We first need to introduce the notion of tetrads. A tetrad is an orthonormal vector basis in terms of which the space-time metric looks locally flat,

${\displaystyle g_{\alpha \beta }=e_{\alpha }^{I}e_{\beta }^{J}\eta _{IJ}}$ 

where ${\displaystyle \eta _{IJ}={\text{diag}}(-1,1,1,1)}$  is the Minkowski metric. The tetrads encode the information about the space-time metric and will be taken as one of the independent variables in the action principle.

Now if one is going to operate on objects that have internal indices one needs to introduce an appropriate derivative (covariant derivative). We introduce an arbitrary covariant derivative via

${\displaystyle {\mathcal {D}}_{\alpha }V_{I}=\partial _{\alpha }V_{I}+{\omega _{\alpha I}}^{J}V_{J}.}$ 

Where ${\displaystyle {\omega _{\alpha I}}^{J}}$  is a spin (Lorentz) connection one-form (the derivative annihilates the Minkowski metric ${\displaystyle \eta _{IJ}}$ ). We define a curvature via

${\displaystyle {\Omega _{\alpha \beta I}}^{J}V_{J}=({\mathcal {D}}_{\alpha }{\mathcal {D}}_{\beta }-{\mathcal {D}}_{\beta }{\mathcal {D}}_{\alpha })V_{I}}$ 

We obtain

${\displaystyle {\Omega _{\alpha \beta }}^{IJ}=2\partial _{[\alpha }{\omega _{\beta ]}}^{IJ}+2{\omega _{[\alpha }}^{IK}{\omega _{\beta ]K}}^{J}}$ .

We introduce the covariant derivative which annihilates the tetrad,

${\displaystyle \nabla _{\alpha }e_{\beta }^{I}=0}$ .

The connection is completely determined by the tetrad. The action of this on the generalized tensor ${\displaystyle V_{\beta }^{I}}$  is given by

${\displaystyle \nabla _{\alpha }V_{\beta }^{I}=\partial _{\alpha }V_{\beta }^{I}-\Gamma _{\alpha \beta }^{\gamma }V_{\gamma }^{I}+{\omega _{\alpha J}}^{I}V_{\beta }^{J}.}$ 

We define a curvature ${\displaystyle {R_{\alpha \beta }}^{IJ}}$  by

${\displaystyle {R_{\alpha \beta I}}^{J}V_{J}=(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha })V_{I}.}$ 

This is easily related to the usual curvature defined by

${\displaystyle {R_{\alpha \beta \gamma }}^{\delta }V_{\delta }=(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha })V_{\gamma }}$ 

via substituting ${\displaystyle V_{\gamma }=V_{I}e_{\gamma }^{I}}$  into this expression (see below for details). One obtains,

${\displaystyle {R_{\alpha \beta \gamma }}^{\delta }=e_{\gamma }^{I}{R_{\alpha \beta I}}^{J}e_{J}^{\delta },\quad R_{\alpha \beta }={R_{\alpha \gamma I}}^{J}e_{\beta }^{I}e_{J}^{\gamma },R={R_{\alpha \beta }}^{IJ}e_{I}^{\alpha }e_{J}^{\beta }}$ 

for the Riemann tensor, Ricci tensor and Ricci scalar respectively.

The Ricci scalar of this curvature can be expressed as ${\displaystyle e_{I}^{\alpha }e_{J}^{\beta }{\Omega _{\alpha \beta }}^{IJ}.}$  The action can be written

${\displaystyle S_{H-P}=\int d^{4}x\;e\;e_{I}^{\alpha }e_{J}^{\beta }{\Omega _{\alpha \beta }}^{IJ}}$ 

where ${\displaystyle e={\sqrt {-g}}}$  but now ${\displaystyle g}$  is a function of the frame field.

We will derive the Einstein equations by varying this action with respect to the tetrad and spin connection as independent quantities.

