Our rst result is the direct analogue of the main theorem of [5], and can be roughly stated as follows: Theorem. Let R be a complex reection of order

Transcription

1 Unfaithful complex hyperbolic triangle groups II: Higher order reections John R. Parker Department of Mathematical Sciences, University of Durham, South Road, Durham DH LE, England. Julien Paupert Department of Mathematics University of Utah 55 South 00 East Salt Lake City, Utah 8, USA. November 7, 007 Abstract We consider symmetric complex hyperbolic triangle groups generated by three complex reections with angle =p. We restrict our attention to those groups where certain words are elliptic. Our goal is to nd necessary conditions for such a group to be discrete. The main application we have in mind is that such groups are candidates for non-arithmetic lattices in SU(,). Introduction In [] Mostow constructed the rst examples of non-arithmetic complex hyperbolic lattices. These lattices were generated by three complex reections R, R and R with the property that there exists a complex hyperbolic isometry J of order so that R j+ = JR j R (here and throughout the paper the indices will be taken mod ). In Mostow's examples the generators R j have order p = ; or 5. Subsequently Deligne{Mostow and Mostow constructed further non-arithmetic lattices as monodromy groups of certain hypergeometric functions, [] and [] (the lattices from [] in dimension were known to Picard who did not consider their arithmetic nature). These lattices are (commensurable with) groups generated by complex reections R j with other values of p; see Mostow [] and Sauter [9]. Subsequently no new non-arithmetic lattices have been constructed. In [5] Parker considered the case p =. That is, he considered complex involutions I, I and I with the property that there is a J of order so that I j+ = JI j J. In particular, he used a theorem of Conway and Jones [] to classify all such groups where I I and I I I are elliptic. Remarkably, when p > nding groups for which R R and R R R are elliptic involves solving the same equation as in the case p =. In this paper we use the solutions to this equation found using [] in [5] in the general case. We describe the conguration space of all groups generated by a complex reection R of order p and a regular elliptic motion J of order. This conguration space is parametrised by the conjugacy class of the product R J, which we represent geometrically in two dierent manners. The rst, following Goldman, Parker, is to consider the trace of R J; this determines the conjugacy class of R J when it is loxodromic, but there is a threefold indetermination when it is elliptic or parabolic. The second manner, following Paupert, is to use the geometric invariants of the conjugacy class, i.e. an angle pair for elliptic isometries and a pair (angle; length) for loxodromic isometries. We will use both parameter spaces in this paper, where we focus on the elliptic case.

2 Our rst result is the direct analogue of the main theorem of [5], and can be roughly stated as follows: Theorem. Let R be a complex reection of order p and J a regular elliptic of order in PU(; ). Suppose that R J and R R = R JR J are elliptic. If the group = hr ; Ji is discrete then one of the following is true: is one of Mostow's lattices. is a subgroup of one of Mostow's lattices. is one of the sporadic groups described below. The sporadic groups correspond to the 8 exceptional solutions from [], which do not depend on p (the groups do change with p of course). We determine for each p > which of these points lie inside our conguration space. One must then analyse each of these groups to decide whether or not it is discrete, if so whether or not it is a lattice and if so whether or not it is arithmetic. We illustrate ways to attack this problem by showing that certain solutions are arithmetic and certain other solutions are non-discrete. We analyse in detail the situation for p =, which can be summarised as follows: Theorem. There are 6 sporadic groups for p =, with the following properties: Four of them x a point in H C. One stabilises a complex line. One is contained in an arithmetic lattice. The ten remaining groups are none of the above. The crucial question is then to determine whether or not the ten remaining groups are discrete. We give a negative answer for four of them, by nding elliptic elements of innite order in the group. Parameters, traces and angles In this section we show how to use a single complex number to parametrise symmetric complex hyperbolic triangle groups generated by three complex reections thorough angle. This generalises the construction in [5] where involutions (that is = ) were considered. We give an alternative description following Paupert [7] where the same space is parametrised by a pair of angles. It is fairly easy to pass between then two parametrisations and we shall use whichever one is best adapted to a particular problem. We then describe how the properties of the group (for example the type of R R and R R R ) vary with. Much of this section follows the relevant parts of [5] rather closely. Otherwise we will try to keep this account as self contained as possible. However, we shall assume a certain amount of background knowledge of complex hyperbolic geometry. For such background material on complex hyperbolic space see [6] and for material on complex hyperbolic triangle groups see [] or [8]. Moreover, many of the ideas we use may be traced back to Mostow's paper [].

3 . Complex reections through angle Let L be a complex line in complex hyperbolic -space HC and write R for the complex reection through angle and xing L. In our applications we shall only consider the case where = =p for an integer p at least. Points of L are xed by R and n, the polar vector of L, is sent by R to e i n. Hence the matrix in U(; ) associated to R has determinant e i. Since we shall be dealing with traces, we want to lift R to a matrix in SU(; ). Hence we multiply the U(; ) matrix by e i =, and take R to be the map given by: R (z) = e i = z + (e i = e i = ) hz; n i hn ; n i n : () We remark that we can add arbitrary multiples of to the angle. Equivalently, since SU(; ) is a triple cover of PU(; ) we could have multiplied R by any power of e i=. Thus in the case where R has order, the in the analogous formula () in [5], the angle was chosen to be and so e i = and e i = become and + respectively. Equivalently, to obtain this we multiply () with = by e i=.. Traces We dene = tr(r J) to be the parameter in our space. We now show how to express in terms of the polar vectors and Hermitian form. We go on to nd the trace of R R in the same way. Our rst lemma and its corollary generalise Lemma. and Corollary.5 of [5] and are proven in a similar manner. Lemma. Let A be any element of SU(; ). Then tr(r A) = e i = tr (A) + (e i = e i = ) ha(n ); n i : hn ; n i Corollary. Let R be a complex involution xing a complex line L with polar vector n. Let J SU(; ) be a regular elliptic map of order. Then := tr(r J) = (e i = e i = ) hj(n ); n i hn ; n i = sin( =)ie i hj(n ); n =6 i : () hn ; n i The following theorem is just a restatement of Theorem 6.. of [6] Proposition. Let be given by (). Then R J is regular elliptic if and only if lies in the interior of the deltoid = z C : jzj 8 Re (z ) + 8jzj 7 0 : () The curve given by the equality in Proposition. is a deltoid. Groups for which R J is regular elliptic correspond to points in the interior of this deltoid. Points on the deltoid correspond to points where R J has repeated eigenvalues, and so is either a complex reection or is parabolic. Since R R R = (R J) we can determine the type of R R R from R J. (Note that it may be that R J is regular elliptic and that R R R is a complex reection.) We now consider R R, and hence by symmetry R R and R R as well. This generalises Lemma.7 of [5] and its proof is analogous.

