Troisième partie · En montant la tourChapitre 14

Les continus

Continua

Read from the draft of 2 October 2026

Plate 14.1Thurston’s congruence link complement at the complex place: an ideal edge and an ideal tetrahedron over the Eisenstein lattice, the eight cusps as the points of

\Proj^1(\F_7)

, and the cusp torus.

14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9

How do the finite objects reappear in real and complex geometry, and which kind of reappearance can each have?

The completions placed the finite geometries at the primes, as links of vertices of buildings. This chapter, on the sixth floor of the tower, finds them in real and complex geometry. There are two ways a finite object can appear in a continuous one: as a configuration of points held in place by a finite group of symmetries, or as a set of classes of an arithmetic configuration modulo a congruence subgroup. The archimedean place, where the congruence that gives a residue field at a prime gives Thurston’s link complement over $\C$ , is of the second kind.

Which kind an object can have is decided by absences: the projective line over $\F_7$ has no configuration in the projective line over $\C$ or in Klein’s plane, only the arithmetic continuum. The chapter then follows the continua that carry the program’s other finite structures: $\Proj^1(\C)$ as the celestial sphere and the Bloch sphere at once, four points of it at the vertices of a regular tetrahedron, and a point of the Cayley plane, with one imaginary unit, in the plane and in its complexification.

The central result · The line life has only an arithmetic continuum in low dimension

The object $\Proj^1(\F_7)$ of $\PSL(2,7)$ has no embedded continuum in $\Proj^1(\C)$ under $\PSL(2,\C)$ , and none in Klein’s plane $\Proj^2(\C)$ under Klein’s representation. It has the arithmetic continuum of the link complement, with $Y=\Proj^1(\C)$ , $L=\PSL(2,\C)$ , $\Lambda=\PSL(2,\mathcal O)$ , $\Lambda'=\Gamma(\mathfrak p)$ and $Z=\Proj^1(K)$ .

Proof

A finite subgroup of $\PSL(2,\C)$ is the image of its preimage in $\SL(2,\C)$ , so by Klein’s classification it is cyclic, dihedral, $A_4$ , $S_4$ or $A_5$ ; none is $\PSL(2,7)$ . In Klein’s plane an orbit of size 8 would have stabilizers of order 21, which form one conjugacy class, the normalizers of the Sylow 7-subgroups. One of them is generated by $g=\mathrm{diag}(\zeta_7^4,\zeta_7^2,\zeta_7)$ , $\zeta_7=e^{2\pi i/7}$ , and the cyclic permutation $h$ of the coordinates, with $hgh^{-1}=g^4$ . The fixed points of $g$ are the three coordinate points, since its eigenvalues are distinct, and $h$ permutes them cyclically. So no point is fixed by the subgroup of order 21.

Status

The definition is the book’s own. The proposition on the line life and the remark that follows it are proved by hand, from Klein’s classification and the absence of faithful real representations below dimension 6, and checked by computation. The identification of the celestial and the Bloch sphere is classical (Penrose and Rindler); the tetrahedron’s three readings were checked in floating point, and their being incarnations of one rigid object of $A_4$ is proved.

The intersection inside $F_4$ is a theorem of Todorov and Dubois-Violette, checked here on Lie algebras in one realization; the Peirce statements, the two complex structures, the two stabilizers of imaginary units and the statements in $\mathfrak e_6$ were computed in floating point, with the gaps between zero and nonzero singular values reported. Their group forms are Krasnov’s, Yokota’s and Boyle’s. Read as gauge algebras, with the sixteen as one family, they belong to the volume, in Chapter XVII.

Deux sortes de continuTwo kinds of continuum

Plate 14.1Thurston’s congruence link complement at the complex place: an ideal edge and an ideal tetrahedron over the Eisenstein lattice, the eight cusps as the points of

\Proj^1(\F_7)

, and the cusp torus.

The word records that the finite object reappears inside a continuous geometry, and the definition says how: an object of a finite group sits in a homogeneous space of a Lie group either as an equivariant configuration of points, or as the classes of an invariant set of an arithmetic group modulo a normal subgroup with quotient the finite group.

