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\topmatter
\title Stability of Lower Central Series of Compact 3-Manifold
Groups
\endtitle
\author T.D. Cochran* and Kent E. Orr**\endauthor
\abstract The {\it length\/} of a group $G$ is the least ordinal
$\a$ such that $G_\a=G_{\a+1}$ where $G_\a$ is the $\a^\supth$ term
of the transfinite lower central series. We begin by establishing
connections between lower central series length and the Parafree
Conjecture, $4$-dimensional topological surgery, and link
concordance. We prove that the length of all surface groups and most
Fuchsian groups is at most $\om$. We show that the length of the
group a Seifert fibration over a base of non-positive even Euler
characteristic is at most $\om$. Our major result is the existence of
closed hyperbolic $3$-manifolds with length at least $2\om$. We
observe that any closed orientable $3$-manifold group has the same
lower central series quotients as a hyperbolic one.
\endabstract
\thanks *partially supported by the National Science
Foundation DMS-9100254, DMS-9400224, DNS-9205540\endgraf
**partially supported by the National Science
Foundation\endthanks
\endtopmatter
\document
\baselineskip=20pt
\sub{\S1. Basic Definitions; History; Main Results; Motivation}
Recall that if $G$ is a group and $\a$ an ordinal, the (transfinite)
{\it lower central series of\/} $G$, $G=G_1\supset
G_2\supset\dots\supset G_\a\supset\dots$, is defined inductively by
$G_1=G$, $G_{\a+1}=[G,G_\a]$ and, if $\a$ is a limit ordinal,
$G_\a=\bigcap_{\b<\a}G_\a$. By cardinality considerations this
series must {\it stabilize\/} and we shall call the least ordinal
$\d$ such that $G_\d=G_{\d+1}$ the {\it length of\/} $G$. Herein the
letter $\d$ is reserved for this least ordinal and consequently
$G_\d$ is also the intersection of all terms of the (transfinite)
lower central series of $G$. If $G_\a=\{1\}$ then $G$ is said to be
{\it$\a$-nilpotent\/}; if $\a$ is finite $G$ is said to be {\it
nilpotent\/}; if $\a=\om$ then $G$ is said to be $\om$-nilpotent or
residually nilpotent; if $\a$ is greater than $\om$ then $G$ is said
to be {\it transfinitely nilpotent\/}. The groups $G/G_\a$ are called
the {\it lower central series quotients of\/} $G$. At the end of this
chapter we try to indicate some of the interrelationships between the
lower central series of finitely presented groups and other important
topics in topology and algebra such as: homology cobordism, the
Parafree Conjecture, link concordance, and the topological
classification of non-simply-connected $4$-manifolds. This paper is
concerned with the questions:
\roster
\item"a)" What lengths are possible for compact $3$-manifold groups?
\newline
and, more generally
\item"b)" How are the quotients $G_\a/G_{\a+1}$ restricted for
compact $3$-manifold groups?
\endroster
Herein we give a lot of new information about a).
The literature about the lower central series length seems sparse. In
1935 W\.~Magnus showed that free groups are $\om$-nilpotent and hence
have length $\om$ (length $2$ if of rank~1) [Ma]. In 1949
A.I\.~Malcev showed that, for any ordinal $\a$, there exists a group
of length $\a$ [Mal]. In 1962 G\.~Baumslag showed that the
fundamental groups of compact surfaces of even Euler characteristic
(except the Klein bottle) are $\om$-nilpotent and hence of length at
most $\om$ [Ba1]. In 1966, P\.~Hall and B\.~Hartley showed that, for
any {\it countable\/} ordinal $\a$, there is a finitely-generated
group of length $\a$ ([HH], see also [H1], [H2]). All of their
examples were non-finitely presented. In 1991 J\.~Levine exhibited
the first finitely-presented group with length greater than $\om$
[L3]. His example has length $\om+1$\footnote"*"{This calculation
is not in [L3] but was carried out by the authors. The interested
reader may write to us for a copy of the calculations.}. The authors
have recently expanded upon this to exhibit large families of
finitely-presented groups with {\it very\/} large length (at least
$2\om$ at present count) and other desirable properties [CO2]. On a
different but related front, in 1969, C\.~Thomas described precisely
those $3$-manifolds whose groups are nilpotent [T]. P\.~Teichner's
recent paper is a good companion to ours [Te].
We now summarize our major results. Probably the most interesting
result is the existence of closed, orientable $3$-manifold groups,
even hyperbolic, which have very large length. This is interesting
given that any hyperbolic $3$-manifold group has a subgroup of finite
index which is $\om$-nilpotent!
\proclaim{Theorem 8.1, 8.2} There exist closed hyperbolic
$3$-manifolds whose fundamental groups have length at least $2\om$
(i.e., greater than $\om+k$ for all integers $k$).
\endproclaim
\newpage
We observe from combining previous work of D\.~Ruberman [R] and
J\.~Stallings [St] that any $3$-manifold has the same lower central
series quotients as a hyperbolic one.
\proclaim{Corollary 7.2} Suppose $M$ is a closed, oriented
$3$-manifold with $\pi_1(M)\cong G$. Then there exists a closed,
oriented hyperbolic $3$-manifold $X$ with $\pi_1(X)\cong P$ and an
epimorphism $f_*$: $P\lra G$ which induces an isomorphism
$P/P_\a\cong G/G_\a$ for each ordinal $\a$. In particular the length
of $G$ equals the length of $P$.
\endproclaim
\bpage
We do not know the precise lengths of these examples but they appear
to be $\om^2$. The existence of these ``pathologies'' makes the
following interesting.
\proclaim{Theorem 3.3} Non-abelian fundamental groups of compact
surfaces have length $\om$ and are $\om$-nilpotent.
\endproclaim
\proclaim{Theorems 5.6 and 5.9} A Fuchsian or planar discontinuous
group (see \S5) whose underlying surface has even Euler characteristic
or non-empty boundary, and is not $S^2$, has length at most $\om$. If
the underlying surface is $S^2$ then the group has length at most
$\om+1$.
\endproclaim
\proclaim{Theorem 6.10} Suppose $G$ is the fundamental group of a
compact $3$-manifold which admits a Seifert fibering with base surface
$X$. If $X$ is neither $S^2$ nor a connected sum of an odd number of
projective planes, then length$(G)\le\om$. If $X$ is $S^2$ then
length$(G)\le\om+1$.
\endproclaim
\bpage
On the other hand it is easy to show that almost all Fuchsian groups
and Seifert fibered groups have length at least $\om$. For example,
\proclaim{Theorem 5.10} If $G$ is a Fuchsian group whose underlying
surface $X$ has $\pi_1(X)$ non-abelian then the length of $G$ is at
least $\om$.
\endproclaim
\proclaim{Propositions 6.11 and 6.12} With $G$ as in 6.10 above, if
$\pi_1(X)$ is not abelian then $G$ has length at least $\om$. If
rank $H_1(G)\ge4$ then $G$ has length at least $\om$.
\endproclaim
\bpage
The reader interested only in the length calculations and not in
connections to other areas of algebra and low-dimensional topology,
should now skip to Chapter~2.
\newpage
\noindent{\bf The relationship between the lower central series of
finitely presented groups}
\noindent{\bf and homology cobordism}
The basic interplay between the homology, in low dimensions, of a
group or space and its lower central series was established by work
of K.T\.~Chen, W.S\.~Massey, J\.~Milnor, J\.~Stallings and W\.~Dwyer
[Ch1, Ch2, Mas, Mi, St, D]. In fact we shall employ the following
theorem of Stallings so often in our present work that we state it
for the readers convenience.
\proclaim{1.1 Stallings' Theorem \rm{[St; p\.~170]}} Let $N\tril G$
be a normal subgroup contained in $G_2$. There is a natural exact
sequence
$$
H_2(G) @>\ \pi_*\ >> H_2(G/N)\lra\f N{[G,N]}\lra0.
$$
If $h:A\lra B$ is a homomorphism inducing an isomorphism on $H_1$
and an epimorphism on $H_2$ then, for any finite $n$, $h$ induces an
isomorphism $A/A_n\cong B/B_n$; $h$ induces an embedding
$A/A_\om\hra B/B_\om$. Moreover if $h$ is surjective it induces an
isomorphism $A/A_\a\cong B/B_\a$ for any ordinal $\a$. (Consequently
$A$ and $B$ have the same length).
\endproclaim
There are similar results for the coefficient groups $\BZ_p$ (which
we shall use in \S8) and $\bbq$ but we shall not state these. As an
easy consequence of 1.1, if $2$ manifolds $M^i$ $i=0$, $1$ are
homology cobordant then their groups $G^i$ are isomorphic modulo any
term of the finite lower central series, i.e.,
$G^0/G^0_n\cong G^1/G^1_n$.
\proclaim{Corollary 1.2} If $M^0$ is homology cobordant to $M^1$ and
the length of $\pi_1(M^0)=G^0$ is finite then the length of
$\pi_1(M^0)$ equals the length of $\pi_1(M^1)=G^1$. Therefore
length$(G^0)\ge\om$ if and only if length$(G^1)\ge\om$.
\endproclaim
\noindent{\bf1.3}\ \ If one defines two groups $G^0$, $G^1$ to be {\it
homology cobordant\/} if there exist maps $\phi_i:G^i\lra H$ to some
third group $H$, which induce isomorphisms on $H_1$ and epimorphisms
on $H_2$ then the following question presents itself.
\mpage
\noindent{\bf1.4 Question}: Is length invariant under homology
cobordism of groups?
An affirmative answer would prove the Parafree Conjecture (see
below) for a very large class of groups. Unfortunately (or
fortunately) we shall herein resolve 1.4 in the negative for general
finitely-presented groups (see 9.1).
\bpage
\noindent{\bf The relationship between the length of 3-manifold
groups and the ``Parafree}
\noindent{\bf Conjecture''}
Recall that a group $G$ is {\it parafree of rank $m$\/} if
$G/G_n\cong F/F_n$, $n\in\BZ_+$, where $F$ is the free group of
rank~$m$, and $G_\om=\{e\}$ [Ba2, Ba3] [Str]. The ``Parafree
Conjecture,'' due to G\.~Baumslag, is: ``A finitely generated
parafree group $G$ has $H_2(G;\BZ)=0$'' [Ki]. A parafree group of
rank~$1$ is easily seen to be free using Stalling's theorem. A
``proof'' of the parafree conjecture appeared in [Gu] but that proof
has been found to have gaps. We remark that the nilpotent completion
of $F$, $\varprojlim(F/F_n)$, is an {\it infinitely\/} generated
parafree group of rank~$m$ whose second homology group is uncountably
generated if $m>1$ [B2]. The parafree conjecture has relations with
link concordance invariants, homology cobordism, the Kervaire
conjecture, Whitehead's Asphericity conjecture and with lengths of
$3$-manifold groups. As regards the latter we have the following. A
group $G$ is called {\it almost parafree\/} if there is some free
group $F$ and homomorphism $F\lra G$ which induces isomorphisms
$F/F_n\lra G/G_n$ for all positive integers $n$ (this is called
weakly parafree in [FT2] and almost parafree in [Ki]). Thus if $G$
is almost parafree then $G/G_\om$ is parafree.
\proclaim{Theorem 1.5} Assume the rank of the abelianization of all
groups is greater than $1$. The following are equivalent.
\roster
\item"$\SP$:" The Parafree Conjecture is true.
\item"$\SA\SP$:" Every finitely generated almost parafree group $G$
has length $\om$ (see {\rm[Ki]}).
\item"$\SE$:" Every finite $\BE$-group has length $\om$. (Recall a
finite $\BE$-group is the fundamental group of a finite $2$-complex
$Y$ such that $H_1(Y)$ is torsion-free and $H_2(Y;\BZ)=\SO$
{\rm[C2]}).
\newpage
\item"$\ST$:" If $G$ is the fundamental group of a closed orientable
$3$-manifold and $G$ is almost parafree, then $G$ has length $\om$.
\endroster
\endproclaim
\sub{Remark} It is interesting that $\ST$, which is restricted to
{\it finitely presented\/} $3$-manifold groups, can imply $\SP$ which
has to do with all finitely {\it generated\/} groups.
\sub{Proof of 1.5} First note that $\SA\SP\Rightarrow\SE$ since every
$\BE$-group is almost parafree [C2; p\.~643]. Suppose $G$ is almost
parafree. Then $G/G_\om$ is finitely generated and parafree. It
follows that $\SP\Rightarrow\SA\SP$ since, by Stallings' Theorem, if
$H_2(G/G_\om)=0$ then $G_\om\cong G_{\om+1}$. Moreover $\ST$ follows
{\it a fortiori\/} from $\SA\SP$. The following will show
$\SE\Rightarrow\SP$.
\proclaim{Lemma 1.6} Given any finitely generated almost parafree
group $G$ there exists a finite $\BE$-group $E$ and an epimorphism
$E\overset f\to\lra G$ which induces an isomorphism $E/E_\om\overset
f\to\lra G/G_\om$.
\endproclaim
\sub{Proof of 1.6} Since $G$ is almost parafree, $H_1(G)\cong\BZ^m$
for some $m$. Choose a generating set $\{x_i, y_j\mid1\le i\le m,
1\le j\le k\}$ for $G$ such that $\{x_i\}$ is a basis for $G/G_2$.
Then, for each $j$ there $y_j=\(\prod^m_{i=1}x^{n_{ij}}_i\)w_j$ for
some $n_{ij}$ where $w_j$ lies in the commutator subgroup of the
free group on $\{x_i,y_j\}$. Let $E=\$,
and $f:E\lra G$ be the obvious epimorphism. Clearly $f$ induces an
isomorphism on $H_1$. One sees by inspection that $E$ is a finite
$\BE$-group. Consider the maps $F\overset g\to\lra E\overset
f\to\thra G$ where $F$ is free on $\{x_i\}$ and $g(x_i)=x_i$.
Therefore $f\circ g$ induces an isomorphism abelianizations and hence
induces epimorphisms $F/F_n\lra G/G_n$ for each $n$. Since $G$ is
almost parafree $G/G_n\cong F/F_n$. Since $F/F_n$ is nilpotent, hence
Hopfian, these maps $F/F_n\lra G/G_n$ induced by $f\circ g$ are
isomorphisms. Similarly $g$ induces isomorphisms $F/F_n\lra E/E_n$.
Therefore $f$ induces isomorphisms $E/E_n\lra G/G_n$. Since $f$ is
surjective it follows that $f$ induces an isomorphism $E/E_\om\cong
G/G_\om$. \qed
\bpage
Applying 1.6 with $G$ a finitely generated parafree group we see
that $H_2(G)\cong H_2(E/E_\om)\cong E_\om/E_{\om+1}$ (since
$H_2(E)=0$) so $\SE\Rightarrow\SP$.
It will suffice now to show that ``not $\SE$'' implies ``not
$\ST$''. Suppose $E$ is a finite $\BE$-group and $\g\in
E_\om-E_{\om+1}$. Let $C=E/\<\g\>$, so $C$ is finitely presented and
$E\overset\pi\to\thra C$ induces an isomorphism on abelianizations and
indeed $E/E_\om\cong C/C_\om$. In particular $C$ is almost parafree.
Moreover the class $[\g]\in H_2(C)$ is not zero (form $K(C,1)$ by
attaching a $2$-cell along $\g$ to the $2$-complex associated to $E$
then adding 3 and higher-dimensional cells as needed) since, indeed,
its image in $H_2(C/C_\om)\cong H_2(E/E_\om)\cong E_\om/E_{\om+1}$ is
not zero (the fact that $[\g]\in H_2(E/E_\om)$ maps to $\g$ under the
Stallings' map is clear, for example, from the proof of Stallings'
theorem in [C3]).
