Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 59 (Introduction to Twistor Theory)

Use the Levi-Civita connection interpretation of the curvature convention. Its Riemann curvature tensor satisfiesThe first antisymmetry comes from the derivative commutator; the second comes from compatibility with the metric tensor. The cyclic identity is the first Bianchi identity, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric Ricci tensor, $\boxed{R_{ab}=R_{ba}}$. The differential Bianchi identity is ${\nabla }_{[a}{R}_{bc]de}=0$.

There is a hypothesis qualification. A metric connection need not be a torsion-free connection. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}{\nabla }_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with torsion tensor.

For example, with a constant Minkowski metric a connection having only ${\Gamma }^1{}_{02}=1$ and ${\Gamma }^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[{\nabla }_0,{\nabla }_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.

Fix the two-component spinor calculus convention ${ϵ}_{01}={ϵ}^{01}=1$, with lowering ${\xi }_A={\xi }^B{ϵ}_{BA}$ and raising ${\xi }^A={ϵ}^{AB}{\xi }_B$. Thus ${ϵ}^{AB}{ϵ}_{AB}=2$; use the same convention for primed indices. The spinor covariant derivative preserves these symplectic forms.

Write $C_{A{A}^{\prime }B{B}^{\prime }}=[{\nabla }_{A{A}^{\prime }},{\nabla }_{B{B}^{\prime }}]$. It is antisymmetric on exchanging the two spacetime index pairs. The tensor-product decomposition of a differential two-form isThis can also be checked directly: symmetrize or antisymmetrize separately in $A,B$ and $A^{\prime },B^{\prime }$. Antisymmetry under simultaneous pair exchange leaves precisely the symmetric–antisymmetric and antisymmetric–symmetric terms. An antisymmetric two-index spinor in a two-dimensional space is a multiple of its symplectic form. Therefore the chiral spinor curvature operators areExchanging $A,B$ in the first expression and relabeling $A^{\prime },B^{\prime }$ shows ${\Delta }_{AB}={\Delta }_{BA}$; the other symmetry follows identically. The projections reconstruct the whole six-component commutator, provingExpanding each projected commutator and relabeling the contracted indices gives the requested derivative expressions:These formulas act tensorially on the field being differentiated, including the covariant derivative's action on the intermediate derivative index. The stated convention removes any ambiguity about signs from contracting alternating spinor forms in the opposite order.

For a torsion-free connection, the covariant Hessian of a scalar is symmetric, so $[{\nabla }_a,{\nabla }_b]f=0$. Moreover, a commutator of two covariant derivatives acts as a derivation on tensor products. Applying the Leibniz rule twice, the terms containing one derivative on each factor cancel between the two orders:Projecting with the parallel symplectic form preserves this identity for the chiral spinor curvature operator. Since its scalar projection obeys ${\Delta }_{AB}f=0$, we obtainThus the chiral spinor curvature operator is algebraic over scalar functions even though its displayed expression uses second derivatives. If torsion tensor is allowed, $[{\nabla }_a,{\nabla }_b]f=-T^c{}_{ab}{\partial }_{c}f$ need not vanish, explaining why metric compatibility alone would not prove the claim.

First use a genuine simple bivector of the specified chirality, with spinor part ${\varphi }^{AB}={\alpha }^A{\alpha }^B$. It is a exterior product: choose independent primed spinors ${\rho }^{A^{\prime }},{\sigma }^{A^{\prime }}$ and take $u^{A{A}^{\prime }}={\alpha }^A{\rho }^{A^{\prime }}$, $v^{B{B}^{\prime }}={\alpha }^B{\sigma }^{B^{\prime }}$. Their antisymmetric product is proportional to $w^{A{A}^{\prime }B{B}^{\prime }}={\alpha }^A{\alpha }^B{ϵ}^{A^{\prime }{B}^{\prime }}$.

For any antisymmetric $w^{ab}$, the Riemann curvature tensor acts on both tensor indices. In the printed curvature convention its complete contraction isbecause the Ricci tensor is symmetric. Substituting the one-chirality bivector and the chiral spinor curvature operator decomposition, the primed-operator term vanishes on contraction with the symmetric ${\alpha }^A{\alpha }^B$. The other term is $2{\Delta }_{AB}({\alpha }^A{\alpha }^B)$. Hence ${\Delta }_{AB}({\alpha }^A{\alpha }^B)=0$.