As a shortcut to performing the calculation we introduce a connection compatible with the tetrad, ${\displaystyle \nabla _{\alpha }e_{\beta }^{I}=0.}$ ^[2] The connection associated with this covariant derivative is completely determined by the tetrad. The difference between the two connections we have introduced is a field ${\displaystyle {C_{\alpha I}}^{J}}$  defined by

${\displaystyle {C_{\alpha I}}^{J}V_{J}=\left(D_{\alpha }-\nabla _{\alpha }\right)V_{I}.}$ 

We can compute the difference between the curvatures of these two covariant derivatives (see below for details),

${\displaystyle {\Omega _{\alpha \beta }}^{IJ}-{R_{\alpha \beta }}^{IJ}=\nabla _{[\alpha }{C_{\beta ]}}^{IJ}+{C_{[\alpha }}^{IM}{C_{\beta ]M}}^{J}}$ 

The reason for this intermediate calculation is that it is easier to compute the variation by reexpressing the action in terms of ${\displaystyle \nabla }$  and ${\displaystyle {C_{\alpha }}^{IJ}}$  and noting that the variation with respect to ${\displaystyle {\omega _{\alpha }}^{IJ}}$  is the same as the variation with respect to ${\displaystyle {C_{\alpha }}^{IJ}}$  (when keeping the tetrad fixed). The action becomes

${\displaystyle S_{H-P}=\int d^{4}x\;e\;e_{I}^{\alpha }e_{J}^{\beta }\left({R_{\alpha \beta }}^{IJ}+\nabla _{[\alpha }{C_{\beta ]}}^{IJ}+{C_{[\alpha }}^{IM}{C_{\beta ]M}}^{J}\right)}$ 

We first vary with respect to ${\displaystyle {C_{\alpha }}^{IJ}}$ . The first term does not depend on ${\displaystyle {C_{\alpha }}^{IJ}}$  so it does not contribute. The second term is a total derivative. The last term yields

${\displaystyle e_{M}^{[a}e_{N}^{b]}\delta _{[I}^{M}\delta _{J]}^{K}{C_{bK}}^{N}=0.}$ 

We show below that this implies that ${\displaystyle {C_{\alpha }}^{IJ}=0}$  as the prefactor ${\displaystyle e_{M}^{[a}e_{N}^{b]}\delta _{[I}^{M}\delta _{J]}^{K}}$  is non-degenerate. This tells us that ${\displaystyle \nabla }$  coincides with ${\displaystyle D}$  when acting on objects with only internal indices. Thus the connection ${\displaystyle D}$  is completely determined by the tetrad and ${\displaystyle \Omega }$  coincides with ${\displaystyle R}$ . To compute the variation with respect to the tetrad we need the variation of ${\displaystyle e=\det e_{\alpha }^{I}}$ . From the standard formula

${\displaystyle \delta \det(a)=\det(a)\left(a^{-1}\right)_{ji}\delta a_{ij}}$ 

we have ${\displaystyle \delta e=ee_{I}^{\alpha }\delta e_{\alpha }^{I}}$ . Or upon using ${\displaystyle \delta \left(e_{\alpha }^{I}e_{I}^{\alpha }\right)=0}$ , this becomes ${\displaystyle \delta e=-ee_{\alpha }^{I}\delta e_{I}^{\alpha }}$ . We compute the second equation by varying with respect to the tetrad,

${\displaystyle {\begin{aligned}\delta S_{H-P}&=\int d^{4}x\;e\left(\left(\delta e_{I}^{\alpha }\right)e_{J}^{\beta }{\Omega _{\alpha \beta }}^{IJ}+e_{I}^{\alpha }\left(\delta e_{J}^{\beta }\right){\Omega _{\alpha \beta }}^{IJ}-e_{\gamma }^{K}\left(\delta e_{K}^{\gamma }\right)e_{I}^{\alpha }e_{J}^{\beta }{\Omega _{\alpha \beta }}^{IJ}\right)\\&=2\int d^{4}x\;e\left(e_{J}^{\beta }{\Omega _{\alpha \beta }}^{IJ}-{1 \over 2}e_{M}^{\gamma }e_{N}^{\delta }e_{\alpha }^{I}{\Omega _{\gamma \delta }}^{MN}\right)\left(\delta e_{I}^{\alpha }\right)\end{aligned}}}$ 

