# How are these two Riemann tensor equations equivalent?

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$$A_{\,;bc}^{a}-A_{\,;cb}^{a}=-R_{\phantom{\mu}dbc}^{a}A^{d}.$$

• 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. Jun 2 '15 at 14:19
• 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. Jun 2 '15 at 15:05
• You cannot use the letter d when multiplying by the metric as its already a dummy index Jun 2 '15 at 15:11
• 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 $\lambda^{a}$. Jun 2 '15 at 15:18
• 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) Jun 2 '15 at 15:27

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