Let $(M, g_M)$ where $M= B \times_f F$ and $g_M=g_B + f^2g_F$, an Einstein warped product manifold (i.e., $Ric_M= \lambda g_M$), with Ricci flat fibermanifold $F$, i.e., $Ric_F=0$. Then $M$ can admit only constant negative Ricci curvature or zero Ricci curvature (i e., $\lambda \le 0$) or $M$ could also have positive constant Ricci curvature (i.e., $\lambda >0$)? In other words, $Ric_F = 0$ necessarily implies $\lambda \le 0$, or can solutions be obtained with $\lambda > 0$?

1$\begingroup$ An obvious result that might interest you, even if it doesn't answer your question, is the following: An Einstein warpedproduct manifold where the base is a Riemannian manifold, independently of dimension, and the ﬁber is Ricciﬂat, we have: $\nabla f^2+[\frac{\lambda (mn)+ R}{m(m1)}]f^2=0$ (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. $\endgroup$– Alexander PigazziniFeb 20 at 21:15
1 Answer
It can not have constant positive Ricci curvature. By BonnetMyers constant positive Ricci curvature implies that $M$ is compact.
If $V$ is a vertical vector then by the formula for Ricci curvature of warped product (page 266 in Besse's book)
$$ Ric(V,V)=Ric_F(V,V) V^2(\frac{\Delta f}{f}+(p1)\frac{\nabla f^2}{f^2}) $$ where $p=\dim F$. If the fiber is Ricci flat then at the point on the base where $f$ achieves minimum (which exists by compactness) it holds that $Ric(V,V)\le 0$

$\begingroup$ Thank you for your answer. You say "Constant positive Ricci curvature implies that $M$ is compact", but you mean that constant positive Ricci curvature implies that $M$ is compact if $Ric_F = 0$, or in general independent of $Ric_F$, to have constant positive Ricci curvature, does $M$ have to be compact? $\endgroup$– exxxit8Feb 20 at 19:59

2$\begingroup$ BonnetMyers theorem says that if $Ric_{M^n}\ge (n1)$ then $diam(M)\le \pi$. This is completely independent of warped products. $\endgroup$ Feb 20 at 20:02

$\begingroup$ Thank you! So if $Ric_F = 0$ then $M$ cannot be compact. But this is also true in the case of semiRiemannian metric and conformal basemanifold $\bar{g_B} = \frac{1} {\phi^2} g_B $? $\endgroup$– exxxit8Feb 20 at 20:10

2$\begingroup$ If by semiRiemannian you mean a metric of arbitrary signature then I don't know if this is possible. Positivity of Ricci does not imply compactness in this case so my argument does not apply. $\endgroup$ Feb 20 at 20:15

$\begingroup$ Thank you so much again dear Professor Kapovitch! So in conclusion the question, regarding my initial question, is independent of the warped product, an Einstein manifold with Riemannian metric to have positive Ricci curvature must be compact, am I right? $\endgroup$– exxxit8Feb 20 at 20:24