FINDING GENERATORS AND RELATIONS FOR
GROUPS ACTING ON THE HYPERBOLIC BALL

arXiv:1701.02452v1 [] 10 Jan 2017

DONALD I. CARTWRIGHT
TIM STEGER
Abstract. In order to enumerate the fake projective planes, as announced
in [2], we found explicit generators and a presentation for each maximal arithmetic subgroup Γ̄ of P U (2, 1) for which the (appropriately normalized) covolume equals 1/N for some integer N ≥ 1. Prasad and Yeung [4, 5] had given a
list of all such Γ̄ (up to equivalence).
The generators were found by a computer search which uses the natural
action of P U (2, 1) on the unit ball B(C2 ) in C2 . Our main results here give
criteria which ensure that the computer search has found sufficiently many
elements of Γ̄ to generate Γ̄, and describes a family of relations amongst the
generating set sufficient to give a presentation of Γ̄.
We give an example illustrating details of how this was done in the case of
a particular Γ̄ (for which N = 864). While there are no fake projective planes
in this case, we exhibit a torsion-free subgroup Π of index N in Γ̄, and give
some properties of the surface Π\B(C2 ).

1. Introduction
Suppose that P is a fake projective plane. Its Euler-characteristic χ(P ) is 3. The
fundamental group Π = π1 (P ) embeds as a cocompact arithmetic lattice subgroup
of P U (2, 1), and so acts on the unit ball B(C2 ) = {(z1 , z2 ) ∈ C2 : |z1 |2 + |z2 |2 < 1}
in C2 , endowed with the hyperbolic metric. Let F be a fundamental domain for
this action. There is a normalization of the hyperbolic volume vol on B(C2 ) and
of the Haar measure µ on P U (2, 1) so that χ(P ) = 3vol(F ) = 3µ(P U (2, 1)/Π). So
µ(P U (2, 1)/Π) = 1. Let Γ̄ ≤ P U (2, 1) be maximal arithmetic, with Π ≤ Γ̄. Then
µ(P U (2, 1)/Γ̄) = 1/N and [Γ̄ : Π] = N for some integer N ≥ 1.
The fundamental group Π of a fake projective plane must also be torsion-free
and Π/[Π, Π] must be finite (see [2] for example).
As announced in [2], we have found all groups Π with these properties, up to
isomorphism. Our method was to find explicit generators and an explicit presentation for each Γ̄, so that the question of finding all subgroups Π of Γ̄ with index N
and the additional required properties, as just mentioned, can be studied.
In Section 2, we give results about finding generators and relations for groups Γ
acting on quite general metric spaces X. The main theorem gives simple conditions
which ensure that a set S of elements of Γ, all of which move a base point 0 by at
most a certain distance r1 , are all the elements of Γ with this property.
In Section 3 we specialize to the case X = B(C2 ), and treat in detail a particular
group Γ. This Γ is one of the maximal arithmetic subgroups Γ̄ ≤ P U (2, 1) listed
in [4, 5] with covolume of the form 1/N , N an integer. In this case N = 864.
Consider the action of Γ on B(C2 ), and let 0 denote the origin in B(C2 ). Two
elements, denoted u and v, generate the stabilizer K of 0 in Γ. Another element b
of Γ was found by a computer search looking for elements g ∈ Γ for which d(g.0, 0)
is small. We use the results of Section 2 to show that the computer search did not
miss any such g, and to get a simple presentation of Γ. It turns out that this Γ is
one of the Deligne-Mostow groups (see Parker [3]).
1

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DONALD I. CARTWRIGHT TIM STEGER

We apply this presentation of Γ to exhibit a torsion-free subgroup Π of index 864.
The abelianization Π/[Π, Π] is Z2 , and so Π is not the fundamental group of a fakeprojective plane. However the ball quotient Π\B(C2 ) is a compact complex surface
with interesting properties, some of which we describe. By a lengthy computer
search not discussed here, we showed that any torsion-free subgroup of index 864
in Γ is conjugate to Π, and so no fake projective planes arise in this context.
In Section 4, we calculate the value of r0 for the example of the previous section.
2. General results
Let Γ be a group acting by isometries on a simply-connected geodesic metric
space X. Let S ⊂ Γ be a finite symmetric generating set for Γ. Fix a point 0 ∈ X,
and suppose that d(0, x) is bounded on the set
FS = {x ∈ X : d(0, x) ≤ d(g.0, x) for all g ∈ S}.

(2.1)

Define
r0 = sup{d(0, x) : x ∈ FS }.
Theorem 2.1. Suppose that there is a number r1 > 2r0 such that
(a) if g ∈ S, then d(g.0, 0) ≤ r1 ,
(b) if g, g ′ ∈ S and d((gg ′ ).0, 0) ≤ r1 , then gg ′ ∈ S.
Then S = {g ∈ Γ : d(g.0, 0) ≤ r1 }.
Corollary 2.1. For Γ, S as in Theorem 2.1, FS is equal to the Dirichlet fundamental domain F = {x ∈ X : d(0, x) ≤ d(g.0, x) for all g ∈ Γ} of Γ centered
at 0.
Proof. Clearly F ⊂ FS . If x ∈ FS \ F , pick a g ∈ Γ so that d(g.0, x) < d(0, x). Now
d(0, x) ≤ r0 because x ∈ FS , and so d(0, g.0) ≤ d(0, x)+d(x, g.0) ≤ 2d(0, x) ≤ 2r0 ≤
r1 , so that g ∈ S, by Theorem 2.1. But then d(0, x) ≤ d(g.0, x), a contradiction. 
Corollary 2.2. Assume Γ, S are as in Theorem 2.1, and that the action of each
g ∈ Γ \ {1} is nontrivial, so that Γ may be regarded as a subgroup of the group C(X)
of continuous maps X → X. Then Γ is discrete for the compact open topology.
Proof. Let V1 = {f ∈ C(X) : f (0) ∈ Br1 (0)}, where Br (x) = {y ∈ X : d(x, y) < r}.
Theorem 2.1 shows that Γ ∩ V1 ⊂ S. For each g ∈ S \ {1}, choose xg ∈ X so that
g.xg 6= xg , let rg = d(g.xg , xg ), and let Vg = {f ∈ C(X) : f (xg ) ∈ Brg (xg )}. Then
g 6∈ Vg . Hence the intersection V of the sets Vg , g ∈ S, is an open neighborhood
of 1 in C(X) such that Γ ∩ V = {1}.

Before starting the proof of Theorem 2.1, we need some lemmas, which have the
same hypotheses as Theorem 2.1.
Lemma 2.1. The group Γ is generated by S0 = {g ∈ S : d(g.0, 0) ≤ 2r0 }.
Proof. As S is finite, there is a δ > 0 so that d(g.0, 0) ≥ 2r0 + δ for all g ∈ S \ S0 .
Let Γ0 denote the subgroup of Γ generated by S0 , and assume that Γ0 $ Γ. Since
Γ is generated by S, there are g ∈ S \ Γ0 , and we choose such a g with d(g.0, 0) as
small as possible. Then d(g.0, 0) > 2r0 , since otherwise g ∈ S0 ⊂ Γ0 . In particular,
g.0 6∈ FS . Since 0 ∈ FS , there is a last point ξ belonging to FS on the geodesic
from 0 to g.0. Choose any point ξ ′ on that geodesic which is outside FS but satisfies
d(ξ, ξ ′ ) < δ/2. As ξ ′ 6∈ FS , there is an h ∈ S such that d(h.0, ξ ′ ) < d(0, ξ ′ ). Hence
d(h.0, g.0) ≤ d(h.0, ξ ′ ) + d(ξ ′ , g.0) < d(0, ξ ′ ) + d(ξ ′ , g.0) = d(0, g.0),

so that d((h−1 g).0, 0) < d(g.0, 0). So h−1 g cannot be in S \ Γ0 , by choice of g. Also,
d(h.0, 0) ≤ d(h.0, ξ ′ ) + d(ξ ′ , 0) < 2d(0, ξ ′ ) ≤ 2(d(0, ξ) + d(ξ, ξ ′ )) < 2r0 + δ.

FINDING GENERATORS AND RELATIONS

3

Hence h ∈ S0 , by definition of δ. Now h−1 g ∈ S by hypothesis (b) above, since
h−1 , g ∈ S and d((h−1 g).0, 0) < d(0, g.0) ≤ r1 . So h−1 g must be in Γ0 . But then
g = h(h−1 g) ∈ Γ0 , contradicting our assumption.

Lemma 2.2. If x ∈ X and if d(0, x) ≤ r0 + ǫ, where 0 < ǫ ≤ (r1 − 2r0 )/2, then
there exists g ∈ S such that g.x ∈ FS , and in particular, d(0, g.x) ≤ r0 .
Proof. Since S is finite, we can choose g ∈ S so that d(0, g.x) is minimal. If g.x ∈
FS , there is nothing to prove, and so assume that g.x 6∈ FS . There exists h ∈ S
such that d(h.0, g.x) < d(0, g.x), and so d(0, (h−1 g).x) = d(h.0, g.x) < d(0, g.x).
By the choice of g, h−1 g cannot be in S, and also d(0, g.x) ≤ d(0, x). So
d(0, (h−1 g).0) ≤ d(0, (h−1 g).x) + d((h−1 g).x, (h−1 g).0) = d(0, (h−1 g).x) + d(x, 0)
< d(0, g.x) + d(x, 0)
≤ 2d(0, x)
≤ 2r0 + 2ǫ ≤ r1 .

