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\topmatter
\title Non-trivial Links and Plats with trivial Gassner Matrices
\endtitle
\author Tim D. Cochran$^{*!}$\endauthor
\thanks *partially supported by the National Science Foundation
\endgraf
!This work was first presented at the regional meeting of the
American Mathematical Society in Denton, Texas, October
1990\endthanks
\abstract We define link-theoretic and plat-theoretic versions of the
Gassner representation and show they are not injective for links of
at least 6~components. In doing so we show how to construct
non-trivial links satisfying all of Joan Birman's conditions
\cite{Bi; p\.~130} for a link which could be the closure of a braid
in the kernel of the Gassner representation. We define a Gassner
matrix for $2n$-plats in such a way to extend the Gassner matrix for
pure $n$-braids. We observe that the classical Gassner
representation for pure braids has a kernel if and only if there is
an element in the kernel which is homotopically essential.
\endabstract
\endtopmatter
\document
\sub{\S0}
Let $B_n$ denote the Artin braid group on ``$n$-strings'' and
$\pb_n$ its normal subgroup consisting of all the pure braids
\cite{Bi, Mo}. These groups have been considerably scrutinized by
both topologists and algebraists \cite{BL}. One question whose answer
has so far eluded us is whether or not the
{\it Gassner representation\/} $G: \pb_n\lra M_{n\x n}(\La)$, into
the group of $n$-by-$n$ matrices over
$\La=\BZ[t^{\pm1}_1,\dots,t^{\pm1}_n]$, is faithful (see \S1)
\cite{Bi; \S3.3} \cite{Ga}. Recently the less discriminating
{\it Burau representation\/} $B: \pb_n\lra M_{n\x n}(\BZ[t^{\pm1}])$
was shown to have a non-trivial kernel for each $n\ge6$ \cite{M, LP}
but these techniques have not yet yielded an element of kernel(G).
This paper is a partial step in that direction.
The connection between pure braids and links is strong. Thus Joan
Birman in 1973 \cite{Bi; pp\.~130} states that if there exists a
non-trivial element $\b\in\pb_n$ such that $G(\b)=I$, then the
$n$-component link $\hat\b$ obtained by ``closing'' the braid would
satisfy the following:
\roster
\item"i)" $\hat\b$ has at least 4 components, each of which is
unknotted and which are pairwise unlinked.
\item"ii)" $\pi_1(S^3-\hat\b)$ can be generated by $n$ meridional
elements.
\item"iii)" a certain specific Alexander matrix for $\hat\b$ is a
matrix of zeros.
\endroster
In particular the Alexander {\it module\/} of $\hat\b$ would be
trivial, that is isomorphic to that of the trivial $n$-component
link. Thus one can search for links which satisfy these criteria.
Birman goes on to remark that ``the only known class of links with
zero Alexander polynomial, the so-called
{\it homology boundary links,\/} cannot satisfy ii) unless the link
is trivial (since an $n$-component link is a homology boundary link
if and only if its group $G$ maps onto $F_n$, the free group of
rank~$n$, so that if $G$ were already a quotient of $F_n$, it would
follow that $G\cong F_n$). In 1981 Jonathan Hillman exhibited
ribbon links with trivial Alexander's module \cite{H1} (see also
\cite{La}). These also cannot satisfy ii) (see \S4). In 1986, the
author discovered new classes of links which have trivial Alexander's
modules and in this paper we show that these actually satisfy all
of Birman's conditions \cite{C1, C2}. If any of these links is a
``{\it pure link\/}'', that is if any of these $n$-component links
is actually isotopic to $\hat\b$ for an $n$-string braid $\b$, then
the Gassner representation is far from faithful. We note that
condition ii) above is the only (pre-Vaughn Jones) general
condition on a link to be a pure link \cite{Bi; p\.~96, Appendix
\#11 p\.~217} and so the links presented herein satisfy it. However
one need only consider the difference between the class of 2-bridge
links and the class of pure 2-braids to see that ii) is still far
from characterizing a pure link. In fact preliminary calculations
using the HOMFLY polynomial (truncated), done for me by Jim Hoste,
have revealed that the examples of Figure~3.3 are {\it not\/} ``pure
links''.
In the present paper we formalize these results in the following
way. Viewing ii) as essentially characterizing the class of
$n$-component {\it $n$-bridge\/} links, we put a group structure
(loosely speaking) on the set of $n$-component $n$-bridge links by
considering the group of $2n$ plats $\SP_{2n}$ in such a way that
$\pb_n$ is a subgroup of $\SP_{2n}$ \cite{Bi; p\.~192}. Then we
extend the classical Gassner representation to a Gassner function
$G: \SP_{2n}\lra M_{n\x n}(\La)$. One then sees that a link
satisfying Birman's conditions will yield an element
$\rho\in\SP_{2n}$ such that $G(\rho)=I$. We then describe how to
construct such examples for $n\ge6$.
