Invertible triads identify zero-cosmological-constant three-dimensional BF gravity

Let $\mathrm{v}\mathrm{o}\mathrm{l}$ be the oriented volume form of $(V,\eta )$. The displayed $J$ satisfies $\eta (J(v)w,z)=\mathrm{v}\mathrm{o}\mathrm{l}(v,w,z)$, so it takes values in $\mathfrak{s}\mathfrak{o}(V,\eta )$ and has zero kernel. Both vector spaces have dimension three: $\eta A$ is an antisymmetric $3\times 3$ matrix for $A\in \mathfrak{s}\mathfrak{o}(V,\eta )$, giving three independent entries. Thus $J$ is an isomorphism. For such $A$, $\mathrm{tr}A=0$ and the alternating-form identity $\mathrm{v}\mathrm{o}\mathrm{l}(Av,w,z)+\mathrm{v}\mathrm{o}\mathrm{l}(v,Aw,z)+\mathrm{v}\mathrm{o}\mathrm{l}(v,w,Az)=(\mathrm{tr}A)\mathrm{v}\mathrm{o}\mathrm{l}(v,w,z)=0$ follows by expanding on an oriented basis. Hence $\eta ([A,J(v)]w,z)=-\mathrm{v}\mathrm{o}\mathrm{l}(v,w,Az)-\mathrm{v}\mathrm{o}\mathrm{l}(v,Aw,z)=\mathrm{v}\mathrm{o}\mathrm{l}(Av,w,z)=\eta (J(Av)w,z),$ which proves $J(Av)=[A,J(v)]$ by nondegeneracy of $\eta$. This also makes the pairing $⟨J(v),J(w)⟩=\eta (v,w)$ invariant and nondegenerate. Since $J$ is constant, the identity gives $D_{\omega }(Je)=J(D_{\omega }e)$ as two-forms, by evaluating on each pair of tangent vectors. Every smooth interior-compact adjoint-valued BF variation $\beta$ equals $J(\delta e)$ for the smooth interior-compact variation $\delta e=J^{-1}\beta$; conversely each $\delta e$ gives such a $\beta$. Thus applying [F2] to $B=Je$ is a variation over the full BF-field space, and gives $F_{\omega }=0$ and $D_{\omega }(Je)=0$. Injectivity of $J$ converts the latter equation to $D_{\omega }e=0$. No invertibility assumption on the coframe field $e$ was used.

On a region where $e$ is a coframe, define a connection on $TU$ by ${\mathrm{\nabla }}_{X}Y=e^{-1}(D_X^{\omega }(eY))$. The Leibniz rule for $D^{\omega }$ gives ${\mathrm{\nabla }}_X(fY)=X(f)Y+f{\mathrm{\nabla }}_{X}Y$. Since $\omega (X)$ is $\eta$-skew, $X{g}_e(Y,Z)=g_e({\mathrm{\nabla }}_{X}Y,Z)+g_e(Y,{\mathrm{\nabla }}_{X}Z)$, so $\mathrm{\nabla }$ is metric-compatible. The equation $D_{\omega }e=0$ gives $e({\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X-[X,Y])=0$; injectivity of $e$ makes the torsion zero.

Metric compatibility and zero torsion imply the Koszul identity $2g_e({\mathrm{\nabla }}_{X}Y,Z)=X{g}_e(Y,Z)+Y{g}_e(Z,X)-Z{g}_e(X,Y)-g_e(X,[Y,Z])+g_e(Y,[Z,X])+g_e(Z,[X,Y])\mathrm{.}$ It follows by adding the metric-compatibility identities for $(X,Y,Z)$ and $(Y,X,Z)$, subtracting the one for $(Z,X,Y)$, and replacing each antisymmetric difference ${\mathrm{\nabla }}_{X}Y-{\mathrm{\nabla }}_{Y}X$ by $[X,Y]$. Nondegeneracy of $g_e$ makes this formula determine ${\mathrm{\nabla }}_{X}Y$ uniquely for every $Z$, so $\mathrm{\nabla }$ is the Levi-Civita connection of $g_e$.

Using the curvature operator defined in the Statement and expanding the vector covariant derivative gives $e(R^{\mathrm{\nabla }}(X,Y)Z)=(D_X^{\omega }{D}_Y^{\omega }-D_Y^{\omega }{D}_X^{\omega }-D_{[X,Y]}^{\omega })(eZ)=F_{\omega }(X,Y)eZ\mathrm{.}$ The derivative terms in $eZ$ cancel; the remaining coefficient is $d\omega (X,Y)+[\omega (X),\omega (Y)]=F_{\omega }(X,Y)$. Thus $F_{\omega }=0$ exactly when $R^{\mathrm{\nabla }}=0$ on a coframe region. For the precise tensor interface in [F3], metric compatibility, applied twice to $g_e(Z,W)$ and subtracting the commutator derivative, gives $g_e(R(X,Y)Z,W)=-g_e(Z,R(X,Y)W)$. Torsion-freeness and the Jacobi identity give $R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0$. Together with antisymmetry in $X,Y$, these identities give pair-exchange symmetry by rearranging the cyclic identity. Therefore $R_{abcd}=g_e(E_a,R(E_c,E_d)E_b)$ is antisymmetric in $a,b$ and in $c,d$, symmetric under pair exchange, and obeys $R_{abcd}+R_{acdb}+R_{adbc}=0$. Its first/third contraction is exactly the ${\mathrm{Ric}}_{bd}$ in the Statement, so [F3] applies with the same convention.

The metric trace of $G=0$ in dimension three is $-\frac{1}{2}\mathrm{Scal}=0$, hence $\mathrm{Scal}=0$ and $\mathrm{Ric}=0$. By [F3], this is equivalent to $R^{\mathrm{\nabla }}=0$. A BF solution on the coframe sector has the Levi-Civita connection by step 3.1; step 4.1 then gives $R^{\mathrm{\nabla }}=0$, hence $G=0$. Conversely, for an invertible $e$ with $G(g_e)=0$, take its Levi-Civita spin connection: for a $V$-valued section $v$ it is determined by $D_X^{\omega }v=e({\mathrm{\nabla }}_X(e^{-1}v))$. This is a connection in the fixed trivialization, and metric compatibility makes its coefficient matrix $\eta$-skew. Torsion-freeness gives $D_{\omega }e=0$; the trace calculation and [F3] give $R^{\mathrm{\nabla }}=0$, and step 4.1 then gives $F_{\omega }=0$. Thus the equivalence includes the necessary spin-connection identification; the Einstein equation alone would not constrain an independently chosen $\omega$. None of these conclusions excludes degenerate BF fields, for which $e^{-1}$ and the induced Lorentz-metric description are unavailable.