Math. Ann. 249, 243-263 (1980)

@ by Springer-Verlag 1980

Conjugacy Relations in Subgroups of the Mapping Class Group and a Group-Theoretic Description of the Rochlin Invariant

Dennis Johnson*

Jet Propulsion Laboratory, 4800 Oak Grove Drive, Pasadena, CA 91103, USA

1. Introduction

Let M be a compact oriented surface of genus g having a single boundary component, ~ f its mapping class group (that is, the group of orientation preserving homeomorphisms of M which are 1 on the boundary rood the subgroup of homeomorphisms which are isotopic to 1 by an isotopy which is pointwise fixed on the boundary), and J the subgroup of J r which acts trivially on HI (M,Z ). Powell has produced (an infinite set of) generators for J of two particularly simple types, namely:

a) a twist on a bounding simple closed curve of M, b) opposite twists on two disjoint homologous simple closed curves of M. Algebraic statements concerning these generators frequently can be translated

into convenient geometric ones ; for example, two such maps of the same type are conjugate in ~ ' iff the defining curves separate M into pieces of the same genus. Our main result here is to give a corresponding recipe for deciding when two such maps are conjugate in J . This recipe will then be extended to conjugacy classification of such maps in fq for any f~ containing J .

Our second goal is to compute certain commutator quotients of J . If ~ is as above, then J/[fq, J ] = X , is abelian. Because of their algebraic simplicity, abelian quotients of J are particulary attractive as objects of study. All known such

quotients are derived from just two • the first (see [J3]) is free abelian of rank (23),

and the second arises from homomorphisms constructed by Birman and Craggs in

~ .(~9/(see [J2] for details). The [BC] and is a Zz-vector space of dimension i = 0 \ ~ /

Birman-Craggs homomorphisms were defined by means of the Rochlin invariant of homology spheres, and this connection with 3-manifold topology points out

~r This paper presents the results of one phase of research carried out at the Jet Propulsion

Laboratory, California Institute of Technology, under Contract NAS 7-100, sponsored by National Aeronautics and Space Administration

0025-5831/80/0249/0243/$04.20

244 D. Johnson

another motivation for the study of (abelian) quotients of J , namely, the possibility of deriving from them invariants for 3-manifolds. To this end, a computation of the universal abelian quotient J /~ ' =X~ would be particularly interesting. In this paper, we use the conjugacy classification scheme to carry out an explicit computation of X , for aj = (9,0 = the subgroup of J# fixing a prescribed quadratic form o) on H a (M, Z2). This gives, as a byproduct, a purely algebraic characterization of the Rochlin invariant (in terms of mapping class group data).

The paper is organized as follows. Section 2 gives a number of preliminary algebraic results we shall need concerning groups with symplectic structure. Section 3 is likewise preliminary in nature, this time dealing with surfaces. Section 4 produces the conjugacy classification results for J or a group ff containing it, and Sect. 5 carries this program out in further detail for f~=(9o,. Finally, Sect. 6 computes J/[(9~,, J ] ; it closes with a corresponding result for the case of a closed surface.

2. Symplectic Spaces

In this section the coefficient ring R will always be Z or Z 2 (everything could be done for an arbitrary modulus m). For R a field, some of the statements below can be found in Artin's book [A1], Chap. 3. We shall be interested in free modules V of finite rank over R which are endowed with a bilinear pairing to R, denoted by x. y, satisfying x .x =0, all xE V. The radical of V is the set rad V of all x~ V such that x . y = 0 for all yE V. The pairing on V will be called primitive if for every linear functional f : V~R satisfying f ( r ad V)= 0 there is an x E V such that f(y)= x.y for all y s V. For R=Z z this condition holds automatically. A primitive pairing with rad V=0 is called sympleetic and in this case we have a self-duality on V (The terms "primitive" and "symplectic" will be applied to V itself when the pairing is clear from context.) It is well known that a symplectic V always has a basis % b i ( i= l . . . . . g) such that ai.ai=b~.bj=O, a~.bj=61j (= Kronecker delta); we call such a basis sympleetic, or an Sp-basis. In particular, the dimension of V is 29 and so always even.

The most common example of a primitive pairing, and the one which concerns us here, is the first homology group of a surface. Let Mg,, (frequently abbreviated herein to M) denote an orientable, oriented surface of genus 9 and having n boundary components 7i (i = 1 .. . . . n). Then H 1 (Mg,,, R) (frequently abbreviated Ht) is a free R-module of dimension 29 + n - 1 and has a natural pairing defined by the intersection of 1-cycles on M. This pairing is always primitive, and the radical is generated by the classes of the boundary curves. In fact, each 7i acquires a preferred orientation from that of M, so determines a homology class % and any n - 1 of the q's form a basis for r a d H 1. In particular, for n = 0 or 1, H 1 is symplectic.

A subspace U of V is symplectic (resp. primitive) if the pairing restricted to U is symplectic (resp. primitive). U is symplectic iff it has a symplectic basis as, bi(i= 1 ... . . k); this set can be extended to a maximal such set of V, say i= 1, ..., 9, which form, then, the basis of a maximal Sp-subspace of V.

Conjugacy Relations in Subgroups 245

Lemma 1. Let V be a module with primitive pairing ; then a) V/rad V acquires a natural induced pairing from that of V and is symplectie. b) I f U is any maximal Sp-subspace of V then the projection of U to V/rad V is

an isomorphism (preserving the pairing) and V= Uq)rad V. c) I f U has Sp-basis ai, bi(i-- 1 . . . . . g) and U' is any other maximal Sp-subspace,

then U' has an Sp-basis of the form ai+el, bi+ fl , where el, f / e r a d V.

Proof. a) and b) are obvious. To prove c), project ai, b i to an Sp-basis of V/rad V; by b) this lifts uniquely to an Sp-basis a'i, b' i of U'. Since a'i and ai have the same image in V/rad E they must differ by something in rad V, and likewise for b i. QED.

I f X is any subset of V, we define X ± = {re Vlv. x = 0, all x~X} ;X ± is a subspace of V containing rad V. Suppose now that U is an Sp-subspace of V with Sp-basis a~, bi(i = 1 .... , k). Extend this to a maximal one, i = 1 . . . . . g, and let c j ( j= 1 . . . . ) be a basis for rad V Clearly a~, bi(i = k + 1 ..... g) and the cj form a basis for U ±, and V= U@ U I. If V is symplectic then so is U I.

If f is an intersection preserving a u t o m o r p h i s m of V which fixes U, that is, f (U) = U, then f ( U j-) = U j. If U is symplectic then the m a p f splits on V= U ® U i into g = f l U and g ± = f l U ~, and we write f = g O g ±. Clearly g, g ± are intersection preserving on U, U ±. Conversely, any pair of intersection preserving au tomor - phisms g,g± of U, U ± define an intersection preserving au tomorph i sm f=g@g± on V which fixes U.

In the remainder of this section we shall restrict V to be symplectic. An au tomorph i sm of V preserving the pairing is called a symplectic map ; the set of all these form a group Sp(V). This group contains certain part icularly simple elements known as transvections. We select a e V and in case R = Z we require a to be primitive, that is, not a non-trivial multiple of any other vector in V. Then the transvection by a is given by T~(x)=x +(a.x)a for all x~ V. It is easily checked that TaeSp(V ) and that T,~(x)=x+r(a.x)a for any integer r. Note that for R = Z z this gives us Ta 2 = 1.

Lemma 2. Suppose a~ V is primitive and that f e S p ( V ) is the identity on a ±. Then f = T~ for some integer r.