In Section 4 we rephrase the question of the faithfulness of the
Gassner representation in terms of Milnor's $\bar\mu$-invariants.
This allows us to see that our examples are, in a sense, the
simplest possible examples. We close with several interesting open
questions from this point of view.
\bpage
\sub{\S1. A Gassner Matrix for Plats}
Let $B_n$ denote the Artin braid group on ``$n$ strings'' and
$\pb_n$ its normal subgroup consisting of all braids whose
corresponding permutation is the identity \cite{Bi, Mo}. Recall
that the braid group has a faithful representation as a group of
right automorphisms of a free group $F=F\axdots$ where if $\hats$
denotes the automorphism corresponding to $\s\in\pb_n$ then
$\hats(x_i)=\la_i x_i\la^{-1}_i$ for some $\la_i\in F$ for
$i=1,\dots,n$ \cite{Bi; Corollary~1.83}. We may assume that the
exponent sum of $x_i$ in $\la_i$ is zero. Let $\La$ denote
$\BZ[t^{\pm1}_1,\dots,t^{\pm1}_n]$, that is, the integral group ring
of the abelianization of $F$. Recall the
{\it Gassner representation\/} $G: \pb_n\lra M_{n\x n}(\La)$ which
is given by $G(\s)=\pi\(\f{\p\hats(x_i)}{\p x_j}\)$ where
$\p/\p x_j$ is Fox's free derivative and $\pi$ is induced by the
homomorphism $\pi: \BZ F\lra\La$ with $\pi(x_i)=t_i$
\cite{Bi; 3.1--3.2}. For the reader's convenience, we recall that
the Fox derivative $D_i=\p/\p x_i$ is determined by the rules:
$D_i(zw)=D_i(z)+zD_i(w)$, $D_i(1)=0$, $D_i(x_j)=\d_{ij}$ and
$D_i(x^{-1}_i)=-x^{-1}_i$.
\midinsert
\vspace{1.2in}
\botcaption{Figure 1}
\endcaption
\endinsert
We define a {\it pure $2n$-plat\/} $\rho$ to be merely a pure
$2n$-braid and denote the group of such as $\SP_{2n}$. Thus the
group $\SP_{2n}$ is ``identical'' to $\pb_{2n}$ and there is the
standard embedding $\pb_n\overset i\to\hra\SP_{2n}=\pb_{2n}$. However
a plat differs from a braid in the definition of its closure. The
{\it closure\/} $L(\b)$ of a {\it pure $n$-braid\/} $\b$ can be
defined to be the {\it oriented\/}, ordered $n$-component link
obtained by closing the braid in the standard fashion and
considering all strings to be oriented upwards. The {\it closure\/}
$L(\rho)$ of a {\it pure $2n$-braid\/} $\rho$ we shall define as
the oriented, ordered $n$-component link obtained by closing the
$2n$-braid as shown in Figure~1, and considering the orientation
induced by allowing the first $n$ strands of the $2n$-braid to go
``up'' and the second $n$ strands to go ``down''. The reader must
be warned that the ``usual'' definition of closing a plat is as in
Figure~2 \cite{Bi}. Our definition is more convenient for our
purposes but is by no means substantively different. The following
are trivial consequences of the definitions.
\midinsert
\vspace{1.8in}
\botcaption{Figure 2}
\endcaption
\endinsert
\proclaim{Proposition 1.1} If $\b$ is a pure $n$-braid then $L(\b)$
is isotopic to $L(i(\b))$ as oriented links.
\endproclaim
\bpage
\proclaim{Proposition 1.2} An $n$-component oriented link $L$ is
isotopic to the closure of a pure plat $\rho\in\SP_{2n}$ if and
only if $L$ is an $n$-component $n$-bridge link.
\endproclaim
\bpage
Thus we would like to think of $\SP_{2n}$ as one way of imposing a
group structure on the set of $n$-component $n$-bridge links in
such a way that $PB_n$ is a natural subgroup.
We shall now extend the notion of Gassner Matrix to elements of
$\SP_{2n}$ in such a way as to retain the strong relationship
between $G(\rho)$ and an Alexander matrix for $L(\rho)$ (see \S2).
