# $4$-dimensional Pseudo-Riemannian manifolds with $R_{ij}=\frac 14 R g_{ij}$

Consider a $$4$$-dimensional Pseudo-Riemannian manifold where $$R_{ij}=\frac 14 Rg_{ij}$$. This would mean that $$G_{ij}=-R_{ij}$$, satisfying the known relation $$G=g^{ij}G_{ij}=-g^{ij}R_{ij}=-R$$ and the EFE could be rewritten as $$-R_{ij}+\Lambda g_{ij}=\kappa T_{ij}$$.

The symmetry between the Ricci tensor and the Einstein tensor is $$R_{ij}=G_{ij}-\frac12 G g_{ij}$$ and $$G_{ij}=R_{ij}-\frac 12 Rg_{ij}$$, which, if $$R_{ij}=\frac 14 Rg_{ij}$$ they both reduce to $$R_{ij}=\frac 14 Rg_{ij}$$.

I would want to know more about this type of manifolds if possible, as I found no information on them. For example, when manifolds have this form, or whether this Einstein tensor is the only Einstein tensor, and (if possible) what happens if it doesn't have this form. Any help would be appreciated. Thanks.

In general, from $${\rm Ric}$$ on $$(M^n,g)$$ we define $$G = {\rm Ric} - \frac{{\rm s}}{2}g \quad \mbox{and}\quad {\rm Ric}_0 = {\rm Ric} - \frac{{\rm s}}{n}g,$$where $$g$$ is the metric and $${\rm s}$$ is the scalar curvature. Those coefficients in front of $$g$$ for each of them have a role. They make the relations $${\rm div}(G) = 0 \quad\mbox{and}\quad {\rm tr}_g({\rm Ric}_0) = 0$$hold. In other words, $$G$$ is defined to be the divergenceless part of $${\rm Ric}$$, while $${\rm Ric}_0$$ is defined to be the traceless part. All the time we'll be using the relations $${\rm div}({\rm Ric}) = {\rm ds}/2$$ (obtained from contractions of Bianchi identities) and $${\rm div}({\rm s}g) = {\rm ds}$$. Some facts:

• If $$n>2$$, then $${\rm Ric} = 0$$ if and only if $$G = 0$$. If $${\rm Ric} = 0$$, then $${\rm s} = 0$$ and so $$G = 0$$. Conversely, if $$G = 0$$, trace it to get $$(1-n/2){\rm s} = 0$$, so that $${\rm s} = 0$$ and $${\rm Ric} = G + {\rm s}g/2 = 0+0 = 0$$.

• If $$n > 2$$, $${\rm Ric} = 0$$ implies $${\rm Ric}_0 = 0$$ by the same argument above. Conversely, if $${\rm Ric}_0 = 0$$, take the divergence of both sides to get that $$(1/2 - 1/n){\rm ds} = 0$$, so that $${\rm ds} = 0$$ and $${\rm s}$$ is a constant. We cannot say anything better.

Also, it is important to note that the Einstein tensor $$G$$ is not directly (at least) related to $$(M^n,g)$$ being an Einstein manifold. For the following reason: if $$(M,g)$$ is Einstein, there is a constant $$\lambda$$ such that $${\rm Ric} = \lambda g$$. Tracing gives that $${\rm s} = \lambda n$$, so $$\lambda = {\rm s}/n$$. Meaning that $$(M,g)$$ is Einstein if and only if $${\rm Ric}_0 = 0$$ and $${\rm s}$$ is a constant (this second condition follows automatically from the previous one if $$n>2$$ by taking $${\rm div}$$ -- this is known as Schur's lemma), and $$G$$ plays no role here.

In general terms, Einstein's field equation is $$G + \Lambda g = T$$, where $$T$$ is the energy-momentum tensor and $$\Lambda$$ is the cosmological constant, which at least in the vacuum case ($$T=0$$) arises as a Lagrange multiplier obtained by trying to optimize the Einstein-Hilbert functional $$\mathscr{S}[g] = \int_M {\rm s}_g\,\nu_g \quad \mbox{subject to}\quad \int_M \nu_g = 1,$$where $${\rm s}_g$$ is the scalar curvature of $$g$$ and $$\nu_g$$ is the volume form induced by $$g$$.

When $$T = 0$$ and $$n=4$$, the situation gets better: tracing the field equation immediately gives that $$\Lambda = {\rm s}/4$$, which is precisely what $$\lambda$$ would be when writing $${\rm Ric} = \lambda g$$ anyway (i.e., $$\Lambda = \lambda = {\rm s}/4$$).

When $$T \neq 0$$ in general, it is not clear whether $$G+\Lambda g =T$$ is the Euler-Lagrange equation for some optimization problem like the above.

I recommend Kühnel's Differential Geometry: Curves-Surfaces-Manifolds (chapter 8) and skimming through The Large Scale Structure of Spacetime by Hawking and Ellis for more details.