It has long been known that general relativity can be formulated by applying a constraint on the BF theory. Plebanski successfully obtained the Riemannian action:
S [ B , ω , λ ] = 1 κ ∫ M [ ( ⋆ B I J + 1 γ B I J ) ∧ F I J ( ω ) + λ I J K L B I J ∧ B K L ] . S[B, \omega, \lambda] = \frac{1}{\kappa}\int_\mathcal{M}\left[\left(^\star\!B^{IJ} + \frac{1}{\gamma}B^{IJ}\right)\wedge F_{IJ}(\omega) + \lambda_{IJKL} B^{IJ}\wedge B^{KL}\right]. S [ B , ω , λ ] = κ 1 ∫ M [ ( ⋆ B I J + γ 1 B I J ) ∧ F I J ( ω ) + λ I J K L B I J ∧ B K L ] .
This can be thought of as a S p i n ( 4 ) \mathrm{Spin}(4) Spin ( 4 ) BF theory and a constraint imposed by the lagrangian multiplier λ \lambda λ . It depends on a s o ( 4 ) \mathfrak{so}(4) so ( 4 ) connection ω \omega ω and a Lie-algebra valued 2-form B B B .
The symmetry of B B B and ∧ \wedge ∧ imposes a constraint on λ \lambda λ as well. Since switching between the indices of B I J B^{IJ} B I J is antisymmetric and that the operation ∧ \wedge ∧ is symmetric, we have the following relation:
λ I J K L = − λ J I K L = − λ I J L K = λ K L I J . \lambda_{IJKL} = -\lambda_{JIKL} = -\lambda_{IJLK} = \lambda_{KLIJ}. λ I J K L = − λ J I K L = − λ I J L K = λ K L I J .
This restrains the dimension of λ \lambda λ down to 21. However, if λ I J K L ∝ ϵ I J K L \lambda_{IJKL}\propto \epsilon_{IJKL} λ I J K L ∝ ϵ I J K L the lagrangian multiplier will give ϵ I J K L B I J ∧ B K L \epsilon_{IJKL}B^{IJ}\wedge B^{KL} ϵ I J K L B I J ∧ B K L , which implies a degenerate metric. Therefore ϵ I J K L λ I J K L = 0 \epsilon^{IJKL}\lambda_{IJKL} = 0 ϵ I J K L λ I J K L = 0 , and that d i m λ = 20 \mathrm{dim}\lambda = 20 dim λ = 20 .
To actually calculate the constraint given by the lagrangian multiplier, we introduce
λ I J K L = Λ I J K L − 1 4 ! Λ M N O P ϵ M N O P ϵ I J K L . \lambda_{IJKL} = \Lambda_{IJKL} - \frac{1}{4!}\Lambda_{MNOP}\epsilon^{MNOP}\epsilon_{IJKL}. λ I J K L = Λ I J K L − 4 ! 1 Λ MNOP ϵ MNOP ϵ I J K L .
We make this substitution because if we assume Λ \Lambda Λ to be the unconstrained lagrangian multiplier and
Λ [ I J K L ] = c ⋅ ϵ I J K L \Lambda_{[IJKL]} = c\cdot\epsilon_{IJKL} Λ [ I J K L ] = c ⋅ ϵ I J K L
we can multiply both sides with ϵ \epsilon ϵ and have
Λ [ I J K L ] ϵ I J K L = c ⋅ ϵ 2 ⟹ c = 1 4 ! Λ [ M N O P ] ϵ M N O P . \Lambda_{[IJKL]}\epsilon^{IJKL} = c\cdot\epsilon^2\implies c = \frac{1}{4!}\Lambda_{[MNOP]}\epsilon^{MNOP}. Λ [ I J K L ] ϵ I J K L = c ⋅ ϵ 2 ⟹ c = 4 ! 1 Λ [ MNOP ] ϵ MNOP .
This gives the derivative of the action with respect to Λ \Lambda Λ :
δ S δ Λ I J K L = B I J ∧ B K L − 1 4 ! ϵ I J K L ϵ M N O P B M N ∧ B O P = 0. \frac{\delta S}{\delta \Lambda_{IJKL}} = B^{IJ}\wedge B^{KL} - \frac{1}{4!}\epsilon^{IJKL}\epsilon_{MNOP}B^{MN}\wedge B^{OP} = 0. δ Λ I J K L δ S = B I J ∧ B K L − 4 ! 1 ϵ I J K L ϵ MNOP B MN ∧ B OP = 0.
which implies
ϵ μ ν ρ σ B μ ν I J B ρ σ K L = 1 4 ! ϵ I J K L ϵ M N O P B μ ν M N B ρ σ O P ϵ μ ν ρ σ . \epsilon^{\mu\nu\rho\sigma} B^{IJ}_{\mu\nu} B^{KL}_{\rho\sigma} = \frac{1}{4!} \epsilon^{IJKL} \epsilon_{MNOP} B^{MN}_{\mu\nu} B^{OP}_{\rho\sigma} \epsilon^{\mu\nu\rho\sigma}. ϵ μν ρ σ B μν I J B ρ σ K L = 4 ! 1 ϵ I J K L ϵ MNOP B μν MN B ρ σ OP ϵ μν ρ σ .
This equation constraints the degrees of freedom of B B B , limiting its dimension down to 16 = 4 × 4 16 = 4\times 4 16 = 4 × 4 . So if we define:
e : = 1 4 ! ϵ M N O P B μ ν M N B ρ σ O P ϵ μ ν ρ σ e := \frac{1}{4!} \epsilon_{MNOP} B^{MN}_{\mu\nu} B^{OP}_{\rho\sigma} \epsilon^{\mu\nu\rho\sigma} e := 4 ! 1 ϵ MNOP B μν MN B ρ σ OP ϵ μν ρ σ
It behaves exactly as the local tetrad. The solution of B B B would be
B = ± ⋆ ( e ∧ e ) and ± e ∧ e . B = \pm ^\star(e\wedge e) \quad\text{and}\quad \pm e\wedge e. B = ± ⋆ ( e ∧ e ) and ± e ∧ e .
Here e e e is the new variable, and the term λ B ∧ B \lambda B\wedge B λ B ∧ B in the action is simply the constraint. So if we put B = ± ⋆ ( e ∧ e ) B = \pm^\star(e\wedge e) B = ± ⋆ ( e ∧ e ) back into the action S S S , we obtain
S [ e , ω ] = 1 κ ∫ M Tr [ ( ⋆ ( e ∧ e ) + 1 γ e ∧ e ) ∧ F ( ω ) ] . S[e, \omega] = \frac{1}{\kappa}\int_\mathcal{M} \operatorname{Tr}\left[\left(^\star(e\wedge e) + \frac{1}{\gamma}e\wedge e\right) \wedge F(\omega)\right]. S [ e , ω ] = κ 1 ∫ M Tr [ ( ⋆ ( e ∧ e ) + γ 1 e ∧ e ) ∧ F ( ω ) ] .
Canonical Analysis
We perform the 3+1 decomposition. We use labels a , b = 1 , 2 , 3 a,b = 1,2,3 a , b = 1 , 2 , 3 as spacial indices on Σ t \Sigma_t Σ t . Introduce
Π a b I J = ⋆ B a b I J + 1 γ B a b I J ⟹ pure spacial; lives on Σ t H a I J = ⋆ B 0 a I J + 1 γ B 0 a I J ⟹ space-time . \begin{aligned}
\Pi_{ab}^{IJ} &= ^\star B_{ab}^{IJ} + \frac{1}{\gamma}B_{ab}^{IJ}\implies \text{pure spacial; lives on }\Sigma_t\\
H_a^{IJ} &= ^\star B_{0a}^{IJ} + \frac{1}{\gamma}B_{0a}^{IJ}\implies \text{space-time}.
\end{aligned} Π ab I J H a I J = ⋆ B ab I J + γ 1 B ab I J ⟹ pure spacial; lives on Σ t = ⋆ B 0 a I J + γ 1 B 0 a I J ⟹ space-time .
