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.


1 Answer 1


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: \begin{equation} \text{Ric}_p(x) = \frac{1}{n-1} \sum_i \langle R(x,z_i)x,z_i \rangle \end{equation} 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 $u$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.