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.

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.

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]

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.

A knot shadow is the curve in the Euclidean plane underlying a knot diagram after its overcrossing and undercrossing information has been erased. At a crossing the two local branches still continue across one another; the drawing does not specify which lies higher in space.

Let $V$ be a vector space over a specified field $F$, and let $v_1,\dots ,v_n\in V$. A linear combination is a finite sumThe coefficients $a_i$ are scalars from that same field; each term uses scalar multiplication, and the terms are joined by vector addition. For coefficients in the real numbers, negative values reverse arrows and fractional values scale them by fractional amounts. The collection of attainable vectors is the linear span. The column picture in Strang's MIT notes interprets solving a linear system as choosing coefficients to reach a target vector.^[ref]

For a concrete calculation over the real numbers, takeThen $2v_1+v_2=(2,2)+(1,-1)=(3,1)=w$. More generally, solving $a+b=x$ and $a-b=y$ givesEvery vector in the Euclidean plane is therefore a linear combination of $v_1,v_2$, so their linear span is all of ${\mathbb{R}}^2$. The target $(3,1)$ has coefficients $(2,1)$; the target $(0,2)$ has coefficients $(1,-1)$, since $(1,1)-(1,-1)=(0,2)$. The displayed formula also makes the coefficients unique. Consequently $(v_1,v_2)$ is an ordered basis, with coordinate vector $[(x,y){]}_{(v_1,v_2)}=((x+y)/2,(x-y)/2{)}^T$.

The payoff is a reusable language for synthesis: linear combinations build vectors from chosen ingredients, linear span decides which targets are reachable, and linear independence decides whether the coefficients are unique. The ingredients can also be polynomials or functions, as long as they belong to a vector space over the specified field.

The Minkowski sum of subsets $A,B$ of a vector space collects the result of vector addition for every possible pair:Think of fixing a point $b$ of $B$, making a translation of the whole of $A$ by $b$, and then taking the union of these copies over every $b\in B$. The points of $B$ specify displacement vectors; they need not be the centre of a shape. Minkowski addition applies to nonconvex sets too. It is different from set union: for $A=B=\{0,1\}$ on the real line, $A\cup B=\{0,1\}$ but $A+B=\{0,1,2\}$.

For axis-aligned rectangles in the Euclidean plane, the coordinate intervals add separately:Every coordinate between $0$ and $a+c$ can be split into one coordinate in $[0,a]$ and another in $[0,c]$, and likewise for the vertical coordinate. Thus a $2\times 1$ rectangle plus a $1\times 2$ rectangle gives a $3\times 3$ square of area $9$, although the two input areas total only $4$. The Minkowski sum adds possible displacements, so its area need not be the sum of input areas.

For closed Euclidean disks $D_r=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le r\}$ centred at the origin, the triangle inequality gives $D_r+D_s\subseteq D_{r+s}$. Conversely, if $r+s>0$, split any $x\in D_{r+s}$ as $x=rx/(r+s)+sx/(r+s)$. Therefore $D_r+D_s=D_{r+s}$, including the case $r=s=0$. Disks of radii $1$ and $2$ give a disk of radius $3$, with areas $\pi ,4\pi ,9\pi$. For disks with centres $u,v$, the resulting centre is $u+v$. These two examples lead to the root-of-area comparison in the Brunn-Minkowski theorem.

Minkowski addition is commutative and associative because vector addition is. Its identity is the singleton $\{0\}$, and $A+\varnothing =\varnothing$. For displacement vectors $u,v$ and $t>0$, translations and uniform dilations satisfywhere $tA=\{ta:a\in A\}$ uses scalar multiplication. A Minkowski sum of convex sets is convex: a line segment between $a_1+b_1$ and $a_2+b_2$ is obtained by adding the corresponding points of the line segments in $A$ and $B$. Closure requires care; see sum of two closed sets need not be closed and sum of a compact set and a closed set.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

On two copies of the real line with their usual topologies, let $U=(1,2)$ and $V=(2,4)$. The coordinate projection maps giveThe first two open sets are unbounded strips; their set intersection is an open rectangle. The point $(1.4,3)$ passes both coordinate tests, whereas $(2,3)$ fails the first because the endpoint $2$ is excluded.

Every point of an open set in the Euclidean plane has such an open rectangle around it inside that open set. Conversely, every open rectangle contains a small Euclidean ball around each of its points. Thus the product topology here is the usual metric topology of the Euclidean plane. With finitely many factors all coordinates may be restricted; on an infinite Cartesian product a basic cylinder set must leave all but finitely many factors free.

In the Euclidean plane, precisely one line through a point outside a given line is parallel to it. This formulation is equivalent to Euclid's fifth postulate within the usual axioms of plane geometry.