Notation. If x 1 . . . . . x n is a set of vectors in V, we denote the subspace they generate by (x 1 . . . . . xn).

Proof of the Lemma. The primitivity of a and the duality on V give us a b ~ V such that a . b = l . Hence U=(a,b) is symplectic, and we find a±=(a)~U i. N o w f (a)=a, so a . f (b )=f (a) . f (b )=a.b , i . e . a . ( f ( b ) - b ) = O . Thus f ( b ) - b ~ a l, say f ( b ) = b + r a + x , where x e U ±. But for any y e U ± we also have f ( y ) = y and get likewise ( f (b ) -b ) . y=O, that is, O=(ra+x).y=x.y. This being true for all y e U ± implies x = 0, since U l is symplectic. Hence f (b) = b + ra. N o w T~(b) = b + ra also, and further, T~, -- 1 on a ±, Since a I + (b) = V, we must have f = 7~. QED.

When R = Z z the Sp-space V m a y be endowed with an addit ional structural feature, namely an Sp-quadratic form (or simply Sp-form). This is by definition a function 09: V--+Z 2 satisfying o(a+b)=o(a)+co(b)+a.b . An iteration of this formula shows easily that o is determined by its values on a basis of V and that these values may be specified arbitrarily. Two Sp-forms co, co' on V,, V' resp. are congruent, written co'-~c0, if there is an i somorphism f : V ~ V ' such that

246 D. Johnson

co'(fx)=co(x), all xe E Note that this implies f x . f y = x . y , so f is a symplectic isomorphism. The map f is called an isometry of (Kco) onto (V',o'). When the respective forms are clear we will say V is congruent to V' and write V-~ V'. The group of all self-isometrics of o, that is, all feSp(V) such that o( fx)=m(x) , all xe V, is the orthogonal group 0o, of co.

If a~, b~ is an Sp-basis of V, the ArJ" invariant of co is given by g

~(o)= Y. ~o(a~) co(b~)~Z~ ; i = 1

it does not depend on the choice of Sp-basis [again, we write simply ~(V) when the form is clear]. Arf showed in [A2] that two Sp-forms are congruent iff they have the same invariant and the dimension of their spaces is the same.

If U is an Sp-subspace of V, then cou = col U is an Sp-form on U, and likewise co~ on U l, Furthermore, writing x~ V as x = u + u ±, we get co(x) = ~o(u) + co(u ±) + u. u ± =cov(u)+co~(u') and hence co is the direct sum cov O)¢o~. Thus the orthogonal splitting of V induced by an Sp-subspace carries over to forms. We clearly have

(V) = ~(U) + ct(U±), implying:

Lemma 3 ("Cancellation Law"). I f cop ~z are forms on V1, V 2 resp. and U1, U z are Sp-subspaces of V 1, V z resp. then Vt ~- V 2 and U a ~ U 2 => U ~ ~- U-~.

Proof dim U~ =dim U~ and ~(U~) = ~(Uz~).

3. Surfaces

We use the following notations and abbreviations: H 1 = H 1 (Mg,,) uses Z or Z 2 coefficients, clear from context if unspecified. We

write " ~ " for "homologous". SCC means "simple closed curve"; BSCC means "bounding simple closed

curve", that is, a null-homologous SCC in M. A BSCC separates M into two pieces, one of which must contain 3M. If V is a BSCC in M, its genus g(y) is the genus of the piece not containing ~M. We use the notation S~, S t for the piece containing t3M and the other piece resp. If g(v)=k then S.~ is an Mk, 1 and S~ is an Mg-k,,+ 1. In particular, a BSCC 7 is maximal if g(y)=g.

For 1, a BSCC the map H i (S~)~H I(M) is injective and intersection preserving. Its image will be denoted by Ur which, since HI(Sy) is symptectic, is an Sp- subspace of dimension 2g(7). The map H 1 (Sr)~HI (M) is not injective (if n > 0); we denote its image by 0r. We have then:

Lemma 4. a) Hi(M)= U~O r b) O~ = U~.

Proof The former statement follows immediately from the use of the Mayer- Vietoris sequence on (M, Se, S~). To deduce the latter, note that any cycle in S,e clearly has zero intersection with any cycle in St, so 0rC U~. Conversely, let x~ U~; by a) we may write x=u+~. Then for any y~ Ur we have x .y=0, i.e. O=u.y + ~ . y = u . y since- ± u~ U~ also. Since Ur is symplectic, we conclude that u=O and hence x = ~ O ~ . QED.

Conjugacy Relations in Subgroups

Vig. l

1" a

247

In particular, let ? be a maximal BSCC ; then S~ is genus zero and its homology is carried by ~?M. We saw in the previous section that (the homology classes o0 the boundary components of M generate radH 1. Thus for any maximal BSCC 7, /]y = radH 1 and U? is a maximal Sp-subspace of H 1. We henceforth denote radH 1 simply by C.

The mapping class group Jg~,, of Mo, . is the group of all orientation preserving homeomorphisms of M0, . which are the identity on OM0, . rood those which are isotopic to 1 by an isotopy which is trivial on the boundary, that is, which leaves the boundary pointwised fixed during the isotopy. This group is generated by twists on SCC's (see [L]). l f~ is an SCC the twist T~ affects any arc e crossing c~ by causing it to turn right, travel once around a, then proceed on as before, as in Fig. I.

Furthermore, the induced map of T~ on H 1 is just the transvection T a, where a is the homology class of c~. In the sequel, we shall frequently confuse a homeomorphism with its mapping class. Also, we make the permanent assump- tion that every homeomorphism is the identity on the boundary.

The homology functor gives a map from J¢0,, to the intersection preserving automorphisms o f H 1 which act trivially on C. In particular, for n =0 or 1 we get a map Jgo.,~Sp(H1), and a classical result states that this map is onto (with either Z or Z 2 coefficients). We shall need the following less well known result:

Lemma 5. Let g be an intersection preserving automorphism of HI(Mg,,) acting trivatty on C; then g is induced by a map in Jt,~,..

A proof will be found in Appendix I. We define Jg,, to be the subgroup of ~/0, . which acts trivally on Hi(M, Z) and

,g~2~ the subgroup acting trivally on HI(M, Z2). Two types of maps of Jo,, are particularly important (see e.g., [P]). If ? is a BSCC of genus k then Ty is in Jo,,, and, by virtue of the existence of a homeomorphism taking 71 to 72 whenever they have the same genus, all genus k maps are conjugate in d¢o, .. Definition. A bounding pair (BP) is an ordered pair (7, 6) of a disjoint homologous SCC's in M, with ? + 0. A BP (7, 6) determines the other type of map of concern to us, namely the map T~ T~- l e J . Note that the map associated to (5, 7) is the inverse of that associated to (7, 6).

A BP separates M into two pieces, one of which must contain 0M. Adapting the notation used for BSCC's to this case, we write S~,~ and Sr.o for these two pieces, the latter containing 0M. S~,a is an Mk, 2 for some k, which we call the genus of (7, 6), written 9(?, 6). As in the BSCC case, all BP maps of the same genus are conjugate in ~'g,,.

