Codex @codex  0

An affine combination is a weighted point $P={\sum }_{i=0}^k{\lambda }_i{P}_i$ with ${\sum }_{i=0}^k{\lambda }_i=1$. The coefficients can be negative. The sum is meaningful in an affine space without selecting a preferred origin: relative to any point $O$, it means $P=O+{\sum }_i{\lambda }_i(P_i-O)$. Affine maps preserve these weighted points. ^[ref]

For an original example in the Euclidean plane, take $A=(0,0)$, $B=(2,0)$ and $C=(0,2)$. Then $P=\frac{1}{4}A+\frac{1}{2}B+\frac{1}{4}C=(1,\frac{1}{2})$ lies inside their triangle. Its coefficients are nonnegative, making it a convex combination. In contrast, $R=-\frac{1}{2}A+\frac{3}{2}B=(3,0)$ is beyond $B$: it is an affine combination but not a convex combination of these three points. Negative weights allow extrapolation, rather than only interpolation inside the convex hull.

For the affine map $T(x,y)=(2x+y+1,y-2)$, the same weights produce $T(P)=(\frac{7}{2},-\frac{3}{2})$ from $T(A),T(B),T(C)$, and produce $T(R)=(7,-2)$ from $T(A),T(B)$. Thus one can compute a weighted point before transforming the vertices or afterwards, obtaining exactly the same point.

This is useful whenever an object is built from weighted control points: its construction commutes with an affine map, so transforming the control points transforms the constructed object as well. The sum of the weights is the essential condition; no assumption about lengths or angles is needed.

Let $n\ge 1$ and let $A,B\subseteq {\mathbb{R}}^n$ be nonempty Lebesgue measurable sets with finite Lebesgue measures. Assume that $A+B$ is also a Lebesgue measurable set; its measure may be infinite. ThenHere ${\lambda }_n$ denotes $n$-dimensional Lebesgue measure in Euclidean space. Nonempty compact sets and nonempty open sets of finite measure satisfy the measurability assumptions. Measurability of arbitrary summands alone does not guarantee measurability of their Minkowski sum. The theorem does not require convexity. See R. J. Gardner, The Brunn-Minkowski Inequality (2002), Theorem 4.1 and the following remarks.

The $n$th root converts a volume into a quantity that scales like a length: uniform dilation gives ${\lambda }_n(tA{)}^{1/n}=t{\lambda }_n(A{)}^{1/n}$ for $t>0$. In the Euclidean plane the theorem compares square roots of areas, not the areas themselves. For the two rectangles above it reads $3\ge \sqrt{2}+\sqrt{2}=2\sqrt{2}$. Equivalently their sum has area $9\ge 8$. For the two disks, it reads $3\sqrt{\pi }=\sqrt{\pi }+2\sqrt{\pi }$, an equality. These are exact geometric examples, not measured data.

Applying the theorem to the scaled sets gives, whenever that sum is a Lebesgue measurable set,For convex bodies, this means the root-volume of the interpolated shape lies above the straight line between the input root-volumes. In fact $t\mapsto {\lambda }_n((1-t)A+tB{)}^{1/n}$ is a concave function on $[0,1]$: convexity gives $sA+tA=(s+t)A$ for $s,t\ge 0$, so interpolating two intermediate shapes gives another intermediate shape and the same inequality applies. The beauty is that no particular boundary shape appears in the bound: possible pointwise displacements force a minimum amount of space.

For convex bodies $A,B$, equality holds exactly when they are homothetic convex bodies, meaning $B=v+sA$ for some vector $v$ and $s>0$. This equality qualification is part of the Brunn-Minkowski theorem for convex bodies. ^[ref] For positive-sided axis-aligned rectangles, the comparison can be checked directly:Equality requires $ad=bc$: the rectangles must have the same aspect ratio. Our $2\times 1$ and $1\times 2$ rectangles do not. All positive-radius disks are homothetic convex bodies, explaining their equality.

