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Spherical symmetry makes moments over a ball isotropic tensors. Odd moments vanish under $x\mapsto -x$, while the second and fourth moments areFor the second moment, a tensor contraction gives $3\alpha =\int r^{2}dV=4\pi a^5/5$. For the fourth, complete index symmetry makes the three coefficients equal and a double tensor contraction gives $15\beta =\int r^{4}dV=4\pi a^7/7$. These identities turn polynomial volume integrals into scalar radial integrals and are useful for inertia and rotational averages.

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 $T^{\prime }=RT{R}^T$.

For a uniform ball of radius $R>0$ and positive mass $M$, let $G$ be its centre of mass. Spherical symmetry gives the second-moment matrix $\int u{u}^{T}dm=(M{R}^2/5)I$, so the central inertia tensor is $(2M{R}^2/5)I$. The tensor parallel-axis theorem gives the displayed shifted tensor. For $X\ne 0$ its eigenvector along $X$ has eigenvalue $2M{R}^2/5$, while the two-dimensional orthogonal complement has repeated eigenvalue $M(2R^2/5+|X{|}^2)$. Thus it is an isotropic tensor exactly at $P=G$. Displacement along a chosen axis does not change that axis's moment; the transverse moments increase by $M|X{|}^2$.

Contract an isotropic tensor with the Kronecker delta or the Levi-Civita symbol, using invariance under proper rotations. The three displayed contractions have ranks one, two and three. The only isotropic vector is zero; an isotropic second-rank tensor is a scalar delta, and an isotropic third-rank tensor is a scalar epsilon. For $T_{ijkl}=\alpha {\delta }_{ij}{\delta }_{kl}+\beta {\delta }_{ik}{\delta }_{jl}+\gamma {\delta }_{il}{\delta }_{jk}$, direct contraction gives $\mu =3\alpha +\beta +\gamma$ and $\nu =\beta -\gamma$.

Here $V_d$ is the volume of the unit ball in $d$-dimensional Euclidean space. Rotational invariance makes the integral an isotropic tensor, while complete symmetry in its four indices equates the three coefficients. Contracting $i=j$ and $k=l$ gives $d(d+2)$ times their common value. The same contraction of the integral is ${\int }_{|x|\le R}|x{|}^4{d}^{d}x=d{V}_d{R}^{d+4}/(d+4)$, proving the formula. For $d=3,R=1$ the coefficient is $4\pi /105$. This is the fourth-order counterpart of the isotropic second moment of a ball.

Under proper rotations a third-rank isotropic tensor is proportional to the Levi-Civita symbol. Its contraction with a field vector is antisymmetric in the two stress indices. A physical symmetric stress therefore vanishes in this linear constitutive model. Under the full orthogonal group an ordinary isotropic third-rank tensor is already zero; the orientation-sensitive Levi-Civita tensor requires the usual pseudotensor distinction.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

An isotropic tensor has unchanged components under every rotation: for rank four, $A_{ijkl}=R_{ia}{R}_{jb}{R}_{kc}{R}_{ld}{A}_{abcd}$. Orthogonality gives $R_{ia}{R}_{ja}={\delta }_{ij}$, so each product of paired Kronecker deltas is invariant. Every linear combination of ${\delta }_{ij}{\delta }_{kl}$, ${\delta }_{ik}{\delta }_{jl}$ and ${\delta }_{il}{\delta }_{jk}$ is therefore isotropic, for arbitrary coefficients.

For the integral, differentiation away from the origin givesAfter multiplication by $x_i{x}_j$, this behaves as $O(r^{-1})$ and is locally integrable in three dimensions. Interpret the integral as the limit with a small central ball removed. A rotation preserves the integration ball and the radial kernel, so the resulting $B$ is an isotropic tensor integral. The symmetries $i\leftrightarrow j$ and $k\leftrightarrow l$ force the two cross-pair coefficients equal:Taking the trace in $k,l$ uses ${\nabla }^2(r^{-1})=0$ off the origin and gives $3A+2B=0$. For the other contraction, the displayed Hessian gives ${\sum }_{ij}{x}_i{x}_j{\partial }_i{\partial }_j(r^{-1})=2/r$. ThusSolving gives $B=2\pi a^2/5$ and $A=-4\pi a^2/15$. Therefore the requested tensor isThis is the quadratically weighted inverse-radius Hessian integral. The origin creates no extra contribution to this weighted integral; even the distributional point term in the unweighted Hessian is annihilated by $x_i{x}_j$.

For an orthogonal change of Cartesian coordinates $x_i^{\prime }=Q_{ij}{x}_j$, an ordinary rank-$n$ tensor transforms byAn isotropic tensor is invariant under such rotations. Orthogonality gives $Q_{ip}{Q}_{jq}{\delta }_{pq}={\delta }_{ij}$, so transforming any of the three products of two Kronecker deltas leaves it unchanged. Their linear combination with arbitrary scalar coefficients is therefore isotropic.