Given a group $C$ as above, we shall construct a $4$-manifold $Y$
whose boundary will be the $3$-manifold counterexample to $\ST$. For
specificity consider a presentation $P$ of $C$ obtained from a
finite presentation for $E$ associated to a $2$-complex with
$H_2=0$, by adding the relation $\g$. Construct a $4$-manifold $Y$
such that $\pi_1(Y)\cong C$ by adding a one-handle to $B^4$ for each
generator of $P$ and a 2-handle for each relator of $P$. The final
2-handle, attached to the null-homologous loop $\g$ should be
attached with framing $\pm1$. Then $H_2(Y)\cong\BZ$ generated by
this last 2-handle and the self-intersection of this class is
$\pm1$. It follows that the map $H_2(Y)\lra H_2(Y,\p Y)$ is an
isomorphism (note $H_2(Y,\p Y)$ is free abelian since $\tor H_2(Y,\p
Y)\cong\tor H^2(Y)\cong\tor H_1(Y)\cong0$). Since $Y$ has no
3-handles, $H_1(Y,\p Y)\cong0$ and hence $H_1(\p Y)\lra H_1(Y)$ is
an isomorphism. Let $G=\pi_1(\p Y)$. Then the inclusion
$i:G\thra C$ is an epimorphism which induces an isomorphism
$G/G_2\cong C/C_2$. Consider a map $F\overset g\to\lra G$ which
induces an isomorphism on abelianizations and hence epimorphisms
$g:F/F_n\lra G/G_n$, and epimorphisms $i\circ g: F/F_n\lra C/C_n$
for all $n$. Since $C/C_n\cong F/F_n$ and $F/F_n$ is Hopfian, the
maps $i\circ g$ are isomorphisms from which it follows that
$F/F_n\cong G/G_n\cong C/C_n$. Therefore $G$ is almost parafree.
Since $i$ is surjective, $G/G_\om\cong C/C_\om$ and hence
$i_*:H_2(G/G_\om)\lra H_2(C/C_\om)$ is an isomorphism. Suppose that
$G$ had length $\om$ so $G_\om\cong G_{\om+1}$ and (from 1.1)
$H_2(G)\lra H_2(G/G_\om)$ is onto. It would follow that the image of
$[\g]$ under the map $H_2(C)\lra H_2(C/C_\om)$ pulls back to a class
$\a\in H_2(G)$ i.e., $i_*\circ\pi_*(\b)=\pi_*([\g])$. This is a
contradiction since $H_2(\p Y)\lra H_2(Y)$ is the zero map so
$H_2(G)\overset i_*\to\lra H_2(C)$ is also the zero map implying
$\pi_*([\g])=i_*\circ\pi_*(\b)=\pi_*\circ i_*(\b)=0$, but we have
previously observed that $\pi_*([\g])\neq0$ in $H_2(C/C_\om)$.
Therefore $G$ is a counterexample to $\ST$ and we have shown ``not
$\SE$'' implies ``not $\ST$''. \qed
\bpage
For a slightly smaller class of groups we can be much more specific
about which $3$-manifold groups are related to the Parafree
Conjecture.
\bpage
\proclaim{Theorem 1.7} For the class $\SS$ of all groups $G$ such
that rank$(H_1(G))=m>1$ and $G$ is normally generated by $m$
elements, the following are equivalent:
\roster
\item"$\SP$:" The Parafree Conjecture is true for groups in $\SS$.
\item"$\SA\SP$:" Every almost parafree group $G$ in $\SS$ has length
$\om$.
\item"$\SR$:" The fundamental group of the exterior of every ribbon
link in $S^3$ has length $\om$. Equivalently the group of the
$0$-surgery along a ribbon link has length $\om$.
\item"$\SE$:" Every finite $\BE$-group $E$ in $\SS$ has length $\om$.
\endroster
\endproclaim
\bpage
\sub{Proof} First note that $\SP\Rightarrow\SA\SP\Rightarrow\SE$
as in 1.5. It is also well known that, for any ribbon link group $G$,
the map $G\thra G/G_\om$ factors through an epimorphism to a finite
$\BE$-group $E$ of rank~$m$ [Hi; p\.~23--25]. It follows from
Stallings' theorem that $G/G_\om\cong E/E_\om$ and hence that
$\SE\Rightarrow\SR$. For if $E_\om\cong E_{\om+1}$ then
$H_2(E/E_\om)\cong0$, implying $H_2(G/G_\om)\cong0$ which implies
$G_\om\cong G_{\om+1}$ as above. Conversely, it is shown by
P\.~Bellis [Be; Thm\.~3.4] that, for any finitely-generated
$\BE$-group in $\SS$, there exists a ribbon link in $S^3$ with group
$G$ and an epimorphism $\phi: G\thra E$ which is an isomorphism on
abelianization. Therefore $G/G_\om\overset\phi\to\cong E/E_\om$ (by
Stallings' theorem) so $H_2(G/G_\om)\cong H_2(E/E_\om)\cong
E_\om/E_{\om+1}$. If the length of $G$ is $\om$ then $H_2(G)\lra
H_2(G/G_\om)$ is surjective. From the naturality of Stallings' exact
sequence with respect to $\phi:G\lra E$ one easily concludes that
$H_2(E/E_\om)\cong0$ and hence that $E$ has length $\om$. Therefore
$\SR\Rightarrow\SE$.
It remains only to show $\SE\Rightarrow\SP$. Suppose $G$ is a
finitely-generated parafree group of rank~$m$ lying in $\SS$,
generated by $\{x_i\mid 1\le i\le\ell\}$. Suppose $G$ is normally
generated by $\{y_i\mid1\le j\le m\}$ where
$x_i=\prod^{n_i}_{k=1}\eta_{ik}y^{\e_{ik}}_{ik}\eta^{-1}_{ik}$. Let
$E\equiv\$ and define $\phi:E\lra G$
by $\phi(x_i)=x_i$, $\phi(y_j)=y_j$. Note $E$ is normally generated
by $m$ elements. The proof is then completed as above in 1.5 and 1.6.
Since the longitudes of a ribbon link lie in $G_\d$, the group of
the $0$-surgery has the same lower central series quotients and
length as that of the ribbon link [\ \ ]. \qed
\bpage
For any integer $N$ there exists a finitely presented group $G$ in
$\SS$ such that $F/F_n\cong G/G_n$ for all $n\le N$ but
length$(G)>\om$ [CO2]. However we are not able to resolve the
Parafree Conjecture at this time.
Note that since every ribbon link is concordant to a trivial link,
every ribbon link group is homology cobordant (in the sense of 1.3)
to a free group. Hence if length were an invariant of general
homology cobordism then $\SR$ would be true, implying the Parafree
Conjecture for groups in $\SS$. In fact it would suffice that
length be invariant of homology cobordism for almost parafree
groups.
\bpage
\noindent{\bf The relationship between the lower central series of
finitely-presented groups}
\noindent{\bf and 4-dimensional topological surgery}
The lower central series has arisen repeatedly in the quest to
identify, in dimension~$4$, for which fundamental groups $G$ the
``basic machinery of manifold theory, the surgery and $5$-dimensional
$s$-cobordism theorems exists in the topological category [FT2;
p\.~1]'' [F1, F2, F3, FQ chapter~6, FT1]. This machinery is known to
exist when $G$ is ``good'' (subexponential growth) [FT1] and
``suspected to fail for non-abelian free groups and hence for a
``random'' group [FT2]''. The general set-up for surgery involves a
compact $4$-manifold $N$ with a degree~1 normal map $f:N\lra X$, and
the ``surgery kernel'' in $H_2(N,\BZ[\pi_1(X)])$. The lower central
series arises in the following way. To perform surgery on
$4$-manifolds one attempts to realize certain $2$-dimensional
homology classes in the kernel by {\it embedded\/} $2$-spheres.
Reducing to a relative version, one has a certain circle $\g$ which
bounds a surface in a $4$-manifold and one hopes to find an embedded
$2$-disk with boundary $\g$ in this relative homology class. All the
successful work in this area has involved approaching this problem
via certain infinite constructs involving either disks (Whitney
disks) or surfaces (gropes, half-gropes, capped gropes); depending on
the treatment ([CF], [FT1], [FQ]). For precise definitions the reader
is referred to [FT2] and [FQ]. However we can give the reader the
basic idea. If $[\g]\in G_2$ then $\g$ bounds a mapped in surface. If
$[\g]\in G_3$ then $\g$ bounds a mapped in surface $S_2$ with
standard symplectic basis $\{a_i, b_i\mid i=1,\dots,g\}$ such that
$a_i$ bounds a mapped in surface $S_{3,i}$ for each $i=1,\dots,g$.
Then $S_1\cup\bigcup^g_{i=1}S_{3,i}$ is called a half-grope of class
$3$ for $\g$. The loop $\g$ bounds a map of a half-grope of class $k$
if and only if $[\g]\in G_k$ [FT2; Lemma~2.1]. In infinite
constructions one often desires an $\infty$-grope. Note that if
$[\g]\in G_\om$, then $[\g]\in G_k$ for each $k$ but this merely
guarantees the existence of a grope of class $k$ for each $k$ and
these might not be related to each other. In fact, $\g$ bounds a map
of an $\infty$-grope if and only if $\g\in G_\d$ [FT2; Lemma~2.2]!
Therefore it would be highly desirable if the length of $G$ were at
most $\om$, for then $G_\om=G_\d$. In \S8 we exhibit the first
examples of compact $3$-manifolds whose groups have length greater
than omega.
We shall now make a specific connection between the Parafree
Conjecture and $4$-dimensional topological surgery. Although this
connection seems provocative, it may not be fundamental. But first,
in order to make the connection we must refer more precisely to the
major result of [FT2]. There $N$ is a compact topological
$4$-manifold, $Y$ is a $\pi_1$-null, codimension~0 submanifold such
that $H_1(Y)\cong H_1(\p Y)$. $N^+$ is the complex obtained from $N$
by attaching $3$-cells to $2$-spheres in $N$ representing a free
basis for the image of $H_2(Y)$. Let $C=\pi_1(Y)$.
\bpage
\noindent{\it Freedman-Teichner Theorem~1.1\/} [FT2; Thm\.~1.1]: Given
$(N,Y)$ as above, if $H_2(C)\lra H_2(C/C_n)$ is the zero map for all $n$ then
there exists a $4$-manifold $N'(\p N'=\p N^+)$ and a simple homotopy
equivalence $h:(N',\p N')\lra(N^+,\p N^+)$ i.e., a manifold structure
on $N^+$.
This theorem is then applied to successfully perform surgery on a
map $f:N\lra X$ with vanishing surgery kernel carried by such a $Y$
(see [FT2; Corollary~1.2]). The power of the theorem, as pointed out
by Freedman and Teichner, is that whereas previous results had relied
on finding a submanifold $Y$ whose second homology is ``spherical'',
that is $H_2(C)=0$, their theorem only requires sphericity modulo
each finite term of the lower central series.
One way in which their hypothesis could be satisfied is if, in fact,
$H_2(C)\lra H_2(C/C_\om)$ is the zero map, a priori stronger than
their condition that $H_2(C)\lra H_2(C/C_n)$ be zero for each finite
$n$. We shall show below that the first condition is in fact
equivalent to their condition if and only if the Parafree Conjecture
is true, and thus is stronger than their condition if and only if
there exists an almost parafree hyperbolic $3$-manifold group of
length greater than $\om$.
\proclaim{Theorem 1.8} Given $4$-manifolds $Y\subset N$ as above
$(\pi_1(Y)\lra\pi_1(N)$ is zero, $H_1(\p Y)\cong H_1(Y))$ such that
$(N,Y)$ satisfies the hypotheses of Freedman-Teichner's Theorem~1.1 in
that $H_2(C)\lra H_2(C/C_n)$ is zero for all finite $n$, but
$H_2(C)\lra H_2(C/C_\om)$ is not zero, there is a counterexample to
the Parafree Conjecture and there is an almost parafree closed
hyperbolic $3$-manifold group which has length greater than $\om$.
Conversely if there exists a counterexample to the Parafree
Conjecture, then there is a pair of $4$-manifolds $(N,Y)$ as above
which satisfies the hypotheses of Freedman-Teichner's Theorem~1.1
that $H_2(C)\lra H_2(C/C_n)$ is zero for all $n$, while
$H_2(C)\lra H_2(C/C_\om)$ is not zero.
\endproclaim
\sub{Proof of 1.8} The condition $H_1(\p Y)\cong H_1(Y))$ is quite
strong. It implies $H_2(Y)$ and $H_1(Y)$ are free abelian and that
$H_2(\p Y)\lra H_2(Y)$ is the zero map [FT2]. Thus there is a map
$F\overset f\to\lra C$ which induces isomorphisms on
abelianizations. Using Dwyers' extension of Stallings' theorem, the
fact that $H_2(C)\lra H_2(C/C_n)$ is zero for each $n$ implies $f$
induces isomorphisms $F/F_n\approx C/C_n$. Hence $C/C_\om$ is a
finitely generated parafree group and $H_2(C/C_\om)\neq0$ since the
map $H_2(C)\lra H_2(C/C_\om)$ was assumed to be non-zero. Thus
$C/C_\om$ is a counterexample to the Parafree Conjecture. By 1.5
``not $\SP$'' implies ``not $\ST$'' as desired.
Conversely if the Parafree Conjecture fails then ``not $\SP$''
implies ``not $\SE$'' as in 1.5. Looking back at the proof of ``not
$\SE$'' implies ``not $\ST$'' in 1.5, we see that we constructed a
compact $4$-manifold $Y$ with almost parafree fundamental group $C$
which satisfies the hypotheses of 1.8. This $Y$ may then be embedded
in $4$-manifolds $N$, such that $\pi_1(Y)\lra\pi_1(N)$ is zero, in
many inequivalent ways. \qed
\bpage
We remark that, while no example of 1.8 is currently known, one can
write down an example $(N,Y)$ which fails the Freedman-Teichner
criterion in that $H_2(C)\lra H_2(C/C_n)$ is non-zero for some $n$.
\bpage
\noindent{\bf The relationship between lower central series and
link concordance}
The relationship between the lower central series of
$G=\pi_1(S^3-L)$ for a link $L$ of $m$ circles in $S^3$ and the
concordance class of $L$ is well established. In the 1950's, Milnor,
building on work of Chen, defined his $\ov\mu$-invariants, a
sequence of increasingly discriminating invariants which inductively
detect whether or not the meridional map $F\lra G$ induces an
isomorphism $F/F_k\cong G/G_k$ for finite $k$. Milnor's invariants
are zero if and only if $G$ is almost parafree or, equivalently, the
longitudes of $L$ lie in $G_\om$. Since the exterior of a concordance
is a homology cobordism between the link exteriors, Stallings'
theorem showed that the quotients $G/G_k$ as well as Milnor's
invariants are invariant under concordance. It is known that {\it
fusions of boundary links\/} have vanishing $\ov\mu$-invariants yet
are not in general concordant to trivial links or even any boundary
link [C1, CO1]. Therefore efforts have turned to finding invariants
``beyond $G_\om$'' [L1, L2, O1, O2, LD, L3, C2]. For example it is
still unknown whether or not the longitudes of a null-concordant link
must lie in $G_\d$ as they do for the trivial link (see
property~$\SR$ above). All of this work is concerned with certain
``universal groups'' to which all link groups map and which,
therefore, can be used to define invariants for all links. The
invariants lie, loosely speaking, in $H_2$ and $H_3$ of these groups.
