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\begin{document}
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\title{Expansions and Quadraticity for Groups}
\author{Dror~Bar-Natan}
\address{
Department of Mathematics\\
University of Toronto\\
Toronto Ontario M5S 2E4\\
Canada
}
\email{\href{mailto:drorbn@math.toronto.edu}{drorbn@math.toronto.edu}}
\urladdr{\url{http://www.math.toronto.edu/~drorbn}}
\subjclass[2010]{57M25, 20E99}
\keywords{
Expansions,
Group Rings,
Groups,
Homomorphic Expansions,
Projectivization,
Quadratic Algebras
}
\thanks{This work was partially supported by NSERC grant RGPIN 262178.}
\date{first edition in future, this edition \today. The
\arXiv{????.????} edition may be older. Electronic version and related
files at \url{\web}.}
\begin{abstract}
\input abstract.tex
\vskip 5mm
This paper is written at two formality levels. Material in ``print'' fonts
is at the fully formal level. {\hand Material in handriting fonts is at
conversation level. At this level I write what I think is true, but at
times it may be imprecise, incomplete, or plain wrong.}
\end{abstract}
\maketitle
\setcounter{tocdepth}{3}
\begin{multicols}{2}
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\tableofcontents
\draftcut
\section{Introduction}
\subsection{The Taylor Expansion}
Before the real thing, which commences in Section~\ref{subsec:Expansions},
we start with a brief reminder on the classical Taylor expansion, which
serves as a motivation for Section~\ref{subsec:Expansions}.
Let $\tilde{R}=C^\infty(V)$ the the algebra of smooth real-valued
functions on some vector space $V$ over $\bbR$, and let $I$ be the ideal
within $\tilde{R}$ of functions that vanish at $0$: $I\coloneqq\{f\in
\tilde{R}\colon f(0)=0\}$. Let $I^0\coloneqq \tilde{R}$, $I^1\coloneqq I$,
and for $n>1$ let $I^n$ be the $n$th power of $I$: the set of all products
of the form $f_1f_2\dots f_n$, where $f_i\in I$ for every $i$. Then $I^n$
is the space of smooth functions that ``vanish at least $n$ times at
$0$'', and hence the quotient $I^n/I^{n+1}$ is ``functions vanishing
$n$ times, while regarding as $0$ functions that vanish more than $n$
times'', which is precisely ``homogeneous polynomials of degree
$n$''. Thus the space
\setcounter{equation}{-1}
\begin{equation} \label{eq:AR}
\hat\calA(\tilde{R})\coloneqq \prod_{n\geq 0}I^n/I^{n+1}
\end{equation}
can be identified as the space of power series on $V$. The ``Taylor
expansion'' is a linear map $Z_T\colon \tilde{R}\to\hat\calA(\tilde{R})$, and one may
show that it is characterised by the following three properties (see
Proposition~\ref{prop:Taylor} on page~\pageref{prop:Taylor}):
\begin{enumerate}[leftmargin=*,labelindent=0pt]
\item {\em $Z_T$ is an expansion:} If $f\in I^n$ then $Z_T(f)$ begins with
$[f]$, the class of $f$ within $I^n/I^{n+1}$. Namely,
\begin{equation} \label{eq:ZTExpansion}
Z_T(f) = (0,\ldots,0,
\underbrace{[f]}_{\mathclap{\text{in degree $n$}}},
\ast,\ast,\ldots),
\end{equation}
where ``$\ast$'' stands for ``something arbitrary''.
\item {\em $Z_T$ is multiplicative:}
\begin{equation} \label{eq:ZTMultiplicative}
Z_T(fg)=Z_T(f)Z_T(g),
\end{equation}
where the product on the left is the pointwise product of functions and
the product on the right is the product of power series.
