THE AMPLE CONE FOR A K3 SURFACE The ample cone or Kähler

19 Pages·2009·1.19 MB·English

THE AMPLE CONE FOR A K3 SURFACE
ARTHURBARAGAR
Abstract. In this paper, we give several pictorial fractal representations of
the ample or Ka¨hler cone for surfaces in a certain class of K3 surfaces. The
classincludessurfacesdescribedbysmooth(2,2,2)formsinP1×P1×P1defined
overasufficientlylargenumberfieldK,whichhavealineparalleltooneofthe
axes,andhavePicardnumberfour. WerelatetheHausdorffdimensionofthis
fractal to the asymptotic growth of orbits of curves under the action of the
surface’sgroupofautomorphisms. WeexperimentallyestimatetheHausdorff
dimensionofthefractaltobe1.296±.010.
TheampleconeorK¨ahlerconeforasurfaceisasignificantandoftencomplicated
geometric object. Though much is known about the ample cone, particularly for
K3surfaces,only a few non-trivialexamples have been explicitly described. These
include the ample cones with a finite number of sides (see [N1] for n = 3, and
[N2, N3] for n 5; the case n = 4 is attributed to Vinberg in an unpublished
≥
work [N1]); the ample cone for a class of K3 surfaces with n = 3 [Ba3]; and the
ample cones for several Kummer surfaces, which are K3 surfaces with n = 20
[V, K-K, Kon]. Though the complexity of the problem generically increases with
n, the problem for K3 surfaces with maximal Picard number (n = 20) appear to
be tractable because of the small size of the transcendental lattice.
In this paper, we introduce accurate pictorial representations of the ample cone
and the associated fractal for surfaces within a class of K3 surfaces with Picard
number n = 4 (see Figures 1, 3, 4, and 5). As far as the author is aware, the
associated fractal has not been studied in any great depth for any ample cone for
whichthe fractalhasa non-integerdimension,except the one in [Ba3]. The fractal
inthatcaseisCantor-like(itisasubsetofS1)andrigorousboundsonitsHausdorff
dimension are calculated in [Ba1]. The Hausdorff dimension of the fractal of this
paper is estimated to be 1.296 .010.
±
Our second main result is to relate the Hausdorff dimension of the fractal to
the growth of the height of curves for an orbit of curves on a surface in this class.
Precisely,letV be asurfacewithinourclassofK3surfacesandlet =Aut(V/K)
A
be its group of automorphisms over a sufficiently large number field K. Let D be
an ample divisor on V and let C be a curve on V. Define
N (C)(t,D)=# C′ (C):C′ D <t .
A { ∈A · }
Here we have abused notation by letting C also represent the divisor class that
′
containsC . TheintersectionC D shouldbethoughtofasalogarithmicheightof
′ ′
·
2000MathematicsSubjectClassification. 14J28,14J50,11D41,11D72,11H56,11G10,37F35,
37D05.
Keywordsandphrases. Fractal,Hausdorffdimension,K3surface,Kleiniangroups,dynamics.
I would like to thank Jeff Lagarias, Bert van Geeman, James McKernan, and the referee for
some valuable conversations and advice. This material is based upon work supported by the
NationalScienceFoundationunderGrantNo.0403686.
1
2 ARTHURBARAGAR
C ,andthequantityN (t,D)shouldbethoughtofasananalogofN (t,h ),
′ (C) (P) D
A A
whichisthenumberofpointsP inan orbitofapointP withWeilheighth (P )
′ D ′
A
bounded by t. See Section 5 of [Ba1] for a detailed comparison. Let
log(N (t,D))
(C)
α= lim A .
t logt
→∞
ForcurvesC withC C >0,this limitexists andis the Hausdorffdimensionofthe
·
fractal associated to the ample cone. Our estimate for α is inferred from a plot of
log(N (t,D)) as a function of logt.
(C)
It sAhould be stressedthat the dynamics studied in this paper are in Pic(V) R,
⊗
andnot onV itself. The richsubjectof dynamics onK3 surfaceshas been studied
by Cantat [C] and McMullen [McM]. Both authors study the dynamics of an
automorphismwhosepullbacktoPic(V)ishyperbolic. OnPic(V) R,thedynamics
⊗
of the pullback is discrete and uninteresting, but on the K3 surface, the orbits of
points under iterations of this map are in places dense and quite fascinating. See
in particular Figure 1 in [C] and Figure 2 in [McM].
