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Every three-dimensional fourth-rank Cartesian tensor invariant under all proper rotations has the displayed form, without assuming index-interchange symmetry. Half-turns about the coordinate axes force every nonzero component to contain each coordinate label an even number of times. The surviving patterns have four equal labels or two equal pairs. Axis permutations identify the latter with $c_{1122}=\alpha$, $c_{1212}=\beta$ and $c_{1221}=\gamma$, while all four-equal components share a value $A$.

Evaluate the component with four copies of $(e_1+e_2)/\sqrt{2}$. Multilinearity gives $(A+\alpha +\beta +\gamma )/2$, which must equal $A$ by rotational invariance. Hence $A=\alpha +\beta +\gamma$, proving the formula. Conversely each pairing of Kronecker deltas is invariant under any orthogonal transformation. For symmetric strain only $\alpha$ and $\beta +\gamma$ matter, giving the two Lamé parameters of isotropic elasticity on tensor components.

For Cartesian tensors, let $R$ be the orthogonal matrix taking old vector components to new ones, $v_i^{\prime }=R_{ij}{v}_j$. A rank-$n$ tensor transforms by one copy of $R$ on each index:Repeated indices are summed according to Einstein notation. In general coordinates a contravariant tensor index instead carries the corresponding Jacobian matrix $\partial x^{\prime i}/\partial x^j$, and a covariant tensor index carries its inverse; in orthonormal bases of Cartesian axes their component laws coincide with the displayed form.

An isotropic tensor has identical components after every proper orthogonal transformation, so $T^{\prime }=T$ for every $R$ with $R^{T}R=I$ and $\det R=1$. Requiring reflection invariance as well is a stronger convention for odd tensor ranks. The Kronecker delta is invariant under all orthogonal transformations, while the Levi-Civita symbol is invariant under proper rotations and acquires a determinant sign under reflections.

Under an orthogonal Cartesian change of axes $Q$, the vector components transform as $v_i^{\prime }=Q_{ia}{v}_a$, while $v_k^{\prime }{v}_k^{\prime }=v_a{v}_a$ and ${\delta }_{ij}=Q_{ia}{Q}_{ja}$. Substitution givesThis is the rank-two Cartesian tensor transformation law. The nonzero assumption ensures that the denominator is positive. In matrix form $P=I-v{v}^T/(v^{T}v)$ is the orthogonal projection matrix onto the orthogonal complement of $v$.

The Cartesian tensor law is $T_{ik}^{\prime }=Q_{ia}{Q}_{kb}{T}_{ab}$. In every complete tensor contraction, summation over a transformed index contracts a pair of $Q$ factors to a Kronecker delta. In detail,Thus all three are invariant under rotations. They are $\mathrm{tr}T$, $\mathrm{tr}(T^2)$ and $\mathrm{tr}(T^3)$. For a real symmetric tensor, orthogonal diagonalization givesExpanding ${\prod }_i(\lambda -{\lambda }_i)$ and using Newton's identities to recover the characteristic polynomial from traces of powers yieldsComparison identifies $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.

Under an orthogonal Cartesian change $x_i^{\prime }=R_{ij}{x}_j$, the chain rule gives ${\partial }_i^{\prime }=(\partial x_j/\partial x_i^{\prime }){\partial }_j=R_{ij}{\partial }_j$. Thus the derivative operator transforms as a rank-one Cartesian tensor. Under a general coordinate change it has the covector transformation law when applied to a scalar; derivatives of vector components then need the usual connection terms. The calculations below use Cartesian coordinates.

The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are Cartesian tensors: transposing $T^{\prime }=RT{R}^T$ gives $(T^{\prime }{)}^T=R{T}^T{R}^T$, so $S^{\prime }=RS{R}^T$ and $A^{\prime }=RA{R}^T$. Hence this is a decomposition into a symmetric tensor and an antisymmetric second-rank tensor, rather than just a decomposition of numerical arrays.