4 Proposition. Let L be a complex line in HC with polar vector n. Let J SU(; ) be a regular elliptic map of order and write L = J(L ). Let R and R = JR J denote the complex reections through with angle xing L and L respectively. Then where is given by (). tr(r R ) = e i = jj + e i = : Corollary.5 If jj > then R R is loxodromic. If jj = cos() then R R has eigenvalues e i =, e i =+i and e i = i. Proof: Each point on L is an e i = eigenvector for R and each point on L is an e i = eigenvector for R. Therefore if z C ; lies on the intersection of L and L we have R R (z) = e i = R (z) = e i = z: Hence z is an e i = eigenvector for R R. If jj > then R R has an eigenvector whose modulus is greater than one. Hence R R is loxodromic. So we assume that jj = cos(). Since R R has an eigenvalue e i = and determinant, we see that its other eigenvalues must be e i =+i and e i = i as claimed.. Angles Conjugacy classes of elliptic isometries in PU(; ) are characterised by an unordered angle pair f ; g with i R=Z. These angles can be dened by noting that an elliptic isometry g belongs to a maximal compact subgroup of PU(; ), which is a group conjugate to U(). Then all elements of U() which are conjugate to g have the same eigenvalues of norm, which we dene to be e i and e i. In concrete terms, a matrix A U(; ) whose associated isometry g PU(; ) is elliptic is semisimple with eigenvalues of norm, say e i, e i and e i. One of these is of negative type in the sense that the associated eigenspace intersects the negative cone in C of the ambient Hermitian form (of signature (; )). Supposing for instance that e i is of negative type, the angle pair of g is then f ; g. It is thus in general not sucient to know the eigenvalues of an elliptic matrix to obtain the angle pair of the corresponding isometry; one must determine which of them is of negative type. This can also be seen in terms of the trace of matrix representatives in SU(; ). Recall from [6] that elliptic isometries have matrix representatives in SU(; ) whose trace lies in the deltoid given in (). Multiplication of A, and hence of its trace, by a cube root of unity corresponds to the same complex hyperbolic isometry in PU(; ). Up to this ambigiuty, the map from angle pairs to traces is given by f ; g 7! e i = i = + e i = i = + e i = i = : This map is three-to-one, except at the exceptional central point f=; =g which is the only preimage of 0. The three preimages in this case correspond to the fact that one of the three eigenvalues corresponds to negative eigenvectors and the other two to positive ones. There are three possible choices and the trace is the same for each of them. Conversely, given a trace = e i + e i + e i i the three preimages of this trace are the three angle pairs f + ; + g; f ; g; f ; g: In order to get these into the parameter space where 0 and 0 one may have to add an integer multiple of to either or both angles and one may have to change their order. The reader can refer to [6], pp. 9{0, for more details on angle pairs and their relation to traces.

5 . The trace parameter space Suppose we are given a symmetrical conguration of three complex lines L, L and L with polar vectors n, n and n. Because they are symmetrical J(n j ) = n j+ where j = ; ; taken mod. Because J preserves the Hermitian form hn j ; n j i is the same positive real number for each j. We normalise n j so that this number is sin( =). Likewise hn j+ ; n j i = hj(n j ); n j i is the same complex number for each j which we dene to be. Using Corollary. we see that That is hj(n j ); n j i = hn j ; n j i e i = e i = = ie i =6 : hn ; n i = hn ; n i = hn ; n i = sin( =); hn ; n i = hn ; n i = hn ; n i = ie i =6 : () All of this works for any Hermitian form of signature (; ). We now make a choice of vectors n, n and n. This determines a Hermitian form. We choose n = 05 ; n = ; n = 0 Together with (), this means that, with this choice, the Hermitian form must be hz; wi = w H z where sin( =) ie i =6 ie i =6 H = ie i =6 sin( =) ie i =6 5 : (5) ie i =6 ie i =6 sin( =) We can immediately write down J and, using (), the reections R j. They are : J = R = R = R = ; (6) 0 0 e i = e i = 0 e i = 0 5 ; (7) 0 0 e i = e i = 0 0 e i = e i = 0 0 e i = 5 ; (8) e i = e i = 0 5 : (9) e i = ei = From this it is clear that the groups = hr ; Ji and = hr ; R ; R i are completely determined up to conjugacy by the parameter. However, not all values of correspond to complex hyperbolic triangle groups. It may be that the Hermitian matrix H does not have signature (; ). We now determine this by nding the eigenvalues of H. In this lemma and throughout the paper we write! = e i= = ( + i p )=. In what follows we will be interested in the case where R J is elliptic. In this case its eigenvalues are unit modulus complex numbers whose product is. We write then as e i, e i and e i i. This means that = e i + e i + e i i. 5

6 Lemma.6 Let H be given by (5) where = e i + e i + e i i. Then the eigenvalues of H are: 8 sin(= + =6 + k=) sin(= + =6 + k=) sin( = = + =6 + k=) for k = 0; ;. Proof: We could solve the characteristic polynomial of H. However, it is easier to observe that eigenvectors for H are 5 ;!! We can immediately write down the eigenvalues as: for k = 0; ;. Subsitituting for gives 5 ;!! 5 : k = sin( =) ie i =6 ki= + ie i =6+ki= k = sin( =) + sin( =6 k=) + sin( =6 k=) + sin( =6 k=): Using elementary trigonometry (and k= = k= k), we see that k = 8 sin(= + =6 + k=) sin(= + =6 + k=) sin( = = + =6 + k=): Corollary.7 When = e i + e i + e i i the matrix H has signature (; ) if and only if 8 sin(= + =) sin(= + =) sin ( + )= + = < 0: (0) Proof: It is easy to check (for example by adding them) that all three eigenvalues cannot be negative. Thus H has signature (; ) if and only if its determinant is negative. Using the trigonometric identity sin() = sin() sin( + =) sin( + =) we see that the product of the three eigenvalues is. 8 sin(= + =) sin(= + =) sin ( + )= + = : Hence H has signature (; ) if and only if this expression is negative. The locus where each eigenvalue is zero corresponds to a line in C. They divide C into seven components which fall into three types: The central triangle. Here all three eigenvalues are positive and H is positive denite. Three innite components that each share a common edge with the central triangle. Here two eigenvalues are positive and one negative. Hence H has signature (; ). This is our parameter space. Three innite components that each only abut the central triangle in a point. Here one eigenvalue is positive and two are negative. These correspond to groups of complex hyperbolic isometries generated by three complex involutions that each x a point. Therefore the parameter space we are interested in, that is those values of satisfying (0), has three components related by multiplication by powers of! = e i=. This corresponds to that fact that J SU(; ) is only dened up to multiplication by a cube root of unity. Hence is only dened up to a cube root of unity. If we factor out by this equivalence, our parameter space becomes one of these components. 6

