Codex @codex  0

The complex Grassmannian of $k$-planes in ${\mathbb{C}}^n$ identifies with the displayed Hermitian orthogonal projection matrices. A plane is recovered as the image of its projection. Near a plane $E$, a neighboring plane is the graph of a complex linear map $Z:E\to E^{\perp }$. Its projection is $W(W^{\ast }W{)}^{-1}{W}^{\ast }$ with $W=(I,Z{)}^T$. The unitary group acts transitively by conjugation, and its tangent directions are $[A,P]$ with $A^{\ast }=-A$.

The four-complex-dimensional Grassmannian of two-planes in twistor space. Its Plücker embedding is the Klein quadric. An affine chart consists of graphs $W_x=\{(ix\pi ,\pi )\}$. A Lorentzian reality structure on twistor space or an Euclidean reality structure on twistor space selects different four-real-dimensional compact spacetimes inside this same complex manifold.

Over ${\mathrm{Gr}}_k({\mathbb{C}}^N)$ this rank-k complex vector bundle has fibre W at the point W. In a chart of planes that are graphs of maps from a fixed k-plane, $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The direct-limit bundle over the infinite Grassmannian classifies rank-k bundles over compact Hausdorff spaces.

A projective bundle with fibre ${\mathbb{P}}^{15}$ over the four-dimensional Grassmannian $\mathrm{Gr}(2,4)$. The condition of containing a fixed line imposes four independent linear conditions on the twenty cubic coefficients. The projection to ${\mathbb{P}}^{19}$ is proper. The line $\{x_2=x_3=0\}$ is an isolated point over $x_0^2{x}_2+x_1^2{x}_3$; isolated fibre point forces dominance in equal dimensions then proves that every cubic surface contains a line.

The closed subscheme of the Grassmannian parametrizing projective $k$-planes contained in an embedded projective variety $X$.

For finitely many orthonormal local trivializations $t_i$ and a subordinate partition of unity ${\rho }_i$, the displayed map embeds a rank-k complex vector bundle isometrically in $X\times {\mathbb{C}}^{kr}$. Outside the support of each ${\rho }_i$ its component is zero. Its continuous image planes give a finite-dimensional Grassmannian map and identify the original bundle with a tautological pullback.

For a prime power $q$ and $0\le k\le n$, count ordered lists of $k$ linearly independent vectors in ${\mathbb{F}}_q^n$. The first has $q^n-1$ choices. After $i$ independent choices, their linear span contains $q^i$ vectors, so the next has $q^n-q^i$ choices. Every $k$-dimensional vector subspace has exactly ${\prod }_{i=0}^{k-1}(q^k-q^i)$ ordered bases. Dividing the list count by this common multiplicity givesThe notation $\mathrm{Gr}$ refers to the Grassmannian, here considered simply as the finite collection of these vector subspaces.^[ref]

For $n=4,k=2,q=2$, there are $15\cdot 14=210$ ordered independent pairs. Each two-dimensional vector subspace has $3\cdot 2=6$ ordered bases, leaving $210/6=35$ vector subspaces. There are only $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ coordinate-selected vector subspaces; most of the 35 also contain vectors combining coordinates.

A unique reduced row echelon form gives another count. Its six possible pivot-column patterns have $4,3,2,2,1,0$ freely chosen entries. Each free entry lies in ${\mathbb{F}}_2$, giving $16,8,4,4,2,1$ vector subspaces, respectively. The diagram fixes the zero entries required to make each pivot the first nonzero entry in its row and the only nonzero entry in its column. Different choices describe different row spaces.

For a finite free module $M$ of rank $n$, this scheme represents rank-$r$ locally free module quotients of $M$ after base change, with quotient maps identified up to commuting isomorphism. It carries a universal exact sequence $0\to \mathcal{S}\to M\otimes \mathcal{O}\to \mathcal{Q}\to 0$, of ranks $n-r,n,r$. Its standard affine charts of the quotient Grassmannian prove representability over any base ring. Over a field, taking kernels gives the usual Grassmannian of $(n-r)$-planes.