The link complement is the example. Let $\zeta=(1+\sqrt{-3})/2$ , $\mathcal O=\Z[\zeta]$ and $\mathfrak p=(2+\zeta)$ , of norm 7. Then $\mathcal O/\mathfrak p\cong\F_7$ and $\PSL(2,\mathcal O)/\Gamma(\mathfrak p)\cong\PSL(2,7)$ ; $\Gamma(\mathfrak p)$ is torsion-free, so $M=\Gamma(\mathfrak p)\backslash\mathbb H^3$ is a hyperbolic manifold of finite volume with deck group $\PSL(2,7)$ ; its eight cusps correspond equivariantly to the points of $\Proj^1(\F_7)$ ; and it is the complement of an eight-component link, tessellated by 28 regular ideal tetrahedra. Its edges are the object of size 28, its faces that of size 56, and its tetrahedra the two objects of size 14.

Definition(Continuum)

Let $X$ be an object of a finite group $G$ . A continuum of $X$ is a homogeneous space $Y$ of a Lie group $L$ together with one of the following. (E) An embedding: an injective homomorphism $\mu\colon G\to L$ and a $G$ -equivariant injective map $X\to Y$ , where $G$ acts on $Y$ through $\mu$ . (A) An arithmetic realization: a discrete subgroup $\Lambda\le L$ , a normal subgroup $\Lambda'$ of $\Lambda$ with $\Lambda/\Lambda'\cong G$ , and a $\Lambda$ -invariant subset $Z\subseteq Y$ such that $\Lambda'\backslash Z\cong X$ as $G$ -sets.

Pas de continu plongéNo embedded continuum

Plate 14.2Klein’s plane: the element

g=\mathrm{diag}(\zeta^4,\zeta^2,\zeta)

of order 7 fixes exactly the three coordinate points, and

h

, with

hgh^{-1}=g^4

, carries them round in a cycle, so the subgroup of order 21 fixes no point and

\Proj^1(\F_7)

has no orbit there.

The line life has no configuration in low dimension. In $\Proj^1(\C)$ the obstruction is Klein’s list: the finite groups of Möbius maps are cyclic, dihedral, $A_4$ , $S_4$ and $A_5$ , and the group of order 168 is none of them. In Klein’s plane the group does act, by its three-dimensional representation, but an orbit of eight points would need a point fixed by a subgroup of order 21, and the element of order seven in it fixes only the three coordinate points, which the element of order three permutes in a cycle.

So the eight points of $\Proj^1(\F_7)$ reappear in a continuum only as classes: the cusps of the link complement, eight classes of points of $\Proj^1(\Q(\sqrt{-3}))$ modulo $\Gamma(\mathfrak p)$ . The same absence keeps the group out of four real dimensions altogether.

Remark(No action on four-dimensional space)

The group of order 168 has no nontrivial homomorphism into $\GL(4,\R)$ : it is simple, so a nontrivial homomorphism is injective, and it has no faithful real representation of dimension less than 6. So it is a subgroup neither of the Lorentz group of $\R^{1,3}$ nor of $\PSL(2,\C)$ . It can meet $\PSL(2,\C)$ only as a quotient of a subgroup, as it does for the arithmetic subgroup $\PSL(2,\mathcal O)$ .

La sphère de Bloch et la sphère célesteThe Bloch sphere and the celestial sphere

Plate 14.3A great circle of

\Proj^1(\C)

, the sphere of null rays: the boost

\mathrm{diag}(e^{s/2},e^{-s/2})

s=1.3

, fixes the null direction of

(1,0)

and carries that of

(1,1)

from

90^\circ

30.5^\circ

from it, a Möbius map that is not an isometry.

Identify Minkowski space $\R^{1,3}$ with the Hermitian $2\times2$ matrices, $(t,x,y,z)\leftrightarrow X=\left(\begin{smallmatrix}t+z&x-iy\\x+iy&t-z\end{smallmatrix}\right)$ , so that $\det X=t^2-x^2-y^2-z^2$ and $A\in\SL(2,\C)$ acts by $X\mapsto AXA^\dagger$ : the double cover of the identity component of the Lorentz group.

The sphere at infinity of $\mathbb H^3=\SL(2,\C)/\mathrm{SU}(2)$ , the space of future unit timelike vectors, is this same $\Proj^1(\C)$ . So the eight cusps of the link complement are eight classes of points of the celestial sphere. Rotations act on it as on the Bloch sphere; boosts act by Möbius maps that are not isometries: the boost $\mathrm{diag}(e^{s/2},e^{-s/2})$ with $s=1.3$ changes the angle between two null directions from $90^\circ$ to $30.5^\circ$ .

