Simplifying the contraction of the riemannian curvature tensor with a timelike projection matrix

Question

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.

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.