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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.

Thus the answer to the global compatibility question is no: ordinary affine Wick rotation is a local analytic continuation, not a globally regular invertible map of the conformally compactified real spacetimes. The global obstruction to Wick rotation of conformal compactifications does not prevent both real slices from being studied inside the same smooth complex Klein quadric.

A representation of the one-arrow quiver is an $n_2\times n_1$ matrix $F$. If $r=\mathrm{rank}F$, changes of basis put it in the form with an $r\times r$ identity block and all other entries zero. Since multiplication by invertible matrices preserves matrix rank,This describes the orbit for every $0\le r\le \min (n_1,n_2)$, including the zero orbit. Its dimension isIndeed, choosing its $r$-dimensional image contributes $r(n_2-r)$ parameters in a Grassmannian, and choosing a surjective map from $k^{n_1}$ onto that image contributes $r{n}_1$. This is a rank orbit of a matrix under left-right multiplication.

For the First Chern class identity, use finite-dimensional classifying maps. A complex line bundle on a compact Hausdorff space embeds in a finite-dimensional trivial vector bundle, so it is a pullback vector bundle of the tautological bundle on some Complex projective space. For the two complex line bundles, the universal tensor product of vector bundles is classified by the mapIts restriction to either factor pulls the tautological bundle back to the tautological bundle on that factor. The degree-two cohomology of the product is the direct sum of the degree-two cohomology of its factors, so these two restrictions determine its First Chern class. Pulling back to $X$ provesThis proves the integral identity, including possible torsion elements in the cohomology of $X$; an argument using only curvature forms would not detect those torsion elements.

The splitting principle for complex vector bundles says that for a vector bundle $E\to X$ there is an iterated projective bundle $p:F(E)\to X$ for which $p^{\ast }$ is injective on integral cohomology, and $p^{\ast }E$ is a direct sum of complex line bundles. The same assertion holds with rational coefficients, and a common such space can split any finite collection of vector bundles. The injectivity follows at each stage from the projective bundle formula for complex vector bundles; a Hermitian metric splits the resulting filtration into a direct sum.

Write $x_i=c_1(L_i)$ for the Chern roots of the pulled-back vector bundle. Define the degree-$2j$ component of its Chern character byThe Fundamental theorem of symmetric polynomials expresses each power-sum symmetric polynomial ${\sum }_i{x}_i^j$ uniquely as an integral polynomial in the elementary symmetric polynomials, which are the Chern classes of $E$. Thus the definition makes sense on $X$ itself, without choosing actual Chern roots there. On a compact Hausdorff space, each vector bundle is classified by a finite-dimensional Grassmannian; its high-degree Chern character components therefore vanish. Hence this defines a class in the direct sum $H^{2\ast }(X;\mathbb{Q})$ even if $X$ has no finite dimension. On more general spaces an infinite Chern character is instead interpreted in the product of the even-degree cohomology groups.

On a common splitting space, the Chern roots of a direct sum are the combined root lists, whereas the First Chern class identity makes the Chern roots of a tensor product of vector bundles the pairwise sums $x_i+y_j$. ConsequentlyThe injectivity in the splitting principle for complex vector bundles descends both identities to $X$. Additivity extends the Chern character to the Grothendieck group by $\mathrm{ch}([E]-[F])=\mathrm{ch}(E)-\mathrm{ch}(F)$. Multiplicativity and $\mathrm{ch}(1)=1$ then show that the Chern character is a unital ring homomorphism.

For Complex K-theory of complex projective space, let $\gamma$ be the tautological bundle, put $t=1-[\gamma ]$, and set $h=-c_1(\gamma )$. The cohomology ring of complex projective space is $\mathbb{Z}[h]/(h^{n+1})$. The CW filtration gives a cofibrationThe K-theory six-term exact sequence and Complex K-theory of a sphere give, inductively,Here $b_n$ is the image of a Bott element from $S^{2n}$, normalized so that $\mathrm{ch}(b_n)=h^n$. This normalization follows from the multiplicative form of Bott periodicity: the Chern character of the degree-two Bott element is an integral generator, and multiplication yields the top-degree generator on $S^{2n}$. In particular, the abelian group $K^0({\mathbb{CP}}^n)$ is a free abelian group of rank $n+1$.

Assume inductively that $1,t,\dots ,t^{n-1}$ form an integral basis on ${\mathbb{CP}}^{n-1}$. Their lifts, together with $b_n$, form an integral basis on ${\mathbb{CP}}^n$. Sincethe Chern characters of this basis are linearly independent over $\mathbb{Q}$: their first nonzero degrees are $0,2,\dots ,2n$. Thus the Chern character is injective on $K^0({\mathbb{CP}}^n)$. By induction $t^n$ restricts to zero on ${\mathbb{CP}}^{n-1}$, so $t^n=a{b}_n$ for an integer $a$. Comparing Chern characters gives $a=1$. Also $\mathrm{ch}(t^{n+1})=0$, so $t^{n+1}=0$. This proves the integral answer, rather than merely its rationalization:The induction begins with ${\mathbb{CP}}^0$ a point and $t=0$.

The points $(x,y)$ are ordered orthonormal two-frames, so $P$ is the Stiefel manifold $V_2({\mathbb{R}}^{n+1})$. Changing an orthonormal basis of the same plane gives the displayed free right $O(2)$-action. Since $O(2)$ is compact, the action is proper, and the quotient is the Grassmannian of unoriented two-planes. The Free proper Lie-group action theorem makesa principal $O(2)$-bundle.

Embed the space $\mathcal{B}$ of complete isotropic flags intoThe incidence conditions $F_i\subset F_{i+1}$ and isotropy equations $⟨F_i,F_i⟩=0$ are closed polynomial conditions. Since the Grassmannian is projective, $\mathcal{B}$ is a projective algebraic variety.

Every complete isotropic flag extends to a symplectic basis. A symplectic change of basis carries any such flag to any other, so $G={\mathrm{Sp}}_{2n}$ acts transitively. The stabilizer $B$ of the standard flag consists of the upper triangular symplectic matrices. It is closed, connected, and solvable. The Lie-Kolchin theorem shows that every connected solvable subgroup fixes a complete flag in $V$; preservation of the symplectic form makes the resulting flag isotropic after taking its first half. Such a subgroup is conjugate into $B$, so $B$ is maximal and hence a Borel subgroup.

Two rectangular matrices are related by invertible left and right multiplication exactly when their ranks agree. The rank-$r$ orbit in ${\mathrm{Mat}}_{b\times a}(k)$ has dimension $r(a+b-r)$: choose its image in a Grassmannian, then a surjective map to that image. These are the representation orbits of the one-arrow quiver.

The flag manifold of a line $\ell$ contained in a two-plane $W$ has projections $(\ell ,W)\mapsto W$ and $(\ell ,W)\mapsto \ell$. Their fibers are respectively ${\mathbb{CP}}^1$ and ${\mathbb{CP}}^2$. This is the flat twistor correspondence: complexified compactified Minkowski space is the Grassmannian of two-planes in twistor space, while projective twistor space is its projective space of lines.