Theorem(One projective line, two spheres) proved

The map $[\psi]\mapsto$ (the ray of $\psi\psi^\dagger$ ) is an $\SL(2,\C)$ -equivariant bijection from $\Proj(\C^2)=\Proj^1(\C)$ onto the set of future null rays, the celestial sphere. For $|\psi|=1$ , $\psi\psi^\dagger=\tfrac12(I+n\cdot\sigma)$ , where $n=(x,y,z)/t$ is the Bloch vector of the pure state $\psi$ . So the celestial sphere and the Bloch sphere are the same sphere $S^2$ . Rotations ( $\mathrm{SU}(2)$ ) act on it as on the Bloch sphere, and boosts act by Möbius maps that are not isometries.

Proof

$\psi\psi^\dagger$ has rank one, hence determinant zero, and trace $|\psi|^2>0$ ; every nonzero null Hermitian matrix of positive trace is of this form. Equivariance: $(A\psi)(A\psi)^\dagger=A\psi\psi^\dagger A^\dagger$ . The formula for $\psi\psi^\dagger$ is the definition of the Bloch vector, and $[\psi_0:\psi_1]\mapsto\psi_1/\psi_0$ is the stereographic coordinate of $n$ , on which $A$ acts by a Möbius map. The rest was checked on random samples.

Quatre effets et quatre directions isotropesFour effects and four null directions

Plate 14.4The four vertices of

K_4

named by the four points

\infty,0,1,2

\Proj^1(\F_3)

: with the four equianharmonic points of

\Proj^1(\C)

they are incarnations of one rigid object of

A_4

, joined by exactly one seam.

Let $r_1,\dots,r_4$ be the vertices of a regular tetrahedron inscribed in the Bloch sphere, and $E_k=\tfrac14(I+r_k\cdot\sigma)$ . These four effects form the qubit’s symmetric informationally complete measurement. Under the identification of the two spheres they become four null vectors $n_k=(1,r_k)$ , linearly independent, with $\sum_kn_k=(4,0,0,0)$ ; and as four points of $\Proj^1(\C)$ they are equianharmonic, every cross-ratio being $e^{\pm i\pi/3}$ .

The three readings are incarnations of one object. The four points of $\Proj^1(\C)$ under the Möbius maps that permute them, the four points of $\Proj^1(\F_3)$ under $\PSL(2,3)\cong A_4$ , and the four vertices of $K_4$ under its rotation group $A_4$ each have a stabilizer of order 3, and $A_4$ has one class of such subgroups, each self-normalizing. So they are incarnations of one rigid object of $A_4$ , and between any two of them there is exactly one seam, whatever the markings, since that class is fixed by every automorphism of $A_4$ .

Proposition(The tetrahedron three times) computed

(1) $\sum_kE_k=I$ , $\operatorname{tr}(E_jE_k)=\tfrac1{12}$ for $j\ne k$ and $\tfrac14$ for $j=k$ , and each $2E_k$ is a pure state. (2) The four effects become the four null vectors $n_k=(1,r_k)$ ; they are linearly independent, $\sum_kn_k=(4,0,0,0)$ , and their Minkowski products are $n_j\cdot n_k=\tfrac43$ for $j\ne k$ and 0 for $j=k$ . (3) As four points of $\Proj^1(\C)$ they are equianharmonic: every cross-ratio is $e^{\pm i\pi/3}$ , and the permutations of them induced by Möbius maps form the alternating group $A_4$ . (4) The four points, the four points of $\Proj^1(\F_3)$ under $\PSL(2,3)\cong A_4$ , and the four vertices of $K_4$ under its rotation group $A_4$ are incarnations of one object of $A_4$ , which is rigid; so between any two of them, once the groups are marked, there is exactly one seam.

Proof

(1)–(3) by direct computation. The Möbius maps inducing a permutation of the points are those preserving the cross-ratio, by sharp 3-transitivity. (4) In each case the stabilizer of a point has order 3; $A_4$ has one class of subgroups of order 3, its four Sylow 3-subgroups, each self-normalizing, so by the stabilizer principle the three sets are incarnations of one rigid object. The class is the only one of its order, so it is invariant under every automorphism of $A_4$ , and the conclusion does not depend on the markings.

