Oct 1, 1993 - As John Millson explained to me, the rst obstruction for deformations in ... Author is very grateful to Nicolaas Kuiper and John Millson for.
DEFORMATIONS OF REPRESENTATIONS OF DISCRETE SUBGROUPS OF SO(3,1) Michael Kapovich October 1, 1993
1. Introduction
Let M be a closed hyperbolic 3-dimensional orbifold (see [T], [Sc] for de nitions), 0 : 1 (M ) ! Isom(H 3 ) be its holonomy representation. Denote the conjugacy class of 0 by [0 ]. In this paper we discuss whether for n = 4 the point [0 ] is isolated in the space
R(1 (M ); n) = Hom(1 (M ); Isom(H n ))=Isom+ (H n ) If [0 ] is isolated, then the corresponding representation is called locally rigid. A suborbifold in M is said to be a virtual ber in a ber bundle over S1 if M admits a nite-sheeted covering p : M0 ! M such that M0 is bered over a circle and a component of the preimage p?1 is a ber of this bration. We start with the following Conjecture 1. The representation 0 is not locally rigid if and only if M contains an incompressible 2-suborbifold which is not a virtual ber in a ber bundle over S1 [Ka 1, Ka 2]. Certainly the case of manifolds is the most interesting and most complicated. The main aim of this paper is to show that Conjecture 1 is not absolutely groundless. First our results deal with re ection orbifolds. In this case we will prove Conjecture 1 and nd that R(1 (M ); 4) is a smooth manifold of dimension (f ? 4), where f is the number of \faces" of the re ection orbifold (Theorem 1). Then we shall consider orbifolds of nite volume and examine the \restriction" map
@T : PR(1(M ); 4) ! PR(1 (T ); 4) where T is an incompressible Euclidean suborbifold corresponding to a \cusp end" of M ; PR(:) mean the space of representations whose restrictions on cyclic parabolic subgroups of 1 (M ) Isom(H 3 ) are induced by conjugations in Isom(H 4 ). Under some conditions the image of this map is 1-dimensional (Example 1 ). The question about deformations of such kind arises naturally if we are trying to construct at conformal structures on manifolds obtained by gluing of two hyperbolic ones along boundary tori [GLT], [Ka1]. Unfortunately, in the general case, the map @T is zero to the rst order (Theorem 3). We prove the \only if" part of Conjecture 1 for in nitely many non-Haken manifolds arising after Dehn surgery on 2-bridge knots (Theorem 2). These are the rst examples of closed hyperbolic 3-manifolds M whose fundamental group are locally rigid in R(1 (M ); 4). Moreover, we prove that for each hyperbolic 2-bridge knot K S3 there exists only one conjugacy class of discrete faithful representations of 1 (S3 ? K ) into Isom(H 4 ) (Section 5).
1
2. Preliminary geometric results.
In this section we collect several elementary facts about geometry of Euclidean spheres in S3. Denote by C the set of all Euclidean spheres in S3 of positive radius. Then C has a natural topology and is a smooth 4-manifold. By Mob(Sn) we denote the group of Moebius transformations acting on Sn. We shall suppose that the hyperbolic 3-space H 3 is realized as a unit ball in R3 R3 = S3; thus Isom(H 3 ) is the group of Moebius transformations of S3 which leave H 3 invariant. Mob+ (Sn) is the subgroup of orientation-preserving Moebius transformations of Sn. Lemma 2.1. Let (1; 2 ; 3; 4 ) 2 C 4 . Then the following trichotomy holds: either (i) T there Tis a sphere 0 2 C orthogonal to all spheres 1 ; 2 ; 3 ; 4 ; T or (ii) 1 2 3 4 3 p , where p is a point; or (iii) spheres 1 ; 2 ; 3 ; 4 are totally geodesic in some metric of constant positive curvature on S3. Corollary 2.1. Let (1; 2 ; 3; 4 ) 2 C 4 . Denote by j the inversion in the sphere j , let ? be the group generated by j . Then the following trichotomy holds: either (i) ? is conjugate in Mob(S3) to a subgroup of Isom(H 