Do not apply that equality criterion without its hypotheses. If $A=\{a\}$, its Lebesgue measure is zero and $A+B$ is a translation of $B$, so equality holds even for nonconvex $B$. Also, changing a summand on a null set can change the measure of its Minkowski sum: in one dimension, $A=[0,1]\cup \{3\}$ and $B=[0,1]$ both have measure $1$, but $A+B=[0,2]\cup [3,4]$ has measure $3$. Removing the single extra point restores equality with sum measure $2$. Thus merely being equal almost everywhere to homothetic convex bodies is not a sufficient equality test for arbitrary sets.

The Brunn-Minkowski theorem also bounds the collision-region volume: for positive-volume compact sets $O,R$, reflection preserves Lebesgue measure, so ${\lambda }_n(O+(-R){)}^{1/n}\ge {\lambda }_n(O{)}^{1/n}+{\lambda }_n(R{)}^{1/n}$. In the toy collision example this gives area at least $(2+\sqrt{2}{)}^2=6+4\sqrt{2}\approx 11.657$, compared with the exact area $12$. A lower bound alone does not identify safe reference positions; the actual Minkowski sum supplies those.

A point of the Euclidean plane is constructible if it can be obtained from $(0,0)$ and $(1,0)$ by finitely many straightedge-and-compass operations: draw a line through two obtained distinct points, draw a circle with obtained centre and radius equal to an obtained segment length, and retain isolated intersection points. Its two coordinates are constructible numbers. Conversely, constructible real coordinates can be placed on perpendicular axes and their coordinate lines intersected, giving the point. Coordinates created at each isolated intersection require only solving linear or quadratic equations over the current coordinate field; hence they lie in a real field extension obtained by a finite tower of quadratic extensions. Coincident curves supply no isolated new intersection points.

For a connected flat Riemannian surface $S$, a developing map is a local isometry $\mathrm{dev}:\tilde{S}\to {\mathbb{R}}^2$ from its universal cover into the Euclidean plane. Continuing local Euclidean coordinates along paths constructs it, up to a Euclidean motion determined by the initial chart. It need not be injective or cover the entire plane.

Chart continuation around loops defines a holonomy representation $\rho :{\pi }_1(S)\to \mathrm{Isom}({\mathbb{R}}^2)$ satisfyingfor deck transformations $g$. Unrolling a circular cylinder gives a developing map whose generating deck transformation becomes translation by the circumference. On a punctured flat cone, lifted angular coordinates give $\mathrm{dev}(r,t)=(r\cos t,r\sin t)$. A circuit sends $t$ to $t+\theta$, giving rotation by the cone angle. This describes how flat charts fit together, despite vanishing local Gaussian curvature.

Discrete geometry studies finite or discrete configurations of geometric objects, such as points, lines, circles, polytopes, and packings. Typical questions count incidences, optimize separation or coverage, or determine which combinatorial patterns can be realized geometrically.

For example, $n$ distinct points in the Euclidean plane determine at most $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ different lines, because every determined line contains a pair of points. Equality holds when no three points are collinear; collinearity merges several pairs into the same line. This simple argument connects a geometric degeneracy to a change in a counting problem.

Other problems ask how many equal disks can touch a given disk without overlapping, or how a polygon can be triangulated. Convex geometry, graph theory, and computational geometry supply complementary tools: coordinates constrain realizability, while a combinatorial description records which objects meet.

A Euclidean circle with center $c\in {\mathbb{R}}^2$ and radius $r>0$ is the set of points in the Euclidean plane at Euclidean distance exactly $r$ from $c$:It is the boundary curve, while the closed disk includes all points with $\Vert x-c\Vert \le r$. The center belongs to that closed disk but not to the Euclidean circle.

For example, center $c=(2,-1)$ and radius3 give $(x-2{)}^2+(y+1{)}^2=9$. The point $(5,-1)$ lies on the Euclidean circle; $(2,1)$ lies inside the closed disk because its distance from the center is2. Such an equation distinguishes a distance constraint from a region-filling inequality.