Also, we decompose the curvature form F μ ν F_{\mu\nu} F μν :
F a b = ∂ [ a ω b ] + [ ω a , ω b ] F 0 a = ∂ 0 ω a − ∂ a ω 0 + [ ω 0 , ω a ] = ω ˙ a − ( ∂ a ω 0 − [ ω a , ω 0 ] ) = : ω ˙ a − D a ω 0 . \begin{aligned}
F_{ab} &= \partial_{[a}\omega_{b]} + [\omega_a, \omega_b]\\
F_{0a} &= \partial_0\omega_a - \partial_a\omega_0 + [\omega_0, \omega_a] = \dot{\omega}_a - (\partial_a\omega_0 - [\omega_a, \omega_0]) =: \dot{\omega}_a - \mathcal{D}_a\omega_0.
\end{aligned} F ab F 0 a = ∂ [ a ω b ] + [ ω a , ω b ] = ∂ 0 ω a − ∂ a ω 0 + [ ω 0 , ω a ] = ω ˙ a − ( ∂ a ω 0 − [ ω a , ω 0 ]) =: ω ˙ a − D a ω 0 .
The action is originally:
S [ Π , H , ω , λ ] = 1 κ ∫ M [ ( ⋆ B I J + 1 γ B I J ) ∧ F I J + λ I J K L B I J ∧ B K L ] S[\Pi, H, \omega, \lambda] = \frac{1}{\kappa}\int_\mathcal{M}\left[\left(^\star B^{IJ} + \frac{1}{\gamma}B^{IJ}\right)\wedge F_{IJ} + \lambda_{IJKL} B^{IJ}\wedge B^{KL}\right] S [ Π , H , ω , λ ] = κ 1 ∫ M [ ( ⋆ B I J + γ 1 B I J ) ∧ F I J + λ I J K L B I J ∧ B K L ]
Substituting the 3+1 decomposed variables:
S = 1 κ ∫ M [ ( Π a b I J ∧ F I J , 0 a ) + ( H a I J ∧ F I J , a b ) + λ I J K L B 0 a I J ∧ B a b K L ] S = \frac{1}{\kappa}\int_\mathcal{M} \left[(\Pi_{ab}^{IJ}\wedge F_{IJ,0a}) + (H_{a}^{IJ}\wedge F_{IJ,ab}) + \lambda_{IJKL}B_{0a}^{IJ}\wedge B_{ab}^{KL}\right] S = κ 1 ∫ M [ ( Π ab I J ∧ F I J , 0 a ) + ( H a I J ∧ F I J , ab ) + λ I J K L B 0 a I J ∧ B ab K L ]
After some algebra, this leads to:
S = 1 κ ∫ M ( Π a b I J ∧ ω ˙ I J , c − Π a b I J ∧ D a ω 0 , I J + H a I J ∧ F I J , a b + λ I J K L { γ 2 γ 2 − 1 ( γ ⋆ H a I J − H a I J ) ∧ γ γ 2 − 1 ( γ Π b c K L − ⋆ Π b c K L ) } ) . \begin{aligned}
S
&= \frac{1}{\kappa}\int_\mathcal{M}\left(\Pi_{ab}^{IJ}\wedge \dot{\omega}_{IJ,c} - \Pi_{ab}^{IJ}\wedge \mathcal{D}_a \omega_{0,IJ} + H_a^{IJ}\wedge F_{IJ,ab} \right. \\
&+ \left.\lambda_{IJKL}\left\{\frac{\gamma^2}{\gamma^2-1}(\gamma^\star H_{a}^{IJ} - H_{a}^{IJ}) \wedge \frac{\gamma}{\gamma^2-1}(\gamma\Pi_{bc}^{KL} - ^\star\Pi_{bc}^{KL})\right\}\right).
\end{aligned} S = κ 1 ∫ M ( Π ab I J ∧ ω ˙ I J , c − Π ab I J ∧ D a ω 0 , I J + H a I J ∧ F I J , ab + λ I J K L { γ 2 − 1 γ 2 ( γ ⋆ H a I J − H a I J ) ∧ γ 2 − 1 γ ( γ Π b c K L − ⋆ Π b c K L ) } ) .
By integrating by parts, the second term gives ω 0 D ω ∧ Π \omega_0 \mathcal{D}_\omega\wedge \Pi ω 0 D ω ∧ Π , and if we redefine λ ′ = γ 2 ( γ 2 − 1 ) 2 λ \lambda' = \frac{\gamma^2}{(\gamma^2-1)^2}\lambda λ ′ = ( γ 2 − 1 ) 2 γ 2 λ , the action is:
S = ∫ d t ∫ Σ t Tr [ Π ∧ ω ˙ + ω 0 D ω ∧ Π + H ∧ F + λ ′ ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) ] . S = \int dt \int_{\Sigma_t} \operatorname{Tr} \left[ \Pi \wedge \dot{\omega} + \omega_0 \mathcal{D}_\omega \wedge \Pi + H \wedge F + \lambda' (\gamma ^\star H - H) \wedge (\gamma ^\star \Pi - \Pi) \right]. S = ∫ d t ∫ Σ t Tr [ Π ∧ ω ˙ + ω 0 D ω ∧ Π + H ∧ F + λ ′ ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) ] .
The canonical pair is ( Π a b I J , ω c K L ) (\Pi^{IJ}_{ab}, \omega^{KL}_c) ( Π ab I J , ω c K L ) . If we introduce
P ( γ , ± ) a b i : = 1 4 ϵ j k i Π a b j k ± 1 2 γ Π a b 0 i ± A c i : = 1 2 ϵ m n i ω c m n ± γ ω c 0 i H ( γ , ± ) a i : = 1 4 ϵ j k i H a j k ± 1 2 γ H a 0 i \begin{aligned}
\mathbb{P}(\gamma, \pm)^i_{ab} &:= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab} \\
{}^{\pm}\mathbb{A}_c^i &:= \frac{1}{2} \epsilon^i_{\ mn} \omega^{mn}_c \pm \gamma \omega^{0i}_c \\
\mathbb{H}(\gamma, \pm)^i_a &:= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a \pm \frac{1}{2\gamma} H^{0i}_a
\end{aligned} P ( γ , ± ) ab i ± A c i H ( γ , ± ) a i := 4 1 ϵ jk i Π ab jk ± 2 γ 1 Π ab 0 i := 2 1 ϵ mn i ω c mn ± γ ω c 0 i := 4 1 ϵ jk i H a jk ± 2 γ 1 H a 0 i
where A \mathbb{A} A is a function of ω \omega ω , its time derivative corresponds to
∂ 0 ± A c i = 1 2 ϵ m n i ∂ 0 ω c m n ± γ ∂ 0 ω c 0 i = 1 2 ϵ m n i ω ˙ c m n ± γ ω ˙ c 0 i = : ± A ˙ c i . \partial_0 {}^{\pm}\mathbb{A}_c^i = \frac{1}{2} \epsilon^i_{\ mn} \partial_0 \omega^{mn}_c \pm \gamma \partial_0 \omega^{0i}_c = \frac{1}{2} \epsilon^i_{\ mn} \dot{\omega}^{mn}_c \pm \gamma \dot{\omega}^{0i}_c =: {}^{\pm}\dot{\mathbb{A}}_c^i. ∂ 0 ± A c i = 2 1 ϵ mn i ∂ 0 ω c mn ± γ ∂ 0 ω c 0 i = 2 1 ϵ mn i ω ˙ c mn ± γ ω ˙ c 0 i =: ± A ˙ c i .
Also notice that
P ( γ , + ) a b i = 1 4 ϵ i j k Π a b j k + 1 2 γ Π a b , i 0 , \mathbb{P}(\gamma, +)^i_{ab} = \frac{1}{4} \epsilon_{ijk} \Pi^{jk}_{ab} + \frac{1}{2\gamma} \Pi^{0}_{ab, i}, P ( γ , + ) ab i = 4 1 ϵ ijk Π ab jk + 2 γ 1 Π ab , i 0 ,
P ( γ , − ) a b i = 1 4 ϵ i j k Π a b j k − 1 2 γ Π a b , i 0 . \mathbb{P}(\gamma, -)^i_{ab} = \frac{1}{4} \epsilon_{ijk} \Pi^{jk}_{ab} - \frac{1}{2\gamma} \Pi^{0}_{ab, i}. P ( γ , − ) ab i = 4 1 ϵ ijk Π ab jk − 2 γ 1 Π ab , i 0 .
