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 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 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.

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.

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 $(T^{\prime }{)}^T=R{T}^T{R}^T=-T^{\prime }$.

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.

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 $w=0$.

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