Repeat with $\alpha +\beta$ and subtract the identities for $\alpha$ and $\beta$. Linearity leaves twice the mixed symmetric product, giving the contracted chiral curvature identityThe polarization works for spinor fields as well as pointwise spinors: the chiral spinor curvature operator is algebraic by the preceding part.

The displayed source expression ${\alpha }_A{\beta }_B{ϵ}_{A^{\prime }{B}^{\prime }}$ needs this clarification. For independent $\alpha ,\beta$ it is not antisymmetric in the spacetime indices. Its antisymmetric part is ${\alpha }_{(A}{\beta }_{B)}{ϵ}_{A^{\prime }{B}^{\prime }}$, which is generally not simple. The proof above uses a genuinely simple bivector first and then polarization, so it establishes the requested conclusion without relying on the incorrect simplicity assertion.

The chiral spinor curvature operator acts algebraically on a two-dimensional spinor space. We may therefore lower its matrix index using the symplectic form and write ${\Delta }_{AB}{\alpha }_C=X_{ABDC}{\alpha }^D$. Symmetry of ${\Delta }_{AB}$ gives symmetry in the first pair of $X$. Preservation of ${ϵ}_{CD}$ says that its action is symplectic; lowering the endomorphism index of a symplectic infinitesimal transformation makes a symmetric two-index tensor. Thus $X_{ABCD}=X_{(AB)(CD)}$.

The additional Riemann curvature tensor symmetry $R_{abcd}=R_{cdab}$ implies $X_{ABCD}=X_{CDAB}$: project both antisymmetric spacetime index pairs onto the unprimed chiral summand. Equivalently this is the symmetry of the unprimed diagonal block of the curvature operator on differential two-forms. The contracted chiral curvature identity is another way of expressing the same three constraints on the initially nine-component pair-symmetric spinor.

Consequently $X$ is a symmetric bilinear form on the three-dimensional space ${\mathrm{Sym}}^{2}S$, with six independent components. Its total symmetrization ${\Psi }_{ABCD}=X_{(ABCD)}$ has five components. The one-dimensional remainder is spanned byIt is symmetric within and between pairs, and $K_{(ABCD)}=0$. The latter follows from the two-dimensional alternating-form identity, or directly by evaluating its components with ${ϵ}_{01}=1$. Hence the chiral Riemann curvature spinor decomposition isA contraction of a totally symmetric spinor with an symplectic form vanishes. Meanwhile the explicit two-term expression for $K$ gives $K_{ABCD}{ϵ}^{AC}{ϵ}^{BD}=3$. Therefore the scalar in the printed normalization isIn particular, $\Psi =X-\Lambda K$ has zero such trace and is the Weyl curvature spinor. The coefficient is fixed by the factor implicit in the printed symmetrization; substituting a differently normalized scalar-curvature term would change the relation between its scalar parameter and this contraction.

In the compact twistor correspondence, a point $p$ of complexified compactified Minkowski space is a two-dimensional subspace $W_p\subset \mathbb{T}$. The correspondence fiber consists of flags $(\ell ,W_p)$ with $\ell$ a one-dimensional subspace of $W_p$. Its image in projective twistor space is precisely $\mathbb{P}(W_p)$, soThis is the twistor line of $p$. In the affine chart the twistor incidence relation gives $W_p=\{(i{p}^{A{A}^{\prime }}{\pi }_{A^{\prime }},{\pi }_{A^{\prime }}):\pi \in {\mathbb{C}}^2\}$. The linear map $\pi \mapsto (ip\pi ,\pi )$ is injective, and projectivizing it gives the same Complex projective space ${\mathbb{CP}}^1$. The subspace argument also covers points at conformal infinity, where that graph chart fails.

A point $q\in \mathbb{PT}$ specifies a one-dimensional subspace ${\ell }_q\subset \mathbb{T}$. Its correspondence fiber consists of two-planes $W$ containing ${\ell }_q$. Passing to the quotient gives a bijection, with inverse taking the inverse image of a line,Since ${\dim }_{\mathbb{C}}(\mathbb{T}/{\ell }_q)=3$, this is an isomorphism of Complex projective spaces. ConsequentlyIt is the compact alpha-plane associated with $q$. In an affine chart with $\pi \ne 0$, the solutions of the twistor incidence relation are $x^{A{A}^{\prime }}=x_0^{A{A}^{\prime }}+u^A{\pi }^{A^{\prime }}$, where $u^A$ supplies two complex coordinates. That affine ${\mathbb{C}}^2$ is only an open subset of the compact ${\mathbb{CP}}^2$; distinguishing them is essential to the printed compact correspondence.

