# Possible sign of scalar curvature for Einstein warped product manifold with Ricci-flat

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 fiber-manifold $$F$$, 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 $$\lambda > 0$$?

• An obvious result that might interest you, even if it doesn't answer your question, is the following: An Einstein warped-product manifold where the base is a Riemannian manifold, independently of dimension, and the ﬁber is 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. Feb 20 at 21:15

It can not have constant positive Ricci curvature. By Bonnet-Myers 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)
$$Ric(V,V)=Ric_F(V,V) -|V|^2(\frac{\Delta f}{f}+(p-1)\frac{|\nabla f|^2}{f^2})$$ 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 $$Ric(V,V)\le 0$$
• 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? Feb 20 at 19:59
• Bonnet-Myers theorem says that if $Ric_{M^n}\ge (n-1)$ then $diam(M)\le \pi$. This is completely independent of warped products. Feb 20 at 20:02
• Thank you! So if $Ric_F = 0$ then $M$ cannot be compact. But this is also true in the case of semi-Riemannian metric and conformal base-manifold $\bar{g_B} = \frac{1} {\phi^2} g_B$? Feb 20 at 20:10