\( \newcommand{\be}{\begin{equation}}
\newcommand{\ee}{\end{equation}} \newcommand{\ba}{\begin{array}}
\newcommand{\ea}{\end{array}} \newcommand{\bea}{\begin{eqnarray}}
\newcommand{\eea}{\end{eqnarray}} \newcommand{\bean}{\begin{eqnarray*}}
\newcommand{\eean}{\end{eqnarray*}} \newcommand{\la}{\label}
\newcommand{\nn}{\nonumber} \newcommand{\half}{{\scriptstyle \frac{1}{2}}}
\newcommand{\third}{{\scriptstyle \frac{1}{3}}}
\newcommand{\bli}[2]{\begin{list}{#1}{\itemsep=0.0cm \topsep=0.0cm
\partopsep=0.0cm #2}} \newcommand{\eli}{\end{list}}
\newtheorem{problem}{Problem}[chapter]
\newcommand{\bprob}{\begin{problem}} \newcommand{\eprob}{\end{problem}}\)
XII. Einstein-Cartan Theory
The field equations in the Einstein-Cartan theory are obtained from
the tetradic action (12.5) by performing independent stationary
variations with respect to the gravitational gauge fields,
i.e. the frame field $\{{{\mathbf{e}}^a}\}$ and the Lorentz
connection $\{ {{\boldsymbol{\omega }}^{ab}}\} $. The variation with
respect to the frame field yields:
\[ {\delta
_{\mathbf{e}}}{S_{{\text{EC}}}} [{\mathbf{e}},{\boldsymbol{\omega
}}] = \frac{1}{{2\kappa }}\int {d{x_4}}\, {\varepsilon
_{abcd}}\delta {{\mathbf{e}}^a} \wedge {{\mathbf{e}}^b} \wedge
\left( {{{\mathbf{R}}^{cd}} + \frac{\Lambda }{3}{{\mathbf{e}}^c}
\wedge {{\mathbf{e}}^d}} \right)\]
The condition that the action vanishes for arbitrary variations
$\delta {{\mathbf{e}}^a}$ gives the equation of motion
\[{{\mathbf{G}}_a}:=\frac{1}{2}{\varepsilon_{abcd}}{{\mathbf{e}}^b}
\wedge \left( {{{\mathbf{R}}^{cd}} + \frac{\Lambda
}{3}{{\mathbf{e}}^c} \wedge {{\mathbf{e}}^d}} \right) = 0\]
which is the Einstein (vacuum) field equation in the tetrad
formalism, including the cosmological constant.
Einstein Tensor
- Expand the curvature 2-form ${{\mathbf{R}}^{cd}}$ in (12.7) to
create the 3-form equation
\[{{\mathbf{G}}_a} = \frac{1}{4}{\varepsilon
_{abcd}}{{\mathbf{e}}^b} \wedge {{\mathbf{e}}^m} \wedge
{{\mathbf{e}}^n} \left( {R_{mn}^{cd} + \frac{\Lambda
}{3}\delta _{mn}^{cd}} \right)\]
- Multiply from the left with the pseudoscalar
${\operatorname{I} _4}$. Then use (C.2) and (B.4) for $p=3$ to
get the 1-form:
\[{*
\mathbf{G}}_a = - \frac{1}{4}\delta _{acd}^{bmn}\left(
{R_{mn}^{cd} + \frac{\Lambda }{3}\delta _{mn}^{cd}}
\right){{\mathbf{e}}_b}\]
- Expand the generalized Kronecker delta's with (B.1) to recover
the tensor form of the Einstein (vacuum) field equations:
\[{*\mathbf{G}}_a = \left( {G_a^b - \Lambda \delta_a^b}
\right){{\mathbf{e}}_b} = 0\]
with the Einstein tensor as defined in Einstein
Tensor. However, the Ricci tensor is not neccesarily symmetric here.