L’espace en creux

Absences, fenêtres, empreintesAbsences, windows, imprints

An absence is a theorem asserting that a collection $\mathcal K$ of structures, given by explicit axioms, has no member with a stated property. It is graded when it comes with a function $\nu\colon\mathcal K\to\R$ (a dimension, a characteristic, a prime, a ratio of orders) and determines the set $W=\nu(\mathcal K)$ of values that occur, its window: where the excluded thing can still happen. A witness is a finite configuration on which the absence can already be checked. For Hurwitz’s theorem on composition algebras, $\nu$ is the dimension and $W=\{1,2,4,8\}$, and a witness that the next step fails is a pair of nonzero sedenions with product zero. For Galois’s theorem, $\mathcal K$ is the pairs $(p,H)$ with $p\ge5$ prime and $H$ of index $p$ in $\PSL(2,p)$, $\nu(p,H)=p$, and $W=\{5,7,11\}$.

The word is meant literally. A seal is cut in negative and leaves a positive figure in the wax: the absence is the cut, the imprint is the figure. The definition classifies pairs of an absence and a theorem, not structures; each entry of the catalogue states the characterizing theorem, and ‘forces’ means only that the theorem uses the absence essentially, not that the imprint is built out of it.

Separating imprints call for one more word. A bridge is refuted when it is proved that its two sides are not incarnations of one object, that no seam exists for the markings in question: a type recurrence proved to be no identification. The points and the lines of the Fano plane give one, which an outer automorphism undoes; the equal orders of $\PSL(3,4)$ and $A_8$ give one that no marking can undo.

La fenêtre de GaloisGalois’s window

In his letter to Auguste Chevalier of 29 May 1832, Galois stated, with an outline for the cases that occur, that $\PSL(2,p)$ admits no action on $p$ points for $p>11$, while for $p=5,7,11$ it does. If the small primes are admitted the window is $\{2,3,5,7,11\}$, since $\PSL(2,2)\cong S_3$ acts on two points and $\PSL(2,3)\cong A_4$ on three. Each exceptional action carries a geometry, and together they are the terminal imprint.

For $p=5$, the action of $\PSL(2,5)$ on the five cosets of $A_4$ and that of $\PSL(2,4)$ on the five points of $\Proj^1(\F_4)$ both have image $A_5$, so $\PSL(2,4)\cong\PSL(2,5)\cong A_5$. For $p=7$ the action builds the Fano plane. For $p=11$ the orbit of size 11 of $\PSL(2,11)$ on five-element subsets is a biplane, a 2-$(11,5,2)$ design whose automorphism group is the image of $\PSL(2,11)$, of order 660. This last survivor has no second life among the linear groups: no other $\PSL(n,q)$ and no alternating group has order 660.

The point stabilizers $A_4$, $S_4$, $A_5$ are the rotation groups of the tetrahedron, the octahedron and the icosahedron, the finite $(2,3,n)$ triangle groups for $n=3,4,5$, and the condition $(p^2-1)/2\in\{12,24,60\}$ says exactly that $p^2-1$ is the order 24, 48 or 120 of one of their double covers in $\SU(2)$: the exceptional actions are cut out by the regular polyhedra. For $p=7$ and $p=11$ the two classes of subgroups of index $p$ are exchanged by conjugation by an element of $\PGL(2,p)$ outside $\PSL(2,p)$, so no rule invariant under $\Aut\PSL(2,7)$ says which seven-point set is the points of the Fano plane and which the lines. This is the source of the marking dependence of Chapter 1.

Le spineur qui ne descend pasThe spinor that does not descend

Klein’s list: every finite subgroup of $\SL(2,\C)$ is conjugate into $\SU(2)$, and is cyclic, binary dihedral, or one of the binary polyhedral groups $2T$, $2O$, $2I$, of orders 24, 48 and 120, the preimages of $A_4$, $S_4$ and $A_5$; so $2I$ is the only non-solvable one. For $p\ge5$ the group $\SL(2,p)$ is perfect, so a nontrivial two-dimensional representation has determinant one and image $2I$, of order $p(p^2-1)$ or $p(p^2-1)/2$; only $p=5$ solves this. The spinor window is $\{3,5\}$, with $\SL(2,3)\cong2T$ and $\SL(2,5)\cong2I$, and in particular $\SL(2,7)$ has no nontrivial two-dimensional representation.

