= First-order three-dimensional gravity
{title2=$T^a=0,\quad R^{ab}=-3\lambda E^a\wedge E^b$}
For a nondegenerate <orthonormal coframe in spacetime> $E^a$ and independent antisymmetric Lorentz <connection 1-form>, the action $\int\epsilon_{abc}(R^{ab}\wedge E^c+\lambda E^a\wedge E^b\wedge E^c)$ has independent coframe and connection variations. Since $\delta R^{ab}=D\delta\omega^{ab}$, integration by parts yields a connection equation $\epsilon_{abc}T^c=0$. The three variations of the cubic coframe term give the coframe equation $\epsilon_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions these are exactly the equations displayed above. The solution has <constant sectional curvature> $K=-3\lambda$. Omitting the factor of three changes the cosmological normalization.
Back to article page
