## Approriate scalling in Yamabe equation

Suppose ${(M,g)}$ is a Riemannian manifold, and ${L_g=\Delta_g -\frac{n-2}{4(n-1)}R_g}$ is the conformal Laplacian. Assume ${u>0}$ satisfies

$\displaystyle L_gu+Ku^p=0$

where ${K}$ is some fixed constant, ${1. Suppose near a point ${x_0\in M}$, there is a coordinates ${x^1,x^2,\cdots, x^n}$. We want to scale the coordinates to ${x^i=\lambda y^i}$,

$\displaystyle g(x)=g_{ij}(x)dx^idx^j=\lambda^2 g_{ij}(\lambda y)dy^idy^j=\lambda^2 \hat{g}(y)$

By the conformal invariance of ${L}$, for any ${\phi}$, we get

$\displaystyle L_{g}(\lambda^{-\frac{n-2}{2}}\phi)=\lambda^{-\frac{n+2}{2}}L_{\hat{g}}(\phi)$

We want to choose ${\phi(y)=\lambda^{\alpha}u(\lambda y)}$ such that

$\displaystyle L_{\hat{g}}(\phi)+K\phi^p=0$

which means

$\displaystyle L_{\hat{g}}(\lambda^{\alpha}u(\lambda y))=\lambda^{\frac{n+2}{2}}L_g (\lambda^{\alpha-\frac{n-2}{2}}u(\lambda y))=-K\lambda^{\alpha+2} (u(\lambda y))^p$

Letting

$\displaystyle \alpha+2=\alpha p$

we get ${\alpha=\frac{2}{p-1}}$.

The above proof may not be right.

Or we should look it more directly

$\displaystyle L_g(u(x))=\lambda^2 L_{\hat{g}}(u(\lambda y))=\lambda^2 Ku(\lambda y)^p$

then

$\displaystyle L_{\hat{g}}(\lambda^\alpha u(\lambda y))=K(\lambda^\alpha u(\lambda y))^p$

with ${\alpha=\frac{2}{p-1}}$.