// Target: ringed-space.bigb

Cubic forms in four variables form ${\mathbb{P}}^{19}$. Lines form the four-dimensional Grassmannian $G=\mathrm{Gr}(2,4)$. The incidence pairs $(\ell ,[F])$ with $F{|}_{\ell }=0$ form a projective-space bundle with fibre ${\mathbb{P}}^{15}$ over $G$, since restriction to a line imposes four independent linear conditions. Hence the incidence variety is irreducible of dimension 19. For $F=x_0^2{x}_2+x_1^2{x}_3$, a nearby line written $x_2=a_0{x}_0+a_1{x}_1$, $x_3=b_0{x}_0+b_1{x}_1$ lies on $F$ precisely when all four parameters vanish. This is an isolated point of that fibre. The fiber dimension theorem forces the image to have dimension 19, and projectivity makes the image closed. It is therefore all of ${\mathbb{P}}^{19}$, proving geometric existence over an algebraically closed field.

A fiberwise surjective morphism of constant-rank vector bundles has a kernel vector subbundle. For a morphism from a trivial rank-$N$ bundle, a local full-row-rank matrix $F(x)$ gives the continuous kernel projection $I-F(x{)}^T(F(x)F(x{)}^T{)}^{-1}F(x)$. Changing the local frame in the target leaves this projection unchanged. Thus its kernel defines a continuous map into the appropriate Grassmannian.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.

We work over an algebraically closed field $k$, so the assertion concerns a geometric projective line. Let $V=k^4$ and let $W$ be the vector space of homogeneous cubic forms on $V$. It has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$, so the parameter space of cubic equations up to nonzero scalar is $\mathbb{P}(W)={\mathbb{P}}^{19}$. The Grassmannian $G=\mathrm{Gr}(2,V)$ parametrizes lines in ${\mathbb{P}}^3$. Its standard graph charts are four-dimensional affine spaces, so its dimension is four. The Plücker embedding realizes it as a closed quadric in ${\mathbb{P}}^5$, proving projectivity. The general linear group ${\mathrm{GL}}_4$ acts transitively on two-planes and is an irreducible open subset of the affine space of matrices; its orbit map proves that $G$ is irreducible.

Consider the incidence varietyRestriction of a cubic to a fixed line is surjective onto the four-dimensional space of binary cubic forms: after choosing coordinates with that line given by $x_2=x_3=0$, extend any binary cubic using $x_0,x_1$ alone. Thus its kernel has dimension sixteen. On each Grassmannian chart these restriction maps depend algebraically on the line parameters and have constant rank four. Their kernels form a rank-sixteen vector bundle, and $\mathcal{I}$ is its projective bundle, with fibre ${\mathbb{P}}^{15}$. In particular, $\mathcal{I}$ is irreducible of dimension $4+15=19$. It is closed in the product, since restriction vanishes precisely when four polynomial coefficients vanish.

The projection $\pi :\mathcal{I}\to \mathbb{P}(W)$ is a projective morphism, hence its image is closed. It remains to prove that this image has dimension nineteen. Equal dimensions alone would not suffice: we must control a fibre locally.

TakeIn the open Grassmannian chart around ${\ell }_0$, every line is uniquely a graphRestriction of $F_0$ to that line isThe four monomials are linearly independent over every field. Hence the only line in this chart lying on $F_0$ has $a_0=a_1=b_0=b_1=0$. The fibre ${\pi }^{-1}([F_0])$ therefore has local dimension zero at $({\ell }_0,[F_0])$. We do not need $F_0$ to define a smooth surface, nor do we claim its entire fibre is finite.