These simplify the canonical trace term Tr [ Π ∧ ω ˙ ] \operatorname{Tr}[\Pi\wedge\dot{\omega}] Tr [ Π ∧ ω ˙ ] :
P + ∧ + A ˙ + P − ∧ − A ˙ = ( Π + Π 0 ) ∧ ( ω ˙ + ω ˙ 0 ) + ( Π − Π 0 ) ∧ ( ω ˙ − ω ˙ 0 ) = 2 Π ∧ ω ˙ + 2 Π 0 ∧ ω ˙ 0 = 1 2 ϵ i j k Π j k ∧ 1 2 ϵ m n i ω ˙ m n + 1 γ Π i 0 ∧ γ ω ˙ i 0 = 1 2 ( δ j m δ k n − δ j n δ k m ) Π j k ∧ ω ˙ m n + Π i 0 ∧ ω ˙ i 0 = 1 2 Π i j ∧ ω ˙ i j + Π i 0 ∧ ω ˙ i 0 = Tr [ Π ∧ ω ˙ ] . \begin{aligned}
\mathbb{P}^{+} \wedge {}^{+}\dot{\mathbb{A}} + \mathbb{P}^{-} \wedge {}^{-}\dot{\mathbb{A}}
&= (\Pi + \Pi^0) \wedge (\dot{\omega} + \dot{\omega}^0) + (\Pi - \Pi^0) \wedge (\dot{\omega} - \dot{\omega}^0) \\
&= 2 \Pi \wedge \dot{\omega} + 2 \Pi^0 \wedge \dot{\omega}^0 \\
&= \frac{1}{2} \epsilon_{ijk} \Pi^{jk} \wedge \frac{1}{2} \epsilon^i_{\, mn} \dot{\omega}^{mn} + \frac{1}{\gamma} \Pi^{0}_i \wedge \gamma \dot{\omega}^{i0} \\
&= \frac{1}{2} (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km}) \Pi^{jk} \wedge \dot{\omega}^{mn} + \Pi^{0}_i \wedge \dot{\omega}^{i0} \\
&= \frac{1}{2} \Pi_{ij} \wedge \dot{\omega}^{ij} + \Pi^0_{i} \wedge \dot{\omega}^{i0} = \operatorname{Tr} [ \Pi \wedge \dot{\omega} ].
\end{aligned} P + ∧ + A ˙ + P − ∧ − A ˙ = ( Π + Π 0 ) ∧ ( ω ˙ + ω ˙ 0 ) + ( Π − Π 0 ) ∧ ( ω ˙ − ω ˙ 0 ) = 2Π ∧ ω ˙ + 2 Π 0 ∧ ω ˙ 0 = 2 1 ϵ ijk Π jk ∧ 2 1 ϵ mn i ω ˙ mn + γ 1 Π i 0 ∧ γ ω ˙ i 0 = 2 1 ( δ jm δ kn − δ jn δ km ) Π jk ∧ ω ˙ mn + Π i 0 ∧ ω ˙ i 0 = 2 1 Π ij ∧ ω ˙ ij + Π i 0 ∧ ω ˙ i 0 = Tr [ Π ∧ ω ˙ ] .
The term ω 0 D ω ∧ Π \omega_0 \mathcal{D}_\omega \wedge \Pi ω 0 D ω ∧ Π acts as a Lagrangian constraint, with ω 0 \omega_0 ω 0 serving as the multiplier. We can express the connection variables as:
+ A + − A 2 = 1 2 ϵ m n i ω c m n and ( + A − − A ) j = 2 γ ω c 0 j . \frac{{}^{+}\mathbb{A} + {}^{-}\mathbb{A}}{2} = \frac{1}{2} \epsilon^i_{\ mn} \omega^{mn}_c \quad \text{and} \quad ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j = 2\gamma \omega^{0j}_c. 2 + A + − A = 2 1 ϵ mn i ω c mn and ( + A − − A ) j = 2 γ ω c 0 j .
Consequently, we can decompose the constraint term:
ω 0 D ω ∧ Π = + A 0 ( D + A P + ) i + − A 0 ( D − A P − ) i . \begin{aligned}
\omega_0 \mathcal{D}_\omega \wedge \Pi &= {}^{+}\mathbb{A}_0 (\mathcal{D}_{{}^{+}\mathbb{A}} \mathbb{P}^{+})_i + {}^{-}\mathbb{A}_0 (\mathcal{D}_{{}^{-}\mathbb{A}} \mathbb{P}^{-})_i.
\end{aligned} ω 0 D ω ∧ Π = + A 0 ( D + A P + ) i + − A 0 ( D − A P − ) i .
Expanding the first term, we find:
+ A 0 ( d P + + + A × P + ) = + A 0 [ d P + + 1 2 ϵ i j k ( ϵ m n j ω m n + γ ω 0 j ) P + k ] = + A 0 [ d P + + 1 2 ϵ i j k ϵ m n j ω m n P + k + γ ϵ i j k ω 0 j P + k ] = + A 0 D ( 1 2 ϵ ω ) P + + + A 0 ϵ i j k + A − − A 2 P + k ∼ η i B i + N i G i . \begin{aligned}
{}^{+}\mathbb{A}_0 \left( d \mathbb{P}^{+} + {}^{+}\mathbb{A} \times \mathbb{P}^{+} \right) &= {}^{+}\mathbb{A}_0 \left[ d \mathbb{P}^{+} + \frac{1}{2}\epsilon_{ijk} \left( \epsilon^j_{\ mn} \omega^{mn} + \gamma \omega^{0j} \right) \mathbb{P}^{+k}\right] \\
&= {}^{+}\mathbb{A}_0 \left[ d \mathbb{P}^{+} + \frac{1}{2} \epsilon_{ijk} \epsilon^j_{\ mn} \omega^{mn} \mathbb{P}^{+k} + \gamma \epsilon_{ijk} \omega^{0j} \mathbb{P}^{+k} \right] \\
&= {}^{+}\mathbb{A}_0 \, \mathcal{D}_{\left(\frac{1}{2}\epsilon \omega\right)} \mathbb{P}^{+} + {}^{+}\mathbb{A}_0 \, \epsilon_{ijk} \frac{{}^{+}\mathbb{A} - {}^{-}\mathbb{A}}{2} \mathbb{P}^{+k} \\
&\sim \eta_i \mathbb{B}^i + N_i \mathbb{G}^i.
\end{aligned} + A 0 ( d P + + + A × P + ) = + A 0 [ d P + + 2 1 ϵ ijk ( ϵ mn j ω mn + γ ω 0 j ) P + k ] = + A 0 [ d P + + 2 1 ϵ ijk ϵ mn j ω mn P + k + γ ϵ ijk ω 0 j P + k ] = + A 0 D ( 2 1 ϵ ω ) P + + + A 0 ϵ ijk 2 + A − − A P + k ∼ η i B i + N i G i .
Here we introduce new constraints B i \mathbb{B}^i B i and G i \mathbb{G}^i G i . Hence, we can rewrite the action as:
S [ P ( γ , ± ) , ± A , λ ] = ∫ d t ∫ Σ t { P + ∧ + A ˙ + P − ∧ − A ˙ + N i G i + η i B i + − A 0 D − A P − + H ∧ F + λ ′ ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) } . \begin{aligned}
S[\mathbb{P}(\gamma, \pm), {}^{\pm}\mathbb{A}, \lambda] &= \int dt \int_{\Sigma_t} \Big\{ \mathbb{P}^{+} \wedge {}^{+}\dot{\mathbb{A}} + \mathbb{P}^{-} \wedge {}^{-}\dot{\mathbb{A}} + N_i \mathbb{G}^i + \eta_i \mathbb{B}^i \\
&+ {}^{-}\mathbb{A}_0 \, \mathcal{D}_{{}^{-}\mathbb{A}} \mathbb{P}^{-} + H \wedge F + \lambda' (\gamma ^\star H - H) \wedge (\gamma ^\star \Pi - \Pi) \Big\}.
