Kg is not finitely generated
Daniel Biss* and Benson Farby
May 20, 2004
1 Introduction
Let g be a closed orientable surface of genus g. The mapping class group
Mod g of g is defined to be the group of isotopy classes of orientation
preserving diffeomorphisms g ! g. Recall that an essential simple closed
curve fl in g is called a bounding curve, or separating curve, if it is null
homologous in g or, equivalently, if fl separates g into two connected com
ponents.
Let K gdenote the subgroup of Mod g generated by the (infinite) collection
of Dehn twists about bounding curves in g. Note that K 1is trivial. It has
been a longstanding problem in the combinatorial topology of surfaces to
determine whether or not the group K gis finitely generated for g 2. For a
discussion of this problem, see, e.g., [Jo1 , Jo3, Bi, Mo1 , Mo3 , Ak ].
McCulloughMiller [MM ] proved that K 2is not finitely generated; Mess
then proved that K2 is in fact an infinite rank free group. Akita proved in [Ak*
* ]
that for all g 2, the rational homology H*(K g; Q) is infinitedimensional
as a vector space over Q. Note that as K g admits a free action on the
Teichmuller space of g, which is contractible and finitedimensional, K ghas
finite cohomological dimension.
For some time it was not known if K g was equal to, or perhaps a finite
index subgroup of, the Torelli group Ig, which is the subgroup of elements
______________________________
*This research was conducted during the period the first author served as a *
*Clay Math
ematics Institute LongTerm Prize Fellow.
ySupported in part by NSF grants DMS9704640 and DMS0244542.
1
of Mod g which act trivially on H1( g; Z). Powell [Po ] proved that K 2 =
I2. Johnson proved in [Jo2 ] that for g 3, K g has infinite index in Ig by
constructing what is now called the Johnson homomorphism, which is the
quotient map in a short exact sequence
1 ! Kg ! Ig ! ^3H=H ! 1
where H = H1( g; Z). Johnson then proved in [Jo3 ] that Ig is finitely gen
erated for all g 3. Our main result is the following.
Theorem 1.1. The group K gis not finitely generated for any g 2.
We will also prove along the way that that the oncepunctured analogue
of K gis not finitely generated.
Theorem 1.1 answers Problem 10 of [Mo2 ], Problem 2.2(i) of [Mo3 ], and
the question/conjecture on page 24 of [Bi]. We would still, however, like
to know the answer to the following question, asked by Morita (see [Mo3 ],
Problem 2.2(ii)).
Question 1.2. Is H1(K g; Z) finitely generated for g 3?
Note that BirmanCraggsJohnson (see, e.g. [BC , Jo1]) and Morita [Mo4 ]
have found large abelian quotients of K g. We would also like to remark
that Morita has discovered (see, e.g., [Mo4 , Mo2 , Mo3 ]) a strong connection
between the algebraic structure of K gand the Casson invariant for homology
3spheres. For example, Morita proved in [Mo4 ] that every integral homology
3sphere can be obtained by gluing two handlebodies along their boundaries
via a map in K g; further, he has been able to express the Casson invariant
as a homomorphism K g ! Z (see, e.g., [Mo1 ]).
Rough outline of the proof. Our proof owes a great intellectual debt
to the paper [MM ] by D. McCullough and A. Miller, where the theorem is
demonstrated in the genus 2 case; indeed we follow the same outline as their
proof.
First, we find an action of K g on the first homology of an abelian cover
Y of g with Galois group Z2g2. While H1(Y ; Z) is infinitely generated, it
is finitely generated as a module over the groupring of the Galois group of
the cover. We view this groupring as the ring Lg of integral Laurent series
2
in 2g  2 variables. This action a priori gives a rather complicated high
dimensional representation of K g. We first project to a Laurent series ring L
in just one variable and then are able to find and quotient out a codimension
two fixed submodule. This reduction to a 2dimensional representation is
crucial for what follows. We then analyze this representation
æ : K g! SL2(L)
The ring L comes equipped with a discrete valuation, and so SL2(L) can
be realized via BruhatTits theory as a group of automorphisms of a certain
simplicial tree. The BassSerre theory of graphs of groups_equivalently, of
groups acting on trees_is especially suited to understanding whether or not
such a group is finitely generated; one such criterion is proven in [MM ]. To
complete the proof, we compute enough about the image of æ to apply this
criterion to show that K gis not finitely generated.
