L’horizon : dessins d’enfants

The horizon: children’s drawings

en chantierRead from the draft of 3 October 2026

Plate Ép.1The drawing of Serre’s triple on the eight points of

\Proj^1(\F_7)

, the cover

X_0(7)\to X(1)

of degree 8: black vertices over

j=0

, white over

j=1728

, the eight edges labelled by the points, and in gold the loop around the cusp

\infty

Ép.1
Ép.2
Ép.3
Ép.4
Ép.5
Ép.6
Ép.7

What would a seam theory of all curves at once look like, and is every coherent symmetry of that whole network arithmetic?

In January 1984 Grothendieck wrote a research programme, the Esquisse d’un programme, and its third and fourth sections propose a seam theory on a grand scale. A drawing on a surface, a dessin d’enfant, is the same object as a finite set with two permutations, as a finite cover of the projective line branched over three points, as a Riemann surface with a meromorphic function, and as an algebraic curve over $\overline\Q$ : by Belyi’s theorem every such curve arises. The absolute Galois group acts on all of these at once.

Klein’s quartic is the textbook case. Its quotient map to the $j$ -line is branched over three points with inertia of orders 2, 3 and 7, and its drawing is the Klein map. The fifteen objects of the group of order 168 are the fifteen quotients of that map, the modular curves of level 7. Galois conjugation acts on them as the outer automorphism acts on the seam table, by the same element that the type law sees in the residues, and it builds the bridge between the points and the lines of the Fano plane that no count can build.

The open question is whether the absolute Galois group equals the Grothendieck–Teichmüller group: whether every coherent symmetry of the whole network of drawings is arithmetic. In seam terms it asks whether every twisted seam of the network is a Galois one. At the level of one finite group the answer is visibly not yes: the finite version of the Grothendieck–Teichmüller group of $\PSL(2,7)$ moves drawings that the relevant Galois elements fix. That does not decide the question, which needs the second level of the tower. The reciprocity law of Chapter 16 is a small, local version of it.

The central result

For $n\ge5$ the outer automorphism group of the profinite fundamental group of $\mathcal M_{0,n}$ is $\widehat{\mathrm{GT}}\times S_n$ . So the twisted seams of the genus-zero network at level two and beyond, up to relabelling the marked points, are exactly the elements of the Grothendieck–Teichmüller group $\widehat{\mathrm{GT}}$ , and $\operatorname{Gal}(\overline\Q/\Q)$ is a closed subgroup of them. Is every twisted seam of the genus-zero network arithmetic? (The question of Drinfeld and Ihara.)

Status

Open, and classical as a question. The finite face is built in the draft: the fifteen objects of 168 as the quotient drawings of the Klein map, with passports, genera and modular names, computed and pairwise distinct; Galois acting on them through the outer automorphism, proved from rigidity, with fields of moduli $\Q$ and $\Q(\sqrt{-7})$ ; the two drawings of degree 7 written exactly over $\Q(\sqrt{-7})$ ; $\widehat{\mathrm{GT}}$ acting as Galois on rigid families, proved; the finite $\mathrm{GT}_1(\PSL(2,7))$ , of order $2^9$ , computed, and shown to move drawings that the commutator subgroup of the Galois group fixes; the collisions of the Klein drawing at 2, 3 and 7, computed, carrying the three types of the type law. The plates of this chapter recompute the drawings, all fifteen passports and the mirror between the two drawings of degree 7.

Still open: the equality of $\operatorname{Gal}(\overline\Q/\Q)$ with $\widehat{\mathrm{GT}}$ ; whether finite shadows that impose the pentagon relation separate the drawings that $\mathrm{GT}_1$ merges, which needs software not yet run; whether the audible twists of $A_6$ are ever Galois; and a projective formalism in which the octahedron over $\F_2$ is recovered. Poonen, Schaefer and Stoll’s solution of $x^2+y^3=z^7$ , by classifying the twisted forms of the Klein quartic, is the horizon’s worked example of twisted seams over the rationals; the draft does not yet treat it.