Our eventual goal is to determine when maps of the above types are conjugate in J . This will be done by associating to each such map a subspace of H~ such that

248 D, Johnson

two of the maps are conjugate in J iff they have the same subspace. When 7 is a BSCC we have already associated a certain subspace of H~ to it, namely, the symplectic subspace U r Extending our analogy with the BSCC case, we now define U.e,~ and 0~,~ to be the images of HI(S~, ~) and HI(S~, ~) in H1(M); the corresponding version of Lemma 4 is then :

Lemma 6. a) HI(S~,6)~HI(M) is injective and so an (intersection preserving) isomorphism to Uy,6. The latter is a primitive subspace of Hi (M) of dimension 29(7, 6)+ 1, and rad U~,~=(c), where c is the homology class of 7.

b) ~,~c~0~,~=(c) and U~,~+ O~.,~=c I. c) U~ o=U~,a and - i _ , , U ~ , ~ - U~,~.

Proof If¢ is a maximal BSCC in S~,~, then U~,e= U~+(c) and this sum is clearly direct ; a) follows from this. A use of the Mayer-Vietoris sequence proves b), and c) follows from b) by an argument similar to that used in Lemma 4.

Note that S~.,o has a natural orientation (derived from that of M) which thereby induces a preferred orientation of), and cs. The homology class c of,2 so oriented is thereby a preferred generator of rad U~,~. Definition. A polarized subspace of H a is a pair (U, c), where U is a primitive subspace of H , with rad U one-dimensional and c is a generator of radU. (If R = Z z the specification of c is superfluous.) Thus a BP (7, 6) gives us a polarized subspace (Ur, a, c); we shall find this the appropriate tool to describe the conjugacy classes of BP maps in ,P'. Note that the polarized subspace of (~, 7) is (U~,~, - c ) ; although Ur, ~ is indifferent to the order of ~, 6, the polarization detects this order.

4. Conjugacy Relations in Subgroups of d/ /~

In the remainder of the paper M will be a fixed surface of type 9, 1 ; the various occurrences of the subscript 9, 1 will be suppressed when clear from context. When Z 2 coefficients are to be emphasized in H 1, we sometimes write H 1 rood 2. Sp(H1) is written simply Sp or Sp mod 2.

If ~ is a subgroup of ~ ' containing J , we shall examine the problem of deciding when two maps of the types we have discussed are conjugate in ~. In particular, we will look at ~ = J , ~=d/~z) and, for ~o an Sp-form on H t mod 2, at ~=(~o,=(by definition) all f ~ ' such that f induces an isometry of ~o on H~ rood 2.

Lemma 7. Let 7,7' be SCC's in M and suppose that T~= T~,, in////; then U~= U /. Similarly,, if (7, 3) and (7', 6') are BP's determining the same map in J , then their polarized subspaces are equal.

The proof will be deferred to Appendix II.

Lemma 8, Let f# be a subgroup of J/¢ containing o¢ and let G~- (~/o¢ be its image in Sp. I f 7, 7' are BSCC's and T~,, T r, are qq-conjugate, then there is a g~G such that g(U~) = U~,, I f (7, ¢5) and (7', 6') give ~-conjugate BP maps, there is a g~ G such that g(Ur,a)= Ur,, ~, and g(c)=c'.

Conjugacy Relations in Subgroups 249

Proof. By hypothesis there is 0~ ff such that 0 TrO-1 = T~(~)= T~,. By Lemma 7, we have Ug~)=U/; but if we let g be the image of 0 in G, then U ~ = H I (0S)=gi l l (S)= g(U~). The proof for BP's is the same.

Corollary. I f T v T,/ cz~ u are conjugate in ~¢ (or J t ) then U~. and U~ are eq at in H 1 (M, Z) (or H 1 (M, Z2) ) ; if two BP maps are conjugate in J (or dd (2)) then their polarized subspaces are likewise equal.

Proof. For f f = J we get G = I , so the subspaces must be equal in Hi(M, Z). For (# = j/eta) we get G = the subgroup of Sp which acts trivially on H I (M, Zz), hence the subspaces must be equal in H l (M, Z2).

We are now ready to give the promised description of conjugacy classes in ~.

t r Theorem 1A. I f y, ~ a e BSCC s, T~ and T 7, are conjugate in J iff U~ = U~, in Hi (M, Z); they are conjugate in ~/¢¢2) iff U~= U~, in HI(M, Zz).

Proof. We have already shown the necessity. To prove sufficiency, we wilt exhibit a map in J (or J t (2)) taking Sy to St,. Since dim U~. = 2g(?0 we have g(7)=g(o/), so S a homeomorphism h such that h(Sr)=S~,, whereby h(U~,)=U~y=U~. Now U~ is symplectic, so the action ofh on H~ splits into the direct sum of symplectic maps g, gX on U~.,/~ = U~, and g is then the induced map of some homeomorphism on S~, which we also denote by g. To get a corresponding homeomorphism on Sr inducin_g gl on U~, note first that Sr has two boundary components, one being 7, so H 1 (S~) splits as U 1 ~ (c), where U 1 is maximally symplectic in H~(S~) and c is the class of 7 (suitably oriented). Furthermore, the projection of H 1 (S~) to U~ has kernel (c) and takes U1 isomorphically onto U¢. Thus we may lift gX to a symplectic map on Uz and then extend it by the identity on (c) to get an intersection preserving map on H~(S~,) which acts trivially on its radical (c). By Lemma 5, this map is induced by a homeomorphism of S~, and we denote this also by g±. Piecing g and g l together along 7 we get a homeomorphism h o of M which fixes S~, and has the same induced map on H a as h. Hence f=hh f f ~ is in ~¢ and f(S,~)=S~e,. In the Z z case, h and h o induce the same map on H~mod2 so hho l~J4 ~z). QED.

Theorem lB. The BP maps T~ T f ~ and T~, T f ~ are conjugate in 3 (or JCl (2~) iff they have the same polarized subspace in H~ (or H a mod 2).

Proof. The necessity has been proved. Conversely, assume equality of the subspaces and again let h be a homeomorphism taking S= S.~,~ to S '= S~,e, and to 7'. Just as in Part A we conclude that h leaves the subspace U = U~,~ invariant. Furthermore, since h(7)= 7' and both ~, 7' have the same homology class c, we get h(c)=c. Thus h is intersection preserving on U=H~(S) and acts trivially on its radical (c), so lifts to a map g on S. By an argument similar to that used in Part A, we may also produce a map g± on S~,~ inducing the same map on U~,~ = U x as does h. Piecing g, g± together along V, (5 gives a map h o agreeing with h on U+ U±=c ± (by Lemma 6) and such that ho(S) = S. Put fo = hhg ~ ; then fo(S) = S' and fo = 1 on e a. By Lemma 2, fo = T[ on H~(M) for some integer r. But T~,(S) = S and T~ induces T~ on H~, so foT~-"=f takes S to S' and = 1 on all of H a, showing fe._g (or ~g~z)). QED.

250 D. J o h n s o n

F i g . 2

F i g . 3

l

,, , j . / , , -

7 b

,~, /~ t

, / L., Y" b'

To complete our classification, we need:

Lemma 9. I f U is any Sp-subspace of Hi(M, Z) then there is a BSCC 7 with U~ = U. I f (U, c) is any polarized subspace of Ha(M, Z) there is a BP(7, 6) having (U, c) as its polarized subspace.

