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Equilateral triangles and the Fano plane
Philippe Caldero and Jérôme Germoni
Abstract. We formulate a definition of equilateral triangles in the complex line that makes
sense over the field with seven elements. Adjacency of these abstract triangles gives rise to the
Heawood graph, which is a way to encode the Fano plane. Through some reformulation, thisgives a geometric construction of the Steiner systems S(2, 3, 7) and S(3, 4, 8). As a consequence, we embed the Heawood graph in a torus, and we derive the exceptional isomorphism
PSL2 (F7 ) ' GL3 (F2 ).

1. EQUILATERAL TRIANGLES. Three points b, c, and d in C form an equilateraltriangle if and only if the ratio (d − b)/(c − b) is −j or −j 2 , where j is a primitive
cubic root of unity. With a (projective) view to extend the notion to other fields, thiscan be written as [∞, b, c, d] ∈ {−j, −j 2 }, where the bracket denotes the cross-ratioof four distinct elements a, b, c, d ∈ P1 (C):[a, b, c, d] =c−a d−b×,d−a c−bwith the usual conventions about infinity: if α ∈ C∗ , then α/0 = ∞, and if α, β, γ, δ ∈C are such that αδ − βγ 6= 0, then (α∞ + β)/(γ∞ + δ) = α/γ .
d

j
1
π
3

b

c
j2
Figure 1. Equilateral triangle in the complex line

Definition. Let K be a field containing a primitive cubic root of unity j . As in C, there
are two of them, the roots of X 2 + X + 1. An unordered triple {b, c, d} (in short: bcd)
of distinct points in K is said to be an equilateral triangle if [∞, b, c, d] ∈ {−j, −j 2 }.
More generally, an equianharmonic quadrangle is a quadruple of distinct points{a, b, c, d} in P1 (K) such as [a, b, c, d] ∈ {−j, −j 2 }.
Remark. Harmonic quadrangles are a projective substitute for the notion of middle
in affine geometry in the following sense: a quadruple {a, b, c, d} is harmonic, i.e.
[a, b, c, d] ∈ {−1, 2, 1/2}, if and only if, when one point is mapped to infinity bya homography, one of the other three is mapped to the middle of the last two ones.
Similarly, a quadrangle is equianharmonic if an only if, when one point is mapped toinfinity, the other three are mapped to an equilateral triangle.

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These two sets of values are remarkable in the fact that, by permuting the four
variables in the cross-ratio λ = [a, b, c, d], one obtains six different values, namely λ,(λ − 1)/λ, 1/(1 − λ), 1/λ, 1 − λ, and λ/(λ − 1), unless the quadrangle is equianharmonic (resp. harmonic), in which case there are only two (resp. three) different
values. In particular, in the definition, the order of points and the choice of a cubicroot j versus the other one j 2 are irrelevant.
Let us recall a few facts about the action of PGL2 (K) on P1 (K) and cross-ratio.
Lemma 1. The group PGL2 (K) acts simply transitively on ordered triples of distinct
points points of P1 (K). Cross-ratio is invariant under PGL2 (K).
If −1 is not a square in K∗ , the group PSL2 (K) acts transitively on unordered
triples. It does not act transitively on ordered triples.
Proof. Let (a, b, c) be a triple of distinct points. The homography h defined by∀z ∈ P1 (K),h(z) =

