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On the six-dimensional exterior square of ${\mathbb{F}}_2^4$, the Pfaffian $Q=x_{12}{x}_{34}+x_{13}{x}_{24}+x_{14}{x}_{23}$ is a nonsingular plus-type quadratic form. The action of $G{L}_4(2)$ preserves it and is faithful: fixing every decomposable exterior line fixes every two-space and then every one-space. The even-weight binary subset module on eight letters, modulo the all-one vector, gives another six-dimensional plus-type space with $Q=\mathrm{wt}(x)/2\mathrm{mod}2$. Coordinate permutations embed $S_8$ faithfully into its orthogonal group. That orthogonal group has order $40320$, while $G{L}_4(2)$ has order $20160$, identifying its index-two image with the alternating group $A_8$.

The isomorphic pair is $A_8\cong PS{L}_4(2)$; $PS{L}_3(4)$ is different. All three have order $20160$, so order alone does not settle the question; explicitly $|PS{L}_3(4)|=(64-1)(64-4)(64-16)/(3\cdot 3)=20160$. We give a concrete realization and an element-order distinction.

On $W={\bigwedge }^2{\mathbb{F}}_2^4$, the Pfaffian isIt is a nonsingular plus-type quadratic form: the three displayed pairs give a hyperbolic decomposition. Change of basis multiplies the Pfaffian by the determinant, which is always one in $G{L}_4(2)$. Hence the exterior-square realization of PSL4 over F2 embeds $G{L}_4(2)$ into $O_6^{+}(2)$. The kernel of a group homomorphism is trivial: an element acting trivially on the exterior square fixes every decomposable two-space, then every one-space as an intersection of two-spaces, and hence is a scalar; over ${\mathbb{F}}_2$ this scalar is one.

For a second model let $E$ be the even-weight vector subspace of ${\mathbb{F}}_2^8$ and quotient by the all-one vector. The dot product descends to a nondegenerate alternating bilinear form on the six-dimensional quotient, andis well defined, since complementing an even subset changes half its weight from $w/2$ to $4-w/2$. It has $(1+70+1)/2=36$ zero vectors, so is plus type. Coordinate permutations give a faithful $S_8$ action: fixing the quotient fixes each weight-two subset, because its complement has weight six, and fixing all two-subsets forces the identity permutation.

From the independently derived symplectic order and quadratic orbit count below,Thus $O_6^{+}(2)\cong S_8$ and the image of $G{L}_4(2)=S{L}_4(2)=PS{L}_4(2)$ is its index-two subgroup $A_8$. The index-two subgroup is unique: a nontrivial group homomorphism $S_8\to C_2$ sends every transposition to the same nonidentity element and is the sign map.

To separate $PS{L}_3(4)$, observe that $A_8$ contains $(12345)(678)$ of order 15. An odd-order projective element of $PS{L}_3(4)$ lifts to a semisimple linear operator, since its order upstairs divides three times an odd order in characteristic two. The possible eigenspace degree patterns are $1+1+1$, $2+1$ and $3$. In the split case all eigenvalues have order dividing three, so the projective element has no order 15. In the $2+1$ case the eigenvalues are $\lambda ,{\lambda }^4,{\lambda }^{-5}$ with $\lambda \in {\mathbb{F}}_{16}^{\times }$; its fifth power is scalar because ${\lambda }^{15}=1$, so its projective order divides five. In the irreducible case the determinant-one torus has order $4^2+4+1=21$ and quotienting its scalar subgroup of order three gives projective order dividing seven. There is therefore no element of order 15 in $PS{L}_3(4)$.

Several small-parameter coincidences connect the finite classical groups to permutation groups. The projective-line action immediately gives $PS{L}_2(2)\cong S_3$, $PS{L}_2(3)\cong A_4$, $PG{L}_2(3)\cong S_4$ and $PS{L}_2(4)\cong A_5$ by faithful actions of degrees three, four and five and their orders. Also $PS{L}_2(5)\cong A_5$: its involutions have Klein-four centralizers, giving five Sylow subgroups; their conjugation action is faithful and embeds its order-60 simple group into $S_5$. Simplicity here can be checked from class sizes $1,15,20,12,12$: no proper nontrivial normal class union has order dividing 60. The same five-subgroup action extends to $PG{L}_2(5)$, with trivial kernel of a group homomorphism and order 120, giving $S_5$. Indeed its kernel intersects $PS{L}_2(5)$ trivially, so centralizes that normal subgroup. Commuting projectively with both upper and lower elementary unipotents forces a representing matrix to be scalar, so this kernel is trivial.

