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

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

We start with a few trivialities. For $g\in G$ we let $\bar{g}\coloneqq
g-1\in\bbQ G$. 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 $\bbQ G$, $\overline{gh} = gh-1 = (g-1)+(h-1)+(g-1)(h-1)
= \bar{g}+\bar{h}+\bar{g}\bar{h}$, and modulo $I^2$ the last term drops
out.\qed
\end{proof}

This proposition implies that if the group $G$ is torsion
then $\calA(G)=0$ justifying Table~\ref{tab:Summary}
line~\ref{cl:torsion}. Indeed if $G$ is torsion and $g\in G$, then $g^k=1$
for some $k$, hence $k\bar{g}=\overline{g^k}=g^k-1=0$, hence $\bar{g}=0$,
hence all the generators of $\calA(G)$ vanish.

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 $1$ is central, $[\bar{x},\bar{y}]$
$= [x,y] = \overline{(x,y)}yx$.
Hence $\overline{(x,y)}yx \in I^2$, hence
$\overline{(x,y)}(yx-1)\in I^3$, 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{Products and Almost-Direct Products} \label{ssec:products}

We aim to prove Proposition~\ref{prop:products}, asserting that
$\calA(G\times H)\cong\calA(G)\otimes\calA(H)$. As we shall see, the
proof revolves around the fact that $\bbQ(G\times H)=(\bbQ G)\otimes(\bbQ
H)$ is twice-filtered. Hence we start with a general fact about
twice-filtered vector spaces.

\begin{figure*}
\centering{\input{figs/lattice.pdf_t}}
\caption{
  Subspaces of $V$ that can be defined using $F'_p$ and $F''_q$ correspond
  to monotone subsets of the lattice $\bbZ_{\geq 0}^2$ and these are
  defined by their ``lower / left boundary lines''. For example, on the
  left $F_{2,1}=F'_2\cap F''_1$ is ``everything above and to the right of the
  solid black line'', and this is the intersection of $F'_2$, ``right of
  the dotted red line labeled $\red F'_2$'', and $F''_1$, ``above the
  dotted blue line labeled $\blue F''_1$''. The right half of this figure
  displays spaces that occurr within the proof of Lemma~\ref{lem:pain}.
} \label{fig:lattice}
\end{figure*}

Suppose a vector space $V$ is twice-filtered. Namely, we have a pair of
filtrations, $V=F'_0\supset F'_1\supset\cdots$ and
$V=F''_0\supset F''_1\supset\cdots$. We write $F_{p,q}\coloneqq F'_p\cap
F''_q$ (see Figure~\ref{fig:lattice}). The associated doubly-graded
space of $V$ is defined by
\[ \gr^2V\coloneqq \bigoplus_{p,q}V_{p,q}
  \coloneqq \bigoplus_{p,q}\frac{F_{p,q}}{F_{p+1,q}+F_{p,q+1}}.
\]
We can define an additional ``diagonal'' filtration on $V$, by setting
$F_n\coloneqq\sum_{p+q=n}F_{p,q}$, and hence a singly-graded
associated space $\gr V=\oplus_nV_n\coloneqq\oplus_nF_n/F_{n+1}$. If
$p+q=n$, then $F_{p,q}\subset F_n$ and $F_{p+1,q}+F_{p,q+1}\subset
F_{n+1}$, and hence there are maps $V_{p,q}\to V_n$, which induce a map
$\alpha\colon\bigoplus_{p+q=n}V_{p,q}\to V_n$.

\begin{lemma} \label{lem:pain} The map $\alpha$ is an isomorphism. \end{lemma}

We need a sub-lemma:

\begin{lemma} \label{lem:D0E0} If $D_1\subset D_0$ and $E_1\subset E_0$
are all subsets of the same vector space, and $D_0\cap E_0\subset D_1$,
then
\[ \frac{D_0+E_0}{D_1+E_1}
  \cong \frac{D_0}{D_1} \oplus \frac{E_0}{E_1+D_0\cap E_0}. \]
\end{lemma}

\begin{proof} Define $\psi\colon \frac{D_0+E_0}{D_1+E_1}
\to \frac{D_0}{D_1} \oplus \frac{E_0}{E_1+D_0\cap E_0}$ by
$[d_0+e_0]\mapsto([d_0],[e_0])$ and verify that this map is well defined:
if $d_0+e_0=d'_0+e'_0$ with $d_0,d'_0\in D_0$ and $e_0,e'_0\in
E_0$, then $d_0-d'_0\in D_0\cap E_0\subset D_1$ and $e_0-e'_0\in
D_0\cap E_0$ so $\psi(d_0+e_0)=\psi(d'_0+e'_0)$, and likewise, easily
$\psi((d_0+e_0)+(d_1+e_1))=\psi(d_0+e_0)$ when $d_1\in D_1$ and $e_1\in
E_1$. The construction of an inverse of $\psi$ is even easier. \qed
\end{proof}

