I'm reading this short piece "Physical and Geometric Interpretations of the Riemann Tensor, Ricci Tensor,and Scalar Curvature" (pdf), and I'm having trouble grasping one the simplifications used in Section 7. Specifically, Equation 9 on page 9.

To give context, the author wants to compute the scalar curvature in "the D − 1 subspace that is orthogonal to a particular [timelike] vector" $t^\mu$. To do this, they contract the Riemannian curvature tensor with what I assume to be the (inverse) metric induced on that D-1 dimensional space: $$ (g^{-1}_{D-1})^{\mu\nu}=g^{\mu\nu} - t^\mu t^\nu \implies R_{D-1} = (g^{\mu\nu} - t^\mu t^\nu)(g^{\alpha\beta} - t^\alpha t^\beta) R_{\mu\alpha\nu\beta}$$

This is simplified in Equation 9 to obtain a contraction with the Einstein tensor. However, I don't know how this simplification is made. I assume the author first expanded

$$(g^{\mu\nu} - t^\mu t^\nu)(g^{\alpha\beta} - t^\alpha t^\beta) R_{\mu\alpha\nu\beta} = (g^{\mu\nu}g^{\alpha\beta}-2g^{\alpha\beta}t^\mu t^\nu + t^\mu t^\nu t^\alpha t^\beta)R_{\mu\alpha\nu\beta}$$

And then contracted $$g^{\mu\nu}g^{\alpha\beta}R_{\mu\alpha\nu\beta} -2g^{\alpha\beta}t^\mu t^\nu R_{\mu\alpha\nu\beta} + t^\mu t^\nu t^\alpha t^\beta R_{\mu\alpha\nu\beta} = R - 2R_{\mu\nu}t^\nu t^\nu + t^\mu t^\nu t^\alpha t^\beta R_{\mu\alpha\nu\beta}$$

The third term is missing in the derivation without a justification. I don't see the symmetry of the projection matrix implying the simplification for simple algebraic reasons, as the metric tensor is also symmetric. If there's anything that I'm missing, nuanced or obvious, I would appreciate any help.

  • $\begingroup$ $R_{\mu\alpha\nu\beta} \ne R_{\mu\alpha , \nu\beta}$ See equation 9 again where the tensor contains derivatives. $\endgroup$
    – joseph h
    May 27, 2023 at 6:00
  • $\begingroup$ I took that as a typo. Earlier on the page the author says "We would have to contract the Riemann tensor, not with $g^{\mu\nu}$, but rather with the projector onto our D−1 space". Plus, in Equation 9 they end up with the Ricci tensor and the Ricci scalar, and that makes sense to me only when the comma is a mistake and the author meant to have the fully lowered Riemann curvature tensor. $\endgroup$
    – Noah M
    May 27, 2023 at 7:37
  • $\begingroup$ Typo’s are always a possibility. I’ll try and have a read of whole article later. $\endgroup$
    – joseph h
    May 27, 2023 at 22:03

1 Answer 1


I came back to the problem, and I found a solution that seems fairly straightforward. Rewriting the last contraction as $$ t^\mu t^\nu t^\alpha t^\beta R_{\mu\alpha\nu\beta} = \langle R(t,t)t,t\rangle$$ makes it clear that it must vanish, from the anti-symmetry of the curvature tensor.

A rearrangement of the contraction such as $$ t^\mu t^\nu t^\alpha t^\beta R_{\mu\alpha\nu\beta} = (t^\mu t^\alpha) (t^\nu t^\beta) R_{\mu\alpha\nu\beta}$$ would also lead to the same conclusion.


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