One gets, after substituting ${\displaystyle {\Omega _{\alpha \beta }}^{IJ}}$  for ${\displaystyle {R_{\alpha \beta }}^{IJ}}$  as given by the previous equation of motion,

${\displaystyle e_{J}^{\gamma }{R_{\alpha \gamma }}^{IJ}-{1 \over 2}{R_{\gamma \delta }}^{MN}e_{M}^{\gamma }e_{N}^{\delta }e_{\alpha }^{I}=0}$ 

which, after multiplication by ${\displaystyle e_{I\beta }}$  just tells us that the Einstein tensor ${\displaystyle R_{\alpha \beta }-{\tfrac {1}{2}}Rg_{\alpha \beta }}$  of the metric defined by the tetrads vanishes. We have therefore proved that the Palatini variation of the action in tetradic form yields the usual Einstein equations.

We change the action by adding a term

${\displaystyle -{1 \over 2\gamma }ee_{I}^{\alpha }e_{J}^{\beta }{\Omega _{\alpha \beta }}^{MN}[\omega ]{\epsilon ^{IJ}}_{MN}}$ 

This modifies the Palatini action to

${\displaystyle S=\int d^{4}x\;e\;e_{I}^{\alpha }e_{J}^{\beta }{P^{IJ}}_{MN}{\Omega _{\alpha \beta }}^{MN}}$ 

where

${\displaystyle {P^{IJ}}_{MN}=\delta _{M}^{[I}\delta _{N}^{J]}-{1 \over 2\gamma }{\epsilon ^{IJ}}_{MN}.}$ 

This action given above is the Holst action, introduced by Holst^[3] and ${\displaystyle \gamma }$  is the Barbero-Immirzi parameter whose role was recognized by Barbero^[4] and Immirizi.^[5] The self dual formulation corresponds to the choice ${\displaystyle \gamma =-i}$ .

It is easy to show these actions give the same equations. However, the case corresponding to ${\displaystyle \gamma =\pm i}$  must be done separately (see article self-dual Palatini action). Assume ${\displaystyle \gamma \not =\pm i}$ , then ${\displaystyle {P^{IJ}}_{MN}}$  has an inverse given by

${\displaystyle {(P^{-1})_{IJ}}^{MN}={\frac {\gamma ^{2}}{\gamma ^{2}+1}}\left(\delta _{I}^{[M}\delta _{J}^{N]}+{\frac {1}{2\gamma }}{\epsilon _{IJ}}^{MN}\right).}$ 

(note this diverges for ${\displaystyle \gamma =\pm i}$ ). As this inverse exists the generalization of the prefactor ${\displaystyle e_{M}^{[a}e_{N}^{b]}\delta _{[I}^{M}\delta _{J]}^{K}}$  will also be non-degenerate and as such equivalent conditions are obtained from variation with respect to the connection. We again obtain ${\displaystyle {C_{\alpha }}^{IJ}=0}$ . While variation with respect to the tetrad yields Einstein's equation plus an additional term. However, this extra term vanishes by symmetries of the Riemann tensor.

Relating usual curvature to the mixed index curvature edit

The usual Riemann curvature tensor ${\displaystyle {R_{\alpha \beta \gamma }}^{\delta }}$  is defined by

${\displaystyle {R_{\alpha \beta \gamma }}^{\delta }V_{\delta }=\left(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha }\right)V_{\gamma }.}$ 

To find the relation to the mixed index curvature tensor let us substitute ${\displaystyle V_{\gamma }=e_{\gamma }^{I}V_{I}}$ 

