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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.

For a smooth one-form $A$, the components ${\partial }_a{A}_b-{\partial }_b{A}_a$ transform as a covariant rank-two tensor. The non-tensorial second derivatives in the ordinary covector derivative under nonlinear coordinate changes are symmetric in $a,b$ and cancel. For a torsion-free connection the same components are ${\nabla }_a{A}_b-{\nabla }_b{A}_a$. This is the coordinate expression of $dA$, without choosing a metric.

In the weak-field, short-wavelength averaging regime, a transverse-traceless tensor strain $H_{ij}$ carries the displayed effective stress-energy tensor in units $c=1$. The average spans several wave periods while remaining small compared with the background curvature scale. The stress is second order in amplitude, so it is absent from a first-order vacuum Linearized Einstein equations calculation. For a plane wave, summing the two equal transverse diagonal contributions gives flux $(16\pi G{)}^{-1}⟨{\dot{H}}_{+}^2+{\dot{H}}_{\times }^2⟩$. This effective averaged stress is not an arbitrary pointwise gravitational energy tensor.

For a slender centreline $\mathbf{X}(\theta )=(A\cos \theta ,B\sin \theta ,b\theta )$, $-\pi \le \theta \le \pi$, let $s^{\prime }(\theta )=(A^2{\sin }^2\theta +B^2{\cos }^2\theta +b^2{)}^{1/2}$ and use local force density $\mathbf{f}=c(2I-\mathbf{tt})\mathbf{v}$, $\mathbf{t}=d\mathbf{X}/ds$, $c=2\pi \mu /\ln (L/R)$. Rotation with angular speed $\Omega$ gives $\mathbf{t}\cdot \mathbf{v}=\Omega AB/s^{\prime }$, henceAxial translation with speed $W$ gives exactly the same coefficient for the torque: $G_z=-cWABb\int d\theta /s^{\prime }$. Indeed $(\mathbf{X}\times \mathbf{t}{)}_z=AB/s^{\prime }$. Both forces and torques here are exerted on the fluid. In the circular case $A=B=a$, this reduces to $F_z/\Omega =G_z/W=-2\pi c{a}^{2}b/\sqrt{a^2+b^2}$. The equality is also a consequence of the Lorentz reciprocal theorem, since the local resistance tensor is symmetric.

A Cartesian tensor invariant under all proper rotations fixing a unit vector $n$ is axisymmetric. Without assuming a symmetric tensor, its general second-order form is the displayed combination of the transverse identity, axial projector and axial antisymmetric part. For $n=e_3$ this isIndeed, imposing $RT{R}^T=T$ for a quarter-turn about $e_3$ forces $T_{11}=T_{22}$, $T_{12}=-T_{21}$ and all mixed axial-transverse entries to vanish. The planar block is a linear combination of the identity and the planar rotation generator, so it commutes with every rotation about the axis. Thus quarter-turn invariance already suffices for a second-order tensor. A symmetric tensor has $\omega =0$, recovering the axisymmetric symmetric rank-two tensor form. The antisymmetric term is allowed because only proper rotations about the fixed axis are imposed.

A tensor derivative of polarization with respect to a sublattice displacement at specified macroscopic electrical conditions. It includes electronic rearrangement and need not equal a formal ionic charge.

Suppose a homogeneous bulk material is unchanged when all spatial coordinates are reversed. Both electric polarization and electric field reverse under this inversion, so a local electric dipole response must obey $P(-E)=-P(E)$. Substituting the nonlinear optical susceptibility expansion givesFor every sufficiently small $E$, all even coefficients vanish. In the full tensor relation, two reversed input electric fields leave their product unchanged while the output electric polarization changes sign, forcing every bulk electric-dipole component of ${\chi }^{(2)}$ to vanish.

The qualification matters: the argument assumes inversion symmetry, a local bulk electric dipole response, and no applied bias that breaks that symmetry. A surface breaks bulk inversion symmetry; spatially varying or magnetic responses also lie outside this local electric-dipole model. Thus detecting second-harmonic generation at a surface or through a weaker magnetic contribution does not contradict the bulk selection rule. Conversely, absence of inversion symmetry permits some quadratic components but does not guarantee a nonzero component in every beam geometry.

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 associative algebras of finite dimension over a field $k$, write a central tensor as $x={\sum }_j{a}_j\otimes b_j$ in a basis of $B$. Commutation with every $a\otimes 1$ forces each $a_j$ into the center of an associative algebra $Z(A)$. Reexpand in a basis of $Z(A)$ and commute with every $1\otimes b$ to force the second coefficients into $Z(B)$. The reverse inclusion follows by multiplying pure tensors.

Let $A,B$ be unital associative algebras over a field, each a simple ring, and $J\ne 0$ a two-sided ideal of their tensor product. Choose $0\ne x={\sum }_{i=1}^n{a}_i\otimes b_i\in J$ of minimal two-factor tensor rank. Both factor families are linearly independent. Since $A{a}_{1}A=A$, choose ${\sum }_j{u}_j{a}_1{v}_j=1$ and replace $x$ by ${\sum }_j(u_j\otimes 1)x(v_j\otimes 1)$. The result has first coefficient $1$, is nonzero by the linear independence of the $b_i$, and retains minimal two-factor tensor rank; its first factor family is therefore still linearly independent. Now apply the same construction in $B$ to make the first second-factor coefficient $1$. The result is again nonzero of minimal two-factor tensor rank, and both its coefficient families are linearly independent.