For completeness, these three products exhaust the rank-four isotropic tensors in three dimensions. Half-turns about coordinate axes force a nonzero component to contain each coordinate label an even number of times. Axis interchanges then leave just the values $X=c_{1111}$, $\alpha =c_{1122}$, $\beta =c_{1212}$ and $\gamma =c_{1221}$. Rotating the first two axes by $\pi /4$ gives $X=\frac{1}{2}X+\frac{1}{2}(\alpha +\beta +\gamma )$, so $X=\alpha +\beta +\gamma$. These component values are exactly those of the three-delta-product expression. Thus isotropy of the elastic medium gives this form for its elasticity tensor.

Contracting with the symmetric strain givesThese are the Lamé parameters. Taking the trace determines the scalar part of the strain:Since ${\sum }_{ij}{e}_{ij}^2=3p^2+{\sum }_{ij}{d}_{ij}^2$, the stored strain energy density isA nonzero traceless test strain proves necessity of $\mu \ge 0$, and a pure scalar strain proves necessity of $3\lambda +2\mu \ge 0$. Conversely these two conditions make both terms nonnegative for every strain. Thus nonnegative isotropic elastic strain energy is equivalent to

In three dimensions, the general rank-two isotropic tensor is $a{\delta }_{ij}$ and the general rank-three tensor invariant under proper rotations is $b{ϵ}_{ijk}$. Here $\delta$ is the Kronecker delta and $ϵ$ the Levi-Civita symbol. The latter is orientation-sensitive: an ordinary third-rank tensor invariant under all orthogonal transformations is zero instead. This distinction does not change the stress conclusion.

For an isotropic material, the constitutive tensor must consequently have $A_{ijk}=b{ϵ}_{ijk}$ in the proper-rotation convention. Hence ${\sigma }_{ij}=b{ϵ}_{ijk}{B}_k=-{\sigma }_{ji}$. Mechanical stress in the model is also symmetric, so ${\sigma }_{ij}=-{\sigma }_{ij}$ and every component vanishes. Thus nonzero stress linear in the field requires anisotropic material coefficients. This is the isotropic linear magnetostriction vanishes restriction, not a claim that all higher-order magnetostriction is impossible in isotropic materials.

An isotropic tensor has unchanged components under every proper orthogonal change of basis: $T_{i_1\dots i_n}=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\dots j_n}$ for every $R\in SO(3)$. Orthogonality gives $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$, so the Kronecker delta is isotropic. The determinant identity gives $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}={ϵ}_{ijk}$, so the Levi-Civita symbol is isotropic too. If reflections are included, this last object is a pseudotensor: it changes sign. The question's isotropy convention therefore refers to proper rotations.

For the sphere moment $T_{i_1\dots i_n}={\int }_{S^2}{\hat{x}}_{i_1}\cdots {\hat{x}}_{i_n}dS$, change variables to $R\hat{x}$. Rotations preserve the sphere and its area measure; inserting the component transformation proves that $T$ is an isotropic tensor. For the second moment, half-turns about coordinate axes kill the off-diagonal components, while rotations interchanging axes make the diagonal entries equal. Thus it is a multiple $a{\delta }_{ij}$. Contracting the indices gives $3a={\int }_{S^2}|\hat{x}{|}^{2}dS=4\pi$. Odd moments vanish by $\hat{x}\mapsto -\hat{x}$, including the third moment. The supplied general fourth-rank form and symmetry under every permutation of the four indices make its three coefficients equal. A double contraction then gives $4\pi =b(9+3+3)$. ThusTo explain the weighted volume integrals, put $v=(1,i,0)$; its bilinear square is $v\cdot v=1+i^2=0$, without complex conjugation. Separating radial and angular factors givesFor $n=2$ the angular factor is $(4\pi /3)v\cdot v=0$. For $n=3$ it is zero by oddness. For $n=4$ it is $(4\pi /15)3(v\cdot v{)}^2=0$. Therefore all three requested integrals vanish, assuming the radial integrals exist. More generally the null-vector cancellation of radial moments follows from rotation about the third axis, which multiplies the integral by $e^{in\varphi }$ without otherwise changing it.