c−a z−b×
c−b z−ais the unique element in PGL2 (K) that maps (a, b, c) to (∞, 0, 1) (if ∞ ∈ {a, b, c},simply erase the corresponding factors). Invariance of cross-ratio is easy to check.
Now, assume −1 is not a square. Then, the homography s defined by s(z) = 1 − z
does not belong to PSL2 (K), so that either h or sh does. Since s maps (∞, 0, 1) to(∞, 1, 0), the unicity claimed above shows that PSL2 (K) is not transitive on ordered
triples. On the other hand, both h and sh map {a, b, c} to {∞, 0, 1} and one of themlies in PSL2 (K). Thus, PSL2 (K) is transitive on unordered triples.
2. EQUIANHARMONIC QUADRANGLES OVER F7 . Let K = F7 be the fieldwith 7 elements. Since 7 − 1 is a multiple of 3, there are two primitive cubic roots ofunity in K, namely j = 2 and j 2 = 4. For example, note that the quadrangles Q3 ={∞, 0, 1, 3} and Q5 = {∞, 0, 1, 5} are equianharmonic.
Lemma 2. There are 28 equianharmonic quadrangles in P1 (F7 ).
Proof. Observe that the cross-ratio is a homography with respect to every variable.
Hence, given three distinct points a, b, c in P1 (F7 ) and λ in P1 (F7 ), there is a unique dsuch that [a, b, c, d] = λ. If λ ∈/ {∞, 0, 1}, the point d is automatically distinct
from a, b, and c. By multiplying the number of triples by the number of admissible values for the cross-ratio, one obtains 8 × 7 × 6 × 2 ordered quadruples and8 × 7 × 6 × 2/4! = 28 equianharmonic quadrangles.
Lemma 3. Equianharmonic quadrangles form a single orbit under PGL2 (F7 ), andtwo orbits under PSL2 (F7 ).
Proof. Let Q = {a, b, c, d} be an equianharmonic quadrangle. Let h be the homography that maps (a, b, c) to (∞, 0, 1). Then, h(d) = [∞, 0, 1, h(d)] = [a, b, c, d].
Since Q is equianharmonic, h maps Q to {∞, 0, 1, 3} or {∞, 0, 1, 5}. Since s : z 7→
1 − z exchanges these two quadrangles, h or sh maps Q to {∞, 0, 1, 3}, which proves
the first claim.
Recall that the cardinality of the orbit of Q3 = {∞, 0, 1, 3} under a group G acting
on the set of quadrangles is |G|/|GQ3 |, where GQ3 is the stabilizer of Q3 . Since
PGL2 (F7 ) acts transitively on the 28 quadrangles, the stabilizer A of Q3 in PGL2 (F7 )has cardinality 336/28 = 12. The group A acts faithfully on Q3 : by lemma 1, if a
homography fixes three points, it is the identity. Since its order is 12, it is isomorphic
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to the alternating group A4 (the unique subgroup of index 2 in the symmetric group).
In fact, using the proof of lemma 1, let us look for homographies that act on Q3 likedouble transpositions: one finds that z 7→ 3/z acts on Q3 as (∞0)(13); that z 7→
(z − 3)/(z − 1) acts as (∞1)(03); and z 7→ (3z − 3)/(z − 3) acts as (∞3)(01).
These are involutions and they commute on Q3 , so that they commute on P1 (F7 ).
The point is that these involutions belong to PSL2 (F7 ). Hence, A ∩ PSL2 (F7 ) contains a subgroup K ' (Z/2Z)2 of order 4; besides, it also contains the order-3 elementz 7→ 1/(1 − z), that permutes {∞, 0, 1} cyclically and fixes 3. Hence, the group A
is included in PSL2 (F7 ), and A is the stabilizer of Q3 in PSL2 (F7 ). Therefore, thePSL2 (F7 )-orbit of Q3 has cardinality 168/12 = 14. Since Q5 = {∞, 0, 1, 5} is in
the same PGL2 (F7 )-orbit as Q3 , its stabilizer is conjugated to A, and the orbit of Q5
has cardinality 14 too.
Lemma 4. The complement of an equianharmonic quadrangle in P1 (F7 ) is equianharmonic. Moreover, both are in the same orbit under PSL2 (F7 ).
Proof. We start with an example: [∞, 0, 1, 3] = 3 = [4, 2, 5, 6]. The homographydefined by h(z) = 5(z − 2)/(z − 4) = (6z + 2)/(4z + 5) maps (4, 2, 5, 6) to(∞, 0, 1, 3), and h belongs to PSL2 (F7 ). This proves the claim for {∞, 0, 1, 3}.
Now, let Q = {a, b, c, d} be an equianharmonic quadrangle. By lemma 3, there isa homography g ∈ PGL2 (F7 ) such that g(Q) = {∞, 0, 1, 3}. Then, the bijection gmaps the complement of Q to {2, 4, 5, 6}, which is equianharmonic. Moreover, g −1 hg
maps the complement of Q to Q. This proves the claim for a general Q.
3. EQUILATERAL TRIANGLES OVER F7 .
Corollary 5. There are 14 equilateral triangles over F7 . They are the following ones:
z+1