\item {\em $Z_T$ is co-multiplicative:} If $g$ is a smooth function of two
variables $x,y\in V$ then its Taylor expansion $Z_T^x(g)$ with respect
to the variable $x$ is a smooth function of the variable $y$ (namely,
each coefficient of each homogeneous polynomial appearing in $Z_T^x(g)$
is a smooth function of $y$). Thus there is an iterated Taylor expansion
$Z_T^{x,y}(g)\coloneqq Z_T^y(Z_T^x(g))$, and it can be interpreted as
taking values in the space of power series in two variables $x,y$. With
all this and with $f\in \tilde{R}$, we have that
\[ Z_T^{x,y}(f(x+y)) = (Z_T(f))(x+y). \]
Alternatively, with $\Box$ denoting the operation $f(x)\mapsto f(x+y)$,
defined on both functions and power series (and doubling the number of
variables in each case), we have that
\begin{equation} \label{eq:ZTCoMultiplicative}
Z_T^{x,y}\circ\Box = \Box\circ Z_T.
\end{equation}
\end{enumerate}
\subsection{General Expansions} \label{subsec:Expansions}
A space ``of power series'' $\hat\calA(R)$ can be defined as in
Equation~\eqref{eq:AR} whenever $R$ is a ring and $I$ is an ideal in
$R$. Yet in this paper we restrict to the case when the ring $R$ is the
group ring of a group $G$.\footnote{So in fact, our motivating example,
the Taylor expansion, is not a special case of our definitions but only
a close associate which is obtained when our definitions are restated
for arbitrary rings. We can make the analogy a bit closer. The algebra
$\tilde{R}=C^\infty(V)$ of functions under pointwise multiplication is
isomorphic, via the Fourier transform and ignoring issues of analysis,
to the algebra $R=C^\infty(V^\ast)$ of functions under convolutions. The
latter algebra $R$ is a continuous version of the group ring of $V^\ast$,
and with this loose identification the definitions in this section match
with the Taylor series example.} So let $G$ be an arbitrary discrete
group whose identity element is denoted $e$. While $G$ is arbitrary,
some groups are more interesting than others, for our purpose. We advise
our readers to inspect the left-most column of Table~\ref{tab:Summary} on
page~\pageref{tab:Summary} to gain a feel for the kind of groups we care
more about. A particularly good example to keep in mind is the pure braid
group $\PB_n$ on $n$ strands, which we discuss in Section~\ref{ssec:PB}.
Let $R\coloneqq\bbQ
G=\{\sum_{i=1}^ka_ig_i\colon a_i\in\bbQ,\,g_i\in G\}$ be the group
ring of $G$ over the rational numbers $\bbQ$, and let $I=I_G$ be the
augmentation ideal of $\bbQ G$:
\[ I=\left\{\sum a_ig_i\colon\sum a_i=0\right\}
=\left\langle g-e\colon g\in G\right\rangle.
\]
We declare that $I^0$ is $R=\bbQ G$ and also consider all higher powers
$I^n$ of $I$.
\begin{definition} The polynomial algebra\footnote{Quillen has a
paper~\cite{Quillen:OnGrOfAGroupRing} devoted to the study of $\calA(G)$,
but he never names that ring beyond ``the associated graded ring of a group
ring''.} $\calA(G)$ of the group $G$ is the direct sum
\[
\calA(G)\coloneqq \bigoplus_{n\geq 0}\calA(G)_n
\coloneqq \bigoplus_{n\geq 0}I^n/I^{n+1}.
\]
The power series algebra $\hat\calA(G)$ of $G$ is the graded completion
of the polynomial algebra $\calA(G)$ of $G$. Thus
\[ \hat\calA(G) = \prod_{n\geq 0}\calA(G)_n = \prod_{n\geq 0}I^n/I^{n+1}. \]
\end{definition}
We note that the product $I^m\otimes I^n\to I^{m+n}$
induced by the product of $R$ descends to a product
$\mu\colon(I^m/I^{m+1})\otimes(I^n/I^{n+1})\to I^{m+n}/I^{m+n+1}$
and hence $\calA(G)$ is in fact a graded algebra over
$\bbQ$. We denote the identity element $[e]\in I^0/I^1$ of $\calA(G)$ by
$1$, and note that there is a map $G\to\calA_1(G)$ by
$g\mapsto\bar{g}\coloneqq[g-e]\in I/I^2$.
Also note that $\calA$ is a functor: a group homomorphism
$\phi\colon G\to H$ induces a morphism $\phi\colon\bbQ G\to\bbQ H$
for which $\phi(I_G)\subset I_H$ and hence $\phi(I_G^n)\subset I_H^n$
for all $n$. Hence we get an induced map $\phi\colon\calA(G)\to\calA(H)$
which is easily seen to be a morphism of graded algebras. Likewise
$\hat\calA$ is a functor too.
