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For a bounded sufficiently regular region, the divergence theorem applied to the vector field $x_i{e}_j$ gives the displayed identity. Contracting with constant vectors gives ${\int }_{\partial \Omega }(b\cdot x)(c\cdot n)dS=|\Omega |b\cdot c$. Contracting with the Kronecker delta or the Levi-Civita symbol respectively gives $\int x\cdot ndS=3|\Omega |$ and $\int x\times ndS=0$ in three dimensions.

A Cartesian basis is an ordered basis of unit vectors along the mutually perpendicular axes of a Cartesian coordinate system. It is an orthonormal basis of a real Euclidean space: $e_i\cdot e_j={\delta }_{ij}$, with ${\delta }_{ij}$ the Kronecker delta. In the plane, the familiar standard basis is $e_1=(1,0{)}^T$, $e_2=(0,1{)}^T$; the vector $p=(2,1{)}^T$ is $2e_1+e_2$. A rotated pair of perpendicular unit vectors is also a Cartesian basis, even though its basis vectors have different standard basis entries.

The advantage is that coordinates can be measured separately by dot products, and the squared norm is the sum of their squares:These are the real Euclidean space instances of the orthonormal basis expansion.^[ref]

An orthogonal basis only requires perpendicular basis vectors; their lengths need not be one. For example, $(2,0{)}^T,(0,3{)}^T$ are orthogonal, but $p=(2,1{)}^T$ has coordinates $(1,1/3{)}^T$, while its dot products with them are $4,3$. For an orthogonal basis $(a_i)$, the coordinate along $a_i$ is $(p\cdot a_i)/(a_i\cdot a_i)$. Dividing each nonzero basis vector by its norm makes it a unit vector without changing its direction. Unit length alone does not imply orthogonality: $(1,0{)}^T$ and $(1,1{)}^T/\sqrt{2}$ are both unit vectors but have dot product $1/\sqrt{2}$.

The examples below show why these distinctions matter: rotating a Cartesian basis changes the coordinate vector while keeping lengths easy to compute; using nonperpendicular basis vectors still gives unique coordinates, but requires the dual basis or a matrix inverse to extract them.

A complex visibility is a calibrated complex correlation of the signals received at a particular interferometric baseline. In a narrow field, with scalar radiation, the far-field approximation for an outgoing source, a narrow frequency channel, negligible noncoplanar phase curvature and a steady sky of mutually incoherent emitters,Here $l,m$ are small dimensionless sky-direction offsets, $I$ is brightness in the chosen channel, and $A$ is the primary beam. This is the Fourier transform of beam-weighted brightness. The two-dimensional formula requires those approximations: a wide field generally needs the third baseline coordinate and curved sky geometry.

For an explicit one-dimensional derivation, set the primary beam to one over the small field, and place independent emitters at angular offsets ${\theta }_j$. Choose a wave phase convention in which, after removing the phase centre's delay, their received complex amplitudes areHere ${\delta }_{jk}$ is the Kronecker delta and angle brackets denote an expected value estimated by time averaging a steady signal. The last equality states mutual incoherence: different emitters have zero averaged cross term, although each emitter's two received copies remain correlated. Multiplying and averaging givesFor continuous brightness this becomes $V(u)=\int I(\theta )e^{-2\pi iu\theta }d\theta$. Real brightness implies the Hermitian symmetry $V(-u)=V(u{)}^{\ast }$; it does not require brightness to be even. In particular, an asymmetric real source usually has a nonzero imaginary complex visibility. At zero baseline, $V(0)={\sum }_j{I}_j$ is the total beam-weighted source strength. Ordinary separated-element correlations need not measure that zero-baseline value.

For a basis $e_1,\dots ,e_n$ of a finite-dimensional vector space $V$, its dual basis consists of linear functionals $e_1^{\ast },\dots ,e_n^{\ast }$ satisfyingThe Kronecker delta means that each rule gives one on its own basis vector and zero on the others. These rules extract coordinates:Applying $e_i^{\ast }$ to a linear combination removes every term except the $i$th. Consequently the dual basis is a basis of the dual space, and every linear functional has expansion $\phi ={\sum }_i\phi (e_i)e_i^{\ast }$. ^[ref].

Coordinate extraction works for a nonorthogonal basis too. With $b_1=(1,0)$ and $b_2=(1,1)$, the dual basis isThe checks $b_1^{\ast }(b_1)=1$, $b_1^{\ast }(b_2)=0$, $b_2^{\ast }(b_1)=0$, and $b_2^{\ast }(b_2)=1$ fix the two rules. For $v=(3,1)$ they give coordinates $(2,1)$, so $v=2b_1+b_2$. The usual dot products with $b_1,b_2$ instead give $3,4$ and do not extract these coordinates: simple projection formulas depend on an orthonormal basis.