1. Background
LetV be analgebraicK3surfacedefinedovera numberfield K. Thatis,V is a
surface with trivial canonical divisor and irregularity equal to zero. Its arithmetic
genus ρ is 1. If C is a smooth curve on V, then the genus of C is given by the
a
adjunction formula,
2g 2=C C.
− ·
The Picard group Pic(V) is an even lattice of dimension n 20. Let =
≤ D
D ,...,D be a basis of Pic(V), so
1 n
{ }
Pic(V)=ZD ZD ZD .
1 2 n
⊕ ⊕···⊕
Let J = [D D ] be the intersection matrix for V with respect to the basis .
i j
· D
By the Hodge index theorem, the signature of J is (1,n 1). That is, J has one
−
positive eigenvalue and n 1 negative eigenvalues.
−
Let =Aut(V/K)bethegroupofautomorphismsonV. Foranautomorphism
A
σ , the pullback σ acts linearly on Pic(V), so can be represented by a matrix
∗
∈A
with integer entries. Since σ preserves intersections, we further have that σ is in
∗
= (Z)= T Sl (Z):TtJT =J .
n
O O { ∈ }
For an ample divisor D, the hypersurface xtJx = D D is a hyperboloid of two
·
sheets, one of which contains D. Let us distinguish this sheet with , and define
H
+ = +(Z)= T :T( )= .
O O { ∈O H H}
Thesurface canbethoughtofasamodelofn 1dimensionalhyperbolicgeometry
H −
imbedded in a Lorentz space, where the Lorentz product is the negative of the
intersectionproduct. Thedistance AB betweentwopointsAandB inthismodel
| |
is given by
H
AtJB =A B = A B cosh(AB )= D Dcosh(AB ),
· || |||| || | | − · | |
where A =√ A A. With this metric, the group of isometries on is +(R),
|| || − · H O
wherethe definition ofthis groupis the obviousanalogof +(Z). For moredetails
O
of this model, we refer the reader to Ratcliff [R].
THE AMPLE CONE FOR A K3 SURFACE 3
Let be the set of effective divisor classes in Pic(V). That is, E if we
E ∈ E
can write E = a C +...+a C with a 0 and C a divisor class that can be
1 1 m m i i
≥
represented with a curve on V.
A divisor D is ample if D E > 0 for all effective divisors E. The ample cone,
·
also known as the K¨ahler cone, is the set
= C Pic(V) R:C E >0 for all E .
K { ∈ ⊗ · ∈E}
The ample cone is convex, and is a subset of the light cone
+ = x Pic(V) R:x x>0,x D >0 ,
L { ∈ ⊗ · · }
where D is any ample divisor. If an irreducible curve has negative self intersection
onV, then it is a smooth rationalcurve andhas self intersection 2. Such a curve
−
is called a 2 curve. If V contains no 2 curves, then = + [Kov]. Let be
2
− − K L E−
the setof nodalclasses;that is, the setof divisorclasses ofsmoothrationalcurves.
Then
= C Pic(V) R:C E >0 for all E .
2
K { ∈ ⊗ · ∈E− }
Furthermore, every plane C x = 0 with C is a face of , so there are no
2
· ∈ E− K
superfluous inequalities in the above description [St].
When the Picard number n is 4, an appropriate (Euclidean) cross section of
the light cone is a sphere. The ample cone is bounded by hyperplanes, each of
which cut out a circle on this sphere. This is one way of describing Figure 1 –
eachcircleinthe patternrepresentsoneofthehyperplanesthatproperlybound .
K
There is, however, a more satisfying interpretation. Each bounding hyperplane of
intersectsthehypersurface ina(hyperbolic)plane. Whenviewedfromapoint
K H
P with P on , we get the Poincar´e sphere representation of , as is further
0 0
− H H
explained in Section 4. Using this interpretation,each circle in Figure 1 represents
a hyperbolic plane, and is the region in the Poincar´esphere bounded by all
K∩H
these planes. The advantage of this interpretation is that it does not depend on
the choice of basis. The different pictures one gets correspond to different choices
of P , so the object is the same – it is only our perspective that has changed.