If $Z(A)=k$, commutation with $a\otimes 1$ gives a tensor in $J$ expressible with at most $n-1$ pure tensors, hence zero by minimality. Independence forces every $c_i$ to be central, hence a scalar multiple of $1$. Independence of $1,c_2,\dots ,c_n$ then forces $n=1$. Thus $1\otimes 1\in J$, proving simplicity of the tensor product without a splitting-field argument.

For a torsion-free connection, the commutator of coordinate covariant derivatives vanishes on a scalar field and acts on a vector field through the displayed Riemann curvature tensor. On a covector field it instead gives $[{\nabla }_{\alpha },{\nabla }_{\beta }]A_{\mu }=-R^{\nu }{}_{\mu \alpha \beta }{A}_{\nu }$. The curvature commutator is a tensor derivation, so each index of a general tensor contributes independently with the corresponding sign. Including the lower derivative index in the second covariant derivative is essential for this identity.

For harmonic electric fields with time convention $e^{i\omega t}$, write $\mathbf{D}={ϵ}^{\ast }(\omega )\mathbf{E}$ for their complex amplitudes, where $\omega$ is angular frequency. The real part ${ϵ}^{\prime }$ describes the in-phase electric displacement field; the minus imaginary part ${ϵ}^{\prime \prime }$ describes the loss component. The sign reverses with the opposite time convention. Complex permittivity can also be a tensor in a linear anisotropic dielectric.

A type-$(1,1)$ tensor identifies with an endomorphism of a finite-dimensional vector space. Composing two such maps produces another type-$(1,1)$ tensor, because $(X,\eta )\mapsto \eta (B(A(X)))$ is a multilinear map. In output-first components the result is $C^a{}_b=B^a{}_c{A}^c{}_b$; the contracted index implements composition, and a basis change conjugates the corresponding matrices.

The tensor relating local current density to electric field. Its inverse, when it exists, is the electrical resistivity tensor; sheet conductivity uses surface current density instead.

For a metric-compatible affine connection, its difference from the Levi-Civita connection is the contorsion tensor. It is determined by the torsion tensor. In derivative-last notation let $T_{ab}{}^c={\Gamma }^c{}_{ab}-{\Gamma }^c{}_{ba}$, the negative of geometric torsion, and lower the last slot with the metric tensor. Then $C_{abc}=\frac{1}{2}(T_{abc}+T_{bca}-T_{cab})$. This is the actual correction added to the Levi-Civita connection; a convention writing $\Gamma =S-K$ has $K=-C$.

If an orthogonal matrix $Q$ satisfies $\dot{Q}=WQ$ for the spin tensor $W^T=-W$, the Jaumann derivative becomes an ordinary derivative in that frame:This follows by differentiating all three factors and using ${\dot{Q}}^T=-Q^{T}W$. It converts a tensor relaxation law using the Jaumann derivative into a componentwise ordinary differential equation.

If a differential operator $L$ is a linear map and has nonzero kernel of a linear map, the equation $LA=J$ determines $A$ only up to a homogeneous solution. Lorentz covariance of $L$ and $J$ therefore does not force independently selected solutions in different inertial frames to obey a particular tensor transformation law. A covariant Green function and consistently transformed homogeneous data do provide such a choice. For the four-potential in Lorenz gauge, $\square A^{\mu }=-{\mu }_0{J}^{\mu }$ still permits $A^{\mu }\mapsto A^{\mu }+{\partial }^{\mu }\chi$ with $\square \chi =0$, a residual Lorenz gauge symmetry. Constant spatial potentials for zero sources are a concrete example.

Use the derivative-index-first velocity gradient $G_{ij}={\partial }_i{u}_j$ and $\omega =G-G^T$. For an objective time derivative the tensor transformation $A^{\ast }=QA{Q}^T$ under $x^{\ast }=Q(t)x+c(t)$ must imply $\mathcal{D}{A}^{\ast }/\mathcal{D}t=Q(\mathcal{D}A/\mathcal{D}t)Q^T$. If $R=\dot{Q}{Q}^T$, then $G^{\ast }=QG{Q}^T-R$ andConsequently the extra terms cancel in $D_{t}A+\frac{1}{2}(\omega A-A\omega )$, proving covariance. With the component-index-first gradient $L_{ij}={\partial }_j{u}_i$, the same physical rate is $D_{t}A-WA+AW$, with spin tensor $W=(L-L^T)/2=-\omega /2$. Changing the gradient convention without changing the commutator sign destroys objectivity.

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.

The deviatoric part of a three-dimensional stress tensor is ${\mathbf{\sigma }}_{\mathrm{dev}}=\mathbf{\sigma }-\mathrm{tr}(\mathbf{\sigma })\mathbf{I}/3$. It is traceless. Adding an isotropic pressure does not change a normal-stress difference.