An alpha-plane is a totally null complex two-plane with tangent vectors $u^A{\pi }^{A^{\prime }}$ for a fixed primed spinor. A beta plane has tangent vectors ${\lambda }^A{u}^{A^{\prime }}$ for a fixed unprimed spinor. The metric tensor pairs two vectors of either family to zero, since the repeated fixed spinor contracts with an symplectic form. These are the two chiral families of totally null planes; labels for their self-duality depend on the orientation convention.

Globally in the twistor correspondence, writewhere $\ell$ is a line and $H$ a three-dimensional hyperplane in twistor space. Both compact families are Complex projective spaces ${\mathbb{CP}}^2$.

For distinct ${\ell }_1,{\ell }_2$, any $W\in A_{{\ell }_1}\cap A_{{\ell }_2}$ must contain their two-dimensional span. Since $W$ itself has dimension two, it must equal that span. Conversely the span belongs to both planes. Thus two distinct compact alpha-planes meet in exactly one point:In affine coordinates, independent ${\pi }_1,{\pi }_2$ determine a unique matrix $x$ from the two incidence equations. If the primed spinors are proportional while the projective twistors are distinct, no finite affine solution exists. The compact intersection still exists, at conformal infinity; saying the planes are disjoint without this affine qualification would be wrong for the printed space $M$.

For the compact alpha-plane $A_{\ell }$ and beta plane $B_H$, an intersection point must satisfy $\ell \subset W\subset H$. If $\ell \not\subset H$, that is impossible. If $\ell \subset H$, intermediate two-planes correspond to lines $W/\ell$ in the two-dimensional quotient $H/\ell$. ThereforeThe nonempty intersection is a projective null geodesic. In an affine chart its tangent must be both $u^A{\pi }^{A^{\prime }}$ and ${\lambda }^A{v}^{A^{\prime }}$; these conditions force a multiple of ${\lambda }^A{\pi }^{A^{\prime }}$. Thus its affine portion is a null line, with the indicated projective compactification.

Use the signature-$(2,2)$ twistor Hermitian formA real Minkowski point $x$ is a Hermitian matrix $x^{A{A}^{\prime }}$. Every vector of its incidence subspace has the form $(ix\pi ,\pi )$. Substitution givesIf the two null geodesics meet at $x$, their null twistors $X,Y$ both belong to this totally isotropic subspace. In addition to their individual nullness, they must obeyFor independent $X,Y$, this says that every point of the projective line $\mathbb{P}\mathrm{span}(X,Y)$ is null: it lies entirely in projective null twistor space. It is the twistor line of their meeting point. In the finite affine setting, distinct null directions at the meeting point make their primed spinors independent; proportional primed spinors would give proportional incident twistors and the same null geodesic.

The meaningful twistor-geometric interpretation of “colinear in $PN$” is that the entire projective line through two distinct projective null twistors lies in projective null twistor space. Mere membership of two points in a projective line would impose no condition, because every pair of distinct points in ${\mathbb{CP}}^3$ spans one.

Indeed, individual nullness givesThis vanishes for every $a,b\in \mathbb{C}$ precisely when $h(X,Y)=0$. Thus $W=\mathrm{span}(X,Y)$ is a maximal totally isotropic two-plane, a real point of Minkowski conformal compactification under the Lorentzian reality structure on twistor space. The two corresponding alpha-planes meet at this unique $W$. Hence their compact intersection is a real spacetime point, and their real null geodesics meet there.

There are two qualifications. The point can be at conformal infinity, so a finite affine meeting is not automatic. If $X,Y$ represent the same projective point instead, their alpha-planes coincide rather than having a unique intersection. If “colinear” is read only as spanning a line in ${\mathbb{CP}}^3$, individual nullness alone does not force a real intersection; the additional orthogonality condition above is indispensable.