Le plan de Cayley et deux sous-groupes maximauxThe Cayley plane and two maximal subgroups

Plate 14.5The derivations of the Albert algebra,

\mathfrak f_4

of dimension 52, and two of its maximal subalgebras:

\mathfrak{spin}(9)

, killing a point

E_{11}

of the Cayley plane, and

\mathfrak s=\mathfrak{su}(3)\oplus\mathfrak{su}(3)

, keeping

\mathfrak h_3(\C)

; they meet in an algebra of dimension 12 that contains the ideal

\mathfrak c

The exceptional group $F_4$ is the automorphism group of the Albert algebra $\mathfrak h_3(\Oct)$ , and $\mathrm{Spin}(9)$ is the stabilizer of a primitive idempotent, a point of the Cayley plane. By the classification of Borel and de Siebenthal, $F_4$ also has a maximal subgroup of full rank, $(\mathrm{SU}(3)\times\mathrm{SU}(3))/\Z_3$ . Todorov and Dubois-Violette found the intersection of the two.

The Lie algebra statement can be checked in one realization. Choose an imaginary unit $e_1$ and let $\C=\R1+\R e_1$ ; then $\mathfrak h_3(\C)$ is a nine-dimensional subalgebra of $\mathfrak h_3(\Oct)$ . The derivations preserving it form $\mathfrak s$ , of dimension 16, equal to its derived algebra; those vanishing on it form an ideal $\mathfrak c$ of dimension 8, and $\mathfrak s/\mathfrak c$ restricts faithfully onto $\mathrm{Der}\,\mathfrak h_3(\C)$ , so $\mathfrak s\cong\mathfrak{su}(3)\oplus\mathfrak{su}(3)$ . Its intersection with $\mathfrak{spin}(9)$ , the derivations killing $E_{11}$ , has dimension 12, a centre of dimension 1 and a derived algebra of dimension 11, and contains $\mathfrak c$ : it is $\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ .

Theorem(Todorov–Dubois-Violette; Todorov–Drenska)

Inside $F_4$ ,

\begin{aligned}\mathrm{Spin}(9)\cap(\mathrm{SU}(3)\times\mathrm{SU}(3))/\Z_3&=S(\mathrm U(2)\times\mathrm U(3))\\&=(\mathrm{SU}(2)\times\mathrm{SU}(3)\times\mathrm U(1))/\Z_6.\end{aligned}

Proof

Cited. The Lie algebra form was checked by linear algebra on the 52-dimensional space of derivations. The identifications use that $\mathfrak f_4$ is compact, so its subalgebras are reductive, and a compact Lie algebra equal to its derived algebra is semisimple; the only compact semisimple Lie algebra of dimension 8 is $\mathfrak{su}(3)$ , and the only one of dimension 11 is $\mathfrak{su}(3)\oplus\mathfrak{su}(2)$ .

Deux structures complexes en un pointTwo complex structures at a point

Plate 14.6At the point

p=E_{11}

: the commutant of

J_{\mathrm{row}}

\mathfrak{spin}(9)

is the intersection of dimension 12, the commutant of

J_{\mathrm{col}}

\mathfrak{su}(2)\oplus\mathfrak{su}(4)

of dimension 18, and commuting with

\omega

cuts the second down to the first.

One point of the Cayley plane already carries the intersection. Let $p=E_{11}$ and let $V=\{X\in\mathfrak h_3(\Oct):p\circ X=\tfrac12X\}$ , the matrices whose only nonzero entries are $x_{12}$ , $x_{13}$ and their mirror images, identified with $\Oct^2$ ; the matrices with $p\circ X=0$ form $\mathfrak h_2(\Oct)$ , whose traceless part is $\R^9$ . These are the Peirce spaces of $p$ . The unit $e_1$ gives two complex structures on $V$ , $J_{\mathrm{row}}(x_{12},x_{13})=(e_1x_{12},e_1x_{13})$ and $J_{\mathrm{col}}(x_{12},x_{13})=(x_{12}e_1,x_{13}e_1)$ ; on the entries of the first column $J_{\mathrm{row}}$ is a right multiplication, so the words left and right depend on the coordinates.