\end{aligned} S [ P ( γ , ± ) , ± A , λ ] = ∫ d t ∫ Σ t { P + ∧ + A ˙ + P − ∧ − A ˙ + N i G i + η i B i + − A 0 D − A P − + H ∧ F + λ ′ ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) } .
The Lagrangian multipliers N , η , λ ′ N, \eta, \lambda' N , η , λ ′ impose the following constraints:
{ B i : = D + A + − A 2 ∧ P ( γ , + ) i ≈ 0 G i : = 1 2 γ ϵ i j k ( + A − − A ) j ∧ P ( γ , + ) k ≈ 0 ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) = 0 ⟹ ϵ μ ν ρ σ B μ ν I J B ρ σ K L − e ϵ I J K L = 0. \begin{cases}
\mathbb{B}^i := \mathcal{D}_{\frac{{}^{+}\mathbb{A} + {}^{-}\mathbb{A}}{2}} \wedge \mathbb{P}(\gamma, +)^i \approx 0 \\
\mathbb{G}^i := \frac{1}{2\gamma} \epsilon_{ijk} ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j \wedge \mathbb{P}(\gamma, +)_k \approx 0 \\
(\gamma^\star H - H)\wedge(\gamma^\star \Pi - \Pi) = 0 \implies \epsilon_{\mu\nu\rho\sigma} B^{IJ}_{\mu\nu} B^{KL}_{\rho\sigma} - e\epsilon^{IJKL} = 0.
\end{cases} ⎩ ⎨ ⎧ B i := D 2 + A + − A ∧ P ( γ , + ) i ≈ 0 G i := 2 γ 1 ϵ ijk ( + A − − A ) j ∧ P ( γ , + ) k ≈ 0 ( γ ⋆ H − H ) ∧ ( γ ⋆ Π − Π ) = 0 ⟹ ϵ μν ρ σ B μν I J B ρ σ K L − e ϵ I J K L = 0.
The index pair [ I J , K L ] [IJ, KL] [ I J , K L ] must be decomposed. We separate it into three cases:
{ I J K L = 0 i 0 j I J K L = i j k l I J K L = 0 i j k . \begin{cases}
IJKL = 0i0j\\
IJKL = ijkl\\
IJKL = 0ijk.
\end{cases} ⎩ ⎨ ⎧ I J K L = 0 i 0 j I J K L = ijk l I J K L = 0 ijk .
For case 1 (I J K L = 0 i 0 j IJKL = 0i0j I J K L = 0 i 0 j ):
B 0 i ∧ B 0 j = 0 ⟹ C H 0 i ∧ C Π 0 j + ( i ↔ j ) = 0 ⟹ ( γ ⋆ H 0 i − H 0 i ) ∧ ( γ ⋆ Π 0 j − Π 0 j ) + ( i ↔ j ) = 0 ⟹ ( γ 2 ϵ j k i H j k − H 0 i ) ∧ ( γ 2 ϵ k l j Π k l − Π 0 j ) + ( i ↔ j ) = 0. \begin{gathered}
B^{0i} \wedge B^{0j} = 0 \\
\implies C^{0i}_H \wedge C^{0j}_\Pi + (i \leftrightarrow j) = 0 \\
\implies (\gamma ^\star H^{0i} - H^{0i}) \wedge (\gamma ^\star \Pi^{0j} - \Pi^{0j}) + (i \leftrightarrow j) = 0 \\
\implies \left(\frac{\gamma}{2}\epsilon^i_{\ jk}H^{jk} - H^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^j_{\ kl}\Pi^{kl} - \Pi^{0j}\right) + (i \leftrightarrow j) = 0.
\end{gathered} B 0 i ∧ B 0 j = 0 ⟹ C H 0 i ∧ C Π 0 j + ( i ↔ j ) = 0 ⟹ ( γ ⋆ H 0 i − H 0 i ) ∧ ( γ ⋆ Π 0 j − Π 0 j ) + ( i ↔ j ) = 0 ⟹ ( 2 γ ϵ jk i H jk − H 0 i ) ∧ ( 2 γ ϵ k l j Π k l − Π 0 j ) + ( i ↔ j ) = 0.
For case 2 (I J K L = i j k l IJKL = ijkl I J K L = ijk l ):
B i j ∧ B k l = 0 ⟹ C H i j ∧ C Π k l + ( i j ↔ k l ) = 0 ⟹ ( γ ⋆ H i j − H i j ) ∧ ( γ ⋆ Π k l − Π k l ) + ( i j ↔ k l ) = 0 ⟹ ϵ i j ( p ( γ 2 ϵ m i j H 0 m − H i j ) ∧ ϵ k l q ) ( γ 2 ϵ n k l Π 0 n − Π k l ) = 0. \begin{gathered}
B^{ij} \wedge B^{kl} = 0 \\
\implies C^{ij}_H \wedge C^{kl}_\Pi + (ij \leftrightarrow kl) = 0 \\
\implies (\gamma ^\star H^{ij} - H^{ij}) \wedge (\gamma ^\star \Pi^{kl} - \Pi^{kl}) + (ij \leftrightarrow kl) = 0 \\
\implies \epsilon^{(p}_{\ \ \ ij}\left(\frac{\gamma}{2}\epsilon^{ij}_{\ \ m} H^{0m} - H^{ij}\right) \wedge \epsilon^{q)}_{\ \ \ kl}\left(\frac{\gamma}{2}\epsilon^{kl}_{\ \ n} \Pi^{0n} - \Pi^{kl}\right) = 0.
\end{gathered} B ij ∧ B k l = 0 ⟹ C H ij ∧ C Π k l + ( ij ↔ k l ) = 0 ⟹ ( γ ⋆ H ij − H ij ) ∧ ( γ ⋆ Π k l − Π k l ) + ( ij ↔ k l ) = 0 ⟹ ϵ ij ( p ( 2 γ ϵ m ij H 0 m − H ij ) ∧ ϵ k l q ) ( 2 γ ϵ n k l Π 0 n − Π k l ) = 0.
For case 3 (I J K L = 0 i j k IJKL = 0ijk I J K L = 0 ijk ):
B 0 i ∧ B j k = ν ϵ 0 i j k ⟹ C H 0 i ∧ C Π j k + ( 0 i ↔ j k ) = ν ϵ 0 i j k ⟹ ( γ ⋆ H 0 i − H 0 i ) ∧ ( γ ⋆ Π j k − Π j k ) + ( 0 i ↔ j k ) = ν ϵ 0 i j k ⟹ ( γ 2 ϵ m n i H m n − H 0 i ) ∧ ( γ 2 ϵ l j k Π 0 l − Π j k ) + ( γ 2 ϵ m n i Π m n − Π 0 i ) ∧ ( γ 2 ϵ l j k H 0 l − H j k ) − ν ϵ 0 i j k = 0. \begin{gathered}
B^{0i} \wedge B^{jk} = \nu \epsilon^{0ijk} \\
\implies C^{0i}_H \wedge C^{jk}_\Pi + (0i \leftrightarrow jk) = \nu \epsilon^{0ijk} \\
\implies (\gamma ^\star H^{0i} - H^{0i}) \wedge (\gamma ^\star \Pi^{jk} - \Pi^{jk}) + (0i \leftrightarrow jk) = \nu \epsilon^{0ijk} \\
\begin{aligned}
\implies &\left(\frac{\gamma}{2}\epsilon^i_{\ mn} H^{mn} - H^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^{jk}_{\ \ l} \Pi^{0l} - \Pi^{jk}\right) +\\
&\left(\frac{\gamma}{2}\epsilon^i_{\ mn}\Pi^{mn} - \Pi^{0i}\right) \wedge \left(\frac{\gamma}{2}\epsilon^{jk}_{\ \ l}H^{0l} - H^{jk}\right) - \nu \epsilon^{0ijk}= 0.