Trois pointsThree points

Plate Ép.1The drawing of Serre’s triple on the eight points of

\Proj^1(\F_7)

, the cover

X_0(7)\to X(1)

of degree 8: black vertices over

j=0

, white over

j=1728

, the eight edges labelled by the points, and in gold the loop around the cusp

\infty

A dessin of degree $n$ is a pair of permutations $(x,y)$ of an $n$ -element set generating a transitive group; with $z=(xy)^{-1}$ its passport is the triple of cycle types and its genus is given by Riemann–Hurwitz, $2g-2=n-c(x)-c(y)-c(z)$ . It is a finite transitive set for the free group on two letters, so in seam language it is an incarnation of an object of its monodromy group. Its other incarnations are covers, surfaces and curves, and each equivalence between them is a seam. For the small dessins below the seams are built: the permutations are computed, a rational function is written down, and the two are matched.

The group of order 168 supplies its dessins through Serre’s triple $x_0=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ , $y_0=\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$ , $z_0=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ , of classes $(2A,3A,7A)$ with $x_0y_0z_0=1$ , the inertia of $X(7)\to X(1)$ over $j=1728$ , 0, $\infty$ . Drawn with $j/1728$ as the Belyi map, black vertices over $j=0$ are the cycles of $y_0$ , white vertices over $j=1728$ the cycles of $x_0$ , and faces the cycles of $z_0$ . On the eight points of $\Proj^1(\F_7)$ the drawing has one face of degree 7 and one of degree 1, the cusp $\infty$ , and genus 0: it is $X_0(7)$ , with $j=(h^2+13h+49)(h^2+5h+1)^3/h$ .

Proposition(The incarnations of a dessin) classical

The following categories are equivalent: dessins; finite connected covers of $\Proj^1(\C)\setminus\{0,1,\infty\}$ ; compact connected Riemann surfaces with a meromorphic function unramified outside 0, 1, $\infty$ ; smooth projective curves over $\C$ with such a function; the same over $\overline\Q$ ; finite extensions of $\overline\Q(t)$ unramified outside $t=0,1,\infty$ . Belyi’s theorem adds that every curve over $\overline\Q$ occurs.

Proof

Covering-space theory, with the fundamental group of the thrice-punctured sphere free on loops around 0 and 1; Riemann’s existence theorem; the algebraicity of compact Riemann surfaces; the invariance of the étale fundamental group under extension of algebraically closed fields of characteristic 0; and the anti-equivalence between curves and function fields.

Le revêtement de type (2, 3, 7)The cover of type (2, 3, 7)

Plate Ép.2The faces of the Klein map are the cusps of

X(7)

, the 24 flexes of Klein’s quartic in eight triangles, each turned by

\tau

, a flex to the other flex on its tangent.

On the group itself the triple gives the regular dessin, the Klein map: 168 edges, 56 black vertices of degree 3, 84 white vertices of degree 2 and 24 faces of degree 7, so $2g-2=168-84-56-24=4$ and the genus is 3. By Hurwitz a surface of genus $g\ge2$ has at most $84(g-1)$ automorphisms, with equality exactly when the quotient has signature $(0;2,3,7)$ , and Klein’s quartic is the only surface of genus 3 with 168. Its 24 faces are the cusps of $X(7)$ ; under the unique equivariant isomorphism with the quartic they are the 24 flexes, and the tangent at the cusp $\infty$ meets the curve again at the cusp $2/7$ .

The same map comes out of Grothendieck’s polyhedra over finite fields. The regular polyhedron $\{3,7\}$ , written by universal formulae in $2\cos(\pi/3)$ and $2\cos(\pi/7)$ , specialized at its exceptional prime 7, where $2\cos(\pi/7)$ reduces to $-2$ , has rotation group $\PSL(2,7)$ : it is the Klein map. The 57 points of the plane over $\F_7$ fall into its orbits $8+21+28$ , with stabilizers $7{:}3$ , $D_8$ and $S_3$ : the sky, the centres of the involutions, and the twenty-eight. The icosahedron does the same at its exceptional primes 2 and 5, returning the two lives of $A_5$ .

