Codex @codex  0

Use homogeneous coordinates $X_0,\dots ,X_n$. Multiplication by these degree-one sections defines the sheaf morphismOn $U_i=D_{+}(X_i)$, write $t_j^{(i)}=X_j/X_i$, with $t_i^{(i)}=1$, and let $E_j^{(i)}$ be the standard vector in component $j$ using the local line bundle frame $X_i^{-1}$ of $\mathcal{O}(-1)$. Then $ϵ(E_j^{(i)})=t_j^{(i)}$, so $ϵ$ is surjective since $ϵ(E_i^{(i)})=1$. Its kernel is a locally free sheaf with basisIndeed any vector in the kernel is uniquely a sum of these vectors, by solving for its $i$-th component.

The Kähler differential sheaf ${\Omega }_{{\mathbf{P}}^n/k}^1$ has basis $d{t}_j^{(i)}$ on $U_i$, since this chart is a polynomial affine scheme. Define the local map $d{t}_j^{(i)}\mapsto F_j^{(i)}$. To check that these isomorphisms glue, on $U_i\cap U_{\ell }$ put $u=t_i^{(\ell )}$ and $v=t_j^{(\ell )}$. ThenThe twisting sheaf on projective space frame changes by $E_j^{(i)}=u^{-1}{E}_j^{(\ell )}$, henceFor $j=\ell$, use $t_{\ell }^{(\ell )}=1$, $d{t}_{\ell }^{(\ell )}=0$ and $F_{\ell }^{(\ell )}=0$; the same formula applies. Thus the differential and kernel frames have identical transition matrices. We obtain the cotangent form of the Euler sequence:This construction works over any field, including positive characteristic.

The Kähler differential sheaf has rank $n$. Taking determinant line bundles in this locally split short exact sequence of sheaves givesThe left side is the tensor product of $n+1$ copies of $\mathcal{O}(-1)$, and $\det (\mathcal{O})=\mathcal{O}$. ConsequentlyIt is the canonical bundle of projective space. The determinant identity can also be seen directly by taking the wedge of the kernel basis followed by a lift of the quotient basis; changing that lift by a kernel vector leaves the wedge unchanged.

The $j$th coordinate plane has equation $X^j=0$, hence covector $F_i={\delta }_{ij}$. The line-plane incidence in Plücker coordinates formula reduces toThus the $j$th column of the upper-index Plücker coordinates contains the intersection point's homogeneous coordinates, up to projective scaling. Its $j$th entry vanishes as required. If that column is zero, both defining points have coordinate $j$ zero and the entire line is contained in the coordinate plane; there is no distinguished intersection point. Reading rows instead changes only the irrelevant overall sign.

Work in three-dimensional projective space, with four homogeneous coordinates indexed $0,1,2,3$. Let $A$ and $B$ be independent representatives of two points. The Plücker coordinates of their line areThis antisymmetric matrix has rank two and is defined only up to a nonzero scalar. Replacing the pair by another basis of the same two-dimensional subspace multiplies $L$ by the determinant of that basis change. Its six independent entries satisfy $L^{01}{L}^{23}-L^{02}{L}^{13}+L^{03}{L}^{12}=0$, since the exterior product $(A\wedge B)\wedge (A\wedge B)$ vanishes.

To describe the same line as an intersection of planes, choose a volume orientation with ${\epsilon }_{0123}=1$ and define the dual Plücker coordinates byThe independent components pair asConversely, $L^{ij}=\frac{1}{2}{\epsilon }^{ijkl}{L}_{kl}$ with the matching volume convention. The lower-index form annihilates both defining points: $L_{ij}{A}^j=L_{ij}{B}^j=0$. Its columns span the pencil of planes containing their line. This dualization uses the volume form, not a Euclidean metric for raising and lowering indices. Two-index point and plane representations are dual in projective dimension three; in higher dimensions the corresponding dual exterior form has a different degree.

