ABSTRACT
Title of Thesis: PROJECTIVE DEFORMATIONS OF TRIANGLE TILINGS
Anton Valerievich Lukyanenko, Master of Arts, 2008
Thesis directed by: Professor William Goldman, Department of Mathematics
A hyperbolic triangle group is the group generated by reflections in the sides
of a triangle in hyperbolic space. For a given hyperbolic triangle group, we find
a one-parameter group of representations into GL(3,R) and associated invariant
cones. We show that the representations are faithful and that the cones are sharp.
We then apply the results of Guichard to approximate the H?older continuity of the
boundaries of the cones. We conjecture that this may be directly calculated by
considering only the Coxeter elements of the triangle group.
PROJECTIVE DEFORMATIONS OF TRIANGLE TILINGS
by
Anton Valerievich Lukyanenko
Thesis submitted to the Faculty of the Graduate School of the
University of Maryland, College Park in partial fulfillment
of the requirements for the degree of
Master of Arts
2008
Advisory Committee:
Professor William Goldman, Chair
Professor Jeffrey Adams
Professor James Schafer
Contents
1 Background 1
1.1 Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Hyperbolic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.4 Coxeter Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Hilbert Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Convex Cones 6
2.1 Tits Cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2.2 Kac-Vinberg Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 H?older Regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.4 ?-Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.5 Guichard?s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 The Generalized Tits Cone 16
3.1 Normalizing Triangle Group Representations . . . . . . . . . . . . . 17
3.2 Constructing the Generalized Tits Cone . . . . . . . . . . . . . . . . 18
3.3 Some Conjectures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4 Coxeter Elements 25
4.1 Infinite Order, Essential . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.2 Minimizing Translation Length in the Hilbert Metric . . . . . . . . . 29
ii
1 Background
1.1 Cones
A cone U in Rn is a domain invariant under positive homotheties:
?u ? U for any u ? U,? > 0.
A cone is sharp (or properly convex, or, in French, saillant) if its closure U does
not contain a full line. It is strictly convex if any codimension-1 subspace of Rn
intersects the boundary ?U of U in at most a ray. The dual U? of a cone U is the
set of linear functionals on Rn that are strictly positive on U.
AspaceX withanactionofagroupGiscalledhomogeneous ifGactstransitively
on X:
for some x ? X,Gx = X.
Relaxing this condition, call X quasi-homogeneous if there exists a compact K
and a group G acting on X such that GK = X. If the quotient X/G is furthermore
Hausdorff, X is called divisible, and G is said to divide X. If G is acting properly
(gK ?K negationslash= ? for only finitely many g) and is torsion-free, then X/G is a manifold.
If G acts properly and is virtually torsion-free (has a subgroup of finite index that
is torsion-free), then X/G is called a (good) orbifold.
The following invariant will allow us to study the boundaries of divisible domains:
Definition 1.1. A transformation g in GL(n,R) is called hyperbolic (or loxodromic)
if its eigenvalues have distinct norms. For such a transformation, let l1 > ... > ln
be the logarithms of the norms of the eigenvalues of g. Define
?g := l1 ?lnl
1 ?ln?1
,
?g := l1 ?lnl
1 ?l2
.
Definition 1.2. For G ?GL(n,R) define:
?G := inf
g?G, g hyperbolic
?g, ?G := sup
g?G, g hyperbolic
?g
It is easy to show that ??1g?1 +??1g = 1, and the same equality holds for groups:
??1G +??1G = 1.
Note that for n = 3, the definitions of ?g and ?g coincide:
?g = ?g = l1 ?l3l
1 ?l2
.
1
1.2 Projective Geometry
The real projective space RPn is the space of lines in Rn+1. Since cones are in-
variant under scaling it is the natural setting for drawing them and analyzing their
properties. For p ? RPn, a nonzero vector in the line p ? Rn+1 gives the homoge-
neous coordinates for p. These are defined up to non-zero scaling. The projection
map Rn+1\{0} ? RPn induces an action of GL(n + 1,R) on RPn. The image of
GL(n+1,R) under this homomorphism into Aut(RPn) is called PGL(n+1,R) and is
isomorphic to SL(n+1,R). We will use this isomorphism to think of PGL(n+1,R)
as a subgroup of GL(n+1,R). The action of PGL(n+1,R) is (n+1)-transitive on
points in general position (no proper subset of the n+1 points is linearly dependent).
RPn is a manifold, with the standard coordinate patch given by dividing out the
last coordinate: 2
64
x1
...
xn+1
3
75mapsto?
2
64
x1/xn+1
...
xn/xn+1
3
75.
Other coordinate patches can be found by dividing out by other coordinates, or by
using the transitivity of the PGL(n+1,R) action.
1.3 Hyperbolic Geometry
Definition 1.3. Let B
0
@
2
4
x
y
z
3
5,
2
4
x
y
z
3
5
1
A = x2 + y2 ? z2 be the Lorentz inner
product on R3. Denote the corresponding norm by bardbl?bardblB and consider the upper
component of the hyperboloid of vectors with norm squared ?1:
Hh =
8
><
>:
2
4
x
y
z
3
5?R3
flfl
flfl
flfl
fl
  
  
  
2
4
x
y
z
3
5
  
  
  
2
B
= ?1 and z > 0
9
>=
>;
By restriction, B induces a positive definite Riemannian metric on Hh. Hh
with this metric is known as the hyperboloid model of the hyperbolic plane H. The
projectivization of Hh into the standard coordinate patch of RP2 is known as the
Klein unit disk HK.
Definition 1.4. The complex projective line CP1 is the set of equivalence classes of
vectors in C2\{0} up to scaling, with the induced automorphism group PGL(2,C).
As with RPn, the standard coordinate patch is given by dividing out the last coor-
dinate: ?
z1
z2
?
mapsto? z1z
2
.
2
We thus get an identification ofCP1 with ?C=C??, the Riemann sphere, on which
GL(2, C) acts by linear fractional transformations:
? a b
c d
?
: z mapsto? az +bcz +d.
Definition 1.5. LetHu = {z ?C| Re(z) > 0}?CP1 be the upper half-plane, with
the Riemannian metric ds2 = dx2+dy2y2 . This is the Poincar?e upper half-plane model
of the hyperbolic plane.
The upper half plane can be transformed to a unit disk using an element of
PGL(2, C). The transformation induces a Riemannian metric on the unit disk,
and we define the disk with this metric to be the Poincar?e unit disk model HD of
hyperbolic space.
These spaces are all isomorphic Riemannian manifolds.
The action GL(2, C) on the Riemann sphere restricts to the action of SL(2, R)
on the upper half-plane and gives all the orientation-preserving automorphisms of
Hu. These are classified by trace:
Lemma 1.6. Let M ? SL(2, R) be a an automorphism of Hu. Then M is, up to
multiplication by ?I, conjugate to one of the following, identified by trace:
?
? 1 0
0 1
?
: z mapsto? z, the identity transformation, with trace 2
?
? 1 1
0 1
?
: z mapsto? z +1, a parabolic transformation, with trace 2
?
? ? 0
0 ??1
?
: z mapsto? ?2z, (? negationslash= 1), a hyperbolic transformation, with trace ? +
??1 > 2
?
? cos? sin?
?sin? cos?
?
, an elliptic transformation, with trace 2cos? < 2
Note that an elliptic transformation is GL(2, C)-conjugate to
? ei? 0
0 e?i?
?
:
z mapsto? e2i?, a rotation fixing the origin and ?.
Proof. If M has real eigenvalues, it can be diagonalized over R to be in one of the
first three forms. In the last case, the eigenvalues are distinct and we get the fourth
form by first diagonalizing overC, and then converting back to the real matrix.