7 .5 Arithmetic When we are discussing the discreteness of hr ; R ; R i below, we shall analyse whether or not the group is arithmetic. In order to do so one must nd a suitable ring containing the matrix entries (after possibly conjugating). Proposition.8 The maps R, R and R may be conjugated within SU(; ) and scaled so that their matrix entries lie in the ring Z[; ; e i ; e i ]. Proof: Conjugating the matrices R, R and R given above by C = diag(e i = ; ; e i = ) these matrices become CR C = e CR C = e CR C = e i = i = i = e i ; 0 0 e i e i ; e i e i e i Hence the group generated by e i = CR j C for j = ; ; consists of matrices whose entries lie in the ring Z[; ; e i ; e i ]. 5 : It is very important to keep track of what happens to H when performing this conjugation. The matrices above preserve the Hermitian form given by (any real multiple of) CH C. So we may take the Hermitian form to be given by: sin( =)CH C = e i e i (e i ) ( e i ) (e i ) e i e i (e i ) 5 : () ( e i ) (e i ) e i e i Hence the entries of sin( =)CH C also lie in the ring Z[; ; e i ; e i ]..6 The angle parameter space In [7], Paupert characterised which angle pairs can arise for the product AB (when it is elliptic), as A and B each vary inside a xed elliptic conjugacy class (i.e. each have a xed angle pair). In the present case, the R (the rst generator) has angles f0; =pg, and J (the second generator) has angles f=; =g. The allowable region in the surface T =S is then a convex quadrilateral (degenerating to a triangle for p = ) with the following properties, which we quote from [7]. The two \totally reducible vertices" V and V (points representing an Abelian group) have respective coordinates f=; = + =pg and f= + =p; =g. These two vertices are joined by a line segment of slope corresponding to the reducible groups which x a point inside HC. Each of these vertices is also the endpoint of a line segment corresponding to the reducible groups which stabilise a complex line. The rst of these segments starts upward at V with slope, bounces o the diagonal and goes on (with slope =) until it reaches the boundary of the square. The second segment starts upward at V with slope = and goes o to the boundary. The last side of the quadrilateral is the resulting portion of the boundary segment fg [0; ]; in all corresponding groups the product R J is parabolic as was proven in [7]. The polygons for p = ; : : : ; ; 0 are 7

8 illustrated in Figure. In what follows we shall assume that R R is elliptic. From Corollary.5, we see that this implies j j. This condition does not have a striaghtforward interpretation in the angle coordinates. In Figure we have drawn the curve corresponding to jj =. When R R R is elliptic and R R is non-loxodromic We restrict our attention to those groups for which R R R is elliptic of nite order and R R is either elliptic of nite order or else parabolic. These are groups for which lies inside or on the deltoid and inside or on the circle. Since they have nite order, the eigenvalues of R R R and R R are all roots of unity. This fact leads to a linear equation in certain cosines of rational multiples of. We nd all solutions to this equation using a theorem of Conway and Jones []. We then go on nd which of these solutions lie in parameter space.. The eigenvalue equation We now investigate when both R R and R R R are elliptic of nite order. In fact our proof will be valid when R R is parabolic as well. We know that, R J (and hence R R R ) is elliptic of nite order if and only if = tr(r J) = e i + e i + e i i ; () where and are rational multiples of. Likewise for R R. In fact we know slightly more. Since the intersection of L and L is an e i = -eigenvector for each of R and R it must be a e i = -eigenvector for R R. From Proposition. we know that tr(r R ) = e i = jj + e i = : Hence the other two eigenvalues of R R are e i =+i and e i = i. (We have taken minus signs in order to make our angles agree with [5].) Thus jj = cos(); () where is a rational multiple of. We solve equations () and () by eliminating. That is, we seek,, rational multiples of so that cos() + = jj = + cos( ) + cos( + ) + cos( ): Rearranging, this becomes = cos() cos( ) cos( + ) cos( ): () In [5] Parker used a theorem of Conway and Jones [] to solve (). Up to complex conjugating 8

9 (so changing the sign of and ) and multiplying by a power of!, the only solutions are: (i) = = = = =6 =6 (ii) = = + = = = = = + = (iii) = = = = = = 5= (iv) =5 =0 =5 7=0 = =0 =0 (v) =5 =0 =5 9=0 = 7=0 7=0 (vi) = =7 =7 6=7 8= = 0= (vii) = =5 =5 =5 =5 =9 =5 (viii) = =5 7=5 =5 9=5 =5 =9 (ix) =7 = =7 = =9 =6 5=6 (x) 5=7 5= 9= 8=7 9=6 =9 =6 (xi) =7 =7 = 7= =6 5=6 =9 We then write down = tr(r J) = e i + e i + e i i and tr (R R R ) = e i + e i + e i i using this table. As indicated earlier, the parameters! j for j = ; correspond to the same group as. Proposition. (Proposition. of [5]) Suppose that R R and R R R are both elliptic of nite order (or possibly R R is parabolic). Up to complex conjugating and multiplying by a power of!, then one of the following is true: (i) = e i= =: m() for some angle ; (ii) = e i= + e i= = e i=6 cos(=) =: s(), for some angle ; (iii) = e i= + e i=6 cos(=) =: (=6; =); (iv) = e i= + e i=6 cos(=5) =: (=6; =5); (v) = e i= + e i=6 cos(=5) =: (=6; =5); (vi) = e i=7 + e i=7 + e 8i=7 =: (=7); (vii) = e i=9 + e i=9 cos(=5) =: (=9; =5); (viii) = e i=9 + e i=9 cos(=5) =: (=9; =5); (ix) = e i=9 + e i=9 cos(=7) =: (=9; =7); (x) = e i=9 + e i=9 cos(=7) =: (=9; =7); (xi) = e i=9 + e i=9 cos(6=7) =: (=9; 6=7). Before proceeding we explain the notation. The possible solutions either lie on one of two curves or are on a nite list of points. The points on the curve from part (i) (for p > ) correspond to Mostow's groups (p; t) (compare []). So we call this the Mostow curve and denote points on it by m(). The points on the curve from part (ii) correspond to subgroups of Mostow's groups, by the results of section 5.. These results generalise the isomorphisms discovered by Sauter in [9], so we call this the Sauter curve and denote points on it by s(). The remaining parts (iii) to (xi) consist of nitely many isolated points which we call sporadic. All the points except for those in part (vi) depend on two angles and so we write them as (; ) = e i + e i cos(). Note 9