\noindent{\em Proof of Lemma~\ref{lem:pain}.} We study ``the part
of the picture where $p\geq s$'' and compare it with ``the part
of the picture where $p\geq s+1$''. Let $s\geq 0$ and set
$F^s_n\coloneqq\sum_{p+q=n,\,p\geq s}F_{p,q}$. Then
\[ \frac{F^s_n}{F^s_{n+1}}
  = \frac{F^{s+1}_n+F_{s,n-s}}{F^{s+1}_{n+1}+F_{s,n-s+1}}
  = \frac{D_0+E_0}{D_1+E_1},
\]
where we denote $D_0\coloneqq F^{s+1}_n$, $D_1\coloneqq F^{s+1}_{n+1}$ (the
Diagonal terms), and $E_0\coloneqq F_{s,n-s}$ and $E_1\coloneqq
F_{s,n-s+1}$ (the Extra terms). Then $D_0\cap E_0=F_{s+1,n-s}\subset D_1$,
and thus by Lemma~\ref{lem:D0E0},
\begin{multline*} \frac{F^s_n}{F^s_{n+1}}
  \cong \frac{D_0}{D_1} \oplus \frac{E_0}{E_1+D_0\cap E_0} \\
  = \frac{F^{s+1}_n}{F^{s+1}_{n+1}}
    \oplus \frac{F_{s,n-s}}{F_{s,n-s+1}+F_{s+1,n-s}}
  = \frac{F^{s+1}_n}{F^{s+1}_{n+1}} \oplus V_{s,n-s}.
\end{multline*}
Hence by induction,
\begin{multline*}
  V_n
  = \frac{F^0_n}{F^0_{n+1}} \cong \frac{F^1_n}{F^1_{n+1}} \oplus V_{0,n} \\
  \cong \ldots \cong V_{n,0}\oplus\ldots\oplus V_{0,n}.
\end{multline*}
We leave it to the reader to verify that the above isomorphism is induced
by the map $\alpha$. \qed

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

Clearly, $I_{GH}=I_G(\bbQ H) + (\bbQ G)I_H$: the ``$\supset$'' inclusion
is obvious, and the ``$\subset$'' inclusion follows from $gh-1 = (g-1)h +
(h-1)$.  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}
and therefore, taking $V=\bbQ G\times H$, $F'_p=I_G^p(\bbQ H)$
and $F''_q=(\bbQ G)I_H^q$ and using notation as in the preceding
discussion, $I_{GH}^n = \sum_{p+q=n}F'_p\cap F''_q = F_n$ and hence
$\calA(G\times H)_n = V_n$. Likewise
\begin{multline*} V_{p,q} =
  \frac{I_G^p\otimes I_H^q}{I_G^{p+1}\otimes I_H^q + I_G^p\otimes I_H^{q+1}} \\
  \cong \frac{I_G^p}{I_G^{p+1}} \otimes \frac{I_H^q}{I_H^{q+1}}
  = \calA(G)_p\otimes\calA(H)_q
\end{multline*}
and thus by Lemma~\ref{lem:pain}, $\calA(G\times H)_n \cong \sum_{p+q=n}
\calA(G)_p\otimes\calA(H)_q$. \qed

The only place in the above proof where we have used the fact that $G$ and
$H$ commute within $G\times H$ 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}). 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
$I_HI_G\subset I_GI_H+I_G^2$ which is a property of almost-direct products
as follows:
\begin{multline*}
  (h-1)(g-1)=(g-1)(h-1)+(g^{h^{-1}}-g)h \\
  \in I_GI_H+I_G^2H.\qquad\qed
\end{multline*}
\end{proof}

MORE.

\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.