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\def\red{\color{red}}

\newcommand{\Ad}{\operatorname{Ad}}
\newcommand{\ad}{\operatorname{ad}}
\newcommand{\bch}{\operatorname{bch}}
\newcommand{\der}{\operatorname{der}}
\newcommand{\diver}{\operatorname{div}}
\newcommand{\mor}{\operatorname{mor}}

\def\act{{\hspace{-1pt}\sslash\hspace{-0.75pt}}}
\def\bbQ{{\mathbb Q}}
\def\calA{{\mathcal A}}
\def\calL{{\mathcal L}}
\def\CW{\text{\it CW}}
\def\FA{\text{\it FA}}
\def\FL{\text{\it FL}}
\def\Loneco{{\calL^{\text{1co}}}}
\def\tbd{\text{\color{red} ?}}
\def\tder{\operatorname{\mathfrak{tder}}}
\def\TAut{\operatorname{TAut}}

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\begin{document}
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{\LARGE\bf Cheat Sheet OneCo}\hfill
\parbox[b]{5in}{\tiny
  \null\hfill\url{http://drorbn.net/AcademicPensieve/2015-04/}
  \newline\null\hfill initiated 14/4/15;
  modified \today, \ampmtime
}

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\vspace{-8mm}

\begin{multicols}{2}\raggedcolumns

{\bf Background.}\ \hfill
$\displaystyle \delta e^\gamma
  = e^\gamma\cdot\left(\delta\gamma\act\frac{1-e^{-\ad\gamma}}{\ad\gamma}\right)
  = \left(\delta\gamma\act\frac{e^{\ad\gamma}-1}{\ad\gamma}\right)\cdot e^\gamma
$

The differential of $\gamma=\bch(\alpha,\beta)$:
\newline\null\hfill
$\displaystyle \delta\gamma\act\frac{1-e^{-\ad\gamma}}{\ad\gamma}
  = \left(\delta\alpha\act\frac{1-e^{-\ad\alpha}}{\ad\alpha}\act e^{-\ad\beta}\right)
  + \left(\delta\beta\act\frac{1-e^{-\ad\beta}}{\ad\beta}\right)
$

{\bf Models.} $\bullet$ In $[x,y]=\delta x$, $xf(y)=f(y+\delta)x$. If $\delta^2=0$, $[x,f(y)] = \delta f'(y)x$.

$\bullet$ In $[x,y]=\delta x+z^2$, $xf(y)=f(y+\delta)x+\frac{z^2}{\delta}(f(y+\delta)-f(y))$. If $\delta^2=0$, $[x,f(y)] = \delta f'(y)x+z^2f'(y)$.

{\bf Deriving Gassner.} $\calL^w$ is $\bbQ\llbracket b_i\rrbracket\langle a_{ij}\rangle$ modulo locality,
$[a_{ij},a_{ik}] = 0$,
$[a_{ik},a_{jk}] = -[a_{ij},a_{jk}] = b_ja_{ik}-b_ia_{jk}$,
and $[a_{ij},a_{ji}] = b_ia_{ji}-b_ja_{ij}$.
Acts on $V = {\bbQ\llbracket b_i\rrbracket\langle x_i=a_{i\infty}\rangle}$ by $[a_{ij},x_i]=0$, $[a_{ij},x_j] = b_ix_j-b_jx_i$. Hence $e^{\ad a_{ij}}x_i=x_i$, $e^{\ad a_{ij}}x_j = e^{b_i}x_j+\frac{b_j}{b_i}(1-e^{b_i})x_i$. Renaming $y_i=x_i/b_i$, $t_i=e^{b_i}$, get $[e^{\ad a_{ij}}]_{y_i,y_j} = \begin{pmatrix} 1 & 1-t_i \\ 0 & t_i \end{pmatrix}$.

{\bf The $\calL^w$ Adjoint representation.} $e^{\ad a_{ij}}$ acts by
\[ a_{kl} \mapsto a_{kl},\qquad a_{ik} \mapsto a_{ik}, \qquad a_{kj} \mapsto e^{-b_i}a_{kj} + \frac{b_k}{b_i}(1-e^{-b_i})a_{ij}, \]
\[ \text{and}\quad a_{jk} \mapsto e^{b_i}a_{jk} + \frac{b_j}{b_i}(1-e^{b_i})a_{ik}\qquad\text{also for $k=i$}. \]

An interpretation?

{\bf 2Dv.} $b$: bracket trace; $c$: cobracket trace; ${\langle b,c\rangle=\delta\in\{0,1\}}$; $\deg b_i = \deg c_j = \deg a_{ij} = \deg\delta = 1$.

$\calA^v$ is $\bbQ\llbracket\delta\rrbracket\FA(b_i,c_j,a_{ij})$ (so $\calL^v = \{f+f^{ij}a_{ij}\}$) modulo locality,
$[a_{ij},a_{ik}] = c_ka_{ij}-c_ja_{ik}$,
$[a_{ik},a_{jk}]=b_ja_{ik}-b_ia_{jk}$,
$[a_{ij},a_{jk}] = (c_j-b_j)a_{ik}+b_ia_{jk}-c_ka_{ij}$,
$[a_{ij},a_{ji}] = ?$,
$[a_{ij},b_i] = -[a_{ij},b_j] = -[a_{ij},c_i] = [a_{ij},c_j] = \delta a_{ij}-b_ic_j$,
$[b_i,c_j]=0$.

\null\hfill$\displaystyle
  a_{ij}f = f^\delta a_{ij} - \frac{b_ic_j}{\delta}(f^\delta-f),
$\hfill$\displaystyle
  [a_{ij},f] = (f^\delta-f)\left(a_{ij}-\frac{b_ic_j}{\delta}\right),
$\hfill\null

with $f^\delta\coloneqq f\act\left({b_i\to b_i+\delta\ b_j\to b_j-\delta \atop c_i\to c_i-\delta\ c_j\to c_j+\delta}\right)$.

{\bf The primitivity condition.} $\displaystyle\ker\left( f+f^{ij}a_{ij} \mapsto \delta f + f^{ij}b_ic_j\right)$. 

{\bf The OneCo Quotient} is $\delta c_i=c_jc_k=0$, so $\Loneco = \left\{(f+f^kc_k)+(f^{ij}+f^{ijk}c_k)a_{ij}\colon f, f^i,f^{ij},f^{ijk}\in\bbQ\llbracket\delta,b_i\rrbracket\right\}$. Then $[a_{ij},f+f^kc_k] = (\delta a_{ij}-b_ic_j) (\partial_if-\partial_jf-f^i+f^j)$.

{\bf Next.} $e^{\ad a_{ij}}(f) = ?$

\vfill{\bf To do.}
$\bullet$ Perhaps I should find a way to highlight the fact that v is a perturbation of w.

\end{multicols}

\end{document}

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