The groups are $\wh F$, the {\it algebraic closure of the free
group\/} $F$, as defined by J\.~Levine [L2] and P\.~Vogel [LD], and
$\ov F$, the {\it algebraic closure of $F$ in its nilpotent
completion\/} $\varprojlim(F/F_n)$, due to Levine [L1]. It was shown
by Levine that $\ov F\cong\wh F/\wh F_\om$ and that
$H_2(\wh F;\BZ)=0$. Thus, using Stallings' exact sequence,
$H_2(\ov F)=\wh F_\om/\wh F_{\om+1}$ so $H_2(\ov F)=0$ if and only
if the length of $\wh F$ is $\om$. $\wh F$ is not very well
understood but Levine showed that it is a direct limit of
higher-dimensional link groups (n.b\. these are always almost
parafree --- see $\SA\SP$ above) so the length of $\wh F$ is
related to the length of such groups and hence to the ``conjectures''
listed above. Moreover it follows from Levine's work that $\ov F$ is
a direct limit of finitely generated parafree groups and hence the
Parafree Conjecture for groups in $\SS$ directly implies $H_2(\ov
F)=0$ and length$(\wh F)=\om$. The ``Strong Parafree Conjecture'',
that a finitely generated parafree group has $H_2=0$ and is of
cohomological dimension at most $2$, implies $H_3(\ov F)=0$ and
hence that invariants associated to $\ov F$ by Levine and Orr always
vanish.
The material is organized as follows.
\widestnumber\item{S3--S5}\roster
\item"\S1" History, Main Results, Motivation
\item"\S2" Elementary Facts and Elementary Examples
\item"\S3--\S5" Lengths of Fuchsian Groups
\item"\S6" Lengths of Groups of Seifert Fibered $3$-Manifolds
\item"\S7" Every $3$-Manifold has the Same Lower
Central Series Quotients as a Hyperbolic $3$-Manifold
\item"\S8" $3$-Manifolds Whose Groups have Length at least $2\om$
\item"\S9" Length and Homology Cobordism
\item"\S10" Questions
\endroster
\vskip.7cm
\sub{\S2. Notation, Elementary Facts, Simple Examples of
Length of $\bold3$-Manifold Groups}
In this paper we use the convention $[a,b]=aba^{-1}b^{-1}$.
The following elementary results will be used throughout.
\proclaim{Proposition 2.1 {\rm(see [MKS; Thm\.~5.1, 5.3])}} Let $a$,
$b$, $c$ be elements of a group $G$. Let $n$ be a positive integer and
$\a$ an ordinal.
\roster
\item"a." If $a\in G_\a$ then $ab\equiv ba\bmod G_{\a+1}$
\item"b." If $[a,b]\in G_\a$ and $[a,c]\in G_\a$ then
$[a,bc]\equiv[a,b][a,c]$ modulo $G_{\a+1}$ and so
$[a,b]^n\equiv[a^n,b]\equiv[a,b^n]$ modulo $G_{\a+1}$.
\endroster
\endproclaim
\proclaim{Proposition 2.2} Let $A$, $B$ be groups with lengths
$\ell(A)$ and $\ell(B)$.
\roster
\item"a." If $f: A\lra B$ is an epimorphism then
$\ell(A)\ge\min\{\om,\ell(B)\}$.
\item"b." If $r: A\lra B$ is a retraction (i.e., there is a
homomorphism $i: B\lra A$ such that $r\circ i$ is the identity) then
$\ell(A)\ge\ell(B)$
\item"c." $(A\x B)_\a\cong A_\a\x B_\a$ and $\ell(A\x B)=
\max\{\ell(A),\ell(B)\}$.
\endroster
\endproclaim
\sub{Remark} b) is false for a general epimorphism since any group
is a quotient of a free group with length $\om$.
\sub{Proof of 2.2} a) Suppose not. Then $\ell(A)=n<\ell(B)$ so there
is some $b\in B_n-B_{n+1}$. One checks that the restriction $f:
A_n\lra B_n$ is onto (this fails if $n$ is not finite), so there is
some $a\in A_n$ such that $f(a)=b$. Since $\ell(A)=n$, $a\in
A_{n+1}$ implying $b\in B_{n+1}$, a contradiction. For b), if $r$ is
a retraction then $r: A_\a\lra B_\a$ is onto for any ordinal $\a$ so
the argument for a) applies. Part c) is clear. \qed
\bpage
The following ideas are important to our construction of elements in
$G_\d$ and $G_\om$.
\sub{Definition 2.3} Let $p$ be a positive integer. An element $g\in
G$ is called a {\it generalized periodic element of period\/} $p$,
or a {\it generalized $p$-element\/} if, for any positive integer
$n$, there is an integer $k=k(n)$ such that $g^{p^k}\in G_n$.
Equivalently, $g$ is of order a power of $p$ in each $G/G_n$ for $n$
finite.
\sub{Example} The group $K$ of the Klein bottle has a presentation
$\__$ and it follows that
$[u,[u,\dots,[u,t]\dots]]=t^{(-2)^n}$ and hence that $t$ is a
generalized $2$-element. The dihedral group $D^n$ of order $2n$ has
a presentation $\____$ and $t^{2^k}\in
D^n_{k+1}$. Thus $t$ is a generalized $2$-element.
\proclaim{Proposition 2.4} If $a$ and $b$ are generalized periodic
elements of $G$ of relatively prime orders $p$, $q$ then $[a,b]\in
G_\om$.
\endproclaim
\sub{Proof of 2.4} We proceed by induction. Suppose $[a,b]\in G_n$.
It suffices to show $[a,b]^{p^k}\in G_{n+1}$ and $[a,b]^{q^t}\in
G_{n+1}$ for some $k$, $t$. Choose $k$, $t$ such that $a^{p^k}\in
G_n$ and $b^{q^t}\in G_n$. By 2.1a, $[a,b]^{p^k}\equiv[a^{p^k},b]$
modulo $G_{n+1}$ and so $[a,b]^{p^k}\in G_{n+1}$. Similarly for
$[a,b]^{q^t}$. \qed
\bpage
\proclaim{Proposition 2.5} Let $a$ and $b$ be periodic elements
(torsion) of the group $G$ of relatively prime orders $p$ and $q$.
Then $[a,b]\in G_\d$. (It suffices that $a^{p^k}$ commutes with $b$
and $b^{q^t}$ commutes with $a$).
\endproclaim
\sub{Proof of 2.5} Suppose $a^{p^k}$ commutes with $b$ and $b^{q^t}$
commutes with $a$. By induction suppose $[a,b]\in G_\a$. Then merely
apply the argument of the proof of 2.4. \qed
\bpage
A few words are in order about the relationship between nilpotence,
transfinite nilpotence and length. If $G_\a=\{e\}$ then clearly the
length of $G$ is at most $\a$. Thus a nilpotent group is of finite
length equal to its nilpotency class and an $\om$-nilpotent group
$(G_\om=\{e\})$ has length at most $\om$ and, in fact, {\it equal\/}
to $\om$ unless it is nilpotent. We shall often use {\it residually
nilpotent\/} (for any $g\in G$, there is a normal subgroup $N$ of
$G$ such that $g\notin N$ and $G/N$ is nilpotent) instead of
{\it$\om$-nilpotent\/} since they are equivalent. But the reader
must be careful not to conclude anything in the other direction, for
there are many groups which are of length~1 (perfect groups) or
length~2 (see below), for example, which are not transfinitely
nilpotent.
\proclaim{Proposition 2.6} If the abelianization of $G\neq e$ is
cyclic then $G$ has length~2.
\endproclaim
\sub{Proof} The surjection $h: G\lra G/G_2$ satisfies the hypothesis
of 1.1 so $G/G_n\cong(G/G_2)/(G/G_2)_n$. Since $G/G_2$ is abelian,
the latter quotient is $G/G_2$ if $n>1$. Therefore $G_n=G_2$. \qed
\bpage
\proclaim{Proposition 2.7} Let $p$ be a prime. The groups $\BZ*\BZ$,
$\BZ*\BZ_{p^k}$, $\BZ_{p^k}*\BZ_{p^m}$ are of length $\om$ and
residually nilpotent.
\endproclaim
\sub{Proof of 2.7} The group $\BZ_{p^k}*\BZ_{p^m}$ is not nilpotent
since a nilpotent group which is generated by a finite number of
elements of finite order must be itself finite which these groups
are clearly not. It follows that $\BZ*\BZ$ and $\BZ*\BZ_{p^k}$ fail
to be nilpotent. It is a theorem of A.I\.~Malcev that these groups
are residually nilpotent ([Mal], see also [Lic]). Alternatively, a
free product of residually finite $p$-groups is residually a finite
$p$-group [MKS; p\.~417]. Since finite $p$-groups are nilpotent,
these groups are residually nilpotent. Hence in each case
$G_\om=\{e\}$ and so length$(G)\le\om$. If the length of $G$ were
$n<\om$ then $G_n=G_\om=\{e\}$ which is a contradiction. \qed
\bpage
The reader is warned that $\BZ_2*\BZ_3$ and $\BZ*\BZ_6$ are not even
transfinitely nilpotent since commutators of elements of order~2
with elements of order~3 lie in $G_\d$ by 2.5. By 2.6,
$\ell(\BZ_2*\BZ_3)=2$ while $\ell(\BZ*\BZ_6)\ge\om$ since it maps
onto $\BZ*\BZ_2$.
\proclaim{Proposition 2.8} The dihedral group $D$ of order~$2^{k+1}$
is $(k+1)$-nilpotent and has length $k+1$. The ``infinite dihedral
group'' $\____$ has length $\om$ and is
residually nilpotent.
\endproclaim
\sub{Proof of 2.8} If $D=\____$ then
there is an split exact sequence $1\lra\BZ_{2^k}\lra
D\lra\BZ_2\lra1$. It can be shown directly (by induction) that
$D_{m+1}\cong2^m\BZ_{2^k}$ so $D_{k+1}\cong\{e\}$ and
$D_k\neq\{e\}$. The infinite dihedral group maps onto any finite
dihedral group so has length at least $\om$ by 2.2a. But the
infinite dihedral group is $\BZ_2*\BZ_2$ hence is residually nilpotent
by 2.7. Therefore it has length $\om$. \qed
\bpage
Now we list examples of $3$-manifold groups of length $1$, $2$, $n$
and $\om$ to give the reader a feeling for the situation.
\roster
\item"(2.10)" A group has length $1$ if and only if it is perfect so
the group of any homology $3$-sphere has length~$1$.
\item"(2.11)" If $G$ is a non-trivial abelian group then $G_2=\{e\}$
so $G$ has length~$2$. Therefore $3$-manifolds with abelian group
such as $S^1\x S^1\x S^1$ have length~$2$. There are many more
examples with length~$2$ however. According to 2.6, if $H_1(M)$ is
cyclic then $\pi_1(M)$ has length~$2$. Examples are knot complements
and their $0$-framed surgeries
\item"(2.12)" To find closed $3$-manifolds with any prescribed finite
length it is convenient to look among those with finite, nilpotent
fundamental groups [T]. For example the generalized quaternion group
$Q^n$ of order $4n$, presented by $\____$ is the group of the Euler class $n$
circle bundle over $\BR P(2)$ with orientable total space. Killing
$u^2$ gives the dihedral group of order $2n$. When $n=2^k$ the
latter is $(k+1)$-nilpotent and has length $k+1$ (see 2.8). Since
$Q^n$ is a $\BZ_2$ central extension of $D^n$, it is easy to see
that $Q^{2^k}$ is $(k+2)$-nilpotent and has length $(k+2)$.
Therefore, for any positive integer $k$ there is a closed,
orientable $3$-manifold with finite fundamental group of length~$k$.
\item"(2.13)" The circle bundle over a torus with Euler class 1 has
the Heisenberg group, $F/F_3$, as its group and hence has length~$3$.
The Euler class $n$ bundles also have nilpotent groups of length~$3$
($2$ if $n=0$).
\item"(2.14)" The length of a non-abelian free group is $\om$ by
[Mal], so if $M=\#^k_{i=1}S^1\x S^2$ for $k\ge2$ then the length of
$\pi_1(M)$ is $\om$. Moreover if $M=M_1\#M_2$ were
$H^1(M_i;\BZ_p)\neq0$ for $i=1$, $2$ then $\pi_1(M)$ maps onto
$\BZ_p*\BZ_p$ and hence has length at least $\om$ by 2.2a and 2.7.
This idea can be formalized as follows.
\endroster
Suppose $M$ is a closed, connected, $3$-manifold and $M=\#M_i$ its
prime decomposition. Let $\wh M$ be obtained by eliminating all $M_i$
which are homology spheres. Then there is a degree one map
$M\lra\wh M$ which induces an isomorphism on $H_1$ so, by Stallings'
theorem we have:
\proclaim{Proposition 2.15} $\pi_1(M)$ and $\pi_1(\wh M)$ have the
same length and isomorphic lower central series quotients.
\endproclaim
Similarly, let $\wh M_p$ be obtained by eliminating all $M_i$ which
are $\BZ_p$-homology spheres. We say $M$ is {\it$\BZ_p$-homology
prime\/} if $\wh M_p$ is prime.
\proclaim{Proposition 2.16} If $M$ is {\it not\/} $\BZ_p$-homology
prime then $\pi_1(M)$ maps onto $\BZ_p*\BZ_p$ so has length at
least $\om$.
\endproclaim
In \S6 we observe that almost all Seifert fibered $3$-manifold
groups have length at least $\om$. Together with 2.16, we begin to
see that it seems clear that ``most'' $3$-manifold groups will have
length at least $\om$, but we cannot formalize this.
\vskip.7cm
\sub{\S3. Surface Groups}
Let $X$ be a compact, connected surface and $G$ its fundamental
group. In this chapter we see that, if $\chi(X)<0$, then $G$ has
length $\om$ and is residually nilpotent. We also give partial results
about when $G$ is residually a finite $p$-group and prove a key
proposition about finite extensions of such groups. These facts will
be crucial in our analysis of the lengths of Fuchsian groups and the
groups of Seifert-fibered $3$-manifolds.
\proclaim{Proposition 3.1} If $H\tril G$ where $H$ is residually a
finite $p$-group ($p$ prime), and $G/H$ is a finite $p$-group then
$G$ is residually a finite $p$-group.
\endproclaim
\sub{Proof} It suffices to find, for an arbitrary non-trivial
$h\in H$, a normal subgroup $N$ of $G$ such that $h\notin N$ and
$G/N$ is a $p$-group. Since $H$ is residually a finite $p$-group,
there is $M\tril H$ of index $p^n$ which does not contain $h$. If
$\{g_i\mid i=1,\dots,p^m\}$ are coset representatives for $H$ in $G$
then let $N=\bigcap^{p^m}_{i=1}M_i$ where $M_i=g_iMg^{-1}_i$. Then
$N$ is normal in $G$ and $H$. Since $h\notin N$, it remains to show
that $G/N$ has order $p^k$. Since the order of $G/H$ is a power of
$p$ it suffices to show that the order of $H/N$ is a power of $p$.
Consider the projections $\pi_i:H/N\lra H/M_i$. Noting that $M_i$ is
of index $p^n$ for all $i$, if $[x]\in H/N$ then $x^{p^n}\in M_i$ so
$x^{p^n}\in N$. Therefore $H/N$ is a $p$-group. \qed
\bpage
We can now combine 3.1 with results of G\. Baumslag to show the
following.