But this implies that h−1 g ∈ S by (b) again, since g, h ∈ S. This is a contradiction,
and so g.x must be in FS .

Remark 1. If in Lemma 2.2 we also assume that ǫ < δ/2, where δ is as in the
proof of Lemma 2.1, then the g in the last lemma can be chosen in S0 . For then
d(0, g.0) ≤ d(0, g.x) + d(g.x, g.0) ≤ 2d(0, x) ≤ 2(r0 + ǫ) < 2r0 + δ.
To prove Theorem 2.1, we consider a g ∈ Γ such that d(g.0, 0) ≤ r1 . By
Lemma 2.1, we can write g = y1 yn · · · yn , where yi ∈ S0 for each i. Since 1 ∈ S0 ,
we may suppose that n is even, and write n = 2m. Moreover, we can assume that
m is odd. Form an equilateral triangle ∆ in the Euclidean plane with horizontal
base whose sides are each divided into m equal segments, marked off by vertices
v0 , v1 , . . . , v3m−1 . We use these to partition ∆ into m2 subtriangles. We illustrate
in the case m = 3:
v3
•
v2 •

• v4
• v5

v1 •
v0 •

•
v8

•
v7

• v6

We define a continuous function ϕ from the boundary of ∆ to X which maps v0
to 0 and vi to (y1 · · · yi ).0 for i = 1, . . . , n, which for i = 1, . . . , n maps the segment
[vi−1 , vi ] to the geodesic from ϕ(vi−1 ) to ϕ(vi ), and which maps the bottom side
of ∆ to the geodesic from g.0 = (y1 · · · yn ).0 to 0.
Because X is simply-connected, we can extend ϕ to a continuous map ϕ : ∆ → X,
where ∆ here refers to the triangle and its interior.
Let ǫ > 0 be as in Lemma 2.2. For a sufficiently large integer r, which we choose
to be odd, by partitioning each of the above subtriangles of ∆ into r2 congruent
subtriangles, we have d(ϕ(t), ϕ(t′ )) < ǫ for all t, t′ in the same (smaller) subtriangle.
We now wish to choose elements x(v) ∈ Γ for each vertex v of each of the
subtriangles, so that
(i) d(x(v).0, ϕ(v)) ≤ r0 for each v not in the interior of the bottom side of ∆,
(ii) d(x(v).0, ϕ(v)) ≤ r1 /2 for the v 6= v0 , vn on the bottom side of ∆.
We shall also define elements y(e) for each directed edge e of each of the subtriangles
in such a way that
(iii) x(w) = x(v)y(e) if e is the edge from v to w.

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DONALD I. CARTWRIGHT TIM STEGER

Note that the y(e)’s are completely determined by the x(v)’s, and the x(v)’s are
completely determined from the y(e)’s provided that one x(v) is specified.
We first choose x(v) for the original vertices v = vi on the left and right sides of ∆
by setting x(v0 ) = 1 and x(vi ) = y1 · · · yi for i = 1, . . . , n. Thus d(x(v).0, ϕ(v)) =
0 ≤ r0 for these v’s. Now if w0 = vi−1 , w1 , . . . , wr = vi are the r + 1 equally spaced
vertices of the edge from vi−1 to vi , we set x(wj ) = x(vi−1 ) for 1 ≤ j < r/2 and
x(wj ) = x(vi ) for r/2 < j ≤ r − 1. Then if 1 ≤ j < r/2, we have
d(x(wj ).0, ϕ(wj )) = d(x(vi−1 ).0, ϕ(wj )) = d(ϕ(vi−1 ), ϕ(wj ))
1
≤ d(ϕ(vi−1 ), ϕ(vi ))
2
1
= d(0, yi .0) ≤ r0 ,
2

(∗)

where the inequality (∗) holds as ϕ on the segment from vi−1 to vi is the geodesic
from ϕ(vi−1 ) to ϕ(vi ). In the same way, d(x(wj ).0, ϕ(wj )) ≤ r0 for r/2 < j ≤ r − 1.
Having chosen x(v) for the v on the left and right sides of ∆, we set y(e) =
x(v)−1 x(w) if e is the edge from v to w on one of those sides. So of the r edges
in the segment from vi−1 to vi , we have y(e) = 1 except for the middle edge, for
which y(e) = yi .
For the vertices v on the bottom side of ∆, we set x(v) = 1 if v is closer to v0
than to vn , and set x(v) = g = y1 · · · yn otherwise, and we set y(e) = x(v)−1 x(w)
if e is the edge from v to w. Since rm is odd, there is no middle vertex on the side,
so there is no ambiguity in the definition of x(v), but there is a middle edge e, and
y(e) = g if e is directed from left to right. For the other edges e′ on the bottom
side, y(e′ ) = 1. If v is a vertex of the bottom side of ∆ closer to v0 than to vn , then
d(x(v).0, ϕ(v)) = d(0, ϕ(v)) ≤

1
r1
d(0, g.0) ≤ ,
2
2

since ϕ on the bottom side is just the geodesic from 0 to g.0. Similarly, if v is a
vertex of the bottom side of ∆ closer to vn , then again d(x(v).0, ϕ(v)) ≤ r1 /2.
We now choose x(v) for vertices which are not on the sides of ∆ as follows. We
successively define x(v) as we move from left to right on a horizontal line. Suppose
that v is a vertex for which x(v) has not yet been chosen, but for which x(v ′ ) has
been chosen for the vertex v ′ immediately to the left of v. Now d(ϕ(v), ϕ(v ′ )) ≤ ǫ
and d(x(v ′ ).0, ϕ(v ′ )) ≤ r0 , so that d(ϕ(v), x(v ′ ).0) ≤ r0 + ǫ. By Lemma 2.2, applied
to x = x(v ′ )−1 ϕ(v), there is a g ∈ S so that d(gx(v ′ )−1 ϕ(v), 0) ≤ r0 . So we set
x(v) = x(v ′ )g −1 . If e is the edge from v ′ to v, we set y(e) = g −1 ∈ S.
We have now defined x(v) for each vertex of the partitioned triangle ∆ so that
(i) and (ii) hold, and these determine y(e) for each directed edge e so that (iii)
holds. We have seen that y(e) ∈ S if e is an edge on the left or right side of ∆ or if
e is a horizontal edge not lying on the bottom side of ∆ whose right hand endpoint
does not lie on the right side of ∆. Also, y(e) = 1 ∈ S for all edges e in the bottom
side of ∆ except the middle one, for which y(e) = g. The theorem will be proved
once we check that y(e) ∈ S for each edge e.
Lemma 2.3. Suppose that v, v ′ and v ′′ are the vertices of a subtriangle in ∆, and
that e, e′ and e′′ are the edges from v to v ′ , v ′ to v ′′ , and v to v ′′ , respectively.
Suppose that y(e) and y(e′ ) are in S. Then y(e′′ ) ∈ S too.
Proof. We have y(e) = x(v)−1 x(v ′ ), y(e′ ) = x(v ′ )−1 x(v ′′ ) and y(e′′ ) = x(v)−1 x(v ′′ ),
and so y(e′′ ) = y(e)y(e′ ). If e′′ is the middle edge of the bottom side of ∆, then
d(y(e′′ ).0, 0) = d(g.0, 0) ≤ r1 by hypothesis, and so y(e′′ ) ∈ S by (b) above. If e′′
is any other edge of the bottom side of ∆, then y(e′′ ) = 1 ∈ S. So we may assume
that e′′ is not on the bottom side of ∆. Hence at most one of v and v ′′ lies on that

FINDING GENERATORS AND RELATIONS

5

bottom side. Hence
d(0, y(e′′ ).0) = d(x(v).0, x(v)y(e′′ ).0)
= d(x(v).0, x(v ′′ ).0)
≤ d(x(v).0, ϕ(v)) + d(ϕ(v), ϕ(v ′′ )) + d(ϕ(v ′′ ), x(v ′′ ).0)

< d(x(v).0, ϕ(v)) + d(ϕ(v ′′ ), x(v ′′ ).0) + ǫ
≤ r0 + r1 /2 + ǫ

(†)

≤ r1 ,

where the inequality (†) holds because at least one of d(x(v).0, ϕ(v)) ≤ r0 and
d(ϕ(v ′′ ), x(v ′′ ).0) ≤ r0 holds, since at most one of v and v ′′ is a vertex of the bottom
side of ∆, and if say v is such a point, then we still have d(x(v).0, ϕ(v)) ≤ r1 /2. 
Conclusion of the proof of Theorem 2.1. We must show that y(e) ∈ S for all edges.
We use Lemma 2.3, working down from the top of ∆, and moving from left to
right. So in the order indicated in the next diagram we work down ∆, finding that
y(e) ∈ S in each case, until we get to the lowest row of triangles.
•
•

e1

e2
•

e4

e5 e6

•

•
e3

•

e7

•
e8

•

e9

•

Then working from the left and from the right, we find that y(e) ∈ S for all the
diagonal edges in the lowest row.
Finally, we get to the middle triangle in the lowest row. For the diagonal edges
e, e′ of that triangle, we have found that y(e), y(e′ ) ∈ S, and so y(e′′ ) ∈ S for the
horizontal edge e′′ of that triangle too, by Lemma 2.3.

