\section{Some Soft Facts about Power Series and Expansions}
\label{sec:soft}

\subsection{$A$-Expansions} Expansions are valued in $\hat\calA(G)$,
a space with a simple general abstract definition. Yet for concrete
groups, it is sometimes difficult to describe $\calA(G)$ in
concrete terms.  It is often the case that it is easy to guess a
graded space $A$ and show that it is ``no smaller'' than $\calA(G)$;
it then remains to affirm that the guess is right --- namely, that
$A\cong\calA(G)$.  In the current section we will see that
we can achieve this affirmation by constructing an $\hatA$-valued
``$A$-expansion''. The first example where this technique is
useful is within the proof of Proposition~\ref{prop:products} in the
next section, where the ``unknown'' is $\calA(G\times H)$ and the
``guess'' is $\calA(G)\otimes\calA(H)$.

\vskip 1mm
\makeatletter\def\thm@space@setup{%
  \thm@preskip=0cm plus 0cm minus 0cm %\thm@postskip=\thm@preskip
}\makeatother
\parpic[r]{$\displaystyle\xymatrix@C=0.6in{
  & \hatA \ar@/_/[d]_{\hat\pi} \\
  \bbQ G \ar[ur]^{Z_A} \ar@{.>}[r]^{\hat\pi\circ Z_A}&
  \hat\calA(G) \ar@{.>}@/_/[u]_{\grh Z_A}
}$}
\begin{definition} (Compare with Definition~\ref{def:Expansion}
and consult the diagram on the right). Let $G$ be a group, and $A$ a
graded vector space along with a degree-respecting surjection $\pi\colon
A\to\calA(G)$ (equivalently, $\hat\pi\colon\hatA\to\hat\calA(G)$, where
$\hatA$ is the graded completion of $A$). An $A$-expansion (for $G$) is
a map $Z_A\colon G\to\hatA$ whose homonymous (uniquely defined) linear
extension $Z_A\colon\bbQ G\to\hatA$ has the property that if $a\in A$ is
if degree $n$ and $f\in I^n$ is such that $\pi a=[f]$ in $I^n/I^{n+1}$,
then $Z_A(f)$ begins with $a$.  Namely,
\[ Z_A(f) = (0,\ldots,0,
    \underbrace{a}_{\mathclap{\text{in degree $n$}}},
    \ast,\ast,\ldots).
\]
\end{definition}

In the language of~\eqref{eq:grgr}, the above definition is equivalent to
the following:
\begin{equation} \label{eq:Agrgr}
  \parbox{\columnwidth-4\parindent}{
    $Z_A\colon\bbQ G\to\hatA$ is a filtration preserving linear map so
    that ${\grh\,Z_A\circ\hat\pi\colon\hatA\to\hatA}$ is the identity map
    of $\hatA$.
  }
\end{equation}

\begin{proposition} \label{prop:AExpansions}
If $Z_A$ is an $A$-expansion for a group $G$, then
$\pi\colon A\xrightarrow{\sim}\calA(G)$ is an isomorphism and $\hat\pi\circ
Z_A$ is an expansion. (Hence finding an $A$-expansion both identifies
$\calA(G)$ and defines an ordinary expansion).
\end{proposition}

\begin{proof} The surjectivity of $\pi$ is given and its injectivity
follows from $\grh\,Z_A\circ\hat\pi=Id$, so it is an isomorphism. Given
this~\eqref{eq:Agrgr} becomes~\eqref{eq:grgr}. \qed
\end{proof}

\begin{example} \label{exa:GenAbel} Let $G$ be an arbitrary Abelian group,
written multiplicatively: $G=\left\langle\{x_\gamma\}_{\gamma\in\Gamma}
\colon \{\prod_\gamma x_\gamma^{a_{r\gamma}}=e\}_{r\in R}\right\rangle$,
where $\{x_\gamma\}$ is a set of commuting generators indexed by some
set $\Gamma$, where $\{\prod_\gamma x_\gamma^{a_{r\gamma}}=e\}$ is a set of
relations indexed by some set $R$, and where $(a_{r\gamma})$ is a matrix
of coefficients in $\bbZ$ having the property that for any any fixed $r\in
R$, $a_{r\gamma}\neq 0$ only for finitely many $\gamma\in\Gamma$.