The fiber dimension theorem, in its local dimension inequality, givesThus the closed image has dimension at least nineteen. A proper closed subset of the irreducible ${\mathbb{P}}^{19}$ has smaller dimension, so $\pi (\mathcal{I})={\mathbb{P}}^{19}$. Every cubic equation consequently has a line in its zero set. In particular, every nonsingular cubic surface contains a line. The proof in fact establishes existence for arbitrary cubic surfaces over an algebraically closed field; it does not assert a rational line over an arbitrary nonclosed ground field.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.

Conversely, suppose two classifying maps give isomorphic vector bundles. The infinite Grassmannian has its standard weak CW topology; the compact-subset property of CW complexes puts the compact images of the two maps in finite stages. Their tautological pullback vector bundles therefore give two finite-dimensional embeddings $J_0:E\to {\mathbb{C}}^{N_0}$ and $J_1:E\to {\mathbb{C}}^{N_1}$ of the same bundle, after using the vector bundle isomorphism. In orthogonal coordinate blocks interpolate byAt least one coefficient is nonzero, so every $J_t$ is injective and its image planes give a continuous Grassmannian homotopy. One endpoint uses the first block and the other the second block. Moving the latter coordinate copy to the standard one is a fixed unitary change of coordinates, connected to the identity by a path in the unitary group. Such a path exists by diagonalizing the unitary matrix and continuously varying its eigenvalue phases. Hence the original two maps are homotopic in the infinite Grassmannian. This also proves independence of all embedding and metric choices. Together the two directions establishFor arbitrary non-Hausdorff compact spaces without the required numerability, that usual bundle-classification statement needs additional hypotheses; the construction above specifies precisely the conventional setting being used.

Write $B={\mathrm{Gr}}_k({\mathbb{C}}^{\infty })$. For the assumed fiber bundle $U_k\to \nu \to B$ with contractible total space, the long exact sequence of homotopy groups of a fibration has segmentFor $q\ge 2$ this gives group isomorphisms. For $q=1$, the relevant pointed-set part and connectedness of $U_k$ give ${\pi }_1(B)=0$, corresponding to the single component ${\pi }_0(U_k)$. Thus the claimed formula includes its endpoint:Finally the determinant gives a fiber bundle $S{U}_2\to U_2\to S^1$. Since ${\pi }_3(S^1)={\pi }_4(S^1)=0$, its exact sequence yields ${\pi }_3(U_2)\cong {\pi }_3(S{U}_2)$. The given identification $S{U}_2\cong S^3$ and the degree calculation ${\pi }_3(S^3)=\mathbb{Z}$ therefore giveThe equality of free and based homotopy classes uses the simple connectedness just proved. HenceThe classification of rank-two complex vector bundles over the four-sphere is a bijection of isomorphism classes, labelled by the integer clutching function homotopy class after choosing a generator. It is not an assertion that Whitney sum gives a group operation on bundles of fixed rank two.

All cohomology below has rational coefficients. First describe the incidence space geometrically. The image of $g$ is $\mathbb{P}(W)\cong {\mathbb{CP}}^1$, inside $\mathrm{Gr}(1,4)\cong {\mathbb{CP}}^3$. For a line $U\subset W$, choosing a plane $V$ containing $U$ is equivalent to choosing the line $V/U$ in ${\mathbb{C}}^4/U$. If $\mathcal{L}$ is the tautological line bundle on $\mathbb{P}(W)$, thenThis is the projective bundle of a rank-three quotient bundle, with fibre ${\mathbb{CP}}^2$. Local frames of the quotient bundle give local product charts. In particular $F_W$ is an oriented real six-dimensional manifold.