3 ); or (ii) ? is conjugate in Mob(S3) to a subgroup of Isom(E 3 ); or (iii) ? is conjugate in Mob(S3) to a subgroup of Isom(S3). Proof of Lemma 2.1. We present the proof that was suggested to the author by N.Kuiper instead of the original one. Consider the sphere S3 as a round sphere in the ane space A 4 RP4 ; respectively Mob(S3) PGL(5; R ). Every sphere 2 C is the intersection of S3 with some ane hyperplane P A 4 ; = (P ). Denote by P 2 RP4 the polar of with respect to S3 (i.e. such point that tangent cone from P to S3 touches S3 at ). Denote by P^ the closure of P in RP4 . Then it is easy to see that (P ) is orthogonal to (Q) i P 2 Q^ (it is sucient to consider rst the case of P 62 A 4 and then apply Mob(S3) PGL(5; R )). Now consider the polars Pj corresponding to j (j = 1; : : : ; 4). Let P^ be the extended hypersubspace in RP4 which passes through these points. Then we have 3 possibilities: T (i) the intersection P TS3 is a sphere of positive radius. This is the desired sphere . (ii) The intersection P S3 is a point 1 2 S3. Then Pj 3 1 , so we have the case (ii) of Lemma. 3 (iii) The intersection T Tabove T is empty. Then P 2 int(S ) and the point P lies in3 the intersection P1 P2 P3 P4 . Applying a projective transformation from Mob(S ) we can map the point P in the center of S3. So, j are \great spheres" in S3 which are totally geodesic in elliptic geometry. Lemma is proved. Remark 1. Independently a generalization of Lemma 2.2 to higher dimensions was proven in [Lu]. Corollary 2.2. Let l and k be two unlinked Euclidean circles in S3; l \ k = ;. Then there exists a Euclidean sphere 0 S3 which is orthogonal to l and k. Proof. We can realize l and k as intersections l = 1 \ 2 and k = 3 \ 4 , so that the case (i) of Lemma 2.2 holds for the collection (1 ; 2 ; 3 ; 4 ): Then the sphere 0 from Lemma 2.2 is orthogonal to l and k. Remark 2. If the sphere 0 (case (i) of Lemma 2.2) is unique (i.e. when j are not orthogonal to a common circle), then it smoothly depends on 2
(1 ; 2 ; 3 ; 4 ) 2 C 4 : Notation. If 1; 2 are spheres in E 3 then by 0 (1 ; 2 ) we shall denote the (external) angle between them. If 1 \2 = ; then we put (1 ; 2 ) = 0. If X Sn then Sp(X ) will denote the round sphere in Sn which contains X and has minimal dimension. Uniqueness of this sphere is evident.
3. Compact re ection orbifolds.
Consider a compact convex nite-sided polyhedron in H 3 , such that every vertex of belongs to precisely 3 edges. Denote the numbers of vertices, edges and faces of by v; e; f respectively. Then we have: 3v = 2e; 2 = v ? e + f: So e = 3f ? 6: Let ? Isom(H 3 ) be the isometry group generated by the re ections in the faces of . Suppose that ? is discrete and is a fundamental polyhedron for it. Then we shall identify the polyhedron with the factor{orbifold H 3 =? . This orbifold is suciently large (i.e. contains an incompressible suborbifold) i is not a tetrahedron [T]. Each (2-dimensional) face i @ is contained in the unique round sphere 0i = Sp(i ) 2 C . According to the Poincare theorem about fundamental polyhedra for discrete groups, the group ? has the presentation
h 1; :::; f : ( j i)n = 1i ij
where nii = 1, Z 3 nij = = (0i ; 0k ) (see [Ma]); the elements j are re ections in the faces j of . Moreover, suppose that we have another con guration (1 ; :::; f ) of spheres in S3, such that
(0i ; 0k ) = (i ; k ) for each i; k. Then the subgroup of SO(4; 1) generated by the re ections in the spheres i is isomorphic to ? . Thus the problem of deforming the representation 0 of ? is equivalent to the problem of deforming the con guration of spheres preserving the angles between the neighboring spheres. Theorem 1. Near the class [I ] of the embedding id : ? ! Isom(H 4 ) the space R(?; 4) is smooth and has dimension f ? 4. 3.1. Proof. De ne r = (1 ; :::; f ) 2 C f . If faces i ; k of have a common edge j then put j = (i ; k ). So, we have the map ~3 = ( 1 ; : : : j : : : ; e ) : C f ! Re