\proclaim{Proposition 3.2} All surface groups are residually finite
$2$-groups and hence residually nilpotent. If $\chi(X)$ is even and
$X$ is not the Klein bottle, then $G$ is residually a finite $p$-group
for any odd prime $p$.
\endproclaim
\sub{Proof} Baumslag has shown that closed surface groups of even
Euler characteristic (except the Klein bottle) are residually free
[Ba1]. Since free groups are residually finite $p$-groups for any $p$
[MKS; p\.~417], the above groups are residually finite $p$-groups for
all primes $p$. Any closed non-orientable surface group has an index
$2$-subgroup which is the group of an orientable surface so,
applying 3.1, such groups are residually finite $2$-groups. Since
finite $p$-groups are nilpotent, all surface groups are residually
nilpotent. \qed
\bpage
\proclaim{Theorem 3.3} Let $X$ be a compact, connected surface and
$G$ its fundamental group. If $\chi(X)<0$ then the length of $G$ is
$\om$. If $X$ is the Klein bottle then the length of $G$ is $\om$.
Otherwise $G$ is abelian and its length is $2$ (or $1$ if $G$ is
trivial).
\endproclaim
\sub{Proof} In the cases where $G$ is not abelian it is easy to see
that $G$ maps onto $\BZ_2*\BZ_2$, the infinite dihedral group, and
hence has length at least $\om$ by 2.2a and 2.7. By 3.2, $G$ has
length at most $\om$. \qed
\bpage
\sub{3.4 Remarks} If $G$ is a group containing a non-trivial
generalized $q$-element $t$, then $G$ is not residually a finite
$p$-group if $(p,q)=1$; for if $\phi:G\to P$ is any epimorphism and
$P$ is a finite $p$-group then $\phi$ factors through $G/G_n$ for
some finite $n$ so $\phi(t)$ is of order $q^k$ and hence is trivial
in $P$. It follows {\it a fortiori\/} that such a group is not
residually free. Therefore the group of the Klein bottle is not
residually a finite $p$ group for odd $p$ since the presentation
$\____$ reveals that $t$ is a non-trivial
generalized $2$-element. The groups of the other surfaces of even
Euler characteristic contain no generalized $p$-elements for any
prime $p$ since they are residually free. The group of $\BR
P(2)\#\BR P(2)\#\BR P(2)$ is not residually free. We do not know if
the odd Euler characteristic surface groups are residually finite
$p$-groups for odd $p$ and this ignorance obstructs the extension of
our results to all Seifert fibered spaces. We suspect that these
surfaces of odd genus greater than $1$ do not have generalized
$p$-elements for any $p$.
\vskip.7cm
\sub{\S4. Free Products of Finite Cyclic Groups}
Any free product $A$ of cyclic groups has length at most $\om$. In
this section we prove this in the case that each factor is {\it
finite\/} cyclic. In the case that the order of every factor is a
power of a single prime $p$, it follows from early work of Malcev
that $A$ is residually nilpotent so $A_\om\cong\{e\}$ [Mal]. But in
all other cases $A$ will fail to be transfinitely nilpotent, so
showing $A_\om\cong A_{\om+1}$ is more difficult.
\sub{Definition} If $G$ is a group, $T(G)$ is the subgroup generated
by the set of elements of the form $[a,b]$ where $a$ and $b$ are of
relatively prime (finite) orders in $G$. Each such element is called
a {\it$T$-element\/}.
\bpage
Note that $T(G)$ is a characteristic subgroup of $G$ and, by 2.5,
is contained in $G_\d$, the intersection of the terms of the lower
central series. Therefore we have $T(G)\subset G_\d\subset G_\om$ in
general.
\sub{Definition} $G$ has Property $T$ if $G_\om=T(G)$.
\bpage
Hence a group with Property $T$ is of length at most $\om$. A
residually nilpotent group has Property $T$ (trivially). A
torsion-free group cannot have Property $T$ unless it is residually
nilpotent. Thus, for example, the fundamental group of the exterior
of a non-trivial knot or non-trivial ribbon link does not have
Property~$T$.
\proclaim{Theorem 4.1} Suppose $A\neq\{e\}$ is a free product of
finite cyclic groups. Then $A$ has Property~$T$. $A$ has length
$\om$ unless the orders of the free factors are pairwise coprime in
which case $A$ has length $2$. Moreover if
$A=\$ $n_i\ge2$, and
$\{p_1,\dots,p_\ell\}$ is the set of distinct primes in the prime
decomposition of $n_1n_2\dots n_m$ and
$n_i=p^{r_{i1}}_1\dots p^{r_{ij}}_j\dots p^{r_{i\ell}}_\ell$,
$1\le i\le m$, $1\le j\le\ell$, then $A_\om$ is normally generated
by $K=\{[c^{k_{ij}}_i,c^{k_{tv}}_t]\mid j\neq v$,
$k_{ij}=n_i/p^{r_{ij}}_j$, $k_{tv}=n_t/p^{r_{tv}}_v\}$ and
$A/A_\om\ovf\x^\ell_{j=1}B_{p_j}$ where
$B_{p_j}=\BZ_{p^{r_{1j}}_j}*\BZ_{p^{r_{2j}}_j}*\dots*\BZ_{p^{r_{mj}}_j}$,
and $\BZ_{p^{r_{ij}}_j}$ is generated by $f(c^{k_{ij}}_i)$,
$k_{ij}=n_i/p^{r_{ij}}_j$.
\endproclaim
\sub{Proof} It is easily seen that we may assume that $A$ is a
{\it finite\/} free product since any element of $A$ lies in such a
finite free product which is a retract of $A$ (see 2.2b). Therefore
suppose $A\cong\$ where $n_i\ge2$, and
$n_i=\prod^\ell_{j=1}p^{r_{ij}}_j$ for $\{p_1,\dots,p_\ell\}$
distinct primes. Note that for each $i$ there is an obvious
epimorphism
$\phi_i:\BZ_{p^{r_{i1}}_1}*\dots*\BZ_{p^{r_{i\ell}}_\ell}\lra
\BZ_{n_i}$ (given by abelianization) and that this map is an
isomorphism on $H_1$. Therefore, letting $\wt
A=\ast^m_{i=1}*^\ell_{j=1}\BZ_{p^{r_{ij}}_j}$ there is an obvious
epimorphism $\phi:\wt A\lra A$ which induces an isomorphism on $H_1$
and an epimorphism on $H_2$ (since $H_2(A)=0$) and hence induces an
isomorphism modulo any term of the lower-central-series. If $\wt A$
is generated by $\a_{ij}$, $1\le i\le m$, $i\le j\le\ell$ then
$\phi(\a_{ij})=c^{k_{ij}}_i$ where $k_{ij}=n_i/p^{r_{ij}}_j$. We can
express $\wt A\cong*^\ell_{j=1}B_j$ where
$B_j\cong*^m_{i=1}\BZ_{p^{r_{ij}}_j}$ generated by $\{\a_{ij}\}$.
Then the kernel of the epimorphism $\pi:\wt
A\lra\times^\ell_{j=1}B_j=B$ is normally generated by the set of
$T$-elements $\{[\a_{ij},\a_{tv}]\mid j\neq v\}$. Examining $\phi_i$
above, one sees that the kernel of $\phi$ is normally generated by
$\{[\a_{ij},\a_{iv}]\mid j\neq v\}$ so that $\pi$ factors through
$\phi$ inducing an epimorphism $f:A\thra\times^\ell_{j=1}B_j$. Since
the kernel of $\pi$ lies in $\wt A_\d$, $\pi$ induces an isomorphism
$\pi:\wt A/\wt A_\d\lra B/B_\d$. Hence $f$ induces an isomorphism
$f:A/A_\d\lra B/B_\d$. But a free product of residually finite
$p$-groups is a residually finite $p$-group hence $B_j$
is residually nilpotent [MKS p\.~417]. It follows that $B$ is
residually nilpotent. Thus $f:A/A_\d\lra B$ is an isomorphism,
$A_\om\cong A_\d$, and $f:A/A_\om\lra\times^\ell_{j=1}B_j$ is an
isomorphism. The kernel of $f$ is normally generated by
$\{[\phi(\a_{ij}),\phi(\a_{tv})]\mid j\neq v\}$ as desired. Since
these are $T$-elements, $A_\om\subset T(A)$ and hence $A_\om=T(A)$.
Hence $A$ has Property~$T$.
Since $A$ has Property~$T$ its length is at most $\om$. If the
orders of free factors are pairwise relatively prime then map from
$A$ to its abelianization is an isomorphism on $H_1$ and an
epimorphism on $H_2$. By Stallings' theorem, then $A$ has the same
length as an abelian group, that is, $2$. Otherwise there is an
epimorphism from $A$ to $\BZ_p*\BZ_p$ for some prime $p$ so
$\ell(A)\ge\ell(\BZ_p*\BZ_p)=\om$ by 2.2a. Thus the length of $A$ is
$\om$. \qed
\vskip1cm
\sub{\S5. Fuchsian and Planar Discontinuous Groups}
In this section we discuss the length of planar discontinuous groups
in preparation for studying the groups of Seifert fibered
$3$-manifold groups.
\newpage
In this paper, for simplicity, a {\it Fuchsian group\/} will be
defined as a group which admits a presentation of one of the
following two forms:
$$
G = \Bigl
\tag"{\bf5.1}"
$$
where $n_i\ge2$.
$$
G = \Bigl\tag"{\bf5.2}"
$$
where $n_i\ge2$ and $g\ge1$.
This is a serious abuse of language although not without precedent
in the literature (e.g\. [JS]). If one sets
$\mu(G)=2-2g-k-\sum^m_{j=1}\(1-\f1{n_i}\)$ in 5.1 (in case of 5.2
replace $2-2g$ by $2-g$). Then it is known that if $\mu(G)<0$, $G$
can be realized as a group of orientation-preserving motions on the
hyperbolic plane, i.e., a true {\it Fuchsian group\/}. If $\mu(G)=0$
then $G$ can be realized as a group of motions on the Euclidean
plane, i.e., a {\it crystallographic group\/}. If $\mu(G)>0$ then $G$
can be realized as a group of motions on the $2$-sphere [ZVC; p\.~122,
p\.~155].
The quotient of $G$ by $\{c_i\}$ is the group of a compact surface
$X$ of genus~$g$ with $k$ boundary components which is orientable in
5.1 and non-orientable in 5.2. This is called the {\it surface
underlying\/} $G$ and adjectives such as {\it closed, orientable,
non-empty boundary, genus\/} and {\it Euler characteristic\/} will be
applied to $G$ but refer to its underlying surface. It will also be
convenient to introduce the {\it signature\/} of $G$,
$(\e,g,n_1,\dots,n_m)$ where $\e=\pm1$ according as $G$ is
orientable or not. Note that free products of cyclic groups
$(k\neq0)$ and surface groups $(m=0)$ are examples of Fuchsian groups
so our work of sections~3 and~4 will not be in vain.
As is standard in treatments of Fuchsian groups, it will be convenient
to define a specific ``capped surface'' $Y$ of which $G$ is the
fundamental group. Given a presentation of form 5.1 or 5.2, let $X$
denote the underlying surface (with boundary) and let $X_0$ be $X$
minus the disjoint union of $m$ open $2$-disks. Now form $Y$ from
$X_0$ by adjoining $m$ $2$-disks to these new $m$ components of
$\p X_0$ via maps of degree $n_i$. Then $\pi_1(Y)\cong G$ and $Y$
shall be called the {\it capped-surface\/} of $G$ (actually
corresponding to the particular presentation). Moreover, borrowing
from the language of orbifolds, we shall use {\it cone point of\/}
$G$ to refer to one of the $c_i$ and the {\it order of the cone
point\/} shall be the corresponding $n_i$.
\proclaim{Theorem 5.3} Let $G$ be a Fuchsian group whose cone points
are each of order a power of a fixed prime $p$.
\roster
\item"i)" if $p=2$, $G$ is a residually finite $2$-group, hence
residually nilpotent, Hopfian and of length at most $\om$.
\item"ii)" if $p\neq2$ and we assume the underlying surface of $G$
has either even Euler characteristic or non-empty boundary, then $G$
is residually nilpotent, Hopfian, of length at most $\om$ and, if
$G$ is not the group of the Klein bottle, is residually a finite
$p$-group.
\endroster
\endproclaim
\sub{5.4 Remarks} Such Fuchsian groups almost always have length at
least $\om$ and hence, in the cases covered by 5.3, almost always
have length precisely $\om$. For, if $g>1$, $G$ maps onto a surface
group of negative Euler characteristic (see 3.3). If $k\ge1$ then $G$
maps onto some $\BZ_p*\BZ_p$ unless $\e=\pm1$, $g=0$, $k=1$ and the
$n_i$ are pairwise coprime; or $\e=-1$, $k=1$, $g=1$ and $m=0$. In the
latter cases $G$ has length $2$. This leaves only the cases $k=0$,
$g\le1$ untreated. If $g=1$, $k=0$ then $G$ maps onto
$H*\$ where $H=\BZ\x\BZ$ if
$\e=+1$ and $H=\BZ_2$ if $\e=-1$. Therefore, if $\e=+1$ $G$ will map
onto $\BZ*\BZ_p$ (and hence have length at least $\om$) unless the
group $\$ is perfect. If $\e=-1$ the
same holds unless this group has $H^1(\ ;\BZ_2)=0$. We shall not
pursue the scattered remaining cases.
\sub{Proof of 5.3} If $G$ has non-empty boundary then $G$ is
isomorphic to a free product of copies of $\BZ$ and $\BZ_{p^{n_i}}$.
Since a free product of residually finite $p$-groups is again a
residually finite $p$-group, we are done in this case [MKS; p\.~417].
We now assume that we are in the closed case. We may also assume
that $G$ is not the group of the Klein bottle since 3.2 handles that
case. The proof is accomplished by induction using 3.1 and the
following Lemma which, in the orientable, positive genus case can be
found in [EEK; Theorem~3.1].
\proclaim{Lemma 5.5 {\rm(compare [H; p\.~119], [EEK; Theorem~3.1])}}
Let $G$ be a closed Fuchsian group with signature
$(\e,g,p^{n_1},\dots,p^{n_m})$, for some prime $p$, which we suppose
is not $(-1,2)$, i.e., $G$ is not the Klein bottle group. Then there
is a finite filtration, $G_k\tril G_{k-1}\tril\dots\tril G_1=G$ where
$G_i/G_{i+1}$ is a finite abelian $p$-group and $G_k$ is the group of
a closed surface which, in case $p$ is odd, has signature
$(\e,g')$ which is not $(-1,2)$, and where $g'\equiv g\bmod2$. (It
follows that $G$ has a normal subgroup $G'$ such that $G/G'$ is a
finite $p$-group and $G'$ is a surface group as above).
\endproclaim
First let us see that 5.5 implies 5.3 for the closed case. If $p=2$,
3.1, 3.2 and 5.5 imply 5.3 directly, since all surface groups are
residually finite $2$-groups.
If $p$ is odd and the Euler characteristic of the underlying surface
of $G$ is even then $g$ is even as is $g_k$. By 3.2, $G_k$ is a
residually finite $p$-group and so, using 3.1 we see that $G$ is a
residually finite $p$-group. Therefore the proof of 5.5 will
complete the proof of 5.3.