For further information on hyperbolic geometry, see [9].
3
1.4 Coxeter Groups
A Coxeter group W on n generators has the presentation
W = ?si flfl (sisj)nij = s2i = 1fi, nij ?N, i,j = 1,...,n.
A Coxeter group with a choice of generators S = {si} is called a Coxeter system.
Coxeter groups with two generators are the dihedral groups, and ones with 3
generators are called triangle groups. In the latter case, we denote the three numbers
nij by p,q,r, and call a triangle group a:
? Spherical Triangle Group if 1p + 1q + 1r > 1
? Euclidean Triangle Group if 1p + 1q + 1r = 1
? Hyperbolic Triangle Group if 1p + 1q + 1r < 1
Equivalently, a (p,q,r) triangle group can be defined as the group generated by
reflections in the sides of a triangle with interior angles pip, piq, pir (see [7]). The above
classification reflects the availability of such triangles in the corresponding geometry.
Given an element w of W, we may write it as words in si in many ways. Define
l(w) to be the length of the shortest word in si that corresponds to w. This extends
to the word metric d(w,wprime) := l(w?1wprime) on W. The word metric on Coxeter groups
has the following essential property:
l(siw) = l(w)+1 or l(w)?1.
For a non-trivial word w, we may always find an si such that siw has length
l(w)?1, which allows inductive arguments.
Given Sprime = {r1,...,rk}?{s1,...,sn}, define
WSprime := ?r1,...,rk?? W
Note that the word metric on WSprime does not necessarily agree with that on all of W
(an element w ? WSprime may have a shorter representation in W).
There are two important classes of elements of W:
? An element w is called essential if it is not conjugate to an element of WSprime for
any proper subset Sprime of the generators.
? An element of the form ?(s1)????(sn), where ? is any permutation of the
generators, is called a Coxeter element. Up to conjugacy there are only two
Coxeter elements in a triangle group: s1s2s3 and its inverse s3s2s1.
We will prove the following fact in Section 4.1:
Theorem 1.7. Coxeter elements of hyperbolic triangle groups have infinite order.
Corollary 1.8. Coxeter elements of hyperbolic triangle groups are essential.
Proof. Any proper subgroup WSprime of a triangle group is of finite order.
4
1.5 Hilbert Metric
Definition 1.9. Let ? be a convex domain, and x,y ? ?. The Hilbert distance
between x and y is
d?(x,y) := 12 inf log [a,x;y,b],
where the infimum is taken over all a and b in ? such that xy ? ab and [a,x;y,b] is
the cross-ratio
[a,x;y,b] := |x?b||y ?a||x?a||y ?b|.
Note that the infimum is attained when a,b ? ??. Due to the projective invari-
ance of the cross-ratio, the Hilbert metric is invariant under the projective automor-
phisms of ?. For a proof of the triangle inequality, see [11].
Straight lines inRPn?? are geodesics in the Hilbert metric. For strictly convex
domains ?, these are the only geodesics, and the Hilbert metric induces a norm on
the tangent spaces of ?.
Example 1. Let ? be the unit ball in Rn. Then dH = dK, the metric on the Klein
model of hyperbolic space.
Proof. Let x,y ? ?. Since ? is homogeneous with respect to its projective auto-
morphisms, we may assume y is the origin, and furthermore rotate such that x lies
on the x-axis, with x-coordinate 0 < a < 1. Thus, we are reduced to the case n = 1.
dH(0,a) = 12 log [?1,0;a,1] = 12 log a+1a?1.
To calculate dK(0,x), we switch to the hyperbola model ofH1 ?R1,1, defined by the
equation x2 ?y2 = ?1 (recall that the metric on hyperbolic space is the restriction
of the Lorentz inner product to this hyperbola). Under the equivalence between the
Klein and hyperbolic models (Figure 1), we associate:
0 ? (0,1), a ? (b,b/a),where b = a?1?a.
To find the distance, we integrate along the hyperbola y = ?x2 +1 in the R1,1
metric pdx2 ?dy2:
dK(0,a) =
Z b
0
s
1?
 dy
dx
?2
dx =
Z b
0
(x2 +1)?1/2dx
Using the substitution x = sinh(u), this reduces to dK(0,a) = sinh?1 b. Plugging
in b,
dK(0,a) = 12 log 1+a1?a = dH(0,a).
5
Klein Model
LightCone
x2Minusy2EqualMinus1
LParen1a,1RParen1
Minus2 Minus1 0 1 2
1
2
Figure 1: The Klein model of hyperbolic 1-space can be viewed as the open segment
between (?1,1) and (1,1). The hyperbola model is given by the hyperbola x2?y2 =
?1. Given a point (a,1) in the Klein model, we identify it with a point in the
hyperbola using a straight line from the origin (dashed line).
2 Convex Cones
2.1 Tits Cone
We now define the Tits cone and discuss its basic properties, basing proofs on
[3, 4, 10].
For this section, fix a Coxeter system (W,S) with S = {s1,...sn} and a basis
{es}s?S for V =Rn. Let B(?,?) be the bilinear form on V defined by:
B(esi,esj) =
(
1 i = j
?cos pinij i negationslash= j .
Definition 2.1. (Tits cone) For each s ? S, define:
Rs(x) := x?2B(x,es)es,
Hs := {x ? V | B(x,es) = 0},
As := {x ? V | B(x,es) > 0},
C :=
\
s?S
As.
Each Rs is a reflection over Hs (see Figure 2). Define the representation ? of W
into GL(3, R) by
?(si) := Rsi
and set G := ?(W). The Tits cone is defined as the image of C under the action of
W:
U :=
[
g?G
gC.
6
As we show below, W acts freely on the images of C, which is a fundamental
domain for the action. The proof stems from the following lemma for the dihedral
case:
Lemma 2.2. Let W = ?a,b flfl (ab)n = a2 = b2 = 1fi, and ? : W ? O(2,R) such that,
in polar coordinates:
? ?(a) is the reflection over the line ? = 0,
? ?(b) is the reflection over ? = pin.
Let C be the region '(r,?) flfl r > 0, 0 < ? < pin)?. Then, for w ? W,
?(w)(C) ? Aa = {(x,y) | x > 0} if and only if l(aw) < l(w).
Proof. We assign each image Cprime of the fundamental domain a word w of minimal
length in {a,b} such that ?(w)(C) = Cprime. C is assigned the null word, its neighbors
are assigned the two words of length 1: a,b. Their neighbors are images of C under
the words ab and ba. To see that those are minimal, note that all words of smaller
length are already taken. Proceeding by induction, a word representing Cprime starts
with a if and only if it is not in Aa (Figure 2).
Hb
Ha
Ab
Aa
C
ea
eb
0
a
b
ab
ba
aba
bab
babaEqualabab
Figure 2: Notation and Lemma 2.2 for D4, in a basis orthogonal with respect to B.
Lemma 2.3. The corners of the Tits cone are dihedral: for any distinct i,j there
is a basis such that Rsi, Rsj, and Ai ?Aj satisfy the conditions in Lemma 2.2.
Proof. Let n = nij = |sisj|. Restricted to the subspace ?ei,ej?, the reflections are
represented by matrices
Ri =
? ?1 2cos pi
n0 1
?
, Rj =
? 1 0
2cos pin ?1
?
.
7
The product
RiRj =
? ?1+4cos2pi
n ?2cos pin
?2cos pin ?1
?
has trace 2cos 2pin and determinant 1. The two eigenvalues are then inverses of each
other, and ? + ??1 = 2cos 2pin < 2. Since this expression is greater than two for ?
real, ? must be imaginary. Thus, under the right basis R1R2 is a rotation of order
n.