10 that complex conjugation sends (; ) to ( ; ) and multiplication by e i= sends (; ) to ( =; ). We use these various values of = tr(r J) to parametrise groups in SU(; ); note that while the values of in the above list do not depend on (the angle of rotation of the reection R ), the corresponding groups do vary with. We will denote by ( ; ) the subgroup of SU(; ) generated by R and J, where the rotation angle of R is and the trace of R J is. Note that there should be no confusion with Mostow's notation (p; t), where p is an integer and t is real. Explicitly, by a simple computation using Mostow's generator R (appropriately normalised in SU(; )), Mostow's group (p; t) is our (=p; e i(=+=p t= =(p)) ).. The points in angle parameters We now list the corresponding points in the angle parameter space. This involves considering the dierences of the eigenvalues and making sure we have a pair f ; g with = 6 6 and = 6 6. For the groups of Mostow and Sauter type the corresponding points in the angle space are piecewise linear curves. We now write these down. Lemma. (i) The point = m() = e i= corresponds to the angle pair f =; g if = [0; ] and f + =; + =g if = [ ; 0]. (ii) The point = s() = e i= + e i= corresponds to the angle pair f5= + ; =g if [ =; =] and f7= ; 5= g if [=; ]. For the sporadic groups, in each case for the angle pair is f + ; g and for it is either f ; g or f + ; + g. These angles may be read o from the table just before Proposition.. f ; g (iii) e i= + e i=6 cos(=) f7=; 5=g e i= + e i=6 cos(=) f=; 5=g (iv) e i= + e i=6 cos(=5) f7=0; =0g e i= + e i=6 cos(=5) f8=5; =0g (v) e i= + e i=6 cos(=5) f9=0; =0g e i= + e i=6 cos(=5) f9=0; =5g (vi) e i=7 + e i=7 + e 8i=7 f=7; 8=7g e i=7 + e i=7 + e 8i=7 f0=7; 8=7g (vii) e i=9 + e i=9 cos(=5) f9=5; 8=5g e i=9 + e i=9 cos(=5) f9=5; =5g (viii) e i=9 + e i=9 cos(=5) f8=5; 7=5g e i=9 + e i=9 cos(=5) f=5; 7=5g (ix) e i=9 + e i=9 cos(=7) f=; 9=g e i=9 + e i=9 cos(=7) f0=7; 9=g (x) e i=9 + e i=9 cos(=7) f7=; 6=7g e i=9 + e i=9 cos(=7) f7=; 9=g (xi) e i=9 + e i=9 cos(6=7) f=7; 5=g e i=9 + e i=9 cos(=7) f=; 5=g 0

11 Figure : The points together with the trace parameter spaces for p = ; : : : ; Figure : The points together with the angle parameter spaces for p = ; : : : ; 0.

12 . Points in parameter space In this section we consider the values of from Proposition. or the corresponding angle pairs and we determine which of them is in the parameter space. In each case, we are only interested in the case where = =p for some integer p >. In each case we know the eigenvalues e i, e i and e i i of R J and we can use the criterion (0) of Corollary.7 to determine for which values of p the form H has signature (; ). Alternatively, we could use the angle pairs and check the linear conditions given in Section.6. We have plotted the points and parameter spaces for p = ; : : : ; 0 in the -plane in Figure and we have plotted the same thing in the angle parameter space in Figure. Proposition. Let R have angle (0; ). (i) Suppose that = m() = e i=. Then H has signature (; ) if and only if sin( + =) cos( = =) > 0: (ii) Suppose that = s() = e i= + e i=. Then H has signature (; ) if and only if sin( = + ) > 0. Proof: We insert the values of and found above into Corollary.7 and simplify. In (i) we have = = and = =6. This leads to 8 sin(= = + =) sin(= + =) sin( = + = + =) < 0: In (ii) we have = = = and = =. This leads to 8 sin( + =) sin( = + = + =) sin( = = + =) < 0 Simplifying, this is equivalent to sin( = + ) cos( ) + > 0: We can ignore the possibility cos( ) = as this only occurs when = e i = e i =, in which case H has a repeated eigenvalue of 0. (By inspection this does not occur in the interval where H has signature (; )).

13 For the sporadic groups we simply work by inspection. The results are given in the following table: (; ) Degenerate (; 0) (iii) = = 5= p p = = 5= = p 7 p = ; 8 p 9 (iv) = =0 =0 p p = = =0 =0 p 9 p = ; 0 p (v) = 7=0 7=0 p p = = 7=0 7=0 p 6 p = p 7 (vi) =7 =7 6=7 p 6 p = 7 p = ; ; p 8 =7 =7 6=7 p p = (vii) =9 =5 =5 p =9 =5 =5 p = ; p = p 5 (viii) =9 =5 =5 p p = =9 =5 =5 p 9 p = ; 0 p = ; p (ix) =9 =6 5=6 p =9 5=6 =6 p = p = p (x) =9 9=6 =6 p p = p = ; ; p =9 =6 9=6 p p = p = (xi) =9 7=6 6=6 p p = =9 6=6 7=6 p 8 p = p = ; p 9 Non-discreteness results There are two simple ways to see that a subgroup of PU(; ) is not discrete: nding an elliptic element of innite order in the group, or nding a subgroup which stabilises a complex line, acting non-discretely on it. Of course, these elementary facts can only be useful if one nds the appropriate element (word) or subgroup. We will present in this section some results of this nature.. Traces for certain words We compute the traces of certain short words in our generators R and J (or R, R, R ). These words seem relevant to us for experimental and/or historical reasons (see [] and [] for instance). These traces can be computed in a straightforward manner from the generators, or by using the formulae from [8]. W tr(w ) R J J R e i = R R J = (R J) JR R = (J R ) e i = + e i = R R e i = + e i = jj R R R R e i = + e i = j j R R R R e i = + e i = j + e i j R R + cos( ) + jj R R R R + cos( ) + j j R R R R + cos( ) + j + e i j R R R = (R J) jj + R R R R R R + jj jj + 6jj + cos( )