By the first part, $V$ is the module $\mathfrak f_4/\mathfrak{spin}(9)$ , the spinor representation of $\mathfrak{spin}(9)$ in $\mathfrak f_4=\mathfrak{spin}(9)\oplus S$ ; Krasnov gives the group versions, $(\mathrm{SU}(3)\times\mathrm{SU}(2)\times\mathrm U(1))/\Z_6$ for the commutant of a right multiplication on both summands and $(\mathrm{SU}(2)\times\mathrm{SU}(4))/\Z_2$ for a left one. Let $\omega$ fix $\C$ and act on its complement, a complex 3-space, as the scalar $e^{2\pi e_1/3}$ . It generates the centre of the $\mathrm{SU}(3)$ of automorphisms of $\Oct$ fixing $e_1$ , the derivations commuting with it form $\mathfrak s$ , and so the intersection is cut from one point and one element of order three:

\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)=\bigl(\mathfrak{su}(4)\oplus\mathfrak{su}(2)\bigr)\cap\mathfrak f_4^{\omega}.

Proposition(Two complex structures at a point) computed

(1) The algebra $\mathfrak{spin}(9)$ preserves $V$ , and $D\mapsto D(p)$ induces an equivariant isomorphism from $\mathfrak f_4/\mathfrak{spin}(9)$ onto $V$ . $\mathfrak{spin}(9)$ acts faithfully on $\R^9$ and preserves the trace form there. (2) The commutant of $J_{\mathrm{row}}$ in $\mathfrak{spin}(9)$ is $\mathfrak{spin}(9)\cap\mathfrak s\cong\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ , of dimension 12. (3) The commutant of $J_{\mathrm{col}}$ in $\mathfrak{spin}(9)$ has dimension 18. It is the stabilizer of the splitting $\R^9=\R^3\oplus\R^6$ , with $\R^3$ the traceless part of $\mathfrak h_2(\C)$ and $\R^6$ the entries $x_{23}\in\C^\perp$ , so it is $\mathfrak{so}(3)\oplus\mathfrak{so}(6)\cong\mathfrak{su}(2)\oplus\mathfrak{su}(4)$ . It contains the algebra of (2), and the ideal $\mathfrak c$ lies in its summand $\mathfrak{so}(6)$ .

Proof

(1) A derivation satisfies $D(p)=D(p\circ p)=2\,p\circ D(p)$ , so $D(p)\in V$ ; the map is linear with kernel $\mathfrak{spin}(9)$ , and $[A,D](p)=A(D(p))$ for $A\in\mathfrak{spin}(9)$ . Both sides have dimension $52-36=16$ . The rest is direct computation; in (3) the commutant preserves $\R^3$ , $\R^6$ and the trace form, so it maps into $\mathfrak{so}(3)\oplus\mathfrak{so}(6)$ , injectively because $\R^9$ is faithful, and both have dimension 18.

Deux stabilisateurs d’unités imaginairesTwo stabilizers of imaginary units

Plate 14.7The table’s units in the cyclic labelling: the 21 maps

x\mapsto ax+b

a\in\{1,2,4\}

, are automorphisms with no signs, all fixing

u=\sum_xe_x

; the three that also fix one unit

e_c

are the maps

x\mapsto a(x-c)+c

The centre of the $\mathrm{SU}(3)$ fixing $e_1$ was cut out by one imaginary unit. Another unit gives another $\mathrm{SU}(3)$ , conjugate to the first, and the two need not be joined by anything the second sees. In the octonions of the table $e_xe_{x+1}=e_{x+3}$ let $u=\sum_xe_x$ , so that $u^2=-7$ , let $v=u/\sqrt7$ , and let $c\in\Z/7$ .

In the vocabulary of the first part, $v$ and $e_c$ are two points of one object of $G_2$ , the sphere $S^6$ , and seams of $G_2$ -sets join them. For the stabilizer $K$ of $v$ they are not joined: $K$ fixes $v$ , while its orbit through $e_c$ has dimension $8-3=5$ . So the bridge between the two units is built for $G_2$ and refuted for $K$ : the status of a bridge depends on the group in which seams are sought. For the points and lines of the Fano plane a twist from outside the group joins what the group keeps apart; here the larger group joins what the smaller one keeps apart, and no new name is needed.

