# Self-dual Palatini action

Ashtekar variables, which were a new canonical formalism of general relativity, raised new hopes for the canonical quantization of general relativity and eventually led to loop quantum gravity. Smolin and others independently discovered that there exists in fact a Lagrangian formulation of the theory by considering the self-dual formulation of the Tetradic Palatini action principle of general relativity.[1][2][3] These proofs were given in terms of spinors. A puerly tensorial proof of the new variables in terms of triads was given by Goldberg[4] and in terms of tetrads by Henneaux et al.[5] Here we in particular fill in details of the proof of results for self-dual variables not given in text books.

## The Palatini action

Main article: spin connection

The Palatini action for general relativity has as its independent variables the tetrad $e_I^\alpha$ and a spin connection $\omega_\alpha^{\;\; IJ}$. Much more details and derivations can be found in the article tetradic Palatini action. The spin connection defines a covariant derivative $D_\alpha$. The space-time metric is recovered from the tetrad by the formula $g_{\alpha \beta} = e^I_\alpha e^J_\beta \eta_{IJ}.$ We define the `curvature' by

$\Omega_{\alpha \beta}^{\;\;\;\; IJ} = \partial_\alpha \omega_\beta^{\;\; IJ} - \partial_\beta \omega_\alpha^{\;\; IJ} + \omega_\alpha^{IK} \omega_{\beta K}^{\;\;\;\; J} - \omega_\beta^{IK} \omega_{\alpha K}^{\;\;\;\; J} \;\;\;\;\; Eq(1).$

The Ricci scalar of this curvature is given by $e_I^\alpha e_J^\beta \Omega_{\alpha \beta}^{\;\;\;\; IJ}$. The Palatini action for general relativity reads

$S = \int d^4 x \; e \;e_I^\alpha e_J^\beta \; \Omega_{\alpha \beta}^{\;\;\;\; IJ} [\omega]$

where $e = \sqrt{-g}$. Variation with respect to the spin connection $\omega_{\alpha \beta}^{\;\;\; IJ}$ implies that the spin connection is determined by the compatibility condition $D_\alpha e_I^\beta = 0$ and hence becomes the usual covariant derivative $\nabla_\alpha$. Hence the connection becomes a function of the tetrads and the curvature $\Omega_{\alpha \beta}^{\;\;\;\; IJ}$ is replaced by the curvature $R_{\alpha \beta}^{\;\;\;\; IJ}$ of $\nabla_\alpha$. Then $e_I^\alpha e_J^\beta R_{\alpha \beta}^{\;\;\;\; IJ}$ is the actual Ricci scalar $R$. Variation with respect to the tetrad gives Einsteins equation $R_{\alpha \beta} - {1 \over 2} g_{\alpha \beta} R = 0$.

## Self-dual variables

### (Anti-)self-dual parts of a tensor

We will need what is called the totally antisymmetry tensor or Levi-Civita symbol, $\epsilon_{IJKL}$. This is equal to either +1 or -1 depending on whether $IJKL$ is either an even or odd permutation of $0123$, respectively, and zero if any two indices take the same value. The internal indices of $\epsilon_{IJKL}$ are raised with the Minkowski metric $\eta^{IJ}$.

Now, given any anti-symmetric tensor $T^{IJ}$, we define its dual as

$*T^{IJ} = {1 \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} T^{ KL}.$

The self-dual part of any tensor $T^{IJ}$ is defined as

$\;^+T^{IJ} := {1 \over 2} \Big( T^{IJ} - {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} T^{KL} \Big)$

with the anti-self-dual part defined as

$\;^-T^{IJ} := {1 \over 2} \Big( T^{IJ} + {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} T^{KL} \Big)$

(the appearance of the imaginary unit $i$ is related to the Minkowski signature as we will see below).

### Tensor decomposition

Now given any anti-symmetric tensor $T^{IJ}$, we can decompose it as

$T^{IJ} = {1 \over 2} (T^{IJ} - {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\;\; IJ} T^{KL}) + {1 \over 2} (T^{IJ} + {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\;\; IJ} T^{KL}) =\;^+T^{IJ} +\;^-T^{IJ}$

where $\;^+T^{IJ}$ and $\;^-T^{IJ}$ are the self-dual and anti-self-dual parts of $T^{IJ}$ respectively. Define the projector onto (anti-)self-dual part of any tensor as

$P^{(\pm)} = {1 \over 2} (1 \mp i *).$

The meaning of these projectors can be made explicit. Let us concentrate of $P^+$,

