# Ngô Quốc Anh

## March 26, 2014

### Two pointwise conformal metrics having the same Ricci tensor must be homothetic

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 13:06

The aim of this note is to recall the following interesting result by X. Xu published  in Proc. AMS in 1992, here.

Theorem. Suppose $(M,g)$ is a compact, oriented Riemannian manifold without boundary of dimension $n \geq 2$. If $\widehat g=e^{2\varphi} g$ and $\text{Ric}(\widehat g)=\text{Ric}(g)$, then $\varphi$ is constant. In other words, two pointwise conformal metrics that have the same Ricci tensor must be homothetic.

A proof for this result is quite simple. First, we recall the following conformal change

$\displaystyle\widehat{\text{Ric}}_{ij}=\text{Ric}_{ij} -(n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big) - \big( \Delta_g \varphi + (n-2) |\nabla \varphi|^2 \big) g_{ij}$

where $\Delta_g u = \text{div}(\nabla u)$. Therefore, if $\text{Ric}(\widehat g)=\text{Ric}(g)$, then we obtain the following fact

$\displaystyle (n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big)+ \big( \Delta_g \varphi + (n-2) |\nabla \varphi|^2 \big) g_{ij}=0.$

However, the term $\nabla_i\varphi \nabla_j\varphi$ appearing in the preceding identity seems to be bad. To avoid it, the author used the following conformal change

$\displaystyle \widehat g = \frac{1}{u^2} g$

for some positive function $u$, i.e. $\varphi = -\log u$ or $u=e^{-\varphi}$. Then we calculate to obtain

$\displaystyle \nabla_i u =-e^\varphi \nabla_i \varphi$

and

$\displaystyle \nabla_i\nabla_j u =-e^{-\varphi}\nabla_i \nabla_j \varphi+ e^{-\varphi}\nabla_i\varphi \nabla_j \varphi.$

Therefore,

$\displaystyle -e^\varphi \text{Hess}(u)=\text{Hess}(\varphi)-d\varphi \otimes d\varphi$

and

$\displaystyle -e^\varphi \Delta u =\Delta \varphi - |\nabla \varphi|^2.$

Hence, the identity $(n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big)+ \big( \Delta_g \varphi + (n-2) |\nabla \varphi|^2 \big) g_{ij}=0$ becomes

$\displaystyle e^\varphi \text{Hess}(u)_{ij}+\frac {e^\varphi}{n-2} \Big(\Delta u -(n-1)\frac{|\nabla u|^2}{u}\Big) g_{ij} =0$

where we also used the following fact

$\displaystyle |\nabla \varphi|^2=\frac{|\nabla u|^2}{u^2}=\frac{e^\varphi}{u}|\nabla u|^2.$

Hence, we have just proved that

$\displaystyle \text{Hess}(u)=-\frac {1}{n-2} \Big(\Delta u -(n-1)\frac{|\nabla u|^2}{u}\Big) g$

and generally that

$\displaystyle\boxed{\widehat{\text{Ric}}_{ij}=\text{Ric}_{ij} +(n-2) e^\varphi \bigg[ \text{Hess}(u) +\Big( \Delta u -(n-1)\frac{|\nabla u|^2}{u} \Big)g \bigg].}$

To go further, the author used the traceless Ricci tensor

$\displaystyle\mathop {{\text{Ric}}}\limits^ \circ =\text{Ric}-\frac{\text{Scal}}{n}g.$

Again, the conformal change rule for $\mathop {{\text{Ric}}}\limits^ \circ$ is given as follows

$\displaystyle {}^{\widehat g}\mathop {{\text{Ric}}}\limits^ \circ{}_{ij}={}^g \mathop {{\text{Ric}}}\limits^ \circ{}_{ij} -(n-2)\big( \text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi \big) +\frac{n-2}{n} \big( \Delta_g \varphi - |\nabla \varphi|^2 \big) g_{ij}.$

Since $\text{Ric}(\widehat g)=\text{Ric}(g)$, it is easy to see that

$\displaystyle {}^{\widehat g}\mathop {{\text{Ric}}}\limits^ \circ{}_{ij}={}^g \mathop {{\text{Ric}}}\limits^ \circ{}_{ij}.$

Therefore, one has

$\displaystyle\text{Hess}(\varphi)_{ij}-\nabla_i\varphi \nabla_j\varphi =\frac{1}{n} \big( \Delta_g \varphi - |\nabla \varphi|^2 \big) g_{ij}.$

Transforming the above identity in terms of $u$ gives

$\displaystyle \text{Hess}(u)=\frac{\Delta u }{n} g.$

This helps us to simplify the former identity as follows

$\displaystyle \frac{\Delta u }{n} g=-\frac {1}{n-2} \Big(\Delta u -(n-1)\frac{|\nabla u|^2}{u}\Big) g$

which is nothing but

$\displaystyle -\frac 2n u\Delta u +|\nabla u|^2=0.$

Integrating this equation over $M$ gives $|\nabla u|=0$ showing that $u$ mus be constant.

In general, if we denote $\sigma=\text{Ric}(\widehat g)-\text{Ric}(g)$ where

$\displaystyle \sigma=(n-2) e^\varphi \bigg[ \text{Hess}(u) +\Big( \Delta u -(n-1)\frac{|\nabla u|^2}{u} \Big)g \bigg],$

we then have

$\displaystyle {}^{\widehat g}\mathop {{\text{Ric}}}\limits^ \circ{} -{}^g \mathop {{\text{Ric}}}\limits^ \circ{} =\sigma - \frac{\text{trace}(\sigma)}{n}g,$

which then implies

$\displaystyle\sigma - \frac{\text{trace}(\sigma)}{n}g=(n-2)e^\varphi \Big[\text{Hess}(u)-\frac{\Delta u}{n}g \Big].$

Hence,

$\displaystyle \frac{\text{trace}(\sigma)}{n}=(n-2)e^\varphi \Big[\frac{n+1}{n} \Delta u-(n-1)\frac{|\nabla u|^2}{u}\Big].$

Thus, we have just shown that

$\displaystyle\frac{\text{trace}(\sigma)}{n(n-2)}=\frac{n+1}{n} \frac{\Delta u}{u}-(n-1)\frac{|\nabla u|^2}{u^2}.$

Integrating both sides over $M$ gives

$\displaystyle \int_M \frac{\text{trace}(\sigma)}{n(n-2)} \leqslant 0.$