$\hat\calA(G)$ sometimes remembers much of the structure of $G$, and
sometimes forgets much of it, as we shall see below. Yet always, for
any group $G$ whatsoever, it makes sense to seek a ``Taylor expansion
for $G$'' --- a map $Z\colon G\to\hat\calA(G)$ satisfying the three
properties that characterize the ordinary Taylor expansion. These are
Definitions \ref{def:Expansion}, \ref{def:ZMultiplicative}, and
\ref{def:ZCoMultiplicative} below.
\begin{definition} \label{def:Expansion}
A map $Z\colon G\to\hat\calA(G)$ is called ``an expansion''\footnote{I
learned this notion from Xiao-Song Lin's~\cite{Lin:Expansions}.}
if its homonymous uniquely defined linear extension $Z\colon\bbQ
G\to\hat\calA(G)$ has the ``universality''\footnote{The origin of the word
``universality'' is in the subject of finite type invariants of
knots~\cite{Bar-Natan:OnVassiliev}, where the analogous property means that
$Z$ is a ``universal finite type invariant''.} property that if
$f\in I^n$ then $Z(f)$ begins with $[f]$, the class of $f$ within
$I^n/I^{n+1}$. Namely, if
\begin{equation} \label{eq:ZExpansion}
Z(f) = (0,\ldots,0,
\underbrace{[f]}_{\mathclap{\text{in degree $n$}}},
\ast,\ast,\ldots),
\end{equation}
where ``$\ast$'' stands for ``something arbitrary''.
\end{definition}
\vskip -2mm\noindent$\overbracket{\hspace{\columnwidth}}$
\noindent{\bf Aside.} One may think of expansions as ``algorithms for
progressively transmitting further and further details of a mathematical
image''. The best way to visualize that is to recall how real-life
pictures are progressively transmitted over slow communication channels.
Think a picture of Brook Taylor, as it would gradually appear in a web
browser connected to the internet over a slow modem:
\vskip 1mm
\noindent\includegraphics[width=0.23\columnwidth]{figs/Taylor-3.png}%
\hfill\includegraphics[width=0.23\columnwidth]{figs/Taylor-4.png}%
\hfill\includegraphics[width=0.23\columnwidth]{figs/Taylor-5.png}%
\hfill\includegraphics[width=0.23\columnwidth]{figs/Taylor-8.png}
Here the space of all pictures plays the role of $R$, and $I^{n+1}$ is
``the information we allow ourselves to forget in the $n$th transmition
step''. So if $P\in R$ is our picture of Brook Taylor, then its projection
to $R/I^n$ is what we see after the $(n-1)$-st transmission step. For
the next transmission step we need to transmit enough to recover the
projection of $P$ to $R/I^{n+1}$, but it would be wastefull to retransmit
what we have sent before, which is in $R/I^n$. So we wish to transmit the
``new'' information in $R/I^{n+1}$: only that part of $R/I^{n+1}$ that
is $0$ in $R/I^n$. That is the kernel of the projection $R/I^{n+1}\to
R/I^n$, which is $I^n/I^{n+1}$. So we need a map $Z=\prod Z_n\colon R\to
\prod I^n/I^{n+1}$ where $Z_n$ is ``the information tansmitted in step
$n$''. Finally, $Z_n$ should have the property that if $P$ is a picture
all of whose details are forgettable until step $n$ (namely, it is in
$I^n$), then nothing about it should be transmitted until step $n$ and
then on step $n$ we must transmit everything about $P$ that is relevant
to step $n$, and that's precisely the class of $P$ in $I^n/I^{n+1}$. This
last sentence is exactly condition~\eqref{eq:ZExpansion}.
When we see a picture gradually appearing on a web browser, it is
precisely because somebody has already chosen an expansion $Z$ for the
space $R$ of all pictures ($Z$ is not unique and there is no canonical
choice for $Z$).
It is often beneficial to try to find an expansion $Z$ that is compatible
with various operations that one may wish to apply to images $P$:
re-colourations, rotations, or the concatanation of several images.