2 Representing Kg on an abelian cover
Consider a standard symplectic basis {a1, . .,.ag, b1, . .,.bg} for H1( g; Z),
where ai . bj = ffii,jand ai . aj = bi . bj = 0. Here and throughout this
article, the symbol . is used to denote the algebraic intersection number of
simple closed curves (or homology classes). By abuse of notation, we will also
sometimes view the ai and bi as elements of ß1( g), considered as relative to
a fixed basepoint.
2.1 The abelian cover
Consider the free abelian group Z2g2 with generators {s2, . .,.sg, t2, . .,.tg}
and the surjection _ : H1( g; Z) ! Z2g2 defined by
_(a1) = _(b1) = 0
_(ai) = si, i 2
_(bi) = ti, i 2
Composing with the Hurewicz map ß1( g) ! H1( g; Z) gives a surjection
' : ß1( g) ! Z2g2; we denote the kernel of ' by K.
3
Let p : Y ! g denote the covering corresponding to the subgroup
K ß1( g). The group Z2g2 then acts on Y by deck transformations.
This action induces an action of Z2g2 on H1(Y ; Z), which is consequently a
1 1
Z s2 , . .,.sg1, t2 , . .,.tg1module. We denote this Laurent series ring by
Lg.
It is rather easy to construct the cover Y explicitly. To this end, con
sider the decomposition of g into two subsurfaces g1,1and 1,1of genus
g  1 and 1, respectively, obtained by cutting along the bounding curve rep
resenting the homotopy class [a1, b1]. Note that the subspace H1( 1,1; Z)
H1( g; Z) is the span of {a1, b1}. Let Y 0denote the universal abelian cover
of g1,1, that is, the cover corresponding to the commutator subgroup of
ß1( g1,1). Since the boundary of g1,1is nullhomologous, it lifts to a col
lection of simple closed boundary curves in Y 0, indexed by the set Z2g2. We
then obtain Y by gluing 1,1to each of these curves along its boundary.
The Lgmodule structure of H1(Y ; Z) can now be read off from this geo
metric description of Y .
Proposition 2.1. The homology group H1(Y ; Z) is a free Lgmodule of rank
2g2
2 + 1 on the following generators: a1, b1, [ai, aj] and [bi, bj] for 2 i <
j g, and [ai, bj] for 2 i, j g, with [ag, bg] excepted. Denote by W the
2g2
free submodule of rank 2  1 obtained by omitting the generators a1 and
b1. We then have p*(a1) = a1, p*(b1) = b1, and p*(c) = 0 for any c 2 W.
Proof.It is a standard fact that the homology of the surface obtained by
sewing in discs along the boundary circles of Y 0is a free Lgmodule on the
generators [ai, aj], [bi, bj], and [ai, bj] (for the sake of normalization, we *
*choose
a single connected fundamental domain X for the action of K on Y and
demand that all these generators be supported in X). Note that the resulting
space is just the universal abelian cover of g1. The element [ag, bg] is
omitted because the relation [a2, b2] . . ...[ag, bg] in ß1( g1) implies that,*
* in
the homology of the cover, [ag, bg] is in the span of the [ai, bi] for 2 i *
*g1.
It is then apparent that the remainder of H1(Y ; Z) is free on the generators
a1 and b1. Finally, the identification of the images of the generators under p*
follows by definition.
We will need to compute the algebraic intersection numbers of certain
curves in Y . To ease the exposition of the next result, it will be convenient *
*to
4
introduce another piece of notation. We denote the set {a2, . .,.ag, b2, . .,.b*
*g}
by {c1, . .,.c2g2} and {s2, . .,.sg, t2, . .,.tg} by {u1, . .,.u2g2}. Thus, H*
*1(Y )
is a free Lgmodule on the generators a1, b1, and [ci, cj] for 1 i < j 2g *
* 2
except i = g and j = 2g.
Proposition 2.2. Suppose i, j, i0, j0 2 {1, . .,.2g  2} with i 6= j and i06= j*
*0.
Assume first that {i, j} \ {i0, j0} = ;. Then there exists ffli,j,i0,j02 {1, 0,*
* 1}
such that
r r2g2
[ci, cj] . u11. .u.2g2[ci0, cj0]=
8
< (1)ri+rj+ri0+rj0ffli,j,i0,j0rk 2 {ffii,k, ffij,k, ffii0,k, ffij0,k} f*
*or all k
= (1)
: 0 otherwise.
