Tensor

A rank-$n$ tensor on a vector space $V$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. Its components transform by one copy of the change-of-basis matrix for each index.

A contravariant vector has coordinate components obeying the displayed transformation law. A tangent vector to a parametrized curve provides the basic example: the chain rule transforms $d{x}^a/ds$ in precisely this way. Pairing it with a covector produces a coordinate-independent scalar. Conversely, if an array pairs invariantly with every contravariant vector, coefficient comparison forces the covector transformation law for that array.

An equality of tensors of the same type. Its tensor component transformation law makes the equality independent of the chosen coordinates. The status of each tensor as dynamical or a fixed background structure is additional information.

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 contravariant tensor transforms with one forward coordinate Jacobian for each upper index. A Lorentz four-vector is the rank-one example; the electromagnetic field tensor with two upper indices is a rank-two example. Lower-index covariant tensors transform with inverse Jacobians, and full contractions yield scalars. Index position specifies a transformation law rather than implying that the numerical components are invariant.

Antisymmetrization over selected tensor indices averages their permutations with the sign of a permutation. Square brackets denote it, for example $T_{[ab]}=(T_{ab}-T_{ba})/2$. It projects onto the alternating part. Antisymmetrized second derivatives of a scalar vanish for a torsion-free connection, but antisymmetrized later derivatives act on a tensor and can produce curvature.

Symmetrization over $r$ selected tensor indices is the average over all $r!$ permutations, with every sign positive. Parentheses denote this operation, for example $T_{(ab)}=(T_{ab}+T_{ba})/2$. It is a projection onto the symmetric part and commutes with a covariant derivative acting on indices outside the parentheses.

A Cartesian second-rank tensor invariant under every proper rotation preserving an unoriented axis $n$ has this form. Rotations about the axis give a planar block $aI+bJ$ and eliminate mixed axial-planar entries. A half-turn about a transverse axis reverses $J$, forcing $b=0$. If only rotations about the oriented axis are imposed, this planar antisymmetric term may survive.

The six-element dihedral group generated by the displayed rotations already forces every invariant second-rank tensor to have the same form as an axially invariant second-rank tensor. The axial threefold rotation eliminates mixed entries and leaves only a planar scalar plus a planar antisymmetric part; the transverse half-turn eliminates the latter. No smaller finite subgroup suffices: cyclic rotation groups retain an axial antisymmetric tensor, while a four-element noncyclic rotation group retains arbitrary diagonal tensors along its three mutually perpendicular half-turn axes.

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.

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.

Under an orthogonal coordinate transformation $R$, a Cartesian pseudotensor has the usual tensor transformation law multiplied by $\det R$. Thus proper rotation matrices act as on ordinary tensors, while an orientation-reversing change contributes an extra minus sign. Interpreted as a rank-three pseudotensor, the Levi-Civita symbol has the same numerical components in every orthonormal frame, because its ordinary three-index transformation already contributes a factor $\det R$.

A symmetric tensor is unchanged by every permutation of its slots. A symmetric second-rank tensor is the rank-two case; symmetry for higher rank requires invariance under every such slot exchange.

For $v=(a,b,0)$ with $r=\sqrt{a^2+b^2}>0$, the symmetric tensor $S=v{v}^T+c{r}^2{e}_3{e}_3^T$ has principal axeswith eigenvalues $r^2,0,c{r}^2$, respectively. This follows directly from the outer product: $v{v}^{T}v=r^{2}v$, while it annihilates vectors perpendicular to $v$. The orthogonal matrix with rows $e_r^T,e_{\theta }^T,e_3^T$ gives $MS{M}^T=\mathrm{diag}(r^2,0,c{r}^2)$.

If $a,b$ depend on position, this is a pointwise eigenframe, not necessarily one fixed Cartesian change of axes throughout space. For $v=0$, the displayed tensor vanishes and every orthonormal basis is a principal basis. Subtracting a scalar multiple of the identity leaves its eigenvectors unchanged and shifts every eigenvalue by the same amount; this is a simple case of coaxial symmetric tensors.

A symmetric tensor invariant under rotations about a fixed nonzero vector $y$ has the form $\alpha I+\beta y{y}^T$. Choose the third coordinate axis parallel to $y$: rotation by a half-turn makes the mixed axial-transverse entries vanish, and all transverse rotations make the transverse block a scalar multiple of the identity. Its two transverse eigenvalues are $\alpha$ and its axial eigenvalue is $\alpha +\beta |y{|}^2$. The trace and the quadratic contraction $y^{T}Ty$ determine these two coefficients. Full isotropy holds exactly when $\beta =0$.

For $|y|=a$ and $n>-1$, this surface integral is finite: near $x=y$ its integrand has size $O(|x-y{|}^{2n})$ and the two-dimensional area element supplies one further power of distance. Rotational covariance and symmetry make it an axisymmetric symmetric rank-two tensor. Put $K=\pi (2a{)}^{2n+2}$. Polar integration about the $y$ axis gives $\mathrm{tr}T=K/(n+1)$ and $y^{T}Ty=a^{2}K/(n+2)$. Solving $3\alpha +\beta a^2=\mathrm{tr}T$ and $\alpha +\beta a^2=y^{T}Ty/a^2$ gives $\alpha =K/[2(n+1)(n+2)]$ and $\beta =\alpha (2n+1)/a^2$. Thus the weighted chord tensor becomes an isotropic tensor exactly at $n=-1/2$.

