About metric and the Ricci curvatrue Recently, I met a question about the relation between $g$ and $-Ric_g$ on the Riemmannian manifold $(M^n, g)$. One said that "without loss of generality that by scaling $g$ in space we have $g \geq - Ric_g$". Is there anyone who can explain it? Thank you!

If $M$ is compact, this is always possible. If we scale $g$ by setting $\hat g = \lambda g$ for some positive constant $\lambda$, then the Ricci tensor (viewed as a symmetric covariant $2$-tensor) doesn't change. So if we choose $\lambda$ larger than the maximum eigenvalue of $-\operatorname{Ric}_g$ on $M$, then we get $\hat g \ge - \operatorname{Ric}_{\hat g}$.

However, if $M$ is noncompact, then the eigenvalues of $-\operatorname{Ric}_g$ might not be bounded above, so this might not be possible.