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


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