Over a field of characteristic zero, degree-$n$ symmetric tensors are spanned by pure powers $a^{\otimes n}$. Inclusion-exclusion expands ${\sum }_{J\subseteq [n]}(-1{)}^{n-|J|}({\sum }_{j\in J}{a}_j{)}^{\otimes n}$ into the sum of all permutation orderings of $a_1\otimes \cdots \otimes a_n$. Every surviving term uses each of the $n$ labels once. This converts an invariant-tensor spanning problem into pure tensor powers.

A second-rank tensor is symmetric when $T_{ij}=T_{ji}$. Every second-rank tensor has the unique decompositioninto symmetric and antisymmetric parts.

Suppose an array $T_{ij}$ is specified in each right-handed orthonormal basis. If $w_i=T_{ij}{v}_j$ transforms as a vector for every vector $v$, then $T$ is a second-order Cartesian tensor under those basis changes. Indeed $v^{\prime }=Rv$, $w^{\prime }=Rw$ imply $T^{\prime }Rv=RTv$ for every $v$, hence $T^{\prime }=RT{R}^T$. The condition on every test vector is essential; one contraction is insufficient.

If contraction of an array with every symmetric second-rank tensor is a vector in every orthonormal frame, its part symmetric in the contracted slots obeys the rank-three tensor transformation law. The difference between its proposed transformation and actual components is symmetric and contracts to zero against every symmetric matrix, so it vanishes. The antisymmetric part is invisible, by the blindness of symmetric contraction tests to antisymmetric arrays, and need not be tensorial.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.

If an array is antisymmetric in every orthonormal basis and its contractions with every antisymmetric second-rank tensor are invariant scalars, it obeys the Cartesian second-rank tensor transformation law. The difference $D=R^{T}A\prime R-A$ is antisymmetric and orthogonal to every antisymmetric test matrix. Choosing the test matrix equal to $D$ gives $D:D=0$, hence $D=0$. The Frobenius inner product is nondegenerate on the antisymmetric subspace.

The Frobenius inner product of a symmetric matrix and an antisymmetric matrix is zero. Thus invariant contractions of an array with all symmetric second-rank tensors test only its symmetric part; an arbitrary basis-dependent antisymmetric part is invisible. A nonzero antisymmetric matrix in one basis and the zero matrix in another passes all such tests with scalar zero but is not a Cartesian second-rank tensor.

A second-rank Cartesian tensor is traceless when the contraction $T_{ii}$ vanishes. For a symmetric second-rank tensor, this is equivalent to its eigenvalues summing to zero. The orientational nematic order parameter is both symmetric and traceless, excluding an isotropic scalar contribution.

For finite sets $X,Y,Z$ and a field $F$, a three-variable tensor $T:X\times Y\times Z\to F$ has slice rank at most $r$ when it is a sum of $r$ terms, each of which separates one variable from the other two. Such terms have one of the formsThe slice rank is the least possible $r$.

A three-variable tensor is diagonal when it vanishes unless its three arguments are equal.

Ifon a finite set $X$, and every $c_x$ is nonzero, then the slice rank of $T$ is $|X|$. The upper bound uses one $x$-slice for each diagonal entry. For the lower bound, restrict any shorter slice decomposition to a common kernel of the coefficient functions in two slice directions; the remaining diagonal matrix has rank larger than the number of slices available in the third direction, a contradiction.

Covariance and contravariance describe how vector and covector components transform when a basis changes.

A covariant tensor is multilinear in vector arguments and transforms with one covariant index for each argument.

The Kronecker delta is ${\delta }_{ij}=1$ when $i=j$ and zero otherwise. It is the component matrix of the identity map in a basis.

A second-rank Cartesian tensor is antisymmetric when its components form a skew-symmetric matrix, $T_{ij}=-T_{ji}$. Orthogonal coordinate changes preserve this property because $T^{\prime }=RT{R}^T$ implies $(T^{\prime }{)}^T=R{T}^T{R}^T=-T^{\prime }$.

In three-dimensional Euclidean space, contracting an antisymmetric second-rank tensor with the Levi-Civita symbol gives an axial vector. The epsilon contraction identity proves the inverse formula: $\frac{1}{2}{ϵ}_{kij}{ϵ}_{klm}{A}_{lm}=\frac{1}{2}(A_{ij}-A_{ji})=A_{ij}$. With this index convention,Thus $A$ is minus the cross-product matrix for $a\times x$. Proper rotations transform $a$ as a vector; an orientation-reversing orthogonal change adds a determinant sign. Stating the index convention prevents a sign ambiguity when converting an antisymmetric tensor into a cross-product operator.

An antisymmetric tensor changes sign under exchange of two indices in a specified subset of its indices. A totally antisymmetric tensor has this property for every pair of indices.