Now, extend ei to an orthonormal basis with respect to B. In this basis, Ri is
reflection over the line Li ?{? = 0}, as desired. In the same basis, Rj must be an
orthogonal reflection such that RiRj is of order n, so it must be reflection over a
line Lj = {? = ?kpin } for some integer k.
Now, the angle ? between ei and ej is given by B:
B(ei,ej) = bardbleibardblB bardblejbardblB cos?
so that
?cos pin = cos?.
This forces ? = ??pi ? pin? (mod 2pi), and since by construction B(ei,Li) =
0,B(ej,Lj) = 0,
? = ?pin.
Conjugating by Ri if necessary, we may assume this sign is positive. Thus, up
to an order-2 rotation Ai ?Aj must be '(r,?) flfl r > 0, 0 < ? < pin?.
Theorem 2.4. (Tits)We use the above notation for the Tits cone. Then
(P) For each w ? W and s ? S, either
wC ? As and l(w) = l(sw)?1, or
wC ? sAs and l(w) = l(sw)+1.
Informally, if wC is on the ?negative? side of the cone with respect to s, then there
is a minimal word equivalent to w that starts with s.
Proof. We split the assertion (P) into smaller statements for an inductive argument:
(Pn) Let w ? W,l(w) = n. For each s ? S, either
wC ? As and l(w) = l(sw)?1, or
wC ? sAs and l(w) = l(sw)+1.
To deal with cases where w doesn?t start with s (for any minimal word equivalent
to w), we need the following assertions:
(Qn) Let w ? W,l(w) = n. For each s,t ? S, there exists u ? Ws,t such that
wC ? uC and l(w) = l(u)+l(u?1w).
8
The intuitive idea is the following: taking s,t ? S, split the Tits cone into sectors
given by the action of the subgroup Ws,t. If wC is in a sector indexed by a word
u ? Ws,t, then w starts with u (there is an equivalent minimal-length word that
starts with u).
P0 and Q0 are trivial. We now prove that Pn and Qn imply Pn+1 and that Pn+1
and Qn imply Qn+1.
Pn and Qn imply Pn+1. Let w ? W with l(w) = n + 1. If l(sw) = l(w) ? 1
(intuitively, w starts with s), then apply Pn to sw and then multiply by s to prove
Pn+1.
If l(sw) > l(w), pick some t such that l(tw) < l(w). Set wprime = tw. Applying
Qn to wprime, we get u ? Ws,t such that wprimeC ? uC and l(wprime) = l(u) ? l(u?1wprime). So
wC = twprimeC ? tuC. We now analyze v = tu. By Lemmas 2.3 and 2.2, Wu,v has
property P. Thus, one of the following is true:
1. v(As ?At) ? As, which implies wC ? vC ? As, or
2. v(As ?At) ? sAs, which would show that wC ? sAs.
Assume by way of contradiction that the latter is true. Then it is also true that
lprime(sv) < lprime(v) (where lprime denotes the word metric in the group Ws,t). We calculate:
l(sw) = l(stwprime) = l((stu)(u?1wprime)) ? lprime(stu)+l(u?1wprime)
< lprime(tu)+l(u?1wprime) = lprime(tu)?lprime(u)+l(wprime) ? l(w).
That contradicts the assumption that l(sw) > l(w).
Pn+1 and Qn imply Qn+1.Lets,t ? S,w ? W withl(w) = n+1. IfwC ? As?At
then we are done with u = 1. Otherwise, choose s to be the letter such that
wC * As. By Pn+1, l(sw) = l(w) ? 1 = n and now Qn applies to sw, giving us
v ? Ws,t such that swC ? v(As ?At) and l(sw) = l(v)+l(v?1sw). We calculate:
l(w) = 1+l(sw) = 1+l(v)+l(v?1) ? l(sv)+l((sv)?1w) ? l(w)
So that the last inequality must be an equality, and Qn+1 is proven using u =
sv.
To summarize, the first assertion is proven by either showing that wC is in the
s-negative side sAs, or by using another generator to reduce the length of the word
and use the inductive hypothesis to show that wC must already be in As. The
second assertion is proven by making one move in the rotation that brings wC to
the sector As ?At and letting induction take care of the remaining rotation.
Corollary 2.5. Let w ? W. Then wC ? C negationslash= ? if and only if w = 1. Thus, the
representation of the Coxeter group in GL(V) is faithful.
Proof. If w negationslash= 1 then there is some s ? S such that l(sw) = l(w)?1. But then by
property (P), wC ? sAs, so wC ?C = ?, a contradiction.
9
Corollary 2.6. The Tits cone U is convex. That is, if x,y ? U, then the segment
xy ? U.
Proof. We may assume without loss of generality that x ? C, the fundamental
domain. Pick a w ? W such that y ? wC. C is convex since it is the intersection of
the half-planes As. Thus, if w = 1, then we are done.
Say l(w) = n and assume inductively that the proposition holds for all shorter
words. Consider the intersection xz = xy ? C. z is in the boundary of C, say in
Hs for some s ? S. But then z ? sC, and the minimal word connecting z and y
has length l(sw) = l(w)?1 by property P. By induction, the segment zy ? U, so
xy ? U.
Considering the restrictions placed on U by the existence of the bilinear form
provide
Theorem2.7. (Vinberg [14, 11])If W is a hyperbolic triangle group, then the closure
of the Tits cone contains no line.
The Tits cone is the domain on which the bilinear form B is positive definite.
By strict convexity, this cannot be all ofR3. B must therefore be of signature (2,1).
We thus have:
Corollary 2.8. The projectivization of the Tits cone for a hyperbolic triangle group
is projectively equivalent to the Klein unit disk.
2.2 Kac-Vinberg Cones
The Tits cone construction was modified by Kac and Vinberg in [8] to produce an
inhomogeneous cone divided by a triangle group.
Let e1,e2,e3 be the standard basis of R3. Define B(?,?) to be a bilinear form,
with Bij = B(ei,ej), such that:
1. det[Bij] < 0.
2. Bii = 2, and the other entries are negative integers.
3. BijBji < 4 for i negationslash= j.
4. B12B23B31 negationslash= B21B32B13.
Example 2. [Bij] =
2
4
2 ?1 ?3
?1 2 ?1
?1 ?1 2
3
5
The rows of [Bij]?1 give vectors li such that B(li,ej) = ?i,j. As with the Tits
cone, define
Ri(v) = v ?B(v,ei)ei
10
Figure 3: The Kac-Vinberg cone for [Bij] in Example 2.
C =R+l1 ?R+l2 ?R+l3
G = ?R1,R2,R3?
U = GC
We refer to U as a Kac-Vinberg cone.
Theorem 2.9. (Theorem 1 in [8]) Let Uprime ?R3 be a quasi-homogeneous sharp cone
whose boundary is twice differentiable except at finitely many points. Then Uprime is
homogeneous, and thus either a triangular cone or an elliptical cone.
Theorem 2.10. (Theorem 2 in [8]) A Kac-Vinberg cone U is not homogeneous.
Proof. Assume by way of contradiction that U is homogeneous. It can be verified
directly that the order of RiRj is nij is given by
4cos2 pin
ij
= BijBji.
If U were a triangular cone, its rotational symmetries RiRj would have to be of
order 3. This would force Bij = ?1 and give a singular matrix.