0
It is clear that if σ , then σ ( )= , so let us define
∗
∈A K K
= T + :T = .
′′
O { ∈O K K}
Ifthereareany 2curvesonV,thenthereexists alargesubsetof + thatcannot
− O
be in . Suppose C C = 2 and C is effective. Let us define the reflection
′′
O · −
through C by
R D =D+(C D)C.
C
·
In the Lorentz space, this is a reflection through the hyperplane C x = 0, or
·
in the hyperbolic geometry , it is reflection through the (hyper)line formed by
H
the intersection of with C x = 0. Note that R is in , since it preserves
C
H · O
intersections. But since R C = C, we have that R .
C C ′′
− 6∈O
Since the canonical divisor on V is trivial and the arithmetic genus is one, the
Riemann-Roch formula on V is
1
l(D)+l( D) D D+2,
− ≥ 2 ·
where l(D) is one more than the dimension of the complete linear system for D.
Hence, if l(D) > 0, then D is effective. In particular, if C C = 2, then either
· −
C or C is effective. Thus R = R for all divisors C Pic(V) with
C C ′′
C C−= 2. Let be the subgroup−of 6∈+Ogenerated by the refle∈ctions through
′
· − O O
4 ARTHURBARAGAR
D
1
Figure 1. A representation of the ample cone. The point D is
1
an accumulation point of circles (one of infinitely many).
2 curves. Note that TR T 1 = R . Hence, is a normal subgroup of +.
C − TC ′
−In [PS-S], Pjatecki˘i-S˘apiroand S˘afarevi˘c show thaOt, for a sufficiently large numOber
field K, the natural map
Φ:Aut(V/K) ′′
→O
σ σ
∗
7→
has a finite kernel and co-kernel, and that = +/ .
O′′ ∼O O′
2. (2,2,2) forms in P1 P1 P1
× ×
The results of this paper depend only on the Picard lattice Pic(V), and applies
toanyK3surfacewiththeintersectionmatrixJ describedbelow. Yettheresultis,
intheauthor’sopinion,muchmoreappealingwhencoupledwithaclassofsurfaces
that have the given Picard lattice. In this section, we describe such a class of K3
surfaces.
Let V be a surface described by a smooth (2,2,2) form in P1 P1 P1. Such
× ×
a surface is the zero locus of a polynomial in three projective variables that is
quadratic in each variable. Let us write
(1) F(X,Y,Z)=X2F (Y,Z)+X X F (Y,Z)+X2F (Y,Z)=0,
0 0 0 1 1 1 2
where X = (X : X ) P1, and the F (Y,Z) are (2,2) forms in P1 P1. Since
0 1 i
∈ ×
smooth (2,2) forms in P1 P1 are elliptic curves, V is fibered by elliptic curves in
×
each of the three directions.
THE AMPLE CONE FOR A K3 SURFACE 5
Let
p : P1 P1 P1 P1
1
× × →
(X,Y,Z) X
7→
be projection onto the X-axis, and similarly define p and p . Let
2 3
Di =p−i 1H
be the pull back of a point H P1 for i=1,2,3.
∈
2.1. The generic case. Generic surfaces in this class have been studied by Wang
[W], Billard [Bi], and the author [Ba2]. Explicit examples (defined over Q and
with full dimension in the moduli space) are given in [B-L]. If V is generic, then
Pic(V)=D Z+D Z+D ZsothePicardnumberisn=3;theintersectionmatrix
1 2 3
with respect to this basis is
0 2 2
2 0 2 ;
 
2 2 0
 
and V contains no 2 curves, so its ample cone is the light cone.
−
An attractive feature of studying smooth (2,2,2) forms in P1 P1 P1 is that
× ×
weget‘forfree’severalautomorphisms. LetusviewEq.1asaquadraticinX with
two roots (say) X and X . We define the map
′
σ : V V
1
→
(X,Y,Z) (X′,Y,Z).
7→
Explicitly,
(F (Y,Z)X +F (Y,Z)X : F (Y,Z)X ) if this is in P1
X′ = (F1(Y,Z)X1+F0(Y,Z)X0 :−F0(Y,Z)X1) otherwise.