A Cartesian isotropic tensor is unchanged by every proper rotation matrix. For a rank-two tensor this means $B^{\prime }=RB{R}^{\mathsf{T}}=B$, or $B_{ij}^{\prime }=R_{ik}{R}_{jl}{B}_{kl}=B_{ij}$. No symmetry of $B$ is assumed. For a quarter-turn about the $z$ axis,Comparing entries yields $B_{11}=B_{22}$, $B_{12}=-B_{21}$, $B_{13}=-B_{23}=-B_{13}$, and $B_{31}=-B_{32}=-B_{31}$. In particular,The remaining matrix has the form $\left(\begin{array}{ccc}b& c& 0\\ -c& b& 0\\ 0& 0& d\end{array}\right)$. A quarter-turn about the $x$ axis givesInvariance now forces $c=0$ and $d=b$, proving that an isotropic second-rank tensor isa scalar multiple of the Kronecker delta.

For the final unheaded conductivity identity, let $P=n{n}^{\mathsf{T}}$, the orthogonal projection onto the unit vector $n$. Then $P^2=P$ and the uniaxial conductivity tensor is $\sigma =\alpha I+\gamma P$. The antisymmetric tensor $D$ sends a vector $v$ to $v\times n$. Applying it twice and using the vector triple product givesConsequentlyThis equals $-\sigma =-\alpha I-\gamma P$ exactly when $(\alpha +\gamma )P=0$. Since $P$ has rank one and is nonzero, the requested parameter isAt this value conductivity vanishes along $n$; it is precisely the singular case excluded in part (ii), so there is no conflict with that part's conclusion.

For a Cartesian second-rank tensor, writeThe first term is a symmetric second-rank tensor and the second is an antisymmetric second-rank tensor. Taking the trace of the symmetric term and subtracting it gives the scalar, symmetric-traceless and axial tensor decomposition:Here $S_{ij}=S_{ji}$ and $S_{ii}=0$. To check reconstruction of the antisymmetric part, contract the Levi-Civita symbols:Thus $P_{ij}=P{\delta }_{ij}+S_{ij}+{ϵ}_{ijk}{A}_k$, and the displayed contractions also prove uniqueness. The vector associated with the antisymmetric part is an axial vector under reflections, and an ordinary vector under proper rotations.

For the specified isotropic tensor law, contract the Kronecker deltas first:Write $T_{ij}=T{\delta }_{ij}+W_{ij}+{ϵ}_{ijk}{V}_k$, where $T=T_{kk}/3$ and $W$ is symmetric and traceless. Transposition preserves its scalar and symmetric-traceless parts and reverses its antisymmetric part. ThereforeUniqueness of the decomposition gives the isotropic elasticity on tensor components relations. Provided their respective coefficients are nonzero, the requested inverses areIf one coefficient vanishes, that strain component has zero stress response: the corresponding stress component must be zero for a solution to exist, and that strain component is not uniquely recoverable. Isotropy by itself does not guarantee invertibility.

If $T_{ij}$ is symmetric then $V=0$, hence $A=0$ and the stress is symmetric. Equivalently, direct transposition gives $P_{ij}-P_{ji}=(\beta -\gamma )(T_{ij}-T_{ji})=0$. Restricting to this symmetric-strain case givesThis final two-parameter expression uses the symmetry assumption from the preceding request. For a general nonsymmetric array, the transpose term cannot be absorbed into the same coefficient: it acts with $\beta +\gamma$ on the symmetric-traceless part and with $\beta -\gamma$ on the antisymmetric part. The stated coefficient $\mu$ is the coefficient of $T_{ij}$ itself; if one instead writes the conventional elastic law with $2\mu$ multiplying the strain, that shear-modulus convention is half this coefficient.

Use the right-handed rotation matrix which sends the positive $x$ direction to the positive $y$ direction:The Cartesian second-rank tensor transformation law givesHere isotropic tensor means invariant under every proper rotation, as is needed for the rank-three assertion involving the Levi-Civita symbol. A passive coordinate rotation through the same angle uses $R^{\mathsf{T}}$ instead; the invariance conclusions are identical.

Setting $S^{\prime }=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the formInvariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf{T}}=\lambda I$ for every orthogonal matrix. The general isotropic second-rank tensor is therefore $S_{ij}=\lambda {\delta }_{ij}$.

An isotropic vector must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus the only isotropic vector is zero. The general isotropic third-rank tensor under proper rotations is $A_{ijk}=\lambda {ϵ}_{ijk}$. If invariance under the whole orthogonal group were required of an ordinary rank-three tensor, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.

The integrand is symmetric in $i,j$, so $T$ is a symmetric tensor. The sphere and its area element are invariant under rotations, and the integrand transforms covariantly when $\mathbf{x},\mathbf{y}$ are rotated together. For fixed $\mathbf{y}$, rotations about its direction therefore leave $T$ unchanged. In axes with $\mathbf{y}$ along the third axis, the mixed transverse-axial entries vanish and the two transverse eigenvalues are equal. Thus it is an axisymmetric symmetric rank-two tensor,where $\alpha ,\beta$ depend on $a,n$ but not on the direction of $\mathbf{y}$. The restriction $n>-1$ makes the integral convergent near $\mathbf{x}=\mathbf{y}$: the integrand is of size distance${}^{2n}$ and the local area element has one further radial power.