$(P^+ T)^{IJ} = ({1 \over 2} (1 - i *) T)^{IJ} = {1 \over 2} (\delta^I_{\; K} \delta^J_{\;\; L} - i {1 \over 2} \epsilon_{KL}^{\;\;\;\;\;\;\; IJ}) T^{KL} = {1 \over 2} (T^{IJ} - {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\;\; IJ} T^{KL}) = \;^+ T^{IJ}.$

Then

$\;^\pm T^{IJ} = (P^{(\pm)} T)^{IJ}.$

### The Lie bracket

An important object is the Lie bracket defined by

$[F , G]^{IJ} := F^{IK} G_K^{\;\; J} - G^{IK} F_K^{\;\; J},$

it appears in the curvature tensor (see the last two terms of $Eq.1$), it also defines the algebraic structure. We have the results (proved below):

$P^{(\pm)} [F , G]^{IJ} = [P^{(\pm)} F, G]^{IJ} = [F , P^{(\pm)} G]^{IJ} = [P^{(\pm)} F , P^{(\pm)} G]^{IJ} \;\;\;\;\; Eq.2$

and

$[F , G] = [P^+ F , P^+ G] + [P^- F , P^- G].$

That is the Lie bracket, which defines an algebra, decomposes into two separate independent parts. We write

$so(1,3)_\mathbb{C} = so (1,3)_\mathbb{C}^+ + so (1,3)_\mathbb{C}^-$

where $so (1,3)_\mathbb{C}^\pm$ contains only the self-dual (anti-self-dual) elements of $so(1,3)_\mathbb{C}$.

## The Self-dual Palatini action

We define the self-dual part, $A_\alpha^{\;\;\; IJ}$, of the connection $\omega_\alpha^{\;\; IJ}$ as

$A_\alpha^{\;\;\; IJ} = {1 \over 2} \big( \omega_\alpha^{\;\; IJ} - {i \over 2} \epsilon_{KL}^{\;\;\;\;\; IJ} \omega^{KL} \big).$

which can be more compactly written

$A_\alpha^{\;\;\; IJ} = (P^+ \omega \big)^{IJ}.$

Define $F_{\alpha \beta}^{\;\; IJ}$ as the curvature of the self-dual connection

$F_{\alpha \beta}^{\;\; IJ } = \partial_\alpha A_\beta^{\;\; IJ} - \partial_\beta A_\alpha^{\;\; IJ} + A_\alpha^{\;\; IK} A_{\beta K}^{\;\;\;\;\; J} - A_\beta^{IK} A_{\alpha K}^{\;\;\;\;\; J}.$

Using $Eq.2$ it is easy to see that the curvature of the self-dual connection is the self-dual part of the curvature of the connection,

$F_{\alpha \beta}^{\;\;\;\; IJ } = \partial_\alpha (P^+ \omega_\beta)^{IJ} - \partial_\beta (P^+ \omega_\alpha)^{IJ} + [P^+ \omega_\alpha , P^+ \omega_\beta]^{IJ}$

$= (P^+ 2 \partial_{[\alpha} \omega_{\beta]})^{IJ} + (P^+ [\omega_\alpha , \omega_\beta])^{IJ}$

$= (P^+ \Omega_{\alpha \beta})^{IJ}.$

The self-dual action is

$S = \int d^4 x \; e \;e_I^\alpha e_J^\beta \; F_{\alpha \beta}^{\;\;\;\; IJ}.$

As the connection is complex we are dealing with complex general relativity and appropriate conditions must be specified to recover the real theory. One can repeat the same calculations done for the Palatini action but now with respect to the self-dual connection $A_\alpha^{\;\; IJ}$. Varying the tetrad field, one obtains a self-dual analog of Einstein's equation:

$\;^+R_{\alpha \beta} - {1 \over 2} g_{\alpha \beta} \;^+R = 0.$

That the curvature of the self-dual connection is the self-dual part of the curvature of the connection helps to simplify the 3+1 formalism (details of the decomposition into the 3+1 formalism are to be given in another article). The resulting Hamiltonian formalism resembles that of a Yang-Mills gauge theory (this does not happen with the 3+1 Palatini formalism which basically collapses down to the usual ADM formalism).

## Derivation of main results for self-dual variables

The results of calculations done here can be found in chapter 3 of notes Ashtekar Variables in Classical Relativity.[6] The method of proof follows that given in section II of The Ashtekar Hamiltonian for General Relativity.[7] We need to establish some results for (anti-)self-dual Lorentzian tensors.

