# Ngô Quốc Anh

## December 21, 2014

### Conformal Changes of the Green function for the conformal Laplacian

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 11:00

Long time ago, I talked about conformal changes for various geometric quantities on a given Riemannian manifold $(M,g)$ of dimension $n$, see this post.

Frequently used in conformal geometry in general, or when solving the prescribed scalar curvature equation in particular, is the conformal Laplacian, defined as follows

$\displaystyle L_g(u) = - \frac{n-1}{4(n-2)}\Delta_g u + \text{Scal}_g u$

where $\text{Scal}_g$ is the scalar curvature of the metric $g$. The operator $L_g$ is conformal in the sense that any change of metric $\widehat g = \varphi ^\frac{4}{n-2}g$ would give the following magic identity

$\displaystyle L_{\widehat g} (u) =\varphi^{-\frac{n+2}{n-2}} L_g (\varphi u).$

Associated to the conformal Laplacian operator $L_g$ is the Green function, if exists, $\mathbb G_{L,g}$. Mathematically, the Green function $\mathbb G_{L,g}$ is defined to be a continuous function

$\mathbb G_{L,g} : M \times M \backslash \{(x,x) : x \in M\} \to \mathbb R$

such that for any $x\in M$, $\mathbb G_{L,g} (x, \cdot) \in L^1(M)$ and for any $u \in C^2(M)$ and any $x \in M$, we have the following representation

$\displaystyle u(x) = \int_M \mathbb G_{L,g}(x,y) L_g(u)(y) dv_g (y).$

In this note, I will show that under the conformal change $\widehat g = \varphi^\frac{4}{n-2} g$, the Green function $\mathbb G_{L,g}$ follows the following magic rule

$\displaystyle \mathbb G_{L,\widehat g}(x,y) = \varphi (x)^{-1} \varphi (y)^{-1} \mathbb G_{L,g} (x,y).$

Obviously, as long as $\mathbb G_{L,g}$ is the Green function for $L_g$, the new function $\mathbb G_{L,\widehat g}$ given by the above formula is well-defined and continuous. Since $\varphi \in C^\infty (M)$, there also holds $\mathbb G_{L,\widehat g} (x, \cdot) \in L^1(M)$. The only thing we need to check is the representation

$\displaystyle u(x) = \int_M \mathbb G_{L,\widehat g}(x,y) L_{\widehat g}(u)(y) dv_{\widehat g} (y).$

for any $u \in C^2(M)$ and any $x \in M$. To see this, by well-known facts, we obtain:

• First the conformal change for $L_{\widehat g}(u)$ as shown above

$\displaystyle L_g (u) =\varphi^{\frac{n+2}{n-2}} L_{\widehat g} (\varphi^{-1} u).$

• Then we have the conformal change for the volume

$\displaystyle dv_g =\varphi^{-\frac{2n}{n-2}} dv_{\widehat g}.$

Hence, using the formula for $\mathbb G_{L,\widehat g}$ given above, we obtain

$\begin{array}{lcl} u(x) \varphi(x) &=& \displaystyle \int_M \mathbb G_{L,g}(x,y) L_g(\varphi u)(y) dv_g (y) \\&=& \displaystyle \int_M \mathbb G_{L,g}(x,y) \varphi (y) ^{\frac{n+2}{n-2}} L_{\widehat g}( u)(y) \varphi (y) ^{-\frac{2n}{n-2}} dv_{\widehat g} (y) \\&=& \displaystyle \int_M \mathbb G_{L,g}(x,y) \varphi (y) ^{-1} L_{\widehat g}( u)(y) dv_{\widehat g} (y).\end{array}$

Hence, we have just proved that

$\begin{array}{lcl} u(x) &=& \displaystyle \int_M \mathbb G_{L,g}(x,y) \varphi(x) ^{-1} \varphi (y)^{-1} L_{\widehat g} (u)(y) dv_{\widehat g} (y) \\&=& \displaystyle \int_M \mathbb G_{L,\widehat g}(x,y) L_{\widehat g}(u)(y) dv_{\widehat g} (y).\end{array}$

Thus, there must hold

$\displaystyle \mathbb G_{L,\widehat g}(x,y) = \varphi (x)^{-1} \varphi (y)^{-1} \mathbb G_{L,g} (x,y)$

as claimed.

It is interesting to note that this type of transformation also holds for the 4th order Paneitz operator $P$ associated with Q-curvature for any dimension $4 \ne n \geqslant 3$. More precisely, there holds

$\displaystyle \mathbb G_{P,\widehat g}(x,y) = \varphi (x)^{-1} \varphi (y)^{-1} \mathbb G_{P,g} (x,y)$

where $\widehat g =\varphi^\frac{4}{n-4} g$. For a precise formulas and definition for 4th order Paneitz operator $P$ as well as its Q-curvature, I refer to this topic.

Finally, the idea of using conformal change for the Green function for the conformal Laplacian $L$ goes back to the work by Lee and Parker in their famous paper about the Yamabe problem published in 1987. The similar identity for the 4th order Paneitz operator $P$ comes from a joint paper by Hang and Yang recently appeared in IMRN.