\sub{Proof of 5.5} We loosely follow [H; p\.~119--120]. Given a
capped surface $Y$ as defined above we define the complexity of $Y$,
$c(Y)$, to be the maximum of the orders of the cone points. We begin
with the capped surface of $G$ with complexity
$(p^{n_1},\dots,p^{n_m})$. If $m=0$ then $G$ is not the Klein bottle
and so $G=G_k$ satisfies the conclusion of 5.5. Otherwise, given such
a $Y$, we shall find a regular covering space $\wt Y$ such that $\wt
Y$ is a capped surface whose cone points are each of order a power
of $p$, $c(\wt Y)k$,
$\phi(a_i)=\phi(b_i)=0$. In both orientable and non-orientable cases
this $\phi$ suffices.
Now suppose $G$ is orientable and there is only one cap of maximal
order $p^{n_1}$. If $g>0$ we can do a preliminary $p$-fold cover of
$Y$ defined by $\phi(a_1)=1$, $\phi(a_i)=\phi(c_j)=\phi(b_k)=0$ if
$i\neq1$. The resulting $\wt Y$ will have $p$ caps of maximal order
and $c(\wt Y)=c(Y)$ and we can then apply the argument above. If
$g=0$ and $m\le2$ then $G$ is cyclic of order a power of $p$ and we
can take $\wt Y$ to be the universal cover of $Y$. If $g=0$ and
$m\ge3$ then map $\phi:G\thra\BZ_{p^{n_2}}$ by $\phi(c_1)=1$,
$\phi(c_2)=-1$, $\phi(c_i)=0$ if $i>2$. This completes all cases when
$X$ is orientable.
Now suppose $X$ is non-orientable with one cap of maximal order
$p^{n_1}$. If $p$ is odd define $\phi:G\thra\BZ_{p^{n_1}}$ by
$\phi(c_1)=1$, $\phi(c_i)=0$ if $i>1$, $\phi(2a_1)=1$ and
$\phi(a_i)=0$ if $i>1$. In case $p$ is $2$, we can do a preliminary
$2$-fold cover $\phi:G\lra\BZ_2$ by $\phi(a_1)=1$, $\phi$ (all
others)$=0$ which results in $\wt Y$ with $2$ caps of maximum order
without increasing complexity and reduces to an earlier case.
The last remark of 5.5 may be shown as follows. The group $G_k$ may
not be normal in $G_{k-2}$ but it is normal in $G_{k-1}$. Redefine
$G_k$ to be the intersection of its conjugacy classes in $G_{k-2}$.
This new $G_k$ is normal in $G_{k-1}$ {\it and\/} $G_{k-2}$, and, by
the proof of 3.1 is of index $p^n$ in $G_{k-2}$ for some $n$. Now
ignore $G_{k-1}$ and consider $G_k\tril G_{k-2}\tril G_{k-3}$ and
proceed as above to shrink $G_k$ to get it normal in $G_{k-3}$.
Eventually we shrink $G_k$ to $G'$ which is normal in $G_1=G$ and of
index a power of $p$. Finally note that we have replaced the
filtration by one where $G_k$ is normal in $G_1$ but the general
remarks above still hold to restrict the topological type of the
surface of which $G_k$ is the fundamental group.
\qed
\bpage
\proclaim{Theorem 5.6} Let $G$ be a Fuchsian group whose underlying
surface is not $S^2$ and either has even Euler characteristic or has
non-empty boundary (excludes precisely a connected sum of an odd
number of projective planes). Then $G$ has Property~$T$, hence
length at most $\om$.
\endproclaim
\sub{Proof of 5.6} We assume the notation of 5.1, 5.2 and our
earlier discussion of capped surfaces. Let
$A=\$ as in 4.1. Let $N$ be the
subgroup of $G$ normally generated by $\{c_i\}$. Let $S$ be
$\pi_1(X)$ where $X$ is the surface underlying $G$. \qed
\bpage
\proclaim{Lemma 5.7} There is a monomorphism $*_{s\in
S}A\overset\phi\to\lra G$ whose image is $N$.
\endproclaim
\sub{Proof of 5.7} Since $X\neq S^2$, $X\neq\BR P^2$, its universal
cover $\wt X$ is a contractible surface with deck translations $S$
and contractible fundamental domain $P$. We shall assume $S$ has a
presentation obtained from 5.1 or 5.2 by setting $c_i=0$ so that we
are implicitly choosing lifts of the generators of $S$ to generators
of $G$. $S$ acts on $\wt X$, tiling $\wt X$ by copies of $P$. Delete
from $\wt X$ the orbit of $m$ disjoint open $2$-disks in $P$ so that
the result is the regular $S$-cover of $X_0$. Then in each translate
of $P$, adjoin to the $i^\supth$ boundary circle a set of $2$-disks
$\{\wt E^s_i\mid s\in S\}$ by maps of degree $n_i$. The resulting
space $\wt Y$ has on it a free, properly discontinuous action of $S$
so $p:\wt Y\lra\wt Y/S\equiv Y$ is a regular cover. Clearly
$N=p_*(\pi_1(\wt Y))$. Moreover $\pi_1(\wt Y)\cong*_{s\in S}A$ in a
natural way. Under this identification, letting $\phi=p_*$, we
complete the proof of 5.7. \qed
\bpage
Consider the exact sequence $1\lra\SA\overset\phi\to\lra
G\overset\pi\to\lra S\lra1$ where $\SA\equiv*_{s\in S}A$. Recall
from 4.1 that $A_\om$ is normally generated by the set
$K=\{[c^{k_{ij}}_i, c^{k_{tv}}_t]\mid j\neq v$,
$k_{ij}=n_i/p^{r_{ij}}_j$, $k_{tv}=n_t/p^{r_{tv}}_v\}$ where
$n_i=\prod^\ell_{j=1}p^{r_{ij}}_j$, $1\le i\le m$. Let $K_s$ be the
corresponding set in the $s^\supth$-copy of $A$ in $\SA$ and let
$K^*$ be the normal closure of $\cup K_s$ in $\SA$. Clearly
$\SA/K^*\ovf*_{s\in S}(B_{p_1}\x\dots\x B_{p_\ell})\equiv\SB$ where
$B_{p_j}$ is as in 4.1. Now we shall force all copies of $B_{p_j}$ to
commute with copies of $B_{p_v}$, $j\neq v$, if we kill all elements
of the form $[b^{s_1}_{p_j},b^{s_2}_{p_v}]$ where $j\neq v$, $s_1\neq
s_2$ and $b^{s_1}_{p_j}$ denotes a generator of one of the free
factors of the $s_1$ copy of $B_{p_j}$ in $\SB$. Recall from 4.1 that
these factors are generated by $\{f((c^{k_{ij}}_i)^{s_1})\mid1\le
i\le m$, $k_{ij}=n_i/p^{r_{ij}}_j\}$. Therefore let
$L=\{[(c^{k_{ij}}_i)^{s_1}, (c^{k_{tv}}_t)^{s_2}]\mid j\neq v$,
$s_1\neq s_2\}$ and $L^*$ be the normal closure of $L$ in $\SA$.
Hence $\SA/L^*K^*\ovf\SB/f(L^*)\cong\x^\ell_{j=1}M_j$ where
$M_j\equiv*_{s\in S}B_{p_j}$. Also observe that $\phi(\cup K_s\cup
L)\sbq G$ is precisely the set of elements of the form
$[s_1c^{k_{ij}}_is^{-1}_1, s_2c^{k_{tv}}_ts^{-1}_2]$ where $j\neq
v$, $k_{ij}=n_i/p^{r_{ij}}_j$, $k_{tv}=n_t/p^{r_{tv}}_v$. Let $\ov
N=\phi(L^*K^*)$. Since $L^*K^*$ is normal in $\SA$ and since
$\cup_sK_s\cup L$ is ``$S$-invariant'', $\ov N$ is normal in $G$.
Since the orders of $c^{k_{ij}}_i$ and $c^{k_{tv}}_t$ are
$p^{r_{ij}}_j$ and $p^{r_{tv}}_v$ respectively, and since $j\neq v$,
$\ov N\subset T(G)$. Let $\ov G=G/\ov N$. It suffices to show
$\ov G_\om\cong\{e\}$ since this will imply that $G_\om\cong T(G)$
and hence that $G$ has Property~$T$.
Note that we have an exact sequence
$1\lra\bigoplus^\ell_{i=1}M_i\overset\phi\to\lra\ov
G\overset\pi\to\lra S\lra1$. Suppose $g\in\ov G_\om$. Since $S$ is
residually nilpotent by 3.2, $g=\phi(x_1,\dots,x_\ell)$. It clearly
suffices to show each $x_i$ is zero. By symmetry it suffices to
consider $x_1$. Consider the projection
$\Psi_1:\x^\ell_{i=1}M_i\lra M_1$, and the commutative diagram below:
$$
\define\toparrow{@>\pretend\phi\haswidth{\text{Clifford}}>>}
\define\piarrow{@>\pretend\pi\haswidth{\text{Clifford}}>>}
\define\plarrow{@>\pretend\ \haswidth{\text{Clifford}}>>}
\split
1\plarrow\x^\ell_{i=1} &M_i\toparrow\ov G\piarrow S\plarrow1\\
\Psi_1 &\Big\downarrow\hskip47pt\Big\downarrow\Psi\hskip33pt\Big\|\\
1\ \ \plarrow\ \ &M_1\plarrow\ov G_1\plarrow S\plarrow1.
\endsplit
$$
One verifies that $\phi(\kernel\Psi_1)$ is normal in $\ov G$ since
$M_1$ is an ``$S$-invariant'' subgroup. It will then suffice to show
$(\ov G_1)_\om=\{e\}$ since then $\Psi(g)=0$ implies
$\Psi_1(x_1,\dots,x_\ell)=x_1=0$. For this purpose it is necessary to
identify $\ov G_1$.
We claim that $\ov G_1\cong\ov G/\<\SS\>$ where
$\SS=\{c^{p^{r_{i1}}_1}_i\mid i=1,\dots,m\}$. The kernel of $\Psi_1$
is normally generated in $\x^\ell_{i=1}M_j$ by $f((c^{k_{ij}}_i)^s)$
where $s\in S$ and $j\neq1$. Therefore the kernel of $\Psi$ in
$\ov G$ is normally generated by $\{c^{k_{ij}}_i\mid 1\le i\le m$,
$j\neq1\}$. Since $k_{ij}=n_i/p^{r_{ij}}_j$ and $j\neq1$,
$p^{r_{i1}}_1$ divides the exponent of $c^{k_{ij}}_i$. Thus kernel
$\Psi_1\subset\__~~$. On the other hand, for fixed $i$,
gcd$\{k_{ij}\mid j\neq1\}=p^{r_{i1}}_1$ so
$\SS\subset\kernel\Psi_1$. We also can see that
$G/\~~~~\cong\ov G/\~~~~$ so we get a presentation for $\ov G_1$ by
adding $\SS$ to the relations of 5.1 or 5.2. We then see immediately
that $\ov G_1$ is a Fuchsian group with underlying surface group $S$
with cone points each of order a power of $p_1$. By 5.3, $\ov G_1$
is residually nilpotent. This completes the proof of 5.6. \qed
\bpage
We can handle the genus zero case but unfortunately it seems to
require special techniques and our conclusion is weaker. We are only
able to show that the length of such a group is at most $\om+1$. Yet
we do not know an example of such a group with length $\om+1$.
\proclaim{Theorem 5.8} Let $G$ be a Fuchsian group with genus zero.
Then $G_\om\sbq T(G)N$ where $N\tril G$ and $[N,G]\subset T(G)$.
\endproclaim
\proclaim{Corollary 5.9} The length of a Fuchsian group of genus
zero is at most $\om+1$.
\endproclaim
\sub{Proof of 5.9} $G_{\om+1}\subset[T(G)N,G]\subset T(G)\subset
G_\d$. Thus $G_{\om+1}=G_\d$.
\sub{Proof of 5.8} $G$ has a presentation $\$.
Our first step will be to prove that we can assume each $n_i$ is a
prime power (the primes will vary). Suppose
$n_1=p^{r_1}_1\dots p^{r_\ell}_\ell$ where the $p_j$ are distinct
primes. Consider the Fuchsian group $\wt
G=\$. We shall define
$\phi:\wt G\lra G$ and show that $\phi:\wt G/T(\wt G)\lra G/T(G)$ is
an isomorphism. Then it is easy to verify that $G$ satisfies the
conclusion of 5.8 if and only if $\wt G$ does. Since $\wt G$ has at
most $(m-1)$ cone points of non-prime-power order, we can continue
this reduction process until we can assume all orders are prime
powers. It remains to define $\phi$. Since gcd$\{n_1/p^{r_j}_j\mid
1\le j\le\ell\}=1$, we may choose integers $x_j$ such that
$\sum^\ell_{j=1}x_j(n_1/p^{r_j}_j)=1$. Let $y_j=x_j(n_1/p^{r_j}_j)$
and note that $(y_j,p_j)=1$ and that $y_j$ is a multiple of
$p^{r_i}_i$ if $i\neq j$. Define $\phi(b_j)=c^{y_j}_1$ and
$\phi(c_i)=c_i$ for $i\ge2$. An inverse $\Psi$ to the induced map
$\phi:\wt G/T(\wt G)\lra G/T(G)$ is constructed by setting
$\Psi(c_1)=b_1\dots b_\ell$ and $\Psi(c_i)=c_i$ if $i>1$. To see that
$\Psi$ is well-defined, note that $[b_i,b_j]\in T(\wt G)$ if $i\neq
j$ so $\Psi(c^{n_1}_1)=(b_1\dots b_\ell)^{n_1}\equiv b^{n_1}_1\dots
b^{n_1}_\ell\equiv0$ in $\wt G/T(\wt G)$. Furthermore
$\Psi\circ\phi(b_j)\equiv b^{y_j}_1\dots b^{y_j}_\ell\equiv b^{y_j}_j$
in $\wt G/T(\wt G)$. Since $y_j\equiv1$ modulo $p^{r_j}_j$, it
follows, with a little checking, that $\Psi$ is $\phi^{-1}$. This
completes the first step of the proof.
We now assume each $n_i=p^{r_i}_i$ for some prime from
$\{p_1,\dots,p_\ell\}$. Re-order and re-index the $c$'s so that
$C_j=\{c_{j1},\dots,c_{jm_j}\}$ is the set of those $c$'s whose
orders are powers of the prime $p_j$. Note that commutators of
elements from distinct $C_j$ are $T$-elements. Let $N$ be the normal
closure of the set $\SS=\{\d_1,\dots,\d_\ell\}$ where
$\d_j=\prod^{m_j}_{k=1}c_{jk}$, the product of the elements of
$C_j$. We claim that $[N,G]\sbq T(G)$. For if $c_k\notin C_j$
then $[\d_j,c_k]\subset T(G)$ since the order of $c_k$ is
relatively prime to $p_j$; and if $c_k\in C_j$ then, using the
relation $\prod^m_{i=1}c_i=1$, we see that $\d_j=\prod_{i\neq j}\d_i$
modulo $T(G)$, implying $[\d_j,c_k]\equiv\Bigl[\prod_{i\neq
j}\d_i,c_k\Bigr]\equiv0$ in $G/T(G)$ since $\d_i$ is a product of
$c$'s which do not lie in $C_j$. It remains only to show $G/T(G)N$ is
residually nilpotent. Merely note that $G/N$ is a free product of
Fuchsian groups $F_1*F_2*\dots*F_\ell$, each of genus zero, such that
$F_j$ has cone points of order various powers of $p_j$. Therefore the
map $G/N\lra G/T(G)N$ factors through $\x^\ell_{j=1}F_j$. By 5.3 and
2.2c the latter group is residually nilpotent, so the map $G\lra
G/T(G)N$ factors through $G/G_\om$. It follows immediately that
$G_\om\subset T(G)N$ as desired. \qed
\proclaim{Theorem 5.10} Let $G$ be a non-trivial Fuchsian group
whose underlying surface is $X$. If $\pi_1(X)$ is not abelian then
length$(G)\ge\om$. If the first betti number of $G$ is at least~$3$
then length$(G)\ge\om$. More precisely, if $X\neq S^2$, $\neq\BR P^2$
then either $G$ $G\cong\BZ$, $G\cong\BZ\x\BZ$ or length$(G)\ge\om$.