Denote 0j = j (r0 ). Let C2 denote the space of spheres orthogonal to the sphere @1 H 3 . Then the groups Mob+ (S3) and Mob+ (S 2 ) act on C f and C2f respectively. Drop the map ~3 to the maps 3 : C f / Mob+ (S3) ! Re and
2 : C2f =Mob+ (S 2 ) ! Re Hence [r0 ] = (2 )?1 ( 0 = ( 01 ; : : : ; 0e )). Moreover, the map 2 is an immersion at the point [r0 ], since H1 (? ; so(3; 1)) = 0 [W]. 3
However, dim[r0 ] C2f =Mob+ (S 3 ) = 3f ? 6 = dim Re , so the map 2 is also a submersion near [r0 ]. Hence the map 3 is a submersion near [r0 ] too. Thus the variety (3 )?1 ( 0 = ( 01 ; : : : ; 0e )) = R(? ; 4) is a manifold of dimension 4f ? e ? 10 = f ? 4 near [r0 ]. Corollary 3.1. The group ? is rigid in SO(4, 1) i the orbifold is not suciently large. 3.2. Now describe the basis of H1 (? ; so(4; 1)). Realize the hyperbolic 3-space H 3 as the upper half-space R3+ = f(x1 ; x2 ; x3 ) : x3 > 0g. Pick an arbitrary sphere 0i as above , center of 0i is a point (x1 ; x2 ; 0). Then de ne the family ti of spheres with the same radius as 0i and center at (x1 ; x2 ; t) , (t 2 R). Let rt (i) = (1 ; :::ti ; :::; f ) 2 C f . For every sphere 0j adjacent to 0i the function (0j ; ti ) has maximum at the point t = 0. So,
d=dt( (0j ; ti))jt=0 = 0 Therefore, the vector d=dt(rt (i))jt=0 is tangent to Hom(? ; SO(4; 1)) C f . By direct calculations it's possible to show that fd=dt(rt (i))jt=0 : i = 1; :::; f ? 4g forms a basis of H1 (? ; so(4; 1)).
Remark 3. The same construction 3.2 works for hyperbolic re ection orbifolds of arbitrary dimension. Thus, if ? SO(n; 1) is any discrete re ection group then dimH1 (?; so(n+ 1; 1)) = maxff ? n ? 1; 0g where f is the number of faces of the fundamental polyhedron of ?.
4. Rigidity of 2-bridge knots.
In this section we will need the following Lemma 4.1. Suppose that G is a nitely generated group, E a G-modulus, R is an element of G so that R = for each 2 E . Denote by : G ! G=hhRii = G0 the natural projection epimorphism, where hhRii is the normal closure of R in G. Let [x] 2 H1 (G; E ) be a class such that the restriction of x to hRi is zero in Z1 (hRi; E ). Then (i) E is G0 -modulus: g0 = g for any g 2 ?1 (g0 ); (ii) there is a class [x0 ] = (x) 2 H1 (G0 ; E ) such that x0 (g0 ) = x(g) for each g0 2 G0 ; g 2 ?1 (g0 ). Proof. De ne x0 as x0 (g0 ) = x(g). Since the action of R on E is trivial, then the de nitions of g0 and x0 don't depend on the choice of g 2 ?1 (g0 ). Let K S3 be a 2-bridge knot (see [R]). Let (p; q) be a pair of coprime integers. Remove from S3 an open regular neighborhood N (K ) of the knot K and denote the resulting manifold with boundary by M1 = M (K ; 1). We shall consider only hyperbolic 2-bridge knots K , i.e. such that int(M (K ; 1)) admits a complete hyperbolic structure. Denote by a simple homotopically nontrivial loop on @M (K ; 1) such that bounds a disc in N (K ); let @M (K ; 1) be a simple homotopically nontrivial loop which is homologically trivial in M (K ; 1). Denote by M(p;q) the manifold obtained from M1 by attaching a solid torus T~ along the boundary so that the loop p q bounds a disc in T~. Suppose that (p; q) are not coprime, and k is their greatest common divisor. Denote by T~(k) the orbifold whose underlying set is D2 S1 and the singular set f0g S1 has order k. Then M(p;q) is the orbifold obtained from M1 by attaching T~(k) so that the loop p=k q=k bounds a disc in T~(k) with one singular point. This procedure is called the generalized Dehn surgery on the knot K ; (p; q) are parameters of the surgery. Remark 4. This de nition is slightly di erent from the standard one. Then for all but nite coprime parameters (p; q) of Dehn surgery on K the resulting manifolds are hyperbolic and are not suciently large [H-T]. For a group F and representation : F ! SO(n; 1) we denote by Adn the corresponding adjoint representation on the Lie algebra so(n; 1). 4