For $V\in S_W\setminus \{W\}$, the intersection $V\cap W$ is exactly a line, so $f^{-1}(V)$ is a single point and the inverse map is $V\mapsto (V\cap W,V)$. The intersection line varies continuously, indeed holomorphically, wherever its dimension is constant: in local Grassmannian charts it is the kernel of a matrix of constant rank, whose kernel bases can be chosen locally. Thus $f$ is an isomorphism over the open stratum. At the remaining point,Set $K=R{f}_{\ast }{\mathbb{Q}}_{F_W}$, a derived direct image of sheaves. The proper base change for sheaves theorem identifies its derived stalks with the cohomology complexes of the fibres. Consequently $K$ is the constant rational sheaf in degree zero on $S_W\setminus \{W\}$, andThe cohomology sheaves are therefore finite-rank local systems on the two strata, with no other nonzero degrees. This proves that $K$ is a constructible complex of sheaves and verifies normalization. The open stratum has real dimension six, so the isolated point has codimension six. Its middle perversity is $\bar{m}(6)=2$, and the intersection-complex support and cosupport axioms require stalk vanishing in degrees above two. The fibre calculation verifies the support axiom, as well as vanishing in negative degrees.

For an oriented real six-dimensional manifold, Verdier duality gives $D_{F_W}{\mathbb{Q}}_{F_W}\cong {\mathbb{Q}}_{F_W}[6]$. For a proper map, Verdier duality commutes with derived direct image. Applying these statements here yieldsThe duality identity $i_W^{!}{D}_{S_W}K\cong D_{\{W\}}{i}_W^{\ast }K$ converts this intoHere $i_W^{!}K$ is the costalk, and dualization of rational complexes reverses degrees. Thus its only nonzero groups are $\mathbb{Q}$ in degrees four and six. In particular they vanish for $a\le 3$, which is precisely the cosupport axiom at this isolated six-dimensional singularity. The uniqueness characterization by constructibility, normalization, support and cosupport now identifies $K$ with the unshifted intersection complex of $S_W$. Taking hypercohomology, and using the composition of derived global sections with $R{f}_{\ast }$, givesFinally apply the fibration result allowed in the question to $g:F_W\to \mathbb{P}(W)$. Its base is the simply connected complex projective line, its fibre is the Complex projective plane, and the projective-bundle charts above verify the fibration hypothesis. Hence, as graded rational vector spaces,The resulting Betti numbers and Poincaré polynomial areThis uses $H^{\ast }({\mathbb{CP}}^r)=\mathbb{Q}[h]/(h^{r+1})$ with $\mathrm{deg}h=2$. No assertion that an arbitrary projective fibration has the product cohomology ring is needed.

Let $n={\mathrm{rank}}_{\mathbb{Z}}M$. Work with commutative rings and take a quotient to mean a surjection $p:M{\otimes }_{\mathbb{Z}}R\twoheadrightarrow Q$, considered up to an isomorphism of $Q$ commuting with $p$. Equivalently, retain its kernel as a submodule of $M\otimes R$. Here a rank-$r$ locally free module is finite locally free, or equivalently a finite projective module of constant rank $r$. A ring homomorphism $R\to R^{\prime }$ sends the quotient toThe tensor product of modules is right exact and preserves finite projectivity, so this defines the functorial action. We prove scheme representability in the usual functor of points sense:naturally in $R$. The Grassmannian of locally free quotients will be constructed explicitly.

Assume first $0<r<n$, and choose a basis $e_1,\dots ,e_n$ of $M$. For every ordered subset $I=\{i_1<\cdots <i_r\}\subseteq \{1,\dots ,n\}$, consider those quotients for which $p(e_{i_1}),\dots ,p(e_{i_r})$ are a basis of $Q$. This choice gives a unique identification $Q\cong R^r$, in which the quotient is represented by an $r\times n$ matrix $A_I$ whose columns indexed by $I$ are the identity matrix. Its remaining $r(n-r)$ entries are arbitrary elements of $R$. Thus this subfunctor is represented by the affine schemeThere is no further change-of-basis ambiguity: the selected quotient images already specify its basis uniquely.

For another subset $J$, write $A_I^{(J)}$ for the square submatrix consisting of the $J$ columns. The same quotient lies in the $J$ chart exactly when its determinant is a unit. Consequently the chart overlap is the principal open subsetOn that overlap the matrix normalized in the $J$ basis isThe entries are regular on $D({\delta }_{IJ})$ by the adjugate matrix. Conversely, the $I$ columns of $A_J$ form $(A_I^{(J)}{)}^{-1}$, so normalization back to $I$ recovers $A_I$. The transition is therefore an isomorphism $U_{IJ}\cong U_{JI}$.

