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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 $(j,k)$ tensor has one fundamental representation factor for every upper index and a dual factor for every lower index. Under $A\in SU(n)$, upper factors transform by $A$ and lower factors by $A^{-1}$; unitarity identifies the latter with conjugate fundamental matrices. Thus complex conjugation exchanges $(j,k)$ and $(k,j)$. Index permutations and contraction with an invariant tensor commute with the action, producing invariant subspaces and decomposition maps.

For a thin layer with essentially uniform tangential electric field, integrating electrical conductivity through its thickness gives the sheet conductivity tensor ${\mathbf{\sigma }}_s$. The two-dimensional form of Ohm's law is $\mathbf{K}={\mathbf{\sigma }}_s{\mathbf{E}}_{\parallel }$, with surface current density $\mathbf{K}$. Its SI unit is $\mathrm{S}=A/V$: amperes per metre divided by volts per metre. When response is the same in every in-plane direction, the sheet conductivity reduces to a scalar $\Sigma$; for uniform bulk electrical conductivity $\sigma$ and thickness $t$, $\Sigma =\sigma t$.

At radius $R$ outside a spherical mass $M$, the Newtonian relative-acceleration tensor has eigenvalues $-g/R,-g/R,2g/R$ with $g=GM/R^2$. It compresses free fall separations tangentially and stretches them radially. The displayed time scale is the inverse square root of the curvature scale controlling this tidal evolution. An ideal circular orbit at that radius has period $2\pi t_{\mathrm{tide}}$, providing a second physical interpretation. For Earth's surface it is about $806$ seconds, while the corresponding ideal low-orbit period is about $84$ minutes. A uniform gravitational acceleration does not determine this relative motion; spatial variation of the field does.

Use the material-first nominal stress $P_{Ii}$ and $F=RU$. The symmetric Biot stress conjugate to the Biot strain tensor is $T^{(1)}=\mathrm{sym}(PR)$. To prove it, set $M=P^T$ and differentiate $F=RU$. Conservation of angular momentum makes $M{F}^T$ symmetric, so the contraction of $R^{T}M$ with $R^T\dot{R}U$ is zero: it is a contraction of a symmetric tensor with a skew tensor after cyclically moving $U$. Thus $M:\dot{F}=\mathrm{sym}(R^{T}M):\dot{U}=T^{(1)}:{\dot{E}}^{(1)}$. No assumption that $PR$ itself is symmetric is needed.

The symmetric cube of a group representation $V$ is the symmetric part of $V^{\otimes 3}$. Over the complex numbers, its character isThe projection onto symmetric tensors averages the six permutations of the three factors. The identity permutation contributes $\chi (g{)}^3$, the three transpositions each contribute $\chi (g)\chi (g^2)$, and the two three-cycles each contribute $\chi (g^3)$. Cubic polynomial invariants are fixed vectors in ${\mathrm{Sym}}^3(V^{\ast })$, whose dimension follows by averaging this character for a finite group.

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.

Three light quarks have a six-dimensional single-quark spin-flavour space ${\mathbb{C}}^3\otimes {\mathbb{C}}^2$. A symmetric spatial ground state and the antisymmetric three-quark colour singlet require the symmetric cube of that space. Its dimension is $\left(\genfrac{}{}{0ex}{}{8}{3}\right)=56$, and restriction to flavour $SU(3)$ and spin $SU(2)$ gives a baryon decuplet of spin $3/2$ and a baryon octet of spin $1/2$. The two mixed-permutation multiplicity spaces pair once in the symmetric sector, explaining why the two flavour-octet copies in the unconstrained tensor cube do not represent two distinct ground-state octets.

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.

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.

Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower trace. The resulting representation has highest weight $p{\omega }_1+q{\omega }_2$. Its dimension isThe subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial inner product, is multiplication by the nonzero polynomial ${\sum }_i{x}_i{y}_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its Weyl dimension formula dimension, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed tensor.

A tensor density of weight $w$ transforms like a tensor with an additional factor $|\det (\partial x/\partial x^{\prime }){|}^w$ under $x\mapsto x^{\prime }$. A scalar density of weight one can be integrated against the coordinate volume element to give an invariant integral. On a spatial hypersurface, $\sqrt{h}$ is such a scalar density; hence ${\pi }^{ij}=\sqrt{h}{p}^{ij}$ is a tensor density when $p^{ij}$ is an ordinary tensor. The Levi-Civita connection extends to densities and satisfies $D_i\sqrt{h}=0$.

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.

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.

After choosing ordered bases, identify a tensor $t={\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ with its matrix $C(t)=(c_{ij})$. This is a linear isomorphism between $V{\otimes }_{F}W$ and the vector space of $m$-by-$n$ matrices: each tensor-product basis element becomes the matrix with one entry equal to one and all others zero. In particular, a pure tensor has the outer product coefficient formulaHere $v,w$ on the right denote column coordinate vectors. The identification depends on the chosen bases, while the tensor itself does not. Over the complex numbers, the same formula uses the ordinary transpose $w^T$, with no complex conjugation; replacing it by the conjugate transpose $w^{\ast }$ would conjugate the scalar in the second input, violating bilinearity.

For unit particle mass, this symmetric third-order tensor transports the pressure tensor in a kinetic fluid description. Its contraction gives the heat flux $q_i=Q_{ijj}/2$. It vanishes for a local Maxwellian, whose odd central velocity moments vanish, but it need not vanish for a general anisotropic or skewed phase-space distribution function.

The torsion of an affine connection is the displayed antisymmetric tensor. It measures the failure of the antisymmetrized covariant derivative to agree with the Lie bracket of vector fields. With derivative-last coefficients ${\nabla }_{\mu }{Y}^{\rho }={\partial }_{\mu }{Y}^{\rho }+{\Gamma }^{\rho }{}_{\nu \mu }{Y}^{\nu }$, its components are ${\mathcal{T}}^{\rho }{}_{\mu \nu }={\Gamma }^{\rho }{}_{\nu \mu }-{\Gamma }^{\rho }{}_{\mu \nu }$. Some conventions use the negative component tensor. Geodesic equations depend on the symmetric connection coefficients, so they alone cannot detect torsion.

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.

In four spacetime dimensions, the trace reversal of a stress-energy tensor has the displayed form, where $T=g^{\mu \nu }{T}_{\mu \nu }$. Its trace is $-T$, so applying trace reversal twice returns the original tensor. Taking the trace of the Einstein field equations without a cosmological constant gives $R=-\kappa T$; substitution yields $R_{\mu \nu }=\kappa {\overline{T}}_{\mu \nu }$. In a general dimension $d>2$, solving for the Ricci tensor instead uses $T_{\mu \nu }-g_{\mu \nu }T/(d-2)$.