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Taking the transpose gives $M^T=(X^T+X)/\sqrt{2}=M$, so $M$ is a symmetric matrix. For $i<j$, $M_{ij}=(X_{ij}+X_{ji})/\sqrt{2}$ has the standard normal distribution, whereas $M_{ii}=\sqrt{2}{X}_{ii}$ has the normal distribution $N(0,2)$. Distinct upper-triangular entries depend on disjoint sets of the original independent random variables, so they remain independent.

Consequently the joint probability density function, relative to ${\prod }_{i\le j}d{m}_{ij}$, isHere $\mathrm{tr}(m^2)={\sum }_i{m}_{ii}^2+2{\sum }_{i<j}{m}_{ij}^2$ because $m$ is a symmetric matrix. This is the Gaussian orthogonal ensemble, with off-diagonal variance one and diagonal variance two; the rescaled matrix $M/\sqrt{n}$ has the corresponding $1/n$ off-diagonal variance.

We will also use its entry covariance. Expanding in the independent entries of $X$, for which $\mathbb{E}[X_{ij}{X}_{kl}]={\delta }_{ik}{\delta }_{jl}$, givesThe symbols ${\delta }_{ab}$ are Kronecker deltas. This formula includes the diagonal case, where the two terms both contribute.

Convolve the identity of part (c) with the inverse denominator $K\ast {\delta }_{-\rho }$ found in part (a). Associativity givesTranslation by a Kronecker delta obeys $(K\ast {\delta }_{\lambda })(\psi )=K(\psi -\lambda )$. Taking the coefficient at $\psi$ therefore yieldsThis is the Kostant multiplicity formula in the paper's convention that $K$ counts negative root partitions. With a partition function counting positive root sums, each argument would have the opposite sign.

The canonical Poisson bracket isThe coordinate derivatives are Kronecker deltas, so direct substitution gives

TakePart (i) proves that Brownian motion solves this martingale problem. For the converse, let $X$ be any continuous adapted process satisfying this martingale problem and $X_0=0$.

Set ${\tau }_n=\inf \{t\ge 0:|X_t|\ge n\}$. Choose a smooth cutoff function ${\chi }_n$ equal to one on a neighbourhood of the closed ball of radius $n$. Apply the martingale problem to ${\chi }_n(x)x_i$ and ${\chi }_n(x)x_i{x}_j$, and stop at ${\tau }_n$. Inside that ball, their Laplace operators are respectively $0$ and $2{\delta }_{ij}$. The optional sampling theorem therefore shows thatare martingales. Continuity makes ${\tau }_n\uparrow \infty$ almost surely. Hence every $X^i$ is a continuous local martingale. The Itô product rule and the fact that a continuous finite-variation local martingale is constant identify its quadratic covariations:Here ${\delta }_{ij}$ is the Kronecker delta.

The Lévy characterization of multidimensional Brownian motion now applies. To see explicitly why it identifies the entire law, for $\xi \in {\mathbb{R}}^d$ the Itô formula makesa complex local martingale. Its absolute value on $[0,T]$ is at most $e^{|\xi {|}^{2}T/2}$, so the bounded local martingale criterion makes it a true martingale there. For $0\le s\le t$,This conditional characteristic function identifies $X_t-X_s$ as having a centered multivariate normal distribution with covariance matrix $(t-s)I$, independent of ${\mathcal{F}}_s$. Together with continuity and $X_0=0$, these are exactly the defining properties of standard $d$-dimensional Brownian motion. Thus the martingale problem for Brownian motion has a unique law.

For an isotropic Newtonian fluid in incompressible flow, the constitutive equation isHere $p$ is pressure, ${\delta }_{ij}$ is the Kronecker delta, $\mu$ is the constant dynamic viscosity, and $E$ is the strain-rate tensor. The more general compressible constitutive equation can include a term proportional to $(\nabla \cdot \mathbf{u}){\delta }_{ij}$; that term vanishes here.

Differentiate the Cauchy stress tensor, retaining the constant dynamic viscosity outside the derivatives:The last equality uses incompressibility. Substitution into the Cauchy momentum equation gives the Navier-Stokes equationThe pressure gradient and the Laplacian of the velocity field therefore give the two contact force densities in this model.

For a source point mass $M$ at the origin, let $\mathbf{R}=R\mathbf{n}$, where $\mathbf{n}$ is an outward unit vector, and let $\mu =GM$ be the source's standard gravitational parameter. Differentiating $g_i=-\mu R_i/R^3$ gives the tidal tensorwhere ${\delta }_{ij}$ is the Kronecker delta and $I$ is the identity matrix. In an orthonormal basis with first axis radial, the tidal tensor isThe radial eigenvalue $2\mu /R^3$ separates nearby freely falling test particles; the two transverse eigenvalues $-\mu /R^3$ bring them together. The farther radial test particle falls inward less rapidly, so its acceleration relative to the nearer one points outward. Transverse gravitational accelerations converge toward the source. Thus an initially resting small cloud stretches radially and squeezes in both transverse directions. These are instantaneous relative accelerations; the subsequent shape also depends on initial velocities and the changing tidal tensor along the cloud's free fall.

