# What does Ricci tensor do with two vectors?

I have found it easier to understand the meaning of a particular tensor if I can find out what does it do if I cancel all its lower indices with vectors in short:

• $g_{ij} u^i v^j$: dot product of $\mathbf u$ and $\mathbf v$.
• $-\Gamma^k_{ij} u^i v^j$ : rate of change of the local coordinates of $\mathbf u$ when parallel transported in the direction of $\mathbf v$.
• $R^l_{ijk} s^i u^j v^k$ : form a parallelogram from $\mathbf u$ and $\mathbf v$ at the given point, parallel transport $\mathbf s$ around it, the resulting vector represents the change of $\mathbf s$ after completing the loop, thus indicating the curvature of the manifold near that the point.

Now doing the same with the Ricci tensor, so $R_{ij} u^i v^j$.

So you supply 2 vectors, and you get a number. What's the meaning of the vectors in this case? What's the meaning of the resulting number?

I read that if the 2 vectors are the same then it tells how does the volume of a ball moved along a geodesic changes. Just like the Christoffel symbols tell you what acceleration you will feel if you go straight (in coordinates) towards a given direction when you supply the same vector twice.

But I'd like to know the more general case when the two vectors isn't the same.

The Ricci tensor obviously a tensor that accepts two vectors and outputs a number. This number represents in some sense the "average" sectional curvature at a given location on the manifold $M$. In GR, we usually use spacetime manifolds.

In his book Riemannian Geometry, Manfredo Do Carmo states the following on page 97:

Let $x = z_n$ be a unit vector in $T_pM$; we take an orthonormal basis $\lbrace z_1,z_2,...,z_{n-1}\rbrace$ of the hyperplane in $T_pM$ orthogonal to $x$ and consider the following averages: $$\text{Ric}_p(x) = \frac{1}{n-1} \sum_i \langle R(x,z_i)x,z_i \rangle$$ for $i = 1,2,...,n-1$

This is a hint of a more formal version of what I said in the first paragraph. Let me add a bit more.

If $u, v$ are tangent vectors at a point $p \in M$, then $$Ric(u,v) = R_{ab} u^a v^b$$ Here, $u^a, v^b$ is abstract index notation. Note, $Ric$ is symmetric, so one way of thinking about $Ric$ is to interpret $Ric(u,u)$ for all tangent vectors $u$ (since this determines $Ric$ in general), and it suffices then to just try to understand this when $||u|| = 1$.

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

So when $e_i, u$ are orthonormal, $\langle R(e_i, u)u, e_i \rangle$ gives the sectional curvature of the plane spanned by $e_i$ and $u$. In particular, $$\frac{1}{n-1} Ric(u, u),$$ when $||u|| = 1$, is an average of the sectional curvatures of 2-planes through $u$.