1 0 2 1 2 0
(cid:26) −
In the generic case, the curves F (Y,Z) = 0 do not intersect, so σ is defined
i 1
everywhere. We similarly define σ and σ . The action of σ on Pic(V) is
2 3 1
1 0 0
−
σ = 2 1 0 ,
1∗  
2 0 1
 
while σ and σ are symmetrically defined. The map σ is reflection across a line
2∗ 3∗ i∗
in H2 with endpoints D and D , where i, j, k is a permutation of 1, 2, and 3.
j k
Thus,thefundamentaldomainfor σ ,σ ,σ isatriplyasymptotictriangle,soby
h 1∗ 2∗ 3∗i
Pjatecki˘i-S˘apiro and S˘afarevi˘c’s result, σ ,σ ,σ has finite index in Aut(V/K).
1 2 3
h i
2.2. A class with Picard number 4. If there exists a point (Q,Q) P1 P1
′
over K¯ such that ∈ ×
F0(Q,Q′)=F1(Q,Q′)=F2(Q,Q′)=0,
then(X,Q,Q)isapointonV forallX P1. Thatis,V containsalineparallelto
′
∈
the X-axis. Since this is a smooth curve of genus zero, it must be a 2 curve. If a
−
surface is generic, modulo this condition, then it has Picard number n=4. These
are the K3 surfaces we study in this paper.
6 ARTHURBARAGAR
AnexampleofsuchaK3surfaceisthesurfaceV givenbyF(X,Y,Z)=0where,
in affine coordinates,
F((x:1),(y :1),(z :1))=f(x,y,z)
=x2(y2+yz+z2+z)+x(y2z2+y2z+z)+(y2z2+y2z+y+z)
This surface includes the line (X,(0 : 1),(0 : 1)), so its Picard number is at least
four. Theequationf(x,1/y,z 1)isequivalentmodulotwotothepolynomialthat
−
describes the surface Y in [B-L], and so V has Picard number four. We can use
2
this example to find infinitely many. Let
G(X,Y,Z)=X2G (Y,Z)+X X G (Y,Z)+X2G (Y,Z)=0
0 0 0 1 1 1 2
be a (2,2,2) form with rational coefficients with odd denominators. The curves
G (Y,Z)=0andG (Y,Z)=0intersectsomewhere,sayatY =QandZ =Q. Let
0 1 ′
us suppose that Q and Q are rational. Though we have imposed some constraints
′
on G, those constraints are of codimension 0 in the moduli space of all (2,2,2)
forms. By making a change of basis, we may assume that Q = Q = 0. After this
′
changeofbasis,letusrequirethatG (Y,Z)havenoconstantterm,soG (0,0)=0.
2 2
This is a one-dimensional constraint. The set of (2,2,2) forms given by
F(X,Y,Z)+2G(X,Y,Z)=0
is ofcodimensionone in the set ofall(2,2,2)forms,and allhavePicardnumber 4.
We have therefore constructed a set of K3 surfaces of full dimension in the moduli
space of all (2,2,2) forms with Picard number 4.
LetV beasurfaceinthisclass. LetussupposethatthelineisparalleltotheX-
axisanddenotethedivisorclassforthislinewithD . Sincethereisonlyonecurve
4
inthisdivisorclass(D D = 2),letusabusenotationslightlyandrepresentthe
4 4
· −
line with D too. Note that the fibers in D and D over Q and Q are each the
4 2 3 ′
union of the line D and a conic.
4
Lemma 2.1. The set = D ,D ,D ,D forms a basis of Pic(V).
1 2 3 4
D { }
Proof. The set is a linearly independent subset of Pic(V), so generatesa sublat-
D
tice of finite index in Pic(V). Let this index be m. It is easy to check that
0 if i=j =4
Di·Dj = 2 if i=j;6 i,j =1,2, or 3
(cid:26) 6
1 if i=1
D D =
4· i 0 if i=2, 3.
(cid:26)
Thus, the intersection matrix J for the sublattice generated by is
D
0 2 2 1
2 0 2 0
J = .
2 2 0 0
1 0 0 2
 − 
Since m2 divides det(J)= 28,weknow m=1 or 2. If m=2, then there exists a
−
nonzerodivisorD Pic(V) that is a linearcombinationof the elements of with
′
∈ D
coefficients 0 or 1/2. Since D Pic(V), we know D D is even and D D Z.
