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Riemann Curvature Tensor Symmetries Proof

  1. Sep 15, 2014 #1
    I am trying to expand $$\varepsilon^{{abcd}} R_{{abcd}}$$ by using four identities of the Riemann curvature tensor:

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

    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:

    \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)

    where I can get rid of the blue or the purple terms using cyclicity (sorry for colors but it'll be a pain to change it), 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. Thank you for any help.
  2. jcsd
  3. Sep 15, 2014 #2


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    Let's see:
    [itex] \epsilon^{abcd} R_{abcd}[/itex]

    By using the symmetries and antisymmetries you won't get anything, because both R and E have the same identities.

    [itex] \epsilon^{abcd} = - \epsilon^{bacd}[/itex]
    [itex] \epsilon^{abcd} = - \epsilon^{abdc}[/itex]
    [itex] \epsilon^{abcd} = \epsilon^{cdab}[/itex]

    So you might have to use the cycling identity:

    [itex]R_{abcd} = - R_{adbc} - R_{acdb}[/itex]

    then look at the contraction:
    LHS: [itex] \epsilon^{abcd} R_{abcd}[/itex]

    [itex] -\epsilon^{abcd} R_{adbc}= -\epsilon^{adbc}R_{adbc} \equiv -\epsilon^{abcd}R_{abcd} [/itex]
    whereas the second term:
    [itex] -\epsilon^{abcd} R_{acdb}= - \epsilon^{acdb}R_{acdb} \equiv -\epsilon^{abcd} R_{abcd} [/itex]

    explain steps: In the first = I have used the symmetries of epsilon.. in the equivelant step I just renamed the summed indices.

    So you have:
    [itex] \epsilon^{abcd} R_{abcd} = - 2 \epsilon^{abcd}R_{abcd}[/itex]
    so it must be zero?

    I hope this helps...Sorry but I don't understand what matrix is the one you have written...
    Last edited: Sep 15, 2014
  4. Sep 17, 2014 #3
    Thank you for the quick reply!

    I'm having trouble understanding the first = sign. How you got from
    $$\varepsilon^{abcd} \overset{\left( i i \right)}{\longrightarrow} -
    \varepsilon^{abdc} \longrightarrow ? \longrightarrow
    \varepsilon^{adbc}$$ As far as I can see from the three identitites you
    can't switch d and b like that. The identities allow either switching the
    first pair of indicies with the second pair (iii) then
    (i) and (ii) allow switching the two indicies
    in the first OR the second pair, respectively. I understand the idea of
    renaming the indicies in the equivalence sign you did afterwards.

    $$ (i) \varepsilon^{{abcd}} = - \varepsilon^{{bacd}}$$
    $$ (ii) \varepsilon^{{abcd}} = - \varepsilon^{{abdc}}$$
    $$ (iii) \varepsilon^{{abcd}} = \varepsilon^{{cdab}}$$
  5. Sep 17, 2014 #4


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    the epsilon is the Levi Civita symbol... it's symmetric under even permutations of its indices and antisymmetric under odd permutations of its indices. Or put in other words, it's a totally antisymmetric thing...that's not the case for R, which is an antisymmetric tensor to its first 2 and last 2 indices and symmetric under the interchange of them two...
    The (i),(ii),(iii) you write and which I referred to is just an illustration of this property and I referred to them because I wanted to tell you that you can't eventually use that to show eg [itex]symmmetric \times antisymmetric =0[/itex], but you had to use the cyclic property of R to work it out.

    i: is an odd permutation of indices (odd=1) a,b
    ii: is again an odd permutation of indices c,d
    iii: is even (4) permutation of indices:
    [itex]\epsilon^{abcd}= -\epsilon^{acbd} = \epsilon^{acdb}= - \epsilon^{cadb} =\epsilon^{cdab}[/itex]
    So [itex] \epsilon^{abcd}= \epsilon^{cdab}[/itex]
    Last edited: Sep 17, 2014
  6. Sep 17, 2014 #5
    Great! I was getting the Levi Civita properties confused with the Riemann Tensor's properties. Thanks again I think I got it.
  7. Sep 17, 2014 #6


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    Gold Member

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