Identity $ \epsilon_{abcd} R^{cd}_{\phantom{cd}mn} = \epsilon_{mncd} R^{cd}_{\phantom{cd}ab}$ in vacuum

Question

starting from \begin{align} \epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \sigma\tau} + \epsilon_{\rho\sigma \xi \kappa} R^{\xi \kappa}_{\phantom{ab} \tau \lambda} + \epsilon_{\rho \tau \xi \kappa} R^{\xi \kappa}_{\phantom{\xi \kappa} \lambda \sigma}=0 \end{align} The paper says, after some simple algebra with the anti-symmetric properties of $\epsilon_{abcd}$ $R_{abcd}$ one can obtain \begin{align} \epsilon_{\rho\lambda \xi \kappa} R^{\xi \kappa}_{\phantom{ab} \sigma \tau} = \epsilon_{\sigma \tau \xi \kappa} R^{\xi \kappa}_{\phantom{ab} \rho \lambda} \end{align}

First my guess was interchange indices $\rho \lambda$ with $\sigma \tau$ with first Bianchi identiy $i.e$ \begin{align} \epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \sigma \tau} = \epsilon_{\rho\sigma \xi \kappa} ( R^{\xi \phantom{ab} \kappa}_{\phantom{a} \tau \lambda}+R^{\xi \phantom{a} \kappa\phantom{b}}_{\phantom{a} \lambda\phantom{b} \tau}) + \epsilon_{\rho \tau \xi \kappa} ( R^{\xi \phantom{ab} \kappa}_{\phantom{a} \lambda \sigma}+R^{\xi \phantom{a} \kappa\phantom{b}}_{\phantom{a} \sigma\phantom{b} \lambda}) \end{align} and \begin{align} \epsilon_{\sigma\tau\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \rho \lambda} = \epsilon_{\sigma\rho \xi \kappa} ( R^{\xi \phantom{ab} \kappa}_{\phantom{a} \lambda\tau}+R^{\xi \phantom{a} \kappa\phantom{b}}_{\phantom{a} \tau\phantom{b} \lambda}) + \epsilon_{\sigma \lambda \xi \kappa} ( R^{\xi \phantom{ab} \kappa}_{\phantom{a} \tau \rho}+R^{\xi \phantom{a} \kappa\phantom{b}}_{\phantom{a} \rho\phantom{b} \tau}) \end{align}

I can see that the first term of $\epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \sigma \tau}$ and $\epsilon_{\sigma\tau\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \rho \lambda}$ are equal but the second part seems not equal to each other (it reduce the same form of ideneity we want to prove)

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The paper says it is simple computation with the anti-symmetric properties of $\epsilon$ and $R$. If you know how to do this can you give me some hint or some idea?

Answer 1

First, operate with $\epsilon^{\sigma\tau\alpha\beta}$ \begin{align} \epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa}_{\phantom{ab} \sigma\tau}\epsilon^{\sigma\tau\alpha\beta} + \epsilon_{\rho\sigma \xi \kappa} R^{\xi \kappa}_{\phantom{ab} \tau \lambda} \epsilon^{\sigma\tau\alpha\beta}+ \epsilon_{\rho \tau \xi \kappa} R^{\xi \kappa}_{\phantom{\xi \kappa} \lambda \sigma}\epsilon^{\sigma\tau\alpha\beta}=0 \end{align} Then you will have \begin{eqnarray} 0&=&\epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\sigma\tau}\epsilon^{\sigma\tau\alpha\beta} + \delta^{\tau\alpha\beta}_{\rho \xi \kappa}R^{\xi \kappa}_{\phantom{ab} \tau \lambda}-\delta^{\sigma\alpha\beta}_{\rho \xi \kappa} R^{\xi \kappa}_{\phantom{\xi \kappa} \lambda \sigma}\\ &=&\epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\sigma\tau}\epsilon^{\sigma\tau\alpha\beta} + 2\delta^{\tau\alpha\beta}_{\rho \xi \kappa}R^{\xi \kappa}_{\phantom{ab} \tau \lambda}\quad(1) \end{eqnarray}

We need $$\delta^{\tau\alpha\beta}_{\rho \xi \kappa} = (\delta^\tau_\rho \delta^\alpha_\xi \delta^\beta_\kappa -\delta^\tau_\rho \delta^\alpha_\kappa \delta^\beta_\xi+\delta^\tau_\xi \delta^\alpha_\rho \delta^\beta_\kappa-\delta^\tau_\xi \delta^\alpha_\kappa \delta^\beta_\rho+\delta^\tau_\kappa \delta^\alpha_\xi \delta^\beta_\rho-\delta^\tau_\kappa \delta^\alpha_\rho \delta^\beta_\xi) $$ Then $$ R^{\xi \kappa}_{\phantom{ab} \tau \lambda}\delta^{\tau\alpha\beta}_{\rho \xi \kappa} = 2 R^{\alpha\beta}{}_{\rho\lambda} $$ By using this eq (1) becomes $$ -4 R^{\alpha\beta}{}_{\rho\lambda} = \epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\sigma\tau}\epsilon^{\sigma\tau\alpha\beta} $$ ($$-Riem= \star Riem \star$$ Note that in this language you need only operate both side with Hodge dual the result is $$\star Riem =Riem \star$$ indeed, the result you want. ) Back to our main method operate both sides with $\epsilon_{\alpha\beta\mu\nu}$ we have $$ -4 \epsilon_{\alpha\beta\mu\nu} R^{\alpha\beta}{}_{\rho\lambda} = \epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\sigma\tau}(-2! (\delta^\sigma_\mu \delta^\tau_\nu-\delta^\sigma_\nu \delta^\tau_\mu)) $$ finally $$ \epsilon_{\mu\nu\alpha\beta} R^{\alpha\beta}{}_{\rho\lambda} =\epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\mu\nu} $$