Locally the spinors are there. A transitive action of $\SL(2,p)$ on $p$ points factors through $\PSL(2,p)$, and its point stabilizer is $2T$, $2O$ or $2I$ for $p=5,7,11$: at each point the stabilizer is a group of spinors acting on $\C^2$. An equivariant bundle over $G/H$ is the same thing as a representation of $H$, and it is equivariantly trivial exactly when that representation is restricted from $G$. So the spin bundle $S_\rho=\SL(2,p)\times_{H_p}\rho$, for a spin representation $\rho$ of the stabilizer, is forced to be nontrivial for $p=7$ and 11. For $2O$ there are two spin representations, $\rho_\pm$, exchanged by $\sqrt2\mapsto-\sqrt2$.

The same counting governs reality. $\PSL(2,7)$ has no faithful real representation of dimension less than 6: its representations 3 and $\bar3$ have character field $\Q(\sqrt{-7})$ and indicator 0, so the realification of 3 is irreducible of dimension 6 with commutant $\C$, a forced complex structure, and $\PSL(2,7)$ is not a group of rotations of $\R^3$. The spin bundle carries a forced quaternionic structure: its sections form $\Ham^7\cong\Ham^3\oplus\Ham^4$, and $6_\pm$ embed $\SL(2,7)$ in the compact symplectic group $\mathrm{Sp}(3)$.

Coïncidences entre groupes simplesCoincidences among the simple groups

Artin determined when two of the simple groups $\PSL(n,q)$, $(n,q)\neq(2,2),(2,3)$, and $A_m$, $m\ge5$, have the same order: only at 60 ($\PSL(2,4)$, $\PSL(2,5)$, $A_5$), 168 ($\PSL(2,7)$, $\PSL(3,2)$), 360 ($\PSL(2,9)$, $A_6$) and 20160 ($\PSL(3,4)$, $\PSL(4,2)$, $A_8$). The groups of each order are isomorphic, with the single exception that $\PSL(3,4)$ is not isomorphic to $\PSL(4,2)\cong A_8$. By computation these are the only coincidences of order among these groups below $10^{40}$.

Two parts are proved directly. The plane and the line meet once: $\PSL(3,q)\cong\PSL(2,r)$ with $r\ge4$ only for $(q,r)=(2,7)$. Every Sylow subgroup of $\PSL(2,r)$ is abelian, except that for odd $r$ the Sylow 2-subgroups are dihedral; the unitriangular group $U_3(q)$ is a nonabelian Sylow subgroup of $\PSL(3,q)$ with centre of order $q$, so it would have to be dihedral, forcing characteristic 2 and then $q=2$, since a dihedral group of order at least 8 has centre of order 2; and $r(r^2-1)=336$ gives $r=7$.

The coincidence at 20160 is not an isomorphism, and element orders show it at once. Inside the Mathieu group $M_{24}$ the two groups are stabilizers: $A_8$ of a pair (octad, point off it), $\PSL(3,4)$ of an ordered triple of points, and both sets have $759\cdot16=12144=24\cdot23\cdot22$ elements. As a bridge between these two objects of $M_{24}$ the coincidence has status type, one group and one size, and it is refuted for every choice of markings, because the stabilizers are not even isomorphic as abstract groups: by the stabilizer principle an object is determined by its stabilizer class, and equal size is not enough.