Theorem(The modular coverings) proved

The covering $X(p)\to X(1)$ is a $G$ -covering, $G=\PSL(2,p)$ , branched over $j=1728,0,\infty$ with canonical inertia generators conjugate to $x_0,y_0,z_0$ ; by rigidity it is the only $G$ -covering of type $(2A,3A,pA)$ , and its genus is 0, 3, 26 for $p=5,7,11$ . As a $G$ -covering it is defined over $\Q(\sqrt{p^*})$ , $p^*=(-1)^{(p-1)/2}p$ , and over no smaller field, and an automorphism of $\overline\Q$ that negates $\sqrt{p^*}$ carries it to the covering of type $(2A,3A,pB)$ , its twist by the outer automorphism.

Proof

The images of $S$ , $TS$ , $T$ fix $i$ , $e^{2\pi i/3}$ and the cusp $\infty$ and generate $G$ ; the genus follows from Riemann–Hurwitz. The classes of the triple are rational over $\Q(\sqrt{p^*})$ , so Serre’s rigidity theorem applies there. An automorphism of $\overline\Q$ moves the canonical inertia generators by a power given by the cyclotomic character, a non-square modulo $p$ when $\sqrt{p^*}$ is negated, so the type becomes $(2A,3A,pB)$ .

Les objets comme dessinsThe objects as drawings

Plate Ép.3The fifteen objects of the group of order 168 as the quotient dessins of the Klein map, each named by its genus: 3 for the regular object, 1 for the objects of sizes 84, 56 and 42 with cyclic stabilizers, and 0 for the other eleven.

For a subgroup $H$ the quotient dessin $D_H$ is $G$ acting on $G/H$ through $x_0$ and $y_0$ : the cover $X(7)/H\to X(1)$ , the modular curve of level 7 attached to $H$ . The fifteen objects of the seam table are exactly the fifteen quotients of the Klein map, each with automorphism group $N_G(H)/H$ , and they are pairwise non-isomorphic. Eleven have genus 0. The regular object is $X(7)$ , of genus 3; the objects with stabilizers $C_2$ , $C_3$ , $C_4$ have genus 1, the last two being the split and non-split Cartan curves $X_{\mathrm{sp}}(7)$ and $X_{\mathrm{ns}}(7)$ ; the cyclotomic object is $X_1(7)$ , the sky $X_0(7)$ , and the object of size 28 is $X_{\mathrm{sp}}^+(7)$ .

The two objects of size 7 are where the group of order 168 first appeared. Galois asked whether the modular equation of degree $p+1$ can be lowered to degree $p$ , that is, whether $\PSL(2,p)$ has a subgroup of index $p$ , and answered that it can for $p=5$ , 7, 11 and no larger prime; for $p=7$ the subgroup is $S_4$ , in two classes, and the seven cosets are the points or the lines of the Fano plane. Their dessins are Galois’s resolvents of degree 7, trees of genus 0 with the same passport. The members of each of the three pairs $(V_4^a,V_4^b)$ , $(A_4^a,A_4^b)$ , $(S_4^a,S_4^b)$ share a passport and are not isomorphic.

Theorem(The seam table as the quotients of the Klein map) computed

The map $H\mapsto D_H$ is a bijection from the fifteen classes of subgroups of $G=\PSL(2,7)$ onto the isomorphism classes of quotient dessins of the Klein map. The automorphism group of $D_H$ is $N_G(H)/H$ . The fifteen dessins are pairwise non-isomorphic, and the two members of each pair $(V_4^a,V_4^b)$ , $(A_4^a,A_4^b)$ , $(S_4^a,S_4^b)$ have the same passport and are not isomorphic.

Proof

A quotient of a regular dessin is determined by a subgroup, and two subgroups give isomorphic dessins exactly when they are conjugate: a bijection $G/H\to G/K$ carrying $x_0$ and $y_0$ to themselves commutes with the group they generate, which is $G$ , so it is a $G$ -isomorphism. The centralizer of the monodromy group is $\Aut_G(G/H)=N_G(H)/H$ by the stabilizer principle. The passports and genera were computed.