′ ′ ′ ′ i
∈ · · ∈
Of the fifteen combinations, only two satisfy these conditions, namely D /2, and
2
D /2. Since D andD representclassesof elliptic curves,they cannotbe reduced
3 2 3
as the sum of two elliptic curves. (cid:3)
THE AMPLE CONE FOR A K3 SURFACE 7
Remark 1. Sincethelatticewithbasis D ,D /2,D ,D iseven,weknowthere
1 2 3 4
{ }
must exist a K3 surface whose Picard lattice is this lattice [Mo].
3. The group of automorphisms
A surface V in our class of K3 surfaces includes the automorphisms σ and σ ,
2 3
as defined in the generic case. The map σ as described above in the generic case
1
is defined in our case for all points on V D . We can extend this map to D as
4 4
\
follows: Without loss of generality, we may assume Q = 0 and Q = 0 (where
1 6 ′1 6
Q = (Q : Q ), Q = (Q : Q )). Let q = Q /Q , q = Q /Q , and consider
0 1 ′ ′0 ′1 0 1 ′ ′0 ′1
(y,z)=(Y /Y ,Z /Z ) in a neighborhood of (q,q ). Let
0 1 0 1 ′
M (z q )=M (y q),
1 ′ 0
− −
and write F , F and F as functions of y and M =(M :M ) P1. Then,
0 1 2 0 1
∈
F (q,M)=0
i
for all M and all i, so (y q) divides F (y,M) for all i. Define G (y,M) so that
i i
−
(y q)G (y,M)=F (y,M).
i i
−
Note that if G (q,M) = 0 for all i and some M, then V has a singularity. Thus,
i
since V is smooth, the equation
X2G (y,M)+X X G (y,M)+X2G (y,M)=0
0 0 0 1 1 1 2
is a quadratic in X for all y in a neighborhoodof q; it is defined on an opensubset
U of V containing D ; and, as before, we may define a map σ on U. The two
4 1′
maps σ and σ agree wherever they are both defined, and piecing them together
1 1′
we get an automorphism σ of V.
1
The following describes how σ acts on Pic(V):
i∗
Theorem 3.1. Let T be the matrix representation of σ in the basis . Then
i i∗ D
1 0 0 0 1 2 0 0 1 0 2 0
−
2 1 0 0 0 1 0 0 0 1 2 1
T1 = 2 0 1 0, T2 =0 −2 1 1 , T3 =0 0 1 0 .
−
 1 0 0 1 0 0 0 1 0 0 0 1
−   −   − 
     
Proof. WecalculateT byconsideringtheintersectionnumbersofthebasiselements
1
in intersected with the elements in the image of under σ . Since σ leaves Y
D D 1∗ 1
and Z fixed, we know σ D =D , σ D =D and σ D =D . Thus,
1∗ 2 2 1∗ 3 3 1∗ 4 4
σ1∗D1·D2 =D1·(σ1∗)−1D2 =D1·σ1∗D2 =D1·D2 =2
σ D D =2
1∗ 2· 1
etc.
Observethat σ2 is the identity, a propertythat wasmade use ofin the above. The
i
only difficult calculation is
σ D D =7.
1∗ 1· 1
Recall that D is the divisor class of the curve
1
F(H,Y,Z)=0
for some fixed H. From Eq. 1 we have
H2F (Y,Z)+H H F (Y,Z)+H2F (Y,Z)=0,
0 0 0 1 1 1 2
8 ARTHURBARAGAR
which is a (2,2) form in P1 P1 (if we think of H as a constant). The divisor D
1
×
intersects σ D if H is a double root of this equation. That is, if
1∗ 1
2H F (Y,Z)+H F (Y,Z)=0,
0 0 1 1
which is another (2,2) form. These two curves intersect at eight points, but one
solutionis(Y,Z)=(Q,Q),forwhichtheaboveargumentthatH isadoublepoint
′
is not valid. Thus,
σ1∗D1·D1 =7.
We therefore get
7 2 2 1
2 0 2 0
T1tJ =JT1 =2 2 0 0 ,
1 0 0 2
 − 
 
from which we derive T .