For the axial vector of an antisymmetric second-rank tensor, contract the two Levi-Civita symbols:ThusWith this convention $Ax=x\times a$, so the displayed matrix is the negative of the cross-product matrix for $a\times x$.

For a Cartesian tensor under a proper orthogonal change $R$, the rank-$n$ law is $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, with repeated indices summed. An isotropic tensor has $T^{\prime }=T$ for every such rotation.

For rank two, $T=RT{R}^T$. A half-turn about a coordinate axis, obtained by applying a quarter-turn twice, reverses each off-diagonal component with exactly one index on that axis. Invariance forces every off-diagonal entry to vanish. A quarter-turn about the third axis exchanges $T_{11}$ and $T_{22}$; quarter-turns about the other axes equate them with $T_{33}$. Thus the isotropic second-rank tensor isConversely $R(\lambda I)R^T=\lambda I$, proving sufficiency.

A two-index Cartesian tensor has components transforming as $T_{ij}^{\prime }=R_{ik}{R}_{jl}{T}_{kl}$ under an orthogonal change of orthonormal basis. The tensor order here counts indices, rather than the matrix rank. Its scalar, symmetric-traceless and axial tensor decomposition follows by settingThen $T_{ij}=Y{\delta }_{ij}+{\Omega }_{ij}+S_{ij}$, with $\Omega$ antisymmetric and $S$ symmetric and traceless. There are three independent antisymmetric components and five independent symmetric traceless components: three off-diagonal entries and two independent diagonal ones in the latter case.

For the symmetric tensor relative to a nonzero vector $A$, put $q=A_i{A}_i>0$, $s=A_i{T}_{ij}{A}_j$, and $t=T_{ii}$. The symmetric tensor decomposition relative to an axis givesIn particular $A_i{B}_i=s/q-s/q=0$. Define the orthogonal projection $P_{ij}={\delta }_{ij}-A_i{A}_j/q$ onto the plane perpendicular to $A$, and setThis tensor is symmetric and kills $A$. Its trace is $t-s/q-2\alpha =0$. To verify the full decomposition, set $u=A/\sqrt{q}$ and split $T$ into its longitudinal component $(u^{T}Tu)u{u}^T$, two mixed components, and $PTP$. The mixed terms are exactly $B{A}^T+A{B}^T$, and $\alpha +\beta q=s/q$ gives the longitudinal coefficient. Thus all requested constraints and the reconstruction hold.

Under an orthogonal change of Cartesian coordinates with $x_i^{\prime }=R_{ia}{x}_a$, a rank-$n$ Cartesian tensor transforms as $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, summing repeated indices. An isotropic tensor is invariant under every rotation. For an isotropic fourth-rank tensor, half-turns about the coordinate axes eliminate components unless each coordinate label occurs an even number of times. Coordinate rotations make the remaining two-pair patterns depend on three constants $c_{1122}=\alpha$, $c_{1212}=\beta$, $c_{1221}=\gamma$. Invariance after rotating the first axis to $(e_1+e_2)/\sqrt{2}$ gives $c_{1111}=\alpha +\beta +\gamma$. HenceThis is a direct proof of the rank-four isotropic decomposition, without assuming any prior interchange symmetry of the four indices. Each displayed delta pairing is itself rotation invariant.

For symmetric strain, contraction gives ${\sigma }_{ij}=\alpha e_{kk}{\delta }_{ij}+(\beta +\gamma )e_{ij}$. Comparing with the Lamé parameters yieldsWrite $p=e_{kk}/3$ and $d_{ij}=e_{ij}-p{\delta }_{ij}$, so $d_{kk}=0$. Since $e_{ij}{e}_{ij}=3p^2+d_{ij}{d}_{ij}$, the strain energy density becomesThe two terms describe volumetric and shear deformation, respectively. Thus nonnegative isotropic elastic strain energy is equivalent toNecessity also follows by taking a nonzero traceless strain and a pure dilation separately. The combination $K=\lambda +2\mu /3$ is the bulk modulus; strict inequalities give positive energy for every nonzero strain.