0
$\begingroup$

The Ricci scalar is the uncontracted version of the Ricci tensor $R=R^{\mu}_{\mu}=g^{\mu\nu}R_{\mu\nu}$. Carrol describes the Ricci scalar as being the trace of the Ricci tensor, but I do not understand what this means. How do you interpret the contraction of the Ricci tensor with the metric as being the trace of the tensor?

He then goes on to say that the Ricci tensor and scalar contain all of the information about traces of the Riemann tensor. What does this mean? How do you interpret this exactly? And why is it important to put all the information about the traces of the Riemann tensor into a single object?

$\endgroup$
2
  • $\begingroup$ What is your understanding of what the trace of a tensor actually is? Why wouldn't it be the contraction of $R^\mu {}_\nu$? $\endgroup$ Commented Nov 22, 2023 at 13:43
  • $\begingroup$ The $4$-index Riemann tensor's symmetries imply there is, to within a sign change, only one way that isn't trivially zero to contract away all of its indices in two pairs. That leads you after one contraction to the Ricci tensor, then after another to the Ricci scalar. $\endgroup$
    – J.G.
    Commented Nov 22, 2023 at 13:55

1 Answer 1

4
$\begingroup$

Go back to the definition of the trace of a matrix in euclidean space. If you choose a basis, in which the linear map $A$ has components $A_i^j$, then the trace is $$TrA=\sum_{i=1}^{d} A_i^i$$ But notice how uncovariant this description is. There seems to be no mention of the metric anywhere. But this is misleading. Recall what $A_i^i$ is. If you let $A$ act on the basis vector $e_i$, you have $$Ae_i=A_i^je_j$$How do you get $A_i^i$ from this?Simple, take the inner product with $e_i$, $$\langle e_i, Ae_i \rangle=A_i^j\langle e_i, e_j \rangle=A_i^i$$ So here you have it. You can rewrite the trace as $$TrA=\sum_{i=1}^N A_i^j \langle e_i, e_j \rangle$$

When you then consider a general metric manifold, the linear map $A$ now does not act on any euclidean space with any metric, but in the tangent space with the metric dictated by the manifold. So you can repeat every step of what I did, if you formally replace $\langle e_i,e_j \rangle$ with $g(e_i,e_j)$ or $g_{ij}$, since the metric gives you the inner product. So the definition of trace on a metric manifold must become $$TrA=\sum_{i=1}^{dimM}A_i^j g_{ij}$$

Finally, starting not with a linear map on and from the tangent space, but with a twice covariant tensor like the Ricci, tensor, and remembering that the inner product of the dual basis is given by $g^{ij}$ instead of $g_{ij}$, you would have arrived at $$TrA=g^{ij}A_{ij}$$ and now hopefully the definition of the Ricci scalar makes sense.

The main point is that the trace of a tensor tells you about the magnitude of its image on repeated copies of the same basis, so to speak. For example, the Ricci tensor tells you about the geodesic(along $\mu$ direction) deviation (along $\nu$) direction about some point in spacetime, so it's a measure of sectional curvature. If you want to know some kind of "average" sectional curvature you must add that curvature in all directions. To do that you have to dot the image with every basis and add it all up. This is what is achieved by contracting with the metric tensor. So the Ricci tensor contains information about the mean sectional curvature of the manifold; in other words, it measures how the volume of objects change along geodesics. Of course all this is meant in a loose sense, but that's the gist of it

$\endgroup$
1
  • 1
    $\begingroup$ Good explanation (+1). $\endgroup$
    – JanG
    Commented Nov 22, 2023 at 16:31

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