If U were an elliptical cone, there would be an inner product (?,?) invariant
under the symmetries of U. It can then be shown that 2(ei,ej) = Bji(ei,ei). Since
the inner product is non-degenerate, and Bij < 0, this implies (ei,ej) negationslash= 0 for all
i,j. The symmetry of the inner product then gives B12B23B31 = B21B32B13, a
contradiction.
Thus, U is not homogeneous, and is not twice differentiable at infinitely many
points.
11
2.3 H?older Regularity
While the boundary of a Kac-Vinberg cone is not twice differentiable, it is more than
just once differentiable. One can extend the notion of nth derivative to non-integer
values of n:
Definition 2.11. Let ? ? (0,1] and f :R?R.
f(?)(x) = limy?x f(x)?f(y)|x?y|? .
A weaker notion of derivative is more useful than the existence of this infinites-
imal quantity. The following definition requires the difference quotient to stay
bounded rather than to converge.
Definition 2.12. Let ? ? (0,1] and f : R ? R. f is ?-H?older if there exists a C
such that for all x,y ?R,
|f(x)?f(y)|? C|x?y|? .
For ??n ? (0,1], f is ?-H?older if the nth derivative f(n) is (??n)-H?older.
We define C? to be smallest constant C fulfilling the above inequality, with
C? = ? if no such C exists.
Example 3. f(x) = |x|? is ?-H?older for ? > 0.
Proof. In all cases, we are reduced to the case ? ? (0,1]. Fix y ?R and define
C(y) = sup
x?R
|f(x)?f(y)|
|x?y|? = supx?R
||x|? ?|y|?|
|x?y|? .
The fraction is not defined for x = y. However, we fill it in with the value 0 by
using l?H?ospital?s rule. Since it increases away from y and approaches 1 as x goes to
??, C(y) = 1. Since this works for any y, f is ?-H?older with constant C? = 1.
Definition 2.13. A curve M in R2 is ?-H?older if it is locally the graph of an ?-
H?older function. For ? ? (1,2] this is equivalent (Lemma 4.2 of [1]) to the condition
that for all compact K ? M there exists CK such that for all p,q ? K
d(q,TpM) ? CKd(p,q)?,
where TpM ? R2 is the tangent line to M at the point p. We use this description
of H?older continuity to define it when M is a submanifold of an arbitrary space
endowed with a distance d.
Thus, while H?older continuity states that the difference quotient |f(x)?f(y)||x?y|? is
bounded on the entire domain of f, for locally H?older functions we require this
condition only as x approaches y. For example, the function f(x) = x2 is locally,
but not globally, 1-H?older since the derivative is locally, but not globally, bounded.
12
Lemma 2.14. (Projective Invariance) Let M ? E ? RPn be a curve in an affine
patch E of projective space, and f a nonsingular projective transformation such that
f(M) ? E. If M is ?-H?older in the metric d given by E, then so is f(M).
Proof. We only have to show the invariance of ?-H?older continuity on compact
subsets of M. Let K be a compact subset of M, and bK a compact neighborhood of
K in E ?f?1(E).
f is bi-Lipschitz on bK: there is a constant c such that for all x,y ? K,
1
cd(x,y) ? d(fx,fy) ? cd(x,y), given by
c = max
(
sup
x,y?K
d(x,y)
d(fx,fy), supx,y?K
d(fx,fy)
d(x,y)
)
.
The ratios are defined everywhere since f is differentiable and invertible, and the
suprema are attained since bK is compact.
Since M is ?-H?older, we have a constant C such that for all x,y ? K,
d(x,TyM) ? Cd(x,y)?.
Extending this inequality using the fact that f is bi-Lipschitz on bK,
1
cd(fx,f(TyM)) ? d(x,TyM) ? Cd(x,y)
? ? C(cd(fx,fy))?.
Since f is a projective transformation, f(TyM) = Tfyf(M), and this reduces to
1
cd(fx,Tfyf(M)) ? C(cd(fx,fy))
?
d(fx,Tfyf(M)) ? (Cc?+1)d(fx,fy)?
So the compact subset K ? f(M) is ?-H?older for any compact K. Thus, f(M)
is ?-H?older.
Definition 2.15. Let M be a submanifold of a metric space. ?M := sup {? ? 2 :
M is ?-H?older}
Lemma 2.16. Let M ? E ?RPn be a submanifold ofRPn contained in a coordinate
patch E. Assume furthermore that M is invariant under a hyperbolic transformation
g, with eigenvalues |?1| > ... > |?n+1| and li = log(|?i|). Then, using the Euclidean
distance given by E,
?M ? ?g = l1 ?ln+1l
1 ?ln
.
13
xMinus
p
gp
TxPlusM
M
xPlus
x0
Figure 4: A g-invariant curve M ? RP2 viewed in a coordinate patch given by g.
The points x+,x0,x? are, respectively, the attracting, saddle, and repelling fixed
points of g.
Proof. We consider the case n = 2. The general case is essentially identical.
Let x+,xo,x? ?RP2 be, respectively, the attracting, saddle, and repelling fixed
points of g. Pick a basis {e1,e2,e3 with e1 ? x?, e2 ? xo, e3 ? x+, so that x+ is the
origin of the standard coordinate patch, and Tx+M is the x-axis (see Figure 4).
For any point p =
? x
y
?
with x negationslash= 0,
limn?? logd(g
n ?p,x+)
n = limn??
log[(?2??11 )2nx2 +(?3??11 )2ny2]
2n
= limn?? log[(?2?
?1
1 )2n(x2 +?3?
?1
2 y2)]
2n
= limn?? 2nlog[?2?
?1
1 ]+log[x+(?3?
?1
2 )2ny2]
2n = l2 ?l1.
limn?? logd(g
n ?p,Tx
+M)
n = limn??
log (?3??11 )ny
n = l3 ?l1.
limn?? logd(g
n ?p,Tx
+M)
logd(gn ?p,x+) =
l1 ?l3
l1 ?l2 = ?g
limn?? logd(g
n ?p,Tx
+M)
logd(gn ?p,x+)?g = 1
limn?? d(g
n ?p,Tx
+M)
d(gn ?p,x+)?g = 1
Thus, if ? > ?g, the last limit would be infinite, so the H?older exponent of M
can be no larger than ?g. Note that, while we computed the H?older exponent in
a coordinate patch not containing all of M, the conclusion is projectively invariant
for a compact neighborhood of x+, and therefore for all of M.
14
Corollary 2.17. Let M be a curve with projective automorphism group G. Then,
?M ? ?G = inf
g?G, g hyperbolic
?g.
Definition 2.18. Let ? ?RPn be a projective domain. Define
?? := sup{? | ?? is ?-H?older}.
The relationship between the automorphism group and H?older exponent estab-
lished in Corollary 2.17 carries over to projective domains.
2.4 ?-Convexity
?-convexity is a useful notion dual to ?-H?older continuity, and is used in key proofs
concerning convex domains.
Definition 2.19. Let M ? E be a manifold in an affine space E with a metric ?.
Then M is ?-convex if for all compact K ? M there is a constant CK such that for
all x,y ? K,
?(x,TyM) ? CK?(x,y)?,
where TyM is the tangent hyperplane to M at y.
As with H?older continuity, given a projective domain ? ?RPn, define
?? = inf{? | ?? is ?-convex}.
Recall that the dual of a cone U ?Rn is the set of linear functionals positive on
U. Viewing a projective domain ? ?RPn?1 as a cone in Rn we have a notion of a
dual projective domain ??. This relates H?older continuity and ?-convexity:
Lemma 2.20. (Benoist, [1])
1
?? +
1
??? = 1.