If we make the extra assumption that the set of values d(g.0, 0), g ∈ Γ, is discrete,
the following result is a consequence of [1, Theorem I.8.10] (applied to the open set
U = {x ∈ X : d(x, 0) < r0 + ǫ} for ǫ > 0 small). We shall sketch a proof along the
lines of that of Theorem 2.1 which does not make that extra assumption.
Theorem 2.2. With the hypotheses of Theorem 2.1, let S0 = {g ∈ S : d(g.0, 0) ≤
2r0 }, as before. Then the set of generators S0 , and the relations g1 g2 g3 = 1, where
the gi are each in S0 , give a presentation of Γ.
Proof. Let S̃0 be a set with a bijection f : s̃ 7→ s from S̃0 to S0 . Let F be the free
group on S̃0 , and denote also by f the induced homomorphism F → Γ. Then f is
surjective by Lemma 2.1. Let ỹ1 , . . . , ỹn ∈ S̃0 , and suppose that g̃ = ỹ1 · · · ỹn ∈ F
is in the kernel of f . We must show that g̃ is in the normal closure H of the set of
elements of F of the form s̃1 s̃2 s̃3 , where s̃i ∈ S̃0 for each i, and s1 s2 s3 = 1 in Γ.
As 1 ∈ S0 , we may assume that n is a multiple 3m of 3, and form a triangle ∆
partitioned into m2 congruent subtriangles, as in the proof of Theorem 2.1. The
vertices are again denoted vi , and we write v3m = v0 , and yi = f (ỹi ) for each i.
We again define a continuous function ϕ from the boundary of ∆ to X which
maps v0 to 0 and vi to (y1 · · · yi ).0 for i = 1, . . . , 3m, and which for i = 1, . . . , 3m
maps the segment [vi−1 , vi ] to the geodesic from ϕ(vi−1 ) to ϕ(vi ). From y1 y2 · · · yn =
1 we see that ϕ(v0 ) = 1 = g = ϕ(v3m ). This time the bottom side of ∆ is mapped
in the same way as the other two sides.

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DONALD I. CARTWRIGHT TIM STEGER

Let ǫ > 0, with ǫ < δ/2, (r1 − 2r0 )/2, and as before, we partition ∆ into subtriangles so that whenever t, t′ are in the same subtriangle, d(ϕ(t), ϕ(t′ )) < ǫ holds.
Using Lemma 2.2 and the remark after it, we can again choose elements x(v) ∈ Γ
for each vertex v of each of the subtriangles, so that d(x(v).0, ϕ(v)) ≤ r0 for each v,
without the complications about the bottom side of ∆ which had to be dealt with
in the proof of Theorem 2.1. If e is the edge from v ′ to v, where v ′ is immediately
to the left of v, and v is not on the right side of ∆, then y(e) = x(v ′ )−1 x(v) ∈ S0 .
We then define elements y(e) for each directed edge e of each of the subtriangles so
that x(w) = x(v)y(e) if e is the edge from v to w. Again using Lemma 2.3, with S
there replaced by S0 , and arguing as in the conclusion of the proof of Theorem 2.1,
we deduce that y(e) ∈ S0 for each edge e.
Of the r edges e in the segment from vi−1 to vi , we have y(e) = 1 except for the
middle edge, for which y(e) = yi . So y(e1 )y(e2 ) · · · y(e3mr ) = 1, where e1 , . . . , e3mr
are the successive edges as we traverse the sides of ∆ in a clockwise direction starting
at v0 , and g̃ = ỹ(e1 ) · · · ỹ(e3mr ).
It is now easy to see that g̃ ∈ H, by using y(e′′ ) = y(e)y(e′ ) in the situation
of (1), so that ỹ(e′′ ) = ỹ(e)ỹ(e′ )h for some h ∈ H. Then in the situation of (2), for
example, y(e)y(e′ ) = y(e)(y(e∗ )y(e′′′ )) = (y(e)y(e∗ ))y(e′′′ ) = y(e′′ )y(e′′′ ),
•

•
e

•
∗

′

e

e

e′′

•

e

(1)

e′

e

′′

′′′

•

•

e

(2)

•

so that ỹ(e)ỹ(e′ ) = ỹ(e)(ỹ(e∗ )ỹ(e′′′ )h) = (ỹ(e)ỹ(e∗ ))ỹ(e′′′ )h = (ỹ(e′′ )h′ )ỹ(e′′′ )h =
ỹ(e′′ )ỹ(e′′′ )h′′ for some h, h′ , h′′ ∈ H. We can, for example, successively use this
device to remove the right hand strip of triangles from ∆, reducing the size of the
triangle being treated, and then repeat this process.

In the next proposition, we use [1, Theorem I.8.10] to show that under extra
hypotheses (satisfied by the example in Section 3), we can omit the g ∈ S0 for
which d(g.0, 0) = 2r0 , and still get a presentation.
Proposition 2.1. Let X = B(C2 ), and suppose that the set of values d(g.0, 0),
g ∈ Γ, is discrete. Assume also the hypotheses of Theorem 2.1. Then the set
S0∗ = {g ∈ S : d(g.0, 0) < 2r0 } is a set of generators of Γ, and the relations
g1 g2 g3 = 1, where the gi are each in S0∗ , give a presentation of Γ.
Proof. For each g ∈ S such that d(g.0, 0) = 2r0 , let m be the midpoint of the
geodesic from 0 to g.0. Let M be the set of these midpoints. Let δ1 > 0 be so small
that 2r0 + 2δ1 < r1 . Since M and S are finite, we can choose a positive δ < δ1 so
that if m, m′ ∈ M and g ∈ S, and if d(g.m, m′ ) < 2δ, then g.m = m′ . So if γ, γ ′ ∈ Γ,
m, m′ ∈ M and B(γ.m, δ) ∩ B(γ ′ .m′ , δ) 6= ∅, then d(γ.0, γ ′ .0) < 2r0 + 2δ < r1 , so
that γ −1 γ ′ ∈ S by Theorem 2.1, and d(m, γ −1 γ ′ .m′ ) < 2δ, so that γ.m = γ ′ .m′ by
choice of δ. Thus if Y is the union of the Γ-orbits of all m ∈ M , then the balls
B(y, δ), y ∈ Y , are pairwise disjoint. Now let X ′ denote the subset of X obtained
by removing all these balls. It follows (because the ambient dimension is > 2) that
X ′ is still simply connected.
Let U = {x ∈ X ′ : d(x, 0) < r0 + δ ′ } of X ′ , for some δ ′ > 0. The proposition will
follow from [1, Theorem I.8.10], applied to this U , once we show that if δ ′ > 0 is
small enough, then any g ∈ Γ such that g(U ) ∩ U 6= ∅ must satisfy d(g.0, 0) < 2r0 .
Clearly d(g.0, 0) < 2r0 + 2δ ′ holds, and because of the discreteness hypothesis, if
δ ′ > 0 is small enough, one even has d(g.0, 0) ≤ 2r0 . Now suppose d(g.0, 0) = 2r0 ,
let m be the midpoint on the geodesic from 0 to g.0, and suppose x ∈ g(U ) ∩ U .
Then d(0, x) < r0 + δ ′ and d(g.0, x) < r0 + δ ′ . Using the CAT(0) property of X,

FINDING GENERATORS AND RELATIONS

7

this shows that d(x, m)2 + r02 ≤ (r0 + δ ′ )2 , so that
q
d(x, m) ≤ (r0 + δ ′ )2 − r02 ,

and this last can be made less than δ if δ ′ is chosen small enough. This contradicts
the hypothesis that x ∈ U ⊂ X ′ .

The fact that X ′ is simply connected could also be used to modify the version
of the proof going through the triangle-shaped simplicial complex.
3. An example
Let ℓ = Q(ζ), where ζ is a primitive 12-th root of unity. Then ζ 4 = ζ 2 − 1, so
that ℓ is a degree 4 extension of Q. Let r = ζ + ζ −1 and k = Q(r). Then r√and ζ 3
are square roots of 3 and −1, respectively, and if ζ = e2πi/12 then r = + 3 and
ζ 3 = i. Let


−r − 1
1
0
1 − r 0 ,
F = 1
(3.1)
0
0
1
and form the group
Γ = {g ∈ M3×3 (Z[ζ]) : g ∗ F g = F }/{ζ ν I : ν = 0, 1, . . . , 11}.

We shall use the results of Section 2 to find a presentation for Γ. Let us first
motivate the choice of this example. Now ι(g) = F −1 g ∗ F defines an involution of
the second kind on the simple algebra M3×3 (ℓ). We can define an algebraic group
G over k so that
G(k) = {g ∈ M3×3 (ℓ) : ι(g)g = I and det(g) = 1}.
For the corresponding adjoint group G,
G(k) = {g ∈ M3×3 (ℓ) : ι(g)g = I}/{tI : t ∈ ℓ and t̄t = 1}.