\end{aligned}
\end{gathered} B 0 i ∧ B jk = ν ϵ 0 ijk ⟹ C H 0 i ∧ C Π jk + ( 0 i ↔ jk ) = ν ϵ 0 ijk ⟹ ( γ ⋆ H 0 i − H 0 i ) ∧ ( γ ⋆ Π jk − Π jk ) + ( 0 i ↔ jk ) = ν ϵ 0 ijk ⟹ ( 2 γ ϵ mn i H mn − H 0 i ) ∧ ( 2 γ ϵ l jk Π 0 l − Π jk ) + ( 2 γ ϵ mn i Π mn − Π 0 i ) ∧ ( 2 γ ϵ l jk H 0 l − H jk ) − ν ϵ 0 ijk = 0.
Recall that we've defined
H ( γ , ± ) a i : = 1 4 ϵ j k i H a j k ± 1 2 γ H a 0 i P ( γ , ± ) a b i : = 1 4 ϵ j k i Π a b j k ± 1 2 γ Π a b 0 i . \begin{aligned}
\mathbb{H}(\gamma, \pm)^i_a &:= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a \pm \frac{1}{2\gamma} H^{0i}_a \\
\mathbb{P}(\gamma, \pm)^i_{ab} &:= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab}.
\end{aligned} H ( γ , ± ) a i P ( γ , ± ) ab i := 4 1 ϵ jk i H a jk ± 2 γ 1 H a 0 i := 4 1 ϵ jk i Π ab jk ± 2 γ 1 Π ab 0 i .
Therefore we can rewrite the constraints as:
B i = D ( + A + − A ) / 2 ∧ P ( γ , + ) i ≈ 0 G i = 1 2 γ ϵ i j k ( + A − − A ) j ∧ P ( γ , + ) k ≈ 0 I i j = H ( γ , − ) ( i ∧ P ( γ , − ) j ) ≈ 0 I I i j = H ( 1 γ , − ) ( i ∧ P ( 1 γ , − ) j ) ≈ 0 I I I i j = H ( γ , − ) i ∧ P ( γ , + ) j + H ( 1 γ , − ) i ∧ P ( γ , − ) j − 1 3 δ i j Tr [ … ] = 0 \begin{aligned}
&\mathbb{B}^i = \mathcal{D}_{({}^{+}\mathbb{A} + {}^{-}\mathbb{A})/2} \wedge \mathbb{P}(\gamma, +)^i \approx 0 \\
&\mathbb{G}^i = \frac{1}{2\gamma} \epsilon^{ijk} ({}^{+}\mathbb{A} - {}^{-}\mathbb{A})_j \wedge \mathbb{P}(\gamma, +)_k \approx 0 \\
&\mathbb{I}^{ij} = \mathbb{H}(\gamma, -)^{(i} \wedge \mathbb{P}(\gamma, -)^{j)} \approx 0 \\
&\mathbb{II}^{ij} = \mathbb{H}\left(\frac{1}{\gamma}, -\right)^{(i} \wedge \mathbb{P}\left(\frac{1}{\gamma}, -\right)^{j)} \approx 0 \\
&\mathbb{III}^{ij} = \mathbb{H}(\gamma, -)^{i} \wedge \mathbb{P}(\gamma, +)^{j} + \mathbb{H}\left(\frac{1}{\gamma}, -\right)^{i} \wedge \mathbb{P}(\gamma, -)^{j} - \frac{1}{3} \delta^{ij} \operatorname{Tr}[\dots] = 0
\end{aligned} B i = D ( + A + − A ) /2 ∧ P ( γ , + ) i ≈ 0 G i = 2 γ 1 ϵ ijk ( + A − − A ) j ∧ P ( γ , + ) k ≈ 0 I ij = H ( γ , − ) ( i ∧ P ( γ , − ) j ) ≈ 0 II ij = H ( γ 1 , − ) ( i ∧ P ( γ 1 , − ) j ) ≈ 0 III ij = H ( γ , − ) i ∧ P ( γ , + ) j + H ( γ 1 , − ) i ∧ P ( γ , − ) j − 3 1 δ ij Tr [ … ] = 0
The last three constraints are derived from the original lagrangian multiplier λ \lambda λ , therefore include 20 constraints in total and are simplicity constraints. In the 3+1 setting, by choosing the time gauge to be n I = ( 1 , 0 , 0 , 0 ) n^I = (1, 0, 0, 0) n I = ( 1 , 0 , 0 , 0 ) , e μ I e^I_\mu e μ I is
{ e 0 0 = N ( lapse ) e a 0 = 0 ( pure spacial, no time-like component ) e 0 i = N i ( shift ) e a i = standard 3-d tetrad \begin{cases}
e^0_0 = N & (\text{lapse}) \\
e^0_a = 0 & (\text{pure spacial, no time-like component}) \\
e^i_0 = N^i & (\text{shift}) \\
e^i_a = \text{standard 3-d tetrad}
\end{cases} ⎩ ⎨ ⎧ e 0 0 = N e a 0 = 0 e 0 i = N i e a i = standard 3-d tetrad ( lapse ) ( pure spacial, no time-like component ) ( shift )
(Indices: I , J ∈ 0 , 1 , 2 , 3 I, J \in 0, 1, 2, 3 I , J ∈ 0 , 1 , 2 , 3 local; μ , ν ∈ 0 , 1 , 2 , 3 \mu, \nu \in 0, 1, 2, 3 μ , ν ∈ 0 , 1 , 2 , 3 global; a , b ∈ 1 , 2 , 3 a,b \in 1, 2, 3 a , b ∈ 1 , 2 , 3 global spacial; i , j ∈ 1 , 2 , 3 i,j \in 1, 2, 3 i , j ∈ 1 , 2 , 3 local spacial).
We consider the solution B = e ∧ e ⟹ B μ ν I J = e μ I e ν J − e μ J e ν I B = e \wedge e \implies B^{IJ}_{\mu\nu} = e^I_\mu e^J_\nu - e^J_\mu e^I_\nu B = e ∧ e ⟹ B μν I J = e μ I e ν J − e μ J e ν I .
Globally spacial components:
B a b 0 i = e a 0 e b i − e b 0 e a i = 0 ⟹ ⋆ B a b 0 i = 1 2 ϵ j k 0 i B a b j k = 1 2 ϵ j k 0 i ( e a j e b k − e b j e a k ) = ϵ j k i e a j e b k B a b j k = e a j e b k − e a k e b j = 2 e a [ j e b k ] ⟹ ⋆ B a b j k = 1 2 ϵ 0 i j k B a b 0 i = 0. \begin{aligned}
B^{0i}_{ab} &= e^0_a e^i_b - e^0_b e^i_a = 0 \implies ^\star B^{0i}_{ab} = \frac{1}{2} \epsilon_{\ \ \ jk}^{0i} B^{jk}_{ab} = \frac{1}{2} \epsilon^{0i}_{\ \ \ jk} (e^j_a e^k_b - e^j_b e^k_a) = \epsilon^{i}_{\ \ jk} e^j_a e^k_b \\
B^{jk}_{ab} &= e^j_a e^k_b - e^k_a e^j_b = 2 e^{[j}_a e^{k]}_b \implies ^\star B^{jk}_{ab} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} B^{0i}_{ab} = 0.
\end{aligned} B ab 0 i B ab jk = e a 0 e b i − e b 0 e a i = 0 ⟹ ⋆ B ab 0 i = 2 1 ϵ jk 0 i B ab jk = 2 1 ϵ jk 0 i ( e a j e b k − e b j e a k ) = ϵ jk i e a j e b k = e a j e b k − e a k e b j = 2 e a [ j e b k ] ⟹ ⋆ B ab jk = 2 1 ϵ 0 i jk B ab 0 i = 0.