The less immediate coincidence $PS{L}_2(9)\cong A_6$ can be exhibited on the ten unordered triple partitions used in Question 4. Let $a=(123)$, $b=(456)$, $w=(12)(36)$, assign $\infty$ to $123\mid 456$ and $0$ to $126\mid 345$, and label the other partitions by applying $a^x{b}^y$ to the zero partition, with $t=x+iy\in {\mathbb{F}}_9$ and $i^2=-1$. Inspection of these nine partitions givesTranslations and inversion generate $PS{L}_2(9)$: conjugating the upper unitriangular matrices by inversion gives the lower ones, and elementary elimination generates $S{L}_2(9)$. Their projective action has order $9(9^2-1)/2=360$. These partition permutations come from even permutations of six letters, whose partition action is faithful as proved below; hence they fill $A_6$.

Finally, on the even-weight module of ${\mathbb{F}}_2^6$ modulo its all-one vector, coordinate permutations preserve a nondegenerate alternating bilinear form in dimension four and act faithfully, again by the weight-two-subset argument. Both $S_6$ and $S{p}_4(2)$ have order 720, proving $S{p}_4(2)\cong S_6$, with commutator subgroup $A_6$. These examples illustrate both the need to quotient group centers and the low-rank exceptions to general simplicity. They do not make equal orders sufficient for group isomorphism, as the two order-20160 examples already demonstrate.

A Berezin integral over independent Grassmann variables selects the coefficient containing every variable. Fix the real orientation by $\int d{\omega }_1\cdots d{\omega }_{2m}{\omega }_{2m}\cdots {\omega }_1=1$. For an antisymmetric matrix $A$, only the order-$m$ term in the exponential survives:Anticommutation collects its top coefficient into the signed sum over pairings that defines the Pfaffian. Reversing the order of $2m$ variables gives $(-1{)}^{m(2m-1)}=(-1{)}^m$, cancelling the prefactor sign. The real Grassmann Gaussian integral is thereforewith this orientation, and a fixed sign times this answer with other orientations. In dimension two, write $A_{12}=a$; the exponential is $1-a{\omega }_1{\omega }_2=1+a{\omega }_2{\omega }_1$, and integration gives $a$. Although $\mathrm{Pf}(A{)}^2=\det A$, writing a positive square root $\sqrt{\det A}$ would give $|a|$ in this test and would lose the correct sign.

For the complex Grassmann Gaussian integral, treat ${\theta }_a$ and ${\bar{\theta }}_a$ as independent integration variables. Every surviving top term contains each $\theta$ and each $\bar{\theta }$ once. Expanding the exponential assigns one matrix entry from every row and column, and anticommutation supplies the permutation sign. Hencewhere $C_n$ is a nonzero constant fixed by the order and normalization of the differentials, independent of $M$. This argument works for singular matrices as well and does not assume that $M$ is Hermitian.

For the explicit $n=2$ check, put $Q={\bar{\theta }}_a{M}_{ab}{\theta }_b$. Moving all the $\theta$ variables before the barred ones givesThe linear and constant terms integrate to zero. Thus the integral is a fixed orientation sign times $M_{11}{M}_{22}-M_{12}{M}_{21}$, exactly the determinant. Choosing the top-monomial integral to be $-1$ gives $C_2=1$.

For the required real-complex compatibility of Grassmann Gaussian integrals, let $u_a={\omega }_a$ and $v_a={\omega }_{n+a}$. Substitution givesIndeed $v^{T}Mu=-u^T{M}^{T}v=u^{T}Mv$ because the variables anticommute and $M^T=-M$. The real quadratic matrix is consequently $A=\mathrm{diag}(M,M)$, of size $2n$. The linear change of Grassmann variables has a constant nonzero Berezin integration Jacobian, so it changes only the allowed normalization factor. For even $n$, the two blocks giveFor odd $n$, $\det M=\det (-M)=(-1{)}^n\det M$ implies $\det M=0$. Each of the separate $u$ and $v$ quadratic exponentials has only even degree and cannot saturate an odd number of variables, so the real integral vanishes too. This covers the odd case without attempting to define a Pfaffian of an odd-dimensional matrix.

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.

For an even-dimensional antisymmetric matrix $M$, put $\theta =(u+iv)/\sqrt{2}$ and $\bar{\theta }=(u-iv)/\sqrt{2}$, with independent Grassmann variables $u,v$. Anticommutation cancels the mixed terms, giving ${\bar{\theta }}^{T}M\theta =(u^{T}Mu+v^{T}Mv)/2$. The constant Grassmann change-of-variables formula changes only the normalization. The two real Pfaffians multiply to $\mathrm{Pf}(M{)}^2=\det M$, agreeing with the complex Grassmann Gaussian integral. In odd dimension $\det M=0$ and both top-degree integrals vanish.

For an antisymmetric matrix $A$, a Berezin integral extracts the top coefficient of the exponential. Expansion into pairings gives the Pfaffian, with a constant sign depending on the order of the differentials. This is a polynomial in the entries of $A$, not the positive square root of its determinant: in dimension two it is proportional to $A_{12}$ and changes sign with that entry. The identity $\mathrm{Pf}(A{)}^2=\det A$ determines its square but does not discard its orientation.