Now k has two archimedean
√ places√v+ and v− , corresponding to the two embeddings
k ֒→ R√mapping r to + 3 and − 3, respectively. The eigenvalues of F are 1 and
−r ± 2. So the form F is definite for v− but not for v+ . Hence
∼ P U (3) and G(kv ) =
∼ P U (2, 1).
G(kv ) =
+

−

Letting Vf denote the set of non-archimedean places of k, if v ∈ Vf , then either
G(kv ) ∼
= P GL(3, kv ) if v splits in ℓ, or G(kv ) ∼
= P UF (3, kv (i)) if v does not split
in ℓ. With a suitable choice of maximal parahorics P v in G(kv ), the following group
Y
Pv
Γ̄ = G(k) ∩
v∈Vf

is one of the maximal arithmetic subgroups of P U (2, 1) whose covolume has the
form 1/N , N an integer. Prasad and Yeung [4, 5] have described all such subgroups,
up to k-equivalence. In this case N = 864. As in [2], lattices can be used to describe
concretely maximal parahorics. We can take P v = {g ∈ G(kv ) : g.xv = xv }, where,
in the cases when v splits in ℓ, xv is the homothety class of the ov -lattice o3v ⊂ kv3 ,
where ov is the valuation ring of kv . When v does not split, xv is the lattice
o3v , where now ov is the valuation ring of kv (i). With this particular choice of
parahorics, Γ̄ is just Γ.
The action of Γ on the unit ball X = B(C2 ) in C2 is described as follows, making
explicit the isomorphism G(kv+ ) ∼
= P U (2, 1). Let






1
0
0
1 0
0
1 0 0
0  , and F0 = 0 1 0  . (3.2)
γ0 = 1 1 − r 0 , Fdiag = 0 1
0
0
1
0 0 1−r
0 0 −1

8

DONALD I. CARTWRIGHT TIM STEGER

Then γ0t Fdiag γ0 = (1 − r)F , so a matrix g is unitary with respect to F if and only if
g ′ = γ0 gγ0−1 is unitary with
to Fdiag . Now let D be the diagonal matrix with
prespect
√
√
diagonal entries 1, 1 and
3 − 1. Taking r = + 3, if g ′ is unitary with respect
to Fdiag , then g̃ = Dg ′ D−1 is unitary with respect to F0 ; that is, g̃ ∈ U (2, 1). If
Z = {ζ ν I : ν = 0, . . . , 11}, for an element gZ of Γ, the action of gZ on B(C2 ) is
given by the usual action of g̃. That is,
 
 ′
z
z
(gZ).(z, w) = (z ′ , w′ ) if g̃ w  = λ w′  for some λ ∈ C.
1
1
Now let u and v be the matrices

 3
ζ + ζ2 − ζ 1 − ζ 0
u = ζ 3 + ζ 2 − 1 ζ − ζ 3 0
0
0
1

ζ3
3
2

and v = ζ + ζ − ζ − 1
0



0 0
1 0 ,
0 1

respectively. They have entries in Z[ζ], are unitary with respect to F , and satisfy
u3 = I, v 4 = I, and (uv)2 = (vu)2 .
They (more precisely, uZ and vZ) generate a subgroup K of Γ of order 288. Magma
shows that an abstract group with presentation hu, v : u3 = v 4 = 1, (uv)2 = (vu)2 i
has order 288, and so K has this presentation.
Let us write simply 0 for the origin (0, 0) ∈ B(C2 ), and g.0 in place of (gZ).(0, 0).

Lemma 3.1. For the action of Γ on X, K is the stabilizer of 0.

Proof. It is easy to see that gZ ∈ Γ fixes 0 if and only if gZ has a matrix representative


g11 g12 0
g21 g22 0 ,
0
0 1
for suitable g√ij ∈ Z[ζ]. Since the 2 × 2 block matrix in the upper left of F is definite
when r = + 3, it is now routine to determine all such gij .

The next step is to find g ∈ Γ \ K for which d(g.0, 0) is small, where d is the
hyperbolic metric on B(C2 ). Now
cosh2 (d(z, w)) =

|1 − hz, wi|2
,
(1 − |z|2 )(1 − |w|2 )

(3.3)

p
(see [1, Page 310] for example) where hz, wi = z1 w̄1 + z2 w̄2 and |z| = |z1 |2 + |z2 |2
for z = (z1 , z2 ) and w = (w1 , w2 ) in B(C2 ).
In particular, writing 0 for the origin in B(C2 ), and using g.0 = (g13 /g33 , g23 /g33 )
and |g13 |2 + |g23 |2 = |g33 |2 − 1 for g = (gij ) ∈ U (2, 1), we see that
cosh2 (d(0, g.0)) = |g33 |2

(3.4)

for g ∈ U (2, 1). Notice that for gZ ∈ Γ, the (3, 3)-entry of g is equal to the
(3, 3)-entry of the g̃ ∈ U (2, 1) defined above, and so (3.4) holds also for gZ ∈ Γ.
The matrix


1
0
0
−ζ 3 − ζ 2 
b = −2ζ 3 − ζ 2 + 2ζ + 2 ζ 3 + ζ 2 − ζ − 1
2
3
ζ +ζ
−ζ − 1
−ζ 3 + ζ + 1

is unitary with respect to F . We shall see below that u, v and b generate Γ, and use
the results of Section 2 to show that some relations they satisfy give a presentation
of Γ. This b was found by a computer search for g ∈ Γ \ K for which d(g.0, 0) is
small.
Notice that d(g.0, 0) is constant on each double coset KgK. Calculations using (3.4) showed that amongst the 288 elements g ∈ Γ of the form bkb, k ∈ K,

FINDING GENERATORS AND RELATIONS

9

there are ten different values of d(g.0, 0). Representatives γj of the 20 double
cosets KgK in which the g ∈ bKb lie were chosen. The smallest few |(γj )33 |2 and
the corresponding d(γj .0, 0) (rounded to 4 decimal places) are as follows:
j

γj

1

1

2

b

3

−1

4

bu
bu

|(γj )33 |2
1

b

−1 −1 −1

v

u

b

√
3+2
√
2 3+4
√
3 3+6

d(γj .0, 0)
0
1.2767
1.6629
1.8778

We use Theorem 2.1 to show that our computer search has not missed any g ∈ Γ for
which d(g.0, 0) is small. Let S = K ∪ Kγ2 K√∪ Kγ3 K, consisting of the g ∈ Γ found
by the computer search to satisfy |g33 |2 ≤ 2 3+4. To verify that S generates Γ, we
need to make a numerical estimate. A somewhat longer direct proof that hSi = Γ
is given in the next section.
Proposition 3.1. For the given S ⊂ Γ, the normalized hyperbolic volume vol(FS )
of FS = {x ∈ B(C2 ) : d(x, 0) ≤ d(x, g.0) for all g ∈ S} satisfies vol(FS ) < 2/864.
The set S generates Γ.
Proof. Standard numerical integration methods show that (up to several decimal
place accuracy) vol(FS ) equals 1/864, but all we need is that vol(FS ) < 2/864. Let
us make explicit the normalization of hyperbolic volume element in B(C2 ) which
makes the formula χ(Γ\B(C2 )) = 3vol(F ) true. For z ∈ B(C2 ), write
 1 + |z| 
1
t = tanh−1 |z| = log
= d(0, z),
2
1 − |z|
Θ = z/|z| ∈ S 1 (C2 ),

dΘ = the usual measure on S 1 (C2 ), having total volume 2π 2 ,
2
dvol(z) = 2 sinh3 (t) cosh(t)dt dΘ.
π
Let Γ′ = hSi. Then the Dirichlet fundamental domains F and F ′ of Γ and Γ′ satisfy
F ⊂ F ′ ⊂ FS . By [4, §8.2, the C11 case], vol(F ) = 1/864. Let M = [Γ : Γ′ ]. Then
vol(F ′ ) = M vol(F ) (and vol(F ′ ) = ∞ if M = ∞), so that M vol(F ) = vol(F ′ ) ≤
vol(FS ) < 2vol(F ) implies that M = 1 and Γ′ = Γ.

Proposition 3.2. For the
√ given S ⊂ Γ, the value of r0 = sup{d(x, 0) : x ∈ FS }
is 21 d(γ3 .0, 0) = 21 cosh−1 ( 3 + 1).
Proof. We defer the proof of this result to the next section.


√
′
′
′
2
One may verify
√ that if g, g2 ∈ S and√gg 6∈ S, then |(gg )33 | ≥ 3 3 + 6. So
if r1 satisfies 2 3 + 4 < cosh (r1 ) < 3 3 + 6, then S satisfies the conditions of
Theorem 2.1.
Remark 2. By Corollary 2.1, FS = F , so that vol(FS ) equals 1/864.
Having verified all the conditions of Theorem 2.1, we now know that S contains
all elements g ∈ Γ satisfying d(g.0, 0) ≤ r1 . The double cosets Kγj K, j = 1, 2, 3,
are symmetric because (buvu)2 v = ζ −1 I, and (bu−1 )4 = I (see below). The sizes of
these Kγj K are 288, 2882/4 and 2882 /3, respectively, adding up to 48, 672, because
{k ∈ K : γ2−1 kγ2 ∈ K} = hvi, and {k ∈ K : γ3−1 kγ3 ∈ K} = hui. Proposition 2.1
gives a presentation of Γ which we now simplify.