${\displaystyle {\begin{aligned}{R_{\alpha \beta \gamma }}^{\delta }V_{\delta }&=\left(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha }\right)V_{\gamma }\\&=\left(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha }\right)\left(e_{\gamma }^{I}V_{I}\right)\\&=e_{\gamma }^{I}\left(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha }\right)V_{I}\\&=e_{\gamma }^{I}{R_{\alpha \beta I}}^{J}e_{J}^{\delta }V_{\delta }\end{aligned}}}$ 

where we have used ${\displaystyle \nabla _{\alpha }e_{\beta }^{I}=0}$ . Since this is true for all ${\displaystyle V_{\delta }}$  we obtain

${\displaystyle {R_{\alpha \beta \gamma }}^{\delta }=e_{\gamma }^{I}{R_{\alpha \beta I}}^{J}e_{J}^{\delta }}$ .

Using this expression we find

${\displaystyle R_{\alpha \beta }={R_{\alpha \gamma \beta }}^{\gamma }={R_{\alpha \gamma I}}^{J}e_{\beta }^{I}e_{J}^{\gamma }.}$ 

Contracting over ${\displaystyle \alpha }$  and ${\displaystyle \beta }$  allows us write the Ricci scalar

${\displaystyle R={R_{\alpha \beta }}^{IJ}e_{I}^{\alpha }e_{J}^{\beta }.}$ 

Difference between curvatures edit

The derivative defined by ${\displaystyle D_{\alpha }V_{I}}$  only knows how to act on internal indices. However, we find it convenient to consider a torsion-free extension to spacetime indices. All calculations will be independent of this choice of extension. Applying ${\displaystyle {\mathcal {D}}_{a}}$  twice on ${\displaystyle V_{I}}$ ,

${\displaystyle {\mathcal {D}}_{\alpha }{\mathcal {D}}_{\beta }V_{I}={\mathcal {D}}_{\alpha }(\nabla _{\beta }V_{I}+{C_{\beta I}}^{J}V_{J})=\nabla _{\alpha }\left(\nabla _{\beta }V_{I}+{C_{\beta I}}^{J}V_{J}\right)+{C_{\alpha I}}^{K}\left(\nabla _{b}V_{K}+{C_{\beta K}}^{J}V_{J}\right)+{\overline {\Gamma }}_{\alpha \beta }^{\gamma }\left(\nabla _{\gamma }V_{I}+{C_{\gamma I}}^{J}V_{J}\right)}$ 

where ${\displaystyle {\overline {\Gamma }}_{\alpha \beta }^{\gamma }}$  is unimportant, we need only note that it is symmetric in ${\displaystyle \alpha }$  and ${\displaystyle \beta }$  as it is torsion-free. Then

${\displaystyle {\begin{aligned}{\Omega _{\alpha \beta I}}^{J}V_{J}&=\left({\mathcal {D}}_{\alpha }{\mathcal {D}}_{\beta }-{\mathcal {D}}_{\beta }{\mathcal {D}}_{\alpha }\right)V_{I}\\&=\left(\nabla _{\alpha }\nabla _{\beta }-\nabla _{\beta }\nabla _{\alpha }\right)V_{I}+\nabla _{\alpha }\left({C_{\beta I}}^{J}V_{J}\right)-\nabla _{\beta }\left({C_{\alpha I}}^{J}V_{J}\right)+{C_{\alpha I}}^{K}\nabla _{\beta }V_{K}-{C_{\beta I}}^{K}\nabla _{\alpha }V_{K}+{C_{\alpha I}}^{K}{C_{\beta K}}^{J}V_{J}-{C_{\beta I}}^{K}{C_{\alpha K}}^{J}V_{J}\\&={R_{\alpha \beta I}}^{J}V_{J}+\left(\nabla _{\alpha }{C_{\beta I}}^{J}-\nabla _{\beta }{C_{\alpha I}}^{J}+{C_{\alpha I}}^{K}{C_{\beta K}}^{J}-{C_{\beta _{I}}}^{K}{C_{\alpha K}}^{J}\right)V_{J}\end{aligned}}}$ 