To vary an antisymmetric tensor without treating its entries as independent physical variables, temporarily allow an unrestricted array $M^{\mu \nu }$ and add ${\lambda }_{\mu \nu }(M^{\mu \nu }+M^{\nu \mu })/2$ to the Lagrangian density, with symmetric Lagrange multiplier $\lambda$. Variation of $\lambda$ enforces antisymmetry. The antisymmetric part of the unrestricted $M$ equation eliminates $\lambda$ and gives the physical Euler-Lagrange field equation. Equivalently, vary only the six independent entries in four spacetime dimensions. Quadratic contractions must be differentiated before applying antisymmetry, or their factors and signs can be lost.

For a real antisymmetric matrix, transpose gives $\mathrm{tr}(A^{2j+1})=\mathrm{tr}((-A{)}^{2j+1})=-\mathrm{tr}(A^{2j+1})$, so this trace vanishes. Every complete pairwise tensor contraction of copies of $A$, using the Euclidean Kronecker delta, splits into closed index cycles. Each cycle is a trace of a power of $A$, up to a sign from index order. An odd total number of copies forces an odd-length cycle, so the whole contraction vanishes. This assertion concerns pairwise index contraction with the metric; inserting additional tensors would be a different claim.

A tensor changes sign whenever two indices are interchanged. In three dimensions, a rank-two antisymmetric tensor is $A_{ij}=a_k{ϵ}_{kij}$, a rank-three one is $A_{ijk}=a{ϵ}_{ijk}$, and every totally antisymmetric tensor of rank greater than three vanishes.

An isotropic tensor is invariant under every proper orthogonal change of basis. A rank-four isotropic tensor has the formRank-five isotropic pseudotensors are formed by multiplying one Kronecker delta by one Levi-Civita symbol and permuting the five indices.

For a uniformly distributed unit vector in three dimensions, $⟨u_i{u}_j{u}_k{u}_l⟩=({\delta }_{ij}{\delta }_{kl}+{\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk})/15$. Contracting indices fixes the coefficient. This identity is useful for incoherent scattering averages and random-axis material models.

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.

In three-dimensional Euclidean space, a rank-three tensor invariant under every proper rotation is a scalar multiple of the Levi-Civita symbol. The distinction between proper rotations and all orthogonal transformations matters: $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}$. A polar tensor of odd rank invariant under every orthogonal transformation must vanish, since the transformation $R=-I$ changes its sign. A nonzero invariant rank-three object under reflections is instead a pseudotensor, whose transformation law includes the extra determinant factor.

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.

The rank-four law $c_{ijkl}=\alpha {\delta }_{ij}{\delta }_{kl}+\beta {\delta }_{ik}{\delta }_{jl}+\gamma {\delta }_{il}{\delta }_{jk}$ sends a strain array to $P_{ij}=\alpha {\delta }_{ij}{T}_{kk}+\beta T_{ij}+\gamma T_{ji}$. Applying the scalar, symmetric-traceless and axial tensor decomposition gives the displayed independent actions on the three components. The inverse requires all relevant coefficients nonzero. Restricting to a symmetric second-rank tensor eliminates the axial part and gives $P_{ij}=\alpha {\delta }_{ij}{T}_{kk}+(\beta +\gamma )T_{ij}$. Without that symmetry restriction the two-parameter form is not generally valid.

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 $\nu =\beta -\gamma$.

A rank-three Cartesian tensor invariant under every proper rotation matrix in three dimensions is a multiple of the Levi-Civita symbol. Half-turns about coordinate axes force every nonzero component to contain each coordinate label once, and quarter-turns make these components alternating. Conversely, the determinant formula $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}$ proves invariance when $\det R=1$. Requiring invariance under reflections as an ordinary tensor forces $b=0$; invariance under proper rotations alone permits nonzero $b$.

A second-rank Cartesian tensor invariant under every proper rotation matrix is a scalar multiple of the Kronecker delta. Half-turns about the coordinate axes kill its off-diagonal entries, and quarter-turns make its diagonal entries equal. A tensor of the form $v_i{w}_j$ is an outer product with matrix rank at most one, so it is isotropic in three dimensions only when it vanishes, equivalently when $v=0$ or $w=0$.

Here $V_d$ is the volume of the unit ball in $d$-dimensional Euclidean space. Rotational symmetry makes the integral an isotropic second-rank tensor $C{\delta }_{ij}$. Taking its trace gives $dC={\int }_{|x|\le R}|x{|}^2{d}^{d}x=d{V}_d{\int }_0^R{r}^{d+1}dr$. This proves the formula. For a constant matrix $M$, contraction gives $\int x^{T}Mx{d}^{d}x=V_d{R}^{d+2}\mathrm{tr}M/(d+2)$. In three dimensions the coefficient is $4\pi R^5/15$.

Einstein notation sums automatically over an index repeated once up and once down, or twice in Euclidean coordinates.

A dummy index is summed over its full specified range. Its name can be changed consistently without changing the expression, as in $A_{ij}{B}_{ij}=A_{ji}{B}_{ji}$. A free index specifies a component and cannot be renamed in only one term of an equation. Relabelling dummy indices is useful for tensor contraction arguments involving a symmetric tensor and an antisymmetric tensor.