Globally timelike components:
B 0 a 0 i = e 0 0 e a i − e a 0 e 0 i = N e a i ⟹ ⋆ B 0 a 0 i = 1 2 ϵ j k 0 i B 0 a j k = 1 2 ϵ j k 0 i ( e 0 j e a k − e 0 k e a j ) = ϵ j k i N j e a k B 0 a j k = e 0 j e a k − e 0 k e a j = 2 N [ j e a k ] ⟹ ⋆ B 0 a j k = 1 2 ϵ 0 i j k B 0 a 0 i = 1 2 ϵ 0 i j k N e a i = N ϵ 0 i j k e a i . \begin{aligned}
B^{0i}_{0a} &= e^0_0 e^i_a - e^0_a e^i_0 = N e^i_a \implies ^\star B^{0i}_{0a} = \frac{1}{2} \epsilon_{\ \ \ jk}^{0i} B^{jk}_{0a} = \frac{1}{2} \epsilon^{0i}_{\ \ jk} (e^j_0 e^k_a - e^k_0 e^j_a) = \epsilon^{i}_{\ \ jk} N^j e^k_a \\
B^{jk}_{0a} &= e^j_0 e^k_a - e^k_0 e^j_a = 2 N^{[j} e^{k]}_a \implies ^\star B^{jk}_{0a} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} B^{0i}_{0a} = \frac{1}{2} \epsilon^{jk}_{\ \ \ 0i} N e^i_a = N \epsilon^{jk}_{\ \ \ 0i} e^{i}_{\ \ a}.
\end{aligned} B 0 a 0 i B 0 a jk = e 0 0 e a i − e a 0 e 0 i = N e a i ⟹ ⋆ B 0 a 0 i = 2 1 ϵ jk 0 i B 0 a jk = 2 1 ϵ jk 0 i ( e 0 j e a k − e 0 k e a j ) = ϵ jk i N j e a k = e 0 j e a k − e 0 k e a j = 2 N [ j e a k ] ⟹ ⋆ B 0 a jk = 2 1 ϵ 0 i jk B 0 a 0 i = 2 1 ϵ 0 i jk N e a i = N ϵ 0 i jk e a i .
Now we calculate the corresponding Π \Pi Π & H H H :
Π a b 0 i = ⋆ B a b 0 i + 1 γ B a b 0 i = ϵ j k i e a j e b k Π a b j k = ⋆ B a b j k + 1 γ B a b j k = 2 γ e a [ j e b k ] H a 0 i = ⋆ B 0 a 0 i + 1 γ B 0 a 0 i = ϵ j k i N j e a k + N γ e a i H a j k = ⋆ B 0 a j k + 1 γ B 0 a j k = N ϵ i j k e a i + 2 γ N [ j e a k ] . \begin{aligned}
\Pi^{0i}_{ab} &= {}^\star B^{0i}_{ab} + \frac{1}{\gamma} B^{0i}_{ab} = \epsilon^{i}_{\ \ jk} e^{j}_{a} e^{k}_{b} \\
\Pi^{jk}_{ab} &= {}^\star B^{jk}_{ab} + \frac{1}{\gamma} B^{jk}_{ab} = \frac{2}{\gamma} e^{[j}_{a} e^{k]}_{b} \\
H^{0i}_{a} &= {}^\star B^{0i}_{0a} + \frac{1}{\gamma} B^{0i}_{0a} = \epsilon^{i}_{\ \ jk} N^{j} e^{k}_{a} + \frac{N}{\gamma} e^{i}_{a} \\
H^{jk}_{a} &= {}^\star B^{jk}_{0a} + \frac{1}{\gamma} B^{jk}_{0a} = N \epsilon_{\ \ i}^{jk} e^{i}_{a} + \frac{2}{\gamma} N^{[j} e^{k]}_{a}.
\end{aligned} Π ab 0 i Π ab jk H a 0 i H a jk = ⋆ B ab 0 i + γ 1 B ab 0 i = ϵ jk i e a j e b k = ⋆ B ab jk + γ 1 B ab jk = γ 2 e a [ j e b k ] = ⋆ B 0 a 0 i + γ 1 B 0 a 0 i = ϵ jk i N j e a k + γ N e a i = ⋆ B 0 a jk + γ 1 B 0 a jk = N ϵ i jk e a i + γ 2 N [ j e a k ] .
Now we can show that
P ( γ , ± ) a b i = 1 4 ϵ j k i Π a b j k ± 1 2 γ Π a b 0 i = 1 4 ϵ j k i 2 γ e [ a j e b ] k ± 1 2 γ ϵ j k i e a j e b k = { P ( γ , − ) a b i = 0 P ( γ , + ) a b i = 1 γ ϵ j k i e a j e b k P ( 1 γ , ± ) a b i = 1 4 ϵ j k i Π a b j k ± γ 2 Π a b 0 i = 1 2 γ ϵ j k i e a j e b k ± γ 2 ϵ j k i e a j e b k = 1 ± γ 2 2 γ P ( γ , + ) a b i H ( γ , − ) a i = 1 4 ϵ j k i H a j k − 1 2 γ H a 0 i = 1 4 ϵ j k i ( N ϵ l j k e a l + 2 γ N [ j e a k ] ) − 1 2 γ ( ϵ j k i N j e a k + N γ e a i ) = 1 2 N δ l i e a l + 1 2 γ ϵ j k i N j e a k − 1 2 γ ϵ j k i N j e a k − N 2 γ 2 e a i = N γ 2 − 1 2 γ 2 e a i H ( 1 γ , − ) a i = 1 4 ϵ j k i H a j k − γ 2 H a 0 i = 1 2 N δ l i e a l + 1 2 γ ϵ j k i N j e a k − γ 2 ( ϵ j k i N j e a k + N γ e a i ) = 1 − γ 2 2 γ ϵ j k i N b e b j e a k = 1 − γ 2 2 N b P ( γ , + ) b a i . \begin{aligned}
\mathbb{P}(\gamma, \pm)^i_{ab} &= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{1}{2\gamma} \Pi^{0i}_{ab} = \frac{1}{4} \epsilon^i_{\ jk} \frac{2}{\gamma} e^j_{[a} e^k_{b]} \pm \frac{1}{2\gamma} \epsilon^i_{\ jk} e^j_a e^k_b =
\begin{cases}
\mathbb{P}(\gamma, -)^i_{ab} = 0 \\
\mathbb{P}(\gamma, +)^i_{ab} = \frac{1}{\gamma} \epsilon^i_{\ jk} e^j_a e^k_b
\end{cases} \\
\mathbb{P}\left(\frac{1}{\gamma}, \pm\right)^i_{ab} &= \frac{1}{4} \epsilon^i_{\ jk} \Pi^{jk}_{ab} \pm \frac{\gamma}{2} \Pi^{0i}_{ab} = \frac{1}{2\gamma} \epsilon^i_{\ jk} e^j_a e^k_b \pm \frac{\gamma}{2} \epsilon^i_{\ jk} e^j_a e^k_b = \frac{1\pm\gamma^2}{2\gamma}\mathbb{P}(\gamma, +)^i_{ab} \\
\mathbb{H}(\gamma, -)^i_a &= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a - \frac{1}{2\gamma} H^{0i}_a = \frac{1}{4} \epsilon^i_{\ jk} \left(N \epsilon^{jk}_{\ \ \ l} e_{a}^l + \frac{2}{\gamma} N^{[j} e^{k]}_a\right) - \frac{1}{2\gamma}\left(\epsilon^{i}_{\ \ jk} N^j e_{a}^k + \frac{N}{\gamma} e^i_a\right) \\
&= \frac{1}{2} N \delta^i_l e^l_a + \frac{1}{2\gamma} \epsilon^i_{\ jk} N^j e^k_a - \frac{1}{2\gamma} \epsilon^{i}_{\ \ jk} N^j e_{a}^k - \frac{N}{2\gamma^2}e^i_a = N \frac{\gamma^2 - 1}{2\gamma^2} e^i_a \\
\mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a &= \frac{1}{4} \epsilon^i_{\ jk} H^{jk}_a - \frac{\gamma}{2} H^{0i}_a = \frac{1}{2} N \delta^i_l e^l_a + \frac{1}{2\gamma} \epsilon^i_{\ jk} N^j e^k_a - \frac{\gamma}{2}\left(\epsilon^{i}_{\ \ jk} N^j e_{a}^k + \frac{N}{\gamma} e^i_a\right) \\
&= \frac{1-\gamma^2}{2\gamma} \epsilon^{i}_{\ \ jk} N^b e^j_b e^k_a = \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba}.
