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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.

The Plücker embedding of $\mathrm{Gr}(2,{\mathbb{C}}^4)$ is a smooth projective quadric in ${\mathbb{CP}}^5$. An antisymmetric $4\times 4$ matrix is nonzero and decomposable precisely when its Pfaffian vanishes, giving the displayed equation. The wedge pairing of two points on this quadric vanishes exactly when the corresponding two-planes intersect. Thus the Klein quadric carries the complex conformal null-separation relation of complexified compactified Minkowski space.

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.

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.

The Plücker embedding maps a $k$-plane to the projective class of the exterior product of a basis. A change of basis multiplies this product by its determinant, so the projective class is well defined. The plane is recovered as the vectors whose exterior product with this element is zero. For two-planes in ${\mathbb{C}}^4$, its image in ${\mathbb{CP}}^5$ is the Klein quadric.