Now, assume that i = i0. Then there exists ffli,j,j02 {1, 0, 1} such that
r r2g2
[ci, cj] . u11. .u.2g2[ci, cj0]=
8
< (1)ri+rj+rj0ffli,j,j0rk 2 {ffii,k, ffij,k, ffii,k, ffij0,k} for a*
*ll k
= (2)
: 0 otherwise.
Lastly,
r r2g2 r r2g2
[ci, cj] . u11. .u.2g2a1 = [ci, cj] . u11. .u.2g2b1 = 0(3)
regardless of the integers rk.
Proof.Equation (3) is clear since the curves in question are apaprently dis
joint. To prove equation (1), notice that the curve representing the cycle
[ci, cj] is a kind of quadrilateral beginning at some basepoint y in the funda
mental domain X, then passing to uiy, followed by uiujy, then ujy, and then
back to the original basepoint y. The curve [ci0, cj0] thus intersects [ci, cj]*
* only
once, at y, but this intersection is not necessarily transverse, so we cannot
determine the value of ffli,j,i0,j0= [ci, cj] . [ci0, cj0] aside from observing*
* that it
lies in the set {1, 0, 1}.
Now, the curve ur11. .u.r2g22g2[ci0, cj0] cannot possibly meet [ci, cj] un*
*less
rk 2 {ffii,k, ffij,k, ffii0,k, ffij0,k} for all k. On the other hand, if the *
*two curves
do meet, then by symmetry, their intersection numbers are determined by
[ci, cj] . [ci0, cj0] as indicated in the statement of the proposition.
5
The verification of equation (2) proceeds in much the same way. The
only subtlety comes in checking the cases ri = 0, in which the curves in
question actually have an entire segment in common. But one can perturb
one of the curves so that they only meet at one endpoint of the segment; the
computation then follows from the usual symmetry.
2.2 The representation
It will be useful for us to consider pointed versions of Ig and K g. We work
with respect to the basepoint x = p(y) 2 g. Denote by I g,*the group
of components of the group of basepointpreserving diffeomorphisms of g
which act trivially on H1( g; Z). öF rgetting the basepoint" clearly gives
a surjective homomorphism Ig,*! Ig. Denote by Kg,*the pullback of the
diagram
Ig,*

" fflfflfflffl
Kg Ø____//Ig
that is, the subgroup of Ig,*generated by twists about bounding curves which
avoid the basepoint. Again, the operation of forgetting the basepoint induces
a surjection Kg,*! Kg.
Recall that K = ß1(Y ). Note that since K is not a characteristic subgroup
of ß1( g), an arbitrary mapping class need not lift to Y . In fact, there are
even elements of Ig which don't lift to Y . However, we have the following.
Proposition 2.3. Each element of Kg,*has a lift to a basepointpreserving
diffeomorphism of Y which is unique up to basepointpreserving isotopy.
Proof.The uniqueness is clear. Moreover, by the universal lifting property
for covering maps, the collection of basepointpreserving mapping classes
that admit such a lift constitutes a subgroup. Thus, we need only verify the
result for Dehn twists about bounding curves, as these generate Kg,*.
So, let C be a bounding curve on g, and denote by tC the twist about
C. Since p is an abelian cover, C lifts to a simple closed curve in Y . Consider
the map etC, which is a simultaneous Dehn twist about all the lifts of C. This
obviously constitutes a lift of tC .
6
These observations are enough to give us our main tool. Henceforth C
will denote an arbitrary bounding curve in g, and eCwill denote a lift of C
to Y . The homology class of eCwill be written
X
c + mp2,...,pg,q2,...,qgsp22. .s.pggtq22. .t.qgga1 + np2,...,pg,q2,...qgsp22*
*. .s.pggtq22. .t.qggb1
where c 2 W (recall W was defined in the statement of Proposition 2.1), the
sum is taken over all integers p2, . .,.pg, q2, . .,.qg, and the m's and n's are
integral coefficients, all but finitely many of which vanish. To simplify the
notation, we will use underlined symbols to refer to (g  1)tuples of objects
indexed by the set {2, . .,.g}. For example, p_will stand for p2, . .,.pg, the
symbol s_will stand for s2, . .,.sg and, crucially, binary operations on under
lined quantities will be performed componentwise, so that s_p_= sp22. .s.pgg.
We are now ready to lift the action of K g,*.
