Cartesian second-rank tensor ($T\prime =RT{R}^T$)

A Cartesian second-rank tensor has component matrix $T$ in each orthonormal basis, with transformation $T\prime =RT{R}^T$ under an orthogonal component change $R$. Both indices transform; this is why assigning arbitrary matrices independently in different bases does not define a tensor. The determinant, trace, and Frobenius inner product with another such tensor are unchanged by these basis transformations.

For a real symmetric Cartesian second-rank tensor $T$ and a nonzero vector $A$, require $B\cdot A=0$ and $C$ symmetric and traceless with $CA=0$. Put $q=A^{T}A$, $s=A^{T}TA$, $t=\mathrm{tr}T$, and $P=I-A{A}^T/q$. The unique decomposition isThe orthogonal projection $P$ isolates the plane perpendicular to $A$. Splitting $I=P+A{A}^T/q$ on both sides of $T$ gives its longitudinal, mixed and transverse blocks, proving reconstruction. The transverse trace fixes $\alpha$, the longitudinal component fixes $\beta$, and the mixed block fixes $B$, proving uniqueness. The remaining transverse block $C$ has two degrees of freedom. Together $\alpha ,\beta ,B,C$ carry $1+1+2+2=6$ components, the number in a symmetric three-by-three matrix.

For a differentiable three-dimensional vector field $F$, split its gradient tensor into symmetric and antisymmetric parts. The symmetric part is $S_{ij}=({\partial }_i{F}_j+{\partial }_j{F}_i)/2$. Contracting the antisymmetric part with the Levi-Civita symbol gives the axial vector $u_k={ϵ}_{kij}{\partial }_i{F}_j/2=(\nabla \times F{)}_k/2$, proving the decomposition. If $\nabla \cdot F=0$, the divergence of the symmetric part is ${\partial }_j{S}_{ij}=\Delta F_i/2$. This lets a vector Laplacian be converted into a boundary integral of a symmetric tensor by the divergence theorem.

A Cartesian second-rank tensor in three dimensions invariant under quarter-turn rotations about all three coordinate axes is a scalar multiple of the identity matrix. Squaring the quarter-turns gives coordinate half-turns; invariance under these forces every off-diagonal component to vanish. The quarter-turns then equate the three diagonal components. This conclusion does not require the original tensor to be symmetric.

Every three-dimensional Cartesian second-rank tensor splits into scalar, symmetric-traceless, and antisymmetric parts. Their coefficients are $P=P_{kk}/3$, $S_{ij}=(P_{ij}+P_{ji})/2-P{\delta }_{ij}$, and $A_k={ϵ}_{kij}{P}_{ij}/2$. The epsilon contraction identity reconstructs the antisymmetric part and proves uniqueness. Under proper rotations $A$ behaves as a vector, but under reflections it is an axial vector.

For a constant orthogonal Cartesian coordinate change $x^{\prime }=Rx$, a Cartesian second-rank tensor obeys $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$. The chain rule gives ${\partial }_j^{\prime }=R_{jm}{\partial }_m$. Consequently ${\partial }_j^{\prime }{T}_{ij}^{\prime }=R_{ik}{R}_{jl}{R}_{jm}{\partial }_m{T}_{kl}=R_{ik}{\partial }_l{T}_{kl}$, the transformation rule for a vector. Under non-Cartesian position-dependent coordinate changes, partial derivatives must be replaced by the appropriate covariant derivative.

A Cartesian second-rank tensor invariant under all rotations about an oriented unit axis $n$ has the displayed form. A half-turn about the axis eliminates the mixed axial-transverse entries, and a quarter-turn makes the transverse block a scalar matrix plus an antisymmetric planar rotation generator. Conversely both transverse terms commute with every rotation about the axis. The antisymmetric term is allowed unless symmetry or an additional axis-reversing rotation is imposed. This differs from an axially invariant second-rank tensor for an unoriented axis.

For a real conductivity tensor invariant under every rotation about a unit vector $n$, the electric current density is $J=a{E}_{\perp }+d{E}_{\parallel }+hn\times E_{\perp }$. Its collinearity with a nonzero real electric field $E$ has three cases. Axial fields are always mapped to an axial current. Purely transverse fields give a collinear current exactly when $h=0$, equivalently when this tensor is symmetric. Fields with nonzero axial and transverse parts require both $h=0$ and $a=d$, making the tensor isotropic. The zero vector is included as a scalar multiple; requiring a nonzero current also requires the relevant scalar conductivity to be nonzero.

To prove necessity, take the dot product of $J=\lambda E$ with $n\times E_{\perp }$. If $E_{\perp }\ne 0$ this gives $h=0$; the transverse and axial equations then respectively give $\lambda =a$ and, when $E_{\parallel }\ne 0$, $\lambda =d$. Reversing these steps proves sufficiency. Rotations about the axis alone permit $h\ne 0$; symmetry under a transverse half-turn would eliminate this antisymmetric term.

A Cartesian second-rank tensor invariant under half-turns about all three coordinate axes is diagonal. Each half-turn has diagonal signs, and each off-diagonal entry changes sign under at least one of them. Invariance therefore kills all off-diagonal entries. It does not make the diagonal entries equal: that requires additional rotational invariance, as for an isotropic second-rank tensor.