\end{aligned} P ( γ , ± ) ab i P ( γ 1 , ± ) ab i H ( γ , − ) a i H ( γ 1 , − ) a i = 4 1 ϵ jk i Π ab jk ± 2 γ 1 Π ab 0 i = 4 1 ϵ jk i γ 2 e [ a j e b ] k ± 2 γ 1 ϵ jk i e a j e b k = { P ( γ , − ) ab i = 0 P ( γ , + ) ab i = γ 1 ϵ jk i e a j e b k = 4 1 ϵ jk i Π ab jk ± 2 γ Π ab 0 i = 2 γ 1 ϵ jk i e a j e b k ± 2 γ ϵ jk i e a j e b k = 2 γ 1 ± γ 2 P ( γ , + ) ab i = 4 1 ϵ jk i H a jk − 2 γ 1 H a 0 i = 4 1 ϵ jk i ( N ϵ l jk e a l + γ 2 N [ j e a k ] ) − 2 γ 1 ( ϵ jk i N j e a k + γ N e a i ) = 2 1 N δ l i e a l + 2 γ 1 ϵ jk i N j e a k − 2 γ 1 ϵ jk i N j e a k − 2 γ 2 N e a i = N 2 γ 2 γ 2 − 1 e a i = 4 1 ϵ jk i H a jk − 2 γ H a 0 i = 2 1 N δ l i e a l + 2 γ 1 ϵ jk i N j e a k − 2 γ ( ϵ jk i N j e a k + γ N e a i ) = 2 γ 1 − γ 2 ϵ jk i N b e b j e a k = 2 1 − γ 2 N b P ( γ , + ) ba i .
Notice that P ( γ , − ) = 0 \mathbb{P}(\gamma, -) = 0 P ( γ , − ) = 0 breaks the internal Lorentz gauge. From the above equations, we can parameterize the variables H a j k H^{jk}_a H a jk and H a 0 i H^{0i}_a H a 0 i purely down to P ( γ , + ) a b i \mathbb{P}(\gamma, +)^i_{ab} P ( γ , + ) ab i and variables N , N i N, N^i N , N i , which we can see explicitly by solving for them:
H ( 1 γ , − ) a i − H ( γ , − ) a i = ( 1 2 γ − γ 2 ) H a 0 i = 1 − γ 2 2 γ H a 0 i (i) γ 2 H ( γ , − ) a i − H ( 1 γ , − ) a i = γ 2 − 1 4 ϵ j k i H a j k (ii) \begin{aligned}
&\mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a - \mathbb{H}(\gamma, -)^i_a = \left(\frac{1}{2\gamma} - \frac{\gamma}{2}\right) H^{0i}_a = \frac{1-\gamma^2}{2\gamma} H^{0i}_a \quad &\text{(i)}\\
&\gamma^2 \mathbb{H}(\gamma, -)^i_a - \mathbb{H}\left(\frac{1}{\gamma}, -\right)^i_a = \frac{\gamma^2 - 1}{4} \epsilon^i_{\ jk} H^{jk}_a \quad &\text{(ii)}
\end{aligned} H ( γ 1 , − ) a i − H ( γ , − ) a i = ( 2 γ 1 − 2 γ ) H a 0 i = 2 γ 1 − γ 2 H a 0 i γ 2 H ( γ , − ) a i − H ( γ 1 , − ) a i = 4 γ 2 − 1 ϵ jk i H a jk (i) (ii)
Plugging in our findings for H ( γ , − ) \mathbb{H}(\gamma, -) H ( γ , − ) and H ( 1 / γ , − ) \mathbb{H}\left(1/\gamma, -\right) H ( 1/ γ , − ) , we get:
(i) ⟹ 1 − γ 2 2 N b P ( γ , + ) b a i + N 1 − γ 2 2 γ 2 e a i = 1 − γ 2 2 γ H a 0 i ⟹ H a 0 i = N γ e a i + γ N b P ( γ , + ) b a i (ii) ⟹ − N 1 − γ 2 2 e a i − 1 − γ 2 2 N b P ( γ , + ) b a i = γ 2 − 1 4 ϵ j k i H a j k ⟹ 1 2 ϵ j k i H a j k = N e a i + N b P ( γ , + ) b a i \begin{aligned}
\text{(i)} &\implies \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba} + N \frac{1 - \gamma^2}{2\gamma^2} e^i_a = \frac{1-\gamma^2}{2\gamma} H^{0i}_a \\
&\implies H^{0i}_a = \frac{N}{\gamma} e^i_a + \gamma N^b \mathbb{P}(\gamma, +)^i_{ba} \\
\text{(ii)} &\implies - N \frac{1 - \gamma^2}{2} e^i_a - \frac{1 - \gamma^2}{2} N^b \mathbb{P}(\gamma, +)^i_{ba} = \frac{\gamma^2 - 1}{4} \epsilon^i_{\ jk} H^{jk}_a \\
&\implies \frac{1}{2} \epsilon^{i}_{\ jk} H^{jk}_a = N e^i_a + N^b \mathbb{P}(\gamma, +)^i_{ba}
\end{aligned} (i) (ii) ⟹ 2 1 − γ 2 N b P ( γ , + ) ba i + N 2 γ 2 1 − γ 2 e a i = 2 γ 1 − γ 2 H a 0 i ⟹ H a 0 i = γ N e a i + γ N b P ( γ , + ) ba i ⟹ − N 2 1 − γ 2 e a i − 2 1 − γ 2 N b P ( γ , + ) ba i = 4 γ 2 − 1 ϵ jk i H a jk ⟹ 2 1 ϵ jk i H a jk = N e a i + N b P ( γ , + ) ba i
Thus, we have parameterized the Plebanski constraints in terms of P ( γ , + ) a b i \mathbb{P}(\gamma, +)^i_{ab} P ( γ , + ) ab i , a scalar lapse N N N , shift vector N a ∈ T ( Σ t ) N^a \in T(\Sigma_t) N a ∈ T ( Σ t ) , and the 3 parameters in the choice of an internal direction n I = ( 1 , 0 , 0 , 0 ) n^I = (1, 0, 0, 0) n I = ( 1 , 0 , 0 , 0 ) . These sum up to 9 + 1 + 3 + 3 = 16 9 + 1 + 3 + 3 = 16 9 + 1 + 3 + 3 = 16 parameters, which is exactly the 16 degrees of freedom in the co-tetrad e μ I e^I_\mu e μ I .
The constraints are "simple" constraints and they must be conserved in time. By arXiv:0902.3416 [gr-qc], this follows from
C i j = e ( j P ˙ ( γ , − ) i ) ∝ N e ( j ∧ [ d e i ) + 1 2 ϵ l m i ) ( + A + − A ) l ∧ e m ] ≈ 0. \mathbb{C}^{ij} = e^{(j} \dot{\mathbb{P}}(\gamma, -)^{i)} \propto N e^{(j}\wedge \left[\mathrm{d}e^{i)} + \frac{1}{2}\epsilon^{i)}_{\ \ \ lm}\left({}^+\mathbb{A} + {}^-\mathbb{A}\right)^l\wedge e^{m} \right]\approx 0. C ij = e ( j P ˙ ( γ , − ) i ) ∝ N e ( j ∧ [ d e i ) + 2 1 ϵ l m i ) ( + A + − A ) l ∧ e m ] ≈ 0.
Together with B i ≈ 0 \mathbb{B}^i \approx 0 B i ≈ 0 , these 9 constraints are equivalent to
I V a i = 1 2 ( + A + − A ) a i − Γ a i ( e ) = 0. \mathbb{IV}^i_a = \frac{1}{2} ({}^{+}\mathbb{A} + {}^{-}\mathbb{A})_a^i - \Gamma_a^i(e) = 0. IV a i = 2 1 ( + A + − A ) a i − Γ a i ( e ) = 0.
Where Γ a i \Gamma_a^i Γ a i is the spin connection compatible with the triad e e e , i.e., the solution to
∂ [ a e b ] i + ϵ j k i Γ [ a j e b ] k = 0. \partial_{[a} e_{b]}^i + \epsilon^i_{\ jk} \Gamma_{[a}^j e_{b]}^k = 0. ∂ [ a e b ] i + ϵ jk i Γ [ a j e b ] k = 0.