Proof. If U has Sp-basis a i, b i (i= 1 . . . . , k) we lift these classes to representative SCC's ~i, fl~ in M satisfying c~i¢~j=fl~c~flj=~ic~fij (i4=j)=0 and ~i intersecting/~i once transversely. There is then a BSCC 7 of genus k which separates these curves

k

from ~M. For example, connect each set ~iufl~ to some point p ~ M - U (a~wfil) by i = l

an arc el, the e~'s being disjoint except at p and disjoint from U (~ufii) except at

their ends. A regular neighborhood of .~. (~i~fl~w~i) has a BSCC boundary ?

filling the requirements. Clearly, U,¢ = U. For a polarized subspace (U, c) with basis al, b~, c, we extend the above set of curves ~, fli to include an oriented SCC 7o representing c such that af~7o=fii¢~'o=0. If now a 7 is chosen, as above, separating the a~, fl[s from 7o, so that 70~S~, then connect 7o to 7 by an arc e in S~, and consider a regular neighborhood of S~w~yo, as in Fig. 2. Its boundary consists of two disjoint homologous SCC's 7~, 72, with 7~ isotopic to 7o, and it is easy to see that the polarized subspace of the BP (7~,72) is (U, +c) (the sign depends on whether e attaches to the "left" or "right" side of 70 ; it is + in the situation depicted in Fig. 2). We may reverse the sign by reversing the order of ~1, 7z. QED.

Remark. The corresponding statements for R = Z 2 follow from the easily proved fact that a subspace of appropriate type in H~(M, Zz) may be lifted to one of the same type in H~(M, Z).

Corollary. The ~¢-conjugacy classes of BSCC maps are in 1 - 1 correspondence with the symplectic subspaces of Hx(M, Z). The J-conjugacy classes of BP maps are in

Conjugacy Relations in Subgroups 251

1 - 1 correspondence with the polarized subspaces of Ht (M, Z). The J/d(2tconjugacy classes of these maps are in 1 - 1 correspondence with the correspondin9 subspaces of Hi (M, Zz).

As an example of the applicat ion of Theorem 1, consider the curves 7, 7' of Fig. 3. Since b' ~- b we have U~ = (a, b) = U<, and so T,~ is conjugate to T./in J . It is easy to see that they are not isotopic.

Let now ff be any subgroup of ~{ containing ~ (resp. ~{tz~) and let G-~ f g / J (re@. ay/~t2)) be its image in Sp (resp. Sp m o d 2).

Theorem 2. For BSCC's 7, 6, T~, and T~ are N-conjugate iff U~ and U a are G-equivalent, that is, 39eG with 9(U,~)=U a. Likewise for BP maps and their polarized U's.

Proof. The necessity was proved in L e m m a 8. Conversely, a g e G taking U~ to U a can be lifted to a h o m e o m o r p h i s m 0e fq, and U ~ ) = g ( U J = U a, so by Theorem 1 T~(,~ = 0T~ 0-1 is conjugate to T~ in ~¢ (resp. ~¢/~2)). But T~. is conjugate to 0T~0-~ in N. QED.

The proof for BP's is the same.

5. Conjugacy Relations in ( ~

Let co be a fixed Sp-form on Hi ( rood 2). If 7 is a BSCC, then we write ~(J=c~(Uj =~(co[UJ and define 7,6 to be congruent if they have the same genus and c~(7)=~(6 ). If T~,, T a are conjugate in (9~,, then the subspaces U 7, U a of H 1 rood2 must be O~-equivalent by T h e o r e m 2 , which implies col U~.~-colU a and thus 9(7) = 9(6) and c~(7)= ~(6). This proves the necessity in the following

Theorem 3A. T~., T~ are (9~-conjugate iff 7, 6 are con#ruent.

Proof (o f sufficiency). Since dim U T = 2 9 ( j = d i m U ~ and e(U~)=c~(Ua), Arf 's theorems tell us that U~ ~- U a. By L e m m a 3 we conclude that U¢ ~- U~- also. Choose isometries e:U<-, Ua and e x'.U,¢~Ua• x', their direct sum f is then an isometry of H 1 rood2 such that f ( U ~ ) = Ua, and the theorem follows now from Theorem 2.

For BP's the si tuation is a bit more complex. Let (7, 6) be a BP and let U,~,a=UoO(c ) with U o maximal symplectic; then c~(Uo) is defined. If al, b i ( i=1 . . . . . k) is an Sp-basis for U o, then by L e m m a 1 any other maximal Sp-subspace U; of U~. a has basis a~+m~c, b~+n~c. Suppose first tha t co(j = co(c) = 0 ; then we get

k

o~( U'o) = ~ co(ai + mic)co(b i + nic) = ~ co(ai)co(bi) = o:( U o) , i = 1 i

since co(c)= 0 and a i . c = b i • c = 0. Thus c~(U0) does not depend on the choice of U o, and we write it simply as e(y, ~). If on the other hand, co(c)= 1, then c~(U~) becomes

~( Uo) + ~ (nlco(ai) + mico(bi) + mlni)co(c) . i

252 D. Johnson

The summation term is just co(~ (nla~ + mibl)). Now for any form on any Sp-space

of dimension 2 we can always find a vector for which co = 1. If n la 1 + rn lbl is such a vector, put rn i = n i = 0 for i > 2 and define U~ by the basis ai + rnic, b i + nlc. U' o is clearly a maximal Sp-subspace of U~,~ and we get c~(U~)= a(Uo)+ 1. Thus e(7~, Y2) is not defined in this case, and we can choose U o in either congruence class.

Definition. (7, 6) is congruent to ()/, 6') if they have the same genus, cg(y) = co(°l'), and, in case this is zero, c~(y, 6 )= a(7', 6').

Theorem 3B. TrT~- 1 and T~,7~71 are conjugate in d) ~ iff (7, 6) is congruent to (7', 6').

Proof. If f~(9o takes the polarized space (U, c) of (?,6) to that of (7',6') I-call it (U', c')] then f(c) = c' and so co(c) = co(c'), i.e. co(y) = co(7')- Further, if U o is maximal Sp- in U, then f (Uo) is maximal Sp- in U', so ~(~, 6)=~(7' , fi') when this is defined.

Suppose conversely that the condit ion is satisfied. If co(7)= co(7')= 1 then we may choose UoC U with ~(Uo)=0, and likewise for U~ C U'; U o and U~ are then congruent. Let e, c' be the classes of 7, Y' and choose d, d'~ U~, U o" resp. such that c .d=c ' . d '= 1. Finally let U 1 = [UoO(C, d)] s and likewise for U'p H 1 is the direct s u m

uo @(c, d)® u , = Uo@(C', d')@ U',.

If now U 1 = 0 then U] = 0 also, and since U o --- U~, the cancellation lemma gives us (c, d)-~(c ', d'), But co(c)= 1, so co(d) = ~(c, d )= cffc', d') = co(d') and thus the map c~c ' , d--*d' is an isometry of(c, d) with (c', d') (it preserves co on a basis). Direct summing this with an isometry of U 0 to U~ gives an isometry f o f H 1 taking (U, c) to (U', c'), and we conclude by Theorem 2 that our maps are conjugate in C~. If U~ 4:0, we proceed as before if o)(d)=co(d'), not ing that in this case we also must have UI~-U'p If co(d)~co(d'), choose a vector x E U l with co(x)=l . Then since x e U 1 C (c, d) ±, we get c- (d + x) = 1 and co(d + x) = 1 + co(d) = o:,(d'), so replace d by d + x and proceed as before. This takes care of the case co(c)=l. When co(c)=eo(c')=O and cffUo)=e(U~), we still have an isometry Uo-+U' o, and if co(d) = ¢o(d') we are again done. If not, replacing d by c + d will accomplish the task, since

co(c + d) = co(c) + co(d) + c. d = co(d) + 1.

This finishes the proof,

6. Commutator Quotients of

If fq is any subgroup of d¢ containing J , then [(~, J ] 9 J ' and so is normal in J , and the quot ient J / [ (# , ~¢] is abelian. We denote this group by X , , and in particular define Xo, = X ~ = J/[(9,~, J ] . Note that if two maps of J are conjugate in N, then they are equal in X~.