Hence:

${\displaystyle {\Omega _{ab}}^{IJ}-{R_{ab}}^{IJ}=2\nabla _{[a}{C_{b]}}^{IJ}+2{C_{[a}}^{IK}{C_{b]K}}^{J}}$ 

Varying the action with respect to the field ${\displaystyle {C_{\alpha }}^{IJ}}$  edit

We would expect ${\displaystyle \nabla _{a}}$  to also annihilate the Minkowski metric ${\displaystyle \eta _{IJ}=e_{\beta I}e_{J}^{\beta }}$ . If we also assume that the covariant derivative ${\displaystyle {\mathcal {D}}_{\alpha }}$  annihilates the Minkowski metric (then said to be torsion-free) we have,

${\displaystyle 0=({\mathcal {D}}_{\alpha }-\nabla _{\alpha })\eta _{IJ}={C_{\alpha I}}^{K}\eta _{KJ}+{C_{aJ}}^{K}\eta _{IK}=C_{\alpha IJ}+C_{\alpha JI}.}$ 

Implying

${\displaystyle C_{\alpha IJ}=C_{\alpha [IJ]}.}$ 

From the last term of the action we have from varying with respect to ${\displaystyle {C_{\alpha I}}^{J},}$ 

${\displaystyle {\begin{aligned}\delta S_{EH}&=\delta \int d^{4}x\;e\;e_{M}^{\gamma }e_{N}^{\beta }{C_{[\gamma }}^{MK}{C_{\beta ]K}}^{N}\\&=\delta \int d^{4}x\;e\;e_{M}^{[\gamma }e_{N}^{\beta ]}{C_{\gamma }}^{MK}{C_{\beta K}}^{N}\\&=\delta \int d^{4}x\;e\;e^{M[\gamma }e_{N}^{\beta ]}{C_{\gamma M}}^{K}{C_{\beta K}}^{N}\\&=\int d^{4}x\;ee^{M[\gamma }e_{N}^{\beta ]}\left(\delta _{\gamma }^{\alpha }\delta _{M}^{I}\delta _{J}^{K}{C_{\beta K}}^{N}+{C_{\gamma M}}^{K}\delta _{\beta }^{\alpha }\delta _{K}^{I}\delta _{J}^{N}\right)\delta {C_{\alpha I}}^{J}\\&=\int d^{4}x\;e\left(e^{I[\alpha }e_{N}^{\beta ]}{C_{\beta J}}^{N}+e^{M[\beta }e_{J}^{\alpha ]}{C_{\beta M}}^{I}\right)\delta {C_{\alpha I}}^{J}\end{aligned}}}$ 

or

${\displaystyle e_{I}^{[\alpha }e_{K}^{\beta ]}{C_{\beta J}}^{K}+e^{K[\beta }e_{J}^{\alpha ]}C_{\beta KI}=0}$ 

or

${\displaystyle {C_{\beta I}}^{K}e_{K}^{[\alpha }e_{J}^{\beta ]}+{C_{\beta J}}^{K}e_{I}^{[\alpha }e_{K}^{\beta ]}=0.}$ 

where we have used ${\displaystyle C_{\beta KI}=-C_{\beta IK}}$ . This can be written more compactly as

${\displaystyle e_{M}^{[\alpha }e_{N}^{\beta ]}\delta _{[I}^{M}\delta _{J]}^{K}{C_{\beta K}}^{N}=0.}$ 

Vanishing of ${\displaystyle {C_{\alpha }}^{IJ}}$  edit

We will show following the reference "Geometrodynamics vs. Connection Dynamics"^[6] that

${\displaystyle {C_{\beta I}}^{K}e_{K}^{[\alpha }e_{J}^{\beta ]}+{C_{\beta J}}^{K}e_{I}^{[\alpha }e_{K}^{\beta ]}=0\quad Eq.1}$ 

implies ${\displaystyle {C_{\alpha I}}^{J}=0.}$  First we define the spacetime tensor field by