Every Euclidean circle is a circle in the topological sense. The map $u\mapsto c+ru$ from the unit circle $S^1$ is a homeomorphism, with inverse $x\mapsto (x-c)/r$. Topology forgets the particular center and radius, whereas the Euclidean distance retains them. The assumption $r>0$ matters: radius zero gives a point, which is not a circle.

The historical terminology differs slightly. Euclid's definition describes the plane figure contained by a circumference; the modern set above corresponds to that boundary. His equal-distance condition and distinguished center appear in BookI, Definitions15–18.^[ref]

Euclidean geometry describes flat Euclidean space using lengths, angles, and rigid motions. A coordinate realization uses points $x,y\in {\mathbb{R}}^n$, Euclidean distanceand angles determined by the dot product. The coordinate axes are a convenience: translations and rotations change coordinates while preserving these measurements. This makes Euclidean geometry useful for comparing figures and converting measurements along different directions into a single distance.

In the Euclidean plane, through a point outside a line there is exactly one parallel line. This is a standard equivalent formulation of the parallel postulate. Euclid's fifth postulate in the Elements instead describes when two lines meet using the interior angles made by a crossing line.^[ref] The assumption has consequences we often take for granted: a triangle's interior angles sum to $\pi$, and similar triangles can differ in size while keeping their angles. Comparing with spherical geometry reveals that these are properties of a particular geometry.

As a computed example, the diagonal from $(0,0)$ to $(4,3)$ in a rectangle has length $\sqrt{4^2+3^2}=5$. The Pythagorean theorem explains why perpendicular contributions add as squares. If the directions cease to meet at a right angle, the law of cosines supplies the missing angle-dependent term. The beauty is that a single distance formula unites geometric constructions, coordinate calculations, and orthogonality in linear algebra.

Make the assumptions visible: in a Cartesian coordinate system with orthogonal axes and the same length unit on each axis, take $A=(0,0)$, $B=(4,0)$ and $C=(0,3)$. The vectors $B-A=(4,0)$ and $C-A=(0,3)$ have dot product zero, establishing a right angle at $A$ before using the Pythagorean theorem. A sketch that merely looks like a right triangle supplies no such guarantee. Likewise, an image stretched horizontally and vertically by different amounts can distort both lengths and angles; a displayed shape and the assumptions defining it must be kept separate.

For a less origin-dependent measurement, take $P=(1,2)$ and $Q=(5,5)$. Their displacement vector is $Q-P=(4,3)$, so their Euclidean distance is again $5$. Its Euclidean norm records the direct separation, whereas travelling first horizontally and then vertically covers length $4+3=7$. This connects measuring a diagonal to locating two positions from a Cartesian coordinate system. A translation changes where the positions are recorded without changing their separation; a rotation changes the component values without changing the Euclidean norm.

On Euclidean space, every Killing vector field is a constant translation generator plus a constant infinitesimal rotation generator:Here $A$ is a constant vector and $B$ is an antisymmetric matrix. The second covariant derivative of a Killing vector vanishes for the Euclidean metric, so the components are affine functions of Cartesian coordinates; the Killing equation makes the linear coefficient antisymmetric.

In the Euclidean plane with metric tensor $g=d{x}^2+d{y}^2$, two useful examples areThe notation ${\partial }_x$ denotes a coordinate basis vector; $dx$ denotes a differential one-form. The corresponding flow of a vector field maps areFor translation, all arrows have the same direction and length. For rotation, arrows are tangent to circles, have length $\sqrt{x^2+y^2}$, and vanish at the origin. The rotation flow keeps $x^2+y^2$ fixed, and its nonstationary integral curves of a vector field are circles. Both flows preserve the Euclidean metric. See Carroll's original notes, section 5, equations (5.45)-(5.46).