Theorem 2. For in nitely many coprime (p; q) the groups 1M(p;q) are locally rigid in
SO(4,1) and moreover H1 (1 M(p;q) ; Adn ) = 0. Proof of Theorem 2. Consider the uniformization M = S3nK = H 3 =?. Then ? = hx; yjxw = wyi, where w = w(x; y), x; y are parabolic elements of ? PSL(2; C ). Denote by A the maximal parabolic subgroup of ? which contains hxi; A = hxi hz i where the elements x; z are represented by the loops and respectively. First consider the case of a \singular" Dehn surgery (r; 0) on the knot K such that in the fundamental group of the hyperbolic orbifold M(r;0) the image of x has the order r. Let r : ? ! ? r = 1 (M(r;0) ) be the holonomy representation; H 3 =?r = M(r;0) . Denote by = fr : ? ! PSL(2; C )g=PSL(2; C ) the collection of conjugacy classes of such representations (where r varies). Lemma 4.2. For every (r; 0)- surgery we have H1 (?r ; Ad4 ) = 0: Proof. The group ?r is generated by two elliptic elements xr = r (x); yr = r (y), which are conjugate in ?r . Consider the group
?r = hxjxr = 1i hyjyr = 1i and the natural projection 'r : ?r ! ?r . Then we have H2 (?r ; Ad4 'r ) = 0 since ?r is almost free. Let [ ] 2 H1 (?r ; Ad4 ) , then [ ] = 'r [ ] 2 H1 (?r ; Ad4 ) is an integrable in nitesimal deformation. Let t : ?r ! SO(4; 1) be a curve tangent to [ ]. Proposition For every t the group t(?r ) is conjugate in SO(4; 1) to a subgroup of SO(3; 1). Proof. The group t(?r ) is generated by two elliptic transformations with the xedpoint sets lt ; kt which are unlinked Euclidean circles in S3. Then the proposition follows from Corollary 2.3. Thus we can nd a coboundary
2 B 1 (?r ; Ad4 'r ) = B 1 (?r ; Ad4 ) such that ? 2 Z 1 (?r ; Ad3 ). However, H1 (?r ; Ad3 ) = 0 by Weil's rigidity theorem [W]. So [ ] = 0. This proves Lemma 2.4.
Corollary 4.2 (i) The restriction map res : H1 (?; Ad4 r ) ! H1 (hxi; Ad4 r ) is injective. (ii) The space H1 (?; Ad4 r ) has dimension 2 . Proof. Consider (i). Let [ ] 2 Ker(res). Then (xm ) = ? Ad4 r (xm ) , where 2 so(4; 1). De ne a cocycles ; in Z1 (?; Ad4 r ) as (g) = ? Ad4 r (g) , (g) = (g) ? (g). The cocycle is cohomologous to and the restriction of to hxi is identically zero. 5
Now we can apply Lemma 4.1 to the projection r : ? ! ?r . Then induces the cocycle r ( ) = 0 2 Z 1 (?r ; Ad4 ) = B 1 (?r ; Ad4 ) (according to Lemma 4.2). Hence for some 2 so(4; 1) we have 0 (g0 ) = ? Ad4 (g0 )( ). However, 0(g0 ) = (g) = ? Ad4 r (g)( ). Therefore, 2 B 1 (?; Ad4 r ) and (i) is proved. Consider (ii). We have the following diagram res0 res3 H1 (?; Ad3 r ) ???! H1 (A; Ad3 r ) ???! H1 (hxi; Ad3 r ) ?? ?? ?? y# y y res1 res2 H1 (?; Ad4 r ) ???! H1 (A; Ad4 r ) ???! H1 (hxi; Ad4 r ) The Abelian group r A is generated by one hyperbolic and one elliptic element; thus its action on R3;1 has no nonzero xed vectors and
0 = 2dimH0 (A; R 3;1 ) = H1 (A; R3;1 ) r
r
Therefore, in the exact sequence 0 = H0 (A; R 3;1 ) ! H0 (A; Ad3 r ) ! r
! H0 (A; Ad4 r ) ! H1(A; R3;1 ) = 0 the homomorphism : H0 (A; Ad3 r ) ! H0 (A; Ad4 r ) is an isomorphism. On another hand: # : H1(hxi; Ad3 r ) = R2 ! H1 (hxi; Ad4 r ) = R4 ! ! H1(hxi; R 3;1 ) = R2 ! 0 r
r
Hence # is a monomorphism with 2-dimensional image. Recall that for H -modulus F the space F H is the set of elements of F xed under the action of H . Now, consider the exact sequence: 0 ! R2 = H1 (hz i; so(3; 1) (x) ) ! r