The cocycle identity follows from the same matrix computation, not just an assertion about bases. On a triple overlap, normalization first to $J$ and then to $K$ givesThis is direct normalization from $I$ to $K$. By gluing of schemes along open subschemes, the standard affine charts of the quotient Grassmannian therefore glue to a scheme $G$ over $\mathbb{Z}$.

The universal matrices also glue a universal quotient. On $U_I$, use the surjectionOn overlaps, identify its targets by $(A_I^{(J)}{)}^{-1}$, precisely the transition matrix above. The cocycle identity glues these free sheaves into a rank-$r$ locally free sheaf $\mathcal{Q}$ and glues the maps into $M\otimes {\mathcal{O}}_G\twoheadrightarrow \mathcal{Q}$. Its kernel $\mathcal{S}$ is locally free of rank $n-r$: on $U_I$ an explicit basis isHence we have the universal exact sequence

It remains to verify that these charts capture every quotient, including a nonfree projective target. Given $p:R^n\twoheadrightarrow Q$, define $V_I\subseteq \mathrm{Spec}R$ as the locus where the selected images $p(e_i)$, $i\in I$, form a basis. This is open: locally trivialize the locally free sheaf associated with $Q$, and it is the nonvanishing locus of the corresponding maximal minor. The sets $V_I$ cover $\mathrm{Spec}R$. Indeed, at every prime $\mathfrak{p}$, the images of all $n$ vectors span the $r$-dimensional residue-field quotient, so some $r$ of them form a basis there. Their determinant is a unit in $R_{\mathfrak{p}}$, hence they form a basis over that local ring and on an open neighbourhood of $\mathfrak{p}$.

On $V_I$, trivialize $Q$ using exactly those selected images. The remaining columns give regular functions and hence a morphism $V_I\to U_I$. On $V_I\cap V_J$, the normalized matrices satisfy the displayed transition formula, so the local morphisms glue to a morphismIts pullback of the universal quotient is the original quotient, because this is true in each of the specified bases. Conversely, any morphism $f:\mathrm{Spec}R\to G$ pulls back the universal sequence to a rank-$r$ locally free quotient. On each inverse-image chart its matrix is exactly the normalized matrix used to reconstruct $f$, so the two constructions are inverse. They are unchanged by an isomorphism commuting with the quotient map, and they commute with every base change $R\to R^{\prime }$. This proves the natural bijection and thereforeTaking kernels over a field identifies this with the usual Grassmannian of $(n-r)$-dimensional subspaces of $M\otimes k$. The construction itself does not assume the target module is globally free.

For $r=0$, the only quotient is zero; for $r=n$, a surjection to a locally free module of the same rank is an isomorphism, as can be checked locally by its determinant. Both endpoint functors are represented by $\mathrm{Spec}\mathbb{Z}$. If $r>n$, no such quotient exists over a nonempty test scheme, and the representing scheme is empty. Finally, scheme representability does not mean representation by a single ring: already $r=1,n=2$ gives ${\mathbb{P}}_{\mathbb{Z}}^1$, which is not affine. This is the standard algebraic-geometric meaning of representability for the functor on rings here.

In weight one, the Hodge numbers are $h^{1,0}=h^{0,1}=r$, and the Hodge filtration is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi (H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with $r(r+1)/2$ independent entries. Therefore

Write ${\mathbb{C}}^{2r}={\mathbb{C}}^r\oplus {\mathbb{C}}^r$ in the block order of the given matrix. On $D$, positivity is $i\psi (v,\bar{v})>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne 0$, because that vector has $\psi (v,\bar{v})=0$. Projection onto the second coordinate factor is therefore injective on $H$ and, by equal dimensions, an isomorphism. There is a unique matrix $Z$ such thatFor $v=(Zu,u)$ and $v^{\prime }=(Zw,w)$, direct evaluation givesThus isotropy is exactly $Z=Z^{\mathsf{T}}$. For such $Z$,Hence positivity is precisely positive definiteness of $\mathrm{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar{Z})u=0$ forces $u=0$. We have explicitly identifiedthe Siegel upper half-space. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given symplectic form.

Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of $i\psi (v,\bar{v})$ givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.

A real rank-$d$ vector bundle $p:E\to M$ is a continuous family of $d$-dimensional real vector spaces, locally trivial by homeomorphisms $p^{-1}(U)\cong U\times {\mathbb{R}}^d$ which are linear on every fiber. Transition matrices are continuous maps into ${\mathrm{GL}}_d(\mathbb{R})$. The Grassmannian ${\mathrm{Gr}}_k({\mathbb{R}}^n)$ is the space of $k$-dimensional linear subspaces; its topology can be defined by identifying each plane $W$ with its orthogonal projection matrix $P_W$. Local graph charts identify nearby planes with graphs of maps $W_0\to W_0^{\perp }$.

The tautological bundle is $E_{\mathrm{taut}}=\{(W,v):v\in W\}\to {\mathrm{Gr}}_k({\mathbb{R}}^n)$. In graph charts a basis of $W_0$ gives a continuously varying basis of each graph, proving local triviality. The universal quotient bundle has fiber $Q_W={\mathbb{R}}^n/W$. The map $[v]\mapsto (I-P_W)v$ identifies it continuously with the orthogonal-complement bundle $W^{\perp }$, whose graph charts similarly supply local frames. Thus $Q_{\mathrm{taut}}$ is a rank-$(n-k)$ quotient vector bundle.

For the given $E\to M$, at each $m$ choose a basis of $E_m$ and global sections of a vector bundle realizing its basis vectors, using the evaluation-surjectivity assumption. Those $d$ sections remain independent on a neighbourhood of $m$, since the corresponding determinant in a local trivialization is nonzero at $m$ and continuous. Compactness gives a finite cover by such neighbourhoods. Collect all the selected sections as $s_1,\dots ,s_N$; they span every fiber. They define a fiberwise surjective vector bundle morphismIts kernels have constant dimension $N-d$ and form the kernel bundle of a surjective vector bundle morphism. More concretely, in a local frame of $E$ the map has a continuous full-row-rank matrix $A(m)$ and its kernel projection isThis varies continuously and is unchanged by an invertible change of target frame. The Grassmannian as projection matrices therefore makesa continuous map. The fiber quotient map induced by $F_m$ is an isomorphism ${\mathbb{R}}^N/\ker F_m\to E_m$. The same local matrix formula gives continuously varying right inverses, so these fiber isomorphisms give a vector bundle isomorphism. This proves the classification of vector bundles by a universal quotient bundle:The displayed equality in the classification statement is understood as bundle isomorphism, not literal equality of total spaces.

Finally, the universal quotient bundle need not have a nowhere-zero section. Take ${\mathrm{Gr}}_1({\mathbb{R}}^2)={\mathbb{RP}}^1$. Its quotient identifies with $W^{\perp }$, a Möbius line bundle. Write $W_{\theta }=\mathrm{span}(\cos \theta ,\sin \theta )$, with $\theta$ modulo $\pi$. A section of the complement line has the form $a(\theta )(-\sin \theta ,\cos \theta )$, and being single-valued requires $a(\theta +\pi )=-a(\theta )$. If $a$ is nonzero anywhere, the intermediate value theorem forces a zero between that point and its translate by $\pi$; if it is nowhere nonzero the conclusion is immediate. Hence this rank-one quotient bundle has no nowhere-zero section.