013
1
z

,

z+1

124

z+1

,

235

z+1

,

346

z+1

,

450

z+1

,

561

,

602


2 126

015
z+1

2 230
z+1

2 341
z+1

2 452
z+1

2 563
z+1

2 605.
z+1

Proof. By lemma 4, one can arrange the 28 equianharmonic quadrangles in 14 complementary pairs. In a given pair, exactly one quadrangle contains ∞. Adding orwithdrawing ∞ gives a one-to-one correspondence between equilateral triangles andequianharmonic quadrangles containing ∞, hence between equilateral triangles andpairsof complementary equianharmonic quadrangles. (E.g.: 013 corresponds to thepair {∞, 0, 1, 3}, {2, 4, 5, 6} .) Hence, the first assertion holds.
To write a list, one starts with 013 and 015. Using invariance of ∞ under affine
transformations, one builds 7 new triangles out of the first two with the translationz 7→ z + 1. These are the rows of the list in the corollary.
Remark. The vertical arrow in the statement
 of corollary 5 has the following meaning. The triangle 013 corresponds to the pair {∞, 0, 1, 3}, {2, 4, 5, 6} . The homog
raphy z 7→ 1/z maps this pair to {∞, 0, 1, 5}, {2, 3, 4, 6} , which corresponds to015. This extends to an action of PGL2 (F7 ) on triangles.
Lemma 6. There is a canonical action of PGL2 (F7 ) on equilateral triangles. All
triangles are in the same PGL2 (F7 )-orbit, but there are two PSL2 (F7 )-orbits described by the lines of corollary 5.
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Proof. By invariance of cross-ratio, a homography in PGL2 (F7 ) maps an equiharmonic quadrangle to another one. Since this action on parts of P1 (F7 ) commutes withtaking the complement, it maps a pair of complementary equianharmonic quadrangles. Hence, the group PGL2 (F7 ) acts on the set of pairs of complementary equianharmonic quadrangles. But there is a one-to-one correspondence between such pairs
and equilateral triangles, one inherits an action of PGL2 (F7 ) on triangles: if abc isa triangle
and g ∈ PGL2 (F7 ), one defines g · abc as the triangle corresponding to
g · {∞, a, b, }, P1 \ {∞, a, b, c} .
Two triangles abc and a0 b0 c0 are in the same PGL2 (F7 )-orbit by lemma 1: thehomography that maps (a, b, c) to (a0 , b0 , c0 ) also maps {∞, a, b, c}, P1 \ {a, b, c}
to {∞, a0 , b0 , c0 }, P1 \ {a0 , b0 , c0 } .
As for PSL2 (F7 )-orbits, lemma 4 implies that the pairs of the form {Q, P1 \ Q},where Q runs over a PSL2 (F7 )-orbit of quadrangles, are orbits of pairs. Hence, bylemma 3, there are two orbits of pairs, corresponding to two orbits of triangles. Since
triangles in the same line in corollary 5 are in the same orbit (simply apply z 7→ z + 1),
the two lines are exactly the two orbits.
E.g.: Let h(z) = 1/z . Then h {∞, 0, 1, 3} = {0, ∞, 1, 5}, so h maps the triangle

013 to 015; moreover, h {∞, 1, 2, 4} = {0, 1, 2, 4} = P1 (F7 ) \ {∞, 3, 5, 6}, sothat h maps the triangle 124 to 356.
Equilateral triangles over F7 as equilateral triangles on a torus. Say that two(equilateral) triangles (over F7 ) are adjacent if they have two vertices in common.
It is easy to check that a triangle in either line of corollary 5 is adjacent to exactly threetriangles, and all three lie on the other line. The upshot is that one can arrange triangles
over F7 as equilateral triangles in the real plane in a periodic manner. By glueing thesides of a fundamental parallelogram, one tiles a torus by 14 triangles (fig. 2).
046

015

245

134

023

346

235

124

013

026

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134

023

126

015

046

235

124

013

126

023
026

126

015
156

046
045

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015
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046
045

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026

245
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023
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045

356
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134
124

156

046
045

356
346

245

026

015
156

046
045

026

356

045

346

245
235

134
124

Figure 2. Triangles over F7 as triangles on a torus

The picture is more appealing when one replaces the adjacency graph, embeddedin the torus, by its dual graph: the meeting points of triangles become 7 hexagons thattile the torus (fig. 3). Since each face touches all the others, this tiling shows that thechromatic number of the torus is at least 7. A polyhedral version of this tiling was
discovered by L. Szilassi [6].
Equilateral triangles and the Fano plane. Forgetting the tilings, let us consider thegraph with 14 vertices labeled by equilateral triangles over F7 , where two vertices are4

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235

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124

Figure 3. Seven hexagons tiling a torus

connected if the corresponding triangles are adjacent. This gives rise to a graph known
as the Heawood graph (see [4]). It is bipartite because a triangle on a line of corollary 5
is adjacent to triangles on the other line.
346