For a concrete counterexample to the weaker reading, take $X=(0,0,1,0)$ and $Y=(1,0,0,1)$ in $({\omega }^0,{\omega }^1,{\pi }_{0^{\prime }},{\pi }_{1^{\prime }})$ coordinates. Each is a null twistor, but $h(X,Y)=1$. Their two incidence equations give $x=-i\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, which is not Hermitian. Thus their alpha-planes meet at a complex point that is not a real Minkowski point.

A real spacetime point $p$ is represented by its twistor line $L_p\cong {\mathbb{CP}}^1$, lying entirely in projective null twistor space. Each point of $L_p$ is a null twistor specifying one real null geodesic through $p$. Their union is the null cone with vertex $p$. Thus the twistor description of the null cone is the family of null geodesics parametrized by the single line $L_p\subset PN$.

Equivalently, the spacetime points on that cone are precisely those whose twistor lines meet $L_p$:For distinct finite points this follows immediately from the incidence equations: a shared nonzero $\pi$ obeys $(q-p{)}^{A{A}^{\prime }}{\pi }_{A^{\prime }}=0$, which is equivalent to $\det (q-p)=0$, the null-separation condition. The flag description extends it to the compact spacetime. Projective null twistors encode whole unoriented null geodesics, so this includes both future and past generators; a time orientation and an affine patch distinguish the two halves. The line itself is the sphere of null directions at the vertex, not a three-dimensional image of the cone embedded pointwise into twistor space.

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.

Given two flags of the same dimensions, choose adapted ordered bases $u_1,\dots ,u_4$ and $v_1,\dots ,v_4$, so their first $d_i$ vectors span their respective $L_i$. The invertible map $A$ sending each $u_j$ to $v_j$ sends the first flag to the second. Let $\delta =\det A\ne 0$. Replace the last target basis vector by ${\delta }^{-1}{v}_4$; this does not alter any subspace in its flag. The corresponding map $g$ has determinant one and still sends every member of the first flag to the corresponding member of the second.

Thus the special linear group acts transitively on every fixed-type flag manifold:The adapted-basis construction is the needed extension of transitivity on nonzero vectors. For an arbitrary subgroup, vector transitivity alone would not justify transitivity on all flags; here the explicit determinant correction proves the stronger claim for the special linear group.

The special linear group acts on the twistor correspondence by $g\cdot (\ell ,W)=(g\ell ,gW)$. Since an invertible linear map preserves inclusion, this is again an incident flag. The two projections commute with that action:This is precisely equivariance: projecting a transformed flag gives the same result as transforming its projection. In particular $g$ sends the twistor line of a spacetime point to the line of the transformed point and sends the alpha-plane of a twistor to that of its transformed twistor. The incidence geometry is preserved, even when a transformation takes a point out of the chosen affine Minkowski chart.

Because $F_1$ is projective twistor space, the printed column represents the line $p=[e_1]$, not a distinguished nonzero vector. Its stabilizer subgroup consists of matrices whose first column is a nonzero multiple of $e_1$:The block form is necessary to preserve the line, and conversely every such determinant-one matrix preserves it. By transitivity, the orbit map $g\mapsto g[e_1]$ is onto $F_1$. Its two values agree precisely when $g^{-1}{g}^{\prime }\in H_p$, equivalently when $g{H}_p=g^{\prime }{H}_p$. This gives the homogeneous space identificationThe dimensions also agree: ${\dim }_{\mathbb{C}}SL(4)=15$, while the free $B$ and $b$ parameters give ${\dim }_{\mathbb{C}}{H}_p=9+3=12$; the quotient has dimension three.

This coset space is not a quotient group, because $H_p$ is not a normal subgroup. For example $h=I+E_{12}\in H_p$. Let $g$ have upper-left block $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and identity on the other two coordinates. Then $g\in SL(4,\mathbb{C})$ but $gh{g}^{-1}=I-E_{21}$ sends $e_1$ to $e_1-e_2$, so it is not in $H_p$. Thus multiplying coset representatives does not define a well-defined group operation.

If instead one required the literal vector $e_1$ to be fixed, one would have $a=1$ and $B\in SL(3,\mathbb{C})$; its quotient would be ${\mathbb{C}}^4\setminus \{0\}$, not $F_1$. The projective reading is required by the printed target quotient and the flag manifold definition.

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.