The same idea explains polynomial interpolation. On polynomials of polynomial degree at most $n$, evaluation at distinct interpolation nodes $x_0,\dots ,x_n$ gives the dual basis to the Lagrange cardinal polynomials, because ${\ell }_i(x_j)={\delta }_{ij}$. Samples are coordinates in this basis, so the reconstruction $p={\sum }_{i}p(x_i){\ell }_i$ is a dual-basis expansion. Differentiation weights from nodal evaluations and factorial monomial basis for polynomial differentiation apply the same principle to different measurements.

In the unscaled variance-one convention, a Gaussian orthogonal ensemble matrix $M$ is a real symmetric matrix whose upper-triangular entries are independent, with $M_{ij}\sim N(0,1)$ for $i<j$ and $M_{ii}\sim N(0,2)$. Its joint probability density function relative to the upper-triangular coordinates is proportional to $\exp [-\mathrm{tr}(M^2)/4]$. The scaled convention uses $M/\sqrt{n}$; the scaling must be stated. If $X$ has independent standard normal entries, $(X+X^T)/\sqrt{2}$ realizes the unscaled convention. Expanding in the entries of $X$ gives the covariance identity $\mathbb{E}[M_{ij}{M}_{kl}]={\delta }_{ik}{\delta }_{jl}+{\delta }_{il}{\delta }_{jk}$, with Kronecker deltas.

The identity tensor is the canonical type-$(1,1)$ tensor that maps each tangent vector to itself. It pairs a covector $\eta$ and vector $V$ as $\eta (V)$, and its components are the Kronecker delta ${\delta }^{\mu }{}_{\nu }$ in every coordinate system.

The impulse response records the response to a unit impulse, with specified initial or boundary conditions. In continuous time the impulse is a Dirac delta function; in discrete time it is the Kronecker delta $\delta [n]$. For a linear time-invariant system with zero initial state, the response to a general input is its convolution with the impulse response. The impulse response of a causal filter is zero before the impulse.

For the five-sample moving average, a discrete impulse produces $h[n]=1/5$ at $n=0,1,2,3,4$ and zero otherwise. The response lasts five samples, so this is a finite impulse response system. Its frequency response is the finite sum of the five delayed phase contributions: one short input experiment determines how every sustained sinusoid is filtered.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

Contract an isotropic tensor with the Kronecker delta or the Levi-Civita symbol, using invariance under proper rotations. The three displayed contractions have ranks one, two and three. The only isotropic vector is zero; an isotropic second-rank tensor is a scalar delta, and an isotropic third-rank tensor is a scalar epsilon. For $T_{ijkl}=\alpha {\delta }_{ij}{\delta }_{kl}+\beta {\delta }_{ik}{\delta }_{jl}+\gamma {\delta }_{il}{\delta }_{jk}$, direct contraction gives $\mu =3\alpha +\beta +\gamma$ and $\nu =\beta -\gamma$.

Every three-dimensional fourth-rank Cartesian tensor invariant under all proper rotations has the displayed form, without assuming index-interchange symmetry. Half-turns about the coordinate axes force every nonzero component to contain each coordinate label an even number of times. The surviving patterns have four equal labels or two equal pairs. Axis permutations identify the latter with $c_{1122}=\alpha$, $c_{1212}=\beta$ and $c_{1221}=\gamma$, while all four-equal components share a value $A$.

Evaluate the component with four copies of $(e_1+e_2)/\sqrt{2}$. Multilinearity gives $(A+\alpha +\beta +\gamma )/2$, which must equal $A$ by rotational invariance. Hence $A=\alpha +\beta +\gamma$, proving the formula. Conversely each pairing of Kronecker deltas is invariant under any orthogonal transformation. For symmetric strain only $\alpha$ and $\beta +\gamma$ matter, giving the two Lamé parameters of isotropic elasticity on tensor components.

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 $w=0$.

A second-rank tensor invariant under every member of the orthogonal group on ${\mathbb{R}}^n$ is a scalar multiple of the Kronecker delta ${\delta }_{ij}$. For a radial weight $f(|x|)$ with absolutely integrable second moments,Reflections force off-diagonal entries to vanish, and coordinate exchanges make all diagonal entries equal. Taking the matrix trace determines their common value. The same proof applies to a ball or a spherical shell centered at the origin.

For $n\ge 3$, invariance under all proper rotations already forces this second-rank form. In two dimensions, proper rotations alone also preserve an antisymmetric tensor; full orthogonal invariance, or the symmetry of $x_i{x}_j$, excludes that extra term. Higher moments use all possible pairings of Kronecker deltas, as in radial tensor moments in three dimensions.