Given $\rho\in\SP_{2n}$, consider $\rho$ as a $2n$-braid to which
is associated an automorphism $\hat\rho(x_i)=\lax$
$i=1,\dots,2n$. Recall that $\hat\rho$ determines the set
$\{\la_i x^n_i\mid n\in\BZ\}$ but not $\la_i$ itself. To clarify the
definition of $\la_i$, we make the following conventions. Choose
$\la_i$ such that the exponent sum of $x_i$ in $\la_i$ equals the
exponent sum of $x_i$ in $\la_{2n-i+1}$ (the latter exponent
{\it is\/} determined by $\hat\rho$), and the exponent sum of
$x_{2n-i+1}$ in $\la_{2n-i+1}$ is the exponent sum of $x_{2n-i+1}$
in $\la_i$. This ensures that, in $\la^{-1}_{2n-i+1}\la_i$, the
exponent sums of $x_i$ and $x_{2n-i+1}$ are zero. Let
$\phi: F\left\lra F\axdots$ be given by
$\phi(x_j)=x_j$ and $\phi(x_{2n-j+1})=x^{-1}_j$ for $1\le j\le n$.
\bpage
\flushpar{\bf Definition 1.3}: The Gassner Matrix $G(\rho)$ of the
pure $2n$-plat $\rho$ is
$$
G_{ij} = \pi\(\f{\p\phi(\la^{-1}_{2n-i+1}\lax\la_{2n-i+1})}
{\p x_j}\)
$$
for $1\le i$, $j\le n$.
\bpage
\proclaim{Proposition 1.4} The function
$G: \SP_{2n}\lra M_{n\x n}(\La)$ extends the Gassner representation
$G: PB_n\lra M_{n\x n}(\La)$.
\endproclaim
\bpage
\flushpar{\bf Proof of 1.4}: If $\b\in PB_n$ with
$\hat\b(x_i)=\a_i x_i\a^{-1}_i$ and $i(\b)=\rho$, then clearly
$\hat\rho(x_i)=x_i$ for $i>n$, and $\la_i$ for $i\le n$ is merely
$\a_i$. Thus, with reference to 1.3, $\la_{2n-i+1}=e$ and
$\phi(\la_i)=\a_i$, implying the result. \qed
\bpage
Unfortunately $G$ does not appear to be a homomorphism and several
attempts at modifying it to be such have failed.
It is not reasonable to expect that $G$ is injective. In fact if
$\b$ is any $n$-braid and $\rho$ is the $2n$-plat of Figure~3, then
$G(\rho)=I$. Note that the closure of $\rho$ is a trivial link. The
appro-
\midinsert
\vspace{1.5in}
\botcaption{Figure 3}
\endcaption
\endinsert
\flushpar priate question is ``Is there a non-trivial
$n$-component $n$-bridge link $L$ such that for some $2n$-plat
representation $\rho$, $G(\rho)=I$?'' This link-theoretic version of
the question of the injectivity of the Gassner representation was
essentially formulated by Birman \cite{Bi; page~130}. In \S2 and \S3
we will see that there are many such links, answering Birman's
question in the negative. The simplest of such can be shown to {\it
not\/} be the closures of $n$-braids so we have not, as of this
writing, been able to decide whether or not the classical Gassner
representation is faithful.
\newpage
\sub{\S2. Relationship of Gassner matrix to the Alexander matrix}
It is well known that the Gassner matrix of a pure braid $\b$ is
equal to the sum of the identity matrix and a certain matrix
presenting the Alexander module $A(L(\b))$ of the closure of $\b$
\cite{Bi~3.10, Mo~13.7}. We shall extend these results to our more
general situation and use these in \S3 to establish the existence of
non-trivial $n$-component $n$-bridge links and pure $2n$-plats with
trivial Gassner matrix.
\bpage
\proclaim{Proposition 2.1} Let $\rho\in\SP_{2n}$ and suppose that
the action of $\rho$ on the free group
$F\left$ is given by $\hat\rho(x_i)=\lax$
$1\le i\le 2n$. Then the fundamental group $\pi_1(S^3-L(\rho))$
admits the presentation:
$$
\left
$$
where $\phi$ is as in \S1.
\endproclaim
\bpage
\flushpar{\bf Proof of 2.1}: Decompose $S^3$ via a $2$-sphere which
slices through the level $\f12$ of the braid. Let $A$, $B$ denote
the resulting $3$-balls. Clearly $\pi_1(A-L(\b))$ has a presentation
$\left$
and $\pi_1(B-L(\rho))$ has a presentation\newline
$\left$.
An application of the Seifert Van-Kampen theorem yields the
presentation
$\left$ for
$\pi_1(S^3-L(\rho))$. Eliminating $z_i$ yields the relations
$\{x_i=x^{-1}_{2n-i+1},\ \lax=\mathbreak
\la_{2n-i+1}x^{-1}_{2n-i+1}
\la^{-1}_{2n-i+1}\mid 1\le i\le n\}$. Eliminating
$\{x_{2n-i+1}\mid 1\le i\le n\}$ yields the desired presentation.