046

134

045

045

124

015

046
126

245

013

156

124
346

235

015

023

245

134

356

026
156

013

023

356
026

235

126

Figure 4. Incidence of triangles: Heawood graph and Fano plane

The Heawood graph is the incidence graph of the Fano plane: one can label vertices
and lines of the Fano plane by triangles, so that adjacency of triangles corresponds toincidence (fig. 4).
Application: two exceptional isomorphisms.
Theorem 7. One has: PSL2 (F7 ) ' GL3 (F2 ).
Proof. One can label vertices of the Fano plane by nonzero vectors in F32 so that thethird vertex on the line containing v and v 0 is labelled by v + v 0 (put for instance thecanonical basis vectors on the vertices of the triangle and use this rule to complete thelabelling). Hence, a permutation f of vertices and lines of the Fano plane preserves,
once completed into a map F32 → F32 by setting f (0) = 0, is additive. However, ona prime field, additivity is equivalent to linearity, so that automorphisms of the Fano
plane are linear automorphisms of F32 . This is in fact a special case of the “fundamental theorem of projective geometry” ([1, Theorem 2.26]), by which any incidence
preserving map is a projective map, i.e. an element in PGL3 (F2 ).
Remark. The action of PGL2 (F7 ) on PSL2 (F7 ) by conjugation embeds the formergroup in Aut(PSL2 (F7 )). On the other hand, the automorphism group Aut(GL3 (F2 ))is the semidirect product of GL3 (F2 ) and Z/2Z acting by g 7→ (g T )−1 . Hence:
PGL2 (F7 ) ' Aut GL3 (F2 ).
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4. STEINER SYSTEMS. Recall a Steiner system with parameters (t, k, n), writtenS(t, k, n), is a set of cardinality n and a collection of k -sets called blocks such thatevery t-set is contained in a unique block.
Proposition 8. An orbit under PSL2 (F7 ) of equianharmonic quadrangles forms an
S(3, 4, 8). An orbit under PSL2 (F7 ) of triangles forms an S(2, 3, 7) (here, the action
is defined in lemma 6).
To be more explicit, recall that a PSL2 (F7 )-orbit of triangles is but a line in corollary 5. Starting from such an orbit, one rebuilds quadrangles by adding ∞ to every
triangle and by adding to the collection the complements of these quadrangles. For
instance, starting from the orbit containing 013, one builds the following S(3, 4, 8):∞013 ∞124 ∞235 ∞346 ∞045 ∞156 ∞026
2456 0356 0146 0125 1236 0234 1345.
Proof. The underlying set of the S(3, 4, 8) is P1 (F7 ); blocks are quadrangles ina given PSL2 (F7 )-orbit. To fix notations, let us consider that of {∞, 0, 1, 3}. Let{a, b, c} be a 3-set. By lemma 1, there exists h ∈ PSL2 (F7 ) that maps {a, b, c}
to {∞, 0, 1}. Then {a, b, c, h−1 (3)} is an equianharmonic quadrangle in the same
PSL2 (F7 )-orbit as {∞, 0, 1, 3}. If {a, b, c} is included in some block {a, b, c, d},then h(d) = [∞, 0, 1, h(d)] = [a, b, c, d] ∈ {3, 5}. Since {∞, 0, 1, 5} is not a block(they are not in the same line in corollary 5, see lemma 6), one has h(d) = 3, which
proves the unicity of a block containing {a, b, c}.
Now, the underlying set of the S(2, 3, 7) is F7 ; blocks are triangles in a fixed
PSL2 (F7 )-orbit, say, that of 013. A block is in particular a triple {a, b, c} such that{∞, a, b, c} is equianharmonic. Hence, given a pair {a, b} in F7 , the unique blockcontaining {a, b} is obtained by erasing ∞ from the unique equianharmonic triangle
in the PSL2 (F7 )-orbit of {∞, 0, 1, 3} that contains {∞, a, b}.
Remark. The passage from quadrangles to triangles follows the classical constructionof an S(t − 1, k − 1, n − 1) out of an S(t, k, n) by selecting blocks containing a fixed
point (here, ∞), then erasing it.
5. ALTERNATIVE PRESENTATIONS. We give two other ways to recover the
Fano plane from PGL2 (F7 ). Proofs of these elementary results appear in [3].
“Nilpotent conic”. The group GL2 (F7 ) acts by conjugation on the space gl2 (F7 ) of
2 × 2 matrices (its Lie algebra). The action factors through PGL2 (F7 ). Since trace
is preserved, the subspace sl2 (F7 ) = ker(tr) is stable. The determinant is a quadratic
form on sl2 (F7 ). Let Q = − det. By the Cayley-Hamilton theorem, the isotropic coneof Q is the set N of nilpotent matrices. Let C be the corresponding conic in P(sl2 (F7 ))defined by the equation Q = 0.
Say a line L in P(sl2 (F7 )) is tangent (resp. secant, resp. an exterior) to C if L ∩ C
is a unique point (resp. a pair of distinct points, resp. empty). A point p outside C is exterior (resp. interior) if p belongs to two (resp. no) tangents. Recall that polarity (withrespect to C ) between a point and a line is the projective counterpart of orthogonality(with respect to Q) between a line and a plane. It turns out that the polar of an exterior(resp. interior) point is a secant (resp. an exterior line). A triangle is self-polar if thepolar of every vertex is the opposite side. Two triangles are adjacent if they share acommon vertex: this makes the set of self-polar triangles into a graph. The action of
PGL2 (F7 ) induces an action on all these geometric objects and on the graph.
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Proposition 9. There are 14 self-polar triangles whose vertices are interior points.
The corresponding graph is the Heawood graph.
This result slightly improves on [5, § 7], in which another quadratic form is used,since here, the Heawood graph comes with a natural action of PGL2 (F7 ).
Elementary 2-subgroups of PSL2 (F7 ). In fact, the projective plane P(sl2 (F7 )) is
the set of involutions in PGL2 (F7 ). For A ∈ GL2 (F7 ), let [A] ∈ PGL2 (F7 ) be thehomography defined by A. Then [A] is an involution in PGL2 (F7 ) if and only if A2
is a scalar matrix. By the Cayley-Hamilton theorem, this means that tr(A) = 0, hence
the claim. Moreover, one can identify involutions in PSL2 (F7 ) and interior points of C .
Since det(A) = (tr(A)2 − tr(A2 ))/2 for every 2 × 2 matrix A, the polar formof Q on sl2 (F7 ) is the bilinear form (A, B) 7→ tr(AB), two elements A, B ∈GL2 (F7 ) ∩ sl2 (F7 ) are orthogonal with respect to Q if and only if [AB] is an involution. This means that two involutions are orthogonal as points in P sl2 (F7 ) if
and only if they commute. In other terms, the vertices of an self-polar triangles form,
together with the neutral elements, a subgroupto (Z/2Z)2 .  isomorphic
 


