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{\LARGE\bf Cheat Sheet OneCo-1611}\hfill
\parbox[b]{5in}{\tiny
  \null\hfill\url{http://drorbn.net/AcademicPensieve/2016-11/}
  \newline\null\hfill
  Restarts \href{http://drorbn.net/AcademicPensieve/Projects/OneCo-1606/}{Projects/OneCo-1606};
  modified \today
}

\vskip -3mm
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\vspace{-8mm}

\begin{multicols*}{2}%\raggedcolumns

\entry{161121b} Control alt-$\Delta$!

\entry{161121a} Roland's 1-co middle $c$ data, from \href{../People/VanDerVeen/nb/1CoMidCStraight10.pdf}{People/\linebreak[0]VanDerVeen/\linebreak[0]1CoMidCStraight10.nb}. At $\{\omega,L,Q,P\}\to\bbO\left(\omega^{-1}e^{L+Q/\omega}(1+\e\omega^{-4}P)\colon acw\right)$, verified at \href{../People/VanDerVeen/nb/NewVdVAlgebraAt17.pdf}{People/\linebreak[0]VanDerVeen/\linebreak[0] NewVdVAlgebraAt17.nb}:
\[ \includegraphics[width=0.85\linewidth]{RolandsData.png} \]

\entry{161112b} Roland's $q\frakg_0$ at \verb$genus0co.nb$: $a=\zeta u$, $\Delta_{jk}(a)=a_j+t_ja_k$, $S(A)=-t^{-1}a$. {\purple Q. How does $\Delta_{q\frakg_0}$ relates to $\Delta_{\frakg_0}$? Is $KV(\frakg_0)$ more than a projection of $KV(\FL)$? Due to $e^{b_1c_2}$?}

\entry{161112a} With $\zeta=\frac{t-1}{b}$, an algebra isomorphism $q\frakg_1\to\frakg_1$ via $a\mapsto\zeta u$, $\e\mapsto-\frac{2\zeta}{t+1}\e$? Roland on 161118: all simplicity advantages in his model arise from the $\zeta$-rescaling and the $cuw\to acw$ PBW change.

\entry{161110} Roland's $q\frakg_1$ at \verb$NewVdVAlgebraAt16.nb$: $[c,u]=u$, $[c,w]=-w$, $[u,w]=(t-1)-\e uw+2\e tc$, $\Delta_{jk}(b,c,u,w)=(b_j+b_k,c_j+c_k, t_ke^{\e c_k}u_j+u_k, e^{\e c_k}w_j+w_k)$, $S(b,c,u,w)=(-b,-c, -t^{-1}ue^{-\e c}, -we^{-\e c})$. Then $a\coloneqq ue^{-\e c}$ so $[c,a]=a$, $[a,w]=(t-1)+\e(t+1)c$, $\Delta_{jk}(a)=t_ka_j+a_ke^{-\e c_j}$, $S(a)=-t^{-1}(ae^{\e c}+\e a)$.

\entry{160920} Roland's $s\frakg_1$: $[w,c]=w$, $[u,c]=-u$, $[w,u]=e^{\epsilon c}$ ($s$ for September).

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{\red 1-Smidgen $sl_2$} Let {\red $\frakg_1$}
be the 4-dimensional Lie algebra $\frakg_1=\langle b,c,u,w\rangle$ over
the ring $R=\bbQ[\e]/(\e^2=0)$, with $b$ central and with \cbox{yellow}{$[w,c]=w$,
$[c,u]=u$, and $[u,w]=b-2\e c$}, with CYBE $r_{ij}=(b_i-\e
c_i)c_j+u_iw_j$ in $\calU(\frakg_1)^{\otimes\{i,j\}}$. Over $\bbQ$,
$\frakg_1$ is a {\red solvable approximation of $sl_2$}: $\frakg_1 \supset
\langle b,u,w,\e b,\e c,\e u,\e w\rangle \supset
\langle b,\e b,\e c,\e u,\e w\rangle \supset 0$.
\hfill\text{\footnotesize(note: $\deg(b,c,u,w,\e)=(1,0,1,0,1)$)}

{\red 0-Smidgen $sl_2$.} Let $\frakg_0$
be $\frakg_1$ at $\e=0$, or $\bbQ\langle
b,c,u,w\rangle/([b,\cdot]=0,\,[c,u]=u,\,[c,w]=-w,\,[u,w]=b$ with
$r_{ij}=b_ic_j+u_iw_j$.
It is $\frakb^\ast\rtimes\frakb$ where $\frakb$ is the 2D Lie algebra
$\bbQ\langle c,w\rangle$ and $(b,u)$ is the dual basis of $(c,w)$.