! R4 = H1(A; Ad3 r ) ! H1 (hxi; Ad3 r )hzi ! 0 However, y acts trivially on H1 (hxi; Ad3 r ) because A is Abelian and H1 (hxi; Ad3 r ) = H1 (hxi; co(2)) where co(2) is the Lie algebra of the centralizer of r (x) and r (z ) in SO(3; 1). This follows that the restriction map res3 is surjective. Therefore, Im(res2 ) = Im(res2 ) = Im(# res3 ) = Im(#) = R2 . Notice also that Ker(res1 ) = Ker(res2 res1) = 0 (accoridng to (i)) and Ker() = 0 (for instance because res0 is injective ). Thus res2 res1 injects H1(?; Ad4 r ) into the image of # which is 2-dimensional. Thus, dimH1 (?; Ad4 r ) 2 On another hand, H1 (?; Ad4 r ) (H1 (?; Ad3 r )) and thus dimH1 (?; Ad4 r ) 2 6
This implies the second assertion of the Corollary. . We continue proof of Theorem 2. The space R(?; 3) has the natural complex structure since we can identify SO(3; 1)+ with PSL(2; C ). Denote by E the projection to R(?; 3) of the set of representations p;q : ? ! ?p;q SO(3; 1) which are the holonomy representations of hyperbolic manifolds M(p;q) . Remark 4. Here and below p and q are coprime integers. Denote by R0 (?; 3) the connected component of R(?; 3) containing 0 . Lemma 4.3 The set E is Zariski dense (over R) in R0 (?; 3). Proof. Step 1. The element 0 (x) is a parabolic element in PSL(2; C ). Take a simply connected neighborhood V of 0 (x) in PSL(2; C ). For each g 2 PSL(2; C ) choose a lift g~ of g to SL(2; C ) and de ne (g) to be the ratio of eigenvalues of g~ (this is well dei ned up to inversion). We can choose V to be so small that the image of does not intersect the set of nonpositive real numbers. Then extend the function to the orbit of V under the conjugation by PSL(2; C ) as: (hgh?1 ) = (g) Finally we put u : [] 7! log(((x))=2) where we choose the branch of logarithm so that log(1) = 0. Then, u[0 ] = 0. The function u([]) is well de ned up to the multiple 1 and there is a way to choose this multiple so that function u is a holomorphic embedding of W into C (see [NZ]). Put E = u(E ) and
U = fjzj=z; z 2 E g The set U is dense on the unit circle (see [NZ]) and the points of E accumulate to zero. Therefore, the set E cannot lie on any real-analytic subset of u(W ). Step 2. Suppose that E is not Zariski-dense and E f ?1 (0) for some nontrivial polynomial f . Then E \ W is contained in the real- analytic set (f u)?1 (0) which is impossible. Corollary 4.3 If E0 is any nite subset of E , then E nE0 is Zariski dense in R0 (?; 3) over R. Lemma 4.4 For in nitely many elements 2 E dim H1 (?; Ad4 ) = 2: Proof. Denote by (L; c ) the so(4; 1)-bundle over M1 with the at connection c associated with the representation Adn (see [JM]). The group cohomology H1 (?; Ad4 ) can be calculated via simplicial cochains of M1 with coecients in the parallel sections of (L; c ) (see [JM]). Thus the spaces of i-chains C i (K ) of the corresponding complex K is nitedimensional. We shall identify C i = C i(K ) for di erent so that the coboundary operators i are linear operators between nite-dimensional spaces
i : C i ! C i+1 which depend algebraically on the parameter . Suppose now that there exists a nite set E0 E such that dim H1 (?; Ad4 ) 3 for every 2 E nE0 . 7
Denote the dimension of C i by Ni . The space Im(0 ) has constant dimension 0 since H0 (?; Ad4 ) = 0 for each [] 2 E . If d = dim Ker(1 ) 3 + 0 for 2 E nE0 then
Im(1 ) = N1 ? d N1 ? 3 ? 0 : Denote by fs ; s = 1; 2; : : : g the complete set of minors of order (N1 ? 2 ? 0) in the matrix 1 . Then X () = 2s () = 0 s1
for every 2 E nE0 . Obviously, () is an algebraic function; however (r ) 6= 0 for every r 2 since dim H1 (?; Ad4 r ) = 2. So, E nE0 is contained in a proper real-algebraic subset of R(?; 3) which contradicts to assertion of Corollary 4.3. Now we can nish the proof of Theorem 2. We have an in nite subset F E such that dim H1 (?, Ad4 ) = 2 for every 2 F . However dim H1 (?; Ad3 ) = 2; so dim H1 (?(p;q) ; Ad4 (p;q) ) = 0 for (p;q) 2 F: Theorem 2 is proved.