We note that the augmentation ideal $I$ of $\bbQ G$ is generated by the
elements $t_\gamma\coloneqq x_\gamma-e$. In terms of these, the relation
$\prod_\gamma x_\gamma^{a_{r\gamma}}=e$ can be re-written as
$\prod_\gamma (e+t_\gamma)^{a_{r\gamma}}=e$,
and using the binomial formula this can be expanded as
\[ \sum_\gamma a_{r\gamma}t_\gamma=0 \quad\text{mod }I^2. \]

We
make a ``guess'' for $\calA(G)$: namely, we let $A$ be the Abelian
algebra over $\bbQ$ with degree 1 generators $\bart_\gamma$ (for each
$\gamma\in\Gamma$ and (additively-written) relations $\sum
a_{r\gamma}\bart_\gamma$ for each $r\in R$.
\end{example}

\subsection{Products and Almost-Direct Products} \label{ssec:products}
We aim to prove Proposition~\ref{prop:products}, asserting that
$\calA(G\times H)\cong\calA(G)\otimes\calA(H)$, using the technique of the
previous section.

\noindent{\em Proof of Proposition~\ref{prop:products}.} Without
further comment we will identify $G$ and $H$ as commuting subgroups of
$GH\coloneqq G\times H$ using the coordinate inclusions, and likewise
$\bbQ G$ and $\bbQ H$ as commuting subalgebras of $\bbQ (GH)=\bbQ
G\otimes\bbQ H$. Let $I_G$, $I_H$, and $I_{GH}$ denote the augmentation
ideals of $\bbQ G$, $\bbQ H$, and $\bbQ(GH)$ respectively. Our
first task is to compare these ideals and their powers.

Clearly, $I_{GH}=I_G(\bbQ H) + (\bbQ G)I_H$: the ``$\supset$'' inclusion
is obvious, and the ``$\subset$'' inclusion follows from $gh-e = (g-e)h +
(h-e)$.  By expanding powers it follows that
\begin{equation} \label{eq:ExpandingPowers}
  I_{GH}^n \!=\! (I_G(\bbQ H)+(\bbQ G)I_H)^n
  \! = \!\!\!\sum_{p+q=n}\!\!\!{I_G^pI_H^q}.
\end{equation}

Let $A=\calA(G)\otimes\calA(H)$, and let $\pi\colon A\to\calA(GH)$
be the composition
\[ \calA(G)\otimes\calA(H) \!\to\! \calA(GH)\otimes\calA(GH)
  \xrightarrow{\mu} \calA(GH)
\]
of the maps induced by the coordinate inclusions and the multiplication map
$\mu$. The map $\pi$ is clearly graded, and it follows
from~\eqref{eq:ExpandingPowers} that it is surjective. Finally, if $Z_G$
and $Z_H$ are expansions for $G$ and $H$ (these exist by
Proposition~\ref{prop:ExpansionsExist}), it is easy to check that
$Z_A\coloneqq Z_G\otimes Z_H$, more precisely defined by $Z_A(gh) =
\sum_{p,q}Z_G(g)_p\otimes Z_H(h)_q$ is an $A$-expansion, where $Z_G(g)_p$
and $Z_H(h)_q$ denote the degree $p$ and degree $q$ of $Z_G(g)$ 
and $Z_H(h)$, respectively. Hence by Proposition~\ref{prop:AExpansions}
$\pi$ is an isomorphism. \qed

The only place in the above proof where we have used the fact that $G$ and
$H$ commute within $GH$ was in Equation~\eqref{eq:ExpandingPowers}.
Indeed, without this commutativity we have that $I_{GH}^n$ is a sum of
$2^n$ products whose factors are $I_G(\bbQ H)$ and $(\bbQ G)I_H$ taken
in an arbitrary order, and we have no way of `sorting' such products to
the form $I_G^pI_H^q$. Yet there is a further situation in which an analog
of Proposition~\ref{prop:products} holds:

\begin{definition} A semi-direct product of groups $G\rtimes H$ is called
``almost-direct'' if the action of $H$ on $G$ descends to the trivial
action of $H$ on the Abelianization of $G$. In other words, if for any
$g\in G$ and $h\in H$, $g^h\equiv g$ modulo $(G,G)$, where $g^h\coloneqq
h^{-1}gh$ and $(G,G)$ denotes the group generated by all commutators of
pairs of elements in $G$.
\end{definition}

\begin{proposition} (Compare \cite[Theorem~3.1]{Papadima:UFTI4Braids} and
\cite[Section 3]{FalkRandell:LCS}). If $G\rtimes H$ is
almost-direct then as vector spaces, $\hat\calA(G\times
H)\cong\hat\calA(G)\hat\otimes\hat\calA(H)$.
\end{proposition}

