Riemannian manifold - scaling the metric

Question

Let $(M,g)$ be a Riemannian manifold. Set $\widetilde{g}=\lambda^2g$. Then ${\rm Ric}(X,Y)=\widetilde{\rm Ric}(X,Y)$. I want to prove $R=\lambda^2\widetilde{R}$.

But $$R=\text{tr} \;{\rm Ric}=\sum_{i}{\rm Ric}(e_i,e_i)=\sum_{i}\widetilde{\rm Ric}(e_i,e_i)=\text{tr}\;\widetilde{\rm Ric}=\widetilde{R}.$$

Where did I go wrong? I am wondering if ${\rm Ric}(X,Y)=\widetilde{\rm Ric}(X,Y)$ is false, but this is stated in my lecture notes.

Answer 1

You can do it in coordinates. You know that $\widetilde{g}_{ij} = \lambda^2g_{ij}$ and $\widetilde{g}^{ij} = \lambda^{-2} g^{ij}$. Then $$R = g^{ij}R_{ij} = \lambda^2 \widetilde{g}^{ij} R_{ij} \stackrel{(\ast)}{=} \lambda^2 \widetilde{g}^{ij} \widetilde{R}_{ij} = \lambda^2 \widetilde{R}, $$where in $(\ast)$ we used that ${\rm Ric} = \widetilde{\rm Ric}$.

Answer 2

$\textbf{Hint:}$ If $(e_i)$ is a orthonormal frame field of $(M,g)$, then you need to rescale $(e_i)$ to make it an orthonormal frame field of $(M,\tilde{g})$.