### Identities for the totally anti-symmetric tensor

Since $\eta_{IJ}$ has signature $(-,+,+,+)$, it follows that

$\epsilon^{IJKL} = - \epsilon_{IJKL} .$

to see this consider,

$\epsilon^{0123} = \eta^{0I} \eta^{1J} \eta^{2K} \eta^{3L} \epsilon_{IJKL}$ $= (-1) (+1) (+1) (+1) \epsilon_{0123} = - \epsilon_{0123} .$

With this definition one can obtain the following identities,

$\epsilon^{IJKO} \epsilon_{LMNO} = - 6 \delta^I_{[L} \delta^J_M \delta^K_{N]} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; Eq.3$

$\epsilon^{IJMN} \epsilon_{KLMN} = - 4 \delta^I_{[K} \delta^J_{L]} = - 2 (\delta^I_K \delta^J_L - \delta^I_L \delta^J_K) \;\;\;\; Eq.4$

(the square brackets denote anti-symmetrizing over the indices).

### Definition of self-dual tensor

It follows from $Eq.4$ that the square of the duality operator is minus the identity,

$** T^{IJ} = {1 \over 4} \epsilon_{KL}^{\;\;\;\;\;\; IJ} \epsilon_{MN}^{\;\;\;\;\;\;\; KL} T^{\;\; MN} = - T^{IJ}$

The minus sign here is due to the minus sign in $Eq.4$, which is in turn due to the Minkowski signature. Had we used Euclidean signature, i.e. $(+,+,+,+)$, instead there would have been a positive sign. We define $S^{IJ}$ to be self-dual if and only if

$*S^{IJ} = i S^{IJ} .$

(with Euclidean signature the self-duality condition would have been $*S^{IJ} = S^{IJ}$). Say $S^{IJ}$ is self-dual, write it as a real and imaginary part,

$S^{IJ} = {1 \over 2} T^{IJ} + i {1 \over 2} U^{IJ} .$

Write the self-dual condition in terms of $U$ and $V$,

$*(T^{IJ} + i U^{IJ}) = {1 \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} (T^{KL} + i U^{KL}) = i (T^{IJ} + i U^{IJ}) .$

Equating real parts we read off

$U^{IJ} = - {1 \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} T^{KL}$

and so

$S^{IJ} = {1 \over 2} (T^{IJ} - {i \over 2} \epsilon_{KL}^{\;\;\;\;\;\; IJ} T^{KL})$

where $T^{IJ}$ is the real part of $2 S^{IJ}$.

### Important lengthy calculation

The following lengthy calculation is important as all the other important formula can easily be derived from it. From the definition of the Lie bracket and with the use of $Eq.3$ we have

$*[F , *G]^{IJ} = {1 \over 2} \epsilon_{MN}^{\;\;\;\;\;\; IJ} (F^{MK} (*G)_K^{\;\;\; N} -(*G)^{MK} F_K^{\;\;\; N})$

$= {1 \over 2} \epsilon_{MN}^{\;\;\;\;\;\;\; IJ} (F^{MK} {1 \over 2} \epsilon_{OPK}^{\;\;\;\;\;\;\;\;\; N} G^{OP} - {1 \over 2} \epsilon_{OP}^{\;\;\;\;\;\; MK} G^{OP} F_K^{\;\;\; N})$

$= {1 \over 4} (\epsilon_{MN}^{\;\;\;\;\;\;\; IJ} \epsilon_{OP}^{\;\;\;\;\;\; KN} + \epsilon_{NM}^{\;\;\;\;\;\;\; IJ} \epsilon_{OP}^{\;\;\;\;\;\; NK}) F^M_{\;\;\; K} G^{OP}$

$= {1 \over 2} \epsilon_{MN}^{\;\;\;\;\;\;\; IJ} \epsilon_{OP}^{\;\;\;\;\;\; KN} F^M_{\;\;\; K} G^{OP}$

$= {1 \over 2} \epsilon^{MIJN} \epsilon_{OPKN} F_M^{\;\;\; K} G^{OP}$

$= - {1 \over 2} \epsilon^{KIJN} \epsilon_{OPMN} F^M_{\;\;\; K} G^{OP}$

$= {1 \over 2} (\delta^K_O \delta^I_P \delta^J_M + \delta^K_M \delta^I_O \delta^J_P + \delta^K_P \delta^I_M \delta^J_O - \delta^K_P \delta^I_O \delta^J_M - \delta^K_M \delta^I_P \delta^J_O - \delta^K_O \delta^I_M \delta^J_P) F^M_{\;\;\; K} G^{OP}$

$= {1 \over 2} (F^J_{\;\;\; K} G^{KI} + F^K_{\;\;\; K} G^{IJ} + F^I_{\;\; K} G^{JK} - F^J_{\;\;\; K} G^{IK} - F^K_{\;\;\; K} G^{JI} - F^I_{\;\; K} G^{KJ})$

$= - F^{IK} G_K^{\;\; \;J} + G^{IK} F_K^{\;\; J}$

$- [F , G]^{IJ}$

That gives the formula

$*[F,*G]^{IJ} = - [F , G]^{IJ} \;\;\;\;\;\; Eq.5.$

which is the starting point for everything else.