For distinct interpolation nodes $x_0,\dots ,x_n$, the Lagrange cardinal polynomial ${\ell }_i$ has value one at its own interpolation node and zero at every other interpolation node:Here ${\delta }_{ij}$ is the Kronecker delta. These Lagrange cardinal polynomials form a basis of the vector space of polynomials of polynomial degree at most $n$: every such polynomial has the unique expansion $p(x)={\sum }_{i}p(x_i){\ell }_i(x)$. The nodal evaluation linear functionals $E_j(p)=p(x_j)$ form the dual basis, since $E_j({\ell }_i)={\delta }_{ij}$.

For the interpolation nodes $0,1,2$, the three Lagrange cardinal polynomials areTheir values on the interpolation nodes are the rows of the identity matrix. In the figure, each colored curve passes through its own unit-height marker and through zero at the other two interpolation nodes. Multiplying the curves by the data $1,2,5$ and adding gives the Lagrange polynomial $x^2+1$.

Since the constant polynomial one interpolates data values all equal to one, uniqueness gives ${\sum }_i{\ell }_i(x)=1$. Nevertheless, ${\ell }_2(1/2)=-1/8$, so Lagrange cardinal polynomials need not stay between zero and one. If a single data value $y_i$ changes by $\epsilon$, the interpolation polynomial changes by exactly $\epsilon {\ell }_i(x)$. Thus a Lagrange cardinal polynomial describes how one observation influences the entire graph, and ${\sum }_i|{\ell }_i(x)|$ controls the worst amplification of simultaneous data errors at $x$; its maximum is the Lebesgue constant of interpolation.

Linear elasticity describes small deformations using a stress that depends linearly on the infinitesimal strain tensor. If $\mathbf{u}$ is the displacement field, thenFor an unstressed isotropic material, the constitutive equation iswhere $\lambda ,\mu$ are the Lamé parameters, ${\delta }_{ij}$ is the Kronecker delta and repeated indices are summed. In particular, $\mu$ is the shear modulus.

The small-displacement-gradient approximation permits superposition: for the same linear material and linear boundary conditions, the sum of two displacement solutions responds to the sum of their applied loads. This makes it possible to calculate bending, vibration and elastic waves from common material coefficients. Small strain alone is not sufficient when rotations are large; a slender elastic rod can turn through a large angle while stretching very little, requiring geometrically nonlinear kinematics.

A linear time-invariant system combines two properties. Linearity means that the output from a linear combination of inputs is the same linear combination of their outputs. Time invariance means that shifting an input by an integer delay shifts its output by exactly that delay. The operation has no preference for a particular time origin.

For discrete time, write the unit impulse as $\delta [n]=1$ at $n=0$ and zero otherwise, a Kronecker delta. Its response $h[n]=T[\delta ][n]$ is the impulse response. A finite input can be decomposed into shifted impulses, givingThe superposition principle supplies the sum and time invariance supplies the shifted copies of $h$. Infinite sums need convergence assumptions; for a bounded input and an impulse response $h$ with absolute summability, the discrete convolution converges absolutely. For a physical system with stored initial energy, this construction describes the response with zero initial state; a separate free response accounts for nonzero initial conditions.

The gain is reuse. Measure or calculate one impulse response, and the convolution predicts the response to every allowed signal without solving a new dynamical problem each time. Alternatively, the frequency response predicts how each oscillatory component changes. ^[ref].

A mixed tensor has $k$ defining indices and $l$ dual indices. The SU(3) action applies $U$ to every upper index and $U^{-1}$ to every lower index. Young symmetrizers select permutation types, while invariant Kronecker deltas take traces and invariant alternating tensors convert an antisymmetric pair of defining indices into a dual index, or conversely. Repeated fundamental and dual tensor rules decompose it with multiplicities: $(p,q)\otimes (1,0)=(p+1,q)\oplus (p-1,q+1)\oplus (p,q-1)$ and the conjugate rule, omitting negative labels.

An orthogonal polynomial family chooses polynomials that do not overlap under a specified inner product. For real polynomials on an interval $(a,b)$, use the weighted inner productAssume $a<b$, the weight $w$ is measurable and positive almost everywhere, and its absolute moments ${\int }_a^b|x{|}^{k}w(x)dx$ are finite for every integer $k\ge 0$. These conditions make the integrals finite and give every nonzero polynomial positive squared norm. A sequence $p_0,p_1,\dots$ is an orthogonal polynomial family when $p_n$ has real coefficients and polynomial degree exactly $n$, and $⟨p_n,p_k{⟩}_w=0$ for $n\ne k$.^[ref]

Here $h_n=\Vert p_n{\Vert }_w^2>0$ and ${\delta }_{nk}$ is the Kronecker delta. Orthogonality does not require $h_n=1$: an orthonormal set is obtained only after replacing $p_n$ by $p_n/\sqrt{h_n}$. Multiplying each member by any nonzero real constant preserves orthogonality, but changes its norm and the corresponding expansion coefficient. The monic orthogonal polynomial convention instead fixes each leading coefficient to one.