La conjugaison comme suture tordueConjugation as a twisted seam

Plate Ép.4The drawings of the seven points and of the seven lines of the Fano plane, side by side: each is the mirror image of the other, as complex conjugation, which exchanges the two resolvents, requires.

Galois acts on the quotients of a rigid regular dessin through the outer automorphism group. For the Klein map an element of $\operatorname{Gal}(\overline\Q/\Q)$ acts as the outer class exactly when it negates $\sqrt{-7}$ ; the quotient dessins whose class $\operatorname{Out}(G)$ fixes have field of moduli $\Q$ , and the three pairs whose classes it exchanges have field of moduli $\Q(\sqrt{-7})$ , and these fields of moduli are fields of definition. The element that does this is the twist of the type law: for Klein’s lattice, $(\sigma,\alpha_\sigma)$ is a twisted Galois symmetry. So the bridge between the points and the lines of the Fano plane, refuted for a fixed marking, is built by Galois conjugation of dessins, by the element the type law sees in the residues.

The two drawings of degree 7 are written down exactly. With the points of order 3 pinned at $x=0,1$ they are $P(x)=k\,x^3(x-1)^3(x-a)$ with $a=(11\mp\sqrt{-7})/8$ ; the dessin of the points of the Fano plane has $a=(5-\alpha)/4$ and that of the lines $a=(5-\bar\alpha)/4$ . Complex conjugation reverses the cyclic order at every vertex, and the two drawings are mirror images. For the three groups of the trinity the Galois-conjugate pairs of quotient dessins are exactly the Gassmann pairs of the seam table: Galois conjugation is the seam that counting cannot build. On other families of $\PSL(2,7)$ the twist exchanging points and lines is realized by Galois elements that act trivially on every character, where the type law cannot see them; the identification of the two Galois elements is a property of the rigid family, the source.

Theorem(Galois acts on quotient dessins through

\mathrm{Out}(G)

) proved

Let $G$ have trivial centre, $C$ a strictly rigid triple of classes, and suppose each power $C^k$ is $\beta_k(C)$ for some $\beta_k\in\Aut(G)$ . Let $X$ be the $G$ -cover of type $C$ branched over $0,1,\infty$ and $D_H$ its quotient by $H$ . For $\sigma\in\operatorname{Gal}(\overline\Q/\Q)$ let $\alpha_\sigma$ be the outer class of $\beta_k$ for the $k$ with $C^k$ the type of $\sigma(X)$ . Then $\sigma(D_H)\cong D_{\alpha_\sigma^{-1}(H)}$ , so Galois acts on the quotient dessins as $\operatorname{Out}(G)$ acts on the classes of subgroups, and the field of moduli of $D_H$ is the fixed field of the $\sigma$ with $\alpha_\sigma(H)$ conjugate to $H$ . For $\PSL(2,p)$ , $p\in\{5,7,11\}$ , and $(2A,3A,pA)$ , $\alpha_\sigma$ is the outer class exactly when $\sigma(\sqrt{p^*})=-\sqrt{p^*}$ , and $\alpha_\sigma$ is the twist of the type law.

Proof

By the branch cycle argument, $\sigma(X)$ is a $G$ -cover branched over $0,1,\infty$ of type a power $C^k=\beta_k(C)$ . The $G$ -covers $(X,\iota\circ\beta^{-1})$ have the types $\beta(C)$ , and by strict rigidity a $G$ -cover of type $\beta(C)$ is unique, so $\sigma(X,\iota)\cong(X,\iota\circ\beta_k^{-1})$ and $\sigma(X/\iota(H))\cong X/\iota(\beta_k^{-1}(H))$ . The power map by a non-square acts on classes and characters as the outer automorphism.