### Derivation of important results

First consider

$*[*F,G]^{IJ} = - *[G,*F]^{IJ} = + [G,F]^{IJ} = - [F , G]^{IJ} .$

where in the first step we have used the anti-symmetry of the Lie bracket to swap $*F$ and $G$, in the second step we used $Eq.5$ and in the last step we used the anti-symmetry of the Lie bracket again. Now using this we obtain

$* (- [F , G]^{IJ}) = *(*[*F,G]^{IJ}) = **[*F,G]^{IJ} = - [*F , G]^{IJ} .$

where we used $** = - 1$ in the third step. So we have then $* [F , G]^{IJ} = [*F , G]^{IJ}$. Similarly we have $* [F , G]^{IJ} = [F , *G]^{IJ}.$

Now if we took $*[F,G]^{IJ} = [*F,G]^{IJ}$ and simply replaced $G$ with $*G$ we would get $*[F,*G]^{IJ} = [*F,*G]^{IJ}$. Combining $- [F , G]^{IJ} = *[F,*G]^{IJ}$ ($Eq.5$) and $*[F,*G]^{IJ} = [*F,*G]^{IJ}$ we obtain

$- [F , G]^{IJ} = [*F,*G]^{IJ} .$

Summarising, we have

$*[F,*G]^{IJ} = - [F , G]^{IJ} = *[*F,G]^{IJ}$

$*[F , G]^{IJ} = [* F , G]^{IJ} = [F , * G]^{IJ} \;\;\;\;\; Eq.6$

$[* F , * G]^{IJ} = - [F , G]^{IJ} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; Eq.7$

Then

$(P^{(\pm)} [F, G])^{IJ} = {1 \over 2} ([F, G]^{IJ} \mp i * [F,G]^{IJ})$

$= {1 \over 2} ([F, G]^{IJ} + [\mp i * F, G]^{IJ})$

$= [P^{(\pm)} F, G]^{IJ} \;\;\;\;\;\;\;\;\;\;\;\; Eq.8$

where we used $Eq.6$ going from the first line to the second line. Similarly we have $(P^{(\pm)} [F , G])^{IJ} = [F , P^{(\pm)} G]^{IJ}$. Now consider $[P^+ F, P^- G]^{IJ}$,

$[P^+ F, P^- G]^{IJ} = {1 \over 4} [(1 - i *) F , (1 + i *) G]^{IJ}$

$= {1 \over 4} [F , G]^{IJ} - {1 \over 4} i [* F , G]^{IJ} + {1 \over 4} i [F , * G]^{IJ} + {1 \over 4} [* F , * G]^{IJ}$

$= {1 \over 4} [F , G]^{IJ} - {1 \over 4} i [* F , G]^{IJ} + {1 \over 4} i [* F , G]^{IJ} - {1 \over 4} [F , G]^{IJ}$

$= 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; Eq.9$

where we have used $Eq.6$ and $Eq.7$ in going from the second line to the third line. Similarly

$[P^- F, P^+ G]^{IJ} = 0 \;\;\;\;\;\;\;\;\;\;\;\;\; Eq.10.$

Starting with $Eq.8$ we have

$(P^{(\pm)} [F, G])^{IJ} = [P^{(\pm)} F, G]^{IJ} = [P^{(\pm)} F, P^{(\pm)} G + P^{(\mp)} G]^{IJ} = [P^{(\pm)} F, P^{(\pm)} G]^{IJ}$

where we have used that any $G$ can be written as a sum of its self-dual and anti-sef-dual parts, i.e. $G = P^{(\pm)} G + P^{(\mp)} G$, and $Eq.9 / Eq.10$.

### Summary of main results

Altogether we have,

$(P^{(\pm)} [F , G])^{IJ} = [P^{(\pm)} F, G]^{IJ} = [F , P^{(\pm)} G]^{IJ} = [P^{(\pm)} F , P^{(\pm)} G]^{IJ}$

which is our main result, already stated above as $Eq.2$. We also have that any bracket splits as

$[F , G]^{IJ} = [P^+ F + P^- F , P^+ G + P^- F]^{IJ}$

$= [P^+ F, P^+ G]^{IJ} + [P^- F , P^- G]^{IJ} .$

into a part that depends only on self-dual Lorentzian tensors and is itself the self-dual part of $[F , G]^{IJ}$, and a part that depends only on anti-self-dual Lorentzian tensors and is the anit-self-dual part of $[F , G]^{IJ}$.