\documentclass[12pt,reqno]{amsart} \def\draft{n}\def\help{n} \usepackage[dvipsnames]{xcolor} \usepackage{ graphicx,dbnsymb,amsmath,amssymb,picins,multicol, mathtools,stmaryrd,enumitem,needspace,makecell,calc,pifont } \usepackage[all]{xy} \usepackage[textwidth=6.5in,textheight=9in,headsep=0.15in,centering]{geometry} % Following http://tex.stackexchange.com/a/847/22475: % Consider inserting "backref" option. \usepackage[breaklinks=true,pagebackref]{hyperref}\hypersetup{colorlinks, linkcolor={green!50!black}, citecolor={green!50!black}, urlcolor=blue } \if\draft y \usepackage[color]{showkeys} \fi %\usepackage{emerald} \usepackage[T1]{fontenc} \input macros.tex \input defs.tex \begin{document} \newdimen\captionwidth\captionwidth=\hsize \setcounter{secnumdepth}{4} \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 May.~18,~2016. 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} \newlength{\quotewidth} \setlength{\quotewidth}{\columnwidth-2\parindent} \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