5. Deformations of nonuniform lattices
5.1. Let ? Isom(H 3 ) is an arbitrary nonuniform lattice (i.e. vol(H 3 =?) < 1 but H 3 =? is not compact). De nition. Let PZ1(?; so(4; 1)) be the subspace of cocycles 2 Z1 (?; so(4; 1)) such that the restriction of to each cyclic parabolic subgroup h i ? is a coboundary in Z1 (h i; so(4; 1)). Then put PH1 (?; so(4; 1)) = PZ1 (?; so(4; 1))=B1 (?; so(4; 1)) The space PH1 (?; so(4; 1)) is called the space of \parabolic cohomology classes" and PZ1 (?; so(4; 1)) is the space of \parabolic cocycles". Theorem 3 For every maximal parabolic subgroup A ? we have
resA : PH1 (?; so(4; 1)) ! PH1(A; so(4; 1)) is identically zero. Proof. The space so(4; 1) admits the Ad? -invariant decomposition so(4; 1) = so(3; 1) V , where V = R3;1 is the Lorentz vector space [JM]. This splitting induces the natural decomposition PH1 (?; so(4; 1)) = PH1 (?; so(3; 1)) PH1 (?; V ): However, PH1 (?; so(3; 1)) = 0 by Calabi{Weil{Garland{Raghunathan Rigidity theorem [GR], [R]. Therefore, projections of every [ ] 2 PH1 (?; so(4; 1)) to PH1 (?; so(3; 1)) and PH1 (A; so(3; 1)) are zero. However, PH1 (A; V ) = 0 since
PH1 (A; so(4; 1)) = PH1 (A; so(3; 1))
So, resA([ ]) = 0.
5.2. Example. Let ? SO(3; 1) be the fundamental group of the complement to any hyperbolic 2-bridge knot (as in the Section 4). Theorem 4 PH1(?; so(4; 1)) = 0. 8
Proof. Suppose that 2 PZ1 (?; so(4; 1)) be a nonzero cocycle. Denote by ? the free group generated by x; y. Then, applying the arguments of Lemma 4.2, we construct a smooth family of representations t of ? into SO(4; 1) such that: (i) 0 = id ; (ii) t (x) and t (y) are all conjugate to x and y for all t ; (iii) the curve t is tangent to the lift of to ? . Each transformation t (x) and t (y) is conjugate in SO(4; 1) to a Euclidean translation. This implies that every 3-dimensional hyperbolic subspace of H 4 which contains 1-dimensional horocycle of t (x) is invariant under t (x); the same is true for t (y). We can assume that t (x) is a Euclidean translation along a line ` in the upper-halfspace model of H 4 . Let P be a Euclidean 3-dimensional subspace in R4 which contains `, horocycle of t (y) and orthogonal to the absolute of H 4 . It follows that t (?) has an invariant 3dimensional hyperbolic hyperplane P \ H 4 . The same arguments as in Lemma 4.2 imply that 2 B1 (?; so(4; 1)). Remark 7. The arguments of the proof show that if [] 2 PR(?; 4), then the class [] contains a representation 1 with image in SO(3; 1). However R.Riley in [Ri] described completely all representations of ? in PSL(2; C ). Any representation which preserves the conjugacy classes of x and z is conjugate to id. Therefore PR(?; 4) consists of a single point. This implies that if [] 2 R(?; 4) is the conjugacy class of a discrete faithful representation then [] = [id]. Indeed, such representation must preserve the conjugacy class of x because (hxi hz i) is conjugate to a lattice in R2 .