L’axiome de FanoFano’s axiom

A complete quadrangle in a projective plane is four points, no three collinear; its six sides meet in pairs of opposite sides at its three diagonal points. Fano’s axiom asserts that the diagonal points of every complete quadrangle are not collinear, and over a division ring $D$ it holds exactly when $\operatorname{char}D\neq2$: normalized to $[e_1],[e_2],[e_3],[e_1+e_2+e_3]$, the quadrangle has diagonal points $[e_1+e_2]$, $[e_1+e_3]$, $[e_2+e_3]$, and a relation among them forces $2a=0$. By computation the diagonal points are collinear for all 7 and 2520 quadrangles of $\mathrm{PG}(2,2)$ and $\mathrm{PG}(2,4)$, and for none of the 234 and 15500 of $\mathrm{PG}(2,3)$ and $\mathrm{PG}(2,5)$.

So the Fano plane embeds in $\mathrm{PG}(2,D)$ exactly when $\operatorname{char}D=2$, and in neither the real nor the complex projective plane. Over the reals there is a stronger absence: by the Sylvester–Gallai theorem every finite set of points not all on one line has a line through exactly two of them, while every line of the Fano plane has three. Among finite planes the failure is rigid: a finite projective plane in which the diagonal points of every quadrangle are collinear is $\mathrm{PG}(2,2^k)$ for some $k$ (Gleason).

In $\mathrm{PG}(2,2)$ the window has its smallest survivor. Read on the complete graph $K_4$ whose vertices are the four points of a quadrangle: the sides are the six edges and the diagonal points are the three perfect matchings, forced onto one line in characteristic 2; $K_4$ with that line is the whole Fano plane. Dually, for a point $c$ the four lines missing $c$ form a quadrilateral whose three diagonal lines, the lines through $c$, are forced to be concurrent, at $c$.

La borne de HurwitzThe Hurwitz bound

Let a finite group $G$ act on a compact Riemann surface $X$ of genus $g$, with quotient of genus $h$ branched with indices $m_1,\dots,m_r$. The Riemann–Hurwitz formula reads $2g-2=|G|\,\mu$, with $\mu(h;m_1,\dots,m_r)=2h-2+\sum(1-1/m_i)$, and the sign of $\mu$ sorts the signatures into three kinds. On the sphere $\mu=-2/|G|<0$, and the signatures with three branch points are $(2,2,n)$, $(2,3,3)$, $(2,3,4)$, $(2,3,5)$, with $2/(-\mu)=2n,12,24,60$: the arithmetic behind Klein’s list. On a torus $\mu=0$. In genus $g\ge2$, $\mu>0$, and the least positive value, $1/42$, gives Hurwitz’s bound $|G|=(2g-2)/\mu\le84(g-1)$.

Equality holds exactly when $X$ is the quotient of the hyperbolic plane by a torsion-free normal subgroup of finite index in the triangle group $\Delta(2,3,7)=\langle x,y\mid x^2=y^3=(xy)^7=1\rangle$. Such a Hurwitz group is perfect, since in the abelianization $x^2=y^3=1$ and $x^7y^7=1$ give $xy=1$, so it is not solvable. A group of order 84 has a normal Sylow 7-subgroup and is solvable, so a Hurwitz group has order at least 168, and no surface of genus 2 attains the bound.

$\Delta(2,3,7)$ has exactly one normal subgroup with quotient $\PSL(2,7)$: there are 336 generating pairs of elements of orders 2 and 3 with product of order 7, and $\Aut\PSL(2,7)$, of order 336, acts freely and transitively on them. So there is exactly one surface of genus 3 with 168 automorphisms, the Klein quartic $x^3y+y^3z+z^3x=0$, and it is the only compact Riemann surface of genus at most 3 on which $\PSL(2,7)$ acts faithfully. One function on either side of zero gives both absences: the regular polyhedra are the spherical $(2,3,n)$, $n\le5$, the Euclidean wall is $n=6$, and the Klein quartic is the first hyperbolic case, $n=7$.

