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For a constant orthogonal Cartesian coordinate change $x_i^{\prime }=Q_{ij}{x}_j$, the tensor component transformation law givesRepeated indices are summed. Each index of an ordinary Cartesian tensor contributes one factor of $Q$. A pseudotensor would have an additional determinant factor under an orientation-reversing transformation; the rules here concern ordinary tensors.

A second-order Cartesian tensor assigns components in each orthonormal coordinate system according toHere $R$ is the matrix taking old vector components to new ones, and repeated indices are summed. This is the tensor component transformation law; an arbitrary array of nine numbers is not a tensor unless its components in different frames obey this rule. No symmetry $T_{ij}=T_{ji}$ is implied by being a second-order tensor.

Let $R$ be an orthogonal change of axes. Since $A$ is a second-order tensor, $A_{ij}^{\prime }=R_{ik}{R}_{j\ell }{A}_{k\ell }$. The scalar-contraction assumption givesfor every tensor $A$. In one fixed frame its components can be any array: in particular, choose one component at a time to be one, all others zero. Equality of the coefficients then forcesMultiplying by $R$ and $R^T$ yieldsThis is exactly the tensor component transformation law. It is the full-contraction version of the quotient theorem for Cartesian tensors. The words “for every” are essential: testing a single $A$, such as the zero tensor or the identity, would not determine all components.

Use the quarter-turn rotation matrixFor a general second-order tensor, direct multiplication givesRequiring $RT{R}^T=T$ gives $T_{11}=T_{22}$, $T_{12}=-T_{21}$, $T_{13}=-T_{23}$ together with $T_{23}=T_{13}$, and $T_{31}=-T_{32}$ together with $T_{32}=T_{31}$. Thus all mixed axial-transverse entries vanish andFor an arbitrary rotation angle $\theta$, the transverse rotation is $Q_{\theta }=\cos \theta I+\sin \theta J$, where $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. The transverse block of $T$ is $\alpha I-\omega J$, so it commutes with $Q_{\theta }$. Therefore $Q_{\theta }(\alpha I-\omega J)Q_{\theta }^T=\alpha I-\omega J$, while the axial component $\beta$ is unchanged.

Hence invariance holds under every rotation about the axis, not just the quarter-turn. This is the general axisymmetric second-order tensor; a symmetric tensor would additionally require $\omega =0$, but that restriction was not assumed.

In orthonormal Cartesian coordinates, a rank-$n$ tensor has components transforming under a change of axes $R$ asRepeated indices are summed. This is the tensor component transformation law; for rank one it is precisely the transformation law of a vector.

Use the vector triple product to write the double cross product asBoth dot products are invariant under an orthogonal change of axes, while $u^{\prime }=Ru$ and ${\Omega }^{\prime }=R\Omega$. ThereforeHence $W$ is a vector. This argument also avoids the sign issue for an individual cross product under an orientation-reversing orthogonal transformation: the double product has two such factors, whose signs cancel.

Apply the symmetric gradient and curl decomposition separately to the electric field and the magnetic field:The corresponding axial parts areThe second equalities use the vacuum Maxwell equations. Their divergence constraints also show that both symmetric gradient tensors are traceless.

An order-$n$ Cartesian tensor has $3^n$ components whose values in two orthonormal Cartesian frames satisfywith repeated indices summed as in Einstein notation. It is an isotropic tensor when its components are unchanged by every proper rotation, $R\in SO(3)$.

For rank two, a half-turn about any coordinate axis changes the sign of each component having exactly one index along that axis. Invariance therefore makes every off-diagonal component zero. Quarter-turns interchanging two coordinate axes then require $T_{11}=T_{22}=T_{33}$. Thus, without assuming symmetry in advance,Conversely, the Kronecker delta is invariant because $R_{ik}{R}_{jk}={\delta }_{ij}$, so every tensor of this form is isotropic.

For rank three the general proper-rotation invariant form iswhere ${ϵ}_{ijk}$ is the Levi-Civita symbol. The convention matters: the isotropic rank-three tensor and reflections distinction says that invariance of a polar tensor under all $R\in O(3)$ would force $b=0$. A nonzero ${ϵ}_{ijk}$ is invariant under proper rotations, and is a pseudotensor under the full orthogonal group. The shell integrals below are invariant under reflections as well, so their odd moments vanish.

Under inversion $\mathbf{x}\mapsto -\mathbf{x}$, the spherical shell and its volume element are unchanged, while the product of three scalar products changes sign. Its integral is therefore its own negative:Equivalently, its rank-three moment tensor is completely symmetric and cannot have a nonzero multiple of the antisymmetric Levi-Civita symbol.

