Chern–Simons boundary geometry and the three-dimensional gravity interface

Define $S_{abcd}$ to be the right-hand side of the displayed formula. Symmetry of $g$ and $\mathrm{Ric}$ gives antisymmetry of $S$ in each index pair and symmetry under exchanging the pairs. Substitution into $S_{abcd}+S_{acdb}+S_{adbc}$ cancels the Ricci and scalar terms pairwise, so $S$ satisfies the first Bianchi identity and is an algebraic curvature tensor. Contracting the first and third indices gives $g^{ac}{S}_{abcd}=3{\mathrm{Ric}}_{bd}-{\mathrm{Ric}}_{bd}-{\mathrm{Ric}}_{bd}+\mathrm{Scal}{g}_{bd}-\frac{1}{2}\mathrm{Scal}(3g_{bd}-g_{bd})={\mathrm{Ric}}_{bd}\mathrm{.}$ Thus $W:=R-S$ is an algebraic curvature tensor with zero Ricci contraction.

In the basis of [A1], an algebraic curvature tensor is represented by a symmetric $3\times 3$ matrix on the basis bivectors $e_0\wedge e_1,e_0\wedge e_2,e_1\wedge e_2$. Write its diagonal entries as $x=W_{0101}$, $y=W_{0202}$, and $z=W_{1212}$. The three diagonal zero-Ricci equations are $x+y=0$, $-x+z=0$, and $-y+z=0$. They imply $2x=0$, hence $x=y=z=0$.

The three off-diagonal zero-Ricci equations are $\mathrm{Ric}(W{)}_{01}=W_{0212}=0$, $\mathrm{Ric}(W{)}_{02}=-W_{0112}=0$, and $\mathrm{Ric}(W{)}_{12}=-W_{0102}=0$. They set the three off-diagonal entries of the same symmetric bivector matrix to zero. Therefore $W=0$, so $R=S$ and the displayed identity holds. If $\mathrm{Ric}=0$, then its scalar contraction is zero and the formula gives $R=0$; conversely $R=0$ implies its Ricci contraction vanishes.

Differentiate the Chern–Simons integrand along $A+s\alpha$. The quadratic term contributes $⟨\alpha \wedge dA+A\wedge d\alpha ⟩$. For the cubic term, the graded bracket and invariant pairing give the cyclic identities $⟨A\wedge [\alpha ,A]⟩=⟨\alpha \wedge [A,A]⟩$ and $⟨A\wedge [A,\alpha ]⟩=⟨\alpha \wedge [A,A]⟩$ for degree-one $A,\alpha$. Thus the three differentiated terms in $\frac{1}{3}⟨A\wedge [A,A]⟩$ are equal, and their average is $⟨\alpha \wedge [A,A]⟩$. The full unintegrated variation is $⟨\alpha \wedge dA+A\wedge d\alpha +\alpha \wedge [A,A]⟩$.

Differentiate $F_{A+s\alpha }$ at $s=0$; bilinearity of the bracket gives $\delta F_A=D_A\alpha$. Thus the BF variation before integration by parts is ${\int }_M⟨\beta \wedge F_A+B\wedge D_A\alpha ⟩$.

The graded Leibniz rule gives $d⟨A\wedge \alpha ⟩=⟨dA\wedge \alpha ⟩-⟨A\wedge d\alpha ⟩$. Since a two-form commutes with a one-form under wedge, $⟨A\wedge d\alpha ⟩=⟨\alpha \wedge dA⟩-d⟨A\wedge \alpha ⟩$. Substitute this into step 1.1, combine $2dA+[A,A]=2F_A$, and apply [F2] to obtain the stated Chern–Simons formula, including its boundary sign.

For variations compactly supported in $\mathrm{int}M$, the boundary term in step 2.1 is zero. If stationarity holds for every such $\alpha$, then ${\int }_M⟨\alpha \wedge F_A⟩=0$ for every local test variation. If $F_A$ were nonzero at some point, nondegeneracy of the pairing would select an algebra element pairing nontrivially with one component; continuity preserves its sign on a smaller chart, and a nonnegative bump supported there would make the integral nonzero. Hence $F_A=0$ on $\mathrm{int}M$.

