Why is the trace of the Riemann curvature tensor useful?

Question

As I understand it, the Ricci curvature tensor is the trace of the Riemann curvature tensor. In other words, \begin{equation} R_{ij} := R_{kij}^{\phantom{kij}k} = g^{km}R_{kijm} \end{equation} But what information does this give me that the Riemann curvature tensor does not?

EDIT:

Based on comments, perhaps another way of phrasing my question is

what information does the Ricci make clear?

or

what is the meaning of the Ricci tensor?

Answer 1

In general relativity, the Ricci tensor contains all the information necessary to characterize matter's effect on curvature. This comes directly from the Einstein equation:

$$R_{ab} + \frac{1}{2} R g_{ab} = 8\pi T_{ab}$$

where $T_{ab}$ is the stress-energy tensor, containing all the information about the density, momentum, pressure, and shear of matter.

Perhaps a better question is this: what information is discarded in the Ricci tensor compared to the Riemann? The answer is the Weyl tensor, which in GR contains information about gravitational waves and tidal deformations. More broadly, the Weyl tensor can be thought of as describing curvature from sources outside the given point, while the Ricci tensor describes curvature originating from the energy, pressure, momentum, and shear stress of matter at the given point.

From a pure mathematics standpoint, I'm not sure why such a decomposition would be useful, but from a physics standpoint, the different roles the Ricci and Weyl tensors play are quite clear.

Answer 2

Here are a few elementary thoughts from a pure Riemannian geometry perspective. (Muphrid's answer gives a nice GR-based perspective, which is obviously a super important application of the Ricci tensor, with which I am much less familiar. I should remark that I don't have much experience with either GR or Einstein manifolds, which is to say I haven't encountered the Ricci tensor much in my life.)

I find the abstract index notation gets in the way of the following explanation a bit, so I'm going to use the following notation: $Ric(\cdot, \cdot)$ is the Ricci tensor; if $v, w$ are tangent vectors at a point $p \in M$, (which would be written in abstract index notation as $v^a, w^a$), then $$ Ric(v, w) = R_{ab} v^a w^b. $$ $Ric$ is symmetric, so one way of thinking about $Ric$ is just to try and interpret $Ric(v,v)$ for all tangent vectors $v$ (since this determines $Ric$ in general), and it suffices then to just try to understand this when $||v|| = 1$.

By definition $Ric$ is the trace of the Riemann curvature tensor on its first and last indices. If $v \in T_pM$ choose an orthonormal basis $e_1, \dots, e_n$ for $T_pM$ with $e_1 = v$; then $$ Ric(v, v) = \sum_{i = 1}^n \langle R(e_i,v)v, e_i \rangle. $$ (If you are only familiar with the abstract index notation, the thing on the right is given by $$ \langle R(v, w) x, y \rangle = g_{\tau d} R_{abc}^{\phantom{abc}\tau} v^a w^b x^c y^d, $$ but I think this gets pretty hard to read with the subscripts $e_i$ sitting around as well.)

Anyway the point is that, when $e_i, v$ are orthonormal, $\langle R(e_i, v)v, e_i \rangle$ gives the sectional curvature of the plane spanned by $e_i$ and $v$. In particular, $$ \frac{1}{n-1} Ric(v, v), $$ when $||v|| = 1$, is an average of the sectional curvatures of 2-planes through $v$. (This same argument is given at the end of Chapter 8 of Lee's book, which I infer might be what you're using since your equation for the Ricci tensor agrees to the letter with the definition on page 124.)

In particular, the condition that $\tfrac{1}{n-1}Ric(v, v) \geq \lambda$ for all unit vectors $v$ is a weaker version of the condition that all sectional curvatures are bounded below by $\lambda$, and similarly for other inequalities; one can think of it as a hypothesis on "average" sectional curvatures starting from $v$.

Among results that use Ricci curvature bounds in their hypotheses, the Bonnet-Myers theorem is a particularly famous one (Theorem 11.8 in Lee). The assumption in that theorem is that when $||v|| = 1$ the Ricci tensor satisfies $$ \frac{1}{n-1} Ric(v, v) \geq \frac{1}{R^2}, $$ which we can now read as saying that for each $v$, the "average" sectional curvature of planes through $v$ is bounded below by $1/R^2$. This is the sectional curvature of a sphere of radius $R$, and so it's appropriate that the conclusion of the theorem is that such a manifold is compact and has diameter less than $\pi R$ (which is the diameter of such a sphere -- "diameter" meaning now the greatest distance between two of its points).

I also am partial to the "Ricci curvature measures the second-order deviation in the volume of small cones" given on Wikipedia as a nice "picture" to keep in my head, but if the proof on Wikipedia doesn't seem to make sense to you, I wouldn't worry about it. (In particular, don't worry about it until you have a strong grasp of Jacobi fields.)