Definition 2.21. Fixing an r0 > 0, we define a function ??? : ? ?R as:
???(x) := inf{?(x,y)|dH(x,y) = r0}sup{?(x,y)?|d
H(x,y) = r0}
,
where dH is the Hilbert metric on ?. The idea is to compare the lengths of the
major and minor axes of Hilbert-metric balls in the ambient metric ?.
Lemma 2.22. (Proposition 15 of [6]) Let ? ? E ?RP2 be a strictly convex domain
in an affine patch and ? ? 2. Then ?? is ?-convex if and only if ??? is bounded
away from 0 on ? for ? the Euclidean metric on E.
15
2.5 Guichard?s Theorem
Since the Hilbert metric is ultimately defined in terms of ??, it is reasonable for the
properties of ?? to be related to those of the metric balls in the Hilbert metric on
?. These can be connected by allowing a circle to approach the boundary of ? and
noting that in the limit the H?older continuity and ?-convexity of the metric ball
must equal those of ??. One can then analyze the action of the group elements on
metric balls in ? to calculate their ?-convexity and, in turn, that of ??. Below, we
elaborate on this approach, taken by Guichard in [6] to prove:
Theorem 2.23. (Guichard [6]) Let ? be a strictly convex quasi-homogeneous do-
main in RPn divided by a torsion-free group G. Then the following equivalent as-
sertions are true:
?? = ?G, ?? = ?G
Proof. We provide a sketch of the proof.
The statements are equivalent by Lemma 2.20 and the same duality for groups
dividing the dual domains. Guichard?s proof focuses on the claim that ?? = ?G.
By Lemma 2.22, ?-convexity of ?? can be verified by bounding the function ???
away from 0 on ?.
To prove the claim, Guichard first defines a notion of (r,epsilon1)-loxodromic elements,
which act on RPn and its exterior algebra in a controlled way. Let Gr,epsilon1 be the
subset of (r,epsilon1)-loxodromic elements of G. Guichard shows that for a given (r,epsilon1),
there exists a compact subset Kr of ? such that Gr,epsilon1Kr = ?. The compactness
of Kr assures that ??? is bounded away from 0 on it for any ?. To show that this
remains true for other points of ?, Guichard writes an arbitrary y ? ? as
y = gx, x ? Kr, g ? Gr,epsilon1.
He then analyzes the difference between ???(x) and ???(y). Assuming that ? ? ?g
for each g, he finds a uniform bound for the ratio of the two numbers. Thus, for
? > ?G, the values of ??? away from the compact Kr are bounded away from 0,
proving that ?? is ?-convex.
Note that any group dividing ? is virtually torsion-free by Selberg?s Lemma in
[13]. Extending the definition of ?g by setting ?g = ? if g is torsion, Theorem
2.23 holds for arbitrary G (note that if G were entirely torsion, it would be finite
and hence not act cocompactly on ?). In particular, the theorem is immediately
applicable to the Kac-Vinberg examples, where G is a Coxeter group.
3 The Generalized Tits Cone
We now generalize the construction of the Tits cone, finding a one-parameter family
of cones for each hyperbolic triangle group. The constructed cones include the Kac-
16
Vinberg cones. Applying Guichard?s result to the deformation space of cones leads
to new conjectures.
3.1 Normalizing Triangle Group Representations
We first normalize the Kac-Vinberg cones of Section 2.2, as well as other triangle
group representations.
Let W be an irreducible hyperbolic triangle group:
W = ?s1,s2,s3 flfl s21 = s22 = s23 = (s1s2)p = (s1s3)q = (s2s3)r = 1fi,2 < p ? q ? r, 1p+1q+1r < 1.
Let ? : W ? GL(R3) be a faithful representation of W such that Ri = ?(si) is
a reflection for each i = 1,2,3. Let Li be the fixed plane, and li the -1-eigenspace
of Ri for each i. For each Ri,Rj, the corresponding fixed lines Li,Lj split R3 into
quadrants. We require furthermore that one of these be a fundamental domain for
the action of the group ?Ri,Rj?.
We now define a normalized form for G = ?(W).
Lemma 3.1. There is a basis for R3 such that R1 and R2 are Euclidean reflections
given by:
R1 =
2
4
1 0 0
0 ?1 0
0 0 1
3
5, R2 =
2
4
cos 2pip sin 2pip 0
sin 2pip ?cos 2pip 0
0 0 1
3
5.
Proof. Let (?,?) denote the standard inner product on R3, and define a new inner
product B by
B(x,y) =
X
g??R1,R2?
(gx,gy)
This bilinear form is still positive-definite and non-degenerate, and also invariant
under the group ?R1,R2?. We adopt it as the inner product on R3. Let e3 be a
unit vector in L1 ?L2 and complete {e3} to an orthonormal basis of R3. Since the
reflections fix e3, in this basis they must be of the form
Ri =
2
4
a b 0
c d 0
e f 1
3
5
Furthermore, they must preserve the inner product B. This is equivalent to R1 and
R2 being symmetric:
Ri =
2
4
a b 0
b d 0
0 0 1
3
5
Note also that each Ri is a reflection, so in each case of trace 1, so a = ?d.
17
Restricting to the plane spanned by the first two basis vectors, we get that
R1 and R2 are orthogonal (Euclidean) reflections in R2 fixing the origin. Up to
conjugation by a rotation, R1 fixes (in cylindrical coordinates) the plane ? = 0, and
R2 fixes ? = ?pip; we may assume the sign is positive. Such transformations are
unique and have the stated matrix representation.
Definition 3.2. Using the basis defined in Lemma 3.1, we use the projection
2
4
x
y
z
3
5mapsto?
? x/z
y/z
?
to view the invariant planes Li in the standard coordinate patch ofRP2. After pro-
jection, L1, L2, and L3 become lines. Now, if L3 intersects L2 inside the coordinate
patch, we rescale the patch so that the Euclidean distance from the origin to the
intersection is 1. We then denote by d the x-coordinate of L1?L3. This may be any
number including ?. In the case that the rescaling was impossible, we set d = 0.
Given this normalization, we say that ? is of characteristic (p,q,r,d). Up to the
choice of ordered generators (a marking), a representation has a unique character-
istic.
3.2 Constructing the Generalized Tits Cone
We now construct a generalized Tits cone for a representation of characteristic
(p,q,r,d).
Definition3.3. LetW beanirreduciblehyperbolictrianglegroupandd ? (cos pip,sec pip).
Define the following reflections:
R1 =
2
4
1 0 0
0 ?1 0
0 0 1
3
5
R2 =
2
4
cos 2pip sin 2pip 0
sin 2pip ?cos 2pip 0
0 0 1
3
5
These reflections satisfy (R1R2)p = I. We define the third reflectionR3 by specifying
its eigenvectors (recall that a reflection has eigenvalues {1,1,?1}). We first require
that
R3
2
4
d
0
1
3
5 =
2
4
d
0
1
3
5, and R3
2
4
cos pip
sin pip
1
3
5 =
2
4
cos pip
sin pip
1
3
5,
18
i.e., R3 fixes the plane spanned by these two vectors. Let the third eigenvector be2
4
x
y
z
3
5, define a change-of-basis matrix M, and impose the condition that the three
eigenvectors are linearly independent:
M =
2
4
d cos pip x
0 sin pip y
1 1 z
3
5, detM = 1
R3 is then given by
R3 = M
2
4
1 0 0
0 1 0
0 0 ?1
3
5M?1.
