# Ngô Quốc Anh

## April 26, 2010

### Conformal invariant operators: Laplacian operators, 2

Filed under: Riemannian geometry — Ngô Quốc Anh @ 19:23

Let me provide another proof of Example 2 in this topic.

Example 2. The conformal Laplacian operator acting on a smooth function $u$ as the following $\displaystyle L_g u= - {\Delta _g}u + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_gu$

is conformal invariant.

The proof relies on the following fact

Proposition. Under a conformal change $g_\omega = e^{2\omega}g$, we have $\displaystyle d{v_{{g_\omega }}} = {e^{n\omega }}d{v_g}$

where $dv_{g_\omega}$ and $dv_g$ are the volume elements with respect to metrics $g_\omega$ and $g$ respectively.

We are now in a position to prove example 2. For simplicity, let us consider the following conformal change $g_\varphi = \varphi^\frac{4}{n-2}g$. The proof is divided into three steps.

Step 1. Showing $\displaystyle {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi}}}u = {\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u$.

Proof. Indeed, for any test function $v\in C_0^\infty(M)$ we have $\displaystyle \int_M {{\varphi ^{ - \frac{{n + 2}}{{n - 2}}}}\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right)vd{v_{{g_\omega }}}} = \int_M {\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right){d{v_g}}}$.

We now use Laplace-de Rham operator for the sake of simplicity. Note that Laplace–de Rham operator is the Laplacian differential operator on sections of the bundle of differential forms on a pseudo-Riemannian manifold. However, the Laplace-de Rham operator is equivalent to the definition of the Laplace–Beltrami operator when acting on a scalar function. Precisely, $\displaystyle \int_M {v{\Delta _g}ud{v_g}} = \int_M {\langle dv,du\rangle d{v_g}}$.

Therefore $\displaystyle\begin{gathered} \int_M {\left( {{\Delta _g}(\varphi u) - {\Delta _g}(\varphi )u} \right)\varphi vd{v_g}} = \int_M {\left( {\left\langle {d(\varphi u),d(\varphi v)} \right\rangle - \left\langle {d(\varphi ),d(\varphi uv)} \right\rangle } \right)d{v_g}} \hfill \\ \qquad= \int_M {{\varphi ^2}\left\langle {d(u),d(v)} \right\rangle d{v_g}} \hfill \\ \qquad= \int_M {{\varphi ^{ - \frac{4}{{n - 2}}}}\left\langle {d(u),d(v)} \right\rangle {\varphi ^{\frac{{2n}}{{n - 2}}}}d{v_g}} \hfill \\\qquad = \int_M {v{\Delta _{{g_\varphi }}}ud{v_g}} . \hfill \\ \end{gathered}$

Step 2. Showing the scalar curvature equation $\displaystyle \Delta_g \varphi + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_{{g_\varphi }}{\varphi ^{\frac{{n + 2}}{{n - 2}}}} = \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi$.

Equivalently, $\displaystyle {L_g}(\varphi ) = \frac{n - 2}{4(n - 1)}{\rm Scal}_{{g_\varphi }}\varphi ^{\frac{{n + 2}}{{n - 2}}}$.

Step 3. Showing $\displaystyle {L_g}(\varphi u) = {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{L_{{g_\varphi }}}(u)$.

Proof. Clearly $\displaystyle\begin{gathered} {L_g}(\varphi u) = - {\Delta _g}(\varphi u) + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi u \hfill \\ \qquad= - {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi }}}u - {\Delta _g}(\varphi )u + \frac{{n - 2}}{{4(n - 1)}}{\rm Scal}_g\varphi u \hfill \\ \qquad= - {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{\Delta _{{g_\varphi }}}u + {L_g}(\varphi )u \hfill \\ \qquad= {\varphi ^{\frac{{n + 2}}{{n - 2}}}}\left[ { - {\Delta _{{g_\varphi }}}u + \frac{n - 2}{4(n - 1)}{\rm Scal}_{{g_\varphi }}u} \right] \hfill \\ \qquad= {\varphi ^{\frac{{n + 2}}{{n - 2}}}}{L_{{g_\varphi }}}(u). \hfill \\ \end{gathered}$

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