A bigon-free dichotomy for pseudoline arrangements, after a note of Grothendieck

Introduction

An arrangement of pseudolines is a finite family of at least three simple closed curves in the real projective plane, none of them separating it, any two of which meet in exactly one point, where they cross, and not all passing through one point [5, 3]. Arrangements of straight lines are the first examples. The vertices are the points where at least two pseudolines meet. We call a pseudoline bigonal when it carries exactly two vertices, so that every other pseudoline meets it at one of these two points; an arrangement is bigon-free when none of its pseudolines is bigonal. Since every pseudoline meets the others, it carries at least one vertex, and it carries exactly one only in a pencil, which is excluded; so bigon-free means that every pseudoline carries at least three vertices.

Four configurations will be used throughout (Figure 1):

$II_4$ four pseudolines in general position, that is, no three through a point;

$II_5$ five pseudolines in general position;

$II'_{3,2}$ five pseudolines $a$, $b$, $c$, $d$, $e$, where $a$, $b$, $c$ pass through a point $p$, the pseudolines $d$, $e$ do not, and $d \cap e$ lies on none of $a$, $b$, $c$; so $\{a,b,c\}$ is the only concurrent triple, and none of the five passes through both $p$ and $d \cap e$;

$III_6$ the complete quadrangle: the six pseudolines joining four points $P_1,\ldots,P_4$ pairwise, each passing through exactly two of them.

A sub-arrangement is a subfamily of at least three pseudolines, with the vertices it determines; we say that $A$ contains a configuration when some sub-arrangement of $A$ is of that type.

The source

The University of Montpellier holds the mathematical papers that Alexandre Grothendieck left to Jean Malgoire, and publishes their facsimiles. Its folder 80, Systèmes de pseudo-droites, is undated in his hand; the archival inventory dates it from 1981 onwards, and an administrative notice of 13 May 1981 used as scrap paper on page 27 agrees. Pages 20 to 28 look for the arrangements that he calls crucial and name them; the four configurations above are among them, with $II_4$, $II_5$, $III_6$ in his notation. Page 24 states, boxed, that every arrangement without bigons is either of type $\mathrm{II}_4$ or contains one of three types, which are $\mathrm{II}_5$, $\mathrm{III}_6$ and a pencil of three with a pencil of two, written $\mathrm{I}'_{3,2}$; we write $\mathrm{II}'_{3,2}$. Page 20 announces the same statement with the three configurations drawn, and page 28 states a lemma on the order in which the pseudolines can be added, of which only a sketch is written.

The notes derive the boxed statement from a theorem on the same page, which, as written, is false (Section 5). We have found no independent proof on the pages, and we have not found the statement in print, in Grünbaum’s monograph [5] or in the survey of Felsner and Goodman [3]. It is elementary, and may be known to specialists. This note records a proof (Theorem 4), the ordering of page 28 (Theorem 6), a consequence (Corollary 7), a computer check, and a formalisation in Lean. The statement is Grothendieck’s; the proofs, the corollary, the check and the formalisation are ours.

The dichotomy

The proof uses one external result, the pseudoline form of the Sylvester–Gallai theorem. A near-pencil is an arrangement in which all pseudolines but one pass through a point; a simple crossing is a vertex on exactly two pseudolines.

The existence of one simple crossing also follows from Melchior’s equality, which holds for pseudolines [2 Th. 1.3]; we return to it in § 6.

We shall often use that two distinct pseudolines meet only once: if a pseudoline passes through two distinct points of another, the two are equal.

Let $Q=\{x,y,z,w\}$ be four pseudolines in general position. Its six vertices are the points $x\cap y$, and two of them are opposite when the two pairs are disjoint, as $x\cap y$ and $z\cap w$; there are three opposite pairs. Two vertices that are not opposite lie on a common pseudoline of $Q$.