Expansions that are compatible with the available operations are what
we call ``Taylor expansions'': see Definition~\ref{def:Taylor} below.
\vskip -2mm\noindent$\underbracket{\hspace{\columnwidth}}$\vskip 2mm
\begin{proposition}[proof below] \label{prop:ExpansionsExist}
Any group $G$ has an expansion (in general, non-unique).
\end{proposition}
Hence the real interest is not in expansions in general, but in expansions
with extra properties as in the definitions that follow.
\noindent{\em Proof of Proposition~\ref{prop:ExpansionsExist}.} For
any natural number $n$ the quotient $I^n/I^{n+1}$ is a linear subspace
of $\bbQ G/I^{n+1}$ and hence there is a (non-unique) projection
$p_n\colon\bbQ G/I^{n+1}\to I^n/I^{n+1}$ which is a one-sided inverse
of the inclusion map. Let $\pi_n\colon\bbQ G\to\bbQ G/I^{n+1}$ be the
quotient map, and for $g\in G$ set $Z(g)\coloneqq \sum_{n=0}^\infty
p_n(\pi_n g)\in\prod I^n/I^{n+1}=\hat\calA(G)$. It is easy to
check that $Z$ is an expansion. \qed
Even before completing the definition of a ``Taylor'' expansion for a
general group, we can already ponder whether a group has ``powerful''
expansions.
\begin{definition} We say that a group $G$ has a faithful expansion if it
has an injective expansion $Z\colon G\to\hat\calA(G)$.
\end{definition}
Proposition~\ref{prop:FaithfulExpansions} in Section~\ref{sec:soft}
implies that if one expansion for a group $G$ is injective, then so is
every other expansion for $G$.
{\red MORE: Same as residually torsion-free nilpotent?}
A summary of what we know about the faithfulness of expansions for
specific groups is in Table~\ref{tab:Summary} on
page~\pageref{tab:Summary}.
Next is the analogue of~\eqref{eq:ZTMultiplicative}:
\begin{definition} \label{def:ZMultiplicative}
An expansion $Z\colon G\to\hat\calA(G)$ is said to be ``multiplicative''
if $Z(g_1g_2)=Z(g_1)Z(g_2)$ for every $g_1,g_2\in G$.
\end{definition}
Before we can state Definition~\ref{def:ZCoMultiplicative}, the analogue
of~\eqref{eq:ZTCoMultiplicative}, we need the following proposition:
\begin{proposition}[Proof in Section~\ref{ssec:products}]
\label{prop:products}
If $G$ and $H$ are groups, then $\calA(G\times
H)\cong\calA(G)\otimes\calA(H)$ and hence $\hat\calA(G\times
H)\cong\hat\calA(G)\hat\otimes\hat\calA(H)$, where everything is
interpreted in the (completed) graded sense:
\[
\hat\calA(G)\hat\otimes\hat\calA(H)
=\prod_n\bigoplus_{p+q=n}\calA(G)_{p}\otimes\calA(H)_{q}.
\]
%(More precisely, $\calA$ is a ``monoidal functor'').
{\red MORE: ``Naturality''.}
\end{proposition}
Now given a group $G$ let $\Box\colon G\to G$ be the ``diagonal''
map $g\mapsto(g,g)$ and let the same symbol $\Box$ also denote the
functorially-induced morphism $\Box\colon\hat\calA(G)\to\hat\calA(G\times
G)\cong\hat\calA(G)\hat\otimes\hat\calA(G)$. The analogue
of~\eqref{eq:ZTCoMultiplicative} is:
\begin{definition} \label{def:ZCoMultiplicative}
An expansion $Z\colon G\to\hat\calA(G)$ is said to be ``co-multiplicative''
if the following diagram is commutative:
\begin{equation} \label{eq:comul}
\vcenter{\vbox{\xymatrix{
G \ar[r]^-\Box \ar[d]_Z &
G\times G \ar[d]^{Z\otimes Z} \\
\hat\calA(G) \ar[r]^-\Box &
\hat\calA(G)\hat\otimes\hat\calA(G).
