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).

We start with the following terminology called “a conformal change“.

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).

1 Comment »

  1. The last line says “which shows that …” Why does the last equation follow? The previous L_{g_\omega} looks quite different.

    Thanks for your nice blog posts.🙂

    Comment by torteloni — December 10, 2015 @ 21:45


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