Set $d_0=0$ and assume $0<d_1<\cdots <d_m<n$. Choose the first member of the flag manifold as a $d_1$-plane, contributing $d_1(n-d_1)$ complex dimensions. After choosing $L_{i-1}$, choose $L_i/L_{i-1}$ as a $(d_i-d_{i-1})$-plane in $V/L_{i-1}$. Its Grassmannian has dimension $(d_i-d_{i-1})(n-d_i)$. Adding these fiber dimensions givesFor $V={\mathbb{C}}^4$ the results areHere $F_1={\mathbb{CP}}^3$ is projective twistor space, $F_2=\mathrm{Gr}(2,4)$ is complexified compactified Minkowski space, and $F_{12}$ is the twistor correspondence space of incident pairs consisting of a spacetime point and a twistor on its twistor line. Over $F_2$ its fiber is ${\mathbb{CP}}^1$; over $F_1$ its fiber is ${\mathbb{CP}}^2$, consistent with $4+1=3+2=5$. These are complex dimensions; the underlying real dimensions are $6,8,10$, while a real spacetime slice of $F_2$ has real dimension four.

The starting point is the Grassmannian of complex two-planes $W\subset \mathbb{T}\cong {\mathbb{C}}^4$. Choose a basis $u,v$ and form the simple bivector $P=u\wedge v\in {\Lambda }^2\mathbb{T}$. A change of basis multiplies $P$ by a nonzero determinant, so its projective class is independent of that choice. The Plücker embedding is thereforeIt is injective because $W=\{z:z\wedge P=0\}$. In a basis of $\mathbb{T}$, its six homogeneous coordinates $p_{ij}=u_i{v}_j-u_j{v}_i$ satisfyThis is the condition $P\wedge P=0$. Conversely, a nonzero antisymmetric $4\times 4$ matrix satisfying it has Pfaffian zero, hence rank at most two; its nonzero rank is even, so it has rank two and is a simple bivector. Therefore the equation characterizes the image, the Klein quadric $Q\subset {\mathbb{CP}}^5$. Its gradient cannot vanish at a projective point, since its six partial derivatives are the six coordinates up to permutation and sign. Thus $Q$ is smooth and has complex dimension four, agreeing with $\mathrm{Gr}(2,4)$.

The Klein quadric is the complexified compactified Minkowski space of the twistor correspondence. A volume form on $\mathbb{T}$ turns $P\wedge P^{\prime }$ into a scalar symmetric bilinear form $B(P,P^{\prime })$. It is zero exactly when the two underlying two-planes have nontrivial intersection: the four spanning vectors are then linearly dependent. In projective twistor space this says that the two twistor lines meet. In the affine graph chart supplied by the twistor incidence relation, it is $\det (x-y)=0$, so it is precisely null separation.

This also determines a conformal geometry on $Q$. At $[P]$, the tangent space is $P^{\perp }/⟨P⟩$ for $B$. The restriction of $B$ descends to a nondegenerate quadratic form on this four-dimensional quotient. Rescaling the representative $P$ changes the identification by a scalar, so its invariant content is a conformal class. In a graph chart its null tangent vectors are the rank-one matrices in $\mathrm{Hom}(W,\mathbb{T}/W)$, equivalently $d{x}^{A{A}^{\prime }}=a^A{b}^{A^{\prime }}$.

The compact complex geometry does not yet specify a real spacetime. The Lorentzian slice comes from the twistor Hermitian form $h$ of signature $(2,2)$: real points are maximal totally isotropic two-planes. In the incidence chart, isotropy of $W_x=\{(ix\pi ,\pi )\}$ is equivalent to $x=x^{†}$, the usual real Minkowski condition. Its twistor lines lie wholly in projective null twistor space. Diagonalizing $h$ as positive and negative two-dimensional blocks also gives a useful global description: an isotropic two-plane projects isomorphically to the positive block, so it is a graph $(z,Uz)$, and isotropy is $U^{†}U=1$. The real compact space is therefore $U(2)$, equivalently $(S^1\times S^3)/{\mathbb{Z}}_2$.

