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For a constant orthogonal matrix $R$, the Laplacian is unchanged by the coordinate transformation $\xi =Rx$. If $u$ is twice continuously differentiable, the chain rule givesbecause ${\sum }_i{R}_{ji}{R}_{ki}={\delta }_{jk}$ by orthogonality of the rows. Here ${\delta }_{jk}$ is the Kronecker delta. In two dimensions, the rotation matrix $\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ with $a^2+b^2=1$ makes the mixed partial derivatives cancel explicitly. The same argument works for reflections and translations, so the Laplacian depends on the Euclidean geometry rather than the orientation or origin of the coordinates.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

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.

In the ideal electric-dipole interaction, a vibrational transition needs a nonzero matrix element of the electric dipole moment. For a diatomic bond stretch, expand the body-axis dipole component as $d(q)=d_e+d_e^{\prime }q+\cdots$. A changing electric dipole moment, rather than a permanent value $d_e$ alone, permits the harmonic fundamental. Keeping only the linear term and harmonic wavefunctions gives $\Delta v=\pm 1$.^[ref]

The derivation is sharper than a picture of a shaking molecule. Orthogonality removes the constant term for $v^{\prime }\ne v$, and creation and annihilation operators giveHere $\delta$ is the Kronecker delta. Thus absorption raises $v$ by one when the dipole derivative and angular factor are nonzero, and its photon satisfiesA nonzero vibrational matrix element is only one factor: rotational electric-dipole selection rules, initial populations, and spectral resolution also constrain visibility.

For an isolated diatomic molecule with identical atomic nuclei, such as molecular hydrogen in its symmetric electronic ground state, stretching preserves zero electric dipole moment. By symmetry, this homonuclear stretch is infrared inactive in the ideal dipole model, although it still vibrates. Other interactions or environments can permit detection.^[ref]