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For an inviscid interface with no surface tension, continuity of normal traction reduces to continuity of pressure. The same condition is used in a sharp Darcy law interface model that neglects capillary pressure and resolved interfacial viscous normal stress. For a general viscous interface, continuity of normal traction balances the full stress tensor; it does not by itself imply continuity of pressure.

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 $\det R$.

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.

On ${\bigwedge }^2{\mathbb{C}}^4$, the invariant Hermitian form and volume tensor define the displayed antilinear map with square one, commuting with SU(4). Its fixed space satisfies $T^{12}=\overline{T^{34}}$, $T^{13}=-\overline{T^{24}}$, and $T^{14}=\overline{T^{23}}$. Three complex choices give six real coordinates, and this fixed space has the original exterior square as its complexification. It is therefore a real representation of dimension six, rather than merely the twelve-dimensional realification of an arbitrary six-dimensional complex representation.

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.

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.

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 $D_i\sqrt{h}=0$.

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.

Traction is the force per unit area transmitted across an oriented surface. For unit normal vector $\mathbf{n}$, the stress tensor gives $\mathbf{t}=\mathbf{\sigma }\mathbf{n}$. Its normal component is normal stress, and its tangential component is shear stress.