Proposition(Two stabilizers of imaginary units) computed

(1) The derivations of $\Oct$ killing $v$ are those commuting with $L_v$ ; they act on the complement of $\langle1,v\rangle$ , a complex 3-space under $L_v$ , as $\mathfrak{su}(3)$ . The same holds for $e_c$ . (2) The two copies of $\mathfrak{su}(3)$ meet in the derivations killing both units, a copy of $\mathfrak{su}(2)$ , and together they generate $\mathfrak g_2$ . (3) The automorphism group $G_2$ is transitive on the unit imaginary octonions, so the stabilizers of $v$ and of $e_c$ are conjugate in $G_2$ ; no element of the stabilizer $K$ of $v$ carries $e_c$ to $\pm v$ . (4) The 21 permutations $e_x\mapsto e_{ax+b}$ , $a\in\{1,2,4\}$ , are automorphisms of $\Oct$ fixing $u$ , and three of them fix $e_c$ .

Proof

(1), (2) and (4) by direct computation. (3) The orbit of $v$ under $G_2$ has dimension $14-8=6$ , so it is open in the sphere $S^6$ of unit imaginary octonions; it is compact, hence closed, and $S^6$ is connected. The elements of $K$ preserve $\langle e_c,v\rangle=1/\sqrt7$ , which is not $\pm1$ .

Un point du plan complexifiéA point of the complexified plane

Plate 14.8The complexified point:

\mathfrak e_6=\mathfrak f_4\oplus iL(J_0)

, of dimension 78, with

\mathfrak{so}(10)

fixing

E_{11}

and

\C^{27}=1\oplus16\oplus10

under it; commuting with

\omega

gives

\mathfrak{su}(3)^{\oplus3}

, and the two commutants in

\mathfrak{so}(10)

have dimensions 15 and 21.

The Cayley plane has a complex sibling with a larger group. Let $\mathfrak h_3(\Oct)^{\C}=\mathfrak h_3(\Oct)\otimes\C$ , with the Hermitian form $\langle X,Y\rangle=\operatorname{tr}(\bar X\circ Y)$ and the cubic norm $N$ , the determinant. Yokota defines the compact group $E_6$ as the $\C$ -linear maps preserving $N$ and $\langle\,,\rangle$ , with Lie algebra $\mathfrak e_6=\mathfrak f_4\oplus i\,L(J_0)$ , where $L_X$ is Jordan multiplication by $X$ and $J_0$ the elements of trace zero. The stabilizer of $E_{11}$ in $E_6$ is $\mathrm{Spin}(10)$ , and Baez’s bioctonionic plane $E_6/((\mathrm{Spin}(10)\times\mathrm U(1))/\Z_4)$ , of real dimension 32, is this complexified plane, its tangent space at a point the $\mathbf{16}$ .

The element $\omega$ acts entrywise. The elements of $\mathfrak e_6$ commuting with it form $\mathfrak e_6^\omega\cong\mathfrak{su}(3)^{\oplus3}$ , of dimension 24; in $\mathfrak{so}(10)$ the commutant of $J_{\mathrm{row}}$ is $\mathfrak k=\mathfrak{so}(10)\cap\mathfrak e_6^\omega\cong\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ , of dimension 15, and that of $J_{\mathrm{col}}$ is the stabilizer of a splitting $W=\R^4\oplus\R^6$ , $\mathfrak{so}(4)\oplus\mathfrak{so}(6)$ , of dimension 21. Their intersections with $\mathfrak f_4$ are the two commutants at the real point, and $\mathfrak k\cap\mathfrak f_4=\mathfrak{spin}(9)\cap\mathfrak f_4^\omega$ , whose $\mathfrak{su}(2)$ is diagonal in the two ideals $\mathfrak{su}(2)$ of $\mathfrak k$ . In groups, $G_{15}=\mathrm C_{\mathrm{Spin}(10)}(J_{\mathrm{col}})\cap E_6^\omega$ , Boyle’s.

Proposition(A point of the complexified plane) computed

(1) The $52+26$ operators of $\mathfrak f_4\oplus iL(J_0)$ span a real Lie algebra of dimension 78, skew-Hermitian for $\langle\,,\rangle$ , preserving $N$ , and acting irreducibly on $\C^{27}$ . (2) The stabilizer of $E_{11}$ in $\mathfrak e_6$ has dimension 45; it preserves the real space $W$ of dimension 10 spanned by $i(E_{22}+E_{33})$ , $E_{22}-E_{33}$ and the real entries $x_{23}$ , and acts on it as all of $\mathfrak{so}(W)$ , so it is $\mathfrak{so}(10)$ . The stabilizer of the line $\C E_{11}$ is $\mathfrak{so}(10)\oplus\R\,iL_{X_0}$ , $X_0=2E_{11}-E_{22}-E_{33}$ . (3) Under $\mathfrak{so}(10)$ , $\C^{27}=\C E_{11}\oplus(V\otimes\C)\oplus(\mathfrak h_2(\Oct)\otimes\C)$ , with irreducible summands of dimensions 1, 16 and 10.

