {e^{ - b\omega }}A({e^{a\omega }}\varphi ), \quad \forall \varphi \in {C^0}(M)$.

Example 1. The Laplace-Beltrami operator is conformal invariant.

Proof. Let us recall from this topic that on $(M,g)$, the Laplace-Beltrami operator is defined to be

$\displaystyle \Delta_g ={\rm div}({\rm grad \left(\cdot\right)})$

and in local coordinates it is given as follows

$\displaystyle\Delta_g = \frac{1}{{\sqrt {\det \left( g\right)} }}\frac{\partial }{{\partial {x^j}}}\left( {\sqrt {\det \left( g\right)} {g^{ij}}\frac{{\partial}}{{\partial {x^i}}}} \right)$.

By a change of metric $g_\omega = e^{2\omega}g$ we need to calculate $\Delta_{g_\omega}$ and then compare with $\Delta_g$.

Indeed, in local coordinates

$\displaystyle {g_\omega } = {e^{2\omega }}{g_{ij}}d{x^i} \otimes d{x^j}$

which implies

$\displaystyle \det ({g_\omega }) = \det ({e^{2\omega }}{g_{ij}}) = {e^{2\omega }}\det (g)$.

Besides,

$\displaystyle {({g_\omega })^{ij}} = {({g_\omega })^{ - 1}} = {({e^{2\omega }}{g_{ij}})^{ - 1}} = \frac{1}{{{e^{2w}}}}{g^{ij}}=e^{-2\omega}g^{ij}$.

Since the tangent vectors $\frac{\partial}{\partial x^i}$ are not affected by conformal change, the rest doesn’t change. Therefore

Example 2. The conformal Laplacian operator is conformal invariant.

Proof. Let us define the so-called conformal Laplacian operator acting on a smooth function $\varphi$ as the following

$\displaystyle L_g (\varphi) = - {\Delta _g}(\varphi) + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi$

where ${\rm Scal}_g$ is nothing but the scalar curvature with respect to metric $g$. We are going to prove under a conformal change $g_\omega = e^{2\omega}g$, the following identity

$\displaystyle {L_{{g_\omega }}}(\varphi ) = {e^{ - \frac{{n + 2}}{2}\omega }}{L_g}\left( {{e^{\frac{{n - 2}}{2}\omega }}\varphi } \right), \quad \forall \varphi \in {C^0}(M)$

holds. Thus if we write

$\displaystyle {e^{\frac{{n - 2}}{2}\omega }} = u$

we then have

$\displaystyle {e^{ - \frac{{n + 2}}{2}\omega }} = {u^{ - \frac{n+2}{{n - 2}}}}$.

It is now clear to see that under the conformal change

$\displaystyle {g_u } = {u^{\frac{4}{{n - 2}}}}g$

we get

$\displaystyle {L_{{g_u}}}(\varphi ) = {u^{ - \frac{{n + 2}}{{n - 2}}}}{L_g}\left( {u\varphi } \right), \quad \forall \varphi \in {C^0}(M)$.

We know from this topic that under a conformal change $g_f=e^fg$

$\displaystyle {\rm Scal}_{{g_f}} = {e^{ - f}}\left( {{\rm Scal}_g - \left( {n - 1} \right)\Delta f - \frac{{\left( {n - 1} \right)\left( {n - 2} \right)}}{4}|\nabla f{|^2}} \right)$.

By using example 1, we firstly have

$\displaystyle\begin{gathered} {L_{{g_\omega }}} = \left( { - {\Delta _{{g_\omega }}} + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_{{g_f}}} \right) \hfill \\ \qquad= - {e^{ - 2\omega }}{\Delta _g} + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_{{g_f}} \hfill \\ \qquad= - {e^{ - 2\omega }}{\Delta _g} + \frac{{n - 2}}{{4(n - 1)}}{e^{ - 2\omega }}\left( {{\rm Scal}_g - 2\left( {n - 1} \right)\Delta \omega - \left( {n - 1} \right)\left( {n - 2} \right)|\nabla \omega {|^2}} \right) \hfill \\ \qquad= {e^{ - 2\omega }}\left[ { - {\Delta _g} + \frac{{n - 2}}{{4(n - 1)}}\left( {{\rm Scal}_g - 2\left( {n - 1} \right)\Delta \omega - \left( {n - 1} \right)\left( {n - 2} \right)|\nabla \omega {|^2}} \right)} \right], \hfill \\ \end{gathered}$

i.e.,

$\displaystyle {L_{{g_\omega }}}(\varphi ) = {e^{ - 2\omega }}\left[ { - {\Delta _g}(\varphi ) + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi - \left( {2\left( {n - 1} \right)\Delta \omega + \left( {n - 1} \right)\left( {n - 2} \right)|\nabla \omega {|^2}} \right)\varphi } \right]$.

Next we have

$\displaystyle\begin{gathered} {L_g}\left( {{e^{\frac{{n - 2}}{2}\omega }}\varphi } \right) = \left( { - {\Delta _g} + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g} \right)\left( {{e^{\frac{{n - 2}}{2}\omega }}\varphi } \right) \hfill \\ \qquad= - {\Delta _g}\left( {{e^{\frac{{n - 2}}{2}\omega }}\varphi } \right) + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g{e^{\frac{{n - 2}}{2}\omega }}\varphi \hfill \\ \qquad= - {\Delta _g}\left( {{e^{\frac{{n - 2}}{2}\omega }}} \right)\varphi - 2\left\langle {\nabla {e^{\frac{{n - 2}}{2}\omega }},\nabla \varphi } \right\rangle - {\Delta _g}(\varphi ){e^{\frac{{n - 2}}{2}\omega }} + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g{e^{\frac{{n - 2}}{2}\omega }}\varphi \hfill \\ \qquad= {e^{\frac{{n - 2}}{2}\omega }}\left[ { - {\Delta _g}(\varphi ) - 2\left\langle {\nabla {e^{\frac{{n - 2}}{2}\omega }},\nabla \varphi } \right\rangle {e^{ - \frac{{n - 2}}{2}\omega }} - {\Delta _g}\left( {{e^{\frac{{n - 2}}{2}\omega }}} \right){e^{ - \frac{{n - 2}}{2}\omega }}\varphi + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi } \right] \hfill \\ \end{gathered}$

which shows that

$\displaystyle {L_{{g_\omega }}}(\varphi ) = {e^{ - \frac{{n + 2}}{2}\omega }}{L_g}\left( {{e^{\frac{{n - 2}}{2}\omega }}\varphi } \right)$.