6. Three examples. Notation. For any orbifold O we shall denote by jOj its underlying set. For a face P of
a polyhedra we shall denote by StP the set o all those faces of which have nonempty intersection with P ; StP = StP ? fP g. We shall suppose that H 3 is realized as a unit ball in R3 . Bending deformations. The following is not the most general description of the \bending", but it's enough for our aims. Suppose that G SO(n +1; 1) is any group which splits as the amalgamated free product G = G1 J G2 so that: (1) G1 and G2 have nite centralizers in SO(n + 1; 1); (2) the centralizer ZJ of the group J in SO(n + 1; 1) is 1-dimensional. Take a nondegenerate curve t in ZJ which contains 1. Then put Gt to be the group generated by G1 and t G2 t?1 . It's easy to see that Gt = t (G), where ft : t 2 [0; 1]g is a continuous curve of homomorphisms of G in SO(n + 1; 1). This curve de nes a nontrivial deformation of the identity representation of G in SO(n + 1; 1). Such deformation is called the \bending in J ". Remark 8. Bending deformations of representations of fundamental groups of hyperbolic manifolds (and orbifolds) of dimension n were constructed by several authors: by W.Thurston [T] for n = 2; in the case of certain re ection groups in H 3 { by B.Apanasov and A.Tetenov [AT]; then in in nitesimal form { by J.Lafontane [L]; and later by C.Kuorouniotis [K]. In the most general form (for graphs of groups) this conception is explained by D.Johnson and J.Millson [JM] (see also [G]). There are examples of B.Apanasov [A] of \pea-pod" groups which admit \stamping" deformations; this construction was generalized by S.Tan [Ta]. For further generalizations see also [KM].
6.1. Example 1. Suppose that we are given a nite-sided convex polyhedron H 3 with the following properties: 9
(a) for some compact face P of all but one faces of StP are orthogonal to a common geodesic plane H 3 ; (b) among the faces in StP there is a face Q2 which enters a cusp made by the faces Q1 ; Q2 ; Q3 ; Q4 where Q3 ; Q4 2 StP . Then Q2 is orthogonal to Q3 ; Q4 . Suppose that is the fundamental polyhedron for the discrete group ? generated by re ection in its faces. Denote by A the group generated by re ections in Q1 ; Q2 ; Q3 ; Q4 . Consider a nontrivial continuous family Sp(Q)2 of Euclidean spheres ( 2 [0; 1]) such that: (i) Sp(Q)02 = Sp(Q)2 ; (ii) Sp(Q)2 is orthogonal to Sp(Q4 ); Sp(Q3 ) and tangent to Sp(Q1 ); (iii) the closed ball in R3 bounded by (Sp(Q)2 ) contains Sp(Q2 ) ; ( 2 [0; 1]). Then (Sp(P ); Sp(Q)2 ) (Sp(P ); Sp(Q2 )). For all suciently small values of there is an elliptic rotation ' 2Mob(S3) around the circle @1 such that: (Sp(P ) ; Sp(Q)2 ) = (Sp(P ); Sp(Q2 )) , where ' Sp(P ) = Sp(P ). De ne the new con guration of spheres Sp( ) , that consists of the same spheres as Sp( ), except of Sp(P ) and Sp(Q2) which are deformed to Sp(P ) and Sp(Q)2 respectively. Then Sp( ) has the same combinatorial type as Sp( ) and the same angles between spheres. The group generated by the re ections in the spheres of Sp( ) de nes the deformation : ? ! G with the following properties: (a) for suciently small values of the representation is discrete and faithful; (b) projection @A of to Hom(A; SO(4; 1))=SO(4; 1) is a nontrivial path; (c) [ ] 2 PR(? ; 3) (see Introduction). P
Φ
2
3
j
3 2
3
α1
α3
2
2 3
3
α2
2
2 3
P
Figure 1: One can generalize this example, however in general case it is rather dicult to determine : whether or not we obtain nontrivial deformations of cusps. 6.2. Example 2. Consider the convex polyhedron in H 3 which is drawn on Figure 1. As usual, if an edge e of is labelled by the integer n then the dihedral angle of at e is =n. Let G be the group generated by re ections in the faces of . First, nd all totally geodesic suborbifolds in . There are only 3 incompressible suborbifolds Di in : @ jDi j = j , i = 1; 2; 3 (see Figure 1). 10
P 2
3 3 3
2
α1
2
2 2
3 3
Φ1+