Proof. If $\ell$ passed through two vertices on a common pseudoline $s$ of $Q$, it would meet $s$ twice and be equal to it. Any three vertices contain two that are not opposite. If $\ell$ passes through no vertex, no three of the five are concurrent. If it passes only through $x\cap y$, then $x,y,\ell$ pass through $p=x\cap y$; $z$ and $w$ miss $p$, since $Q$ is in general position; and $z\cap w$ is on neither $x$ nor $y$, nor on $\ell$, which passes through no other vertex of $Q$. □

A pseudoline through two opposite vertices of $Q$ is a diagonal of $Q$. Two diagonals through the same opposite pair meet twice, so they are equal: $Q$ has at most three diagonals in $\mathcal{A}$.

Proof. Let $\mathcal{A}$ be bigon-free, $Q=\{x,y,z,w\}\subseteq\mathcal{A}$ in general position (Lemma 2), and suppose that $\mathcal{A}$ contains neither $II_5$ nor $II'_{3,2}$. By Lemma 3, every pseudoline of $\mathcal{A}\setminus Q$ is a diagonal of $Q$. If there is none, $\mathcal{A}=Q$ is $II_4$. If there is exactly one, $\delta$ through $x\cap y$ and $z\cap w$, then $\delta$ meets every other pseudoline of $\mathcal{A}=Q\cup\{\delta\}$ at one of these two points, and is bigonal. If there are at least two, say $\delta_1$ through $P_1=x\cap y$ and $P_2=z\cap w$, and $\delta_2$ through $P_3=x\cap z$ and $P_4=y\cap w$, then the six pseudolines $x,y,z,w,\delta_1,\delta_2$ pass through the pairs $\{P_1,P_3\}$, $\{P_1,P_4\}$, $\{P_2,P_3\}$, $\{P_2,P_4\}$, $\{P_1,P_2\}$, $\{P_3,P_4\}$, and through no other $P_i$, since $Q$ is in general position and each diagonal meets $Q$ only at its two vertices. They form a $III_6$. □

None of the four alternatives can be omitted. $\mathrm{II}_4$, $\mathrm{II}_5$ and $\mathrm{II}'_{3,2}$ are bigon-free (in $\mathrm{II}'_{3,2}$, each of $a$, $b$, $c$ carries $p$ and its points on $d$ and $e$, and each of $d$, $e$ carries $d \cap e$ and three more), and so is $\mathrm{III}_6$, each of whose sides carries two of the $P_i$ and its point on the opposite side. And $\mathrm{III}_6$ contains neither $\mathrm{II}_5$ nor $\mathrm{II}'_{3,2}$: deleting the side $P_1P_2$ leaves two concurrent triples, at $P_3$ and at $P_4$, and every five sides arise this way.

Proof. A bigon-free arrangement of five pseudolines is not $\mathrm{II}_4$ and cannot contain a $\mathrm{III}_6$, so by Theorem 4 it contains, and therefore is, a $\mathrm{II}_5$ or a $\mathrm{II}'_{3,2}$; both are bigon-free. □

The ordering of page 28

Page 28 asks for a numbering $D_1,\ldots,D_n$ of the pseudolines, starting from a $\mathrm{II}_4$ or a $\mathrm{III}_6$, such that no $D_i$ is bigonal in the partial arrangement $\{D_1,\ldots,D_i\}$. We prove a stronger form.

Proof. Adding a pseudoline never removes a vertex from the pseudolines already present. So it suffices that each pseudoline, when it is added, meets the earlier ones in at least three distinct points.

A pseudoline $\ell$ outside a $\mathrm{III}_6$ meets its six sides in at least three points. Indeed, a point lies on at most three sides, and only the $P_i$ lie on three, since any three sides include two with a common $P_i$, and these meet only there. Two of the $P_i$ share a side, so two points lie on at most five sides. If $\mathcal{A}$ contains a $\mathrm{III}_6$, start from it and add the other pseudolines in any order.

