===== recycled on Sun Mar 22 12:11:25 EDT 2015 by drorbn on Debian-1207 ======


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 doubly filtered. Hence we start with some general facts about
doubly-filtered vector spaces.

A double-filtration on a vector space $V$ is a collection of subspaces
$F_{p,q}$ of $V$, with $p,q$ non-negative integers, such that $V=F_{0,0}$
and for every $p,q$, $F_{p,q}\supset F_{p+1,q}$ and $F_{p,q}\supset
F_{p,q+1}$. The associated doubly-graded space of $V$ is defined by
\[ \gr^2V\coloneqq \bigoplus_{p,q}V_{p,q}
  \coloneqq \bigoplus_{p,q}\frac{F_{p,q}}{F_{p+1,q}+F_{p,q+1}}.
\]
A double-filtration also defines a single filtration, by setting
$F_n\coloneqq\sum_{p+q=n}F_{p,q}$, and hence a singly-graded
associated space $\gr V=\oplus_nF_n/F_{n+1}$. 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
$\alpha_n\colon\bigoplus_{p+q=n}V_{p,q}\to V_n$.

\begin{lemma} The map $\alpha_n$ is an isomorphism. \end{lemma}

\begin{proof} The surjectivity of $\alpha_n$ is obvious. For $s\geq 0$, let
$\alpha_n^s$ be the restriction of $\alpha_n$ to the sum
$V_n^s\coloneqq\bigoplus_{p+q=n,\,p<s}V_{p,q}$. It is clear that
$\alpha_n^{n+1} = \alpha_n$ and that $\alpha_n^0$ is injective, so
let us assume that $\alpha_n^s$ is injective and prove the same for
$\alpha_n^{s+1}$. MORE.
\end{proof}

===== recycled on Sun Apr  5 12:54:26 EDT 2015 by drorbn on Debian-1207 ======


\begin{proof} The surjectivity of $\alpha$ is obvious. For $s\geq 0$, let
$\alpha_s$ be the restriction of $\alpha$ to the sum
$V_n^s\coloneqq\bigoplus_{p+q=n,\,p\leq s}V_{p,q}$. It is clear that
$\alpha_n = \alpha$ and that $\alpha_{-1}$ is injective, so
let us assume that $\alpha_{s-1}$ is injective and prove the same for
$\alpha_s$. Indeed, suppose $v=(v_{<s},v_s)\in V_n^s=V_n^{s-1}\oplus
V_{s,n-s}$ is such that $\alpha_sv=0$. Then there are
$u_{<s}\in\bigoplus_{p+q=n,\,p<s}F_{p,q}$ and $u_s\in F_{s,n-s}$ that
project to $v_{<s}$ and to $v_s$, and we know that their sum projects to
the kernel of $\alpha_s$, and so $u=u_{<s}+u_s\in
F_{n+1}=\sum_{p+q=n+1}F_{p,q}$
MORE.
\end{proof}

===== recycled on Sun Apr  5 12:54:29 EDT 2015 by drorbn on Debian-1207 ======


\begin{proof} We construct the inverse $\beta$ of $\alpha$
\end{proof}

===== recycled on Sun Apr  5 12:54:29 EDT 2015 by drorbn on Debian-1207 ======


\begin{proof} The surjectivity of $\alpha_n$ follows from
$\sum_{p+q=n}F_{p,q} = F_n$. 
\end{proof}

===== recycled on Fri May  8 16:26:04 EDT 2015 by drorbn on Debian-1207 ======


equality of maps $G\to\calA(G)\otimes\calA(G)$ holds true:
\[ (Z\otimes Z)\circ\Box = \Box\circ Z. \]

===== recycled on Wed Oct  7 10:29:37 EDT 2015 by drorbn on Debian-1207 ======


For any $\Gamma$, the polynomial algebra $\calA \coloneqq
\calA(G(\Gamma))$ is generated by the elements $\{\barx=[x-1]\colon x\in
V\}$ in $\calA_1=I/I^2$.  But modulo $I^2$, the edge relation
\[ x\xrightarrow{z^{\pm 1}}y \colon\qquad x=y^{z^{\pm 1}} \]
becomes $\barx=\bary$ (see )

===== recycled on Sat Feb 20 10:07:05 EST 2016 by drorbn on Ubuntu-1404 ======


\begin{multline*}
  \left(
    w\xrightarrow{y}x\xrightarrow{w}y\xrightarrow{x}z
  \right) \mapsto \\
  \left\langle
    w,x,y,z \colon w=x^y,\, x=y^w,\, y=z^x
  \right\rangle.
\end{multline*}

===== recycled on Fri Mar  4 16:08:19 EST 2016 by drorbn on Ubuntu-1404 ======

MKS: in $\langle a,b\rangle/(a^2=1)$, $(((a,b),a),(a,b))=1$.

===== recycled on Fri Mar  4 16:08:58 EST 2016 by drorbn on Ubuntu-1404 ======

We prefer not to do so to allow us to keep all edges in a tree oriented
towards the root.

===== recycled on Fri Mar  4 16:11:40 EST 2016 by drorbn on Ubuntu-1404 ======

and if all edges in $\Gamma$ are directed towards some root vertex

===== recycled on Fri Mar  4 16:11:48 EST 2016 by drorbn on Ubuntu-1404 ======

and if all edges in $\Gamma$ are directed towards some root vertex
$x_0\in V$.