For constant tangent vectors ${\alpha }_1,{\alpha }_2$ on the affine space ${\mathcal{A}}_{\mathrm{\Sigma }}$, differentiating the potential current ${\theta }_A(\alpha )=-\lambda ⟨A\wedge \alpha ⟩$ from step 2.1 gives $D_{{\alpha }_1}\mathrm{\Theta }({\alpha }_2)=-\lambda {\int }_{\mathrm{\Sigma }}⟨{\alpha }_1\wedge {\alpha }_2⟩$ and $D_{{\alpha }_2}\mathrm{\Theta }({\alpha }_1)=-\lambda {\int }_{\mathrm{\Sigma }}⟨{\alpha }_2\wedge {\alpha }_1⟩$. The field-space exterior derivative is their difference; antisymmetry of the wedge gives ${\mathrm{\Omega }}_{\mathrm{\Sigma }}=-2\lambda {\int }_{\mathrm{\Sigma }}⟨{\alpha }_1\wedge {\alpha }_2⟩$. Its coefficients are constant, so its field-space exterior derivative is zero.

Invariance of the pairing makes $D_A$ obey the graded Leibniz rule on paired forms. Therefore $d⟨B\wedge \alpha ⟩=⟨D_{A}B\wedge \alpha ⟩-⟨B\wedge D_A\alpha ⟩$. As in step 2.1, the two-form $D_{A}B$ commutes under wedge with $\alpha$. Apply [F2] to obtain the stated BF formula. For variations supported in the interior the boundary integral vanishes. Independent compactly supported $\beta$ and $\alpha$ then test $F_A$ and $D_{A}B$ separately; the local nondegenerate-pairing bump argument in step 3.1 shows each must vanish.

At a flat $a$, $D_a^2ϵ=[F_a,ϵ]=0$, so $D_aϵ$ is tangent to the flat locus. For a tangent $\alpha$, $D_a\alpha =0$. Invariance of the pairing yields $d⟨ϵ,\alpha ⟩=⟨D_aϵ\wedge \alpha ⟩+⟨ϵ,D_a\alpha ⟩$. Integrating and applying [F2] on $\mathrm{\Sigma }$ gives ${\int }_{\mathrm{\Sigma }}⟨D_aϵ\wedge \alpha ⟩={\int }_{\mathrm{\partial }\mathrm{\Sigma }}⟨ϵ,\alpha ⟩$, which proves the gauge-contraction formula. If $ϵ{\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Sigma }}=0$, this contraction vanishes for every tangent $\alpha$. If $ϵ$ is fixed under variation, direct differentiation of $Q_{ϵ}(a)=-2\lambda {\int }_{\mathrm{\partial }\mathrm{\Sigma }}⟨ϵ,a⟩$ gives exactly the same contraction.

For $\lambda \mathrm{\ne }0$, nonzero boundary parameters can in fact give nonzero contractions. Take $\mathfrak{g}=\mathbb{R}$ with pairing multiplication, the unit disk $\mathrm{\Sigma }\subset {\mathbb{R}}^2$ with its standard orientation, $a=0$, $ϵ(x,y)=x$, and $\alpha =dy$. Then $F_a=0$, $D_a\alpha =d(dy)=0$, and $D_aϵ=dx$. Polar coordinates give ${\int }_{\mathrm{\Sigma }}dx\wedge dy={\int }_0^{2\pi }{\int }_0^{1}rdrd\theta =\pi$, so ${\mathrm{\Omega }}_{\mathrm{\Sigma }}(D_aϵ,\alpha )=-2\lambda \pi \mathrm{\ne }0$ when $\lambda \mathrm{\ne }0$. This establishes the stated possibility without asserting that every nonzero boundary parameter has a nonzero charge. If $\lambda =0$, the formulas for ${\mathrm{\Theta }}_{\mathrm{\Sigma }}$, ${\mathrm{\Omega }}_{\mathrm{\Sigma }}$ and $Q_{ϵ}$ all have the common factor $\lambda$ and vanish; hence every gauge direction is null in this case.

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.