# Spaces of Constant Curvature and the Ricci Tensor

Airsteve0
Hi all, I was just interested in verification of a concept. If we are given the full Riemann tensor in the form which implies constant curvature (i.e. lambda multiplying metric components) does this imply that the Ricci tensor vanishes? The question stems from why the vacuum equations cannot be constructed in dimensions less than 4. Thanks for any clarification!

Staff Emeritus
Gold Member
If we are given the full Riemann tensor in the form which implies constant curvature (i.e. lambda multiplying metric components)

This doesn't sound right to me. Could you write out the equation you have in mind? You seem to be confusing the existence of a cosmological constant term with constant curvature. Do you mean Riemann tensor, or Ricci tensor? If you really mean the Riemann tensor, then there are 4 indices, which can't be equated to the 2 indices of the metric.

Are you specifically interested in the Riemannian case, or semi-Riemannian as well? E.g., are you talking about applications such as cosmology, where the curvature of the spacetime is clearly not constant, but the spatial geometry of a hypersurface of constant t is a constant-curvature space?

TrickyDicky
Hi all, I was just interested in verification of a concept. If we are given the full Riemann tensor in the form which implies constant curvature (i.e. lambda multiplying metric components) does this imply that the Ricci tensor vanishes? The question stems from why the vacuum equations cannot be constructed in dimensions less than 4. Thanks for any clarification!

Solutions with constant curvature are vacuum solutions of the EFE, either with, or without cosmology constant(like the static ones such as Schwarzschild's), since you seem to be talking about those with lambda i.e you said"( lambda multiplying metric components)", then obviously the Ricci tensor doesn't vanish:it is lambda times the metric.

TrickyDicky
The question stems from why the vacuum equations cannot be constructed in dimensions less than 4. Thanks for any clarification!

It is not so much that they can be constructed but the vanishing of the Ricci tensor in less than 4-dim is trivial in the sense that the Riemann tensor reduces to the Ricci tensor (or to the Ricci scalar in two dim.) and making the Riemann tensor zero gets you a flat solution, wich is not GR anymore.

Staff Emeritus
Gold Member
Solutions with constant curvature are vacuum solutions of the EFE

This is incorrect. For example, Einstein's static cosmology (i.e., dust plus a fine-tuned cosmological constant) has constant curvature but is not a vacuum solution.

TrickyDicky
This is incorrect. For example, Einstein's static cosmology (i.e., dust plus a fine-tuned cosmological constant) has constant curvature but is not a vacuum solution.

Indeed, I should have said that constant curvature solutions have Ricci tensor of the form:metric times a constant k, and that vacuum solutions also have this form, with k=0 for the ones without cosmological constant, and k=lambda for the ones with.
So my phrasing was backwards.

andrien
In dimension<4,einstein eqn. are trivial(without cosmological constant).