Proposition 2.4. The operation which associates to an element of Kg,*, the
action of its lift to Y on H1(Y ; Z) gives rise to a representation
eæ: Kg,*! GL 1+(2g2(Lg)
2 )
Proof.We must check that eætakes composition to multiplication and that
its image respects the Lgaction on H1(Y ; Z). The former condition follows
from the uniqueness up to isotopy of lifts; the latter holds because for any
bounding curve C in g, the set of all lifts of C to Y is Lginvariant.
2.3 Reducing dimension
The representation eæ is highdimensional; we instead would like to work
with a 2dimensional representation. We will achieve this by proving that eæ
contains a large subrepresentation, namely W , that we will be able to ignore.
In order to do this we first need to analyze the image under eæof a twist about
a bounding curve.
Proposition 2.5. Let C be a bounding curve on g and 1 i < j 2g  2.
Then ~æ(tC )([ci, cj]) = [ci, cj] + d where d can be written as a sum of terms
each of which is divisible by (uk  1)(ul 1) for some 1 k < l 2g  2.
7
Proof.We first assume that C~ = [ci0, cj0]. Recall now that if fi = {fik} is a
family of mutually disjoint and nonisotopic simple closed curves on a surface,
and if ff is another simple closed curve, then the homology class of the twist
tfi(ff) of ff about fi is
X
[tfi(ff)] = [ff] + (ff . fik)[fik] (4)
k
Now, if {i, j} = {i0, j0}, then of course ~æ(tC )([ci, cj]) = [ci, cj]. If, *
*instead,
{i, j} \ {i0, j0} = ;, then equation (1) tells us that
~æ(tC )([ci, cj]) = [ci, cj] + ffli,j,i0,j0(ui 1)(uj  1)(u1i0 1)(u1j0 1*
*)[ci0, cj0]
which is of the desired form if we set k = i and l = j. Lastly, suppose that
{i, j} \ {i0, j0} contains a single element, say without loss of generality i =*
* i0.
Then equation (2) gives us
1 1
~æ(tC )([ci, cj])=[ci, cj] + ffli,j,,j0ui + 1  ui(uj  1) uj0  1 [ci0, cj0]
1
= [ci, cj] + ffli,j,,j0u1j0ui  1 + ui (uj  1)(uj0 1)[ci0, *
*cj0]
which again gives us what we want, with k = j and l = j0.
The general case follows from this calculation by the linearity present in
equation (4) along with the vanishing of equation (3).
We now use Proposition 2.5 to find a substantially smaller representation
of Kg,*. Denote by L the Laurent series ring Z[t, t1], and define : Lg ! L
by (si) = 1 for 2 i g and
(
t if i = 2
(ti) =
1 if 3 i g
The homomorphism induces a homomorphism
^ : GL n(Lg) ! GL n(L)
for any n 1. Now define ^æ: Kg,*! GL 1+(2g2(L) by
2 )
æ^= ^ O eæ
This is of course the same as the representation obtained by tensoring
H1(Y ; Z) over Lg with L. Recall that W H1(Y ; Z) was defined in the
statement of Proposition 2.1.
8
Corollary 2.6. The representation æ^ becomes trivial upon restriction to
W Lg L.
Proof.Proposition 2.5 guarantees us that for any bounding curve C and any
1 i < j 2g  2, we have eæ(tC )([ci, cj]) = [ci, cj] + d, where d is a sum *
*of
terms each of which is divisible by (uk1)(ul1) for some 1 k < l 2g2.
Since k 6= l, at least one of uk and ul is not equal to t2, so it must be the
case that each of the summands of d vanishes when we tensor with L. Thus,
^æ(tC )([ci, cj]) = [ci, cj]. The desired result then follows from the fact tha*
*t the
tC generate Kg,*and the [ci, cj] generate W Lg L.
We are now able to define the representation that will actually allow us to
prove our result. Since the image of ^æ: Kg,*! GL 1+(2g2(L) fixes W Lg L,
2 )
we may pass to a quotient representation
~ ~
H1(Y ; Z) Lg L
æ~: Kg,*! Aut L _______________ GL 2(L)
W Lg L
The first thing we will need to know about ~æis the following.
Proposition 2.7. The image of ~æis actually contained in SL2 (L) rather
than GL 2(L). Moreover, for a bounding curve C on g, we have
_ P P !