\qed
\bpage
Recall that the {\it Alexander matrix\/} associated to the
presentation $\left$ is defined by
$A_{ij}=\pi\(\f{\p r_i}{\p x_j}\)$ \cite{F; p\.~100} and, in the
case that $\left\cong\pi_1(S^3-L(\rho))$, presents
the Alexander's module $A(\rho)$ of $L(\rho)$ \cite{Cr; \S3}.
\bpage
\proclaim{Proposition 2.2} If $\rho\in\SP_{2n}$ then $G(\rho)-I$ is
the Alexander matrix of $L(\rho)$ associated to the presentation of
2.1.
\endproclaim
\bpage
\flushpar{\bf Proof}: Note that $r_i=\g_ix^{-1}_i$ where
$\g_i=\phi\(\la^{-1}_{2n-i+1}\lax\la_{2n-i+1}\)$. Then
$\f{\p r_i}{\p x_j}=\f{\p\g_i}{\p x_j}+\g_i\f{\p x^{-1}_i}{\p x_j}=
\f{\p\g_i}{\p x_j}-\g_ix^{-1}_i\d_{ij}$ where $\d_{ij}=1$ only when
$i=j$. Since $\pi(\g_i)=1\cd e=1$ in $\La$,
$A_{ij}=\pi\(\f{\p\g_i}{\p x_j}\)-\d_{ij}=G_{ij}-\d_{ij}$, as
desired. \qed
\bpage
Since the triviality of the Gassner matrix is now seen to entail
the Alexander matrix of a specific presentation being the zero
matrix, we investigate when the latter may occur. First we observe
that this is only a property of the link, not of the particular
$n$-bridge presentation.
\bpage
\proclaim{Proposition 2.3} The Alexander matrix for
$\left<\xdots\mid r_1,\dots,r_n\right>$ is a zero matrix if and
only if {\it any\/} $n\x n$ Alexander matrix presenting the Alexander
module of the associated group is a zero matrix.
\endproclaim
\bpage
\flushpar{\bf Proof of 2.3}: Let
$\pi\cong\left<\xdots\mid r_1,\dots,r_n\right>$. By definition
there is a short exact sequence
$0\lra\text{image}(\wh A)\overset i\to\hra\La^n\overset f\to\lra
A\lra 0$ where $\wh A: \La^n\lra\La^n$ is given by multiplication
by the Alexander matrix and $A$ is the Alexander module
\cite{Tr; \S3}. If for one presentation $\wh A=0$ then clearly
$A\cong\La^n$. But the isomorphism class of the Alexander module is
an invariant of the isomorphism class of the group $\pi$ so for any
presentation of the group, $A\cong\La^n$. But $f$ is then an
epimorphism between finitely-generated free modules of identical
rank over a Noetherian ring, hence an isomorphism. Thus
image~$(\wh A)=0$ and $A_{ij}$ is always a matrix of zeros. \qed.
\bpage
\proclaim{Proposition 2.4} Suppose the exponent sum of $x_i$ in
$w_i$ is zero for all $1\le i\le n$. Then the Alexander matrix of
$\left$ is zero if and only if
$w_i\in F''$ (the second derived) for each $i$.
\endproclaim
\bpage
\flushpar{\bf Proof of 2.4}: The proof of Theorem 3.16 of
\cite{Bi} suffices. \qed
\bpage
The following extends \cite{Bi; Thm\.~3.16} and
\cite{CL; Thm\.~3.10} to our present situation.
\proclaim{Theorem 2.5} Suppose $L$ is an $n$-component link which
admits an $n$-bridge presentation or suppose that $L=L(\rho)$ is
the closure of a pure $2n$-plat $\rho$. Then the Gassner matrix of
$L$ (equivalently of $\rho$) is the identity if and only if the
longitudes of $L$ lie in the second derived subgroup of
$\pi_1(S^3-L)$ (equivalently if and only if
$w_i=\phi(\la^{-1}_{2n-i+1}\la_i)$ lies in $F''$ for all
$1\le i\le n$).