10

DONALD I. CARTWRIGHT TIM STEGER

Proposition 3.3. A presentation of Γ is given by the generators u, v and b and
the relations
u3 = v 4 = b3 = 1, (uv)2 = (vu)2 , vb = bv, (buv)3 = (buvu)2 v = 1.

(3.5)

Proof. By Proposition 2.1, the set S0∗ = {g ∈ Γ : d(g.0, 0) < 2r0 } of generators,
and the relations gi gj gk = 1, where g1 , g2 , g3 ∈ S0∗ , give a presentation of Γ. By
Theorem 2.1, S0∗ is the union of the two double cosets Kγ1 K = K and Kγ2 K. So
these relations have the form (k1′ γi1 k1′′ )(k2′ γi2 k2′′ )(k3′ γi3 k3′′ ) = 1, where kν′ , kν′′ ∈ K
and i1 , i2 , i3 ∈ {1, 2}. Using the known presentation of K, and cyclic permutations,
the relations of the form γi1 k1 γi2 k2 γi3 k3 = 1, where 1 ≤ i1 ≤ i2 , i3 ≤ 2, are
sufficient to give a presentation. After finding a word in u and v for each such ki ,
we obtained a list of words in u, v and b coming from these relations. Magma’s
routine Simplify can be used to complete the proof, but it is not hard to see this
more directly as follows. When i1 = 1, we need only include in the list of words
all those coming from the relations of the form (i): bkb = k ′ , for k, k ′ ∈ K. When
i1 = 2 (so that i2 = i3 = 2 too), we need only include, for each k1 ∈ K such that
bk1 b ∈ KbK, a single pair (k2 , k3 ) such that bk1 bk2 bk3 = 1, since the other such
relations follow from this one and the relations (i). The only relations of the form (i)
are the relations bv ν uvub = v ν−1 u−1 v −1 u−1 , ν = 0, 1, 2, 3, which follow from the
relations vb = bv and (buvu)2 v = 1 in (3.5). Next, matrix calculations show that
there are 40 elements k ∈ K such that bkb ∈ KbK, giving 40 relations bkb = k ′ bk ′′ .
We need only show that all of these are deducible from the relations given in (3.5).
Using bv = vb, any equation bkb = k ′ bk ′′ gives equations bv i kv j b = v i k ′ bk ′′ v j . So
we needed only deduce from (3.5) the five relations
b1b = uvuvbuvu,
buvub = (vu)−2 ,

bub = ubu,

buvu−1 b = v −1 u−1 v −1 bu−1 ,

and buvu−1 vub = v 2 u−1 v −1 u−1 bu−1 v −1 u−1 .

Firstly, (buvu)2 v = 1 and (vu)2 = (uv)2 imply that b−1 = vuvubuvu, and this
and b3 = 1 imply the first relation. The relations (bvu)3 = 1 and bv = vb imply
that bub = (vubvuv)−1 , and this and b−1 = vuvubuvu give bub = ubu. To get the
third relation, use vb = bv to see that v(buvu−1 b)u equals
b(vuvu)ubu = b(vu)2 bub = b(vu)2 b(vu)2 (vu)−2 ub = v(uv)−2 u = u−1 v −1 b.
The fourth relation is immediate from (buvu)2 v = 1 and bv = vb. Finally, from
(uv)2 = (vu)2 and our formula for b−1 we have
(uvu)−1 b−1 (uvu) = (uvu)−1 (uvuvbuvu)(uvu) = vbuvu2 vu = vbk
for k = uvu−1 vu, using u3 = 1. So vbk has order 3. Hence bkb = v −1 (kvbv)−1 v −1 ,
and the fifth relation easily follows.

As mentioned above, (bu−1 )4 = 1. By Proposition 3.3, this is a consequence of
the relations in (3.5). Explicitly,
(bu−1 )4 = bu(ubu)(ubu)(ubu)u = bu(bub)(bub)(bub)u = b(ubu)bbubb(ubu)
= b(bub)bbubb(bub)
= bbuuub
= 1.
Let us record here the connection between Γ and a Deligne-Mostow group whose
presentation (see Parker [3]) is
Γ3,4 = hJ, R1 , A1 : J 3 = R13 = A41 = 1, A1 = (JR1−1 J)2 , A1 R1 = R1 A1 i.

Using the fact that the orbifold Euler characteristics of Γ3,4 \B(C2 ) and Γ\B(C2 )
are both equal to 1/288, several experts, including John Parker, Sai-Kee Yeung and
Martin Deraux, knew that Γ and Γ3,4 are isomorphic. The following proof based on

FINDING GENERATORS AND RELATIONS

11

presentations is a slight modification of one communicated to us by John Parker.
It influenced our choice of the generators u and v for K.
Proposition 3.4. There is an isomorphism ψ : Γ → Γ3,4 such that
ψ(u) = JR1 J −1 ,

Its inverse satisfies ψ

−1

ψ(v) = A1 ,

(J) = buv, ψ

−1

and

(R1 ) = b and ψ

ψ(b) = R1 .
−1

(3.6)

(A1 ) = v.

Proof. Setting R2 = JR1 J −1 , we have R1 R2 A1 = J and A1 JR2 = JR1−1 J. So
(ψ(u)ψ(v))2 = (R1−1 J)2 = R1−1 ·A1 JR2 = A1 R1−1 JR2 = A1 R2 A1 R2 = (ψ(v)ψ(u))2 .

Next, ψ(b)ψ(u)ψ(v) = R1 R2 A1 = J, which implies that (ψ(b)ψ(u)ψ(v))3 = 1. Now
(ψ(b)ψ(u)ψ(v)ψ(u))2 ψ(v) = (R1 R2 A1 R2 )2 A1 = (JR2 )2 A1 = JR2 JR1−1 J = 1. So
there is a homomorphism ψ : Γ → Γ3,4 satisfying (3.6). We similarly check that
we have a homomorphism ψ̃ : Γ3,4 → Γ mapping J, R1 and A1 to buv, b and v,
respectively, and that ψ and ψ̃ are mutually inverse.


We now exhibit a torsion-free subgroup of Γ having index 864. It has three
generators, all in KbK. The elements of K are most neatly expressed if we use not
only the generators u and v, but also j = (uv)2 , which is the diagonal matrix with
diagonal entries ζ, ζ and 1, and which generates the center of K.
Lemma 3.2. The non-trivial elements of finite order in Γ have order dividing 24.
(i) Any element of order 2 is conjugate to one of v 2 , j 6 or (bu−1 )2 .
(ii) Any element of order 3 is conjugate to one of u, j 4 , uj 4 , buv, or their
inverses.
Proof. By [1, Corollary II.2.8(1)] for example, any g ∈ U (2, 1) of finite order fixes
at least one point of B(C2 ), and so in particular this holds for any g ∈ Γ of finite
order. Conjugating g, we may assume that the fixed point is in the fundamental
domain F of Γ, and so d(g.0, 0) ≤ 2r0 . Thus g lies in K ∪ Kγ2 K ∪ Kγ3 K, and so
is conjugate to an element of K ∪ γ2 K ∪ γ3 K. Checking these 864 elements, we
see that g must have order dividing 24. After listing the elements of order 2 and 3
amongst them, routine calculations verify (i) and (ii).

Proposition 3.5. The elements
a1 = vuv −1 j 4 buvj 2 ,

a2 = v 2 ubuv −1 uv 2 j

a3 = u−1 v 2 uj 9 bv −1 uv −1 j 8
∼ Z2 .
generate a torsion-free subgroup Π of 864, for which Π/[Π, Π] =
and

Proof. Using our presentation of Γ, Magma’s Index command verifies that Π has
index 864 in Γ, and the AbelianQuotientInvariants command verifies that it has
abelianization Z2 .
We now check that Π is torsion-free. Suppose that Π contains an element π 6= 1
of finite order. By Lemma 3.2, we can assume that π has order 2 or 3. So for one
of the elements t listed in (i) and (ii) of Lemma 3.2, there is a g ∈ Γ such that
gtg −1 ∈ Π. Using Index, one verifies that the 864 elements bµ k, µ = 0, 1, 2, k ∈ K,
form a transversal for Π in Γ. So we may assume that g = bµ k. But now Index
verifies that none of the elements bµ kt(bµ k)−1 is in Γ.