Furthermore, to satisfy the triangle group relations, require:
(R1R3)q = I ?? tr(R1R3) = 2cos 2piq +1
(R2R3)r = I ?? tr(R2R3) = 2cos 2pir +1
Together with detM = 1, this gives three linear equations in x,y,z. When
d /?{cos pip,sec pip},
we can solve for {x,y,z} and complete the definition of R3. We then set ?(si) =
Ri. This gives a representation of the triangle group in GL(3,R). We will show
(Corollary 3.8) that this representation is faithful, so ? has characteristic (p,q,r,d).
Following the notation of the Tits cone, define C to be the cone
C =R+
2
4
0
0
1
3
5?R+
2
4
cos pip
sin pip
1
3
5?R+
2
4
d
0
1
3
5.
For each i, consider the complement of Li, the +1-eigenspace of Ri, and let Ai
be the component containing C (e.g., A1 is the half-space y > 0).
Lastly, define
G = ?(W), U =
[
g?G
gC.
The constraints on d ensure that the corners of the triangular cone C look like
Figure 2:
19
Figure 5: Generalized Tits cone for the (4,4,4) triangle group with d increasing from
.8 to 1.4 in increments of .1.
20
Lemma 3.4. In the above notation, for distinct i,j ? {1,2,3}, n = |RiRj|, there
exists a basis of R3 such that
Ri =
2
4
1 0 0
0 ?1 0
0 0 1
3
5 and Rj =
2
4
cos 2pin sin 2pin 0
sin 2pin ?cos 2pin 0
0 0 1
3
5.
Furthermore,
Ai ?Aj =R+
2
4
1
0
0
3
5?R+
2
4
cos pip
sin pip
1
3
5?R
2
4
0
0
1
3
5.
Proof. The lemma is true by construction for i = 1,j = 2. We focus on the case
i = 1,j = 3.
Lemma 3.1 shows that Ri,Rj can be written in the desired form.
We now approach the normalization in a different way to prove the second as-
sertion. We first conjugate by the affine translation
2
4
1 0 ?d
0 1 0
0 0 1
3
5 so that R3 now
fixes
2
4
0
0
1
3
5. Note that R1 remains unchanged.
Say M is a matrix that conjugates the new R1 and R3 to the desired form. Then
M must commute with R1 and have
2
4
0
0
1
3
5 as an eigenvector (since M
2
4
0
0
1
3
5 has to
be fixed by R3 and also stay in the fixed eigenspace of R1. This gives us that M
must preserve the planes
span
0
@
2
4
0
0
1
3
5,
2
4
1
0
0
3
5
1
A, span
0
@
2
4
0
0
1
3
5,
2
4
0
1
0
3
5
1
A,
These divide R3 into quadrants, which M must permute.
Now, sinced > cos pip, weknowthatA1?A3 iscontainedinoneofthesequadrants.
This remains true for M(A1 ?A3). Given R1 and R3 in normalized form, there are
four choices for A1 ? A3. Two of these are not contained in a quadrant, and the
other two differ by an order-2 rotation that commutes with the reflections. We may
pick the one we want.
The other cases follow immediately, with the use of d < sec pip to prove the cases
i,j ?{2,3}.
Corollary 3.5. The construction of the Tits cone fails for d /? (cos pip,sec pip).
21
Proof. While we can construct cones outside this interval, the corners are not acute
and the corresponding space U does not have nice combinatorial properties. In
particular, the quotient space U/G is an orbifold if and only if d is in the specified
range.
Corollary 3.6. The generalized Tits cone with the origin removed, U\{0}, is open.
Note that we have excluded the case of ideal hyperbolic triangle groups, where
RiRj is of infinite order for some i,j from this discussion. In these cases, U\{0} is
not open and the theory of quasi-homogeneous domains does not apply.
Theorem 3.7. We use the above notation for the generalized Tits cone. Let w ? W.
Then
(P) For each w ? W and s ? S, either
wC ? As and l(w) = l(sw)?1, or
wC ? sAs and l(w) = l(sw)+1.
Informally, if wC is on the ?negative? side of the cone with respect to s, then there
is a minimal word equivalent to w that starts with s.
Proof. Lemma 3.4 and Lemma 2.2 give us Property (P) for each group Wij. The
proof for W is then identical to Theorem 2.4, except for the use of Lemma 3.4
instead of Lemma 2.3.
Corollary 3.8. The representation of W given by ?(si) = Ri is faithful, and that
the generalized Tits cone is convex. The proof is the same as in Section 2.1 for the
standard Tits cone.
The proof of Theorem 2.7, which states that the Tits cone contains no full lines,
relies on the bilinear form on the Tits cone. However, a generalized Tits cone is
given by an symmetric bilinear form on R3 only if it is an elliptical cone. We must
therefore reprove the result for the generalized case. We first define some notation
and prove a lemma:
Definition 3.9. For a generalized Tits cone U, Define ? to be the projection of
U\{0} onto the sphere S2, and E the upper hemisphere z > 0 of S2.
Lemma 3.10. Let U1,U2 be two generalized Tits cones for a given triangle group
(the last parameter, d, may vary). Then the corresponding spaces ?1 and ?2 are
homeomorphic.
Proof. The closures of the generating cones C project to triangles on S2, which are
homeomorphic by stretching the side of length d and imposing a linear isomophism
on the rest of the simplex. Since the rest of ?i is defined by the faithful G-action, the
homeomorphism of triangles extends to an homeomorphism of the spaces ?i.
22
Theorem 3.11. Let G be a hyperbolic triangle group. Then the closure of the
generalized Tits cone U contains no full lines.
Proof. If ? ? E, then we are done, so assume otherwise.
First, consider the case where ? intersects the equator of S2, so there is a point
p ? ???E. Since ? is open, it contains an open neighborhood around p, so there
is a point q of ? that is in the lower hemisphere. The action of ?R1,R2? provides
more points in the lower hemisphere. Since
2
4
0
0
1
3
5 ? ?, we have by convexity that
the generalized Tits cone U = R3 and ? = S2. However, since G is hyperbolic,
its Tits cone does not contain a line by Theorem 2.7, so its projection ?prime into S2
lies entirely in the upper hemisphere E and is contractible. By Lemma 3.10, ? is
homeomorphic to ?prime is impossible.
The remaining case is that ? is contained entirely in E but has a point in the
closure of E. By applying the reflections R1,R2 to the point and invoking convexity,
? = E, so ? = E. Since G leaves ? invariant, it must also leave ?? invariant, so
all elements of G are in the affine group Aff(2,R). The Tits cone provides a faithful
linear representation of G, so G is virtually torsion free by Selberg?s Lemma ([13]),
i.e. there is a torsion-free Gprime ? G of finite index. Since G acts properly, so does
Gprime, and thus ?/Gprime is a manifold, with an induced affine structure. We can also
find a hyperbolic structure on ?/Gprime by means of the G-equivariant homeomorphism
(Lemma 3.10) between ? and the corresponding ?prime for the Tits cone. Since ?/Gprime
has a hyperbolic structure, it has genus greater than one. However, Benzecri showed
in [2] that there is no affine structure on such a manifold.
Consider the projection ? of U into S2 (or, equivalently, into RP2). We may
now apply Theorems 1.1, 5.1 and Fact 5.4 of [1].
Corollary 3.12. Since W is a hyperbolic group,
1. ? is strictly convex (i.e. its boundary does not contain a segment).
2. ?? is once differentiable.
3. All non-torsion elements of G are hyperbolic (have distinct real eigenvalues).
4. The action of G on ?? is minimal.
5. The set { fixed boundary points of g | g ? G non-torsion} is dense in ?????.
Thus, for a given hyperbolic triangle group W, we have constructed a deforma-
tion space of cones with the same combinatorial properties as the Tits cone. We
now apply this to the study of H?older regularity.