\endproclaim
\bpage
\flushpar{\bf Proof of 2.5}: Suppose the Gassner matrix of $\rho$
is the identity. Then by 2.2 the associated Alexander matrix for
$L(\rho)$ is the zero matrix. By \cite{C2, Thm\.~3.1},
$\pi_1(S^3-L(\rho))\equiv\pi$ has a presentation of the form
$\left$ where $v_i$ represents an
i$^{\text{th}}$ longitude $l_i$ of the link. By 2.3 the Alexander
matrix associated to {\it this\/} presentation is the zero matrix
and by 2.4 it follows that $v_i\in F''$ for each $i$ (since it is
well known that if $l_i\notin F'$ then some linking numbers are
non-zero and the Alexander matrix cannot be zero \cite{Bi;
Cor\.~3.14.1}.) It follows immediately that $l_i\in\pi''$. On the
other hand, by 2.1, $\pi$ has a presentation $\left$ where $w_i=\phi(\la^{-1}_{2n-i+1}\la_i)$ so by 2.4
$w_i\in F''$. For the other implication, suppose that the longitudes
of $l_i$ lie in $\pi''$. It follows that $F/F''\approxeq\pi/\pi''$
and hence that the $v_i\in F''$ (see presentation above) \cite{C2,
see proof of 3.2 from 3.1}. Apply 2.4 to see that the Alexander
matrix associated to $\left$ is the zero
matrix then apply 2.3 and 2.2 to yield the desired ``triviality'' of
the Gassner matrix. \qed
\newpage
\sub{\S3. $n$-component $n$-bridge links with trivial Gassner
matrix}
We review a construction of \cite{C2, \S1}.
\midinsert
\vspace{2.4in}
\botcaption{Figure 4}
\endcaption
\endinsert
Let $A$ be the set of letters $\{x_i\mid i=1,\dots,n\}$. A
{\it 1-bracket\/} is an element of $A$. An {\it m-bracket\/} is a
symbol $(\b_1,\b_2)$ where $\b_1$ is a k-bracket $k\ge 1$ and
$\b_2$ is an $m-k$ bracket $(m-k\ge 1)$. We say that the
{\it weight\/} of an m-bracket is $m$. To each pair of brackets
$\{\b_1,\b_2\}$ we shall associate an unoriented link with
$n_1+n_2$ components (where $w(\b_1)=$~weight~$(\b_1)=n_1$ and
$w(\b_2)=n_2$). Begin with the Hopf link with the components labelled
by $\b_1$, $\b_2$ respectively. If $w(\b_1)\neq1$ then
$\b_1=(\b_{11}, \b_{12})$ and we describe a procedure for replacing
that component by two components labelled by $\b_{11}$ and $\b_{12}$
respectively. Continuing this procedure (applied to $\b_2$ also)
until all labels have weight 1 yields a link with $w(\b_1)+w(\b_2)$
components. The procedure is shown in Figure~4 and is usually
called ``Bing-Doubling''. There is a choice of clasp which is not
important to our present considerations. Any such link
$L(\b)=L(\{\b_1,\b_2\})$ is clearly an $n$-component $n$-bridge link
where $n=w(\b_1)+w(\b_2)$.
Define a {\it generalized simple bracket\/}, GSB, to be any bracket
of the form $(x_i,\a)$ or $(\a,x_i)$ where $\a$ is a GSB of lesser
weight, and a GSB of weight~1 is merely any 1-bracket. Thus all
m-brackets are GSB for $m=1$, $2$, $3$ but $((x_1,x_2), (x_3,x_4))$
is not a GSB. In \cite{C2} the following criterion was derived for
when the longitudes of $L(\{\b_1,\b_2\})$ lie in $G''$.
\bpage
\proclaim{Theorem 3.2} {\rm(2.2 of \cite{C2})} Suppose $(\b_1,\b_2)$
contains no repeated letters, $w(\b_i)>1$ for $i=1$, $2$, and at
least one of $\b_1$, $\b_2$ is not a generalized simple bracket.
Then the $n$-component $n$-bridge link $L(\{\b_1,\b_2\})$ described
above has the property that its longitudes lie in $\pi''$.
\endproclaim
\topinsert
\vspace{3.4in}
\botcaption{Figure 5}
\endcaption
\endinsert
\bpage
\flushpar{\bf Examples}: The examples of minimal weight are several
6-component links corresponding to $\b_1=(x_1,x_2)$
$\b_2=((x_3,x_4), (x_5,x_6))$ which are isotopic to those obtained
from the Borromean Rings by ``Bing-Doubling'' each component and
assigning orientations arbitrarily. Several distinct ordered,
oriented links arise in such a way. One is shown in Figure~5.
More generally let $\b_1=(x_1,x_2)$ and $\b_2=((x_3,x_4),\s)$ where
$\s$ is any bracket of weight $(n-4)$, $n\ge 6$, whose underlying
sequence is $x_5x_6\dots x_n$ or any permutation of that sequence.
This link is shown in Figure~6.
