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\begin{document} \latintext
\newgeometry{textwidth=8in,textheight=10.5in}

{\LARGE{\bf Cheat Sheet CU and QU}\hfill
\parbox[b]{2.5in}{\tiny
  \null\hfill\url{http://drorbn.net/AcademicPensieve/2019-10}
  \newline\null\hfill
  modified \today.
}}

\vskip -3mm
\rule{\textwidth}{1pt}
\vspace{-8mm}

\begin{multicols}{2}

{\bf\red Our Algebras.} Let
$sl_{2+}^\epsilon \coloneqq L\langle y,b,a,x\rangle$ subject to ${[a,x]=x}$,
$[b,y]=-\epsilon y$, $[a,b]=0$, $[a,y]=-y$, $[b,x]=\epsilon x$, and $[x,y]=\epsilon a+b$. So
$t\coloneqq\epsilon a-b$ is central and if $\exists \epsilon^{-1}$, $sl_{2+}^\epsilon/\langle t\rangle\cong sl_2$.

Indeed if $\epsilon$ is invertible, the map
\[ \phi_\epsilon\colon sl_{2+}^\epsilon \to sl_{2+}^1, \quad (y,b,a,x) \mapsto (\epsilon y, \epsilon b, a, x), \]
is an isomorphism of Lie algebras, and $sl_{2+}^1/\langle t\rangle \cong sl_{2+}^1/(a=b) \cong L\langle y,a,x\rangle/([a,x]=x,[a,y]=-y,[x,y]=2a) \cong sl_2$.

$U$ is either $CU=\calU(sl_{2+}^\epsilon)\llbracket\hbar\rrbracket$ or
$QU=\calU_\hbar(sl_{2+}^\epsilon) = A\langle y,b,a,x\rangle\llbracket\hbar\rrbracket$ with
$[a,x]=x$, $[b,y]=-\epsilon y$, $[a,b]=0$, $[a,y]=-y$, $[b,x]=\epsilon
x$, and $xy-qyx=(1-AB)/\hbar$, where $q=\bbe^{\hbar\epsilon}$,
$A=\bbe^{-\hbar\epsilon a}$, and $B=\bbe^{-\hbar b}$. Set also
$T=A^{-1}B=\bbe^{\hbar t}$.

{\red\bf The Quantum Leap.} Also decree that in $QU$,
\[ \Delta(y,b,a,x) = (y_1+B_1y_2, b_1+b_2, a_1+a_2, x_1+A_1x_2), \]
\[ S(y,b,a,x) = (-B^{-1}y, -b, -a, -A^{-1}x),\]
\vskip 3pt
and $R=\sum\hbar^{j+k}y^kb^j\otimes a^jx^k/j![k]_q! = 1 + \hbar r + O(\hbar^2)$, where $r = y\otimes x+b\otimes a$ satisfies CYBE, $[r_{12},r_{13}] + [r_{12},r_{23}] + [r_{13},r_{23}] = 0$.

$\phi_\epsilon$ extends to $QU$ with $\phi_\epsilon(\hbar)=\hbar/\epsilon$ and $R$ is $\phi_\epsilon$-invariant.

{\red\bf The $s,t$ Alternative.} With $s=\epsilon a+b$ (and $t=\epsilon a-b$), we have that $sl_{2+}^\epsilon$ is
\[ L\langle t,y,s,x\rangle
  \left/\left(
    [t,\cdot]=0,\, [s,x]=2\epsilon x,\, [s,y]=-2\epsilon y,\, [x,y]=s
  \right)\right.,
\]
with $r = y\otimes x+(s-t)\otimes(s+t)/4\epsilon \sim y\otimes x+(s\otimes s+s\otimes t-t\otimes s)/4\epsilon$.

{\red\bf The $d,t$ Alternative.} With $d=s/\epsilon$, this becomes
\[ L\langle t,y,d,x\rangle
  \left/\left(
    [t,\cdot]=0,\, [d,x]=2x,\, [d,y]=-2y,\, [x,y]=\epsilon d
  \right)\right.,
\]
with $r \sim y\otimes x+(\epsilon d\otimes d+d\otimes t-t\otimes d)/4$.

\end{multicols}

\end{document}

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