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)

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?

  • 1
    $\begingroup$ can you refer to the paper in question? $\endgroup$
    – Bruce Lee
    Commented Feb 10, 2016 at 9:57
  • $\begingroup$ The paper i refer is arxiv 1006.3168v2 $\endgroup$
    – phy_math
    Commented Feb 10, 2016 at 13:05

1 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 $$ \epsilon_{\mu\nu\alpha\beta} R^{\alpha\beta}{}_{\rho\lambda} =\epsilon_{\rho\lambda\xi \kappa} R^{\xi \kappa} {}_{\mu\nu} $$


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