## Conformally invariant Laplacian

On a compact manifold ${(M^n,g)}$, ${n\geq 3}$,

$\displaystyle L_g=-\Delta_g+\frac{n-2}{4(n-1)}R_g$

is called conformal laplacian operator. This follows from the following fact. Suppose ${\tilde{g}=\phi^{\frac{4}{n-2}}g}$, then

$\displaystyle L_{\tilde{g}}(f)=\phi^{-\frac{n+2}{n-2}}L_g(\phi f), \quad\forall\,f\in C^\infty(M)$

Proof: Firstly suppose ${f>0}$, define ${\bar{g}=(\phi f)^{\frac{4}{n-2}}g}$, then

$\displaystyle L_{\bar{g}}(\phi f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(\phi f)^{\frac{n+2}{n-2}}$

on the other hand ${\bar{g}=f^{\frac{4}{n-2}}\tilde{g}}$

$\displaystyle L_{\tilde{g}}(f)=\frac{n-2}{4(n-1)}R_{\bar{g}}(f)^{\frac{n+2}{n-2}}$

So we proved the equality. For general ${f\in C^\infty(M)}$, ${\exists\, C>0}$ such that ${f+C>0}$. By the special case

$\displaystyle L_{\tilde{g}}(f+C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (f+C))$

$\displaystyle L_{\tilde{g}}(C)=\phi^{-\frac{n+2}{n-2}}L_g(\phi (C))$

Thus the general case is also true. $\Box$