Otherwise, let $Q \subseteq\mathcal{A}$ be in general position (Lemma 2). By Lemma 3, a pseudoline outside $Q$ meets $Q$ in at least three points unless it is a diagonal of $Q$, and $\mathcal{A}$ has at most one diagonal $\delta$ of $Q$, since two would form a $\mathrm{III}_6$ with $Q$. Start from $Q$, add the pseudolines other than $\delta$ in any order, and add $\delta$ last: then all its vertices are present, and there are at least three since $\mathcal{A}$ is bigon-free. □

Arrangements avoiding the three configurations

Proof. A near-pencil contains none of them: it has a single vertex on more than two pseudolines, while $\mathrm{III}_6$ has four, and any five of its pseudolines include at least four through the centre, while $\mathrm{II}_5$ has no concurrent triple and $\mathrm{II}'_{3,2}$ has two pseudolines off its triple point.

Conversely, let $\mathcal{A}$ contain none of the three. It is not $\mathrm{II}_4$, since $n \ge6$, so by Theorem 4 it has a bigonal pseudoline $D$, with vertices $P$ and $Q$. Every other pseudoline passes through exactly one of them, since $D$ is the only pseudoline through both. Let $s$ and $t$ be the numbers of pseudolines other than $D$ through $P$ and through $Q$, so $s+t=n-1 \ge5$. If $s \ge3$ and $t \ge2$, three pseudolines $a,b,c \ne D$ through $P$ and two $d,e \ne D$ through $Q$ form a $\mathrm{II}'_{3,2}$; $d \cap e=Q$ lies on none of $a,b,c$, and $P$ on neither $d$ nor $e$. The case $s \ge2$, $t \ge3$ is symmetric. Since $s+t \ge5$, the remaining cases are $\min(s,t) \le1$. If $t=1$, the $s+1=n-1$ pseudolines through $P$ (with $D$) and the one other through $Q$ form a near-pencil, and similarly if $s=1$. The case $t=0$ is excluded, since then all pseudolines pass through $P$. □

For $n=4$ the arrangements avoiding the three configurations are $\mathrm{II}_4$ and the near-pencil, and for $n=5$ the near-pencil and the arrangement of two pencils of three sharing a pseudoline; these small cases agree with the enumeration below.

The statement of page 24 as written

The theorem from which page 24 derives the boxed statement assumes that there are no two vertices such that every pseudoline passes through one of them, and that the arrangement contains no $\mathrm{II}_{5}$, and concludes that it is a complete quadrangle $\mathrm{III}_{6}$ or a complete quadrilateral with its three diagonals, $\mathrm{IV}_{7}$. Six straight lines already refute it.

Example 8. Let $A=(0,0)$, $B=(1,0)$, $C=(0,1)$ in the affine plane, and take the six lines

the last three passing through $A$, $B$ and $C$. Writing a line $u_{0}x+u_{1}y+u_{2}z=0$ of the projective plane as an integer triple, three lines are concurrent if and only if the determinant of their triples vanishes. Of the twenty triples, exactly three vanish, $\{AB,CA,a\}$, $\{AB,BC,b\}$ and $\{BC,CA,c\}$, concurrent at $A$, $B$, $C$. Two vertices carry at most five of the six lines, since two of the triple points share a side; every five lines contain a concurrent triple; and there are three triple points, not the four of $\mathrm{III}_{6}$, and six lines, not the seven of $\mathrm{IV}_{7}$ (Figure 5).

The example contains a $\mathrm{II}'_{3,2}$ (the five lines $AB$, $CA$, $a$, $b$, $c$ have the single concurrent triple $\{AB,CA,a\}$), so it does not contradict Theorem 4. Nor does $\mathrm{IV}_{7}$, which contains a $\mathrm{II}'_{3,2}$ as well: two sides $x$, $y$ of the quadrilateral and the three diagonals have the single concurrent triple formed by $x$, $y$ and the diagonal through $x\cap y$ (the three diagonals are not concurrent, since the Fano matroid is not orientable [1]).

The theorem of page 24 becomes true once $\mathrm{II}'_{3,2}$ is excluded together with $\mathrm{II}_{5}$, and $\mathrm{IV}_{7}$ is then dropped from its conclusion. Its hypothesis implies that the arrangement is bigon-free, since the two vertices of a bigonal pseudoline would be met by every pseudoline. The proof of Theorem 4 then shows that the arrangement consists of four pseudolines $Q$ in general position and some of their diagonals. With no diagonal, or one, two vertices of $Q$ meet every pseudoline; with three, it is $\mathrm{IV}_{7}$, which contains a $\mathrm{II}'_{3,2}$. So it has exactly two diagonals and is a $\mathrm{III}_{6}$.