We conclude this section by mentioning some other properties of Π.
Let us first note that Π cannot be lifted to a subgroup of SU (2, 1). The determinants of a1 , a2 and a3 are ζ 3 , ζ 3 and −1, respectively, and so the aν could be replaced
by ζ −1 ω i1 a1 , ζ −1 ω i2 a2 and −ω i3 a3 , where ω = e2πi/3 and i1 , i2 , i3 ∈ Z, to obtain
−3 3 −1 −1 −1
−1
3
−4
generators with determinant 1. But a−3
I is
2 a3 a1 a2 a3 a2 a3 a1 a2 a1 a3 a1 = ζ
unchanged by any choice of the integers i1 , i2 and i3 , as the number of aν ’s appearing in the product on the left is equal to the number of aν−1 ’s, for each ν. So we get

12

DONALD I. CARTWRIGHT TIM STEGER

a relation in P U (2, 1) but not in SU (2, 1). It was found using Magma’s Rewrite
command, which derives a presentation of Π from that of Γ̄.
Magma shows that the normalizer of Π in Γ contains Π as a subgroup of index 3,
and is generated by Π and j 4 . One may verify that
3
j 4 a1 j −4 = ζ 3 a3 a−3
2 a3 a1 ,

j 4 a2 j −4 = ζ −1 a3−1 ,

and

−1
−1 −1
2 −1 −1
j 4 a3 j −4 = ζ −1 a−1
1 a2 a1 a2 a1 a2 a1 a3 a1 a2 a1 .

Let us show that j 4 induces a non-trivial action on Π/[Π, Π]. By Proposition 3.5,
there is an isomorphism ϕ : Π/[Π, Π] → Z2 , so we have a surjective homomorphism
−1
−3 3
3
3
f : Π → Π/[Π, Π] ∼
= Z2 . Using the relation a22 a−1
1 a2 a1 a3 a1 a2 a3 a1 a3 a1 = ζ I, we
2
∼
see that 3f (a1 ) − 2f (a2 ) + 7f (a3 ) = (0, 0), and since Π/[Π, Π] = Z , this must be
the only condition on the f (aν ). So we can choose the isomorphism ϕ so that f
maps a1 , a2 and a3 to (1, 3), (−2, 1) and (−1, −1), respectively, and then


0 −1
4
−4
f (π) = (m, n)
=⇒ f (j πj ) = (m, n)
for all π ∈ Π.
(3.7)
1 −1

Next consider the ball quotient X = Π\B(C2 ). Now j 4 induces an automorphism
of X. Let us show that this automorphism has precisely 9 fixed points.

Proposition 3.6. The automorphism of X induced by j 4 has exactly 9 fixed points.
These are the three points Π(bµ .0), µ = 0, 1, −1, and six points Π(hi .z0 ), where
hi ∈ Γ for i = 1, . . . , 6, and where z0 ∈ B(C2 ) is the unique fixed point of buv.
Proof. If Π(j 4 .z) = Πz, then πj 4 .z = z for some π ∈ Π. This implies that πj 4
has finite order. It cannot be trivial, since Π is torsion-free. If π ∈ Π, then
π ′ = (πj 4 )3 = (π)(j 4 πj 8 )(j 8 πj 4 ) is also in Π. Since the possible orders of the
elements of Γ are the divisors of 24, if πj 4 has finite order, then 1 = (πj 4 )24 = (π ′ )8 ,
so π ′ must be 1, so that (πj 4 )3 must be 1. So πj 4 must have order 3. So for one
of the eight elements t listed in Lemma 3.2(ii), πj 4 = gtg −1 for some g ∈ Γ. Thus
gtg −1 j −4 ∈ Π. Since the elements bµ k, µ = 0, 1, −1 and k ∈ K, form a set of coset
representatives for Π in Γ, and since j 4 normalizes Π, we can assume that g = bµ k
for some µ and k.
When t = j 4 , we have bµ ktk −1 b−µ j −4 = bµ j 4 b−µ j −4 , independent of k. We find
that these three elements are in Π. Explicitly, bµ j 4 b−µ j −4 = πµ for
−3 −1
−1
2
π0 = 1, π1 = ζ −4 a2 a−2
1 a3 a1 and π−1 = a2 a1 a3 a1 .
µ

(3.8)
4

and these equations mean that the three points Π(b .0) are fixed by j .
Write πj 4 = gtg −1 for some g ∈ Γ, where t is one of the eight elements t listed
in Lemma 3.2(ii). In the notation of (3.8), and writing g = π ′ bµ k, where π ′ ∈ Π,
µ ∈ {0, 1, −1}, and k ∈ K, we get
πj 4 = π ′ bµ ktk −1 b−µ π ′

−1

= π ′ bµ ktk −1 (j −4 b−µ πµ j 4 )π ′

−1

= π ′ (bµ k)(tj −4 )(bµ k)−1 (πµ j 4 π ′

−1 −4

j

)j 4 .

So (bµ k)(tj −4 )(bµ k)−1 is in Π, and therefore either t = j 4 or tj −4 has infinite
order. In particular, apart from t = j 4 , our t cannot be in K, and so must be buv
or (buv)−1 .
We find that bµ ktk −1 b−µ j −4 ∈ Π never occurs when t = (buv)−1 . For t = buv,
we find that bµ ktk −1 b−µ j −4 ∈ Π for only 18 pairs (µ, k). This means that j 4 fixes
Π(bµ k.z0 ) for these 18 (µ, k)’s. If (µ, k) satisfies bµ ktk −1 b−µ j −4 ∈ Π, then so does
(µ, kj 4 ), since we can write bµ j 4 = πµ j 4 bµ for some πµ ∈ Π, as we have just seen.
Moreover, Π(bµ kj 4 .z0 ) = Π(bµ k.z0 ), since kj 4 = j 4 k and so
Π(bµ kj 4 .z0 ) = Π(πµ j 4 bµ k.z0 ) = Π(j 4 bµ k.z0 ) = Π(bµ k.z0 ).

FINDING GENERATORS AND RELATIONS

13

So we need only consider six of the (µ, k)’s, and correspondingly setting
h1 = b−1 vuj 3 ,

h2 = u−1 vj,

h3 = buv 2 j 2 ,

h4 = b−1 v 2 uj 3 ,

h5 = vj 2 ,

h6 = bvu−1 v,

−4
we have hi (buv)h−1
= πi′ ∈ Π for i = 1, . . . , 6; explicitly,
i j

π1′ = ζ 4 a22 a1 a33 ,

π2′ = j 8 a1 j 4 ,

−1
π3′ = ζ 2 j 8 a1 a32 j 4 a2 a1 a−2
2 a1 .

π4′ = ζ −5 a33 a21 a33 ,

−1 8
π5′ = ζ −1 j 4 a−1
1 a2 j ,

π6′ = ζa2 a−1
1 .

The six points Π(hi .z0 ) are distinct, as we see by checking that (a) the nontrivial
′
g ∈ Γ fixing z0 are just (buv)±1 , and (b) (bµ k ′ )(buv)ǫ (bµ k)−1 is not in Π for
ǫ = 0, 1, 2, when (µ′ , k ′ ) and (µ, k) in the above list of six pairs are distinct.

Finally, we show that Π is a congruence subgroup of Γ.
The prime 3 ramifies in Q(ζ) (as does 2), and F9 = Z[ζ]/rZ[ζ] is a field of order 9.
Let ρ : Z[ζ] → F9 be the natural map, and write i for ρ(ζ). Then i2 = −1, and
F9 = F3 (i). Applying ρ to matrix entries, we map Γ to a group of matrices over F9 ,
modulo hii. The image ρ(g) of any g ∈ M3×3 (Z[ζ]) unitary with respect to the F
of (3.1) is unitary with respect to ρ(F ), and so if we conjugate by C = ρ(γ0 ), where
γ0 is as in (3.2), then ρ′ (g) = Cρ(g)C −1 is unitary in the “usual” way.
So ρ′ maps Γ to the group P U (3, F9 ) of unitary matrices with entries in F3 (i),
modulo scalars. This map is surjective. In fact, ρ′ (Γ1 ) = P U (3, F9 ), where Γ1
is the normal index 3 subgroup of Γ consisting of the gZ ∈ Γ having a matrix
representative g of determinant 1. One may check that Γ1 = hv, bu−1 , u−1 bi, and
that hρ′ (v), ρ′ (bu−1 ), ρ′ (u−1 b)i = P U (3, F9 ).
The given generators a1 , a2 and a3 of Π have determinants ζ 3 , ζ 3 and −1, respectively, and so Π ⊂ Γ1 . Now −ζa2 and −a1 a2 are mapped by ρ′ to the matrices




−i −i − 1
i
i
−i
i+1
i−1
−1  , and M = −i − 1
i
−i  ,
R= 1
i−1
0
i−1
i
−i − 1
i
respectively, which satisfy R7 = I, M 3 = I and M RM −1 = R2 . Moreover, −a3
is mapped to R−1 . Hence Π is mapped onto the subgroup hR, M i of P U (3, F9 ),
which has order 21. Now |P U (3, F9 )| = 6048 = 288 × 21, and so the conditions on
a gZ ∈ Γ to be in Π are that gZ ∈ Γ1 and that ρ′ (g) ∈ hR, M i.
4. Calculation of r0 .
For any symmetric set S ⊂ U (2, 1), the following lemma simplifies the description
of the set FS defined in (2.1) in the case X = B(C2 ).
Lemma 4.1. If g ∈ U (2, 1) and z = (z1 , z2 ) ∈ B(C2 ), then d(0, z) ≤ d(0, g.z) if
and only if |g3,1 z1 + g3,2 z2 + g3,3 | ≥ 1.