1 + ni_,jmp,qs_p_i_t_q_j_ mi_,jmp,qs_p_i_t_q_j_
æ~(tC ) = P _ _ _pi qj P _ _ _piqj (5)
ni_,j_np_,q_s___t___1  mi_,j_np_,q_s___t___
Furthermore, ~ædescends to a representation
æ : Kg ! SL2(L)
Proof.Observe that the statement that the image of ~ælies in SL2(L) rather
than GL 2(L) follows formally from equation (5), so it suffices to verify that
equality. To establish that, we compute before projecting to L via by
simply expanding out the summations
X
(etC)*(a1) a1 + (a1 . s_i_t_j_eC)s_i_t_j_[Ce] (mod W )
and X
(etC)*(b1) b1 + (b1 . s_i_t_j_eC)s_i_t_j_[Ce] (mod W )
9
using the formulas
a1 . s_i_t_j_eC= ni_,j_
and
b1 . s_i_t_j_eC= mi_,j_
To verify the last statement, consider an element j of Kg,*that lies in the
kernel of the projection Kg,*! Kg. Denote by ejthe basepointpreserving lift
of j to Y. Since j is isotopic to the identity once we forget basepoints, ejmust
be isotopic to a diffeomorphism covering the identity map on g. Thus, we
must have an equation
_ !
(s_p_t_q_)0
^æ(j) = pq
0 (s__t__)
But in order for this to lie in SL2, it must be the identity matrix, so ^æfacto*
*rs
through the quotient Kg of Kg,*.
3 Amalgamated products and infinite gener
ation
Denote by H the image of the homomorphism æ : Kg ! SL2(L). Our goal
is to prove that H is not finitely generated. We now describe how we will do
this.
Consider the inclusion SL2 (L) SL2(Q[t, t1]). The field Q(t) obtained
by adjoining a free variable t to the rational numbers is equipped with a
discrete valuation and contains Q[t, t1], so one can apply the construc
tion of BruhatTitsSerre to find a (locally infinite) simplicial tree on which
SL2(Q[t, t1]) acts by isometries. The BassSerre theory of groups acting
on trees can then be applied (see [BM ], x5) to express SL2 (Q[t, t1]) as an
amalgamated product:
SL2(Q[t, t1]) ~=A *U B (6)
where A = SL2(Q[t]),
_ ! _ !
t1 0 t 0
B = A
0 1 0 1
10
and U = A \ B.
This decomposition allows one to apply the theory of graphs of groups to
obtain the following criterion, which is Proposition 5 in [MM ].
Proposition 3.1 (Criterion for infinite generation). Let A *U B be
an amalgamated product, and let H be any subgroup. Suppose there exist
elements Mk 2 A\U and Nk 2 B\U such that
1. MkNkM1k2 H; and
2. (H \ A)MkU 6= (H \ A)MlU whenever k 6= l.
Then H is not finitely generated.
We apply Proposition 3.1 to the situation above, with SL2 (Q[t, t1]) ~=
A *U B and with H = æ(Kg). Our goal now is to find matrices Mk and Nk
satisfying the desired criterion.
3.1 The elements Mk and N
For a positive integer k, we let
_ !
1 0
Mk = 2 SL2(L)
k 1
We also set _ !
1 t  2 + t1
N = 2 SL2(L)
0 1
We now verify that the first hypothesis of Proposition 3.1 holds in our
case; here we are taking Nk = N for all k.
Proposition 3.2. For each k 1, the matrix MkNM1klies in H.
Proof.First of all, consider the simple closed bounding curve C shown in
Figure 1. The figure is drawn so that the homology of the leftmost handle
of g is spanned by {a1, b1}. We now lift C to Y ; this is shown in Figure 2.
Here, each octogan with a handle coming out of it corresponds to a single
fundamental domain for the Z2g2action on Y ; we have drawn the two lifts
of C that meet the fundamental domain X. In general, of course, the base
11
Figure 1: The curve C
Figure 2: The two lifts of C that meet b1.
is a 4(g  1)gon; the figure corresponds to the case g = 3. It is clear that
no lifts of C meet a1, so that æ(tC )(a1) = a1. Moreover, by twisting b1 about
the two curves shown in Figure 2, one sees that
æ(tC )(b1) = b1 + (t2  2 + t12)a1 = b1 + (t  2 + t1)a1
and therefore that æ(tC ) = N.
12
Secondly, since b1 is in the kernel of the map ß1( g) ! Z2g2, the twist
tb1 lifts to Y . Denoting by T the simultaneous twist about all the lifts of b1
to Y , we see that Mk = T*k. Set C0 = tkb1(C). We then have
k
MkNM1k = T*k etC*T*
= etC0*
the last equality following from the general formula ftaf1 = tf(a), where f
is any mapping class and ta any Dehn twist. Since C0 bounds in g, we see
that MkNM1k= æ(tC0).