Verification

Enumeration

All arrangements of $n \le9$ pseudolines were enumerated, simple or not, as reorientation classes of rank-3 oriented matroids whose underlying matroid is simple [1], and the simple ones up to $n = 10$. The method is single-element extension with pruning by the three-term Grassmann–Plücker relations, and a canonical form computed on the flag graph of the cell decomposition of the sphere. The counts of classes, $1, 2, 4, 17, 143, 4890, 461053$ for $n = 3,\ldots,9$, and $4382$ and $312356$ simple ones for $n = 9, 10$, agree with the known values [4]. All $466094$ bigon-free arrangements with $n \le9$ ($461049$ of them with $n = 9$), and all $312356$ simple ones with $n = 10$, satisfy Theorem 4 and the strong ordering of Theorem 6. A deletion-closed search of the arrangements containing none of the three configurations finds only the near-pencils for $6 \le n \le16$, as Corollary 7 predicts. The hypothesis is weak: at $n = 9$, all but four arrangements are bigon-free, and all but $4382$ of the bigon-free ones (the simple arrangements) were found to contain a $\mathrm{II}'_{3,2}$.

Formalisation

Theorem 4 is formalised in Lean 4 with the Mathlib library, as part of the Lean appendix of [6], for a combinatorial model of arrangements that page 60 of the folder describes: a map, topological or combinatorial, whose underlying surface is a real projective plane. A map of a possibly non-orientable surface is a generalised map [8]: a finite set of flags with three fixed-point-free involutions $\alpha_0,\alpha_1,\alpha_2$, with $\alpha_0\alpha_2$ a fixed-point-free involution; vertices, edges and faces are the orbits of $\langle\alpha_1,\alpha_2\rangle$, $\langle\alpha_0,\alpha_2\rangle$, $\langle\alpha_0,\alpha_1\rangle$, and the Euler characteristic is $\chi= V - E + F$. A combinatorial arrangement labels the flags by pseudolines, constantly on edges, so that each pseudoline is connected and passes exactly once through each of its vertices, every vertex lies on at least two pseudolines, two pseudolines meet at exactly one vertex, and $\chi= 1$. The axioms do not require the pseudolines to go straight at a vertex, so the formal theorem is slightly more general than the statement for maps. Figure 6 shows the three involutions near an edge.

Vector (PDF)

The formal proof follows § 2, except for Theorem 1, which it replaces by Melchior’s counting [9], [2 Th. 1.3]: the skeleton of a bigon-free map has no loop and no multiple edge, so every face has at least three sides; with $\chi= 1$ and $E = \sum_v \nu(v)$, where $\nu(v)$ is the number of pseudolines through $v$, this gives $\sum_v (3 - \nu(v)) \ge3$, hence a simple crossing. The rest is incidence geometry. The proof uses only Lean’s standard axioms. It is registered in the Palomar Registry as entry PALOMAR-2026-11-07-000003, from the repository Commutator-IO/grothendieck-project-palomar (theorem Grothendieck.Folder80Maps2.PseudolineMap.map_dichotomy). The ordering (Theorem 6), Corollary 7, and the passage from a topological arrangement to its map are not formalised.

Statement on the use of artificial intelligence

The transcription of folder 80 used here, the search for a proof, the enumeration program, the Lean formalisation and a first draft of this text were produced with language models (Claude Opus 5 and Opus 5.5, Anthropic), under the author’s direction. The author checked the proofs, the computations and the references, and takes full responsibility for the content. The formal proof is checked mechanically by Lean. The transcriptions and the enumeration program are in the public repository Commutator-IO/grothendieck-project (directory computations/bigon).

Acknowledgements

The facsimiles are published by the University of Montpellier, grothendieck.umontpellier.fr.

Commutator, Paris, France

Email address: [email protected]