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If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

DefineThey sum to $t_{ij}$, with $s_{ji}=s_{ij}$ and $a_{ji}=-a_{ij}$. Interchanging two slots of a tensor still produces a tensor, so these are a symmetric second-rank tensor and an antisymmetric second-rank tensor, not merely symmetric and antisymmetric arrays in a selected frame.

For uniqueness, if $t=s+a=s^{\prime }+a^{\prime }$, then $s-s^{\prime }=a^{\prime }-a$. The common difference is both symmetric and antisymmetric, so each of its components equals its own negative and vanishes over $\mathbb{R}$. Thus the decomposition is unique. These are the symmetric and antisymmetric parts of a matrix, with their tensor transformation law preserved.

Take the four-divergence of the Proca equation. Commuting partial derivatives and using that $F^{ab}$ is an antisymmetric second-rank tensor giveConsequently, for $m\ne 0$, ${\partial }_a{A}^a=0$. This is the Lorenz constraint in Proca theory: an equation of motion enforces it, rather than a choice of gauge fixing. It leaves $(\square -m^2)A^a=0$. At $m=0$ the divergence argument supplies no such constraint; the massless gauge symmetry requires a separate treatment.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$, $\hslash =c=1$, and $\square ={\partial }_a{\partial }^a=-{\partial }_t^2+{\nabla }^2$. The Proca field has nonzero mass $m$. Vary its action, using the fact that the gauge field strength is an antisymmetric second-rank tensor:The integration by parts discards a boundary term, with the variation fixed at the boundary. Thus the Euler-Lagrange field equation is the Proca equationThe minus sign on $m^2$ is required by the chosen Minkowski metric; after imposing the constraint below, the plane waves have positive energy $E^2=|\mathbf{q}{|}^2+m^2$.

True. Since $A$ and $A^{\prime }$ are antisymmetric, the matrix $D=R^T{A}^{\prime }R-A$ is antisymmetric. Invariance of contractions with every antisymmetric second-rank tensor gives $D:B=0$ for every antisymmetric test matrix $B$. Such a matrix can be freely prescribed in one basis and then transformed to define a valid test tensor. Choose $B=D$ in that basis. Thenso $D=0$ and $A^{\prime }=RA{R}^T$. Together with the given antisymmetry, this proves that $A$ is an antisymmetric second-rank tensor. The antisymmetric contraction test for an antisymmetric tensor works because the Frobenius inner product is nondegenerate on the antisymmetric subspace.

Writing $a_k{x}_k=\mathbf{a}\cdot \mathbf{x}$ and differentiating givesIts symmetric second-rank tensor part and antisymmetric second-rank tensor part areTaking the trace makes the two symmetric terms cancel, so the divergence isContracting the antisymmetric part with the Levi-Civita symbol gives the curl

A second-rank tensor is antisymmetric whenUnder an orthogonal Cartesian coordinate change with matrix $R$,Thereforeafter relabelling the dummy indices. Hence the antisymmetric second-rank tensor property is independent of Cartesian coordinates.