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In three dimensions, ${ϵ}_{ijk}$ is zero when indices repeat and is the sign of the permutation $(i,j,k)$ otherwise.

For smooth vector fields $F,G$,andBoth follow by contracting two Levi-Civita symbols and applying the product rule.

The Laplacian takes the divergence of a scalar field's gradient. For a real-valued function $f$ on an open region of Euclidean space, with continuous partial derivatives through order two, its expression in a Cartesian coordinate system isThus the Laplacian, also called the Laplace operator, adds the coordinate second derivatives; equivalently it is the matrix trace of the Hessian matrix. Its output is another scalar field.^[ref]

Three original examples distinguish shape from this sum. The bowl $f(x,y)=x^2+y^2$ has $\Delta f=2+2=4$. The saddle $g(x,y)=x^2-y^2$ instead has $\Delta g=2-2=0$: opposite second derivatives cancel. It is a harmonic function, although its graph of a function is not a plane. For the Gaussian function $h(x,y)=e^{-r^2}$, with $r^2=x^2+y^2$, differentiation givesIts Laplacian is negative inside the circle of radius one, zero on that circle, and positive outside it. The Gaussian function itself is positive throughout: its value and its Laplacian describe different properties.

The sum does not favor a particular orientation. Under a change of Cartesian coordinate system represented by an orthogonal matrix $Q$, the Hessian matrix becomes $Q^{\mathsf{T}}(D^{2}f)Q$. The cyclic property of the trace and $Q{Q}^{\mathsf{T}}=I$ preserve its matrix trace. This makes the Laplacian useful when the behavior depends on spatial differences equally in every direction, rather than on the arbitrary choice of axes.

For a vector field, the Euclidean Laplacian acts separately on components expressed in a fixed Cartesian coordinate system. Components in a position-dependent basis require differentiating that basis too; applying a scalar formula separately to polar components omits those terms.^[ref] The scalar formula in Laplacian in polar coordinates is valid away from $r=0$; smoothness at the origin must be handled in coordinates that remain regular there.

The curl of $F$ has components

For a continuously differentiable vector field in three-dimensional Euclidean space, the component of its curl along a unit normal vector is the limiting circulation per unit area around a small loop in the perpendicular plane. The loop direction and the normal obey the right-hand rule. Stokes theorem gives the corresponding relation for a finite surface.

For the rigidly rotating velocity field $G=(-\Omega y,\Omega x,0)$, direct differentiation gives $\nabla \times G=(0,0,2\Omega )$. A circle of radius $R$ has tangential speed $\Omega R$ and circumference $2\pi R$, so its circulation is $2\pi \Omega R^2$, twice the angular velocity times the enclosed area. The factor of two distinguishes vorticity from rigid-body angular velocity.

Curved integral curves alone do not demonstrate nonzero curl. The field $H=(x,-y,0)$ has zero curl, yet its trajectories satisfy $x=x_0{e}^t$, $y=y_0{e}^{-t}$ and hence $xy=x_0{y}_0$: away from the axes they are hyperbolas. The curl measures local circulation of the field, not the curvature of an individual trajectory.

For $F=(F_1,F_2,F_3)$ in Cartesian coordinates,

For a continuously differentiable vector field, the divergence is the limiting outward flux per unit volume through a small closed surface. Opposite faces of a small rectangular box almost cancel; the remaining differences in the three components give the sum of partial derivatives above. The divergence theorem extends this local balance to a whole region.

For $F=(x,y,0)$, the divergence is $1+1+0=2$. In a planar unit square, the right edge contributes outward flux $1$ and the top contributes $1$, while the left and bottom contribute zero. The total $2$ equals the integral of the planar divergence over area $1$.

If $F$ is a velocity field, positive divergence means local volume expansion, not automatically creation of mass. The continuity equation gives $D\rho /Dt=-\rho \nabla \cdot F$, where $D/Dt$ is the material derivative. For this field, a uniform initial density evolves as $\rho (t)=\rho (0)e^{-2t}$ while material volumes grow as $e^{2t}$. Their product stays constant.

If a helicity density evolves as a sum of directional derivatives along divergence-free fields tangent to the boundary, its volume integral is conserved because both terms become vanishing boundary fluxes.

A vector field assigns a vector to every point of its domain.

Along a geodesic $\gamma$ with tangent $T$, the index form on endpoint-vanishing vector fields is

If failure of a property has a finite witness, one can often axiomatize the property by a universal sentence forbidding each possible finite witness. Compactness supplies finite witnesses when failure is expressed by inconsistency of an associated theory.

