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Use the convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$, with $C_a{}^b{}_c=-C_c{}^b{}_a$. Expanding the Jacobi identity on three basis elements givesso each coefficient satisfiesAntisymmetry in the input indices makes this equivalent toThe vertical bars exclude the summed index from antisymmetrization.

In dimension three choose a nonzero contravariant volume form ${ϵ}^{abc}$ on ${\mathfrak{g}}^{\ast }$ and its dual ${ϵ}_{abc}$, normalized to have component $1$ on an oriented basis. The three-dimensional Lie algebra structure decomposition starts withThe contraction identifies an antisymmetric pair of indices with one dual index. Every such matrix splits uniquely into its symmetric part $n^{bd}=A^{(bd)}$ and antisymmetric part. The latter has the unique representationThus $n$ is a symmetric contravariant tensor and $a$ a covector, andUsing ${ϵ}_{acd}{ϵ}^{efd}={\delta }_a^e{\delta }_c^f-{\delta }_a^f{\delta }_c^e$ gives the inverse formulaThe factors $1/2$ follow from the unnormalized contraction in this particular question.

Only the alternating triple $(e_1,e_2,e_3)$ has to be checked. Substitution of the inverse formula into its Jacobiator givesOne can see the cancellation explicitly by using an auxiliary positive inner product and diagonalizing the symmetric matrix $n$ by an orientation-preserving orthogonal change of basis. If its diagonal entries are $n_1,n_2,n_3$, the brackets areIn their Jacobiator, the terms involving two $n$'s or two $a$'s cancel and the remaining terms are $-\frac{1}{2}(n_1{a}_1{e}_1+n_2{a}_2{e}_2+n_3{a}_3{e}_3)$. Returning to the original basis proves the displayed formula. ThereforeThis imposes three algebraic conditions on the nine independent entries of the original antisymmetric structure constants; if $a\ne 0$, it must lie in the kernel of $n$.