Cartesian tensor ($T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$)

A Cartesian tensor has components in each orthonormal basis of a Euclidean space, related by the displayed rule under an orthogonal component change $R$. Each index contributes one transformation matrix. The Euclidean inner product identifies upper and lower indices in orthonormal coordinates, so the distinction between covariant and contravariant components does not change this rule. An isotropic tensor has unchanged components under all proper rotations; requiring invariance under reflections as well excludes the ordinary odd-parity cases. A Cartesian second-rank tensor is the matrix case, with transformation $T^{\prime }=RT{R}^T$.

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.