La face localeThe local face

Plate Ép.5The collision objects of the Klein dessin’s reductions on the fifteen objects:

A_4^a

(\alpha)

and

A_4^b

(\bar\alpha)

, the split prime 2;

S_3

at the inert prime 3; the sky,

7{:}3

, at the ramified prime 7.

The field of moduli of a three-point cover is unramified outside the primes dividing its monodromy group, here 2, 3 and 7. Reduce the Klein quartic with its action at a prime over each of them. In characteristic 0 the special points form three orbits, the 24 cusps, the 56 points of order 3 and the 84 points of order 2; in each reduction two of them collide, and the orbit they collide onto is an object of the seam table. At 2 the points of orders 2 and 3 collide onto 14 points with stabilizer $A_4$ , of class $A_4^a$ at the prime $(\alpha)$ and $A_4^b$ at $(\bar\alpha)$ . At 3 they collide onto 28 points with stabilizer $S_3$ . At 7, on Serre’s model $v^2=u^7-u$ , the cusps and the points of order 3 collide onto the 8 points of the sky. At a prime over 13 the picture of characteristic 0 survives.

The three local types of the type law appear on these collision objects. At the split prime 2 the two primes see the two members of an outer pair and the twist relates the two residues, a duality. At the inert prime 3 the collision object is fixed by the outer automorphism, but the trace of $z\mapsto z+1$ reduces into $\F_9$ and not into $\F_3$ , so the twist is realized only semilinearly, through the Frobenius of $\F_9$ . At the ramified prime 7 the collision object is the sky, and the twist is realized by an automorphism of the reduction, $(u,v)\mapsto(3u,\sqrt3\,v)$ over $\F_{49}$ .

Theorem(Collisions at the primes of 168) computed

In the reduction of the Klein dessin the special points form exactly two orbits. At $p=2$ : the 24 cusps, and one orbit of 14 points with stabilizer $A_4$ , of class $A_4^a$ at $(\alpha)$ and $A_4^b$ at $(\bar\alpha)$ . At $p=3$ : the 24 cusps, and 28 points with stabilizer $S_3$ . At $p=7$ : the 84 points of order 2, and the 8 points $v=0$ , $u\in\Proj^1(\F_7)$ , with stabilizer $7{:}3$ . At a prime over 13 every element fixes as many points as in characteristic 0.

Le groupe de Grothendieck–TeichmüllerThe Grothendieck–Teichmüller group

Plate Ép.6On the fifteen quotient dessins of the Klein map the orbits of

\widehat{\mathrm{GT}}

are the Galois orbits: the three pairs that the conjugation of

\sqrt{-7}

exchanges, lit, and nine dessins each fixed.

Drinfeld’s group $\widehat{\mathrm{GT}}$ is the group of pairs $(\lambda,f)$ , $\lambda\in\widehat\Z^\times$ and $f$ in the closed commutator subgroup of the profinite free group on $x$ and $y$ , for which $x\mapsto x^\lambda$ , $y\mapsto f^{-1}y^\lambda f$ is an automorphism satisfying three relations: two on the free group itself, the first level of the Teichmüller tower, and the pentagon on the pure braid group on five strands, the second level. It acts on dessins, and $\sigma\mapsto(\chi(\sigma),f_\sigma)$ embeds the absolute Galois group in it compatibly with the two actions; injectivity rests on Belyi’s theorem. On the quotients of a rigid regular dessin, such as the Klein map, every element of $\widehat{\mathrm{GT}}$ acts as a Galois element does, so on the fifteen quotient dessins the $\widehat{\mathrm{GT}}$ -orbits are the Galois orbits, the outer pairs. The non-rigid objects of the seam table are no test of $\widehat{\mathrm{GT}}$ against Galois: what matters is the rigidity of the triple, not of the object.