14 We use this to prove non-discreteness results, some valid for any value of p like the following, and some depending on the value of p (see last section for p = ). Lemma. Let R, R and R be given by (7), (8) and (9). If j j > then R R R R is loxodromic. If j j then R R R R is elliptic with eigenvalues e i =, e i =+i0 and e i = i0 where j j = cos 0. Proof: A short computation shows that the trace of R R R R is tr(r R R R ) = ei = j j + e i = : Since all points of L (the mirror of R ) and of R (L ) (the mirror of R R R ) are e i = eigenvectors for R and R R R respectively, we see that their intersection is an e i = eigenvector for R R R R. If this point is outside H C then R R R R is loxodromic. On the other hand, if this point is inside complex hyperbolic space then R R R R is elliptic with eigenvalues e i =, e i =+i0 and e i = i0 where j j = cos 0. We now list j j for the values of given in Proposition.. It is clear that complex conjugating or multiplying by a cube root of unity does not aect this quantity. We also give the value of 0 or indicate that R R R R is loxodromic. jj j j 0 (i) e i= = + cos() (ii) e i= + e i= + cos() + cos() (iii) e i= + e i=6 cos(=) = = (iv) e i= + e i=6 cos(=5) + cos(=5) =5 = (v) e i= + e i=6 cos(=5) + cos(=5) =5 = (vi) e i=7 + e i=7 + e 8i=7 = = (vii) e i=9 + e i=9 cos(=5) = + cos(=5) =5 (viii) e i=9 + e i=9 cos(=5) = + cos(=5) =5 (ix) e i=9 + e i=9 cos(=7) + cos(=7) =7 + cos(=7) loxodromic (x) e i=9 + e i=9 cos(=7) + cos(5=7) 5=7 + cos(=7) irrational (xi) e i=9 + e i=9 cos(6=7) + cos(=7) =7 + cos(6=7) irrational In case (ix) j j > and so R R R R is loxodromic. In the last two cases j j is less than. Therefore, using Lemma. we see that R R R R is elliptic and we nd the angle 0 satisfying + cos( 0 ) = j j = + cos(=7) and + cos(6=7), respectively. Using the theorem of Conway and Jones (Theorem 7 of []), we see that 0 is not a rational multiple if and hence R R R R is elliptic of innite order. Therefore, for these values of, the group hr ; Ji is not discrete. This proves: Corollary. Suppose that = e i=9 +e i=9 cos(=7) or = e i=9 +e i=9 cos(6=7). Then R R R R is elliptic of innite order. Hence when examining possible candidates for discrete groups it suces to consider the values of given in parts (i) to (ix) of Proposition..

15 . Triangle subgroups We will systematically analyse triangle subgroups in order to nd conditions for discreteness. By triangle subgroups we mean subgroups which x a point in CP ; depending on the position of that point, the subgroup will stabilise a complex line (if the point is outside of H C ), x a point in H C or on its boundary. The two-generator group in question then acts as a hyperbolic, spherical, or Euclidean triangle group respectively, and the angles of the triangle action are obtained from the eigenvalues of the generators, see Proposition. below. We use the lists which answer the classical question of plane geometry: given three angles, and, when is the group generated by the reections in the sides of an (; ; ) triangle discrete? In the case of a spherical triangle, the list is due to Schwarz ([0]), see table. below; for hyperbolic triangles, it is essentially due to Knapp ([8]), see Proposition. below. In the Euclidean case there are the three obvious triangles (=; =; =), (=; =; =) and (=; =; =6), and only one other, namely (=; =6; =6) (see []). We start by listing the Schwarz triangles, i.e. the spherical triangles answering the above question. The possible triangles are arranged as in []: colunar triangles appear together on one line, in increasing order of size. Schwarz triangles (=; =; p=q) (=; =; =); (=; =; =); (=; =; =) (=; =; =); (=; =; =) (=; =; =); (=; =; =); (=; =; =); (=; =; =) (=; =; =); (=; =; =); (=; =; =) (=; =; =5); (=; =; =5); (=; =; =5); (=; =; =5) (=5; =; =); (=5; =; =); (=5; =; =) (=; =5; =5); (=; =5; =5); (=; =5; =5) (=; =5; =5); (=; =5; =5); (=; =5; =5); (=; =5; =5) (=5; =; =5); (=5; =; =5); (=5; =; =5); (=5; =; =5) (=5; =5; =5); (=5; =5; =5) (=; =; =5); (=; =; =5); (=; =; =5) (=5; =5; =5); (=5; =5; =5) (=; =5; =); (=; =5; =); (=; =5; =); (=; =5; =) (=5; =5; =); (=5; =5; =); (=5; =5; =) Proposition. (Lemma. of Klimenko and Sakuma [7]) Suppose that the group generated by reections in the sides of a hyperbolic triangle is discrete. Then the angles of the triangle appear on the following list: (i) =p, =q, =r where =p + =q + =r < ; (ii) =p, =p, =r where r is odd and =p + =r < =; 5

16 (iii) =p, =, =p where p 7 is odd; (iv) =p, =, =p where p 7 is not divisible by ; (v) =p, =p, =p where p 7 is odd; (vi) =p, =p, =p where p 7 is odd; (vii) =7, =, =7. The case when two of the angles are =p and =q was proved by Knapp in Theorem. of [8]. This list is the same as that above without case (vi). Knapp's theorem was rediscovered by Mostow and appears as Theorem.7 of [] except that Mostow missed case (vii). Given a group generated by reections in the sides of a triangle (hyperbolic, Euclidean or spherical) we can consider the index group of holomorphic motions. The product of a pair of reections in sides that make an angle is an elliptic rotation through. If we represent this map as a matrix in SU() or SU(; ) the eigenvalues are e i. This group of holomorphic motions is generated by A and B where A, B, AB are elliptic rotations corresponding to products of pairs of reections. Proposition. For j = ; suppose that B j is a complex reection in SU(; ) with eigenvalues e i j=, e i j=, e i j= and mirror L j. Then L \ L corresponds to an e i = i = eigenvector of B B. Suppose that B B is elliptic and its other eigenvalues are e i =6+i =6i =. Then on the orthogonal complement of L \ L the group hb ; B i acts as the holomorphic subgroup of a ( =; =; =) triangle group. Proof: We conjugate so that L \ L corresponds to the vector (; 0; 0) T. Then B and B are block diagonal matrices in S U()U(; ) or S U()U(). Scaling, we may suppose that they lie in U() SU(; ) or U() SU(). Then the action on the orthogonal complement of L \ L is given by the unimodular matrices in the lower right hand block. These block matrices have eigenvalues e i =, e i =, e i = for B, B, B B (up to maybe scaling by ). The result follows from the remarks above. For example, consider R and R. Their eigenvalues are e i =, e i =, e i =. Scaling, we see that the eigenvalues of e i =6 R and e i =6 R are e i =, e i =, and e i =. The eigenvalues of R R are e i =+i, e i = i, and e i =, so that the eigenvalues of e i = R R are e i, e i, and e i = ( e i = )( e i = ). Therefore hr ; R i acts on the orthognal complement of L \ L as the holomorphic subgroup of a ( =; =; ) triangle group. Similarly, R and R R R have a common eigenvector L \R (L ). When R R R R is elliptic the group hr ; R R R i acts on the orthogonal complement of L \R (L ) as a ( =; =; 0 ) triangle group. We can apply the result of Klimenko and Sakuma, Proposition. (in fact we only need the earlier version of Knapp [8]) to eliminate some of the cases, using the values for and 0 given in Table.. This gives the following: Proposition.5 Suppose that = =p. (i) If p > and or = e i= + e i=6 cos(=5) then hr ; R i is not discrete. (ii) If p 6= ; 5 and or = e i=9 + e i=9 cos(=5) then hr ; R R R i is not discrete. 6

