Ricci Tensor

Relevant parts to questions...

• Defined as the contraction $R_{ij}=R_{irj}^r$ of the Riemann-Christoffel Tensor.
• Rank 2 (covariant), with $3^2=9$ components in 3D.
• $R_{ij}=0$ in Euclidean space (since $R_{irj}^r=0$ there).
• Distinct from Ricci’s Theorem (which is the statement $g_{ik,\ell }=0$).

The Ricci Tensor is the second-rank tensor obtained by Contraction the Riemann-Christoffel Tensor over its contravariant index and one covariant index:

$$\boxed{R_{ij}=R_{irj}^r}$$

It collapses the full four-index curvature tensor into a simpler two-index object, while still carrying nontrivial geometric information.

Properties

• Rank 2, covariant, so $R_{ij}$ has $3^2=9$ components in 3D (and $n^2$ in $n$-D).
• Tensor - it is a contraction of a tensor, so by Contraction it is itself a tensor.
• Symmetric::$R_{ij}=R_{ji}$ (for the Levi-Civita connection - the specific ${\Gamma }_{jk}^i$ defined from the metric).
• Vanishes in Euclidean space::since $R_{ijk}^r=0$ everywhere in a flat space, all of its contractions vanish as well. $R_{ij}=0$ is a weaker condition than $R_{ijk}^r=0$.

Relation to the Riemann-Christoffel Tensor

The full rank-4 Riemann-Christoffel Tensor $R_{ijk}^r$ has $3^4=81$ components in 3D. The Ricci tensor is a summary - a partial invariant built by contracting.

$$R_{ij}=R_{irj}^r=\sum _{r=1}^n{R}_{irj}^r$$

In 2D, the Ricci tensor (and scalar curvature) carries all curvature information - the Riemann-Christoffel tensor has only one independent component. In higher dimensions, $R_{ij}$ loses information: spaces can be Ricci-flat ($R_{ij}=0$) without being Euclidean.

Applications

1. Einstein’s field equations in general relativity relate the Ricci tensor and scalar curvature $R=g^{ij}{R}_{ij}$ to the stress-energy tensor: $R_{ij}-\frac{1}{2}R{g}_{ij}+\Lambda g_{ij}=\kappa T_{ij}$.
2. Characterising geometry - manifolds of constant Ricci curvature (Einstein manifolds) are a major object of study in differential geometry.
3. Simplified curvature checks - if $R_{ij}\ne 0$, the space is not Euclidean (a quicker test than computing the full Riemann-Christoffel tensor).

Ricci-flat ≠ Euclidean (in 4D+)

In dimensions $\ge 4$, a space with $R_{ij}=0$ can still have non-zero $R_{ijk}^r$. Such spaces are called Ricci-flat - famous examples include the Schwarzschild spacetime (a vacuum solution of GR), which is curved even though its Ricci tensor vanishes.

Contract Riemann-Christoffel over one upper and one lower index; what remains measures “average” curvature.