Bianchi Identity using null tetrad

Question

I'm currently looking at the Newman-Penrose Formalism, and trying to understand where there sets of equations come from. For that, I need to know how I can write the second Bianchi identity for the Riemann tensor using the tetrad null frame $\lbrace e_{a},e_{b},e_{c},e_{d}\rbrace$. In other words, given that

$$g(e_{a},e_{b})=\left(\begin{array}{cccc}0 & -1 & 0 & 0 \\-1 & 0 & 0 & 0 \\0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\end{array}\right)$$

and $\Gamma_{abc}=g(\nabla_{e_{a}}e_{b},e_{c})$, I would like to write down

$$R_{abcd;e}+R_{abde;c}+R_{abec;d}=0$$

using the given null frame (and not the usual local coordinates expression). Can anyone help?

This is not homework.

Answer 1

In tetrad formalisms you don't want the Christoffel symbols $\Gamma_{abc}$, you want the connection 1-forms. For a given basis 1-form $e^a$, $d(e^a)$ is a 2-form. Now it must be that $$d(e^a) = \omega^a{}_b \wedge e_b.$$ for a matrix of 1-forms $\omega^a{}_b$. Actually, it's usually better to think of $\omega^a{}_b$ as a matrix-valued one-form. In fact, it takes values only in $\mathfrak{so}(1,3)$, the Lie algebra of the Lorentz group. More concretely, that means that $\omega^a{}_b$ is antisymmetric with respect to the Minkowski metric.

Anyway, in this formalism, the Riemann curvature is an $\mathfrak{so}(1,3)$-valued 2-form given by $$R^a{}_b = d\omega^a{}_b + \omega^a{}_c \wedge \omega^c{}_b.$$ The statement of the Bianchi identity is $$dR^a{}_b + \omega^a{}_c \wedge R^c{}_b + R^a{}_c \wedge \omega^c{}_b = 0.$$

You will find this material in Sections 14.5 and 14.6 of Gravitation by Misner, Thorne, and Wheeler.

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The above does not connect to the Newman-Penrose formalism since it doesn't mention spinors. To obtain the NP formalism, you must use dyads, the spinor lifts of null tetrads. In this formalism the connection 1-form is a $2\times 2$ antisymmetric complex matrix, so there are $3\cdot 4=12$ components, each of which can be a given its own Greek letter. The Newman-Penrose equations are now obtained by writing out in dyad components the equations above, representing tetrad derivatives by $D, \Delta, \delta, \overline{\delta}$.

I have never seen this done in detail as I imagine the calculations are horribly tedious and boring.

Answer 2

The method I will shown here may be not full option but may be helpful \begin{eqnarray} D_\Gamma R^{\mu \nu} &=& 0 \;,\\ \mathrm d R^{\mu \nu} + \Gamma^\mu{}_\lambda \wedge R^{\lambda \nu} + \Gamma^ \nu{}_\gamma \wedge R^{\mu \gamma} &=&0\;,\\ \partial_{[\beta} R^{\mu \nu}{}_{{\rho\sigma}]} +\Gamma_{[\beta|}{}^\mu{}_\lambda R^{\lambda \nu}{}_{|{\rho\sigma}]}+ \Gamma_{[\beta|}{}^ \nu{}_\lambda R^{\mu \lambda }{}_{|{\rho\sigma}]} -\Gamma_{[\beta}{}^\gamma{}_{\rho |} R^{\mu \nu}{}_{\gamma | \sigma]} - \Gamma_{[\beta}{}^\gamma{}_\rho R^{\mu \nu}{}_{\rho ] \gamma} &=&0\;,\\ \therefore \nabla_{[\beta} R^{\mu \nu}{}_{{\rho\sigma}]} =0\;. \end{eqnarray} Note that My $\Gamma$ use a different convention $$\Gamma_{cab}=g(\nabla_{e_{a}}e_{b},e_{c})\;.$$ So $$\Gamma_c{}^a{}_b = \Gamma_{(c}{}^a{}_{b)}$$