17 5 The two curves in parameter space In this section we determine exactly which points on the two curves from Proposition. correspond to discrete groups. The rst curve corresponds to Mostow's original groups from [], where sucient conditions for discreteness were found; these sucient conditions were generalised in [] (to a condition which Mostow calls INT, which also contains the groups from []). A necessary condition, using triangle subgroups, was then found in [], leaving only nine groups in dimension and one in dimension not covered by either of these criteria. Mostow proves in [] that all but three of these are discrete; the last three cases were shown to be discrete by Sauter in [9]. See [] or [9] for more details on the history of this question. Determining which points of the second curve (or Sauter curve) correspond to discrete groups uses the result that every such group is a subgroup of a group on the Mostow curve, see section 5. below. It then remains to discard the possibility that a discrete group on the Sauter curve could be a subgroup (of innite index) of a non-discrete group on the Mostow curve (theorem 5.8). This is done by a careful analysis of triangle subgroups. 5. Sauter groups are subgroups of Mostow groups In this section we prove that each group on the Sauter curve is isomorphic to a subgroup of a group on the Mostow curve, but for a dierent value of, the rotation angle of R : Proposition 5. (; e i = + e i = ) is isomorphic to a subgroup of ( ; e i= ). Recall that ( ; ) denotes the group generated by R and J, where is the rotation angle of R and is the trace of R J. Proof: We begin by dening some elements of hr ; Ji, with as rotation angle of R. These denitions follow Sauter. A A = R R J = J (J R ) J = e i = e i = e i = jj e i = 0 e i = 0 = JR R = (J R ) ; A = J R R J = J(J R ) J The result follows by noting that the subgroup ha ; Ji of hr ; Ji corresponds to the required parameters. Namely, if = e i= then A is a complex reection with angle (as can be seen by its trace), and A J has trace e i = ( jj ) + e i = = e i = + e i =. Note that we have replaced J with J, which is conjugate to J (as in Sauter's isomorphisms). 5 ; In fact we can say more: Lemma 5. ha ; A ; A i is a normal subgroup of hr ; Ji. Proof: If jj = then R i R j R i = R j R i R j. Using this and R = JR J, we nd R A R = R R R JR = R R R R J = A ; R A R = R JR R R = R R J J R R J JR R = A A A ; R A R = R J R R J R = JR R = A : (compare the identities in Section 7 of []). Moreover, from the denition of A ; A ; A we have: JA i J = A i+ (taking the index i mod. ). The lemma follows. 7

18 When = e i=, Sauter considers the map R 7! A, R 7! A, R 7! A. That is, he considers the group for which the trace of the generator is = e i= + e i= and the parameter is e i = + e i =. This is a point on the curve of the second type. In fact, as Sauter is only interested in the case of Mostow groups, he only considers this map in the case when = =, Theorem 6. of [9]. In this case e i = + e i = = e i=9 which is an intersection point of the curves of Mostow and Sauter types (see section 5.5). The special case of groups on the Mostow curve with = (or equivalently! or!) was considered by Livne in [9]. Such groups have signature (; ) when < =. In [] Parker showed that Livne's groups contain subgroups that are triangle groups generated by involutions. This is another special case of Proposition 5.. Conversely, if R, R and R have the form (7), (8), (9) then the map satises Q = e i= e i =+i= 0 e i = e i= + e i= QR Q = R ; QR Q = R ; QR Q = R R R : Therefore hr ; R ; R i is a normal subgroup of hq; R ; Ji. 5. The Mostow curves Proposition 5. Suppose that = =p and = e i=. Then the following subgroups of hr ; Ji have a common xed vector and on its orthogonal complement they act as (the holomorphic subgroup of) a triangle group as follows: (i) hr ; R i acts as a ( =; =; =) triangle group; (ii) hr ; R R R i acts as a ( =; =; ) triangle group; (iii) hr ; R R R i acts as a ( =; =; ) triangle group; (iv) ha ; A i = hr (v) ha ; A A A i = hr group. R J; JR R R i acts as a (=; =; ) triangle group; J; (R R ) i acts as a (=; = = =; = =) triangle Proof: (i) The eigenvalues of R and R are e i =, e i =, e i =, and those of R R are e i =+i=, e i = i=, e i =. The result follows from Proposition.. (ii) The eigenvalues of R and R R R are e i =, e i =, e i =. The trace of R R R R is e i = e i = cos(). Therefore its eigenvalues are e i =+i, e i = i, e i =. The result follows from Proposition.. (iii) This is similar to (ii) except that tr(r R R R ) = e i = + e i = cos( + ) = e i = e i = cos( ): (iv) The eigenvalues of A and A are e i=, e i=, e i=. The trace of A A is e i= e i= cos( ). Hence its eigenvalues are e i=, e i +i= and e i +i=. The result is again similar. (v) Finally, the eigenvalues of A are e i=, e i=, e i= and those of A A A = (A J ) = (R R ) are e i, e i, e i. The eigenvalues of A A = (A )(A A A ) are e i=, e i +i=, e i +i=. The result follows from Proposition.. 5 Combining Proposition 5. with Proposition. we obtain 8