1 0
0 3
1 −3
3 −3
E.g.: The group K given by the classes of
,
,
,
.
0 1
1 0
1 −1
1 −3
From subgroups to equilateral triangles. A subgroup of PSL2 (F7 ) isomorphic to(Z/2Z)2 decomposes P1 (F7 ) into two orbits of cardinal 4 that turn out to be equianharmonic. Since all of them are conjugated under PGL2 (F7 ), it is enough (and easy)
to check it in a single example. This procedure induces a one-to-one correspondence
between self-polar triangles in P(sl2 (F7 )) and equilateral triangles in F7 .
E.g.: The orbits of the subgroup K defined above are {∞, 0, 1, 3} and {2, 4, 5, 6}. To
check this, compute the action of all elements of K on ∞ and take the complement.
Conclusion. In this article, we havec seen several geometric realizations of combinatorial structures related to the Fano plane–the Heawood graph, a tiling of a torus byseven hexagons, the Steiner systems S(2, 3, 7) and S(3, 4, 8). It is amusing that somekind of special triangles is involved in each construction.
REFERENCES
1. E. Artin, Geometric algebra, Reprint of the 1957 original. Wiley Classics Library. A Wiley-Interscience
Publication. John Wiley & Sons, Inc., New York, 1988. x+214 pp.
2. E. Brown and N. Loehr, Why is PSL(2, 7) ∼= GL(3, 2)? This Monthly 116 (2009), No. 8, 727–732.
3. Ph. Caldero, J. Germoni, Histoires hédonistes de groupes et de géométrie, Calvage et Mounet, Paris, 2014.
4. H. S. M. Coxeter, Self-dual configurations and regular graphs. Bull. Amer. Math. Soc. 56 (1950), 413–455.
5. H. S. M. Coxeter, My graph. Proc. London Math. Soc. (3) 46 (1983), no. 1, 117–136.
6. L. Szilassi, Les toroı̈des réguliers. [Regular toroids] Dual French-English text. Structural Topology No. 13(1986), 69–80.
PHILIPPE CALDERO (caldero@math.univ-lyon1.fr) received his PhD in representation theory in 1993
from Paris VI University. He is an associate professor in Lyon, France.
Institut Camille Jordan, Université Claude Bernard Lyon 1, Villeurbanne, France
caldero@math.univ-lyon1.fr
JÉRÔME GERMONI (germoni@math.univ-lyon1.fr) received his PhD in representation theory in 1997
from Strasbourg University. He is an associate professor in Lyon, France.
Institut Camille Jordan, Université Claude Bernard Lyon 1, Villeurbanne, France
germoni@math.univ-lyon1.fr

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