Proof. Since U (2, 1) acts transitively on B(C2 ), we may write z = h.0 for some
h ∈ U (2, 1). So by (3.4), d(0, z) ≤ d(0, g.z) if and only if |h3,3 | ≤ |(gh)3,3 |. Since
zν = hν,3 /h3,3 for ν = 1, 2, we have (gh)3,3 = (g3,1 z1 + g3,2 z2 + g3,3 )h3,3 , and the
result follows.

−1
Now
√ ∪ Kbu bK ⊂ Γ be as in Section 3. Write r
√ let Γ and S = K ∪ KbK
for + 3. For 1 < ρ < (r + 1) 2, let Uρ denote the union of the 12 open discs
in C of radius 1 with centers ρζ λ , λ = 0, 1, . . . , 11. Let Bρ denote the bounded
component of C \ Uρ . The conditions on ρ ensure that Bρ exists. See the diagram
below.
√
Let B1 and B2 denote Bρ for ρ = (r + 1)/ 2 and ρ = r + 1, respectively. In
the diagram, ρ′ and ρ′′ are the two solutions t > 0 of |teiπ/12 − ρ| = 1. When

14

DONALD I. CARTWRIGHT TIM STEGER

√
ρ = (r + 1)/√ 2, we have ρ′ = 1 √
and ρ′′ = r + 1. When ρ = r + 1, we have
ρ′ = (r + 1)/ 2 and ρ′′ = r(r + 1)/ 2.
Write κ for the square root of r − 1.

Lemma 4.2. Let (w1 , w2 ) ∈ C2 . Then (w1 , w2 ) ∈ FS if and only if
√
√
(i) u1 w1 + u2 w2 ∈ B1 for each of the pairs (u1 , u2 ) = ( r + 1, 0), (0, r + 1)
and (κ−1 e−iπ/12 , κ−1 ζ 3ν e−iπ/12 ) for ν = 0, 1, 2, 3, and
(ii) u1 w1 + u2 w2 ∈ B2 for each of the pairs (u1 , u2 ) = (κ−1 , κ−1 (ζ + 1)ζ 1+3ν )
and (κ−1 (ζ + 1)ζ 1+3ν , κ−1 ) for ν = 0, 1, 2, 3,
√
in which case, |w1 |, |w2 | ≤ 1/ r + 1.
ρeiπ/6
•
′

0
•

Bρ

•ρ
•

′′
•ρ

eiπ/12

iπ/12

e

ρ−1

•

ρ

Proof. Given w = (w1 , w2 ) ∈ B(C2 ), to verify that w ∈ FS , we must show that
d(0, w) ≤ d(0, (bk).w) and that d(0, w) ≤ d(0, (bu−1 bk).w) for all k ∈ K. Since b
commutes with v, and bu−1 b commutes with u, we must check 288/4 + 288/3 = 168
conditions.
Let γ0 be as in (3.2), and let D, as before, be the diagonal matrix with diagonal
entries 1, 1 and κ. The g ∈ U (2, 1) to which we apply Lemma 4.1 are the matrices
(bk)˜= Dγ0 bkγ0−1 D−1 and (bu−1 bk)˜= Dγ0 bu−1 bkγ0−1 D−1 , where k ∈ K. Now
(bk)˜3i = κ(γ0 bkγ0−1 )3i

for

i = 1, 2,

and (bk)˜33 = (γ0 bkγ0−1 )33 ,

and similarly with b replaced by bu−1 b. Note also that for λ ∈ Z,

(γ0 bkj λ γ0−1 )3i = (γ0 bkγ0−1 )3i ζ λ for i = 1, 2, and (γ0 bkj λ γ0−1 )33 = (γ0 bkγ0−1 )33 ,

and similarly with b replaced by bu−1 b. So the conditions for w = (w1 , w2 ) ∈ FS
to hold have the form
|κ(g31 w1 + g32 w2 )ζ λ + g33 | ≥ 1 for λ = 0, . . . , 11,

for 6 matrices g of the form γ0 bkγ0−1 , and 8 matrices g of the form γ0 bu−1 bkγ0−1 .
If g = γ0 bkγ0−1 , then g33 = (ζ + 1)/ζ, and if g = γ0 bu−1 bkγ0−1 , then g33 = r + 1.
By taking k = uv 2 u−1 j, j −2 , and uv 2−ν j 3(ν−1) , for ν = 0, 1, 2, 3, respectively, we
get from g = γ0 bkγ0−1 the triples (g31 , g32 , g33 ) equal to ((ζ + 1)ζ −1 , 0, (ζ + 1)ζ −1 ),
(0, (ζ + 1)ζ −1 , (ζ + 1)ζ −1 ) and ((r + 1)ζ −1 /2, (r + 1)ζ −1 ζ 3ν /2, (ζ + 1)ζ −1 ). Using
√ eiπ/12 and κ(r + 1)/2 = κ−1 , and replacing λ by 6 − λ, we see that
ζ + 1 = r+1
2

the conditions coming from the six g√of the form γ0 bkγ0−1 are just the conditions
u1 w1 + u2 w2 6∈ Uρ for
√ ρ = (r + 1)/ 2 for the six√(u1 , u2 ) listed in (i). Taking
the case (u1 , u2 ) = ( r + 1,
√0), if u1 w1 + u2 w2 = r + 1 w1 is in the unbounded
component of √
C \ Uρ , then r + 1 |w1 | ≥ r + 1 (since ρ′′ equals r + 1 in this
√ case),
2
,
w
)
∈
B(C
).
So
r + 1 w1
and so |w1 | ≥ r + 1 > 1, which is impossible for (w
√1 2
′
is in the bounded component
B
,
and
so
|w
|
≤
1/
r
+
1
(since
ρ
equals
1
in this
ρ
1
√
case). Similarly, |w2 | ≤ 1/ r + 1 for all (w1 , w2 ) ∈ FS .
By taking k = v 1−ν j 3ν−1 , and k = vu−1 v 2+ν j 9 , for ν = 0, 1, 2, 3, respectively,
we get from g = γ0 bu−1 bkγ0−1 the triples (g31 , g32 , g33 ) equal to ((r + 1)/2, (r +
1)(ζ + 1)ζ 1+3ν /2, r + 1) and ((r + 1)(ζ + 1)ζ 1+3ν /2, (r + 1)/2, r + 1), ν = 0, 1, 2, 3.
Replacing λ by 6 − λ, we see that the conditions coming from the eight g of the

FINDING GENERATORS AND RELATIONS

15

form γ0 bu−1 bkγ0−1 are just the conditions u1 w1 +√u2 w2 6∈ Uρ for ρ = r + 1 for the
eight (u1 , u2 ) listed in (ii). Using |w1 |, |w2 | ≤ 1/ r + 1 for (w1 , w2 ) ∈ FS , we see
that u1 w1 + u2 w2 is in the bounded component Bρ of C \ Uρ in each case.

So calculation of r0 in this case is equivalent to calculation of the maximum
value ρ0 , say, of |w| on the set of w = (w1 , w2 )∈ C2 satisfying the conditions (i)
1+ρ0
. As we have seen, |w1 |, |w2 | ≤
and (ii) in Lemma 4.2, and r0 = 12 log 1−ρ
0
√
√
1/ r + 1 for (w1 , w2 ) ∈ FS . So FS is compact, and ρ0 ≤ r − 1.
We can now show that the value of r0 is 12 d(γ3 .0, 0) = 21 cosh−1 (r + 1), where
γ3 = bu−1 b. We first prove that this is a lower bound for r0 .
Lemma 4.3. For the above S, we have r0 ≥ 12 d(γ3 .0, 0) =
p
is, ρ0 ≥ (r − 1) r/2.

1
2

cosh−1 (r + 1). That

Proof. Consider the geodesic [0, γ3 .0] from 0 to γ3 .0, and let m be the point
on [0, γ3 .0] equidistant between 0 and γ3 .0. Let us show that m ∈ FS . If m 6∈ FS ,
there is a g ∈ S so that d(g.0, m) < d(0, m). Now g.0 6= 0, so that g 6∈ K. Also,
d(g.0, 0) ≤ d(g.0, m) + d(m, 0) < 2d(m, 0) = d(γ3 .0, 0),

and so g 6∈ Kγ3 K. So g must be in Kγ2 K = KbK. Since m ∈ [0, γ3 .0],

d(g.0, γ3 .0) ≤ d(g.0, m) + d(m, γ3 .0) < d(0, m) + d(m, γ3 .0) = d(0, γ3 .0),

so that g −1 γ3 ∈ K ∪ KbK. Now g −1 γ3 6∈ K, since otherwise g.0 = γ3 .0, so that m
is closer to γ3 .0 than to 0. Since KbK is symmetric, we have γ3−1 g ∈ KbK. Thus
g must be in G = KbK ∩ γ3 KbK. One may verify that G = (huibK) ∪ (huib−1 K).
Since u is in K and commutes with γ3 , it fixes [0, γ3 .0], and so d(g.0, m) is constant
on both double cosets huibK and huib−1 K. Note that uγ3−1 = γ3 u−1 is an element
in Γ of order 2 which interchanges 0 and γ3 .0, and so fixes m. The map f : g 7→
uγ3−1 g is an involution of G, and d(f (g).0, m) = d(g.0, m). Also, f (b) = ub−1 u, so
that f interchanges the two double cosets, and so d(g.0, m) is constant on G. So to
show the result, we need only check that
not hold. Now
√ d(b.0, m) < d(0, m) does √
γ3 .0 = (z1 , z2 ) for z1 = −(r − 1)ζ 2 /(2 r − 1) and z2 = (i +p1)ζ 2 /(2 r − 1). Write
mt =p(tz1 , tz2 ) for 0 ≤ t ≤ 1. Then m = mt for t = (1 − 1 − |γ3 .0|2 )/|γ3 .0|2 =
(1 − 1 − r/2 )/(r/2) = r − 1. Some routine calculations show that b.m = m, and