1
For T , we note that σ D = D and σ D = D , so the only difficult intersec-
2 2∗ 1 1 2∗ 3 3
tions to calculate areσ D D , σ D D , andσ D D . We calculate σ D D
2∗ 2· 2 2∗ 2· 4 2∗ 4· 4 2∗ 2· 2
inthesamewaythatwecalculatedtheintersectionσ D D above,onlythistime
1∗ 1· 1
no intersections are discarded, so σ D D =8.
2∗ 2· 2
We calculate σ D D by noting that the curve F(X,Y,Q) is the union of the
2∗ 2· 4 ′
line D and the conic σ D . Thus
4 2 4
D =D +σ D .
3 4 2∗ 4
Hence
σ D D =σ D D σ D σ D
2∗ 2· 4 2∗ 2· 3− 2∗ 2· 2∗ 4
=2+0=2.
We also use this observation to calculate σ D D :
2∗ 4· 4
0=D D =(D +σ D ) (D +σ D )
3· 3 4 2∗ 4 · 4 2∗ 4
= 2+2σ D D 2.
− 2∗ 4· 4−
Thus,
0 2 2 1
2 8 2 2
T2tJ =JT2 =2 2 0 0,
1 2 0 2
 
from which we derive T . The matrix T is found in asymmetric way. (cid:3)
2 3
The elements T , T , and T are all in + (as they should) but the group
1 2 3
O
T ,T ,T is not all of +. There is, of course, the reflection R through D ,
h 1 2 3i O D4 4
and the map S that switches D and D , which are
2 3
1 0 0 0 1 0 0 0
0 1 0 0 0 0 1 0
RD4 =0 0 1 0 , and S =0 1 0 0.
1 0 0 1 0 0 0 1
 −   
   
THE AMPLE CONE FOR A K3 SURFACE 9
There is also another matrix
1 8 8 0
0 1 0 0
T4 =0 −0 1 0,
−
0 4 4 1
 
 
which was found by trial and error. Note that T =ST S.
3 2
Lemma 3.2. The maps S and T are in .
4 ′′
O
Proof. Any map T + can be written as a product T = T T where T
′ ′′ ′ ′
and T (by P∈jatOecki˘i-S˘apiro and S˘afarevi˘c’s result). The map T is in ∈ Oif
′′ ′′ ′′
∈ O O
and only if T =I. Note that T = for any T that is not the identity.
′ ′ ′ ′
K∩K ∅ ∈O
Thus, to show that T , it is enough to find an ample D such that TD is also
′′
∈O
ample.
The divisor D =D +D +D is ample, and SD =D, so S .
1 2 3 ′′
∈O
Since the divisor D represents an elliptic fibration, it is irreducible. Since D
1 4
represents a 2 curve, it is irreducible. Let D = 2D +D . The intersection of
′ 1 4
−
D with its irreducible components arenon-negative,so the intersectionofD with
′ ′
any effective divisoris non-negative. Hence, D is in the closure of the ample cone.
′
Since D D =0, it is on the boundary of the ample cone. Thus, since the ample
1 1
·
coneisconvex,thedivisorD =D +D =3D +D iseitherintheampleconeor
′′ ′ 1 1 4
ontheboundaryoftheamplecone. Ifitisontheboundaryoftheamplecone,then
there exists a 2 curve whose intersection with D and 2D +D are both 0. It is
1 1 4
−
not difficult to verify that no such 2 curve exists, so D is not on the boundary
′′
of the ample cone, and is therefore−ample. Finally, T D =D , so T . (cid:3)
4 ′′ ′′ 4 ′′
∈O
Remark 2. Thereexistsanautomorphismσ Aut(V)suchthatσ =T . Itcan
4 ∈ 4∗ 4
bedescribedasfollows: ForapointP V,letE betheellipticcurveinthedivisor
∈
class D such that P E. Let O be the unique point of intersection between E
1 E
∈
and D . Define σ (P) = P, where P is the additive inverse of P with respect
4 4
− −
to addition on the elliptic curve E with zero at O . Then σ is an automorphism
E 4
on V. For every E in the class D , σ fixes E, and since σ (O ) = O , it also
1 4 4 E E
fixes D . Thus D and D are both eigenvectors of σ . This is enough to narrow
4 1 4 4∗
down the possibilities for σ to T , ST , S, and the identity. In [B-McK] we show
4∗ 4 4
σ =T .
4∗ 4
This alsoshowsthat,forthe classofK3surfacesdescribedinSection2,wemay
take K =Q. That is, Q is a sufficiently large number field.
