Calculation of double dual of Riemann tensor

1. Jul 6, 2012

shichao116

Hi all, I encounter a technical problem about tensor calculation when studying general relativity. I think it should be proper to post it here.

Riemann curvature tensor has Bianchi identity: $R^\alpha$$_{[\beta\gamma\delta;\epsilon]}=0$

Now given double (Hodge)dual of Riemann tensor: G = *R*, in component form:
$G^{\alpha\beta}$$_{\gamma\delta}=1/2\epsilon^{\alpha\beta\mu\nu}R_{\mu\nu}$$^{\rho\sigma}1/2\epsilon_{\rho\sigma\gamma\delta}$

Show that the Bianchi identity can be simply written in terms of divergence of G as
$\nabla\cdot G=0$.

In component form:
$G_{\alpha\beta\gamma}$$^{\delta}$$_{;\delta}=0$

PS: $$\nabla$$ and ";" represent covariant derivative in abstract and component form respectively.

I've never done such calculation and is overwhelmed by so much super- and subscripts. Can anyone show me step by step how to get the final answer from the beginning? Thanks very much.

Last edited: Jul 6, 2012
2. Jul 6, 2012

Muphrid

Heh, this would be really quite easy with some geometric algebra (or at least an exterior derivative). Still, have you tried simply going with the brute force approach and applying the product rule to the expression for G?

3. Jul 6, 2012

shichao116

Hi Muphrid, I tried to use brute force, but I don't know how to deal with the two Levi-Civita symbol, because as you might see, the divergence actually sum with one of the Levi-Civita's subscript on the RHS.

Would mind show me how to do it if it's not too lengthy? Thanks very much.

4. Jul 6, 2012

Staff: Mentor

The Einstein tensor G is the *contracted* double dual of the Riemann tensor. So you need to contract two of the indices after taking the double dual, so G has only two indices, not four. Then you take the covariant divergence on the second index.

5. Jul 6, 2012

shichao116

Hi Peter, this is not Einstein tensor. You obtain Einstein tensor by contracting two indices of "this" G. This is exercise 13.11 of book "Gravitation".

6. Jul 6, 2012

Staff: Mentor

Ah, I see. I'll check my copy of the book when I get a chance. MTW does have a lot of exercises involving "index gymnastics".

7. Jul 6, 2012

robphy

You'll probably need identities like

$$\epsilon_{abmn}\epsilon^{pqcd} =(-1)^\sigma\delta_{abmn}^{pqcd} =(-1)^\sigma (4!)\delta_{[a\vphantom{b}}^{p} \delta_{b\vphantom{]}}^q \delta_{m\vphantom{]}}^{c\vphantom{p}} \delta_{n]}^{d\vphantom{p}},$$
where $\sigma$ is the number of negative signs in the metric tensor,
and
$$\delta_{abmn}^{pqcd} =\delta_{ab}^{pq}\delta_{mn\vphantom{b}}^{cd} +\delta_{mn\vphantom{b}}^{pq}\delta_{ab}^{cd} +(2!)^2\delta_{[m[a}^{pq}\delta_{b]n]}^{cd\vphantom{p}},$$
where $\delta_{ab}^{pq}=(2!)\delta_{[a}^p\delta_{b]}^q =(2!)\delta_{a\vphantom{b}}^{[p}\delta_{b}^{q]}$.

8. Jul 6, 2012

Mentz114

Isn't the action of the Levi-Civita's here

Gαβγδ = 1/2ϵαβμνRμνρσ1/2ϵρσγδ

anti-symmetrizing two pairs of indices so that

Gαβγδ = (1/4)R[αβ][γδ]

which seems to go part of the way.

Last edited: Jul 6, 2012