# Fiber manifold Ricci flat, physical meaning

In a warped-product spacetime, what a physical meaning we have for Ricci-flat Fiber?

I'll explain.. it is well known that a Ricci-flat spacetime means that the cosmological constant need not vanish, but if is the Fiber to be Ricci-flat and not spacetime, what can it mean?

• Can you give an example? I'm thinking of a tangent bundle as a simple example but not sure if I understand "Ricci-flat Fiber" as it has been a long while. For any manifold, regardless of Ricci, the tangent spaces are R^N, completely flat. That is not to say that moving across the entire bundle is trivial as mapping one T(p) to another T(q) is non-trival. So, are you saying that each Fiber is "flat" or that the bundle is flat even though the base space is not (or is that even possible)? – ggcg Dec 31 '18 at 19:33
• I'm saying that $M=B \times _f F$ with the metric $\bar{g}=g + f^2 \ddot{g}$ and where $(B,g)$ is the Base manifold and $(F, \ddot{g})$ is the Fiber manifold, in my question $F$ is a Ricci flat manifold. – exxxit8 Dec 31 '18 at 19:50
• That helps, but now are you asking what are the consequences of this type of space-time manifold on the CC? You say the F is NOT part of space-time so how do you see it coming into the picture? Is it a Gauge field manifold? Or, is it some additional compactified portion of space (as in string theory)? In the former case the two are treated separately, in the latter the extra metric components might produce an effect, but seeing as how they are flat the effect may vanish. – ggcg Dec 31 '18 at 21:17
• so if I consider a warped product spacetime without consider Gauge manifold or the Fiber manifold an additional compactified space, the flatness of the Fiber manifold has no influence on the spacetime structure, is that so? – exxxit8 Jan 2 '19 at 8:08
• Are you asking in the context of Kaluza-Klein theories? For example, in string theory compactifications on Calabi-Yau manifolds are often considered. Calabi-Yau manifold is a compact Kähler Ricci-flat manifold. The physical motivation for the construct is that this allows unbroken $\mathcal{N}=1$ supersymmetry in the effective $D=4$ theories. – A.V.S. Jan 2 '19 at 9:02