Introduce the second moment tensor $M_{ij}={\int }_V{x}_i{x}_j{d}^{3}x$. The spherical shell is preserved by every rotation, so this is an isotropic tensor, $M_{ij}=A{\delta }_{ij}$. Its trace determines the constant:Contracting with the components of the two unit vectors givesThis is the second-order case of radial tensor moments in three dimensions.

A pseudotensor has an additional determinant factor under an orthogonal transformation. Thus its rank-three transformation law isThe contracted expression $L_{ia}{L}_{jb}{L}_{kc}{ϵ}_{abc}$ is fully antisymmetric in $i,j,k$, so it equals a scalar multiple of ${ϵ}_{ijk}$. For $(i,j,k)=(1,2,3)$ it is precisely the determinant expansion of $L$, fixing that scalar to $\det L$. ThereforeThe Levi-Civita symbol consequently has the same numerical entries in every orthonormal frame when interpreted as a pseudotensor. With an ordinary rank-three tensor transformation law alone it would instead acquire a minus sign under an orientation-reversing transformation.

Under the specified passive change of orthonormal basis, each Cartesian tensor index acquires one copy of $L$. Thus a tensor of order $n$ transforms asRepeated old-basis indices are summed from one to three. Orthogonality of the new basis gives $L{L}^T=I$. Taking determinants yields $(\det L{)}^2=1$, hence $\boxed{\det L=+1\text{ or }-1}$. The positive case preserves orientation and the negative case reverses it.

For a Cartesian change of components $x_i^{\prime }=R_{ij}{x}_j$, a rank-$n$ tensor transforms according towith repeated indices summed. An isotropic tensor has unchanged components under every proper rotation matrix.

For rank two write $T^{\prime }=RT{R}^T$. Invariance under quarter-turns implies invariance under their squares, the coordinate half-turnsEach off-diagonal entry changes sign under at least one of these, so all off-diagonal entries vanish. A quarter-turn about the third axis then interchanges the first two diagonal entries, making them equal; a quarter-turn about the first axis equates the other two. ThereforeConversely $R(\lambda I)R^T=\lambda I$ for every rotation. This proves the isotropic second-rank tensor classification without assuming that the original tensor was symmetric. Also $\lambda =T_{kk}/3$ is a scalar, since the trace is rotation-invariant.

The required contractions areSubstituting into the axis decomposition of a symmetric matrix givesIndeed $A^{T}B=-4+4=0$, $CA=0$ and $\mathrm{tr}C=-15+3+12=0$. Adding $25I+5A{A}^T+B{A}^T+A{B}^T+C$ reconstructs the given tensor.

Under an orthogonal matrix $R$, the Levi-Civita symbol satisfiesUse this twice, and use $R_{mc}{R}_{mf}={\delta }_{cf}$ for the contracted index. It follows thatTherefore $\boxed{T_{ijkl}^{\prime }=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{T}_{abcd}=T_{ijkl}}$: it is a fourth-rank isotropic tensor. Although a single alternating symbol is a pseudotensor under reflections, the two determinant factors multiply to one, so this product is an ordinary tensor even for improper orthogonal transformations.

Write the inverse tensor asInsert this into $({\delta }_{ik}+\lambda X_{ik})Y_{kj}={\delta }_{ij}$ and equate powers of $\lambda$. One obtains successivelyThus the Neumann series givesEvery contraction is an ordinary ordered matrix product. No transpose or symmetry assumption is used. Multiplying the truncated series on either side by $I+\lambda X$ leaves $I-{\lambda }^4{X}^4$, which verifies the error order. For a fixed finite matrix this expansion converges whenever $|\lambda |\Vert X\Vert <1$ in a submultiplicative matrix norm.

For a Cartesian coordinate change $x_i^{\prime }=R_{ij}{x}_j$ with $R$ an orthogonal matrix, a rank-$n$ tensor obeyswith repeated indices summed according to Einstein notation. A tensor is isotropic if $T_{i_1\cdots i_n}^{\prime }=T_{i_1\cdots i_n}$ for every proper rotation matrix $R$. Under proper rotations the Levi-Civita symbol is an isotropic third-rank tensor; if reflections are included, it is a pseudotensor. The convention using proper rotations is the one relevant to the rank-three classification here.

