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


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