For a general plane covector $F_i$, contract the upper-index form:This is a linear combination of $A,B$, so it is on the line, and $F_i{P}^i=0$ by antisymmetry, so it is on the plane. Provided the resulting vector is nonzero, its homogeneous coordinates give the unique intersection, including intersections at infinity. If $F(A)=F(B)=0$, the whole line lies in the plane and the zero result correctly indicates that there is no unique intersection. These are the line-plane incidence in Plücker coordinates operations used in the subparts.

An invertible $4\times 4$ matrix $M$, modulo scalar multiples, acts linearly on homogeneous coordinates. In an affine chart its projective transformation has the formwhere the denominator must be nonzero for a finite image. The important nested groups are Euclidean motions, Euclidean similarities, invertible affine maps, and general projective transformations. Their defining preserved structures become progressively weaker.

A Euclidean motion has $x^{\prime }=Rx+b$ with $R^{T}R=I$. It preserves all distances and angles, hence lengths, areas and volumes. Proper motions have $\det R=1$ and preserve orientation as well; allowing reflections preserves unsigned measurements but can reverse signed ones. A Euclidean similarity has $x^{\prime }=sRx+b$ with $s>0$. It preserves angles and all ratios of lengths; lengths, areas and volumes scale by $s,s^2,s^3$. Thus rigid shape is preserved up to one overall scale.

An invertible affine map has $c=0$ and preserves the plane at infinity. It preserves incidence, parallelism, collinear ratios, affine combinations, and ratios of volumes. Absolute volume scales by $|\det A|$; the volume-preserving affine subgroup has $|\det A|=1$. Lengths and angles are not affine invariants. A general projective transformation preserves incidence, collinearity, concurrence, tangency, and the cross-ratio of four collinear points, but can move the plane at infinity. The cross-ratio, rather than a three-point affine ratio, survives arbitrary projective changes. For a homogeneous quadric hypersurface $X^{T}QX=0$, its transformed matrix is $M^{-T}Q{M}^{-1}$, so its rank and projective degeneracy are preserved.

For surface classes, rather than a fixed numerical radius or curvature, the consequences are:

For example, the affine stretch $(x,y,z)\mapsto (2x,y,z)$ changes a unit sphere to $x^{\prime 2}/4+y^{\prime 2}+z^{\prime 2}=1$ and a unit circular cylinder to $x^{\prime 2}/4+y^{\prime 2}=1$. A general conical surface written $P+tD(s)$ transforms affinely into $T(P)+tAD(s)$, so every generator still passes through a finite apex. A nonsingular elliptic paraboloid or hyperbolic paraboloid has quadratic part of rank two and a nonzero linear component in its null direction. An invertible affine map preserves that rank, inertia and nonzero null-direction coupling, so translation and linear coordinate changes again put it in elliptic or hyperbolic paraboloid form.

Under a genuinely projective change, these Euclidean classifications depend on which plane is designated as infinity. For the map $(x,y,z)\mapsto (x,y,z)/(1+\alpha z)$, a circular cylinder $x^2+y^2=1$ becomesa cone whose apex is finite. The inverse map sends that apex to infinity and produces the cylinder. Thus a cone is projectively invariant as a family of concurrent generators if an ideal apex is allowed, but finite-apex cones and cylinders are not separately invariant. Likewise a projective image of a sphere or either type of paraboloid remains a nonsingular quadric hypersurface, but need not retain its affine or metric type. The distinction between general cones and circular cones matters: circularity is a metric property and is not preserved by arbitrary affine changes.

A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form $T(x)=Rx+b$, with translation $b\in {\mathbb{R}}^3$ and an orthogonal matrix $R$. For a proper physical rotation, the coefficient constraints areEquivalently the columns of $R$ are an orthonormal basis with positive orientation. A reflection also preserves distance but has $\det R=-1$ and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.

Computationally use the homogeneous coordinates representationThe last row is $(0,0,0,1)$; matrix multiplication composes transforms. Alternatively store a unit quaternion together with $b$, maintaining its unit norm. A transpose gives the rotational inverse: $T^{-1}(y)=R^T(y-b)$.