It was shown by the au thor in [J1] that, when the genus of M ( = Mg, 1) is > 3, the BP's of genus 1 generate J ; this then also applies t oXe . The situation for Xo, is particularly simple, for by Theorem 3B there are just three (9,~-conjugacy classes of

Conjugacy Relations in Subgroups 253

( ~ ........... ~,

Fig. 4

such maps: those with co(y)= 1, those with co(7)=0 and c~( 7, 8)=0, and those with co(7)=0 and e(y, 6)=1. Let these classes project to the elements v, w0, w I of X,~ respectively; these 3 elements then generate X,o. We shall calculate X~ explicitly below; to begin with, we exhibit some additional relations between v, w o and w~.

The following statement, of a classical nature, admits a direct geometric proof: If 7, 6 are oriented, homologous SCC's and not - 0 , then there is an f e J with

f(7) = & We need only a special case of this result, so rather than prove the above, we

prove:

Lemma 10. Let (3,fl') be a BP; then there is an f e J such that f(¢/)=/~'.

Proof. Let c~ be an SCC intersecting [J,//' each once transversely. A neighborhood of ~w//w/?' looks like Fig. 4, and since [3~-fl ', we have y--- 7'-"~0. Ire ' is as shown in Fig. 4, then c~ -- c( so T~, T~- 1 is in J . It is easily verified that T,, T~- ~(3) is isotopic to fl'. QED.

Lemma 11. Let (7, 6) be a BP and suppose that T~ef¢ for some integer r. Then i~-~)'e[~,~¢]. Proof. Let f e ~ be such that f l7)=6; then f T J ' - l = TI~)= T a and

f T ~ ' f - I = T ~ ~, so (T~Ta-1)'=T~*Ta-*=[T~*,U]e[fq, J ] .

C o r o l l a r y 1. J / [ ~ ' , J ] = 0.

Proof. Here f f = ~ / , and T r is always in ft. A different proof was given in [J1].

C o r o l l a r y 2 . v = 2 w 0 = 2 w 1 = 0 in Xo,( 9 ~ 3).

Proof. Since, for any 7, T~ 2 = 1 in Sp mod 2, that is, T~. z ~ J[(2) C (9~, we get (T~T~- ~)2 is always in [ (9 , J ] . Writing the group operation in Xo additively, we see then that 2 x = 0 for all xeX,o. To show that v=0, recall that it is represented by a map ~T~- ~ with co(*;)= 1. The Corollary then follows from Lemma 11 and the following

Lemma t2. For c in H 1 mod 2, T~E O~,/ff co(c) = 1.

Proof.

co( T~(x)) = co(x + (c. x)c) = o)(x) + (c . x)co(c) + (c-x) 2 = o)(x) + (1 + co(c)Xc- x).

tf co(¢)= 1, we get just co(x) for any x, so T~e 0o,. If ogc) =0, choose x so that c . x = 1 and get co(T~(x))= t+co(x)=t=co(x). QED.

254 D. Johnson

$1 e2 ~ ¢

Fig. 5a--¢ ~¢)

We have now shown that X,o has two generators Wo, w~, each of order at most 2. Our next objective is to show that w 0 =0. We shall need the elements s, t of X~, where s is the image inX,o of T~, ~, being a BSCC of genus 1 with ~(y)=0, and t the image of a genus 2 BSCC map with ~=0 . We now invoke the relation on twists exploited in [J1]. For the surface depicted in Fig. 5a (an M0~,) we have the relation T~TaT: = T~I T,T,3T~4. Each T,~ obviously commutes with all the other maps. If now we glue the Ml . l ' s S~ and S z to ~2, e3 respectively, as in (b), then the surface obtained may be imbedded into an Mo. ~(g > 3) in such a way that ~ +0. We have then a"~4~fl~-el d¢O and g(a, e4)=g([~ , e l ) = l , g(e2)=g(e3)= 1, 9(7)=2. If the relation is written

we see that it gives a relation among maps in J . We may further assume that c9 restricted to each of the S~'s has zero Arf

invariant. To see this, let % b i be an Sp-basis of HI(S~)C HI(M), and extend these four elements to an Sp-basis of Ht(M). Define a form co' by co'(a~)=co'(b,)=O for i = 1, 2, and extend the definitions to the remaining basis elements so as to insure a(o ') = a(os) (possible since g > 3). Then co' is congruent to co, so let fE ~ transform o to ¢o'; if we transform our original imbedding by f, with a'~ =f(a~), etc., we have then co( a'i) = o( b'i) = O.

For such an imbedding of Fig. 5b, we have ~(~2) = ~(e3) = ~()') =0 , so T~ = T~3 = s in Xo and T~=t. Turning to the BP@I, fi), we see that its surface has maximal BSCC ~2, so U~ = HI(S 0 is a maximal Sp-subspace of U~,,~, and U~, has zero Arf invariant. The same applies to (e 4, ~); but since e~ ~g4, the reduction of these two BP maps to X,o must both be either v [when co(q)= 1] or w 0 [when co(e~) = 0 ] and our relation reduces to:

t=2s+2(v or Wo)=0 (since everything in X~, has order <2) .

Finally we apply the relation to Fig. 5c. This can again be imbedded in Mo, ~ for g >3, and we have g(a, ~4)=g(fl, el)=g(),, ~3) = 1, 9@2)=2, with the relation being written

By an argument similar to the one used above, we may assume that co(et)=co(rh) = a~(e3)=c~013)=0. Since U,.,, =(e3, t/a; q ) and a(e 3, q3)=0, T~T~ i reduces to w o

Conjugacy Relations in Subgroups 255

in X,o. Likewise T~T~ -1 reduces to w o. U,,~, has boundary element e4~_el +%, so co(e4)=0. The remaining generators of U,,~4 are, e.g., ~1 and r/1 +r/3 , both with co = 0. Thus T~T~21 also reduces to w o. Finally, ~2 is of genus 2 and has c~(e2)= 0, so we get the relation:

t=3Wo=W o .

We have already shown that t = 0 in Xco, so w o = 0 also. X~ has now been shown to be either 0 or Z 2 and generated by w 1. The final

step is then

Theorem4. Xo,~-Z2 for 9>3. The unique isomorphism is qiven by the Birman- Craggs homomorphism O,o-

Remarks. For the definition of ~, see [J2]. These homomorphisms from J to Z 2 were originally defined for closed surfaces and o9 of zero Arf invariant by Birman and Craggs in [BC] (using a different notation). The definition uses the Rochlin invariant of a homology 3-sphere; in fact, given k~3 t, a homology 3-sphere is constructed using k, and Q(k) is just its Rochlin invariant. The relation to quadratic forms and extension to open surfaces and Arf invariant 1 forms is found in [J23. It was shown there that for any o9, Qco maps J onto Z 2.

Proof of the Theorem. We need only show that 0~ actually defines a homomor- phism of X~ to Z2, that is, to show that Q,o is zero on [(9o; J ]. Let g~(9,o, k ~ J . Then o~,(gkg- lk - i) =O,o(gko- t )_ eo,(k) ' since 0o is homomorphic on J . But by the Corollary to Theorem 2 of [J2], ~o(okg-1)= Q~o(k) whenever ge 6~,o. QED.