The first four Legendre polynomials, named for Adrien-Marie Legendre, use $w=1$ on $[-1,1]$ and the normalization $P_n(1)=1$:Their orthogonality and squared norms areThus their Gram matrix has a positive diagonal and zero entries elsewhere, although their function graphs cross and overlap on the page. Orthogonality measures an integral of products, rather than separation of function graphs.^[ref]

For these four polynomials, a short verification expands each product and uses ${\int }_{-1}^1{x}^{2j+1}dx=0$ and ${\int }_{-1}^1{x}^{2j}dx=2/(2j+1)$. For example, $⟨P_0,P_2⟩=(3(2/3)-2)/2=0$ and $⟨P_1,P_3⟩=(5(2/5)-3(2/3))/2=0$. The same calculation gives the diagonal entries $2,2/3,2/5,2/7$. More generally, apply the Gram-Schmidt process to the monomial basis $1,x,x^2,\dots$: at step $n$, subtract projections onto the earlier polynomials. The result retains its leading term, has degree $n$, and has positive norm, proving existence without a differential equation. Different degrees also give linear independence, so $p_0,\dots ,p_m$ form a basis of the polynomials of degree at most $m$.

This basis turns least-squares solutions for functions into independent scalar calculations, as the decoupled coefficients of orthogonal polynomial projections show. In a spectral method, increasing the degree enlarges the approximation space while keeping earlier projection coefficients fixed. Weighted integrals also connect the same orthogonal polynomials to Gaussian quadrature: its nodes are polynomial roots of a member of the family. The projection problem minimizes an integrated squared error; Gaussian quadrature uses the family to choose evaluation points for an integral.

An orthonormal family of orthogonal polynomials also gives each member unit norm in the chosen weighted inner product. Its pairwise inner products are the Kronecker delta. Dividing every nonzero orthogonal polynomial by its norm produces such a family; the sign or complex phase still requires a convention, such as a positive real leading coefficient.

The Boltzmann factor factorizes over bonds. Define the spin-chain transfer matrix byHere ${\delta }_{ab}$ is the Kronecker delta. Summing the product of these matrix elements over all configurations, and closing the final bond by periodic boundary conditions, gives the canonical partition functionHence, writing $I$ for the identity matrix and $P$ for the all-ones matrix,Every element of $P^2$ is a sum of $Q$ ones, while every diagonal element of $P$ is one. ThusMore explicitly, for $u=(1,\dots ,1{)}^T$, $Pu=Qu$. If ${\sum }_a{v}_a=0$, then $Pv=0$. These complementary eigenspaces have dimensions $1$ and $Q-1$. Equivalently, $P/Q$ is the orthogonal projector onto the linear span of $u$. The eigenvalues and their eigenvalue multiplicities are thereforeSince the two matrices share these eigenvectors, the exact finite-ring canonical partition function isBoth terms matter in the finite-ring low-temperature limit: discarding the second would lose the correct ground state degeneracy factor and introduce a spurious single-wall contribution.

Take $L\ge 2$, so neighbouring operators act at distinct sites. The identityexpresses the Hamiltonian operator of this quantum XY chain in spin ladder operators. Substituting the Schwinger boson representation gives the requested quartic expression:Operators at different sites commute, allowing normal ordering without a correction here. For the first quartic term, the inverse discrete Fourier transforms giveHere ${\delta }^{(L)}$ is one when its two momenta agree modulo $2\pi$ and zero otherwise; it expresses lattice momentum conservation. The other quartic term has the same ordered operators with phase $e^{i(p-r)}$. On the support of this Kronecker delta, $p-r=t-q$ modulo $2\pi$, so the two phases sum to $2\cos (q-t)$. Thus the Fourier vertex of a Schwinger boson chain isAn equivalent three-sum expression sets $t=p+q-r$ modulo $2\pi$, so its coefficient is $\cos (r-p)$. The factor $1/L$ comes from four factors $L^{-1/2}$ and one site sum, while the factor of two from the two exchange terms cancels the original $1/2$. The formula is Hermitian: conjugation exchanges $p$ with $t$ and $q$ with $r$, and the coefficient is unchanged after imposing lattice momentum conservation. It also preserves each local total boson number in its real-space form, hence preserves the physical constrained subspace.