# Riemann Curvature Tensor Symmetries Proof

I am trying to expand $$\varepsilon^{{abcd}} R_{{abcd}}$$ by using four identities of the Riemann curvature tensor:

Symmetry $$R_{{abcd}} = R_{{cdab}}$$ Antisymmetry first pair of indicies $$R_{{abcd}} = - R_{{bacd}}$$ Antisymmetry last pair of indicies $$R_{{abcd}} = - R_{{abdc}}$$ Cyclicity $$R_{{abcd}} + R_{{adbc}} + R_{{acdb}} = 0$$

From what I understand, the terms should cancel out and I should end up with is $$\varepsilon^{{abcd}} R_{{abcd}} = 0.$$

What I ended up with was this mess:

$$\begin{array}{l} \varepsilon^{{abcd}} R_{{abcd}} = R_{\left[ {abcd} \right]} = \frac{1}{4!} \left( \underset{- R_{{dcab}}}{\underset{+ R_{{cdab}}}{\underset{- R_{{abdc}}}{\underset{{\color{dark green} + R_{{badc}}}}{\underset{- {\color{red} {\color{black} R_{{bacd}}}}}{{\color{blue} R_{{abcd}}}}}}}} + \underset{{\color{magenta} - R_{{adbc}}}}{\underset{{\color{red} + R_{{cbad}}}}{\underset{- R_{{cbda}}}{\underset{+ R_{{bcda}}}{{\color{magenta} {\color{black} R_{{dabc}}}}}}}} + \underset{- R_{{abdc}}}{\underset{+ R_{{dcba}}}{\underset{- R_{{cdba}}}{\underset{- R_{{dcab}}}{\underset{{\color{dark green} + R_{{badc}}}}{\underset{- R_{{bacd}}}{\underset{+ R_{{abcd}}}{R_{{cdab}}}}}}}}} + \underset{+ R_{{dabc}}}{\underset{- R_{{cbda}}}{\underset{{\color{red} + R_{{cbad}}}}{\underset{- {\color{blue} {\color{black} R_{{bcad}}}}}{{R_{{bcda}}}}}}} - \underset{- R_{{bdca}}}{\underset{{\color{blue} + R_{{acdb}}}}{\underset{+ R_{{dbca}}}{\underset{- R_{{dbac}}}{\underset{+ R_{{bdac}}}{R_{{acbd}}}}}}} - \underset{{\color{blue} + R_{{adbc}}}}{\underset{- R_{{bcda}}}{\underset{+ {\color{blue} {\color{black} R_{{bcad}}}}_{}}{\underset{{\color{red} - R_{{cbad}}}}{R_{{adcb}}}}}} - \underset{{\color{black} + R_{{abcd}}}}{\underset{- R_{{bacd}}}{\underset{{\color{dark green} + R_{{badc}}}}{R_{{abdc}}}}} - \underset{{\color{magenta} + R_{{abcd}}}}{\underset{- R_{{abdc}}}{\underset{{\color{red} {\color{dark green} + R_{{badc}}}}}{\underset{- {\color{red} {\color{black} R_{{bacd}}}}}{R_{{cdba}}}}}} \right) \end{array}$$

where I can get rid of the blue or the purple terms using cyclicity, but I'm stuck because I cant see how I can get all the terms to cancel. The main problem seems to be is that the last term in the cyclicity identity $$\left( R_{{acdb}} \right)$$ can only be acquired from the 5th term $$\left( R_{{acbd}} \right)$$ in the expression i have. After I get rid of 6 terms with cyclicity I was thinking I could get of what remains with some symmetry relationship. Am I going down the wrong path here? Do I need another relationship? Carroll in Introduction to General Relativity'' says in eq 3.83 that all I have to do is expand the expression for $$R_{\left[ {abcd} \right]}$$ and mess with the indicies using the 4 identities to proove that it reduces to zero.

To increase visibility I posted the same question on http://www.physicsforums.com/showthread.php?p=4852429#post4852429, but haven't gotten any replies yet.

## 2 Answers

I think that it is only necessary to use the cyclic identity. Contracting both sides with the Levi-Civita, we should have $$0 = (R_{abcd} + R_{adbc} + R_{acdb}) \varepsilon^{abcd} \tag{1}.$$ Let $S = R_{abcd}\varepsilon^{abcd}$. Then $R_{adbc}\varepsilon^{abcd} = -R_{adbc}\varepsilon^{acbd} = R_{adbc}\varepsilon^{adbc} = S$ where the last step is renaming the dummy indices. With the same argument the third term is also equal to $S$, so we (1) says that $3S = 0$.

First, by definition of $\varepsilon$ $$\varepsilon^{{abcd}} R_{{abcd}} = \varepsilon^{{abcd}} R_{{a[bcd]}} =\text{any symmatrization of the indices of R}$$ But \begin{eqnarray}R_{{a[bcd]}} &=&\frac{1}{3!}\bigg\{ R_{abcd}-R_{abdc}+R_{adbc}-R_{adcb}+R_{acdb} -R_{acbd} \bigg\}\;,\\ &=& \frac 1 3 \bigg\{ R_{abcd}+R_{adbc} +R_{acdb}\bigg\}\;,\\ &=& 0\;.\texttt{(by the first Bianchi identity)} \end{eqnarray} So, $$\varepsilon^{{abcd}} R_{{abcd}} = \varepsilon^{{abcd}} R_{{a[bcd]}} =\varepsilon^{{abcd}}\times 0=0\;.\quad\quad\quad\blacksquare$$