Graduate Hodge operator on 4th rank tensor

[binbagsss]

I am reading a article that defines a L, R acting Hodge dual denoted by * as:

$$R^{*\alpha \mu \kappa \rho}=\epsilon^{\alpha \mu \nu \lambda}R_{\nu \lambda \kappa \rho},$$

contracting over the first two indices (and similar definition for #*^R^{\alpha \mu \kappa \rho} # contracting over the last two indices.

MAIN QUESTION
They have that #(*)^2=-1. # However I compute the following, can someone please tell me what I am doing wrong? Thanks

$$R*_{ab}^{kp}=\epsilon_{ab\alpha\mu}R^{*\alpha \mu \kappa \rho}

=\epsilon_{ab \alpha \mu} \epsilon^{\alpha \mu \nu \lambda}R_{\nu \lambda \kappa \rho}$$$$=2 \delta^{a}_{\nu} \delta^b_{\lambda}-2\delta^b_{\nu}\delta^a_{\lambda}R_{\nu\lambda \kappa \rho}

=2 R_{ab\kappa \rho}-2_{ba\kappa \rho} = 4 R_{ab \kappa \rho} $$

 

[ergospherical]

When acting on a $p$-form, and with a Lorentzian signature, $*^2 = - (-1)^{p(n-p)}$. To show it use the identity $\epsilon^{a \dots c \dots} \epsilon_{b \dots c \dots} = -p! (n-p)! \delta^{a}_{[b} \dots \delta^{\dots}_{\dots]}$. I think that's what you're trying to do in your working but it's so messy & so many typos I can't follow it.

 

[binbagsss]

When acting on a $p$-form, and with a Lorentzian signature, $*^2 = - (-1)^{p(n-p)}$. To show it use the identity $\epsilon^{a \dots c \dots} \epsilon_{b \dots c \dots} = -p! (n-p)! \delta^{a \dots}_{[b \dots]}$. I think that's what you're trying to do in your working but it's so messy I can't follow it.

Yeh, apologies was just trying to edit the format now- need to familiarise myself with the phantom command for index spacing (just need some chicken first, too much running).

I don't have a background in differential forms, but I think a tensor can only be written as a p-form if it is antisymmetric- so here when you say 'p-form' this is referring to considering the first two indices and last two indices separately, I assume?

I have started with identity:

$\epsilon^{\alpha \mu \nu \lambda}\epsilon_{\beta \gamma \kappa \rho}=-4!\delta^{[\alpha_{\beta}\delta^{\mu}_{\gamma}\delta^{\nu}_{\kappa}\delta^{\lambda}]_{\rho}$And then it can be shown that for two contracted indices:

$$\epsilon^{\alpha \mu \nu \lambda} \epsilon_{\lambda \nu \kappa \rho}=2\delta^{\mu}_{\kappa}\delta^{\alpha}_{\rho}-2\delta^{\alpha}_{\kappa}\delta^{\mu}_{\rho} $$

 

[ergospherical]

You can Hodge over either pair of the Riemann indices. Since it's effectively a 2-form it might be easier to rename it e.g. $R_{abcd} \equiv A_{ab}$ for fixed $c$, $d$. Then\begin{align*}
{(*^2 A)}_{c_1 c_2} = \frac{1}{2} \epsilon_{c_1 c_2 a_1 a_2} {(*A)}^{a_1 a_2} = \frac{1}{4} \epsilon_{c_1 c_2 a_1 a_2} \epsilon^{a_1 a_2 b_1 b_2} A_{b_1 b_2}
\end{align*}Now use $\epsilon_{c_1 c_2 a_1 a_2} \epsilon^{a_1 a_2 b_1 b_2} = -4 \delta^{b_1}_{[c_1} \dots \delta^{b_2}_{c_2]} = -4(\delta^{b_1}_{c_1} \delta^{b_2}_{c_2} - \delta^{b_1}_{c_2} \delta^{b_2}_{c_1})$