All of the links which arise as in Theorem 3.2 are non-trivial, in
fact none is link homotopic to a trivial link because the
Bing-Double of any homotopically essential link is homotopically
essential \cite{C1; Corollary 8.2}. This reference also shows
the equivalent fact that Milnor's invariant $\ov\mu(i_1i_2\dots
i_n)$, for some permutation of $\{1,\dots,n\}$, will be $\pm1$ for
the $n$-component links of Theorem~3.2. In summary we have shown:
\bpage
\proclaim{Theorem 3.5} Suppose $\rho$ is a pure $2n$-plat arising
from the Brunnian $n$-component $n$-bridge link $L(\{\b_1,\b_2\})$
as in Theorem~3.2. Then $\rho$ and $L(\{\b_1,\b_2\})$ are non-trivial
(even homotopically essential) but have trivial Gassner matrices. If
any such $L(\{\b_1,\b_2\})$ were the closure of a pure $n$-braid
then the Gassner representation on pure $n$-braids would have a
non-trivial kernel. In particular each of these links satisfies all
the conditions i)--v) of \cite{Bi, p\.~130}.
\endproclaim
\midinsert
\vspace{3in}
\botcaption{Figure 6}
\endcaption
\endinsert
%This space is inserted to force section 4 to begin on a new page
\midinsert
\vspace{3.8in}
%\botcaption{Figure 6}
%\endcaption
\endinsert
\newpage
\sub{\S4. Relations with Milnor's $\bar\mu$-invariants and Massey
products}
In this section we translate the question of the faithfulness of
the classical Gassner representation to the language of some
well-studied link invariants, Milnor's $\bar\mu$-invariants. This
allows us to see that examples of the sort suggested by Theorem~3.5
would yield (if they could be realized by pure links) the simplest
possible (from this other point of view) examples of braids in the
kernel of the classical Gassner representation, and {\it are\/} the
simplest examples for our {\it extended\/} Gassner representation.
Many aspects of this translation to $\bar\mu$-invariants were known
to Jon Hillman \cite{Hi2} and perhaps others, but they are
certainly not as generally known as they deserve to be. Also,
Birman's question 20 \cite{Bi; p\.~218} translates the faithfulness
question to a purely algebraic question about elements of the free
group which is intimately related to the present discussion.
Let us briefly recall the ideas and notation behind Milnor's
$\bar\mu$-invariants for an ordered, oriented, $n$-component link
$L$ in $S^3$. Given a positive integer $k$, called the weight, and
a sequence $I=i_1\dots i_k$ of elements of the set $\{1,\dots,n\}$
Milnor's invariant $\bar\mu(i_1\dots i_k)$ for $L$ is an integer
(well defined modulo the ideal generated by certain
$\bar\mu$-invariants of lesser weight). These invariants of weight
$k$ detect precisely if the longitudes of $L$ lie in the
k$^{\text{th}}$ term of the lower central series $G_k$ of the link
group $G=\pi_1(S^3-L)$.
Our first result says that for the class of $n$-bridge
$n$-component links, the property of being {\it trivial\/} is
equivalent to the vanishing of all the $\bar\mu$-invariants. This
would certainly be false for more general links since, for example,
the $\bar\mu$-invariants of any boundary link are zero.
\bpage
\proclaim{Theorem 4.1} Suppose $\rho\in\SP_{2n}$ (or $\s\in\pb_n$).
Then the following are equivalent.
\roster
\item"a)" $L(\rho)$ (or $L(\s)$) is a trivial $n$-component link
\item"b)" $G=\pi_1(S^3-L(\rho))$ (or $L(\s)$) is free of rank~$n$
\item"c)" the longitudes of $L(\rho)$ (or $L(\s)$) are trivial in
$G$
\item"d)" the longitudes of $L(\rho)$ (or $L(\s)$) lie in $G_k$ for
each $k\in\BZ_+$
\item"e)" Milnor's invariants $\bar\mu(i_1\dots i_k)$ vanish for $L$
for any $I=i_1\dots i_k$
\item"f)" all Massey products in $S^3-L$ are trivial
\item"g)" $\s$ is the trivial braid
\endroster
\endproclaim
\bpage
\flushpar{\bf Proof of 4.1}: The following implications are trivial:
$g\Rightarrow a\Rightarrow c\Rightarrow d\Leftrightarrow e$. The
following are well-known $a\Leftrightarrow b\Leftrightarrow c$,
$e\Leftrightarrow f$ \cite{Po, Tu}. It suffices therefore to show
$d\Rightarrow b$ and, in the case of braids, $b\Rightarrow g$.
Suppose $d$ holds for $L(\rho)$ (or $L(\s)$). By \cite{C2,
Thm\.~3.1}, $G$ has a presentation of the form
$$ where $w_i$ represents an
i$^{\text{th}}$ longitude of $L$. We shall establish, by induction,
that $w_i\in\bigcap^\infty_{k=1}F_k$, the intersection of the terms
of the lower-central series of $F\ngth$. Since this
intersection is well-known to be trivial, $b$ would follow.