Take the polar axis along $\mathbf{y}$. Because $|\mathbf{x}|=|\mathbf{y}|=a$,The area element is $a^2\sin \theta d\theta d\varphi$. Hence with $s=1-\cos \theta$,

For the second scalar integral, take the trace. It contracts the two chord-vector factors into their squared length:Put $K=\pi (2a{)}^{2n+2}$. The tensor form gives $3\alpha +\beta a^2=K/(n+1)$, while its axial contraction gives $\alpha +\beta a^2=K/(n+2)$. Solving yields the weighted chord tensor of a sphere coefficientsBecause $a>0$, this is an isotropic tensor exactly when $\beta =0$. Therefore $n=-1/2$ is the unique permitted isotropic exponent.

Here isotropic tensors are invariant under proper rotation matrices, $R^{T}R=I$ and $\det R=1$. The isotropic second-rank tensor and isotropic third-rank tensor forms in three dimensions arewith scalar coefficients $A,B$. One can see completeness directly. A half-turn about each coordinate axis kills every off-diagonal component of a rank-two tensor, and quarter-turns make its three diagonal components equal. For rank three, the half-turns require that each coordinate label occur an odd number of times in a nonzero component. Thus only permutations of $(1,2,3)$ survive. Quarter-turns interchange two labels while reversing one sign, forcing the surviving components to alternate with the parity of a permutation. This gives a multiple of the Levi-Civita symbol.

The tensor transformation law verifies invariance of these forms:The first equality uses orthogonality and the second uses the determinant formula. If reflections were also required while using the ordinary tensor transformation law, the rank-three coefficient would have to vanish. The proper-rotation convention is the one under which $B{ϵ}_{ijk}$ is an isotropic tensor.

The ball is unchanged by any rotation matrix, and rotating coordinates has Jacobian determinant one. Thus changing variables $\mathbf{y}=R\mathbf{x}$ givesThese moments are therefore isotropic tensor integrals. For the Cartesian moments of a three-dimensional ball, the second moment is $\alpha {\delta }_{ij}$. Taking its tensor contraction givesThe third moment vanishes: $\mathbf{x}\mapsto -\mathbf{x}$ preserves the ball and changes the sign of its integrand. Equivalently, its symmetry under index interchange excludes a nonzero multiple of the antisymmetric Levi-Civita symbol.

The fourth moment is symmetric under every permutation of its four indices. The general rank-four isotropic tensor form consequently has equal coefficients, yieldingFor example, comparing $T_{1122},T_{1212},T_{1221}$ shows that the three coefficients agree. Contracting $i=j$ and $k=l$ givesFinally the vector triple product identity gives $\mathbf{x}\times (\mathbf{\Omega }\times \mathbf{x})=r^2\mathbf{\Omega }-(\mathbf{\Omega }\cdot \mathbf{x})\mathbf{x}$. The first term integrates to $3\alpha \mathbf{\Omega }$ and the second to $\alpha \mathbf{\Omega }$, using the second moment. Hence

A Cartesian second-rank tensor is an isotropic tensor if its components are unchanged under every proper rotation matrix $R$:The Kronecker delta is isotropic because $R_{ik}{R}_{jk}={\delta }_{ij}$, the orthogonality relation $R{R}^T=I$.

No symmetry of $T$ needs to be assumed. A half-turn about the first axis has matrix $\mathrm{diag}(1,-1,-1)$; invariance under it forces $T_{12},T_{13},T_{21},T_{31}$ to vanish. A half-turn about the second axis additionally kills $T_{23}$ and $T_{32}$. Thus $T=\mathrm{diag}(d_1,d_2,d_3)$. Each half-turn is the square of an allowed quarter-turn. A quarter-turn about the third axis interchanges $d_1,d_2$, so they are equal. A quarter-turn about the second axis interchanges $d_1,d_3$, so all three are equal. Consequently the complete family of isotropic second-rank tensors isConversely, every member of this family is invariant because $R(\lambda I)R^T=\lambda I$.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

In three dimensions, the most general isotropic tensors of ranks zero through three areThe scalar and zero vector are unchanged by rotations. For every $R\in SO(3)$, $R_{ia}{R}_{jb}{\delta }_{ab}={\delta }_{ij}$ and $R_{ia}{R}_{jb}{R}_{kc}{\epsilon }_{abc}=(\det R){\epsilon }_{ijk}={\epsilon }_{ijk}$, proving isotropy.

A tensor is isotropic tensor if its components are unchanged under every proper orthogonal change of basis. The general isotropic rank-four tensor iswhere $a,b,c$ are scalars.