23
0.8 0.9 1.0 1.1 1.2 1.3 1.4 d
1
2
3
4
5
?d
Figure 6: ?g for various g of length at most 10, as the parameter d for the generalized
Tits cone varies. p=q=r=4. The thick graphs correspond to the Coxeter elements
R1R2R3 and R3R2R1.
3.3 Some Conjectures
Since for a hyperbolic group W a generalized Tits cone U is convex, we may apply
Guichard?s Theorem 2.23. Denote by ? the projectivization of U, i.e. the image of
U\{0} under the projection map
2
4
x
y
z
3
5mapsto?
? x/z
y/z
?
,
viewed as a coordinate patch of RP2. Since U is strictly convex, the map is defined
on all of U\{0}. By Theorem 2.23 the H?older coefficient of ?? equals the infimum
of the Guichard alphas for the non-torsion elements of G:
?? = inf
g?G,|g|=?
?g = inf
g?G,|g|=?
l3 ?l1
l3 ?l2,
where l1 > l2 > l3 are the logs of the absolute values of the eigenvectors of g.
The explicit construction of the generalized Tits cones allowed us to apply com-
putational methods to the calculation of H?older continuity, leading to the three
conjectures.
Figure 6 shows the graphs for ?g of various elements of the (4,4,4) triangle
group as the deformation parameter d changes. The highlighted curves represent
the Coxeter elements. Based on this graph (and similar further experimental data),
we make the following conjectures:
24
Conjecture 3.13. In the notation for the generalized Tits cone,
?? = min{?R1R2R3,?R3R2R1}
Attempts to prove the conjecture are complicated by the infinite (or at least
prohibitively large) number of equivalence classes of elements, up to conjugacy and
powers. Furthermore, the graph shows that the ?g lines intersect at multiple points,
so a monotonicity result cannot be established.
Conjecture 3.14. Using the notation for the generalized Tits cone with p = 4,q =
4,r = 4, there exists d ? (cos pip,sec pip) such that ?R1R2R3 = ? for any ? ? (1,2).
Corollary 3.15. Let ? ? (1,2). There exist quasi-homogeneous domains ?1,?2
with
??1 < ?, ??2 > ?.
If Conjecture 3.13 is true, ?1 and ?2 can be chosen so that these are equalities.
Proof. Assuming Conjecture 3.14, we find the Tits cone U and corresponding pro-
jective domain ? such that ?R1R2R3 = ?. The first part of the conjecture is then
given by Guichard?s Theorem:
?? = inf
g?G, |g|=?
?g ? ?R1R2R3
The second part is given by duality:
1
?? +
1
??? = 1.
If ?? = min{?R1R2R3,?R3R2R1}, then it is continuous as a function of d. Since
it can be arbitrarily close to 1 and 2, it can be anywhere in the range.
4 Coxeter Elements
We conclude by proving two important properties of Coxeter elements in hyperbolic
triangle groups. We also note that Coxeter elements are critical in other situations,
such as the proof of the Goldman-Parker conjecture ([12, 5]) which considers repre-
sentations of ideal triangle groups in the isometries of complex hyperbolic space.
4.1 Infinite Order, Essential
We now prove Theorem 1.7.
Recall that hyperbolic automorphisms of H have infinite order and have trace
greater than two. To prove Theorem 1.7, we will show that there is an action of a
triangle group on the upper half-planeHu such that the square of a Coxeter element
acts by a hyperbolic transformation.
25
The Tits cone construction shows that W can be embedded into Isom(H) as a
group generated by reflections. We consider a faithful representation of W into the
isometries of Hu:
G = ?R1,R2,R3 flfl (R1R2)p = (R2R3)q = (R1R3)r = R2i = Ifi? PSL(2,R)
Lemma 4.1. Set A = R1R2, B = R2R3, AB = R1R3. Then the trace of the square
of a Coxeter element,
tr(ABA?1B?1) = tr(R1R3R2R1R3R2) = tr(R1R3R2)2,
is given by Goldman?s favorite polynomial K(x,y,z):
tr(ABA?1B?1) = K(x,y,z) = x2 +y2 +z2 ?xyz ?2
where x = tr(A),y = tr(B),z = tr(AB).
Proof. Given X ? SL(2, R), note that by the Cauchy-Schwartz Theorem (a matrix
satisfies its minimal polynomial),
X2 ?tr(X)X +I = 0
X ?tr(X)I +X?1 = 0
Multiplying through by another matrix Y ? SL(2, R),
XY ?tr(X)Y +X?1Y = 0
Taking traces,
tr(XY)?tr(X)tr(Y)+tr(X?1Y) = 0
tr(XY) = tr(X)tr(Y)?tr(X?1Y)
Also note that in SL(2,R) trace is invariant under inversion as well as conjugation.
We now apply the above trace expansion to ABA?1B?1, taking X = A and Y =
BA?1B?1:
tr(ABA?1B?1) = tr(A)tr(BA?1B?1)?tr(A?1BA?1B?1)
= tr(A)2 ?tr(A?1BA?1B?1)
Splitting again with X = A?1B, Y = A?1B?1,
tr(ABA?1B?1) = tr(A)2 ?[tr(A?1B)tr(A?1B?1)?tr(B?1AA?1B?1)]
= tr(A)2 ?tr(AB)[tr(A?1B)]+tr(B2)
= tr(A)2 ?tr(AB)[tr(A)tr(B)?tr(AB)]+[tr(B)tr(B)?tr(I)]
= tr(A)2 +tr(AB)2 ?tr(A)tr(B)tr(AB)+tr(B)2 ?2
= tr(A)2 +tr(B)2 +tr(AB)2 ?tr(A)tr(B)tr(AB)?2
= x2 +y2 +z2 ?xyz ?2
26
Now, A is the product of reflections in two lines meeting at angle pip, so it is
a rotation by angle 2pip . It must therefore have trace ?2cos(pip). Likewise, B has
trace ?2cos(piq), and AB has trace ?2cos(pir). We are free to chose signs for two
of these, but the third is determined once we make the choice. We choose tr(A) =
?2cos(pip), tr(B) = ?2cos(piq) and delay the proof (Lemma 4.3) that this forces
tr(AB) = ?2cos(piq).
Lemma 4.2. K(x,y,z) > 2 for a hyperbolic group.
Proof. Making the above choices for the traces of A, B, and AB, Goldman?s favorite
polynomial becomes
K(p,q,r) = 4[(cos(pip))2 +(cos(piq))2 +(cos(pir))2 +2cos(pip)(cos(piq))(cos(pir))? 12]
We derive this from the cosine angle-sum formula. To ease the notation, define
P = cos(pip),Q = cos(piq),R = cos(pi ? pip ? piq) = ?cos(pip + piq).
If we at the same time had R = cos(pir), this would correspond to the Euclidean
angle case.
cos
 pi
p +
pi
q
?
= cos pip cos piq ?sin pip sin piq
= cos pip cos piq ?
q
1?cos2pip
q
1?cos2piq
?R = PQ?
p
1?P2
p
1?Q2p
1?P2
p
1?Q2 = PQ+R
(1?P2)(1?Q2) = (PQ+R)2
1?P2 ?Q2 +P2Q2 = P2Q2 +2PQR+R2
P2 +Q2 +R2 ?2PQR? 12 = 12
So that in the Euclidean case we would have K(x,y,z) = 2, either the identity or a
parabolic element.