3.2 Distinctness of double cosets
The rest of this paper is devoted to proving the following.
Proposition 3.3. With the notation as above, we have
(H \ A)MkU 6= (H \ A)MlU
for all k 6= l.
Given Proposition 3.3, whose proof we present in the next section, we are
now able to establish our main result, Theorem 1.1.
Proof of Theorem 1.1. We apply Proposition 3.1 to the subgroup H =
æ(Kg) of SL2 (Q[t, t1]) ~= A *U B, with Mk and Nk = N as above. First
observe that Mk 2 A\U since t does not divide k, and N 2 B\U since t 
2 + t1 62 Q[t]. Therefore, in light of Propositions 3.2 and 3.3, Proposition 3*
*.1
implies that H is not finitely generated. As Kg surjects onto H, it is not
finitely generated.
Note that since Kg,*surjects onto Kg, it follows that Kg,*is also not
finitely generated.
4 The proof of Proposition 3.3
In this section we prove Proposition 3.3. In order to do this we will prove
that the elements of H = æ(Kg) are of a very special form. To state this
precisely, we will need the following.
13
Definition 4.1 (Balanced polynomials). Let f 2 Z[u11, . .,.un1] be a
Laurent polynomial in n variables over the integers. We say that f is balanced
if
1. f(1, 1, . .,.1) = 0; and
2. for all ntuples (i1, . .,.in) 2 Zn, the coefficients of ui11. .u.innand
ui11. .u.innin f are equal.
Parallel with a crucial observation of McCulloughMiller [MM ], we have
the following.
Proposition 4.2. Each element of H has the form
_ !
1 + P Q1
Q2 1  P
where P, Q1, and Q2 are balanced.
Proof.Recall the map : Lg ! L above. We begin by fixing an element
T = æ(tC ) 2 H and writing
_ !
(1 + R1) (S1)
T =
(S2) (1  R2)
where now the Ri and Si lie in Lg. Equation (5) gives us expressions for
R1, R2, S1, and S2 in terms of the m and n coefficients. Now, since the twist
tC lies in the Torelli group Ig, we have
p* etC* (a1) = p*(a1) = a1.
But equation (5) tells us that
p* etC* (a1) = a1 + R1(1, 1, . .,.1)a1 + S2(1, 1, . .,.1)b1
and so R1(1, 1, . .,.1) = S2(1, 1, . .,.1) = 0. A similar analysis of b1 allows
us to conclude that R2(1, 1, . .,.1) = S1(1, 1, . .,.1) = 0.
It follows via formal manipulations from equation (5) that S1 and S2 also
satisfy the other criterion for balancedness. We now turn our attention to
R1 and R2. Notice that for all p_and q_, we have
0 = Ce . s_p_t_q_
X X
= mi_,j_ni_+p_,j_+q_ ni_,j_mi_+p_,j_+q_
i_,j_ i_,j_
14
From this, it follows that R1 = R2, from which one can deduce formally that
R1 is balanced.
Since it is clear that takes balanced polynomials to balanced polyno
mials, we have the desired property for elements of the form æ(tC ). But the
set of elements of SL2(L) of the desired form is evidently a subgroup, so the
result follows since the tC generate Kg.
Following Lemma 7 in [MM ], we will now see how Proposition 3.3 follows
rather formally from Proposition 4.2.
Proof of Proposition 3.3.Suppose, that the Mk and Mlare in the same dou
ble coset, that is, that we have a matrix equation
_ ! _ ! _ ! _ !
1 0 1 + P Q1 1 0 u v
= (7)
k 1 Q2 1  P l 1 wt z
with _ !
1 + P Q1
2 H \ A
Q2 1  P
and _ !
u v
2 U
wt z
By Proposition 4.2, we know that P, Q1, and Q2 are balanced. By the
definition of A, they also lie in Q[t]. Thus, they are constant and hence
vanish. Therefore, setting t = 0 in equation (7) gives
_ ! _ ! _ ! _ !
1 0 1 0 u(0) v(0) u(0) v(0)
= =
k 1 l 1 0 z(0) lu(0) lv(0) + z(0)
which obviously implies that k = l.
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Dept. of Mathematics, University of Chicago
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Email: daniel@math.uchicago.edu, farb@math.uchicago.edu
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