4. Visualizing the group
′′
O
As noted earlier, the hypersurface is a three dimensional hyperbolic space.
H
In this section, we describe how to project to the Poincar´e ball and upper half
H
plane models ofH3,andthroughthese models,we investigatethe tilings generated
by the actions of + and on .
′′
O O H
The matrix J has three negativeeigenvaluesandonepositive eigenvalue. Let us
denote them with a2, a2, a2, and a2. Let Q be the matrix that diagonalizes
− 1 − 2 − 3 4
J so that
J = QtAtJ AQ,
0
−
10 ARTHURBARAGAR
where
a 0 0 0 1 0 0 0
1
0 a 0 0 0 1 0 0
A=0 02 a 0 and J0 =0 0 1 0 .
3
0 0 0 a4 0 0 0 −1
   
The surface + described by ytJ y = 1 with y > 0 is the usual Lorentz model
0 4
V −
of H3. The map y = λAQx sends to +, where λ=(√D D) 1. The stereo-
−
± H V ·
graphic projection of + to the hypersurface y =0 through the point (0,0,0, 1)
4
V −
is an isomorphism of + to the Poincar´e ball model of H3. This projection distin-
V
guishes a point, the point P on where P = λ 1Q 1A 1(0,0,0,1). The point
0 0 − − −
H
P is the point “closest to the eye,” and is mapped to the center of the Poincar´e
0
ball. Byapplyinganisometryto beforeprojecting,wecanmakeachoiceforP .
0
H
Throughoutthis paper,we makethe choiceP =D +D +D , whichis anample
0 1 2 3
divisor.
ThePoincar´eballcanbeunwrappedtogivethePoincar´eupperhalfspacemodel
for H3. This map distinguishes a point P on ∂ , the boundary of , which is
∞ H H
sent to the point at infinity. The point P can be represented with a point on the
light cone x x =0. In the upper half sp∞ace model (with z > 0), planes in H3 are
·
modeled with planes and half spheres that are perpendicular to the plane z = 0.
This model is often representedby its boundary,the planez =0 together with the
point at infinity, on which planes in H3 appear as circles or lines. The rest of the
figures in this paper are these types of representations of H3.
The map T is a reflection through a plane in . That plane is the intersection
1
H
of with the hyperplane (2, 2, 2,1) x=0. When mapped to the plane z =0,
H − − ·
as described above with the point P =D , its boundary is a circle Γ , as shown
1 1
∞
in Figure 2. The plane includes the points (divisors) D and D , which are on the
2 3
light cone x x=0, which represents ∂ .
· H
ThemapSisalsoareflectionthroughaplanein . Thatplaneistheintersection
H
of with the hyperplane (0,1, 1,0) x =0, and is represented by the line Γ in
S
H − ·
Figure 2. It includes the points D , D +D , and P =(1,1,1, 2).
1 1 4 1
−
The maps T , T , andT areeachrotationsby π aboutparallellines (parallelin
2 3 4
),eachwithoneendpointatD ,andtheotherendpointatD ,D ,andD +D ,
1 3 2 1 4
H
respectively. The map T rotates everything on one side of the plane represented
2
by Γ to the other side. The map T maps everything from on one side of Γ to
2 3 3
the other.
ThemapR isreflectionthroughtheplanerepresentedbyΓ . InFigure2,the
D4 ′1
planes Γ and Γ are the planes perpendicular to T D and T D , respectively, so
′2 ′3 2 4 3 4
represent the reflections T R T and T R T , respectively.
3 D4 3 2 D4 2
Wealsorepresent,withΓ ,theplanethroughwhichT T T S =T T ST reflects.
5 2 4 3 2 4 2
We can now see enough to prove something.
Let = T ,T ,T ,S,R ,andlet betheregioninH3boundedbytheplanes
G h 1 2 4 D4i F
representedby Γ , Γ , and Γ in Figure 2, and above the hemispheres represented
2 5 S
by Γ , Γ , and Γ .
′1 ′2 1
Lemma 4.1. The group has finite index in +.
G O
Proof. The region is a fundamental domain for . Since is a subgroup of +,
F G G O
a fundamental domain for + lies within . Since has finite volume, the group
has finite index in +. O F F (cid:3)
G O