Therefore if $X$ is neither $S^2$ nor a connected sum of an odd
number of projective planes then either the length of $G$ is
precisely $\om$ or $G\cong\BZ$ or $G\cong\BZ\x\BZ$ or
$G\cong*\BZ_{n_i}$ where $(n_i,n_j)=1$, in which cases the length of
$G$ is $2$.
\endproclaim
\sub{Proof of 5.10} $G$ maps onto $\pi_1(X)$ so the first claim of
5.10 follows from 2.2a and 3.3. If $\b^1(G)\ge3$ then $\pi_1(X)$
cannot be abelian so the second claim of 5.10 follows. Now suppose
$X\neq S^2$ and $X\neq\BR P^2$ and $\pi_1(X)$ is abelian. Then $X$
is either an annulus, a Moebius band, a $2$-disk or $S^1\x S^1$. If
$X$ is an annulus or a Moebius band then $G$ maps onto some
$\BZ*\BZ_p$ unless $m=0$ in which case $G\cong\BZ$. If $X=D^2$ then
$G$ maps onto some $\BZ_p*\BZ_p$ unless $G=*\BZ_{n_i}$ where
$(n_i,n_j)=1$. If $X=S^1\x S^1$ then either $G=\BZ\x\BZ$ or $G$ maps
onto the group $\$ for some prime $p$, which can
be shown to have infinite length. For the last claim of 5.10, apply
the above and 5.6. \qed
\vskip.7cm
\sub{\S6. Seifert Fibered ${\boldkey3}$-Manifold Groups}
In this section we show that the fundamental group of a compact
Seifert fibered $3$-manifold with a base surface of even Euler
characteristic which is not $S^2$ (or with non-empty boundary) has
length at most $\om$. If the base surface is $S^2$ we can only show
the length is at most $\om+1$. In the case that base surface is closed
and of odd Euler characteristic we can say nothing only because of our
ignorance concerning those surface groups! We would expect that {\it
all\/} Seifert fibered $3$-manifold groups have length at most $\om$.
Conversely, at the end of this section we observe that the length of
most Seifert fibered groups is at least $\om$, in particular if the
first betti number is at least $4$ then this is true.
\proclaim{Theorem 6.1} Let $G$ be a central extension $1\lra
N\overset i\to\lra G\overset\pi\to\lra P\lra1$ of a group $P$ where
$P$ has Property~$T$ (see \S5). Then $G_\om=(N\cap G_\om)G_\d$, and
$G_{\om+1}=G_\d$. More generally, upon omitting the word ``central''
above, it is sufficient that $T(P)$ is the normal closure of
$\{[\pi(\a_i), \pi(\b_i)]\}$ where $[N,\b_i]=[\a_i,N]=\{e\}$.
\endproclaim
\sub{Proof} We identify $N$ with $i(N)$. Suppose $g\in G_\om$, so
$\pi(g)\in P_\om=T(P)$. It follows that
$\pi(g)=\prod^k_{i=1}\eta_i[a_i,b_i]\eta^{-1}_i$ where
$a_i$, $b_i$ are of relatively prime finite orders, say $p_i$ and
$q_i$ in $P$. Choose elements $\xi_i$, $\a_i$, $\b_i$ of $G$ in the
inverse image of $\eta_i$, $a_i$, $b_i$ respectively. Let
$h=\prod^k_{i=1}\xi_i[\a_i,\b_i]\xi^{-1}_i$ so $g=nh$ for some
$n\in N$. It will suffice to show that, for any ordinal $\g$,
$[\a_i,\b_i]\in G_\g$. Since $\pi(\a_i)$ is of order $p_i$,
$\a^{p_i}_i$ lies in the center of $G$ and similarly for
$\b^{q_i}_i$. Then apply 2.5 and we are done. \qed
\bpage
\proclaim{Theorem 6.2} Suppose $G$ is the fundamental group of a
compact $3$-manifold which has an orientable Seifert fibration (the
circle fibers can be continuously oriented) over a base surface $X$
which satisfies one of the following:
\roster
\item"i)" $\p X\neq\emptyset$
\item"ii)" $X$ is closed and orientable
\item"iii)" $X$ is closed, non-orientable with even genus
\endroster
In cases {\rm i)} and {\rm iii)} $G$ has length at most $\om$. In case
{\rm ii)} $G$ has length at most $\om+1$, and, if $X\neq S^2$,
$G_\om/G_{\om+1}$ is a (possibly trivial) cyclic group generated by a
power of the regular fiber.
\endproclaim
\sub{Proof of 6.2} Let $N$ be the subgroup generated by a regular
fiber $t$. It is well known that $G$ has a presentation of one of
the following forms [H]:
$$
\align
G &=\Bigl
\endalign
$$
$$
\align
G &=\Bigl
\endalign
$$
where $n_i\ge2$ and $(n_i,s_i)=1$.
Thus there is an exact sequence $1\lra N\lra G\overset\pi\to\lra
P\lra1$ where $P$ is a Fuchsian group with underlying surface $X$,
satisfying the hypotheses of 5.6 except in the case that $X$ is
$S^2$. The latter case will be handled separately. Therefore $P$ has
Property~$T$ and 6.1 applies to $G$ to show that
$G_\om=\G_{\om+1}$ for some integer $N$. In cases i) and iii)
it is easy to see from 6.3 or 6.4 that $t$ represents an element of
infinite order in the abelianization so $t^N\in G_\om$ implies $N=0$
and consequently $G_\om\cong G_{\om+1}$.
Now assume $X=S^2$. Then we have the exact sequence $1\lra N\lra
G\overset\pi\to\lra
P\lra1$ where 5.8 applies to $P$ to say that $P_\om\subset T(P)M$
where $M\tril P$ and $[M,P]\subset T(P)$.
It suffices to show that, if $g\in G_\om$, then the element
$[c_1,g]$ lies in $G_\d$ (since $G_{\om+1}$ is normally generated by
$\{[c^{\pm1}_i,g]$, $[t^{\pm1}_i,g]\mid g\in G_\om\}$). We proceed
by induction. Recall from 5.8 that $M\tril P$ is the normal closure of
$\{\d_1,\dots,\d_k\}$ where
$\d_j=\prod^m_{j=1}c^{n_ix_{ij}/p^{r_{ij}}_j}$ where
$\sum_{j=1}n_ix_{ij}/p^{r_{ij}}_j=1$ and
$n_i=\prod^\ell_{j=1}p^{r_{ij}}_j$. Therefore, proceeding as in the
non-zero genus case, we can choose $n\in N$, $\tl m\in G$ such that
$g=nh\tl m$ where $h\in G_\d$ and $\pi(\tl m)\in M$. Furthermore we
can choose $\tl m$ to be in the normal closure in $G$ of
$\{\d_1,\dots,\d_k\}$ viewed as elements of $G$. Therefore it will
suffice to show $[c_1,\d_j]\in G_\d$ for fixed $j$. Assume
$[c_1,\d_j]\in G_\b$ for {\it all\/} $j$. The proof will be complete
if we show $[c_1,\d_j]\in G_{\b+1}$ for any {\it fixed\/} $j$.
\proclaim{Lemma 6.5} $[c^q_1,\d_j]\in G_\d$ whenever $q$ is an
integer dividing $n_1$ such that $(n_1/q,p_j)=1$.
\endproclaim
\sub{Proof of 6.5} Consider
$\bigl[c^q_1,c^{n_ix_{ij}/p^{r_{ij}}_j}_i\bigr]$. Since $c^q_1$ is of
order $n_1/q$ in $G$ modulo its center, and similarly
$c^{n_ix_{ij}/p^{r_{ij}}_i}_i$ is of order $p^{r_{ij}}_j$, the
commutator lies in $G_\d$ by 2.5. It follows that $[c^q_1,\d_j]$ lies
in $G_\d$ since $\d_j$ is a product of such powers of $c_i$'s (use
2.16). \qed
\bpage
It follows from 6.5 (by taking $q=p^{r_{1j}}_j$) that
$[c_1,\d_j]^{p^{r_{1j}}_j}\in G_{\b+1}$. To conclude that
$[c_1,\d_j]\in G_{\b+1}$ it will now suffice to show $[c_1,\d_j]^q\in
G_{\b+1}$ where $q=n_1/p^{r_{1j}}_j$ since $(q,p^{r_{1j}}_j)=1$.
\proclaim{Lemma 6.6} $\prod^\ell_{j=1}\d_j\in G_\d N$.
\endproclaim
\sub{Proof of 6.6} Recall that
$\prod^\ell_{j=1}\d_j=\prod^\ell_{j=1}\prod^m_{i=1}
c^{n_ix_{ij}/p^{r_{ij}}_j}_i$. The factors of $\d_j$ commute in
$G/G_\d$ with the factors of $\d_k$ if $j\neq k$, since the
commutator of two such factors is of the form $[a,b]$ where
$a^{p^n_j}\in N$, $b^{p^m_k}\in N$, so $[a,b]\in G_\d$ by 2.5.
Therefore
$\prod^\ell_{j=1}\d_j\equiv\prod^m_{i=1}\prod^\ell_{j=1}
c^{n_ix_{ij}/p^{r_{ij}}_j}_i$ in $G/G_\d$. Since
$\sum^\ell_{j=1}n_ix_{ij}/p^{r_{ij}}_j=1$,
$\prod^\ell_{j=1}\d_j\equiv c_1\dots c_m\equiv t^b$ in
$G/G_\d$, completing the proof. \qed
\bpage
Using 6.6, we see that
$[c_1,\d_j]^q\equiv[c_1,\d_1\dots\hat\d_j\dots\d_\ell]^q$ modulo
$G_{\b+1}$. Since we assumed $[c_1,\d_k]\in G_\b$ for all $k$, the
latter term is congruent to $\(\prod_{k\neq j}[c_1,\d_k]\)^q$ and
hence $\prod_{k\neq j}[c^q_1,\d_k]$ modulo $G_{\b+1}$ (using 2.1).
Since $(n_1/q,p_k)=(p^{r_{1j}}_j,p_k)=1$, we can apply 6.5 to conclude
that $[c^q_1,\d_k]\in G_\d$ and hence that $[c_1,\d_j]^q\in G_{\b+1}$,
thus finishing the proof that $G_{\om+1}=G_\d$ in the genus zero,
closed, case of Theorem~6.2. This completes the proof of 6.2. \qed
\bpage
In general the fundamental group of a compact Seifert fibered
$3$-manifold has one of the following forms.
$$
\align
&\Bigl
\endalign
$$
$$
\align
&\Bigl
\endalign
$$
\proclaim{Theorem 6.9} Suppose $G$ is the fundamental group of a
compact $3$-manifold which admits a Seifert fibration whose base
surface is not $S^2$ or a connected sum of an odd number of
$\BR P(2)$'s. Then no non-zero power of the class represented by the
regular fiber lies in $G_\om$. (The result holds except when the base
is $S^2$ or $\BR P(2)$ but we do not need this stronger fact.)
\endproclaim
\sub{Proof of 6.9} The proof breaks into several cases but in each
case we shall construct a homomorphism $\phi:G\lra H$ where $H$ is
residually nilpotent $(H_\om=\{e\})$ and $\phi(t)$ is of infinite
order. Clearly this is sufficient. Throughout, with regard to the
presentations 6.7, 6.8, let $x=\prod^m_{i=1}n_i$, let
$\b_i=s_i\(\prod_{j\neq i}n_j\)$, and let $\e=\ex+\sum^m_{i=1}\b_i$.
First we consider the cases in which there is an orientable Seifert
fibering (i.e., all $\e_i$, $\eta_i$, $\th_i$ are $+1$). If the base
surface has non-empty boundary or is non-orientable then the regular
fiber $t$ can easily be seen to be of infinite order in the
abelianization (as was noted in the proof of 6.2). Therefore we need
only consider the case that the base surface is orientable with
genus at least one and $G$ has a presentation as in 6.3. Let
$H=\$ and let $\phi(a_1)=a$,
$\phi(b_1)=b$, $\phi(a_i)=\phi(b_i)=0$ for $i>1$, $\phi(t)=t^x$,
$\phi(c_i)=t^{\b_i}$. Note that $H$ is one of the ``Heisenberg
groups'' of 2.14 and is nilpotent. Since $H$ is the fundamental
group of an orientable circle bundle over the torus, it is
torsion-free.
Now consider the case of a non-orientable Seifert fibration over a
closed non-orientable base surface of genus at least two (since $\BR
P(2)$ is excluded). Here we have a presentation as in 6.8 with
$k=0$. Without loss of generality we may assume $\e_i=-1$ for
$1\le i\le r$
where we use the $+$ sign if the number of $\e_i$ equal to $-1$ is
odd. Let $\phi(t)=t^x$, $\phi(c_i)=t^{\b_i}$ and $\phi(a_1)=a$. If
the number of $\e_i$ equal to $-1$ is odd then let $\phi(a_2)=a$,
$\phi(a_3)=a^{-1}$, $\phi(a_4)=a$, $\phi(a_5)=a^{-1}$, et~cetera,
ending with $\phi(a_r)=a^{-1}$, and let $\phi(a_g)=b$ and
$\phi(a_i)=0$ for $r~~*\lra H\overset\pi\to\lra K\lra 1$ where
$K=\$ is the group of the Klein Bottle, there is
an exact sequence $1\lra\\lra\wt H\lra\BZ\x\BZ\lra 1$ since the
kernel of induced map $K\lra\BZ_2\x\BZ_2$ is the group of a torus.
Moreover $t$ is central in $\wt H$ since $\wt H$ is generated by
$a^2$, $b^2$, $[a,b]$, $ba^2b^{-1}$ and $ab^2a^{-1}$. Therefore $\wt
H$ is one of the Heisenberg groups as discussed above. By 3.1 it
will suffice to show that the Heisenberg group
$\$ is residually a finite
$2$-group. For this it suffices, for any integer $n$, to find a
quotient which is a finite $2$-group in which $t^n$ survives. But
$\$ is
such a group.
Next consider the case of a non-orientable Seifert fibration with a
closed orientable base surface (of genus at least~$1$). Here we have
a presentation as in 6.7 with $k=0$ and $g\ge1$. We can assume
$\e_1=-1$. Then let $H=\$ and let $\phi(c_i)=t^{\b_i}$,
$\phi(t)=t^x$, $\phi(a_1)=a$, $\phi(b_1)=b$ and otherwise
$\phi(a_i)=0$ if $\e_i=+1$ or $\phi(a_i)=a$ if $\e_i=-1$ and
$\phi(b_i)=0$ if $\eta_i=+1$ or $\phi(b_i)=a$ if $\eta_i=-1$. It is
easy to see that $t$ is of infinite order in $H$ by calculating, for
example, that $t$ is of infinite order in the abelianization of the
kernel of the epimorphism $H\lra\BZ_2$ given by sending $a$ and $b$
to $1$. We see that $H$ is residually nilpotent as follows. The
commutator subgroup $H_2$ is normally generated by $[a,b]$, $[a,t]$
and $[b,t]$ and hence normally generated by $t^r$, where $r=1$ if
$e$ is odd, and $r=2$ if $e$ is even. Thus $H_2$ is the subgroup
generated by $t^r$. It follows by induction that $H_n$ is the
infinite cyclic subgroup $(r2^{n-1})\BZ\sbq\BZ$ generated by
$t^{r2^{n-1}}$. Hence $H_\om=\{e\}$.