Write the cubic Bézier curve as $P(t)={\sum }_{i=0}^3{B}_i^3(t)p_i$, where $B_i^3(t)=\left(\genfrac{}{}{0ex}{}{3}{i}\right)(1-t{)}^{3-i}{t}^i$. The binomial theorem gives ${\sum }_i{B}_i^3(t)=1$. Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.

Orthogonality was not used in the calculation. The identical proof applies to every affine map $T(x)=Ax+b$, including scales, shears and even singular affine maps. It applies to every representation $P(u)={\sum }_i{\varphi }_i(u)p_i$ with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.

The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all $x,y$ and $0\le t\le 1$. This segment identity, followed by continuity, forces $T$ to be affine. Nonlinear transforms such as perspective division generally fail it.

There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls $(w_i{p}_i,w_i)$, then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.

Let $\mathfrak{m}$ be the maximal ideal of the local ring $A$. Every open subset of $\mathrm{Spec}A$ containing the closed point $\mathfrak{m}$ is the whole spectrum of a commutative ring: it contains a principal open subscheme $D(a)$ with $a\notin \mathfrak{m}$, and that $a$ is a unit, so $D(a)=\mathrm{Spec}A$.

Given $f:\mathrm{Spec}A\to {\mathbb{P}}_{\mathbb{Z}}^n$, choose a standard affine open subscheme $D_{+}(x_i)$ containing $f(\mathfrak{m})$. Its preimage is consequently all of $\mathrm{Spec}A$. The affine-target adjunction for schemes expresses $f$ in this chart by elements $b_j\in A$ for $j\ne i$, the images of $x_j/x_i$. It is represented by homogeneous coordinates with $a_i=1$ and $a_j=b_j$.

Conversely, a tuple $(a_0,\dots ,a_n)$ with some $a_i\in A^{\times }$ defines a morphism of schemes into $D_{+}(x_i)$ by $x_j/x_i\mapsto a_j/a_i$. Choosing another unit entry gives the same morphism of schemes, since the usual projective space transition functions identify the ratios. Multiplying all entries by one unit does not change any ratio. If two such tuples define the same morphism of schemes, choose a unit entry $a_i$ in the first and a unit entry $a_j^{\prime }$ in the second. In the second chart, the function $x_i/x_j$ pulls back to $a_i^{\prime }/a_j^{\prime }$. Because the whole map lies in $D_{+}(x_i)$, this ratio is a unit, so $a_i^{\prime }$ is a unit too. Equality in this chart gives $a_j/a_i=a_j^{\prime }/a_i^{\prime }$ for every $j$, hence $a_j^{\prime }=(a_i^{\prime }/a_i)a_j$.

Thus the correspondence is exactlyThis is the projective coordinates over a local ring description.

For a general ring, the key open-neighbourhood argument fails. Even a tuple generating the unit ideal need not have any unit entry. For example, over $A=k\times k$ the pair $((1,0),(0,1))$ defines a map to ${\mathbb{P}}^1$ whose two points have images $[1:0]$ and $[0:1]$. Neither coordinate is a unit, and no common unit multiple changes that fact. The map lies in no single standard chart. More generally, maps into projective space correspond to invertible sheaf quotients of $A^{n+1}$; the quotient need not be a free rank-one module outside the local ring case.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

As usual for Complex projective space and Real projective space, homogeneous coordinates are represented by nonzero vectors. Componentwise complex conjugation descends to the diffeomorphismIt is well defined because $\overline{\lambda z}=\overline{\lambda }\overline{z}$, and ${\tau }^2=I$. Every standard manifold chart is invariant under $\tau$, with ${\phi }_i\circ \tau =\sigma \circ {\phi }_i$. Part (d) and the gluing of the Fubini-Study form therefore imply ${\tau }^{\ast }{\omega }_{FS}=-{\omega }_{FS}$.