Now, since pip + piq + pir < pi, cos(pir) > R. We now want to prove the following
statement:
P2 +Q2 +(cos(pir))2 ?2PQ(cos(pir))? 12 > P2 +Q2 +R2 ?2PQR? 12
(cos(pir))2 ?2PQ(cos(pir)) > R2 ?2PQR
27
Figure 7: A triangle in the hyperbolic plane (left) can be deformed to the ideal case
(right).
To do this, we show that cos(?)2 ?2PQ(cos(?)) is decreasing with respect to ? on
the interval (cos(pi ? pip ? piq),0). The derivative is negative on this interval since
?sin(?)(cos(?)?2PQ) < 0 iff (cos(?)?2PQ) > 0),
and we have
cos(?) >cos(pi ? pip ? piq)?cos(pip)cos(piq) =
?cos(pip + piq)?cos(pip)cos(piq) = sin(pip)sin(pip) ? 0
Thus, pir > pi ? pip ? piq implies cos(pir) > R, which in turn implies
(cos(pir))2 ?2PQ(cos(pir)) > R2 ?2PQR,
and we have
P2 +Q2 +R2 ?2PQR? 12 > 12.
Then, the trace of ABA?1B?1 is greater than two, so ABA?1B?1 must be a
hyperbolic transformation. Up to conjugation, it is the square of a Coxeter element
g, so g has infinite order. The inverse g?1 also has infinite order, and represents the
other conjugacy class of Coxeter elements in the triangle group.
We still have to decide how to assign signs to the traces of A, B, and AB.
Lemma 4.3. If tr(A) = ?2cos(pip), tr(B) = ?2cos(piq), and tr(AB) = ?cos(pir),
then tr(AB) = cos(pir).
Proof. We first consider the limiting case p = ?, q = ?, r = ?. A,B,AB are then
parabolic elements. Without loss of generality,
A =
? ?1 1
0 ?1
?
,B =
? ?1 0
b ?1
?
28
The product is then
AB =
? 1+b ?1
?b 1
?
,
which has trace 2+b = ?cos(pi/?) = ?2.
If we choose tr(AB) = 2, then b = 0, making B projectively equivalent to the
identity. Since it?s the product of two different reflections, this is impossible. We
must therefore have a negative trace.
Now, recall that A = R1R2, B = R2R3, where R1,R2,R3 are reflections over
hyperbolic lines defining a triangle with interior angles pip, piq, pir (Figure 7). If we
move the lines, the corresponding transformations vary continuously. If we move
them so that the endpoints of the triangle move toward the boundary, the interior
angles decrease, so the trace of AB approaches ?2 through a monotone function.
Since the interior angle starts off acute, it stays acute, so the trace of AB is never
0. Thus, tr(AB) < 0 in the case we are interested in.
4.2 Minimizing Translation Length in the Hilbert Metric
Lemma 4.4. (McMullen, [11]) Let ? be a strictly convex domain in RP2, and
g ? PGL(3, R) an automorphism of ? with eigenvalues |?1| > |?2| > |?3| and
li = log(|?i|). Then,
inf
p??
dH(p,gp) = l1 ?l32
Proof. Let x+,xo,x? ? RP2 be, respectively, the attracting, saddle, and repelling
fixed points of g. Pick a basis {e1,e2,e3 with e1 ? x?, e2 ? xo, e3 ? x+, so that x+
is the origin of the standard coordinate patch, and Tx+M is the x-axis (as in Figure
4). g acts on the coordinate patch by the linear transformation
? ?
2??11 0
0 ?3??11
?
.
We let p =
? 0
y
?
and calculate the translation length (restricting to the positive
y-axis for the second equality):
d(p,gp) = d
 ? 0
y
?
,
? 0
?3??11 y
??
= 12 log[?,y;?3??11 y,0]
= 12 log
flfl
flfl(y ?0)(?3??11 y ??)
(y ??)(?3??11 y ?0)
flfl
flfl
= 12 log
flfl
flfl y
??3??11 y
flfl
flfl
= l1 ?l32
29
Note that the above calculations technically require taking limits instead of
dividing infinities, but the result is the same.
Now, if p is not on the y-axis, we may project segment ???p,gp?? into the positive
y-axis. Projections decrease the value of the cross-ratio, and thus the distance
between p and gp is at least as large as if p had been on the y-axis in the first place.
Thus, p is moved by a distance of at least l1?l32 .
Theorem 4.5. (McMullen, [11]) Coxeter elements minimize translation length in
the generalized Tits cone ? among the essential elements of W.
inf
w?W,w essential
 
inf
p??
dH(p,gp)
?
= min
w?W,w Coxeter
 
inf
p??
dH(p,gp)
?
Proof. We say a curve in ? represents an element w ? W if it connects a point p
with its image wp. The translation length of an element is the infimum over the
lengths of all curves ? that represent it. Since the geodesics in the Hilbert metric
are the straight segments, we may restrict the infimum to the ? that are segments.
We now show that any straight segment ? representing w can be modified with-
out changing its length to represent a Coxeter element.
Pick a basepoint p, which is without loss of generality contained in C. Consider
the straight segment? connectingpand wp. Using the inductive process in Corollary
2.6, we may build a minimal representation si1 ...sik of w such that a ? goes from
p to wp, first crosses the side of the fundamental region corresponding to si1, then
si1si2, etc.
Now, since w is essential, all generators of W must appear in si1 ...sik, so some
Coxeter element g appears as a subword of w. These combinatorial properties of
the group are reflected in the generalized Tits cone.
We now modify ?: it is only allowed to cross a wall (of the image of the funda-
mental domain) when the wall corresponds to a letter of the Coxeter element we are
trying to create. If it doesn?t, ? reflects off of the wall (the modification to ? can be
made precise by considering orbifold covering spaces). For example, if w = s1s3s1s2,
we would allow it to cross the side corresponding to s1 only once. As a result, the
modified ? now represents a Coxeter element, but has the same length.
30
References
[1] Y. Benoist. Convexes divisibles I. Tata Inst. Fund. Res. Stud. Math., 17:339?
374, 2004.
[2] J. P. Benzecri. Vari?et?es localement affines. Sem. Topologie et Gom. Diff., Ch.
Ehresmann, 7, 1958-60.
[3] N. Bourbaki. Lie Groups and Lie Algebras Chapters 4-6. Springer, 1968.
[4] M. Davis. The Geometry and Topology of Coxeter Groups. Princeton University
Press, 2008.
[5] W. Goldman and J. Parker. Complex hyperbolic ideal triangle groups. J. Reine
Angew. Math., 425:71?86, 1992.
[6] O. Guichard. On Holder regularity of divisible convex subsets. Ergodic Theory
Dyn. Syst., 25(6):1857?1880, 2005.
[7] W. Moser H. Coxeter. Generators and Relations for Discrete Groups. Springer-
Verlag, 1972.
[8] V. Kac and E. Vinberg. Quasi-homogeneous cones. Math. Notes, 1(3):347?354,
1967.
[9] E. Katok. Fuchsian groups. University of Chicago Press, 1992.
[10] J. L. Koszul. Lectures on hyperbolic Coxeter groups. Univ. Notre Dame, 1967.
[11] C. T. McMullen. Coxeter groups, Salem numbers and the Hilbert metric. Publ.
Math. Inst. Hautes tudes Sci., 95:151?183, 2002.
[12] R. Schwartz. A better proof of the Goldman-Parker conjecture. Geom. Topol.,
9:1539?1601, 2005.
[13] A. Selberg. Contributions to Function Theory (Internat. Colloq. Function The-
ory, Bombay), pages 147?164, 1960. Tata Institute of Fundamental Research,
Bombay.
[14] E. Vinberg. Discrete groups that are generated by reflections. Math USSR Izv.,
5:1083?1119, 1971.
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