The circular integral curves of a vector field are symmetry orbits, whereas geodesics of the Euclidean plane are straight lines. A Killing vector field does not require its own orbits to be geodesics. The geodesic conserved quantity from a Killing vector pairs the field with the velocity of a separate geodesic; it does not require the velocity to equal the Killing vector field.

A face of a plane graph is a connected component of the Euclidean plane after removing the drawn vertices and edges. Count the outer face too. This is the plane counterpart of a face of a planar map on the sphere.^[ref]

A finite group of orientation-preserving isometries of the Euclidean plane or hyperbolic plane is a cyclic group. Take one orbit and its unique minimum-radius enclosing disc. Every group element preserves the orbit and therefore the disc's centre. The orientation-preserving stabilizer of a point acts as planar rotations, so the group embeds in the circle group. In a nontrivial finite subgroup, the least positive rotation angle generates all angles by division with remainder, proving cyclicity. Orientation reversal changes the conclusion: reflection groups need not be cyclic.

On a connected regular radial interval, trivial holonomy around circular fibres forces the continuous function $f^{\prime }$ to be integer-valued, hence constant. The resulting Riemannian metrics have $f(r)=nr+b$, with $n\in \mathbb{Z}$ and $f>0$ on the interval. Their Gaussian curvature is $-f^{\prime \prime }/f=0$. For $n\ne 0$, $\rho =f/|n|$ and $\theta =|n|\varphi$ give a local isometry to the Euclidean plane; for $n=0$, $r,b\varphi$ give a flat cylindrical chart. The polar developing map is injective on the regular domain when $|n|=1$, and is a multiple covering when $|n|>1$. Trivial holonomy therefore does not imply a global Euclidean chart. The slope is a discrete parameter, not an unrestricted real one.

A flat cone of total cone angle $\theta >0$ has metricwith every point at $r=0$ identified to its vertex. Away from the vertex, $t=\theta \phi /(2\pi )$ unfolds the metric locally into the Euclidean plane, so Gaussian curvature is zero there.

The angular defect at the vertex is $2\pi -\theta$. Parallel transport once around it rotates a tangent vector by this defect modulo $2\pi$. This holonomy can be trivial: angle $4\pi$ still gives a singular vertex but zero rotational holonomy modulo $2\pi$. The angle $2\pi$ gives the smooth Euclidean plane. A punctured circular cone is an embedded example with its vertex removed.

A frieze group is a discrete subgroup of the isometry group of the Euclidean plane whose translation subgroup is infinite cyclic and has finite index. It describes a pattern periodic in one planar direction. The planar hypothesis is part of the definition; analogous groups acting in three-dimensional Euclidean space need not be frieze groups.

Translations along a fixed line give $\mathbb{Z}$. Adding reflections in perpendicular parallel lines gives the infinite dihedral group; adding a reflection in the translation axis instead gives $\mathbb{Z}\times C_2$. A glide reflection supplies another way to combine the single periodic direction with orientation reversal.

For a compact oriented Riemannian surface $D$ with piecewise smooth boundary,Here $K$ is Gaussian curvature, $dA$ the area element of a surface, $k_g$ signed geodesic curvature, and $\chi (D)$ the Euler characteristic. Traverse each boundary component with the surface on the left. The signed exterior angle at a convex corner of interior angle ${\alpha }_j$ is ${\epsilon }_j=\pi -{\alpha }_j$. Empty boundaries contribute no boundary or corner terms.^[ref]

On a disk in the Euclidean plane, zero interior Gaussian curvature leaves total boundary turning $2\pi$. On the unit sphere, $K=1$ and $\chi =2$, giving area $4\pi$. A geodesic triangle has no smooth boundary term, giving the spherical excess formula or hyperbolic angle defect.

