\section*{Recycling}

\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