La composition s’arrête à huit, la géométrie au planComposition stops at eight, geometry at the plane

A composition algebra over $\R$ is a real algebra with identity and a positive definite form $N$ with $N(xy)=N(x)N(y)$; it has no zero divisors. The Cayley–Dickson double of an algebra with involution is $A\oplus A$ with $(a,b)(c,d)=(ac-\bar db,\ da+b\bar c)$, $\overline{(a,b)}=(\bar a,-b)$ and $N(a,b)=N(a)+N(b)$, and from $\R$ it gives $\C$, $\Ham$, $\Oct$ and the sedenions $\mathbb S$, of dimensions 1,2,4,8,16. By the doubling lemma the double of a composition algebra is a composition algebra exactly when the algebra is associative, is associative exactly when the algebra is commutative and associative, and is commutative exactly when the involution is trivial. So each doubling loses a property, as the lemma predicts, and the last loss has a witness: in the convention above, $(e_1+e_{10})(e_4-e_{15})=0$.

The same window recurs. The associative real division algebras of finite dimension are $\R$, $\C$, $\Ham$ (Frobenius); every finite-dimensional real division algebra, normed or not, has dimension 1, 2, 4 or 8 (Bott and Milnor, Kervaire), which also follows from Adams’s theorem that maps $S^{2n-1}\to S^n$ of Hopf invariant one exist only for $n=2,4,8$.

The octonions coordinatize a projective plane but no projective space of higher dimension, and the stop is forced twice. Synthetically: in dimension at least 3 Desargues’s theorem holds and the coordinates form an associative division ring (Veblen and Young), while a plane is Moufang exactly when it is coordinatized by an alternative division ring (Moufang; Bruck and Kleinfeld; Kleinfeld). Algebraically: the Hermitian matrices $\mathfrak h_n(\Oct)$ with $x\circ y=\tfrac12(xy+yx)$ form a Jordan algebra exactly when $n\le3$ (Jordan, von Neumann and Wigner; Albert), and the failure has a small witness in $\mathfrak h_4(\Oct)$, using three octonion units that do not associate. The 27-dimensional $\mathfrak h_3(\Oct)$ has automorphism group $F_4$, of dimension 52, and its primitive idempotents form the Cayley plane $F_4/\Spin(9)$, of dimension 16. Along the chain, associativity allows projective spaces of every dimension, alternativity without associativity exactly the plane, and the sedenions, which are not alternative, no plane at all.

La dimension trois dans l’espace de HilbertDimension three in Hilbert space

In a real or complex Hilbert space of finite dimension $d\ge2$, a context is an orthonormal basis, a frame function is a nonnegative function on rays summing to 1 on every context, and a valuation is a frame function with values in $\{0,1\}$, choosing exactly one ray from each context. For $d\ge3$ every frame function is $e\mapsto\langle e,\rho e\rangle$ for a density operator $\rho$ (Gleason), and there is no valuation (Kochen and Specker). The second follows from the first, dimension by dimension: a valuation would be a frame function, hence continuous on the connected unit sphere, with the two values 0 and 1. In dimension three a witness is Peres’s 33 rays, with 72 orthogonal pairs and 16 orthogonal triples, on which an exhaustive search finds no admissible assignment.

Both theorems fail for $d=2$, for one reason. On the qubit the contexts are the antipodal pairs $\{n,-n\}$ of the Bloch sphere, pairwise disjoint, so $f(n)=(1+n_z^3)/2$ is a frame function not of trace form, and choosing in each pair the point whose first nonzero coordinate among $(n_z,n_y,n_x)$ is positive is a valuation. The gap closes when projections are replaced by effects: for every $d\ge2$ a generalized probability measure on effects is $E\mapsto\operatorname{tr}(\rho E)$ for a unique density operator (Busch).

In dimension four the absence has a witness that can be checked by hand (Cabello, Estebaranz and García-Alcaine): 18 rays of $\R^4$ with coordinates in $\{0,\pm1\}$, forming 9 contexts, every ray in exactly two of them. A valuation would put one 1 in each of the 9 contexts, an odd total, while counting each ray twice, an even total. A search over all $2^{18}$ assignments confirms that none is a valuation.

Comment les absences dépendent les unes des autresHow the absences depend on one another