Now consider the case of a non-orientable Seifert fibration with a
non-orientable base surface with non-empty boundary. Let
$H=\bigl$, and let $\phi(c_i)=t^{\b_i}$,
$\phi(t)=t^x$, $\phi(a_i)=a_i$, $\phi(d_j)=d_j$ if $j\neq k$ and
$\phi(d_k)=\(t^\e\prod_{j\neq k}d_j\)^{-1}\prod a^2_i$. Then $H$ is
clearly a semi-direct product of a free abelian group with $\BZ$
(generated by $t$) and $H_n=2^{n-1}\BZ\sbq\BZ$ by [L4; 3.1] or by
direct calculation. Thus $H$ is residually nilpotent.
Finally we have the case of a non-orientable Seifert fibration with
an orientable base surface with non-empty boundary. Then the
procedure just above works (solve for $d_k$ and let all $a_i$, $b_i$,
$d_i$ commute) to yield a map to a semi-direct product as above.
Note that $g=0$, $k=1$ is not possible for a non-orientable Seifert
fibration.
This completes the proof of 6.9. \qed
\proclaim{Theorem 6.10} The fundamental group $G$ of a compact
Seifert fibered $3$-manifold whose base surface is neither $S^2$ nor a
connected sum of an odd number of projective planes has length at
most $\om$. If the base surface is $S^2$, $G$ has length at most
$\om+1$.
\endproclaim
\sub{Proof of 6.10} $G$ is the extension of a Fuchsian group $P$ by
the cyclic subgroup $N$ generated by the regular fiber $t$. From the
proof of 5.6 we see that $T(P)$ is the normal closure of a set of
elements $[\pi(\a_i),\pi(\b_i)]$ where $\a_i$ and $\b_i$ are
conjugates of a power of some $c_j$. But any conjugate of $c_j$
commutes with $t$ since $c_j$ commutes with $t$. Hence, by 6.1,
$G_\om=(N\cap G_\om)G_\d$. By 6.9 $N\cap G_\om=\{e\}$ so
$G_\om=G_\d$ and $G$ has length at most $\om$.
If the base surface is $S^2$ then the Seifert fibration must have
orientable fibers and hence is covered by 6.2. \qed
\bpage
On the other hand, almost all Seifert fibered $3$-manifold groups
have length at least $\om$.
\proclaim{Proposition 6.11} If $\pi_1(X)$ is not abelian then $G$, the
group of a Seifert fibered $3$-manifold, has length at least $\om$.
\endproclaim
\sub{Proof} There is an epimorphism $G\lra\pi_1(X)$ so by 2.2a and
3.3, length$(G)\ge\om$.
\proclaim{Proposition 6.12} If $M$ is Seifert fibered and rank
$(H_1(M))\ge4$ then length $(\pi_1(M))\ge\om$.
\endproclaim
\sub{Proof} From 6.3 and 6.4 we see that rank $(H_1(M))\le2g+k+1$ if
$X$ is orientable and $g+k+1$ if $X$ is non-orientable. By 6.11 we
may assume $X$ is $\BR P(2)$, a Moebius band, an annulus, a torus,
$D^2$ or $S^2$. But in these cases rank $(H_1(M))\le3$. \qed
\proclaim{Theorem 6.13} If $G$ is the fundamental group of a compact
Seifert fibered $3$-manifold with base surface $X$ such that
$\pi_1(X)$ is not abelian and $X$ is not a connected sum of an odd
number of projective planes, then $G$ has length $\om$.
\endproclaim
\vskip.7cm
\sub\nofrills{\S7. $\bold3$-Manifold Groups Have the Same Lower
Central Series as Hyperbolic}
\ \ {\bf $\bold3$-Manifold Groups}
In this section we will show that, for any closed, oriented
$3$-manifold there is a hyperbolic $3$-manifold (in particular a prime
$3$-manifold with torsion-free fundamental group) whose groups have
the same length and the same lower central series quotients. This is
very similar in spirit to earlier work of Robert Myers and Chuck
Livingston on homology cobordisms between $3$-manifolds [My2] [Li1].
In our situation, however, an homology cobordism to a hyperbolic
$3$-manifold is not sufficient since such only preserves the
isomorphism type of $G/G_\a$ for finite values of $\a$, and therefore
does not preserve length (see 9.1). The stronger result needed here
is implied by a theorem of D\.~Ruberman [R; Theorem~2.6 and
Corollary~2.7]. Recent work of Akio Kawauchi on ``almost identical
imitations'' is also closely related [K1, K2]. Ruberman's result
proves even more than we need so we include our own very similar
(but shorter) proof.
\proclaim{Theorem 7.1 \rm{(D\.~Ruberman [R; Theorem~2.6])}} Let $M$ be
a closed, connected, oriented $3$-manifold. There exists a closed,
connected, oriented hyperbolic $3$-manifold $X$ and a degree one map
$f$: $X\lra M$ which induces isomorphisms on all integral homology
groups.
\endproclaim
\proclaim{Corollary 7.2} Suppose $M$ is a closed, connected, oriented
$3$-manifold with $\pi_1(M)\cong G$. Then there is a closed,
connected, oriented hyperbolic $3$-manifold $X$ with $\pi_1(X)\cong
P$ and an epimorphism $f_*$: $P\lra G$ which induces an isomorphism
$f_\a$: $P/P_\a\lra G/G_\a$ for each ordinal $\a$. The length of $P$
is equal to the length of $G$.
\endproclaim
\sub{Proof of 7.2} Using 7.1, the degree one map induces an
epimorphism $f_*$: $P\lra G$ which satisfies the hypotheses of
Stalling's theorem (1.1). It follows that each $f_\a$ is an
isomorphism. It then follows easily from this that the lengths are
equal. \qed
\bpage
\sub{Proof of 7.1} Let $T$ be a ``standard trivial'' embedding of
the wedge $W$ of $k$ circles into $S^3$. A tame embedding
$g:W\lra S^3$ is called {\it characteristic\/} if there is an
orientation-preserving homeomorphism $\phi:S^3\lra S^3$ such that
$\phi\circ g=T$ and $\phi(S^3-$image~$g)=S^3-$image~$T$. In other
words there is a degree one map rel boundary from $E(g)$ (the
exterior of image$(g)$) to $E(T)$. It is well known that $g$ is
characteristic if (and only if) $g$ defines a {\it boundary
tangle,\/} one such that the $k$ component ``knots'' $g(S^1)$ bound
pairwise disjoint (except at wedge point) embedded, oriented
surfaces in $S^3$ [CS, Mil]. The theorem will follow from the
existence of characteristic, simple, Haken tangles. By this we mean
a characteristic embedding $g$ whose exterior $N=E(g)$ is
irreducible, with incompressible boundary, such that every properly
embedded incompressible torus or annulus is boundary parallel.
\proclaim{Lemma 7.3} For any $k\ge1$ there exists a characteristic
tangle $g$ whose exterior $N$ is a simple, Haken $3$-manifold (with
incompressible boundary).
\endproclaim
Assuming 7.3, choose a Heegard splitting of genus~$k$ with $k\ge2$
for $M$. Identifying $E(T)$ with the handlebody of genus~$k$ and $M$
with $E(T)\cup_\psi E(T)$ for some homeomorphism $\psi$ of
$\p E(T)$, there is a degree one map $f:E(g)\cup_\psi E(g)\lra
E(T)\cup_\psi E(T)\equiv M$ which is an isomorphism on homology.
Moreover $X=E(g)\cup_\psi E(g)$ is a simple Haken manifold [My1;
Lemma~3.3] which is hyperbolic by W\.~Thurston [Mo].
\sub{Proof of 7.3} Fix $k\ge2$. Let $h:W\lra S^3$ be a $2k$-string
simple Haken tangle (see [My2; Prop\.~4.1 and proof of 5.2]). Let
$F'$ denote $S^1\x S^1$ with an open disk deleted. Choose an
identification $\phi$ of $\(\natural^k_{i=1}F'\)\x[0,1]$ with a
regular neighborhood of image$(h)$. Let $F$ be the complement in $F'$
of a small open collar of $\p F'$, and let $F_i$ be a copy of $F$
lying in the $i^\supth$ copy of $F$ in $\natural^k_{i=1}F'$. Let $N$
be the manifold obtained from $E(h)$ by identifying $\phi(F_i\x\{0\})$
with $\phi(F_i\x\{1\})$ for $i=1,\dots,k$. Note that $N=E(g)$ where
$g:\bigvee^k_{i=1}S^1\lra S^3$ and $g$ restricted to the $i^\supth$
circle is $h$ restricted to $\phi(\p F_i\x\{1/2\})$. Since
$\{\phi(\p F_1)$, $\phi(\p F_2),\dots,\phi(\p F_k)\}$ is a boundary
link, $g$ is a boundary tangle and hence characteristic. It remains
only to show that $N$ is simple, irreducible and has incompressible
boundary. This is a standard glueing result [My3; Lemma~2.1]. \qed
\bpage
This completes the proof of Theorem 7.1.
\vskip.7cm
\sub{\S8. $\bold3$-Manifold Groups With Length Greater than
$\bold\om$}
Let $K$ be the fundamental group of the Klein bottle. We shall show
that, for any odd integer $q$, $C=K*\BZ_q$ has length at least
$2\om$. This demonstrates that there are closed, orientable
$3$-manifolds whose groups have length at least $2\om$. For there is
a twisted interval bundle over the Klein bottle whose total space is
an orientable $3$-manifold $Y$ with boundary a torus. Letting $X$ be
the ``double'' of $Y$ along its boundary observe, that $Y$ is a
retract of $X$ and hence $\pi_1(Y)\cong K$ is a retract of
$\pi_1(X)$. Therefore $C$ is a retract of $\pi_1(X)*\BZ_q$ and so
the length of $\pi_1(X)*\BZ_q$ is at least $2\om$ by 2.2 and 8.1. But
this group is the fundamental group of the closed, orientable
$3$-manifold $X\#L(q,1)$.
\proclaim{Theorem 8.1} Let $K=\*__$ be the
Klein bottle group and $q$ be an odd positive integer. Then
$C=K*\BZ_q$ has length at least $2\om$. If $K'$ has $K$ as retract
then $K'*\BZ_q$ has length at least $2\om$.
\endproclaim
\proclaim{Corollary 8.2} There exist closed, hyperbolic $3$-manifold
groups of length at least $2\om$.
\endproclaim
\sub{Proof of 8.2} The argument in the first paragraph above shows
that the group of $X\#L(q,1)$ has length at least $2\om$. Then apply
Corollary~7.2.
\sub{Proof of 8.1} We can assume $q$ is prime since if $q=mn$ then
$K*\BZ_m$ is a retract of $K*\BZ_{mn}$. The proof is accomplished in
the following stages. We consider the presentation $C=\____$.
\proclaim{Lemma 8.3} $C_\om$ is the normal subgroup generated by
$[s,t]$.
\endproclaim
The following enables us to find a presentation for $C/C_{\om+k}$ for
any $k\in\BZ_+$.
\proclaim{Lemma 8.4} If $A\tril C$ is the subgroup {\it normally\/}
generated by $\{a_\a\mid\a\in J\}$ and $B\tril C$ is the subgroup
{\it generated\/} by $\{b_\b\mid\b\in J'\}$ then $[A,B]$ is the
subgroup normally generated by $\{[a_\a,b_\b]\mid\a\in J, \b\in J'\}$.
\endproclaim
\proclaim{Lemma 8.5} $[s,t]$ is of order $q$ in $H_2(C/C_\om)$.
\endproclaim
\proclaim{Lemma 8.6} $H_2(C)\lra H_2(C/C_{\om+k})$ is not surjective.
\endproclaim
Lemma 8.6 implies Theorem 8.1 since by Stallings' theorem the
cokernel of $\pi_*:H_2(C)\mathbreak
\lra H_2(C/C_{\om+k})$ is
$C_{\om+k}/C_{\om+k+1}$. It follows that the length of $C$ is
greater than $\om+k$ for all $k\in\BZ_+$, and hence at least $2\om$.
Lemma~8.5 is not strictly necessary but is much easier to prove than
8.6 and is enough to show that the length of $C$ is greater than
$\om$, since $H_2(C)\cong H_2(K)\op H_2(\BZ_q)\cong0$ and so
$H_2(C/C_\om)\cong C_\om/C_{\om+1}$ by Stallings' theorem.
\sub{Proof of 8.3} Since $[u,t^k]=t^{(-2)^k}$,
$[u,[u,\dots,[u,t]]\dots]=t^{(-2)^n}$ so $t^{2^n}\in C_{n+1}$ (in
fact it lies in $K_{n+1}$). Thus $t$ is a generalized
$2$-element because it has order a power of $2$ in each $C/C_j$,
$j\in\BZ_+$. By 2.4, $[s,t]\in C_\om$. Now it suffices to
show that $B=C/\<[s,t]\>$ is residually nilpotent. The group $B$ has
presentation $\____$. Note that the subgroup $J$ generated by $t$ is
normal and infinite cyclic. Moreover there is a semi-direct product
decomposition $B=J\rtimes(\BZ*\BZ_q)$ where $\BZ*\BZ_q=\____$. $J$ is a $\BZ[\BZ*\BZ_q]$ module where $u_*(t^k)=t^{-k}$ and
$s_*(t^k)=t^k$. Levine has shown that, since $\BZ*\BZ_q$ is
residually nilpotent (see 2.7), $B_\om\cong\bigcap^\infty_{j=1}I^jJ$
where $I$ is the augmentation ideal of $\BZ[\BZ*\BZ_q]$ [L4]. Suppose
$w=w(u,s)$ is an element of $\BZ*\BZ_q$. Since $s$ acts trivially on
$J$, the action $w_*$ is equal to $u^m_*$ where $m$ is the exponent
sum of $u$ in $w$. Thus $IJ=I'J$ where $I'$ is the augmentation
ideal of $J$ as a $\BZ[u,u^{-1}]$ module. Similarly $I^jJ=(I')^jJ$
and the latter is easily seen to be $2^jJ$. Therefore
$B_\om=\bigcap^\infty_{j=1}2^jJ=\{e\}$ as desired. \qed
\bpage
\sub{Proof of 8.4} Since $[A,B]$ is normally generated by elements
of the form $[x,y]$ where $x\in A$, $y\in B$. We know
$x=\Pi\eta_ia^{\pm1}_i\eta^{-1}_i$. Using the identities
$[ab,c]=a[b,c]a^{-1}[a,c]$ and $[a,bc]=[a,b]b[a,c]b^{-1}$ we see
that $[x,y]$ is a product of conjugates of elements of the form
$[b^{\pm1},\eta a^{\pm1}\eta^{-1}]$ and their inverses, where $b$ is
a generator of $B$ and $a$ is a normal generator of $A$. Now use the
identity $[b,\eta
a\eta^{-1}]=\g[b,a]\g^{-1}\eta[[\eta^{-1},b],a]\eta^{-1}$ where
$\g=b\eta b^{-1}$ and then expand the triple commutator by using
$[ab,c]=a[b,c]a^{-1}[a,c]$ and the fact that $[\eta^{-1},b]$ is a
product of generators of $B$ (or their inverses). This shows that
$[A,B]$ is normally generated by elements of the form
$[a^{\pm1}_\a,b^{\pm1}_\b]$. Since $[x^{\pm1},y^{-1}]$ is conjugate
to $[x^{\mp1},y]$ and $[x^{-1},y]=yx^{-1}y^{-1}[x,y]^{-1}yxy^{-1}$,
the result follows. \qed
\bpage
\sub{Proof of 8.5} Let $B=C/C_\om$ as in 8.3 above. Let $D$ be the
normal subgroup of $B$ generated by $\{s,t\}$ so there is a split
short exact sequence
$$
1\lra D\overset f\to\lra B\overset\pi\to\thra\BZ\lra1
$$
where $\BZ$ is generated by $u$. A presentation for $D$ is
$\$ where $f(s_i)=u^isu^{-i}$ and $f(t)=t$. Thus
$D\cong\BZ\x\(*^\infty_{i=-\infty}\BZ_q\)$ generated by $t$ and
$\{s_i\}$ and $H_2(D)\cong\x^\infty_{i=-\infty}\BZ_q$
generated by $\{[s_i,t^{\e_i}]\mid i\in\BZ\}$ which we denote
$\{e_i\mid i\in\BZ\}$. There is then an exact sequence:
$$
H_2(D)@> u_*-\id>> H_2(D)@>\ f_*>> H_2(B)
@>\ \p_*>> H_1(D)@>u_*-\id_*>> H_1(D)\lra .
$$
Since $u_*(t)=t^{-1}$ and $u_*(s_i)=s_{i+1}$,
$u_*([s_i,t^{\e_i}])=[s_{i+1},t^{\e_{i+1}}]$. Thus
$u_*(e_i)=e_{i+1}$ and so $H_2(D)/\$ is $\BZ_q$ generated
by $[s,t]=e_0$. It follows from the exact sequence that
$H_2(B)\cong\BZ_q$ generated by $[s,t]$. \qed
\bpage
\proclaim{Lemma 8.6} $H_2(C)\lra H_2(C/C_{\om+k})$ is not surjective
(so the length of $C$ is at least $2\om$).