One finite group at a time, Guillot’s finite group $\mathrm{GT}(G)$ sees only the two relations of the first level. For $\PSL(2,7)$ its part $\mathrm{GT}_1$ with $\lambda=1$ has order $2^9$ , and it moves regular dessins that the commutator subgroup of the Galois group fixes, for the types $(4,4,3)$ , $(4,3,3)$ , $(4,4,4)$ , $(7,3,4)$ and $(7,7,4)$ . This does not bear on whether $\widehat{\mathrm{GT}}$ is the Galois group, because the maps from $\widehat{\mathrm{GT}}$ to the finite groups need not be onto. It is a finite instance of Grothendieck’s two-level principle: coherence at level one, imposed on one finite group, does not force arithmeticity.

Proposition(The finite group over-counts) computed

$|\mathrm{GT}_1(G)|=1$ , $2^9$ , $2^{48}$ for $G=A_5$ , $\PSL(2,7)$ , $\PSL(2,11)$ . Every marked regular dessin alone in its class triple is fixed; among them is the Klein map. The image of the commutator subgroup $\operatorname{Gal}(\overline\Q/\Q)'$ in $\mathrm{GT}_1(\PSL(2,7))$ is a proper subgroup: the two marked regular dessins of type $(4,4,3)$ , of genus 15, are exchanged by $\mathrm{GT}_1$ , but their quotients by $S_4^a$ have the abelian field of moduli $\Q(\sqrt7)$ , so $\operatorname{Gal}(\overline\Q/\Q)'$ fixes them.

Gal et GTGal and GT

Plate Ép.7A twisted seam that Galois supplies: the Frobenius at 2 as a twisting element, of power 2, and made equivariant as

\tau

, of power 4, on the flexes of Klein’s quartic and on the coordinatizations of the Fano plane by

\F_8

The question at the scale of all curves can be put in the language of this book. For each $n\ge4$ the finite étale covers of $\mathcal M_{0,n}$ over $\overline\Q$ form a Galois category, joined to the others by the maps that forget points and by the symmetric group; for $n=4$ its objects are the dessins. A twisted seam of the whole network at level $n$ is an outer automorphism of the profinite fundamental group, and from level two on these are the elements of $\widehat{\mathrm{GT}}$ , with the absolute Galois group among them. At level one alone the arithmetic statement is false: the outer automorphisms of the free group include $x\mapsto xy$ , $y\mapsto y$ , which changes passports, as Galois never does.

The Galois groups the draft has seen act on the network of 168 are small: $\operatorname{Gal}(\Q(\sqrt{-7})/\Q)$ through the outer automorphism, and the cyclotomic groups above it through the automorphism groups of objects, where $\sigma_2$ , corrected by its twisting element, is the twist $\tau$ of the flexes. Next: whether finite shadows that impose the pentagon relation separate the dessins of $\PSL(2,7)$ that $\mathrm{GT}_1$ merges, and whether the audible twists of $A_6$ are ever Galois. On the horizon, Poonen, Schaefer and Stoll found the primitive integer solutions of $x^2+y^3=z^7$ by classifying the twisted forms of the Klein quartic that the equation produces: twisted seams over the rationals, which the draft does not yet treat.

Open questionopen

Is every twisted seam of the genus-zero network arithmetic, $\operatorname{Gal}(\overline\Q/\Q)=\widehat{\mathrm{GT}}$ ? At the finite level of 168: does imposing the pentagon relation on finite quotients separate the regular dessins of $\PSL(2,7)$ that $\mathrm{GT}_1$ moves and Galois fixes?

What would settle the local question for 168 is the list of coherent symmetries of the network of its fifteen drawings with their seams, at the second level of the tower as well as the first, compared element by element with the Galois action; what would settle the global one is the equality of the absolute Galois group with $\widehat{\mathrm{GT}}$ , which is open. The Esquisse gives seam theory a single profinite group whose finite quotients carry every regular dessin, a tower whose coherent twists form $\widehat{\mathrm{GT}}$ , and the anabelian programme; seam theory gives the Esquisse a finite, computable network in which these statements can be tested, the type law as the local face of the Galois action, and the marks, which name what local data can and cannot hear.

So far seam theory mostly re-reads known mathematics. If the network-level statements of this fourth part are true, they would be the first results that cannot be read off a single classical source, and each needs a literature check before anyone calls it new.