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Treat the orthonormal coframe in spacetime and the Lorentz connection 1-form as independent variables, with nondegenerate coframe and ${\omega }_{ab}=-{\omega }_{ba}$. Fix the curvature conventionThe second expression is Cartan's first structure equation; the first is Cartan's second structure equation. Use compactly supported variations, or boundary data that make the variation's surface term vanish.

For a connection variation, $\delta R^{ab}=D\delta {\omega }^{ab}$, where $D$ is the covariant exterior derivative. Since $D{ϵ}_{abc}=0$, its graded product rule givesFor a coframe variation, relabeling the dummy indices makes the three variations of the cubic term identical. HenceThe independent variations therefore give ${ϵ}_{abc}{T}^c=0$ and ${ϵ}_{abc}(R^{ab}+3\lambda E^a\wedge E^b)=0$. In three dimensions the alternating symbol identifies an antisymmetric index pair with a single index, so these equations are equivalent toThese are the equations of first-order three-dimensional gravity. The first selects the Levi-Civita connection; the second fixes constant sectional curvature $K=-3\lambda$. The factor of three follows from varying the coefficient exactly as printed.

On a static coordinate patch where $f>0$, choose the Lorentz frame metric ${\eta }_{ab}=\mathrm{diag}(-1,1,1)$ and coframePut $A=f^{\prime }/(2\sqrt{f})$ and $B=\sqrt{f}/r$. Exterior differentiation gives $d{E}^0=-A{E}^0\wedge E^1$, $d{E}^1=0$, and $d{E}^2=B{E}^1\wedge E^2$. Solving the equation of vanishing torsion forms givesHere antisymmetry is in the two raised or two lowered frame indices; mixed time-space components have equal, rather than opposite, signs. This is the Lorentzian connection-form antisymmetry needed to retain the curvature signs.

Using Cartan's second structure equation, the nonzero curvature 2-forms areFor example, $d(A{E}^0)=-(\sqrt{f}{A}^{\prime }+A^2)E^0\wedge E^1=-f^{\prime \prime }{E}^0\wedge E^1/2$, while the $02$ form is the product ${\omega }^0{}_1\wedge {\omega }^{12}=-AB{E}^0\wedge E^2$. Thus the field equations require $f^{\prime \prime }/2=3\lambda$ and $f^{\prime }/(2r)=3\lambda$. Integrating the latter gives $f=C+3\lambda r^2$; the former is then satisfied automatically. The origin condition fixes $C=1$, yieldingAs a sign and normalization check, the Ricci tensor is $R_{\mu \nu }=-6\lambda g_{\mu \nu }$ and the scalar curvature is $R=-18\lambda$. The coframe calculation applies initially in the region with $f>0$; if $\lambda <0$, $f$ vanishes at $r=1/\sqrt{-3\lambda }$, delimiting that static chart without changing the derived function.