Let $(M,g)=(N,\ddot{g})\times f(B,\bar{g})$ be an Einstein warped-product manifold (i.e. $Ric=\lambda g$) where $f:N \rightarrow (0, \infty)$ (positive scalar function) and with $g= \ddot{g}+f^2 \bar{g}$.


If $(B, \bar{g})$ is Ricci flat, being $(M, g)$ an Einstein manifold, this means that $(M, g)$ must be only Ricci flat or not? Or $(M, g)$ can be Einstein (not Ricci flat) even though $(B, \bar{g})$ is Ricci flat?



Einstein warped-product manifold with flat fiber
When we have an Einstein warped-product manifold where the base is a Riemannian manifold, independently of dimension, and the ﬁber is a Ricci-ﬂat, we have: $|\nabla f|^2+[\frac{\lambda (m-n)+ R}{m(m-1)}]f^2=0$ (with $n$ and $m$ the dimension of the base and the fiber, respectively and $R$ is the scalar curvature of the base).
Then, either $R$ $\leq$ $\lambda (n − m)$ or $f$ is trivial.
This means that the Einstein warped-product manifold will not necessarily be flat even though the fiber is Ricci-flat (for details see, for example, <https://www.researchgate.net/publication/319151263_On_the_structure_of_Einstein_warped_product_semi-Riemannian_manifolds>).