For an ordered basis $(x_i)$ of a Lie algebra $\mathfrak{g}$, the ordered monomials $x_1^{a_1}\cdots x_n^{a_n}$ form a basis of $U(\mathfrak{g})$. In particular, a triangular decomposition $\mathfrak{g}={\mathfrak{n}}^{-}\oplus \mathfrak{t}\oplus {\mathfrak{n}}^{+}$ gives $U(\mathfrak{g})\cong U({\mathfrak{n}}^{-})\otimes U(\mathfrak{t})\otimes U({\mathfrak{n}}^{+})$ as vector spaces.

A time-reversal operator $T$ intertwines forward and backward evolution:It must be antiunitary in ordinary quantum mechanics. Its conjugate linearity sends $i$ to $-i$, allowing it to commute with a time-reversal-invariant Hamiltonian rather than reversing the sign of its energy.

In the Heisenberg picture, states are fixed and an observable evolves asFor a time-independent Schrödinger-picture observable,

For $F⊣G$, the right adjoint $G$ is full and faithful exactly when the counit $FG\to 1$ is an isomorphism. Dually, $F$ is full and faithful exactly when the unit $1\to GF$ is an isomorphism.

Every unipotent algebraic subgroup of ${\mathrm{GL}}_n$ is conjugate to a subgroup of the upper unitriangular group. The filtration by vanishing initial superdiagonals then proves that every unipotent algebraic group is a nilpotent group.

Every finite union-closed family other than $\{\varnothing \}$ has an element belonging to at leastof its members. The proof applies the binary entropy product inequality coordinate by coordinate to the union of two independent uniform members of the family.

For $x,y\in [0,1]$, the binary entropy and the golden ratio $\phi$ satisfy

Let $X\ge 0$ have density $g$, let $U$ be independent and uniform on $[0,1]$, and setThen

For independent $U_1,U_2\sim \mathrm{Uniform}(0,1)$, the sum has triangular densityIts upper tail for $1\le c\le 2$ is $(2-c{)}^2/2$.

An ultrametric space is a metric space whose metric satisfies the ultrametric inequality. The rational numbers with $d_p(x,y)=|x-y{|}_p$, and their P-adic number completion, are examples. Their geometry organizes points into nested clusters rather than measuring separation along a number line.

For integers $a,b$ and $k\ge 0$, the definition of the P-adic valuation givesThus a closed ball of radius $p^{-k}$ in the integers is a congruence class modulo $p^k$. In the p-adic integers the same closed ball is $a+p^k{\mathbb{Z}}_p$. More shared low-order digits mean a smaller p-adic absolute value of the difference.

The binary tree first splits the sixteen integers into even and odd congruence classes, then refines modulo $4$, $8$, and $16$. Each node fixes one more low-order digit. The leaves $5$ and $13$ share the class $5$ modulo $8$, then separate modulo $16$, so their distance is exactly $1/8$. The leaves are deliberately ordered by their low-order digits; horizontal spacing and edge lengths are layout choices, not the metric. The singleton leaves describe this finite sample; a congruence class modulo $16$ in all p-adic integers remains infinite.

The ultrametric inequality also explains the unusual behavior of balls. Every point of an open ball can serve as its center: if $d(a,b)<r$, then $d(a,x)<r$ if and only if $d(b,x)<r$. Two open balls that intersect are nested, since a point of their intersection serves as a common center and their radii can be compared. The same nesting rule holds for closed balls and for a pair consisting of one open ball and one closed ball.

Every open ball is a clopen set. For a point $y$ outside $B(a,r)$, the open ball $B(y,d(a,y))$ stays outside $B(a,r)$: if $d(y,z)<d(a,y)$, the ultrametric inequality forces $d(a,z)=d(a,y)\ge r$. Hence the complement is an open set. Also every closed ball of positive radius is an open set, because $B(b,r)$ lies in the closed ball of radius $r$ around $a$ whenever $d(a,b)\le r$. A radius-zero closed ball is a singleton and need not be an open set. Distinct points can therefore be separated by a clopen set, proving that an ultrametric space is totally disconnected. These ball properties turn distance thresholds into a hierarchy of compatible clusters. ^[ref]

Quantifier elimination for dense linear orders shows that the one-types over $\mathbb{N}\subset \mathbb{Q}$ are determined by cuts: equality to a natural number, one of the intervals between consecutive natural numbers, the ray below zero, or the cut above every natural number. Only the last type is non-isolated.