19 Corollary 5. Suppose that = =p and = e i=. If hr ; Ji is discrete then either (i) = =q, = =p =q = =r or (ii) = = =q and = 6=p = =r where p is odd. Suppose that p, q and r are integers (or posssibly ) in modulus at least so that =p+=q+=r = = then up to permutation p, q and r take one of the values in this table: p 5 6 q r p=(p 6) 0 q=(q 6) r=(r 6) = 5= 7= 8= 0= 0=7 8 5 The following theorem is a restatement of work of Mostow [] and Sauter [9]. Theorem 5.5 Suppose that = =p and = e i= where n max ; n < < min + ; o The group hr ; Ji corresponding to these parameters is discrete if and only if one of the following is true: (i) p = ; ; 5; 6; 7; 8; 9; 0; ; 8 or and = =q, = =r for some integers q, r (possibly innity); (ii) p = 5; 7 or 9 and = =p; (iii) p = 5; 7 or 9 and = 6=p; (iv) p = 5; ; ; 0 or and = =; (v) p = 5; ; ; 0 or and = = =p. Theorem 6. of Sauter [9] shows that the group from (v) with = =p and = = for p = 5; ;, 0 or is isomorphic to the group from (i) with = = and = =p. This isomorphism is given by identifying A with R and J with J. Theorem 6. of Sauter [9] shows that the group from (ii) with = =p, = =p is isomorphic to the group from (i) with = =p and =. This isomorphism is slightly more complicated and involves sending a square root of A to R and J to J. The groups from (ii) and (iii) are isomorphic as are the groups from (iv) and (v) via the map that xes R but sends J to J. 5. The Sauter curves Proposition 5.6 Suppose that = =p and = e i= + e i=. Then the following subgroups of hr ; Ji have a common xed vector and on its orthogonal complement they act as (the holomorphic subgroup of) a triangle group as follows: (i) hr ; R i acts as a ( =; =; ) triangle group; (ii) hr ; R R i acts as a ( =; = = =; = =) triangle group. 9

20 Proof: When = e i= +e i= then R R R = (R J) has one eigenvalue e i and a repeated eigenvalue e i. Also, R R has trace tr(r R ) = e i = jj + e i = = e i = cos() + e i = : Hence R R has eigenvalues e i =, e i =+i and e i = i. Then n = L \ L is a common eigenvector of R and R R R, and hence of R R. The eigenvalues are e i =, e i and e i = i respectively. Thus on the orthogonal complement of n the group hr ; R R i acts as a ( =; = = =; = =) triangle group. We again use Proposition. to eliminate all but nitely many points. Corollary 5.7 Suppose that = =p and = e i= +e i=. If hr ; Ji is discrete then = =q where either (i) = =p =q = =r and = 6=q = =s or (ii) p = and = = = =q = =s. Proof: Using = =p in hr ; R i and Proposition., we see that either = =q or =p and in the latter case p is odd. Assume that = =q. Then hr ; R R i acts as a (=p; = =p =q; = =q) triangle group. From Proposition. we see that either = =p =q or = =q has the form =r. The result follows from the table above. Note that in (ii) we have =s = = =q = (= = =q) and, when s is not divisible by we are in case (iv) of Proposition.. If = =p then hr ; R R i acts as a (=p; = =p; = 6=p) triangle group. By inspection we see that the only possible values of p satisfying Proposition. are p = 6; 8; 9; 0; ; ; 8. Of these, only p = 9 is odd. We now eliminate this case. Suppose that = =9 and = =9. That is: We calculate (R R ) 6 = R = e i=7 e i=7 cos(=9) cos(=9) 0 e i= e i=7 e i=9 cos(=9) e 6i=7 cos(=9) + 0 e 8i=7 cos(=9) + e i=9 cos(=9) e 8i=9 The eigenvalues of (R R ) 6 are e 6i=9, e 8i=9, e 8i=9. Thus (R R ) 6 is a complex reection with angle 8=, that is =. Also tr (R R ) 6 J = e 6i=7 cos(=9) + : This trace does not appear on our list of possible values of. Therefore hr ; Ji is not discrete. Indeed, we may calculate that tr (R R ) 6 (R R ) 6 = e 8i=9 cos(=9) cos(=9) + e 6i=9 : 0 5 : 5 :

21 Now cos(=9) cos(=9) lies in ( ; ), but is not equal to twice the cosine of a rational multiple of (for example this follows from the theorem of Conway and Jones). Hence (R R ) 6 (R R ) 6 is elliptic of innite order. We now consider the points we have not eliminated by Corollary 5.7. By inspection, we see that each of these groups is a subgroup of one of the discrete groups from Theorem 5.5. Therefore we have proven: Theorem 5.8 Suppose that = =p and = e i= + e i= where =p < < =p. The group hr ; Ji is discrete if and only if = =q where either (i) q = ; ; 5; 6; 7; 8; 9; 0; ; 8 or and =p =q = =r for some integer r; (ii) p = and q = 5; ; ; 0 or. Note that (i) includes the case where p = and r = q (where q > 5), which is Proposition.5 of [5], and the case where p = and q = ; ; 5; 6; 7; 8; 9; 0; ; 8 or. 5. An example of a discrete group on the Sauter curve In x5.. of [6] Paupert considers the group with p = and angle parameters (5=; =). Writing R J as a unimodular matrix, these angle parameters translate to eigenvalues, e i= and e i=. (The xed point of R J corresponds to the e i= -eigenspace.) Hence the trace of R J is = + e i= + e i= =. This appears on the list of discrete groups. In this case: R = R = R = e i=9 e i=9 0 e i= e i=9 e i=9 0 0 e i=9 e i=9 0 0 e i=9 e i= e i=9 0 e i=9 e i=9 We can easily check that an e i= -eigenvector of R J is e 5i=9 e 5i=9 which is negative. We now give another, more direct, way to see that this group is discrete. Using Proposition.8 we see that hr ; R ; R i may be conjugated in PU(; ) so that their matrix entries lie in Z e i=, the Eisenstein integers. Since this is a discrete subring of C the group is automatically discrete. 5.5 The two curves are related for p = In section 5. we have seen that each group on the Sauter curve is a subgroup of a group on the Mostow curve, but for a dierent value of (the angle of rotation of the generators). In the special case where = =, we can apply this twice to see that each group on the Sauter curve is a 5 5 ; 5 ; 5 :