}}}
\end{equation}
This amounts to saying that for every $g\in G$, $\zeta\coloneqq Z(G)$ is
group-like, namely, it satisfies $\Box\zeta=\zeta\otimes\zeta$.
\end{definition}
Finally, we come to the definition of ``Taylor'':
\begin{definition} \label{def:Taylor} We say that a group $G$ is
``Taylor'' if it has a Taylor expansion --- an $\hat\calA(G)$-valued
multiplicative and co-multiplicative expansion.
\end{definition}
{\red MORE: Really, $\Box$ makes $\calA$ into a bialgebra and we seek
a bialgebra morphism.}
A summary of what we know about the Taylor property for specific groups
is in Table~\ref{tab:Summary} on
page~\pageref{tab:Summary}.
\begin{remark} As a matter of convenience, we fixed our ground ring to
be $\bbQ$, though our definitions make sense over arbitrary ground rings.
In practice, raw expansions and multiplicative often exist also over
$\bbZ$, yet Taylor expansions often require characteristic $0$. See
e.g.\ Remark~\ref{rem:LOFZ}.
\end{remark}
\vskip -2mm\noindent$\overbracket{\hspace{\columnwidth}}$
\noindent{\bf Aside.} Why Care About Expansions? Groups are sometimes
complicated. It is sometimes difficult to decide if a group element $g$
is trivial or not. Given an expansion $Z$, compute $Z(g)$ and (at least
if $Z$ is faithful) the question is susceptable to a degree-by-degree
study, where often at least the low degrees are easy. It is sometimes
difficult to decide if a certain equation, written within a group $G$,
has solution. E.g., is $g\in G$ a square of some $h\in G$? Likewise,
given $g_1$ and $g_2$, is there $h$ such that $g_2=h^{-1}g_1h$? Namely,
are $g_1$ and $g_2$ conjugate? Given a multiplicative $Z$ such questions
become similarly susceptable to a degree-by-degree study.
\vskip -2mm\noindent$\underbracket{\hspace{\columnwidth}}$\vskip 2mm
\subsection{Quadraticity} If an expansion is to be useful, we must
understand its target space, $\hat\calA(G)$. Clearly, as every element
in $I^n$ is a product of elements in $I$, the degree $n$ piece
$\calA(G)_n=I^n/I^{n+1}$ is generated by products of elements in
$\calA(G)_1=I/I^2$. So $\calA(G)$ is generated by the degree $1$
elements in it. $\calA(G)$ is especially simple if all the relations
between its generators are in degree $2$.
\begin{definition} Following~\cite{Lee:VirtualIsQuadratic},
we say that the group $G$ is ``quadratic'' if the
$\bbQ$-algebra $\calA(G)$ is a quadratic
algebra~\cite{PolishchukPositselski:QuadraticAlgebras}. Namely, if
$\calA(G)$ is the algebra freely generated by
$\calA(G)_1$ modulo the ideal generated by the kernel of the multiplication
map $\mu_{11}\colon\calA(G)_1\otimes\calA(G)_1\to\calA(G)_2$:
\[ \calA(G) \!=\!
\left\langle \frac{I}{I^2} \right\rangle\!\left/\!\left\langle
\!\ker\left(\frac{I}{I^2}\!\otimes\!\frac{I}{I^2}\to \frac{I^2}{I^3}\right)
\right\rangle\right..
\]
\end{definition}
A summary of what we know about the quadraticity of specific groups
is in Table~\ref{tab:Summary} on
page~\pageref{tab:Summary}.