Proof

By direct computation, as for $\mathfrak f_4$ ; the norm is checked through its derivative along random complex points, and irreducibility through commutants of complex dimension 1.

Deux copies, une seule autoconjuguéeTwo copies, only one self-conjugate

Plate 14.9The sixteen eigenvalues of the second copy’s centre on

V\otimes\C

y

up to a common factor:

\tfrac16

six times,

\tfrac13

three times,

-\tfrac23

three times,

-\tfrac12

twice, 1 and 0. Reflected through zero they do not return to themselves, so the representation is not self-conjugate.

The difference between $\mathfrak f_4$ and $\mathfrak e_6$ is seen on the $\mathbf{16}$ , through its complex conjugate. A complex representation is self-conjugate when it is isomorphic to its complex conjugate. The copy of $\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ inside $\mathfrak f_4$ acts on $V\otimes\C$ as the complexification of its action on $V$ , so that representation is self-conjugate.

A second copy lives in $\mathfrak e_6$ outside $\mathfrak f_4$ . The vectors of $V\otimes\C$ killed by $\mathfrak c$ form a complex space of dimension 4; for each of the two ideals $\mathfrak{su}(2)$ of $\mathfrak k$ , those killed by $\mathfrak c$ and by it form a space of dimension 2 with no nonzero real vector, and the stabilizer in $\mathfrak k$ of such a vector is a subalgebra $\mathfrak c\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ of dimension 12, not contained in $\mathfrak f_4$ . Its centre acts on $V\otimes\C$ with eigenvalues $iy$ , and the multiset of the $y$ is not symmetric under $y\mapsto-y$ ; since conjugation replaces $iy$ by $-iy$ , this representation is not self-conjugate.

Proposition(Two copies of

\mathfrak{su}(3)\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)

) computed

(1) The algebra $\mathfrak{spin}(9)\cap\mathfrak f_4^\omega$ acts on $V\otimes\C$ as the complexification of its action on $V$ , so this representation is self-conjugate. (2) The vectors of $V\otimes\C$ killed by $\mathfrak c$ form a complex space of dimension 4. For each of the two ideals $\mathfrak{su}(2)$ of $\mathfrak k$ , the vectors killed by $\mathfrak c$ and by it form a space of dimension 2 containing no nonzero real vector. (3) The stabilizer in $\mathfrak k$ of such a vector is a subalgebra $\mathfrak c\oplus\mathfrak{su}(2)\oplus\mathfrak u(1)$ of dimension 12, not contained in $\mathfrak f_4$ . Its centre acts on $V\otimes\C$ with eigenvalues $iy$ , where the sixteen numbers $y$ are proportional to $\tfrac16$ (six times), $\tfrac13$ (three times), $-\tfrac23$ (three times), $-\tfrac12$ (twice), 1 and 0. They are not symmetric under $y\mapsto-y$ , so this representation is not self-conjugate.

Proof

(1) The complexification of a real representation is self-conjugate. (2) and (3) by direct computation. The multiset of eigenvalues of the centre is an invariant of the representation, and conjugation replaces $iy$ by $-iy$ .

Floor six is the continuum completing the finite geometries: at the complex place the link complement carries the line life arithmetically, and no configuration can; one sphere is both the celestial and the Bloch sphere; and one point of the Cayley plane with one imaginary unit carries the intersection of two maximal subgroups of $F_4$ , whose second copy in $\mathfrak e_6$ is not self-conjugate.

The next chapter assembles the tower: the theorem that joins each floor to the one below it, and seven objects followed from floor to floor.

Introduced here: continuum
Also in this chapter: objectincarnationseamrigid objectbridgestatusrefutedabsencecompletion
Objects: the skythe twenty-eightthe object of size 56the object of size 14, class athe object of size 14, class b
In the volume: IThe Founding SentenceXThe Finite Celestial SphereXIIIThe Level-Seven ShadowXVIIOne Point of the Cayley Plane