Under a proper rotation matrix $R$, the product of coordinates transforms as a rank-$n$ tensor. The unit ball is unchanged by the substitution $\mathbf{y}=R\mathbf{x}$, whose Jacobian determinant is one. HenceThus $S$ is an isotropic tensor. An isotropic second-rank tensor has the form $S_{ij}=A{\delta }_{ij}$. Taking the trace and using spherical coordinates,Therefore the isotropic second moment of a ball isAn isotropic third-rank tensor has the form $B{ϵ}_{ijk}$, but $S_{ijk}$ is symmetric in every pair of indices, forcing $B=0$. Alternatively, $\mathbf{x}\mapsto -\mathbf{x}$ makes its integrand odd on the ball. Thus

For a local electric-dipole response of order $n$, writeIts susceptibility is a rank-$(n+1)$ tensor. Initially it has $3^{n+1}$ components in three dimensions, before symmetries and frequency constraints are applied. nonlinear-susceptibility tensor symmetry involves several logically different restrictions.

Neumann's principle requires invariance under the material point group. For an operation $R$,For inversion, $R=-I$, so ${\chi }^{(n)}=(-1{)}^{n+1}{\chi }^{(n)}$. Thus all even orders vanish in a centrosymmetric material's local bulk electric-dipole response. In particular ${\chi }^{(2)}=0$ there, while ${\chi }^{(3)}$ is permitted. Surface symmetry breaking, an applied bias or nonlocal magnetic/quadrupole response lies outside that statement. A noncentrosymmetric crystal permits some second-order components, but other point-group operations can still eliminate a chosen component.

Intrinsic permutation symmetry follows from interchanging commuting input field factors, provided the associated frequency labels are interchanged too. At second order,For equal input frequencies this makes the last two spatial indices symmetric, reducing 27 components to 18 before material symmetries. At unequal frequencies it does not identify the two index arrangements with the arguments held fixed. Reality of the time-domain polarization separately givescausality imposes dispersion relations; neither arbitrary real coefficients nor full index symmetry should be inferred merely from the word lossless.

Kleinman symmetry is a stronger approximation far from relevant resonances when frequency dispersion can be neglected. A frequency-independent effective interaction energy makes its derivatives interchangeable, allowing full permutation of output and input spatial indices. At second order a fully symmetric rank-three tensor has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ components. Close to resonance or across a dispersive band, that reduction can fail even when absorption at the measured frequencies is small. The original analysis distinguishes high-resonance electronic response from strongly dispersive lattice mechanisms and identifies additional symmetry constraints on the former.^[ref]

For an isotropic fourth-rank susceptibility,If the relevant input slots are fully interchangeable, the coefficients coincide and ${\chi }_{xxxx}^{(3)}=3{\chi }_{xxyy}^{(3)}$. With distinct or mixed-sign frequency slots, frequency arguments must be followed and isotropy alone leaves the three displayed structures. These distinctions explain why tensor contractions, pump polarizations and operating frequencies must be specified before comparing effective self- and cross-phase coefficients.

Use coordinates $(ct,x,y,z)$ and put $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. For an observer moving at $+v\hat{\mathbf{x}}$, the coordinate Lorentz boost isThe given electromagnetic field tensor transforms as $F^{\prime }=\Lambda F{\Lambda }^{\mathsf{T}}$. For example,whileThe $01$ component is unchanged because its boost determinant is ${\gamma }^2(1-{\beta }^2)=1$. The $23$ component is unchanged because neither index lies in the boost plane. Reading off all six components givesThe signs follow the stated positive observer velocity and the supplied tensor convention. In vector form for a boost in any direction, ${\mathbf{E}}_{\parallel }^{\prime }={\mathbf{E}}_{\parallel }$, ${\mathbf{E}}_{\perp }^{\prime }=\gamma ({\mathbf{E}}_{\perp }+\mathbf{v}\times \mathbf{B})$, with analogous ${\mathbf{B}}_{\perp }^{\prime }=\gamma ({\mathbf{B}}_{\perp }-\mathbf{v}\times \mathbf{E}/c^2)$.

Fix coordinates $x^{\mu }=(ct,x,y,z)$ and use the following convention for the contravariant electromagnetic field tensor:For $O^{\prime }$ moving with velocity $+v\hat{\mathbf{x}}$ relative to $O$, the Lorentz transformation hasThe rank-two tensor transformation rule isFor example,Similarly, $F^{\prime 03}=-\gamma (E_z+v{B}_y)/c$ and $F^{\prime 13}=\gamma (B_y+v{E}_z/c^2)$. The $01$ component is unchanged because ${\gamma }^2(1-{\beta }^2)=1$, and the $23$ component is unchanged since neither index lies in the boost plane. Reading off the six field components givesAn alternative overall sign convention for $F^{\mu \nu }$ gives the same physical Lorentz transformation of electromagnetic fields when its identifications with $\mathbf{E}$ and $\mathbf{B}$ are used consistently.