There is a curious anomaly in the above proof. Until the final invocation of the homomorphism 0~,, everything has been proved using only algebra and 2-dimensional topology, and by these means alone we concluded that Xo = 0 or Z 2. The statement X,o 4= 0 is a purely algebraic one (definitions for everything in sight can be given which avoid even references to 2-D topology), and yet we have found no other way to exhibit a non-zero element of it (that is, an f¢ [¢o , J ] ) but through the use of the homomorphisms ~,, which involve the at least implicit construction of a 4-manifold. A more direct construction of e,o, or at least a proof that X~,4:0, without the use of 4-manifolds or homology 3-spheres, would be of interest.

By way of justifying the title of this paper, we also point out that the group X~, gives us a group-theoretic definition of the Rochlin invariant of a homology 3-sphere W. In fact, using the constructions of [J2], we represent W as a generalized Heegaard decomposition with gluing map k e ~ ; if co is the self- intersection form of the Heegaard surface, then the Rochlin invariant of W may be defined as the image of k in X,, = Z 2 (see [J2]/ 'or further details). This definition is not, however, computationatly effective, since as mentioned above we have no general method for deciding when an element k e J is in Ida,o, J ] .

It is not hard to show that 0 0 o ~ = 1 in Spmod2, that is, ~ (9o,=J/¢(2). Thus all co al l ¢o

we have [j1(2~ j ] C ~ [(9,o, J ] . The latter is, by Theorem 4, precisely ~ , Ker Qco, all co

that is, the subgroup c# found in [BC], Sect. 4, and [J2].

256 D. Johnson

Conjecture. cK = [~¢¢(z), j ] (g >= 3).

The truth of the above would show X a~2> to be the space of cubic Boolean polynomials found in [J2-], Theorem 4.

It is possible to use Theorem 4 to find the corresponding quotient X~ °~a for a

closed surface M o. We have an exact sequence P 0 - - * K - + J g , 1 - - - - - ~ J g - - ~ 0 , w h e r e p is

induced by an inclusion map Mg, I ~ M o and K is generated by: a) a twist T~ on the boundary of M0,,. b) BP maps of genus g - 1. See, for example, [B], pp. 156-160 a. The map p

• " closed induces a surjectlon X~,~X~, , and it is easy to see that its kernel is precisely the image of K in X,o. It was shown in [BC] and [J2] that for any BSCCT, Qo,(T.,,)= ~(y), and that, for any BP (7, 3), 0,,,(T~T~-1) is 0 if o9(7)= 1 and is ~(y, 6) i f o9(7) = 0. Note that ~(c~)= 7(co), so e,o(~)= c~(og). If c is the class of y in the BP (7, 3) and Ur,o= Uo@(c ) where U 0 is maximal Sp-with basis a i, b i (i= 1, ..., g - 1 ) , we extend to a full Sp-basis ai, b~, c, d of H 1 and obtain ~(og)=~(Uo)+oo(c)~o(d ). If ~o(c) = 0 this gives ~(co) = c~(U0) = ~(7, 6). Thus the image of K in X~, = Z z is precisely the subgroup generated by ~(co), and we have shown:

Theorem 5. For a closed surJace of genus g >= 3, X,o = 0 if ~(~) = 1 and X,~ = Z z if c~(e9) =0, with isomorphism given by Oo,.

Appendix I

We prove Lemma 5 in this appendix. Let 7 be a maximal BSCC of M = M g , , ; HI(M ) then splits into U~(~C, and Ur is symplectic. Writing any x e H I ( M ) as a

/ k

column vector lut, with uE U~, csC, the fact that g = 1 on C implies that g may be \ J c

wr,tt oi , eform and in eo is nterse t'o

preserving, we find that for u, vs U~,

u. v = g(u). g(v) = (A(u) + b(u)). (A(v) + b(v)) = A(u). A(v),

since b(u), b(v)s C = r a d H , . Thus A is a symplectic automorphism of U~=HI(S~). By the classical results there is a homeomorphism of S~ inducing A on U~, and if we extend this homeomorphism by the identity to all of M, we find its induced

0 1 map to be (O ~)' Now (A ~ ) = ( O 1)(b ~ ) s o we really need only show that

(lb O1) is induced by a homeomorphism of M for any map b: U ~ C . The matrices

(~ 10)forma free abelian group with generators(lb, j 0) 1 ' where for i = t . . . . . 2g, e~ is

a basis of Ur, and for j = 1 . . . . . n - 1, cj is a basis of C given by the classes of the boundary curves 7~ of M, and finally, b~ is the map of Uy to C taking e~ to cj and

1 Using the relation illustrated in Fig. 5c for g = 2 , we can show that the BP maps alone suffice to generate K

Conjugacy Relations in Subgroups 257

( . ) (b)

Fig. 6a and b

(1 01) by some homeomor - the remaining e's to zero. Thus it suffices to induce bi j

phism. For example, let {ei} be a symplectic basis, say % b~; we show how to get a map sending al to a~ +cj and fixing the other basis elements.

The classes a~, b~ may be represented in M by a "canonical set" of oriented SCC's % fli satisfying c~ i c~ ~j =/~i c~ ¢/j = 0, c~ i ~/~j = 0 for i =l=j, and c~ i,/~i intersecting once transversely. This collection of curves does not separate M, so let 2 be an arc from/~1 to 7: with its interior disjoint from all the curves %/~, ?k; see Fig. 6a. Let ~,ez be the two components of the boundary of a regular neighborhood of /~t u 2 u ?j. Application of T~ T~- t has no effect on any of the curves %/~, ?k except cq, which goes to e'x as shown in Fig. 6b. The class a'~ of ~'1 is clearly a~+c~. QED.

Appendix II

In this Appendix we will show that isotopic BSCC or BP maps have the same (polarizedt subspaces. The proof will be broken into a number of lemmas. First, we need a

Definition. If f is an automorphism of nl(M) (= n) then Fix(f) is the subgroup of elements of n left fixed by f , and ~(f) is the projection of Fix(f) into Hi(M, Z) I=H1).

The base point for n will always be chosen on 0M; this choice permits us to conclude (as in [J3], Sect. 2) that any homeomorphism f of M gives a well defined automorphism of n (also denoted by f ) which depends only on its mapping class. Our method of proof is to show that for f the automorphism of n induced by T~, 7 a BSCC, we get 4,(f)= [?e, and for f induced by the BP map T~T~ -1, we get • ( f )= t]_: ~. The equality of T~, _T./then implies F ix( f )=Fix( f ' ) and q~(f)= ~(f'), that is, U, I = U,. But then U~ = 0~ = t]~ = U~,. The same argument for BP's shows that U~,~ = U./,~,, since by Lemma 6, U ~ = UT, ~.

Lemma A. Let X, Y be groups with respective automorphisms f, g. Let X + Y denote the free product of X and Y and f + g the automorphism of X + Y induced by f, g. Then Fix( f+9) is generated by Fix(f) CX CX + Y and Fix(g) C YCX + Y.

Proof. Every element z of X + Y has a unique expression of the form z = x l y l x z . . . y . (n>l) , where x~#1 for i>1 and y i ~ l for i<n. Then (f+g)(z)

258

~1 al ~ql a 2 ~2

F ig . 7a---e

Fig . 8

q' I/

(b,

~t

. . . . G . . . . o I /

~J

D. J o h n s o n

7 k

• .. G,, P - ' ® t /

¢,

= f ( x O g ( Y l ) ... 9(Y,) is also of this form, so ( f + g ) ( z ) = z=~f(x i )=xi , g(Yi)=Y;, all i. QED.

Definition. An element a of a group G iscalled simple if the only solutions of x k = a (xe G, ke Z) are those for which k = + 1. In other words, a is "not a power". Note in particular that this implies a 4:1. If G is free, then we know that the only elements commuting with a simple element are its powers (see, e.g., [MKS], p. 42, Problem 6).