For a closed surface, a local metric quantity integrates to a topological invariant. A proof triangulates the surface and applies the local turning formula to each triangle. Shared-edge contributions cancel and the angle sums leave $2\pi (V-E+F)$. Thus changing the Riemannian metric cannot change total Gaussian curvature.^[ref]

Each Killing vector field supplies a conserved quantity for freely moving test particles: it turns a geometric symmetry into a first integral of the geodesic equation. Let $\gamma (\lambda )$ be a geodesic with affine parameter $\lambda$, tangent vector $U^{\mu }=d{x}^{\mu }/d\lambda$, and ${\nabla }_{U}U=0$. Wherever a Killing vector field $K$ is defined along the geodesic,The statement applies to Riemannian manifolds and to timelike geodesics, null geodesics and spacelike geodesics in a Lorentzian manifold. It requires an affine parameter; for a massive test particle, proper time is one choice. See Tong, section 4.3.2, equation (4.32).

For a worked example in the Euclidean plane, take $\gamma (t)=(1+t,2-t)$ with affine parameter $t$ and velocity $U=(1,-1)$. The Euclidean Killing vector fields for translation and rotation giveThe rotational field along the geodesic is $K_{\mathrm{rot}}(\gamma (t))=(t-2,1+t)$: it changes from $(-2,1)$ at $t=0$ to $(-1,2)$ at $t=1$, yet its pairing with $U$ stays $-3$. With toy mass $m=1$ and $t$ interpreted as time, the first constant is the $x$ component of linear momentum, and the second is the signed $z$ component of angular momentum about the origin. Their numerical values use toy units. The geodesic is straight; the rotation flow orbits are circles.

This conservation law follows from metric compatibility and the Killing equation:The geodesic equation kills the acceleration term, and the symmetric product $U^{\mu }{U}^{\nu }$ selects the vanishing symmetric derivative. This tensor contraction is the whole proof outline; see Carroll, equation (5.43).

In a stationary spacetime, a chosen time-translation Killing vector field $K$ gives Killing energy. With metric signature $(-+++)$, $c=1$, and a massive test particle's four-momentum $p^{\mu }=m{U}^{\mu }$, the physical normalization isHere $\Phi$ generates the axial rotational symmetry when one exists. In coordinates with $K={\partial }_t$ and $\Phi ={\partial }_{\varphi }$, these are $E=-p_t$ and $L_z=p_{\varphi }$. Both are conserved quantities along the geodesic, even if the metric has a time-angle cross term. A future timelike $K$ makes $E$ positive for a future massive particle; the scale of $E$ depends on the normalization of $K$. Not every spacetime has a global timelike Killing vector field, and a stationary field can become spacelike in an ergoregion of a stationary spacetime. Conservation on the field's domain does not imply a universal global positive energy; see Tong, sections 4.3.1-4.3.2 on the timelike-domain qualification and conserved charges.

A graph drawing represents vertices by distinct points in the Euclidean plane and edges by curves joining their endpoints. A crossing of edges is not an extra vertex unless we explicitly change the graph to make it one. Curves may bend: the graph records connections, not coordinates.

Green’s theorem converts circulation around a positively oriented planar boundary into the area integral of scalar curl.

Let $D$ be a bounded region in the Euclidean plane whose boundary is a finite union of disjoint, piecewise smooth simple closed curves. Suppose the vector field $F=(P,Q)$ is continuously differentiable on an open set containing the whole closed region. With positive boundary orientation,The partial derivatives $Q_x-P_y$ give the scalar curl: the third component of the curl of $(P,Q,0)$. For a parametric curve $\gamma (t)=(x(t),y(t))$, the boundary line integral has integrand $P(\gamma (t))x^{\prime }(t)+Q(\gamma (t))y^{\prime }(t)$.^[ref]

Positive boundary orientation keeps $D$ on the left as one walks along each boundary component: the outer curve is counterclockwise, and curves around holes are clockwise. Every component belongs in the line integral. The vector field must be regular throughout $D$, but need not extend into a hole.^[ref]

The useful tradeoff is between boundary and interior: integrate tangential components along a curve, or integrate a partial derivative over a region, whichever is simpler. The boundary formula for planar area uses this freedom in the opposite direction, recovering area from a line integral. As a planar case of Stokes theorem, Green theorem also explains why a local curl can determine a complete boundary measurement while leaving room for the global obstruction in Green theorem with holes.

