# 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

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

## December 13, 2013

### Norm of some 2-tensors involving the Ricci curvature tensor

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 4:09

Following the previous note where we discuss about the norm of the tensor $\text{Ric}-\frac 1n \overline{\text{Scal}}g$ in terms of the trace-free Ricci tensor, i.e. the following identity

$\displaystyle \boxed{{\left| {{\text{Ric}} - \dfrac{\overline{\text{Scal}}}{n}g} \right|^2} = |\mathop {{\text{Ric}}}\limits^ \circ {|^2} + \dfrac{1}{n}{(\text{Scal} - \overline{\text{Scal}})^2}}$

holds. Today, we derive some further identites involving the Ricci tensor $\text{Ric}$. As we have already seen from the previous notes, the trace-free Ricci tensor is given by

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

First, we calculate $\text{Ric}-\frac gn \text{Scal}$. For simplicity, we denote by $R$ and $S$ the Ricci tensor and the scalar curvature respectively. By definition, we have

$\begin{array}{lcl} \displaystyle {\left| {R - \frac{g}{n}S} \right|^2} &=& \displaystyle {g^{im}}{g^{jn}}{\left( {R - \frac{g}{n}S} \right)_{ij}}{\left( {R - \frac{g}{n}S} \right)_{mn}} \hfill \\ &=& \displaystyle {g^{im}}{g^{jn}}\left( {{R_{ij}}{R_{mn}} - \frac{S}{n}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{S^2}}}{{{n^2}}}{g_{ij}}{g_{mn}}} \right) \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}{g^{im}}{g^{jn}}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{S^2}}}{{{n^2}}}{g^{im}}\underbrace {{g^{jn}}{g_{ij}}}_{\delta _i^n}{g_{mn}} \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}{g^{im}}\underbrace {{g^{jn}}{g_{ij}}}_{\delta _i^n}{R_{mn}} - \frac{S}{n}{g^{im}}\underbrace {{g^{jn}}{g_{mn}}}_{\delta _m^j}{R_{ij}} + \frac{{{S^2}}}{{{n^2}}}\underbrace {{g^{im}}{g_{mi}}}_n \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}\underbrace {{g^{nm}}{R_{mn}}}_S - \frac{S}{n}\underbrace {{g^{ij}}{R_{ij}}}_S + \frac{{{S^2}}}{n} \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{{{S^2}}}{n}. \end{array}$