If M is an open surface, rc is free and has a set of free generators (a "canonical basis") represented by SCC's % fl, ( i= 1 .. . . . 9), where all the curves are disjoint except at the base point, and are arranged there as in Fig. 7a. Figure 7b shows ak, flk on the form of the surface we shall use here. If we compute the product

k

I-[ r~,, fl,], i = 1

we find it to be represented by the curve 7k of Fig. 7c. Let now 7 be a BSCC of genus k, as illustrated in Fig. 8. Using the basis ei, fli as shown in Fig. 7b, we find:

T~ (ai) = y[ 1 o~i7 k I i <- k

and T~ leaves other basis elements unchanged. If we let X be the subgroup of n (freely) generated by % fli(i < k) and Y the subgroup generated by el, fli( i > k), then n = X + Y and T~ = f + 1 Y, where f = T~IX = conjugation by y[ i eX. We shall apply Lemma A to conclude:

Lemma B. ¢(T~)= Uy.

Conjugacy Relat ions in Subgroups 259

~t ~k÷l

Fig. 9a and b

"t

Cb)

Proof. By Lemma A, Fix(7;) is generated by Fix(f) and Y. But Fix ( f )= {xe X I), ~ l XTk = X}. Since X is free and 7ke X is simple, we have Fix(f) is just the cyclic subgroup generated by ~k" Thus Fix(T~) is generated by ?'k and ei, fli(i> k). Their images in H 1 are 0 and a~, b~(i>k), where the latter is a symplectic basis for 0 r. QED.

This completes the proof of Lemma 7 for BSCC maps. We turn now to the case of a BP(7, 6), as illustrated in Fig. 9a; we have shown

the basis elements ek+ 1, flk+i only. If d is the element of ~ illustrated in Fig. 9b, then we find that

~-1 where 7k is as before. It is shown in [J3], Sect. 4, that the d =7~k + ifik+ 1 k+ 1, action of T~T~ -1 on the basis ~, fli is:

c~i, fli conjugated by d for i< k

(Zk+ l"+~kO~k+ 1

fi~+ 1 and oq, f l i ( i>k+ 1) unchanged.

If we let XCzr be generated by ~i, f l i ( i~k+l ) and Y by ai, f l i ( i>k+l ) then Lemma A tells us that ~(T~T;1) is generated by the corresponding homology classes ai, b~(i > k + 1) and the image of Fix (f) in H t, where f = T~, T~ 1 IX. To show that ~(7~T~- 1)= 0~,~ =(bk+ a, ak+ 2,'", bg) it suffices then to show that the image of Fix(f) in H 1 is just the cyclic subgroup (bk+ 1)" We shall need the following:

Lenuna C. Let F be freely generated by xl , . . . , x,, y, z, and put P = the subgroup generated by the xi's, Q = the subgroup generated by P and y. Let h be an automorphism of F given by q ~ y q y - 1 for q6Q and z~pz , p some element of P. Then Fix(h) is generated by y and z - lp- lyz.

Before proving this lemma, we show how to use it to obtain our desired result. The group X, generated by a i, ill, "'"ak+l' ilk+l, is also easily seen to be freely generated by ~l .. . . . ~k + l, d. If we put ( a 1 . . . . . ilk) = (Xi,..., X2k)' d = y, ak+ 1 = Z,

k

~'k = 17I [~,/~,] = P , i = i

we see that f = TTT j- llX is of the form specified in Lemma C, and conclude that Fix(f) is generated by y =d and z - l p - l y z = a~-+li7 / 1 d a,+ 1. But

d -1 ~))k~k+ 1ilk+ lOCk+ I '

260 D. Johnson

and hence z - l p - l y z = f l k + r The images of d, flk+l in H 1 are bo th just bk+ i; this proves then tha t ¢~(T~T;-l)= 0~,~,

Proof of Lemma C. It is easy to see that h does fix y and z - l p - l y z , so Fix(h) contains the group they generate; we must only show the converse. N o w every element of F can be written in the form qlz"lq2z"2...qr+ l, where q~e Q and e i = + 1. Such an expression will be called a strin9 ; its length is r. Fur thermore , every string can be reduced to one satisfying the addit ional conditions e~_ 1 = - e i ~ q ~ + 1 for i = 2 . . . . . r. Such a string is called reduced, and every element of F has a unique such expression. Let us examine the effect of the m a p h on a string at the locale ... z~'-~qS'.. . . We have 4 cases (with suitable al terations for the first and last q's):

a) el-~ = ~ = 1: we get . . , zq~z...goes to ... pzyq~y-lpz... ; thus q~ is replaced by YqiY- 1 p.

b) e~_ 1 = - ei = 1 : . . . zqiz- i ... ~ . . . pzyq~y- i z - l p - 1 ..., so q~ is replaced by yq~y- ~. c) e i _ l = - ~ i = - l : . . . z - l q F . . . ~ . . . z - l p - l y q i y - l p z .... so qi is replaced by

p- a YqiY- 1 p. d) ~i- 1 ~ ~ - 1 :... z - lqiz- ~ ...--~... z - lp- iyqiy- l z - ~p- ~ .... so qi is replaced by

p- i yqiy- 1

An examinat ion of cases b), c) shows immediately that if the original string is reduced then so is the resulting one, and furthermore, it has the same length and sequence of e~'s. If then a~ Fix(h) is expressed as a reduced string, then its sequence of q~'s must be left unaltered by the above substi tutions a)-d). We shall prove the l emma now by induction on the length r, assuming it to hold for all reduced strings of length r - 2 .

To begin the induction, if r =0 , we have a = q l and h(a)= YqxY- ~= q~ implies that qa = S , since y is simple. I f r = 1 we have a =qlz~lq2. Suppose for example that e 1 = 1. Then h(a)=yqly-lpzyq2y - 1 =q~zq2 ' so yq~y-lp must equal ql.

Sublemma D. a) y q y - l p = q has no solutions q~(2. b) p - l y q y - l = q has no solutions q6Q.

Proof Let q =f i xy k, where x is a reduced word in x 1 . . . . . x,, y and neither begins nor ends with a power of y. I f q is a solution of a), we must have yJ+ lxyk-lp=yJxyk. If k4~ 1, then f i+lxyk- lp is a reduced word, since p~P does not involve y. This is impossible, since one side of the equality begins with yJ+ 1 and the other with y~. I f k = t we get xp = y - lxy, i.e., p = x - ly - ~xy. The right side is a reduced word involving y, again impossible; this proves a). b) follows f rom a) by inverting and replacing q -1 by q.

Par t a) o f the sub lemma shows that qazq2 can never be fixed by h, and par t b) shows analogously that q~z-lq2 can never be fixed; hence, for r = l , L e m m a C holds vacuously.

Note that, in the same way, the sub lemma shows that el cannot be + 1, and also disallows cases a) and d). Thus for r > 2 , a must begin qiz - lqzz .... and h(a) will begin yqiy- iz -~p-~yq2y-~pz ... . We must then have yqay - l=q i , P- ~Yq2Y- ~P = qz. It is easy to see tha t p - ly is simple (p does not involve y) and so we conclude that qa=Y/, q 2 = ( p - l y ) k for some j , k ~ Z , whence z- iqzz=(z- ~p- ~yz) ~. Hence a' =q~z~...q,+ l has length r - 2 and is also fixed by h.

C o n j u g a c y Rela t ions in S u b g r o u p s 261

By hypothesis, it is in the group generated by y and z-~p-~yz, and so then is a. QED.