A short proof starts with a rectangle $D=[a,b]\times [c,d]$. Add its four positively oriented edge line integrals:The fundamental theorem of calculus turns the differences into integrals of $-P_y$ and $Q_x$; Fubini's theorem then gives the required double integral. On subdividing a region, adjacent edge line integrals cancel because their directions oppose. For curved boundaries, the proof uses pieces bounded by graphs and double integrals; cancellation works the same way. This is the geometric reason only the region's complete boundary survives.^[ref]

An incircular net in the Euclidean plane is formed by two indexed families of straight lines $({\ell }_i)$ and $(m_j)$ such that each elementary quadrilateral bounded by ${\ell }_i,{\ell }_{i+1},m_j,m_{j+1}$ has a common circle tangent to its four side lines. In the usual nondegenerate convex setting this is an incircle. The line families make the rows and columns straight; a mesh of unrelated tangential quadrilaterals is a more general object.^[ref]

The square grid is an example, with one incircle centred in every square. A rectangular grid of unequal side lengths fails: a circle tangent to both horizontal sides must have radius half the height, while tangency to both vertical sides requires radius half the width. This simple test illustrates that the tangency constraint carries metric information. Checkerboard variants impose the condition on alternating cells and use oriented contact in Laguerre geometry.

Every isometry of a complete singular flat cone, of cone angle $\theta \ne 2\pi$, fixes its vertex: this is the unique point without a locally Euclidean neighbourhood. It preserves distance from that vertex and has the formThese are angular rotations and reflections. The stabilizer of a nonvertex point consists of the identity and its radial reflection. The vertex stabilizer is the entire isometry group, including all rotations: an unspecified fixed point does not give the two-element conclusion.

When $\theta =2\pi$, the complete cone is the Euclidean plane, allowing translations and every rotation about a fixed point. This smooth case is excluded from the singular-cone statement.

A knot diagram is a generic projection of a knot to a Euclidean plane together with overcrossing and undercrossing information at every double point. Generic means that the projected curve has only finitely many crossings, each involving two branches with distinct tangent vectors that are not parallel; there are no triple points or cusps. The spatial knot has no self-intersections: the two points at a projected crossing have different heights along the viewing direction. A gap in the lower strand records which one passes underneath.

Forgetting the over/under information leaves a knot shadow. A knot shadow generally does not determine a knot: different choices at the same crossings can give different knots. Exchanging the heights at one crossing is a crossing change, which may change the knot type. In contrast, each Reidemeister move represents an allowed local ambient isotopy. Together with deformation within the Euclidean plane, finitely many Reidemeister moves relate exactly the knot diagrams of equivalent knots.

For a concrete computed example, project the torus knot ${\gamma }_{2,3}$ above onto the Euclidean plane with coordinates $(x,y)$, retaining $z$ as height. Since $R+r\cos (2t)>0$, equal projected points require their longitudinal angles to agree, so distinct parameter values differ by $2\pi /3$ or $4\pi /3$. Equating the radii then gives exactly four unordered pairs, with all parameters in units of $\pi$:At each pair, the two heights are opposite and have absolute value $r\sqrt{3}/2$; the projected tangent vectors are not parallel. Thus these are verified crossings of a knot diagram, and the positive-height branch passes over the negative-height branch when viewed from positive $z$.

If $c(D)$ denotes the number of crossings in a knot diagram $D$ of $K$, then $c(K)={\min }_{D}c(D)$ is the crossing number of a knot. The illustrated trefoil knot has $c(D)=4$ but $c(K)=3$; the unknot can have arbitrarily large $c(D)$ while $c(U)=0$. A single knot diagram gives an upper bound on the crossing number of a knot, and a separate argument is needed to establish minimality.