\subsection{Some Basic Examples}
\input table.tex
\subsubsection{The Infinite Cyclic Group $\bbZ$} As our first
example we take the group $\bbZ$, which we write in multiplicative
notation: $G = \langle x\rangle = \{x^k\colon k\in\bbZ\}$. Then
$\bbQ G=\bbQ[x,x^{-1}]$ can be identified as the ring of Laurent
polynomials in a variable $x$. The augmentation ideal $I$
of $\bbQ G$ is the ideal $\langle\tilx\rangle$ generated by the
single element $\tilx=x-1$; indeed, clearly $\langle \tilx\rangle\subset
I$, and the identities $x^k-1=(x^{k-1}+x^{k-2}+\dots+1)\tilx$ and
$x^{-k}-1=-(x^{-1}+x^{-2}+\dots+x^{-k})\tilx$ (for positive $k$) prove that
$x^k-1\in\langle \tilx\rangle$ for any $k$, and therefore $I\subset\langle
\tilx\rangle$. Therefore $\calA(G) = \bigoplus\langle \tilx\rangle^n/\langle
\tilx\rangle^{n+1} = \bigoplus \langle \tilx^n\rangle/\langle \tilx^{n+1}\rangle =
\bbQ[\barx]$ is a polynomial ring in one variable $\barx$, where $\barx$
is the class of $\tilx$ in $I/I^2$ (and $\barx^n$ is the class of $\tilx^n$
in $I^n/I^{n+1}$). Thus $\hat\calA(G)=\bbQ\llbracket\barx\rrbracket$
is the ring of power series in $\barx$.
A homomorphic expansion $Z\colon G\to\bbQ\llbracket\barx\rrbracket$
is determined by its value $\xi = Z(x)$ on the generator $x$
of $G$. Condition~\eqref{eq:ZExpansion} is satisfied iff $\xi =
1+\barx+O(\barx)^2$ (note that such $\xi$'s are always invertible
in $\bbQ\llbracket\barx\rrbracket$, so $Z(x^{-1})=\xi^{-1}$ makes sense).
Indeed,~\eqref{eq:ZExpansion} with $n = 0,1$ and $f = 1,t$ forces the
first two coefficients of $\xi$ to be as stated, and if $\xi$ is
as stated and $f=t^n\in I^n$ then $Z(f) = Z((x-1)^n) = (\xi-1)^n =
(\barx+O(\barx)^2)^n = \barx^n+O(\barx)^{n+1}$, sufficiently
proving~\eqref{eq:ZExpansion}. Quite clearly, $Z$ is faithful.
We leave it to the reader to verify that the map
$\Box\colon\bbQ\llbracket\barx\rrbracket = \hat\calA(G) \to
\hat\calA(G)\hat\otimes\hat\calA(G) = \bbQ\llbracket\barx_1,\barx_2\rrbracket$
is the substitution $\barx\to\barx_1+\barx_2$. Therefore the commutativity
of~\eqref{eq:comul} starting with $x\in G$ in the upper left corner is the
equation $\xi(\barx_1+\barx_2) = \xi(\barx_1)\xi(\barx_2)$, whose
unique solution within power series satisfying our initial condition is
$\xi(\barx)=\exp(\barx)$. Therefore there is a unique Taylor expansion
for $G=\langle x\rangle$ and it is given by $Z(x^k) = e^{k\barx}$.
\subsubsection{Abelian Groups} \label{sssec:AbelianGroups}
Similar analysis shows that if $G=\bbZ^m=\langle x_1,\ldots,x_m\colon
x_ix_j=x_jx_i\rangle$, then $\bbQ G=\bbQ[x_i,x_i^{-1}]$
is the ring of Laurent polynomials in $m$ variables, the
augmentation ideal $I=\langle \tilx_1,\ldots,\tilx_m\rangle$
is generated by the $m$ elements $\tilx_i=x_i-1$, and
$\hat\calA(G)=\bbQ\llbracket\barx_1,\ldots,\barx_m\rrbracket$ is the
ring of power series in the variables $\barx_i$, the images of $\tilx_i$
in $I/I^2$. There is a unique Taylor expansion for $G$ and it is given
by $Z(x_i)=e^{\barx_i}$. This expansion is faithful.
If $G$ is finitely generated Abelian, then it is of the form $\bbZ^m\times
T$, for some torsion group $T$. By Proposition~\ref{prop:products},
$\hat\calA(G)=\hat\calA(\bbZ^m)\hat\otimes\hat\calA(T)$. By
Corollary~\ref{cor:Torsion} below we find that $\calA(T)=0$. Hence there is
a unique Taylor expansion for $G$, it is given by
$Z(x_i)=e^{\barx_i}$ with $x_i$ and $\barx_i$ as before, and it is faithful
iff the torsion part $T$ is trivial.
The case of general Abelian groups serves as Example~\ref{exa:GenAbel}.
