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Poisson in A Relativist's Toolkit defines the Riemann tensor as$$A_{\,;\alpha\beta}^{\mu}-A_{\,;\beta\alpha}^{\mu}=-R_{\phantom{\mu}\nu\alpha\beta}^{\mu}A^{\nu}.$$

Foster and Nightingale's A Short Course in General Relativity give$$\lambda_{a;bc}-\lambda_{a;cb}=R_{\phantom{\mu}abc}^{d}\lambda_{d}.$$ How can I show these two equations are equivalent? I've relabelled Poisson's using Latin indices as$$A_{\,;bc}^{a}-A_{\,;cb}^{a}=-R_{\phantom{\mu}dbc}^{a}A^{d}.$$

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  • $\begingroup$ First off you need to multiply by a metric tensor to move the summation to an upper index in the second equation, then use the property of R to be antisymmetric in the first two indices to switch the order. $\endgroup$
    – Triatticus
    Commented Jun 2, 2015 at 14:19
  • $\begingroup$ So I multiply by $g^{ad}$ to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}=R_{\phantom{\mu}abc}^{d}\lambda^{a}.$$ Then use the Riemann tensor symmetries to give$$\lambda_{;bc}^{d}-\lambda_{;cb}^{d}= -R_{\phantom{\mu}dbc}^{a}\lambda^{a}.$$ But it's still not the same as Poisson's equation. $\endgroup$
    – Peter4075
    Commented Jun 2, 2015 at 15:05
  • $\begingroup$ You cannot use the letter d when multiplying by the metric as its already a dummy index $\endgroup$
    – Triatticus
    Commented Jun 2, 2015 at 15:11
  • $\begingroup$ OK, so what metric do I multiply by? If I try $g^{ax}$ it does weird things to the rhs (ie I have two upper indices in the Riemann tensor). And I'm still left with $\lambda^{a}$. $\endgroup$
    – Peter4075
    Commented Jun 2, 2015 at 15:18
  • $\begingroup$ Well the index on labda doesnt change with multiplication by the metric, you then reorder the first two indices and like the answer to this question says swap the upper and lower dummy indices (implicite use of the metric) $\endgroup$
    – Triatticus
    Commented Jun 2, 2015 at 15:27

1 Answer 1

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Using the anti-simmetry of the two first indices of the Riemann tensor and the fact that $ x^a y_a = x_a y^a $.

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