Ashtekar variables

Remarks

The constraints of general relativity become: the Gauss constraint $D_a E^a_i = \partial_a E^a_i + \epsilon_{ijk} A^j_a E^{ak} = 0$, the diffeomorphism constraint $E^a_i F^i_{ab} = 0$, and the Hamiltonian constraint, where $F$ is the curvature of $A$. For $\gamma = i$ and $\Lambda = 0$ the Hamiltonian constraint is the polynomial $\epsilon^{ijk} E^a_i E^b_j F_{kab} = 0$. Example: on flat $\R^3$ at rest, with $q_{ab} = \delta_{ab}$, $K_{ab} = 0$ and the constant triad $e^a_i = \delta^a_i$, one gets $E^a_i = \delta^a_i$ and $A^i_a = 0$, which satisfies all constraints. Non-example: the Levi-Civita connection of $q$ alone is not an Ashtekar–Barbero connection when $K_{ab} \ne 0$, since it carries no information on the extrinsic curvature.