Suppose, by induction that $w_i\in F_k$ for each $i$. Since $l_i$,
the i$^{\text{th}}$ longitude, lies in $G_{k+1}$, $w_i$ must lie in
the normal subgroup generated by $F_{k+1}$ and the relations
$[x_i,w_i]$. But since $[x_i,w_i]\in F_{k+1}$, it follows that
$w_i\in F_{k+1}$. This completes the inductive step establishing
$d\Rightarrow b$. Now suppose $\s$ is a braid such that $G$ is
free. Since $G$ has a presentation $$ where $\hat\s(x_i)=\la_ix_i\la^{-1}_i$ and the
exponent sum of $x_i$ in $\la_i$ is zero (a result of Artin
\cite{Bi; 2.2}), it follows that $[x_i,\la_i]=e$ for each $i$ (free
groups are Hopfian). The restriction on the exponent sum then
implies $\la_i=e$ for all $i$ which implies $\hat\s$ is the identity
endomorphism. It follows immediately that $\s$ is trivial \cite{Bi;
1.8.3}. \qed
\bpage
This is to be contrasted with our 2.5 and Birman's 3.16 \cite{Bi}
which say that the Gassner Matrix is trivial if and only if the
longitudes of the associated link lie in $G''=[G_2,G_2]$. Thus the
translation of the question of the faithfulness of the Gassner
representation to links is ``If the Alexander module of a link is
trivial, does it have vanishing $\bar\mu$-invariants?''. It is
known that the Alexander module controls {\it some\/}
$\bar\mu$-invariants. For example, for a 2-component link, if the
Alexander polynomial is trivial then each
$\bar\mu(111\dots12\dots2)$ is zero \cite{Sm}. In fact Lorenzo
Traldi has worked out a presentation matrix for the Alexander
module in terms of the $\bar\mu$-invariants \cite{Tr}. Each entry
of this matrix involves a certain ``symmetric sum'' of
$\bar\mu$-invariants so that if the Alexander matrix is the zero
matrix, one gets systems of equations. Linear dependencies among the
$\bar\mu$-invariants provide further equations. Using these facts
Traldi achieved some limited but interesting results about the
vanishing of $\bar\mu$-invariants. Nonetheless the author settled
the long-standing question in \cite{C2, C1} by exhibiting many
links with trivial Alexander module and {\it non-vanishing\/}
$\bar\mu$-invariants. Thus many $\bar\mu$-invariants are simply
{\it not\/} controlled by the Alexander module. Therefore the
answer to the Gassner question translated to arbitrary links is
decidedly ``No''. Links are much more non-abelian than the
Alexander module can capture. However these examples are in the
category of {\it arbitrary\/} links. Therefore the question
remaining is
\sub{Question 4.2} If $L$ is an {\it $n$-bridge\/} $n$-component
link (respectively the closure of a pure $2n$-plat, respectively the
closure of a pure $n$-braid) what restrictions are placed on the
$\bar\mu$-invariants?
Theorem 3.5 shows that in the category of {\it $2n$-plats\/},
$\bar\mu(123\dots n)$ (or any permutation) is {\bf not} controlled
by the Alexander module so the analogue of the Gassner
representation here has a ``kernel''. Indeed these are the simplest
possible examples within this class as the following shows. Recall
that a Brunnian link (or braid) is one for which any proper sublink
(proper subbraid) is trivial.
\bpage
\proclaim{Proposition 4.3} Suppose $L$ is a non-trivial
$n$-component link which is the closure of the plat
$\rho\in\SP_{2n}$ whose Gassner matrix is trivial, and suppose $n$
is minimal with respect to these properties. Then $L$ is a Brunnian
link (if $L=L(\b)$ then $\b$ is a Brunnian braid \cite{Bi;
Theorem~3.18}) and so all $\bar\mu$-invariants of weight less than
$n$ vanish as do all $\bar\mu$-invariants $\bar\mu(I)$ where $I$
fails to contain each element of $\{1,\dots,n\}$ at least once.
Therefore the $\bar\mu$-invariants of least weight which might fail
to vanish are $\bar\mu(I)$ where $I$ is a permutation of the
sequence $12\dots n$.
\endproclaim
\bpage
\flushpar{\bf Proof of 4.3}: Suppose $G(\rho)=I$. By 2.5 the
longitudes of $L$ lie in the second derived $G''$. Suppose $J$ is a
proper sublink of $L$. It is easy to see that the longitudes of $J$
lie in its second derived group so, by 2.5 and minimality, $J$ is a
trivial link. But Milnor himself showed that $\bar\mu(I)$ depends
only on the sublink $J$ consisting of those components of $L$ whose
corresponding index appears in $I$. Therefore if some index is
missing from $I$, $\bar\mu(I)$ is zero since $J$ is trivial. \qed
\bpage
Moreover, Traldi has shown that these first-possible non-vanishing
invariants $\{\bar\mu(i_1\dots i_n) \mathbreak
\mid i_1\dots i_n$ a permutation
of $1\dots n\}$ vanish when the Alexander Module is trivial in the
cases of fewer than 6 components \cite{Tr; p\.~423 last paragraph}.