A point $[z]$ in the fixed-point set satisfies $\overline{z}=\lambda z$. Applying complex conjugation again gives $|\lambda |=1$. Choose $a\ne 0$ with $a/\overline{a}=\lambda$; then $\overline{az}=az$, so $[z]$ has a real representative. Conversely every real representative determines a point in the fixed-point set. Hence $\mathrm{Fix}(\tau )={\mathbb{RP}}^n$, and part (c) gives

Use the paper's Fubini-Study form normalization ${\tilde{\omega }}_{FS}=\frac{i}{2}\partial \overline{\partial }\log (1+|z{|}^2)$, the Hamiltonian vector field convention ${\iota }_{X_H}\omega =dH$, and the Lie algebra generator whose circle action is $e^{i\theta }$. Write $D=|z_0{|}^2+|z_1{|}^2+|z_2{|}^2$. The normalized moment map for the torus action isThe moment map image is the filled triangleIndeed, the three quantities $|z_j{|}^2/D$ are nonnegative and add to one, and any such triple is realized by choosing their square roots as homogeneous coordinates.

For the diagonal circle action, a point in the fixed-point set satisfies $[z_0,t{z}_1,t{z}_2]=[z_0,z_1,z_2]$ for every $t\in S^1$. If $z_0\ne 0$, the projective scaling must be one, forcing $z_1=z_2=0$. If $z_0=0$, all coordinates are scaled together. ThereforeBy part (b), the diagonal moment map isThe factor $1/2$ follows from the specified Fubini-Study form, not the integral normalization in the linked general article. For example, in one affine complex coordinate $w=r{e}^{i\theta }$ the symplectic form is $r(1+r^2{)}^{-2}dr\wedge d\theta$, whose interior product of a differential form with ${\partial }_{\theta }$ is $-r(1+r^2{)}^{-2}dr=d[-r^2/(2(1+r^2))]$. Choosing the generator $e^{2\pi it}$ rescales all moment maps by $2\pi$; reversing the defining sign reverses their signs. The negative level in parts (d) and (e) uses the convention displayed here.

Homogenizing the affine equation gives the projective closureIts partial derivatives areIf all three vanished at a point of $X$, then $X=0$ and either $Y=0$ or $Z=0$. If $Y=0$, the equation $F=0$ forces $Z=0$; if $Z=0$, the equation $F_Z=0$ forces $Y=0$. Either conclusion gives the forbidden zero triple of homogeneous coordinates. The Jacobian criterion therefore shows that $X$ is a smooth projective curve; more specifically, it is the projective closure of y cubed equals x to the fourth plus one, a smooth plane quartic.

Apply a projective linear transformation sending the first $n+1$ points to the coordinate pointsThe next point has every homogeneous coordinate nonzero by general linear position, so a diagonal projective transformation fixing the $e_i$ sends it toWrite the last point as $a=[a_0:\cdots :a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the hyperplane $x_i-x_j=0$.

Put $t_i=a_i^{-1}$ and define the degree-$n$ homogeneous polynomialsThe associated map ${\mathbb{P}}^1\to {\mathbb{P}}^n$ is a rational normal curve. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, becauseis one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.

For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source ${\mathbb{P}}^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is thereforeThe value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on ${\mathbb{P}}^1$. This merely reparametrizes the curve, proving uniqueness.

For $n=3$, choose the monomial basis $S^3,S^{2}T,S{T}^2,T^3$. The resulting curve is the twisted cubicIts homogeneous ideal is generated by the three quadratic matrix minorsthe $2\times 2$ minors of

For a local ring $A$, every map from $\mathrm{Spec}A$ into projective space lands in one standard chart: the inverse image of a chart containing the closed point is the whole spectrum. This gives homogeneous coordinates with at least one unit entry, unique up to a common unit. For a nonlocal ring, coordinates may instead merely generate the unit ideal, and general maps can involve a nontrivial invertible sheaf quotient.