# Scalar curvature of warped-product manifolds - intuition

Let $(M, g) = (N_1, g_1) \times f(N_2, g_2)$ be an Einstein warped-product manifold, with metric $g=g_1+f^2g_2$.

What does it mean if you choose the scalar curvature, of its base-manifold $(N_1, g_1)$, equal to a multiple of the $f$ warping function?

Does it has any geometrical or physical meaning?

• One cannot simply choose the scalar curvature of the base manifold. Given a random function $f$ they may not be a metric on the base realizing this as the scalar curvature (think e.g. surfaces, where the Gauss-Bonnet theorem gives restrictions). – Danu Dec 10 '17 at 17:55
• @Danu, thank you for your answer, yes I know, maybe I expressed myself badly ... I consider that the choice of $f$ as scalar curvature (or its multiple) is also a correct choice for the metric... and in this eventuality, what does it mean? Has any geometrical or physical meaning? – Alexander Pigazzini Dec 10 '17 at 18:39