22 subgroup of a group on the Mostow curve, for the same value of. Moreover, Theorem 5.8 tells us that the Sauter subgroup in question is discrete only if the larger Mostow group is discrete. In other words, in terms of discrete groups, the two curves are the same; this is visible on Figure. The precise statement is the following: Proposition 5.9 (=; e i= + e i= ) is isomorphic to a subgroup of (=; e i= ). Proof: We have seen in section 5. that (; e i = + e i = ) is isomorphic to a subgroup of ( ; e i= ). Applying this with = =, we obtain that (; e i=9 + e i=9 ) is isomorphic to a subgroup of (=; e i= ). Now the point is that e i=9 + e i=9 = e i=9 = e 8i=9, so that the former group (of Sauter type) is also of Mostow type, and we can therefore repeat the process. This tells us that (8=; e i= + e i= ) is isomorphic to a subgroup of (=; e i= ). Note that 8= = = mod.. In fact we can say more and identify the corresponding subgroups in terms of generators: h(r R ) ; Ji is a subgroup of ha ; Ji, which is in turn a subgroup of hr ; Ji (see section 5.), and these various subgroups correspond to the values of the parameters described above. Indeed, when p = and jj =, we have tr((r R ) ) = e 6i=9 + e 8i=9. Therefore (R R ) is a complex reection with angle e i=9 = e i=. We also have tr((r R ) J) = + cos(=). When = e i= then this trace is e i= + e i= as required. 6 Sporadic groups generated by complex reections of order 6. Which sporadic points are in the parameter space? This was determined earlier for all values of p (see Table.). following: H has signature (; ) for the following sporadic values of : For p =, the situation is the e i= + e i=6 cos(=); e i= + e i=6 cos(=5); e i= + e i=6 cos(=5); e i= + e i=6 cos(=); e i= + e i=6 cos(=5); e i= + e i=6 cos(=5); e i=9 + e i=9 cos(=5); e i=9 + e i=9 cos(=5); e i=9 + e i=9 cos(=7); e i=9 + e i=9 cos(6=7); e i=7 + e i=7 + e 8i=7 : H is degenerate (signature (,0) or (,), see below) for = e i=9 + e i=9 cos() where = =5; =5; =7; =7 or 6=7. The corresponding ve points are on the boundary of our parameter space. H is positive denite when = e i=9 + e i=9 cos(=7) and = e i=7 + e i=7 + e 8i=7. The corresponding two points are outside of our parameter space. 6. Reducible sporadic groups These correspond to = e i=9 + e i=9 cos() where = =5; =5; =7; =7 or 6=7. These groups have a common eigenvector (; ; ) T ; as in Proposition. we analyse their action on the orthogonal complement of this vector. In this case, normalising R for convenience as e i=9 R, the eigenvalues on this complement are e i= and e i= for R, e i= and e i= for J, and e i and e i for R J. Therefore the group hr ; Ji acts as (the holomorphic subgroup of) a (=; =; ) triangle group. For = =5 or =5, this is in fact a (nite) (; ; 5) triangle group;

23 Figure : The sporadic points together with the trace parameter space for p = Figure : The sporadic points (crosses) together with the angle parameter space for p =. We have also included all discrete points (diamonds) on the two curves.

24 R J R J R J R J R 5 5 J R 7 7 J Figure 5: Schematic picture of triangle group action when = e i=9 + e i=9 cos() for = =5 (); =5 (); =7 (). for = =7, it is a discrete hyperbolic (; ; 7) triangle group. For = =7 or 6=7, it is a nondiscrete spherical triangle group; in fact we already know that these two groups are non-discrete from Corollary., as R R R R is elliptic of innite order. In Figure 5 we draw a schematic picture of the triangle group action induced by the generators in the three discrete cases. 6. Non-discrete irreducible sporadic groups The trace of R R is When = = we have tr(r R ) = cos( ) + jj + : tr(r R ) = jj = + cos() where is given in the table just before Proposition.. Evaluating in each case we can nd whether R R is loxodromic, parabolic or elliptic and, in the latter case, whether or not it has nite order. This enables us to eliminate one pair of complex conjugate (see also Proposition.5): Lemma 6. Suppose that = = and = e i= + e i=6 cos(=5) or its complex conjugate. The R R is elliptic of innite order. Proof: We know that tr(r R ) = jj = + cos (=5) = + cos(=5): Since R R has real trace it must have eigenvalues e i, e i and. Hence its trace is + cos(). In other words, cos() = cos(=5): Using the theorem of Conway and Jones, Theorem 7 of [], we see that is not a rational multiple of. This proves the result.

25 The proof of the following lemma is similar to the proof of Lemma., knowing that: tr(r R R R ) = e i = e i + e i = : Lemma 6. Let R, R and R be given by (7), (8) and (9). If j + e i j > then R R R R is loxodromic. If j j then R R R R is elliptic with eigenvalues e i =, e i =+i and e i = i where j + e i j = cos. Corollary 6. Suppose that = = and = e i=9 + e i=9 cos(=5). Then R R R R is elliptic of innite order. Proof: When = = and = e i=9 + e i=9 cos(=5) we see that j + e i j = 6 p 5: Hence j + e i j = cos where is not a rational multiple of. 6. Arithmeticity of sporadic groups We begin by applying Proposition.8 to show that when = e i=7 + e i=7 + e 8i=7 the group hr ; R ; R i is contained in an arithmetic lattice. In particular, putting = = into the expression in equation () for sin(=)ch C where C = diag(e i=9 ; ; e i=9 ) we obtain: p CH C = ( + i p )= ( + i p )= ( i p )= ( + i p )= ( i p )= ( i p )= Proposition 6. Let R, R and R be complex reections with angle = = so that the group hr ; R ; R i has parameter = e i=7 + e i=7 + e 8i=7 = ( i p 7)=. Then hr ; R ; R i is contained in an arithmetic lattice and hence is discrete. Proof: We have e i = ( + i p )= and = ( i p 7)=. Both of these are algebraic integers in the eld Q p ; i p. In particular, using Proposition.8, we can ensure that the matrix entries of all elements of hr ; R ; R i are algebraic integers in Q p ; i p. This eld is a totally imaginary quadratic extension of the totally real number eld Q p. The only nontrivial Galois conjugation in Q p sends p to p. This is compatible with g, the Galois conjugation in Q p ; i p xing ( +i p )= and sending = ( i p 7)= to = ( +i p 7)=. The matrix p CH C has entries in the ring of integers of Q p ; i p and signature (; ). The Galois conjugation g sends p H to p CH C. But we know that the point lies outside our parameter space and so corresponds to a group which has signature (; 0). Therefore using standard arguments (for example [0] or Proposition. of [5]) we see that this group is arithmetic and hence discrete. 5 : There are ten more sporadic groups with signature (; ). The goal of the rest of this section is to show that none of them are contained in an arithmetic lattice. Proposition 6.5 Let R, R and R be complex reections with angle = = so that the group hr ; R ; R i has parameter 6= e i=7 + e i=7 + e 8i=7 = ( i p 7)=. Then hr ; R ; R i is not contained in an arithmetic lattice. 5