Outside the source, the matrix trace is $2\mu /R^3-\mu /R^3-\mu /R^3=0$. This matches $\mathrm{tr}T=-{\nabla }^2\Phi =0$ in a region with zero mass density, by the Poisson equation for Newtonian gravity. Zero matrix trace means no leading change of a small initially resting cloud's volume at order $t^2$, not that each separation stays fixed. Inside matter, the same sign convention gives $\mathrm{tr}T=-4\pi G\rho$ instead.

An electrostatic interaction involving an electric quadrupole moment, the second-order term in a multipole expansion. About a specified origin, use the conventionHere $q_a$ is an electric charge, ${\mathbf{r}}_a$ its position, and ${\delta }_{ij}$ the Kronecker delta. Far from the electric charges, the quadrupole contribution to electrostatic potential iswhere ${\epsilon }_0$ is vacuum permittivity. Thus a point electric charge has quadrupole interaction potential energy proportional to $r^{-3}$ at fixed orientation. A second electric quadrupole moment couples to second derivatives of this electrostatic potential, giving quadrupole–quadrupole potential energy proportional to $r^{-5}$ at fixed orientations. These are far-field terms, rather than the whole intermolecular force at close contact.

Benzene has no permanent electric dipole moment by symmetry, but its electric quadrupole moment need not vanish. Its electrostatic interaction therefore depends on the relative molecular orientations even without permanent electric dipole moments. Published electric quadrupole moment conventions can differ by a factor of two; the tensor convention above must be matched before inserting tabulated values.^[ref]

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

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.

For the outward unit normal on a three-dimensional sphere, the off-diagonal integrals vanish by reflection symmetry. The diagonal integrals are equal by rotational symmetry, and their sum is ${\int }_{S^2}|\mathbf{n}{|}^{2}d\Omega =4\pi$. This proves the displayed matrix identity, where ${\delta }_{ij}$ is the Kronecker delta. Contracting it with a constant vector gives ${\int }_{S^2}(\mathbf{U}\cdot \mathbf{n})\mathbf{n}d\Omega =(4\pi /3)\mathbf{U}$, useful for integrating far-field stress forces.

The Smagorinsky model approximates the deviatoric subgrid-scale stress using an eddy viscosity determined by the resolved strain-rate tensor:Here ${\bar{S}}_{ij}=({\partial }_j{\bar{u}}_i+{\partial }_i{\bar{u}}_j)/2$, $\Delta$ is the filter length, $C_s$ is dimensionless, and repeated indices use the Einstein summation convention. The Kronecker delta selects the isotropic part. The construction estimates a missing effect from resolved deformation rather than calculating each unresolved eddy.^[ref]

For simple shear flow $\bar{u}=(\gamma y,0,0)$, the only nonzero strain components are ${\bar{S}}_{12}={\bar{S}}_{21}=\gamma /2$, giving $|\bar{S}|=|\gamma |$. Taking $C_s=0.1$, $\Delta =0.02\mathrm{m}$ and $\gamma =10{\mathrm{s}}^{-1}$ yieldsThese are illustrative parameter choices, not a universal fitted constant. The dimensions follow from ${\Delta }^2|\bar{S}|$.

For an incompressible flow, the resolved-to-unresolved kinetic energy transfer isA nonnegative eddy viscosity thus removes resolved kinetic energy and cannot represent local turbulent backscatter. Coefficients and near-wall treatment require physical and numerical judgment.

Joseph Smagorinsky's 1963 atmospheric-circulation experiment used a lateral exchange coefficient based on horizontal deformation and a length scale. The modern three-dimensional expression above extends that strain-dependent modeling idea; it should not be mistaken for the complete original atmospheric model.^[ref]

A subgrid-scale model represents the influence of unresolved motion on the quantities retained in a large eddy simulation. Its need follows directly from the nonlinear velocity product in the Navier-Stokes equations: averaging a product generally differs from multiplying the averages.

Let a bar denote a spatial averaging filter. With constant mass density $\rho$, constant kinematic viscosity $\nu$, and a time-independent translation-invariant filter on an unbounded or periodic domain, filtering the incompressible momentum equation giveswhere repeated indices use the Einstein summation convention, $p$ is pressure, and the subgrid-scale stress in this sign convention isThe assumed filter commutes with the displayed derivatives. The definition of ${\tau }_{ij}$ is exact; a closure must approximate it using available resolved information.^[ref]

For an elementary example, average two equally weighted values of one velocity component, $a$ and $-a$. The mean is zero but the mean square is $a^2$, so the corresponding stress component is $a^2$. A resolved mean alone has lost the information needed to reconstruct that covariance.

The Smagorinsky model closes the deviatoric part with an eddy viscosity:where ${\delta }_{ij}$ is the Kronecker delta, ${\bar{S}}_{ij}=({\partial }_j{\bar{u}}_i+{\partial }_i{\bar{u}}_j)/2$ is the resolved strain-rate tensor, $C_s$ a dimensionless model coefficient and $\Delta$ a characteristic filter width. The isotropic stress can be included in a modified pressure. This strain-dependent viscosity has its historical origin in Joseph Smagorinsky's atmospheric circulation modelling.^[ref][ref]

For an incompressible flow, the model transfers resolved kinetic energy out at the rate per unit massA fixed nonnegative eddy viscosity therefore permits dissipation but excludes local turbulent backscatter. A dynamic subgrid-scale model can instead estimate its coefficient from resolved stresses at two filter levels. This addresses a modelling choice, rather than changing the exact definition of the unresolved stress.^[ref]

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]