\endproclaim
\sub{Proof of 8.6} Let $D'$ be the {\it normal\/} subgroup of
$B=C/C_{\om+k}$ generated by $\{s,t\}$. From 8.3 and 8.4 we know
that $C_{\om+k}$ is the normal closure of
$\{[\d_1, [\d_2,\dots, [\d_k,[s,t]]\dots]]\mid\d_i\in\{u, s, t\}\}$.
There is an exact sequence $1\lra D'\lra B\lra\BZ\lra1$ and $D'$ has
a presentation $\$ where
the latter relations are rewritten using
$ut_i\bar u=t^{-1}_{i+1}$ and $us_iu^{-1}=s_{i+1}$ (so here
$s_i=u^isu^{-i}$, $t_i=u^itu^{-i}=t^{(-1)i}$). We now analyze
$H_2(D',\BZ_q)$. Let $F=$ the free group on $\{s_i,t\mid i\in\BZ\}$,
and $R$ denote the relation subgroup of $D'$ so $D'=F/R$. It will be
more convenient to work with the mod~$q$ lower central series [St].
This differs from the ordinary lower central series in that, for
non-limit ordinals, $G_{\a+1,q}\equiv G\#G_{\a,q}$ where, in
general, for any subgroup $U$ of $G$, $G\#U$ is the subgroup
generated by all elements of the form $[g,u]v^p$ for $g\in G$\ \ $u$,
$v\in U$. Since $R\sbq F_{2,q}$ there is a natural homomorphism
$\phi:R/R\#F\lra F_{2,q}/F_{3,q}$. Note that if $x\in F_m$, then
$[u,x]=(ux\bar u)\bar x$ lies in $F_m$ also. Therefore, the relation
$[\d^i_1,\dots[s_i,t_i]]$ lies in $F_{3,q}$ unless {\it each\/} $\d_i$
is $u$. And the relation $[u,[u,\dots[s_i,t_i]]$ equals
$\prod^k_{i=1}(u-\id)_*([s_i,t_i])$ in the abelian group
$F_{2,q}/F_{3,q}$. It follows that the image of $\phi$ is generated by
$\{s^q_i\}$ and $\{(u-\id)^k_*([s_i,t_i])\mid i\in\BZ\}$. We have the
long exact sequence with coefficients in $\BZ_q$:
$$
H_2(D')@> u_*-\id>> H_2(D')@>\ i_*>> H_2(B)\lra H_1(D')\lra H_1(D')
\lra .
$$
By [St; Thm\.~2.1], $H_2(D')\cong R/R\#F$, so the cokernel of
$(u_*-\id)$ maps, via $\phi$, onto the image of $\phi$ modulo the
image of $\phi\circ(u_*-\id)$. The image of $\phi\circ(u_*-\id)$ is
the subgroup generated by $\{s^q_{i+1}-s^q_i, (u-\id)^{k+1}_*
([s_i,t_i])\mid i\in\BZ\}$. Since $t_i=t^{(-1)^i}_i$ and
$[s_i,t^{-1}]=-[s_i,t]$ in $F_{2,q}/F_{3,q}$, all the $t_i$'s in the
above may be replaced by $t$'s. Thus it suffices to show that the
subgroup generated by $\{(u-\id)^k_*([s_i,t])\}$ modulo the subgroup
generated by $\{(u-\id)^{k+1}_*([s_i,t])\}$ is non-trivial for each
fixed $k\in\BZ_+$. It is well known that $F_{2,q}/F_{3,q}$ is a
$\BZ_q$-vector space with basis
$\SS=\{s^q_i,t^q,[s_i,t],[s_i,s_j]\mid i>j\}$. It is an easy
exercise that the following is also a basis: (here let
$\a_i=[s_i,t]$)
$\SS'=\{s^q_i,t^q,[s_i,s_j],\a_0,\a_1-\a_0,\dots,(u-\id)^k_*\a_0,
(u-\id)^{k+1}_*\a_i\}$. Since image of $\phi$ contains
$(u-\id)^k_*\a_0$, for example, it is non-trivial even upon quotient
by $\{(u-\id)^{k+1}_*\a_i\}$. In fact, $F_{2,q}/F_{3,q}$ modulo
$\{(u-\id)^{k+1}_*\a_i, s^q_{i+1}-s^q_i\}$ has basis
$\SS''=\{s^q_0,t^q,\a_0,\a_1-\a_0,\dots,(u-\id)^k_*\a_0\}\cong
\BZ^{k+3}_q$. The elements $(u-\id)^k_*\a_0$ and $s^q_0$ are basis
elements. It follows that the image of $H_2(D')@>i_*>>H_2(B)$ maps
onto $\BZ_q\x\BZ_q$ with $[u,[u,\dots[u,[s,t]]]$ mapping to $(1,0)$
and $s^q$ mapping to $(0,1)$. Therefore $H_2(B;\BZ_q)$ is of $\BZ_q$
rank at least~2. Since $H_2(C;\BZ_q)\cong\BZ_q$, the map
$\pi_*:H_2(C)\lra H_2(B)$ is not surjective. Since $\pi:C\lra B$ is
an isomorphism on $H_1$, by the naturality of the Universal
Coefficient Theorem, it follows that $H_2(C)\lra H_2(B)$ is not
surjective on integral homology. \qed
\bpage
This completes the proof of 8.6 and of Theorem 8.1.
There are other $3$-manifolds with very long fundamental groups. Let
$X_p$ be the oriented $3$-manifold obtained by $(p,p,0)$ Dehn
surgery on the Borromean Rings as shown in Figure~8.7
\midinsert
\vspace{1in}
\botcaption{figure 8.7}\endcaption
\endinsert
\proclaim{Theorem 8.8} If $(q,p)=1$ then $X_p\#L(q,\ell)$ (or
connected sum of $X_p$ with any manifold with $\pi_1$ equal to
$\BZ_q$) has a fundamental group whose length is at least $2\om$.
\endproclaim
We shall not prove this theorem since its proof is similar to but
much harder than that of 8.1. Letting $s$ generate the $\BZ_q$ as
before, first one shows that $C=A*\BZ_q=\pi_1(X_p)*\pi_1(L(q,1))$
has a presentation
$$
\.
$$
Then one shows by induction that $x$ and $y$ are generalized
$p$-elements, specifically that $x^{p^k}$ and $y^{p^k}$ lie in
$A_{k+1}$ where $A=\pi_1(X_p)$. It follows that $[x,s]$ and $[y,s]$
lie in $G_\om$. One then shows that $G/\<[x,s],[y,s]\>$ is
$\om$-nilpotent by observing that the latter group is a semidirect
product $\BZ\x\BZ\rtimes(\BZ*\BZ_q)$ where $\BZ\x\BZ=\$ and
calculating that the action is nilpotent. The final (messy) step is
to show that the cokernel of $H_2(G ; \BZ_q)\lra H_2(G/G_{\om+k} ;
\BZ_q)$ maps {\it onto\/} $\BZ_q\x\BZ_q$ by examining the exact
sequences related to the subgroup $D$ (of infinite cyclic index) of
$G/G_{\om+k}$ which is normally generated by $\{x,y,s\}$. The
interested reader may write to the authors for a copy of these
calculations.
Finally, replacing $X_p$ of 8.8 by $(p,0)$ surgery on a Whitehead link
also works although a complete proof has been written down only for
$p=4$.
\vskip.4cm
\sub{\S9. Length of the lower central series and Homology Cobordism}
We shall say that two finitely-presented groups $G$ and $G'$ are
{\it homology cobordant\/} if there exists a third
finitely-presented group $H$ and homomorphisms $f:G\lra H$,
$f':G'\lra H$ which induce isomorphisms on $H_1$ and epimorphisms on
$H_2$. Then if two compact manifolds are homology cobordant in the
usual sense, their fundamental groups will be homology cobordant. If
we also insist that the images of $f$ and $f'$ normally generate
$H$, two groups are homology cobordant if and only if they have
isomorphic {\it algebraic closures\/}, in the sense of J\.~Levine
([L2], see especially Propositions~5 and~6). As mentioned in \S1,
Stallings' theorem implies that the isomorphism type of $G/G_n$, $n$
finite, is an invariant of homology cobordism. It follows that the
length of $G$ is an invariant of homology cobordism if the length of
$G$ is finite. If it were true that length were an invariant of
homology cobordism, then all slice link exteriors would have groups
of length $\om$ since any slice link exterior is homology cobordant
to a trivial link exterior. It would follow that the Parafree
Conjecture is true for a large class of groups (see 1.5). However we
shall observe below that length is {\it not\/} an invariant of
homology cobordism. In fact an example was already found in [L3;
\S6] although it was not explicitly pointed out. Since the known
examples will fail to be ``weakly parafree'' ($F/F_n\cong G/G_n$ for
all finite $n$ (see \S1)), this leaves open the possibility that
length might be invariant under homology cobordism of weakly
parafree groups for the simple reason that all such groups may have
length at most $\om$.
\proclaim{Proposition 9.1 \rm{(compare [L3; Section 6])}} There exist
finitely presented groups $B$ and $C$ and a homomorphism $f:B\lra C$
whose image normally generates $C$ and which induces an isomorphism
on $H_1$ and an epimorphism on $H_2$ but where length$(C)>\om$ and
length$(B)=\om$.
\endproclaim
\sub{Proof} Let $B=\____$,
$C=\____$ and define $f(u)=u$,
$f(s)=s$ and $f(t)=t^3$. Since $H_1(B)$ and $H_1(C)$ are
$\BZ\x\BZ\x\BZ_2$ where $t$ is an element of order~2, $f$ induces an
isomorphism on $H_1$. We claim that $H_2(C)\cong\BZ$ generated by
$[s,t^3]$. To see this let $D$ be the normal subgroup generated by
$\{s,t\}$. Then $D$ has presentation $\$ where $i\in\BZ$, $s_i=u^isu^{-i}$ and $\e_i=(-1)^i$, so
$H_2(D)$ is free abelian on the set $\{\a_i\mid i\in\BZ\}$ where
$\a_i=[s_i,t^{3\e_i}]$. There is an exact sequence as below
$$
\lra H_2(D)@> u_*-\id>> H_2(D)\lra H_2(C)\lra H_1(D)
@> u_*-\id>> H_1(D)\lra .
$$
Since $u_*(\a_i)=u[s_i,t^{3\e_i}]u^{-1}=\a_{i+1}$, the cokernel of
$u_*-\id$ on $H_2$ is $\BZ$, generated by $\a_0=[s,t^3]$. It is
easily seen that $u_*-\id$ is injective on $H_1(D)$ so
$H_2(C)\cong\BZ$ generated by $[s,t^3]$. By an entirely analogous
argument, $H_2(B)\cong\BZ$ generated by $[s,t]$. Since
$f([s,t])=[s,t^3]$, $f$ induces an isomorphism on $H_2$.
But note that $B$ is isomorphic to the quotient of $C$ by the normal
subgroup generated by $[s,t]$. We claim that the kernel of this map
$\pi:C\lra B$ is $C_\om$. First, $[s,t]\in C_\om$ since, assuming
$[s,t]\in C_n$, we see that $[s,t]^3\equiv[s,t^3]\equiv0$ modulo
$C_{n+1}$ and $[s,t]^{2^n}\equiv[s,t^{2^n}]\equiv0$ modulo $C_{n+1}$
since $t$ is a generalized $2$-element. Moreover $B$ is residually
nilpotent since it has a semidirect product decomposition
$1\lra\BZ\lra B\lra\BZ*\BZ\lra1$ where $\BZ$ is generated by $t$. It
follows that $B_\om=\{e\}$ just as in the proof of 8.3. Therefore
the length of $B$ is $\om$ ($B$ is not nilpotent since it maps onto
$\BZ*\BZ$).
Finally, by Stallings' theorem, $C_\om/C_{\om+1}$ is isomorphic to
the cokernel of $H_2(C)\overset\pi_*\to\lra H_2(B)$. But
$\pi_*([s,t^3])=[s,t^3]\equiv[s,t]^3$ in $H_2(B)$ so
$C_\om/C_{\om+1}\cong\BZ_3$. Hence the length of $C$ is at least
$\om+1$. \qed
\bpage
It would take more work to find an example which could be realized
by an homology cobordism of $3$-manifolds but this should not be
difficult.
\vskip.4cm
\sub{\S10. Questions and Problems}
\roster
\item"1." Prove that any planar discontinuous group has length at
most $\om$. (see 5.6 and 5.9)
\item"2." Prove that the fundamental group of a compact Seifert
fibered $3$-manifold has length at most $\om$. (see 6.10).
\item"3." What lengths are possible for fundamental groups of
compact $3$-manifolds? Are there restrictions on $G_\a/G_{\a+1}$?
\item"4." Does the fundamental group of the exterior of a ribbon
link in $S^3$ have length $\om$? (see 1.7).
\endroster
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\endref
\endRefs
\baselineskip=10pt
\vskip.7cm
\line{Tim D. Cochran\hfil Kent E. Orr}
\line{Mathematics Department\hfil Mathematics Department}
\line{Rice University\ \ P.O.Box 1892\hfil Indiana
University}
\line{Houston, Texas\hfil Bloomington, Indiana}
\line{77251--1892\hfil 47405}
\end
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