Reconstruction of a Weyl tensor from its electric and magnetic parts: symmetries, trace-freeness and recovery

(All component families.) Expanding the definition with [L1] gives $C_{0a0b}=E_{ab}$, $C_{0abc}=B_{ap}{ϵ}_{pbc}$, $C_{ab0c}=B_{cp}{ϵ}_{pab}$ and $C_{abcd}={\delta }_{ac}{E}_{bd}+{\delta }_{bd}{E}_{ac}-{\delta }_{ad}{E}_{bc}-{\delta }_{bc}{E}_{ad}$; components with repeated zeros follow from antisymmetry. Antisymmetry of each pair and pair interchange now follow from symmetric $E,B$. For the cyclic identity, families with two zeros cancel by symmetric $E$, while purely spatial families cancel directly in the four electric terms. For one-zero families, the only nontrivial fully alternating sum is $B_{ap}{ϵ}_{pbc}+B_{bp}{ϵ}_{pca}+B_{cp}{ϵ}_{pab}=({\mathrm{tr}}_{\delta }B){ϵ}_{abc}=0$; sums with repeated spatial indices cancel by symmetry of $B$. Thus the full cyclic identity holds.

(Every trace and recovery.) The Ricci-type contraction gives ${\gamma }^{ac}{C}_{a0c0}=-{\mathrm{tr}}_{\delta }E=0$, ${\gamma }^{ac}{C}_{a0cb}=0$ by symmetric $B$ against epsilon, and ${\gamma }^{ac}{C}_{aicj}=E_{ij}-{\sum }_a{C}_{aiaj}=-({\mathrm{tr}}_{\delta }E){\delta }_{ij}=0$. Pair interchange and the two antisymmetries reduce every cross-pair trace to these, while within-pair traces vanish by antisymmetry. The electric recovery is already in step 1.1. For the magnetic recovery, raising two spatial indices introduces no sign and $C_{0a0b}^{\ast }=\frac{1}{2}{\sum }_{c,d}{C}_{0acd}{ϵ}_{0bcd}=\frac{1}{2}{B}_{ap}{ϵ}_{pcd}{ϵ}_{bcd}=B_{ab}$ by [L2]. This proves (a),(b).

(Full inverse and bijection.) Conversely let $W$ have the algebraic Weyl symmetries and zero traces. Its electric parts are symmetric and tracefree by pair interchange and contraction; For the magnetic parts, $B_{ab}=\frac{1}{2}{ϵ}_{bcd}{W}_{0acd}$; contracting $a=b$ gives the cyclic sum $W_{0123}+W_{0231}+W_{0312}=0$, and contracting ${ϵ}_{eab}{B}_{ab}$ gives a Ricci-type trace of $W$, hence zero. In three dimensions this latter vanishing is exactly symmetry of $B$, so $B$ is symmetric and tracefree. The epsilon inverse in [L2] recovers $W_{0abc}=B_{ap}{ϵ}_{pbc}$ from $B$. Pair interchange fixes $W_{ab0c}$. The remaining spatial algebraic curvature has six independent components: identify its antisymmetric pairs with the three basis pairs $(23),(31),(12)$, giving a symmetric $3\times 3$ matrix. Its spatial Ricci contraction is determined by $W_{0a0b}=E_{ab}$ because the full trace is zero. This contraction is injective on the six spatial components: in dimension three its three diagonal equations determine $W_{1212},W_{1313},W_{2323}$, and its three off-diagonal equations determine the remaining three pair-matrix entries. The explicit spatial formula in step 1.1 has that contraction, so it is the unique spatial tensor. Hence every component equals the reconstruction, which is an inverse to the electric-magnetic map. Linearity and step 2.1 prove the claimed bijection, with no additional curvature variables.