Thus the example of Figure~5 is the simplest possible example of this
phenomenon.
The reader will recall that these $\bar\mu$-invariants (with no
repeated indices) vanish precisely when the link is link-homotopic
to the trivial link (see for example \cite{HL}). The connection to
link homotopy is fascinating. For example we can argue that these
homotopy invariants are the only ones we need concern ourselves
with.
\bpage
\proclaim{Proposition 4.4} Suppose that the Gassner representation
for pure braids has a non-trivial kernel. Then there exists a pure
braid in the kernel of the Gassner representation which is
homotopically essential.
\endproclaim
\bpage
\flushpar{\bf Proof of 4.4}: Let $\b$ be a non-trivial pure braid
such that $G(\b)=I$ and such that $\b$ is of minimal braid
index~$n$. By 4.1 there is some sequence $I=i_1i_2\dots i_k$ of
letters from $\{1,\dots,n\}$ such that $\bar\mu(I)$ is non-zero for
$L(\b)$ and such that all $\bar\mu$-invariants of weight less than
$k$ vanish. Suppose that there are $n_j$ occurences of the letter
$j$ in $I$. Form a new braid $\b'$ by replacing the j$^{\text{th}}$
strand by $n_j$ parallel strands. Since $G(\b)=I$, all linking
numbers are zero, so that the obvious choice of parallel is
0-framed. Milnor showed that the $\bar\mu$-invariants of $L(\b')$
are closely related to those of $L(\b)$ \cite{Mi; Thm\.~7}. In
particular if $I'$ is the sequence obtained from $I$ by replacing
the occurences of the letter $j$ by $\{j_1,\dots,j_{n_j}\}$ then
$\bar\mu(I')$ for $L(\b')$ equals $\bar\mu(I)$ for $L(\b)$. If $J$
is any sequence with {\bf no repeated indices} for the link $L(\b')$
whose weight is less than that of $I'$, application of the same
theorem reveals that $\bar\mu(J)$ is equal to a $\bar\mu$-invariant
of weight less than $k$ evaluated on $L(\b)$, and is thus zero. It
follows that $L(\b')$ is homotopically non-trivial \cite{Mi; \S5}.
Moreover, by 2.5, to show $G(\b')=I$, it suffices to show the
longitudes of $L(\b')$ lie in the second derived of its link group.
But it is an easy exercise to see that if the longitudes of $L(\b)$
lie in the second derived of its group then the same will be true
of any parallel link. \qed
\bpage
Note that the braid $\b'$ of 4.4 is certainly not Brunnian since
$\b$ is a proper sub-braid.
The question of which $\bar\mu$-invariants may be realized as the
first non-vanishing $\bar\mu$-invariants of a {\it pure\/} link, is
fascinating, even without considering the Alexander module. My
first intuition was that only the homotopy invariants could be
realized but this is only true for $n=2$. In fact Deborah Goldsmith
long ago characterized what braids were null-homotopic (see
\cite{G}). In fact I do not know {\it any\/} restrictions on the
$\bar\mu$-invariants of a {\it pure\/} 3-component link.
\newpage
\sub{\S5. Questions}
\roster
\item"1." Which $\bar\mu$-invariants can be realized (as the
first non-vanishing $\bar\mu$-invariant) by pure links, that is
links which are closures of pure braids? More generally what are
restrictions on the $\bar\mu$-invariants of a pure link?
\item"2." The construction of \S3 uses a link-theoretic
construction, Bing-Doubling, to create links with desired
$\bar\mu$-invariants. The set of plats is closed under this
operation but not the set of pure links. Is there a braid-theoretic
operation which can be used to create braids with desired
$\bar\mu$-invariants? For example it follows from recent work of
T\.~Stanford, X.S\.~Lin and D\.~Bar-Natan, that a braid lying in the
k$^{\text{th}}$ term of the lower central series of the pure braid
group will have vanishing $\bar\mu$-invariants up to a high weight
depending on $k$ \cite{BN, Li2, St}. It should not be so difficult
to make this more explicit.
\item"3." Is there a non-trivial $2n$-plat with trivial Gassner
matrix for $n=4$ and $n=5$?
\item"4." What is the inverse image of the identity for the Gassner
function on $\SP_{2n}$?
\endroster
\newpage
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\endRefs
\vskip.9cm
\rightline{Tim D. Cochran}
\rightline{cochran\@rice.edu}
\rightline{Mathematics Department}
\rightline{Rice University, P.O. Box 1892}
\rightline{Houston, Texas 77251}
\end