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


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


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


\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].


\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.


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