We can also calculate the Poisson bracket:
{ I V a i ( x ) , P ( γ , − ) b c j ( y ) } = 1 2 ϵ a b c δ i j δ ( x − y ) . \{ \mathbb{IV}^i_a(x), \mathbb{P}(\gamma, -)^j_{bc}(y) \} = \frac{1}{2} \epsilon_{abc} \delta^{ij} \delta(x-y). { IV a i ( x ) , P ( γ , − ) b c j ( y )} = 2 1 ϵ ab c δ ij δ ( x − y ) .
Which makes them secondary constraints. By setting them strongly to zero, we define K a i = ( + A a i − − A a i ) / 2 K_a^i = ({}^{+}\mathbb{A}^i_a - {}^{-}\mathbb{A}^i_a)/2 K a i = ( + A a i − − A a i ) /2 , such that:
+ A a i = Γ a i + γ K a i = : A a i . {}^{+}\mathbb{A}_a^i = \Gamma_a^i + \gamma K_a^i =: A_a^i. + A a i = Γ a i + γ K a i =: A a i .
The action is then reduced to:
S [ P ( γ , + ) , + A , N , N i ] = 1 κ ∫ d t ∫ Σ ( P ( γ , + ) i ∧ + A ˙ i + N i D A ∧ P ( γ , + ) i + ϵ j k i N j P ( γ , + ) k ∧ ( γ F 0 i + F j k ϵ i j k ) + N e i ∧ ( F 0 i γ + F j k ϵ i j k ) ) . \begin{aligned}
S[\mathbb{P}(\gamma, +), {}^{+}\mathbb{A}, N, N_i]
&= \frac{1}{\kappa} \int dt \int_{\Sigma} \Big( \mathbb{P}(\gamma, +)_i \wedge {}^+\dot{\mathbb{A}}^i \\
&+ N_i \mathcal{D}_{A} \wedge \mathbb{P}(\gamma, +)^i + \epsilon^i_{\ jk} N^j \mathbb{P}(\gamma, +)^k \wedge (\gamma F_{0i} + F^{jk}\epsilon_{ijk}) \\
&+ Ne^i \wedge (\frac{F_{0i}}{\gamma} + F^{jk}\epsilon_{ijk} )\Big).
\end{aligned} S [ P ( γ , + ) , + A , N , N i ] = κ 1 ∫ d t ∫ Σ ( P ( γ , + ) i ∧ + A ˙ i + N i D A ∧ P ( γ , + ) i + ϵ jk i N j P ( γ , + ) k ∧ ( γ F 0 i + F jk ϵ ijk ) + N e i ∧ ( γ F 0 i + F jk ϵ ijk ) ) .
This is the standard Hamiltonian formulation of general relativity in terms of SU(2) connection variables A a i A_a^i A a i . If we now introduce the densitized triad
E i a = γ ϵ a b c P ( γ , + ) b c i , E^a_i = \gamma \epsilon^{abc} \mathbb{P}(\gamma, +)_{bc}^i, E i a = γ ϵ ab c P ( γ , + ) b c i ,
The Poisson bracket between E i a E^a_i E i a and A b j A^j_b A b j is now
{ E i a ( x ) , A b j ( y ) } = κ δ b a δ i j δ ( x − y ) , { E i a ( x ) , E j b ( y ) } = { A a i ( x ) , A b j ( y ) } = 0. \{E^a_i(x), A^j_b(y)\} = \kappa \delta^a_b \delta^j_i \delta(x-y), \quad \{E^a_i(x), E^b_j(y)\} = \{A^i_a(x), A^j_b(y)\} = 0. { E i a ( x ) , A b j ( y )} = κ δ b a δ i j δ ( x − y ) , { E i a ( x ) , E j b ( y )} = { A a i ( x ) , A b j ( y )} = 0.
And the action takes the familiar Ashtekar-Barbero form:
S [ E , A , N , N i , N i ] = 1 γ κ ∫ d t ∫ Σ t d 3 x [ E i a A ˙ a i − N b V b ( E i a , A a i ) − N S ( E i a , A a i ) − N i G i ( E i a , A a i ) ] \begin{aligned}
S[E, A, N, N^i, N_i] &= \frac{1}{\gamma\kappa} \int dt \int_{\Sigma_t} d^3x \Big[ E^a_i \dot{A}^i_a - N^b V_b(E^a_i, A^i_a) \\
&- N S(E^a_i, A^i_a) - N^i G_i(E^a_i, A^i_a) \Big]
\end{aligned} S [ E , A , N , N i , N i ] = γκ 1 ∫ d t ∫ Σ t d 3 x [ E i a A ˙ a i − N b V b ( E i a , A a i ) − NS ( E i a , A a i ) − N i G i ( E i a , A a i ) ]
With the constraints
{ V b = E j a F a b j − ( 1 + γ 2 ) K b i G i S = E i a E j b ∣ det E ∣ ( ϵ k i j F a b k − 2 ( 1 + γ 2 ) K [ a i K b ] j ) G i = D a E i a \begin{cases}
V_b = E^a_j F^j_{ab} - (1+\gamma^2)K^i_b G_i \\
S = \frac{E^a_i E^b_j}{\sqrt{|\det E|}} \left( \epsilon^{ij}_{\ \ k} F^k_{ab} - 2(1+\gamma^2)K^i_{[a} K^j_{b]} \right) \\
G_i = \mathcal{D}_a E^a_i
\end{cases} ⎩ ⎨ ⎧ V b = E j a F ab j − ( 1 + γ 2 ) K b i G i S = ∣ d e t E ∣ E i a E j b ( ϵ k ij F ab k − 2 ( 1 + γ 2 ) K [ a i K b ] j ) G i = D a E i a
These three constraints completely dictate both the symmetries and the dynamics of General Relativity. Because it is a fully constrained system, its Hamiltonian is simply a linear combination of these constraints. Physically, they can be interpreted as follows:
1. The Gauss Constraint (G i ≈ 0 G_i \approx 0 G i ≈ 0 )
This is the non-Abelian equivalent of Gauss's Law from Yang-Mills theory (∇ ⋅ E ⃗ = 0 \nabla \cdot \vec{E} = 0 ∇ ⋅ E = 0 ). Here, E i a E^a_i E i a acts as the electric field conjugate to the connection A a i A_a^i A a i . It generates local S U ( 2 ) SU(2) S U ( 2 ) internal gauge rotations. By setting G i ≈ 0 G_i \approx 0 G i ≈ 0 , we ensure that the physical state of the universe is independent of how we locally orient our internal 3D reference frame (the target space of the triads). Picking a different internal S U ( 2 ) SU(2) S U ( 2 ) basis at any point in space doesn't change the physical geometry.
2. The Vector Constraint (V b ≈ 0 V_b \approx 0 V b ≈ 0 )
In the action, this constraint is multiplied by the shift vector N b N^b N b . It is the generator of spatial diffeomorphisms upon the 3D slice Σ \Sigma Σ . Setting V b ≈ 0 V_b \approx 0 V b ≈ 0 enforces the principle that coordinates are arbitrary—if you stretch, twist, or remap the spatial coordinate grid on Σ \Sigma Σ , the underlying physical geometry remains exactly the same. The E j a F a b j E^a_j F^j_{ab} E j a F ab j term essentially comes from the Lie derivative of the connection along the spatial slice, while the − ( 1 + γ 2 ) K b i G i - (1+\gamma^2)K^i_b G_i − ( 1 + γ 2 ) K b i G i part is a proportional correction term.
3. The Scalar Constraint (S ≈ 0 S \approx 0 S ≈ 0 )
In the action, this is multiplied by the scalar lapse function N N N . This constraint generates "time evolution," which in General Relativity corresponds to deforming the spatial slice Σ t \Sigma_t Σ t forward in the normal direction into the 4D spacetime bulk. Because this must also vanish (S ≈ 0 S \approx 0 S ≈ 0 ), it embodies the fact that there is no absolute background time in GR. This is the core dynamical equation of General Relativity.