\begin{proof} It is enough to re-prove Equation~\eqref{eq:ExpandingPowers}
in the present case. The inclusion $\supset$ is trivial, and so following
the discussion above it is sufficient to show that in an arbitrary ordered
product of factors each of which is $\bbQ G$, $\bbQ H$, $I_G$, or $I_H$
we can sort the $\bbQ G$ and $I_G$ factors to the left and all the $\bbQ
H$ and $I_H$ factors to the right by a series of $\subset$ inclusions,
without decreasing the total number of $I_G$ / $I_H$ factors appearing.
For this we use the equalities / inclusions $HG=GH$, $(\bbQ H)I_G=I_G(\bbQ
H)$, and $I_H(\bbQ G)\subset I_{GH}=I_G(\bbQ H)+I_H$ which are easily shown
to hold in an arbitrary semi-direct product, and the inclusion
$I_HI_G\subset I_GI_H+I_G^2H$ which is a property of almost-direct products
as follows:
\[ (h-e)(g-e)=(g-e)(h-e)+(g^{h^{-1}}-g)h \]
\hfill\hfill\hfill$\displaystyle \in I_GI_H+I_G^2H.$ \qed
\end{proof}

MORE: a description of $\calA(GH)$ as an algebra.

\subsection{More About the Polynomial Algebra $\calA(G)$}
\label{ssec:Trivialities}

We start with a few trivialities. For $g\in G$ we let $\bar{g}\coloneqq
g-e\in\bbQ G$. It is clear that $\bar{g}\in I$ and that elements of the
form $\bar{g}$ generate $I$. And as $I/I^2$ generates $\calA(G)$, the
classes of the $\bar{g}$'s in $I/I^2$ generate $\calA(G)$.

\begin{proposition} In $I/I^2$, $\overline{gh}=\bar{g}+\bar{h}$ for any
$g,h\in G$. In particular, $\overline{g^k}=k\bar{g}$ for any $k\in\bbZ$.
\end{proposition}

\begin{proof} In $\bbQ G$, $\overline{gh} = gh-e = (g-e)+(h-e)+(g-e)(h-e)
= \bar{g}+\bar{h}+\bar{g}\bar{h}$, and modulo $I^2$ the last term drops
out.\qed
\end{proof}

\begin{corollary} \label{cor:Torsion}
If a group $G$ is torsion then $\calA(G)=0$ (justifying
Table~\ref{tab:Summary} line~\ref{cl:torsion}).
\end{corollary}

\begin{proof} Indeed if $G$ is torsion and $g\in G$, then
$g^k=e$ for some $k$, hence $k\bar{g}=\overline{g^k}=g^k-e=0$, hence
$\bar{g}=0$, hence all the generators of $\calA(G)$ vanish. \qed
\end{proof}

If $x$ and $y$ are elements of a group, we denote their group-commutator by
$(x,y)\coloneqq xyx^{-1}y^{-1}$. If $a$ and $b$ are elements of an algebra,
we denote their algebra-commutator by $[a,b]=ab-ba$. In our context, these
two notions are compatible:

\begin{proposition} If $x,y\in G$, then $\overline{(x,y)}\in I^2$ and in
$\calA(G)_2=I^2/I^3$, $\overline{(x,y)} = [\bar{x},\bar{y}]$.
\end{proposition}

\begin{proof} In $\bbQ G$ and since $e$ is central, $[\bar{x},\bar{y}]$
$= [x,y] = \overline{(x,y)}yx$.
Hence $\overline{(x,y)}yx \in I^2$, hence
$\overline{(x,y)}(yx-e)\in I^3$, hence modulo $I^3$, $\overline{(x,y)} =
\overline{(x,y)}yx = [\bar{x},\bar{y}]$. \qed
\end{proof}

The above proposition has a stronger variant:

\begin{proposition} \label{prop:StrongerCommutators} If $x,y\in G$ are
such that $\bar{x}\in I^m$ and $\bar{y}\in I^n$, then $\overline{(x,y)}\in
I^{m+n}$ and in $\calA(G)_{m+n}$, $\overline{(x,y)} = [\bar{x},\bar{y}]$.
\end{proposition}

\begin{proof} Same proof, with $I^2$ replaced with $I^{m+n}$ and $I^3$
with $I^{m+n+1}$. \qed \end{proof}

\subsection{More About Expansions $Z\colon G\to\hat\calA(G)$}

MORE Existance in the plain case, uniqueness, $A$-expansions.

MORE: Something about GT/GRT.