Let $G$ be the finite group and choose a point $p$. Its orbit $E=\{gp:g\in G\}$ is a nonempty finite set. Every group element permutes $E$ and maps a closed metric disc onto a disc of the same radius. By the permitted uniqueness of the smallest enclosing disc, it fixes that disc and therefore its centre $o$. Thus $G$ lies in the orientation-preserving stabilizer of one point.

In the Euclidean plane, translate $o$ to zero. Every isometry fixing zero is an orthogonal linear map: distance preservation gives preservation of inner products by the polarization identity, and its values on an orthonormal basis then determine the linear map. Orientation preservation gives determinant one, so it is a rotation. Consequently $G$ embeds in the circle group of rotation angles.

For the hyperbolic plane, use this precise standard fact about its isometries: after placing a point at the origin of the Poincare disc model, every orientation-preserving isometry fixing it has the form $z\mapsto e^{i\theta }z$. Equivalently, its point stabilizer is the rotation group, and its action on directions at the point is faithful. This is a general hyperbolic-plane fact allowed by the question. It again identifies $G$ with a finite subgroup of the circle group.

Finally, a nontrivial finite rotation group has a least positive angle $\theta$ in $[0,2\pi )$. Divide any other angle $\varphi$ by $\theta$: the remainder $\varphi -k\theta$ is another group angle in $[0,\theta )$, so it must be zero. Similarly the remainder of $2\pi$ is zero, and $\theta =2\pi /m$ for an integer $m$. The rotation through $\theta$ generates the group. The trivial group is cyclic as well. Therefore both finite orientation-preserving plane isometry groups are cyclic. This proves the finite orientation-preserving plane isometry groups are cyclic result using the same enclosing-disc argument in both geometries.

A plane crystallographic group is a discrete subgroup of the Euclidean plane's isometry group with compact quotient; equivalently a compact set has translates covering the plane. We prove the crystallographic restriction theorem without assuming its conclusion. The orientation-preserving subgroup $G^{+}$ has index at most two and still has a cocompact group action. Its subgroup $T$ of translations is discrete as an additive subgroup of ${\mathbb{R}}^2$.

Suppose $T$ were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in $G^{+}$ is a translation. Thus $G^{+}$ would be an abelian group. Every member would then commute with the given nontrivial rotation $g$ and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then $G$ would be finite, contradicting cocompactness on the unbounded plane. Hence $T$ contains a nonzero vector $v$.

If $g$ is a half-turn its order is already two. Otherwise its linear part $R$ rotates $v$ to a linearly independent vector $Rv$, and conjugating translations shows that $Rv\in T$. Thus $T$ spans the plane and is a euclidean lattice. A discrete additive subgroup meets every bounded closed set in finitely many points: an accumulating sequence would give nonzero differences tending to zero, contradicting isolation of the identity. To see the euclidean lattice assertion directly, choose a shortest nonzero translation $v$. Translations on its line are integer multiples of $v$. Reducing all parallel components modulo $v$ shows that nonzero perpendicular components are bounded away from zero, since otherwise infinitely many translation vectors would accumulate in a compact rectangle. Choose a vector $w$ with the smallest positive perpendicular component; reducing by $w$ and then $v$ proves $T=\mathbb{Z}v+\mathbb{Z}w$.

Conjugation by $g$ preserves this euclidean lattice. In its euclidean lattice basis, $R$ is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also $2\cos \theta$, where $\theta$ is the rotation angle. Since $g$ is nontrivial,These values give angles of order $2,3,4,6$ respectively. Hence

For $n\ne 0$, put $\rho =f(r)/|n|$ and $\theta =|n|\varphi$. Then $d{\rho }^2=d{r}^2$ andThese give a local isometry wherever $\rho >0$. For $n=0$, $f=b>0$, and the local coordinates $X=r$, $Y=b\varphi$ similarly give the Euclidean plane metric tensor. Equivalently, the Gaussian curvature of the original Riemannian metric is $-f^{\prime \prime }/f=0$ throughout this family.

A local isometry need not be a global isometry. If $|n|=1$, the displayed polar map is injective modulo $\varphi \sim \varphi +2\pi$ and identifies the regular domain with a Euclidean annulus, possibly a punctured disc or the punctured Euclidean plane, according to the radial interval. If $|n|>1$, it is a $|n|$-fold covering of that annulus: distinct angles differing by $2\pi /|n|$ have the same image. This developing map cannot be made injective by a different choice of Euclidean coordinates, since two local Euclidean isometries on a connected flat domain differ by a rigid motion once their value and differential are fixed at one point. For $n=0$, the domain is a flat cylinder; its circular curves are closed geodesics, whereas the Euclidean plane has no nonconstant closed geodesic.

All these metrics are locally Euclidean; they are not all globally Euclidean. Even the regular $|n|=1$ domain is an annular region rather than the entire Euclidean plane. A regular axis can be added without a conical singularity only when $|n|=1$; with the full radial range this completion becomes the Euclidean plane. Trivial holonomy alone therefore does not supply a global injective Euclidean chart.