We have now shown that q)(TTT ~ 1)= O~,a, and hence that if T~T;X= T~,T~,' then U~,o = U./,~,. To complete the proof of Lemma 7 for BP maps, we must show the equality of not just the above subspaces, but also of their corresponding polarized subspaces. In other words, we must also show that e = c', these being the homology classes corresponding to ~, 7'. In [J3], Sect. 5, a homomorphism ~ : J ~ H 1 is defined such that t(T~T~ ~)= 2kc, where k is the genus of (~, 6). From this we conclude that 2kc = 2k'c'. Since c, c' are the homology classes of SCC's, they must be primitive; also, k, k' are positive. These facts imply that c = c', finishing the proof of Lemma 7.

Appendix III

In this appendix we describe an alternate and sometimes more useful repre- sentation of the "subspace invariants" U by means of the exterior powers A k of H r Recall that AkHI is the quotient of the kth tensor power of H, by the subspace generated by all products x ~ ®. . . ® x k in which two of the xi's are equal ; the image of x 1 ®. . . ®x k in AkH1 is denoted by x 1/x x 2 A ... ^ x k. In AaH1, for example, we have generating elements of the form x ^ y A z , and relations of the form x A y A x = O , x A y A z = - - X A Z A y . Let now al, b i be an Sp-basis of i l l ; then the

g

element O= ~ ai,~b~ of A2H, does not depend on the choice of Sp-basis. If i = l

UCH~ is an Sp-subspace with Sp-basis % bi( i=l , . . . ,k) we likewise have an k

element Or= ~. ai/x b ~ A 2 H I depending on U alone. i = l

Suppose (U,c) is a polarized subspace of H 1, so U = U o O ( c ), where U o is maximally symplectic in U. Let U 0 have Sp-basis % bi(i= 1,..., k); by Lemma 1, any other maximal symplectic subspace U~ of U has an Sp-basis of the form

a i + m~c, b~ + n~c(m~, n~e R = Z or Z2).

We get, then,

k

Ov~ = ~ (a i + mic ) A (b i + nic ) i = l

k k

= ~ a i A bi + Y" (nla ~ - mibi) A c = Ovo + x t, c, i = 1 i = 1

k

where x = ~ (n~a~-mibi). Thus Ovo^c=Ov6AceAaH1 is independent of the i = l

choice of both U 0 and its basis : it depends on (U, c) alone. We call this element z(v,~) ; note that Z(v,-c)= -T(v,c).

Since x. x = 0 for x • H , and intersection is bilinear we get a natural homomor- phism j:A2V---+R given by x A y - ~ x . y . We can use J to define similar maps A 3 V ~ V and A4V-->AZV, also denoted by J, as follows:

J(x A y/x z) = (x. y)z + (y. z)x + (z" x)y

262 D. Johnson

and similarly

J ( W A X A y A z ) = ~ [ (W.X)yAz+(y .Z )WAX] . cyclic p e r m s

of x y z

It is easily seen that these maps are well defined; they are symplectic versions of the "contract ions" of tensor calculus.

Lemma, I f U CH 1 is symplectic of dimension 2k then xE U iff J(O v/~ x) = ( k - 1)x. I f (U,c) is a polarized subspace of dimension 2 k + l then x e U iff J(v(v,~ ) ̂ x ) - - ( k - 1)c ^ x.

Proof Let a~, bi(i= 1 . . . . ,k) be an Sp-basis of U. We find forj<_k:

J (OvAa j )=J [ ~ a i A b i A a j ) . )

But for i + j , J(a~ A b~ A a~) = a~ and thus J(O~: A aj) = (k - 1)aj. The same holds for b~, so we get J ( O v A x ) = ( k - 1 ) x for all xE U. If y~ U ±, then J(a~Ab~ A y ) = y and so J(O v ^ y)=ky. Hence for any z e H , , J(O v ^ z ) - ( k - 1)z is the projection of z into U ±, and so is zero iff z~ U.

If (U, c) is polarized, choose a maximal Sp-subspace U o of U and a basis ai, b i ( i=l . . . . ,k) of U o. Let d~U~ be such that e . d = l ; then U ~ = U o ~ ( c , d ) is symplectic, and H~ = U@(d)~U~. For a~ ( j<k) we get

J(z A aj)= J(O~o A c A aj)=(~OJ(ai A bi A c A aj)"

But J(a i/x b, ^ c A a j) is easily seen to be c/~ a j, so we get ( k - 1)c A ai; likewise, J (z A b j) = (k - 1) c ^ bj. Since J (r ^ c) = J (0) = 0 = (k - 1) c/x c, the claimed equality does indeed hold for all x~U. For x~U-~ we find J ( a i A b i A c / x x ) = c / ~ x , so J(z /~ x) = k(c ^ x). Finally, J(z ^ d) = k(c ^ d) + Ovo. If now x = u + rd + uJf(r~ R), we get :

J (z A x ) - ( k - 1)c A x =r(Ovo +c A d)+c A u-~

= rOvo + c ^ (rd + u~). Now OVo involves basis elements of H, which are entirely distinct from those found in c ^ (rd + u~), so J (z A x ) - (k - 1) c/~ x = 0 iff r = 0 and c ^ u,l = 0. Likewise c and u~ involve distinct basis elements, so c A u~ = 0 iff u~ = 0. The equality holds then iff x = u ~ U, QED.

Corollary 1. The symplectic subspaces U, U' are equal iff O~ = 0 v, ; the polarized subspaces (U, c), U', c') are equal iff zw,~)= z~v,,~, ).

If "; is a BSCC we write Ov, e AZH~ simply as 0~, defining similarly zr,~ for a BP (%fi) 2 Then we have:

Corollary 2. T~, T~ are conjugate in J iff Or = O~ ; T~,T~ ' and T~, T~, 1 are conjugate in

2 It is shown in [J3] (see Lemmas 3 and 4B) that there is a homomorphism from J to the additive group A3H1 which takes the value ~r,~ on T~T~-l

Conjugacy Relations in Subgroups 263

Acknowledgements. I would like to thank both Joan Birman and Robion Kirby for their interest in this work as well as for useful conversations.

References

[AI] [A2]

[B]

[J1]

[J2]

[J3]

EL3

[MKS]

[P]

Artin, E. : Geometric algebra. New York : Interscience 1957 Aft, C. : Untersuchungen fiber quadratische Formen in K6rpern der Charakteristik 2. Crelles Math. J. 183, 148 167 (194I) Birman, J.: Braids, links, and mapping class groups. Annals of Mathematical Studies. Princeton: Princeton University Press 1975 group of homeomorphisms of a closed, oriented 2-manifold. Trans. Amer. Math. Soc. 237, 283-309 (1978) Johnson, D. : Homeomorphisms of a surface which act trivially on homology. Proc. Amer. Math. Soc. 75, 1t9-125 (1979) Johnson, D. : Quadratic forms and the Birman-Craggs homomorphisms. Trans. Amer. Math. Soc. (to appear) Johnson, D.: An abelian quotient of the mapping class group J0. Math. Ann. 249, 225-242 (1980) Lickorisch, W.B.R. : A representation of orientable combinatorial 3-manifolds. Ann. of Math. 76, 531-540 (1962) Magnus, W., Karass, A., Solitar, D. : Combinatorial group theory. New York: Interscience 1966 Powell, J. : Two theorems on the mapping class group of surfaces. Proc. Amer. Math. Soc. 68, 347-350 (1978)

Received November 13, 1978; in revised form August 6, 1979