so d(b.0, m) = d(0, m). So m ∈ FS , and r0 ≥ d(0, m) = 12 d(0, γ3 .0).
Proposition 4.1. If (w1 , w2 ) ∈ FS , then |w1 |2 + |w2 |2 ≤ 2r − 3 = 2r (r − 1)2 .
Proof. Let FS ∗ = {z ∈ B(C2 ) : d(0, z) ≤ d(g.0, z) for all g ∈ S ∗ } for S ∗ = K∪KbK.
Since S ∗ ⊂ S, we have FS ⊂ FS ∗ . We shall in fact show that |w1 |2 + |w2 |2 ≤ 2r − 3
for (w1 , w2 ) ∈ FS ∗ .
Suppose that w = (w1 , w2 ) ∈ FS ∗ and |w1 |2 + |w2 |2 is maximal. Then d(0, w) =
d(0, g.w) for some g ∈ KbK. Replacing w by k.w for some k ∈ K, if necessary,
we may suppose that g = buv 2 u−1 j 7 . Let g̃ = Dγ0 gγ0−1 D−1 ∈ U (2, 1) for this g.
−1
Since g̃31 = −κ(ζ + 1)ζ −1 , g̃32 = 0 and
4.1 shows that
√
√ g̃33 = (ζ + 1)ζ , Lemma
r
−
1
w
−
1|.
Since
r
+
1
w1 ∈ B1 , this
1 = |g̃31 w1 +
g̃
w
+
g̃
|
=
|ζ
+
1|
|
1
33
√ 32 2
means that r + 1 w1 is on the rightmost arc of the boundary of B1 . That is,
1
1
−√
e−iθ for some θ ∈ [−π/12, π/12].
w1 = √
r−1
r+1

Fixing w1 , we see from Lemma 4.2(i) that w2 must lie on or outside
√ various circles,
including, for ν = 0, 1, 2, 3 and ǫ = ±, the circles Cν,ǫ of radius r − 1 and center
√
 √

αν,ǫ = −iν w1 − r + 1 eǫiπ/12 = iν eǫiπ/4 + e−iθ / r + 1 .
In the following diagram, we have taken θ = π/24:

16

DONALD I. CARTWRIGHT TIM STEGER
C1,+

C1,−

C2,−
C0,+
p

•

0
C2,+

P

q

C0,−
Q
C3,−

C3,+

Let Uǫ (θ) denote the union of the four open discs bounded by the circles Cν,ǫ ,
ν = 0, 1, 2, 3. Using 0 < cos(θ+ǫπ/4) < 1, we see that C\Uǫ (θ) has two components,
the bounded one containing 0. So the complement of U (θ) = U+ (θ) ∪ U− (θ) has
a bounded component containing 0. The set U (θ) is obviously invariant under the
rotations z 7→ iλ z, λ = 0, 1, 2, 3. It is also invariant under the reflection Rν,ǫ in the
line through 0 and αν,ǫ , for each ν and ǫ. For if 0 6= α ∈ C, the reflection in the
line through 0 and α is the map Rα : z 7→ αz̄/ᾱ. It is then easy to check that
Rν,ǫ (αν ′ ,ǫ′ ) = αν ′′ ,ǫ′

for ν ′′ = 2ν − ν ′ + (ǫ − ǫ′ )/2 (mod 4).

Let p and P be the points of intersection of C0,+ and C0,− . Then R0,− (C0,+ ) = C3,+
and R0,− (C0,− ) = C0,− , so that R0,− (p) and R0,− (P ) are the points q and Q of
intersection of C0,− and C3,+ . In particular, |q| = |p| and |Q| = |q|.
It is now clear that |z| ≤ |p| for all z in the bounded component of C \ U (θ), and
that |z| ≥ |P | for all z in the unbounded component.
We now evaluate |p| and |P |. Let z ∈ C0,+ ∩ C0,− . Then
z + w1 −

√
√
√


r + 1 eiπ/12 = r − 1 = z + w1 − r + 1 e−iπ/12 .

For α, β ∈ C with β 6∈ R, |α √
− β| = |α − β̄| if and only if α ∈ R. So z + w1 must be
real, and writing z + w1 = t r + 1, with t ∈ R, we have
√
√
√
r − 1 = t r + 1 − r + 1 eiπ/12 ,
so that
r+1
t2 − √ t + r − 1 = 0.
2
√
√
The solutions of this are t = (r − 1)/ 2 and t = 2. Taking the smaller of these,
√
r − 1√
r + 1 = r − 1.
p + w1 = √
2
√ √
√
√
So |p| = |√r √
− 1 − w1 |. Taking
instead
√
√ t = 2, we see that |P | = | √2 r + 1 − w1 |.
So |P | ≥ 2 r + 1 − 1/ r + 1 > 1/ r + 1, and therefore |z| > 1/ r + 1 for all z
in the unbounded component of C \ U (θ).

FINDING GENERATORS AND RELATIONS

17

that w2 is in the bounded component of C \ U (θ), and
So (w1 , w2 ) ∈ FS ∗ implies
√
therefore |w2 | ≤ |p| = | r − 1 − w1 |. Thus
√
|w1 |2 + |w2 |2 ≤ |w1 |2 + | r − 1 − w1 |2
√
= r − 1 + 2|w1 |2 − 2 r − 1 Re(w1 )
 1

√
1
=r−1+2
+
− 2 cos θ
r−1 r+1

 1
√
1
−√
cos θ
−2 r−1 √
r−1
r+1
√
= 3(r − 1) − r(r − 1) 2 cos θ
√ r+1
≤ 3(r − 1) − r(r − 1) 2 √
8
= 2r − 3.


We conclude by giving a direct proof that the set S generates Γ. The following
lemma uses a modification of an argument shown to us by Gopal Prasad.
√
Lemma 4.4. If g ∈ Γ \ K, then d(g.0, 0) ≥ d(b.0, 0) = cosh−1 ( r + 2 ).

Proof. By considering the (3, 3)-entry of g ∗ F g − F = 0, we see that

|g13 |2 + |g13 − (r − 1)g23 |2 = (r − 1) |g33 |2 − 1 .

(4.1)

Write α = g13 , β = g13 − (r − 1)g23 and γ = g33 . By hypothesis, g.0 6= 0, and so
g13 6= 0 or g23 6= 0. Hence

α, β, γ ∈ Z[ζ], |α|2 + |β|2 = (r − 1) |γ|2 − 1 , and α, β are not both 0. (4.2)

We claim that under conditions (4.2), |γ|2 ≥ r+2 must hold. Writing α = a0 +a1 ζ +
a2 ζ 2 +a3 ζ 3 , we have |α|2 = P (α)+rQ(α), where P (α) = a20 +a0 a2 +a22 +a21 +a1 a3 +a23
and Q(α) = a0 a1 + a1 a2 + a2 a3 . Our hypothesis is that
P (α) + P (β) + P (γ) = 3Q(γ) + 1

and P (γ) = Q(α) + Q(β) + Q(γ) + 1, (4.3)

and we want to show that P (γ) + rQ(γ) ≥ 2 + r. Now P is a positive definite form,
and γ 6= 0 and either α 6= 0 or β 6= 0. So the left hand side of the first equation
in (4.3) is at least 2, so that Q(γ) > 0. Since Q(γ) ∈ Z, we have Q(γ) ≥ 1. So all
we have to do is show that P (γ) ≥ 2. Now Q(γ) 6= 0 implies that a0 a1 , a1 a2 or
a2 a3 6= 0. If a0 a1 6= 0, then P (α) = (a2 + a0 /2)2 + 3a20 /4 + (a3 + a1 /2)2 + 3a21 /4 ≥
3(a20 + a21 )/4 ≥ 3/2 > 1, so that P (α) ≥ 2. The other two cases are similar.

The following result implies that Γ is generated by K and b.
Lemma 4.5. If g ∈ Γ \ K, then there exists k ∈ K so that d(bkg.0, 0) < d(g.0, 0).

Proof. If there is no such k ∈ K, then d(s.(g.0), 0) ≥ d(g.0, 0) for all s ∈ S ∗ , and
4.1, and so d(g.0, 0) ≤ r0 .
so g.0 ∈ FS ∗ , in the notation of the proof of Proposition
√
But by Lemma 4.4, d(g.0, 0) ≥ d(b.0, 0) = cosh−1 ( r + 2 ), which is greater than
−1
1
(r + 1) = r0 .

2 cosh
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Sci. Paris, I 348 (2010) 11–13.
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