${\displaystyle S_{\alpha \beta \gamma }:=C_{\alpha IJ}e_{\beta }^{I}e_{\gamma }^{J}.}$ 

Then the condition ${\displaystyle C_{\alpha IJ}=C_{\alpha [IJ]}}$  is equivalent to ${\displaystyle S_{\alpha \beta \gamma }=S_{\alpha [\beta \gamma ]}}$ . Contracting Eq. 1 with ${\displaystyle e_{\alpha }^{I}e_{\gamma }^{J}}$  one calculates that

${\displaystyle {C_{\beta J}}^{I}e_{\gamma }^{J}e_{I}^{\beta }=0.}$ 

As ${\displaystyle {S_{\alpha \beta }}^{\gamma }={C_{\alpha I}}^{J}e_{\beta }^{I}e_{J}^{\gamma },}$  we have ${\displaystyle {S_{\beta \gamma }}^{\beta }=0.}$  We write it as

${\displaystyle ({C_{\beta I}}^{J}e_{J}^{\beta })e_{\gamma }^{I}=0,}$ 

and as ${\displaystyle e_{\alpha }^{I}}$  are invertible this implies

${\displaystyle {C_{\beta I}}^{J}e_{J}^{\beta }=0.}$ 

Thus the terms ${\displaystyle {C_{\beta I}}^{K}e_{K}^{\beta }e_{J}^{\alpha },}$  and ${\displaystyle {C_{\beta J}}^{K}e_{I}^{\alpha }e_{K}^{\beta }}$  of Eq. 1 both vanish and Eq. 1 reduces to

${\displaystyle {C_{\beta I}}^{K}e_{K}^{\alpha }e_{J}^{\beta }-{C_{\beta J}}^{K}e_{I}^{\beta }e_{K}^{\alpha }=0.}$ 

If we now contract this with ${\displaystyle e_{\gamma }^{I}e_{\delta }^{J}}$ , we get

${\displaystyle {\begin{aligned}0&=\left({C_{\beta I}}^{K}e_{K}^{\alpha }e_{J}^{\beta }-{C_{\beta J}}^{K}e_{I}^{\beta }e_{K}^{\alpha }\right)e_{\gamma }^{I}e_{\delta }^{J}\\&={C_{\beta I}}^{K}e_{K}^{\alpha }e_{\gamma }^{I}\delta _{\delta }^{\beta }-{C_{\beta J}}^{K}\delta _{\gamma }^{\beta }e_{K}^{\alpha }e_{\delta }^{J}\\&={C_{\delta I}}^{K}e_{\gamma }^{I}e_{K}^{\alpha }-{C_{\gamma J}}^{K}e_{\delta }^{J}e_{K}^{\alpha }\end{aligned}}}$ 

or

${\displaystyle {S_{\gamma \delta }}^{\alpha }={S_{(\gamma \delta )}}^{\alpha }.}$ 

Since we have ${\displaystyle S_{\alpha \beta \gamma }=S_{\alpha [\beta \gamma ]}}$  and ${\displaystyle S_{\alpha \beta \gamma }=S_{(\alpha \beta )\gamma }}$ , we can successively interchange the first two and then last two indices with appropriate sign change each time to obtain,

${\displaystyle S_{\alpha \beta \gamma }=S_{\beta \alpha \gamma }=-S_{\beta \gamma \alpha }=-S_{\gamma \beta \alpha }=S_{\gamma \alpha \beta }=S_{\alpha \gamma \beta }=-S_{\alpha \beta \gamma }}$ 

Implying

${\displaystyle S_{\alpha \beta \gamma }=0,}$ 

or

${\displaystyle C_{\alpha IJ}e_{\beta }^{I}e_{\gamma }^{J}=0,}$ 

and since the ${\displaystyle e_{\alpha }^{I}}$  are invertible, we get ${\displaystyle C_{\alpha IJ}=0}$ . This is the desired result.

• Lanczos tensor
• Weyl tensor