\subsubsection{Free Groups} \label{sssec:FreeGroups}
We leave it to the reader to verify that if $G$ is a free group on some
set of generators $\{x_\gamma\}_{\gamma\in\Gamma}$ then $\calA(G)$ is
the free associative algebra generated by $\bar{x_\gamma}=x_\gamma-e$ in
$I/I^2$. One noteworthy expansion for $G$ is the Magnus expansion $Z_M$,
the multiplicative extension of $x_\gamma\mapsto 1+\bar{x_\gamma}$ and
$x_\gamma^{-1}\mapsto 1-\bar{x_\gamma}+\bar{x_\gamma}^2-\ldots$. It
is faithful (e.g.~\cite{MagnusKarrassSolitar:CGT}), it is
defined over $\bbZ$, but it is not Taylor. Another noteworthy
expansion is the exponential expansion $Z_E$, the multiplicative
extension of $x_\gamma^{\pm 1}\mapsto\exp(\pm\bar{x_\gamma})$. By
Proposition~\cite{prop:FaithfulExpansions} and the faithfulness of
$Z_M$, the exponential expansion is also faithful. $Z_E$ is Taylor,
though it is only defined over $\bbQ$. $Z_E$ is {\em not} the unique
Taylor expansion for $G$ --- we leave it to the reader to verify
that if $\{f_\gamma\}_{\gamma\in\Gamma}$ are Lie power series in the
generators of $\calA(G)$ which vanish in degree 1, then $x_\gamma^{\pm
1}\mapsto\exp(\pm(\bar{x_\gamma}+f_\gamma))$ always defines a (faithful)
Taylor expansion for $G$.
\subsection{The First Sophisticated Example: Pure Braids and Vassiliev
Invariants} \label{ssec:PB}
As an illustrative example, we discuss the pure braid group on $m$
strands, $\PB_m$. Our purpose is only to highlight the results; the
proofs are merely cited.
\parpic[r]{\input{figs/PureBraid.pdf_t}}
The group $\PB_m$ is the fundamental group of the configuration space
$C_m=\{z\in\bbC^m\colon i\neq j\Rightarrow z_i\neq z_j\}$ of $m$ distinct
points in the plane. A pure braid can be visualized as on the right.
The group $\PB_m$ is generated by the love-behind-the-bars elements
$\{\sigma_{ij}\colon 1\leq ij$ by declaring $t_{ij}=t_{ji}$). The
relations of $\calA(\PB_m)$ can be derived from the relations of
$\PB_m$, and come out to be $[t_{ij},t_{kl}]=0$ whenever $i,j,k,l$
are distinct, and $[t_{ij}+t_{ik},t_{jk}]=0$ whenever $i,j,k$ are
distinct. Hence $\calA(\PB_m)$ is the well-known ``Drinfel'd-Kohno
algebra'' \cite{Drinfeld:QuasiHopf, Drinfeld:GalQQ, Kohno:MonRep,
Kohno:LinearRepresentations}.
Note that the augmentation ideal $I$ is always generated by differences of
group elements, and that any two pure braids differ by finitely many
``crossing changes'' $\overcrossing\leftrightarrow\undercrossing$. Hence
for $\PB_m$, the ideal $I$ is generated by differences
$\doublepoint\coloneqq\overcrossing-\undercrossing$ as in the theory of
finite type (Vassiliev) invariants (e.g., \cite{Bar-Natan:OnVassiliev,
Bar-Natan:Braids}). Hence $I^n$ is generated by ``$n$-singular pure
braids'' (pure braids with $n$ double points $\doublepoint$), hence
$(\bbQ\PB_m/I^{n+1})^\ast$ is precisely the space of type $n$ invariants,
and a little further inspection of the definitions shows that an expansion
for $\PB_m$ is precisely what is called within the language of finite type
invariants ``a Universal Finite Type Invariant (UFTI) for $\PB_m$''.
It is well known that a multiplicative and co-multiplicative (i.e., Taylor)
expansion / UFTI $Z$ exists for $\PB_m$. However I still don't know a
simple group-theoretic proof of that fact. Indeed the simplest
formula I know for such a $Z$ is
\[
Z(\gamma)= \hspace{-10mm}\sumint_{\substack{
n\geq 0 \\
0