Figure 2: (i) Suppose D1 is totally geodesic; then we split along D1 and consider the \upper half" +1 (Figure 2). Then +1 contains an incompressible Euclidean rectangle suborbifold, that is impossible. So, 1 cannot be the boundary of the underlying set of any totally geodesic suborbifold in . (ii) Consider D2 . According to Andreev's theorem (see [T]) there exists a convex polyhedron +2 H 3 as on Figure 3. The face D2 @ j+2 j is a rectangle symmetric under rotation of order 2 around the axis ` . Let ?2 = (+2) ; then ?2 [ +2 is a convex polyhedron isometric to the initial one . So D2 is a totally geodesic suborbifold. (iii) The same arguments imply that the orbifold D3 is totally geodesic. Thus the orbifold has exactly two totally geodesic 2-dimensional suborbifolds D2 ; D3 . Fundamental groups of D2 ; D3 have 1-dimensional centralizers in Isom(H 4 ). The corresponding bending deformations 2 ; 3 of the group G in Isom(H 4 ) de ne classes 20 ; 30 which span the 2-dimensional space H1 (G; so(4; 1)). On another hand, deformation space R(G; 4) is a smooth 2-dimensional surface. This shows that bending deformations in intersecting surfaces can span a plane tangent to a smooth surface in the representation variety. This result shows the striking di erence between lattices in Isom(H 3 ) and Isom(H n ) (n > 3) since in higher dimensions there are examples [JM] when a linear combination of two bending cocycles is not tangent to any smooth curve in the representation variety. 6.3. Example 3. Our next aim is to construct an example of group H which does not admit bending deformations at all, while it is possible to deform H since its fundamental polyhedron has six faces. Change only one dihedral angle of the polyhedron : instead of the angle =3 at the edge j we consider the angle =5. Denote the new polyhedron by . (i) The same arguments as in Example 2 imply that 1 cannot be boundary of a totally geodesic suborbifold D1 of . Next notice that both 2 ; 3 don't intersect the edge j . 11
2 3 3 2
3
2
2 3
2
α2
2 2
l
Φ2+
Figure 3: (ii) Consider the curve 2 @ j j and suppose that 2 = @ jE2 j; E2 is a totally geodesic suborbifold. Then split along E2 and obtain two parts : +2 j and ?2 which does not contain j . The hyperbolic polyhedron ?2 is isometric to ?2 . Then the rectangles D2 and E2 are also isometric. Fix the polyhedron +2, the face E2 @ j +2j and denote the faces in StE2 by Qi (i = 1; : : : ; 4) so that Q1 j . The remaining face in @ +2 nStE2 will be denoted by S . Then:
S meets Qi (i = 1; 2; 3) by the angles =2; =3; =2 ,
(a)
Sp(S ) is orthogonal to @ H 3 .
(b) The sphere Sp(S ) with the properties (a), (b) is unique up to the re ection 2 in E2 ( 2 preserves Q1 ). Thus, (a) &(b) , the angle between S and Q1 is equal to =5 . However, we can consider Sp(Qi ) as spheres which contain faces of the polyhedron +2 (since D2 is isometric to E2 ). Let R be the remaining face of @ j+2 jnStD2 . Then R has the same properties (a), (b), however the angle between R and Q1 is equal to =3. This contradiction implies that E2 cannot be a totally geodesic suborbifold in . (iii) The same arguments as above are valid for the curve 3 . So the curves k (k = 1; 2; 3) cannot be boundaries of underlying sets of totally geodesic suborbifolds of . Hence does not contain totally geodesic suborbifolds at all. Let H be the discrete group generated by re ections in faces of . Then H does not admit bending deformations, however Theorem 1 implies that the deformation space R(H; 4) is 2-dimensional. 6.4. Conjecture 2. For every cocompact discrete subgroup G Isom(H 3 ) the variety R(G; 4) is smooth at the point [id]. 12
Remark 9. As John Millson explained to me, the rst obstruction for deformations in this case is always zero because it belongs to the subspace H2 (G; so(3; 1)) of H2 (G; so(4; 1)), however H2 (G; so(3; 1)) = H1 (G; so(3; 1)) = 0.
Acknowledgements. Author is very grateful to Nicolaas Kuiper and John Millson for interesting discussions and to the referee of this paper for helpful remarks. A considerable part of this text was written during the author's stay in IHES [Ka 3]. This work was also supported by NSF grant numbers 8505550 and 8902619 administered through MSRI and University of Maryland at College Park.
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