The same result is especially transparent in coordinates of the Plücker embedding after a complex linear change of coordinates on ${\Lambda }^2\mathbb{T}$. The real Lorentzian cone has equationIts nonzero real generators represent points. Normalize the lengths of its positive and negative parts to one; real projectivization identifies their simultaneous antipodes. ConsequentlyThis is the Projective Minkowski compactification. Its universal cover is $\mathbb{R}\times S^3$, with deck generator $(s,v)\mapsto (s+\pi ,-v)$, so its fundamental group is $\mathbb{Z}$. The projective version includes periodic identifications; an unwrapped Einstein-cylinder or Penrose diagram description uses the appropriate cover and affine domain instead.

For example, use homogeneous cone coordinates $[U,V,t,\mathbf{x}]$ with quadratic equation $2UV+t^2-|\mathbf{x}{|}^2=0$. The affine patch $U\ne 0$ sets $U=1$ and then $V=-(t^2-|\mathbf{x}{|}^2)/2$. The complement $U=0$ is the null cone $t^2-|\mathbf{x}{|}^2=0$ at infinity. It is much larger than a single point: different null directions retain different limits there.

For the Euclidean slice, choose an antilinear quaternionic structure on a complex vector space $J$ on twistor space, for exampleThe Euclidean reality structure on twistor space selects $J$-invariant complex two-planes. Such a plane is a quaternionic line in ${\mathbb{H}}^2$, givingThere are no fixed projective twistors for $J$: if $JZ=cZ$, applying $J$ again would give $-Z=|c{|}^{2}Z$. Every projective twistor instead lies on the unique invariant line $\mathbb{P}\mathrm{span}(Z,JZ)$. Thus the Euclidean correspondence becomes a fibration ${\mathbb{CP}}^3\to S^4$ with fiber ${\mathbb{CP}}^1$, in contrast to the Lorentzian correspondence where only null twistors represent real null geodesics. Distinct Euclidean real points have disjoint twistor lines, since the line through any $Z$ and $JZ$ is unique. This reflects the absence of nonzero real null vectors for a positive-definite metric.

On the Klein quadric the Euclidean real cone has signature $(1,5)$; normalize its timelike coordinate to one and obtain the unit sphere $S^4$. Its affine Euclidean chart omits one point. Inverse stereographic projection explicitly gives the round conformal metricwhich extends smoothly across that point. This proves the one-point conformal compactification of Euclidean space. The distinction between the Lorentzian Hermitian and Euclidean quaternionic reality structures is also described in Adamo's lectures on twistor theory; the global statements here follow from the explicit models just derived.

Finally, Wick rotation $t\mapsto it$ compares affine real slices through their complexification. Locally it changes $d{t}^2-|d\mathbf{x}{|}^2$ to $-d{\tau }^2-|d\mathbf{x}{|}^2$, the negative of the Euclidean metric; the overall sign does not change its null cone in the complexified setting. It does not identify the two real conformal compactifications globally. The Lorentzian projective space has fundamental group $\mathbb{Z}$, while $S^4$ is simply connected. Its causal universal cover $\mathbb{R}\times S^3$ is noncompact, so passing to that cover does not repair the mismatch with the compact Euclidean sphere.

There is also a direct algebraic obstruction. In the homogeneous coordinates above the affine Wick substitution extends only as the rational mapSubstitution verifies that the image satisfies the same complex quadric equation. For $U=0$ and $t\ne 0$ it collapses the point to $[0,1,0,0]$, while at some infinity points every output coordinate vanishes. In particular it is indeterminate at $[0,1,0,0]$. This is a genuine failure of holomorphic extension: the affine paths $(t,\mathbf{x})=(0,(s,0,0))$ and $(s,(is,0,0))$, as $s\to \infty$, both tend to that input point. Their Wick images tend respectively to $[0,1,0,0]$ and $[0,0,1,(1,0,0)]$, which are distinct points of the complex quadric.