{\red How did these arise?} $sl_2=\frakb^+\oplus\frakb^-/\frakh\eqqcolon
sl_2^+/\frakh$, where $\frakb^+=\langle c,w\rangle/[w,c]=w$ is a Lie
bialgebra with $\delta\colon\frakb^+\to\frakb^+\otimes\frakb^+$ by
$\delta\colon(c,w)\mapsto(0,c\wedge w)$. Going back,
$sl_2^+=\calD(\frakb^+) = (\frakb^+)^\ast\oplus\frakb^+ = \langle
b,u,c,w\rangle/\cdots$. {\red Idea.} Replace $\delta\to\e\delta$ over
$\bbQ[\e]/(\e^{k+1}=0)$. At $k=0$, get $\frakg_0$. At $k=1$,
get $[w,c]=w$, $[w,b']=-\e w$, $[c,u]=u$, $[b',u]=-\e u$,
$[b',c]=0$, and $[u,w]=b'-\e c$. Now note that $b'+\e c$ is
central, so switch to $b\coloneqq b'+\e c$. This is $\frakg_1$.

\rule{\linewidth}{1pt}

{\red Lemma.} $R_{ij} \!=\! e^{b_ic_j+u_iw_j}
\!=\! \bbO\left(
  \exp\left(b_ic_j+\frac{e^{b_i}-1}{b_i}u_iw_j\right)
    |i\colon\!u_i,\,j\colon\!c_jw_j
\right).$

{\red The Big $\frakg_0$ Lemma.} With $[c,u]=u,\,[c,w]=-w,\,[u,w]=b$,
\begin{enumerate}[leftmargin=*,labelindent=0pt]

\item $\bbO(e^{\gamma c+\beta u}|uc) = \bbO(e^{\gamma c+e^{-\gamma}\beta u}|cu)$, meaning $e^{\beta u}e^{\gamma c} = e^{\gamma c}e^{e^{-\gamma}\beta u}$.
    \newline Proof. $u\cdot\gamma c = \gamma \cdot(cu-u)$ \ldots

\item $\bbO(e^{\gamma c+\alpha w}|wc) = \bbO(e^{\gamma c+e^{\gamma}\alpha w}|cw)$

\item $\bbO(e^{\alpha w+\beta u}|wu) =
     \bbO(e^{-b\alpha\beta+\alpha w+\beta u}|uw)$

\item $\bbO(e^{\delta uw}|wu)e^{\beta u}
     = e^{\nu\beta u}\bbO(e^{\delta uw}|wu)$,
     with \cbox{yellow}{$\nu=(1+b\delta)^{-1}$}
\newline{\footnotesize
  (a. expand and crunch.\hfill b. use $w=b\hat{x}$, $u=\partial_x$.
  \hfill c. use ``scatter and glow''.)}

\item $\bbO(e^{\delta uw}|wu) =
     \bbO(\nu e^{\nu\delta uw}|uw)$

\item $\bbO(e^{\beta u+\alpha w+\delta uw}|wu)
     = \bbO(\nu
       e^{-b\nu\alpha\beta+\nu\alpha w+\nu\beta u+\nu\delta uw}
     |uw)$

\end{enumerate}

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{\red The Big $\frakg_1$ Lemma.}
\[
  \bbO\left(e^{\alpha w+\beta u+\delta uw}|wu\right) = \bbO\left(\nu
  (1+\e\nu\Lambda) e^{\nu(-b\alpha\beta+\alpha w+\beta u+\delta uw)}|cuw\right)
\]

Here $\Lambda$ is for {\greektext L'ogos}, ``a principle of order and knowledge'', a balanced
quartic in $\alpha$, $\beta$, $c$, $u$, and $w$:
\begin{align*} \Lambda = &
  - b\nu\left(\nu^2\alpha^2\beta^2+4\delta\nu\alpha\beta+2\delta^2\right)/2
  - \delta\nu^3(3b\delta+2)\beta^2u^2/2 \\
& - b\delta^4\nu^3u^2w^2/2
  - \delta^2\nu^3(2b\delta+1)\beta u^2w \\
& - \nu^2(2b\delta+1)(\nu\alpha\beta+2\delta)\beta u
  - 2b\delta^2\nu^2(\nu\alpha\beta+\delta)uw \\
& + \delta\nu^3(b\delta+2)\alpha^2w^2/2
  + 2(\nu\alpha\beta+\delta)c
  + 2\delta\nu\beta cu
  + 2\delta^2\nu cuw \\
& + 2\delta\nu \alpha cw
  + \delta^2\nu^3\alpha uw^2
  + \nu^2(\nu\alpha\beta+2\delta)\alpha w.
\end{align*}

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\entry{160801a} Roland's $sm_qsl(2)$ formulas, \bbs{VanDerVeen}{160731}{180741}, with $q=e^\e$, $t=e^b$: $b$ central, $[w,c]=w$, $[c,u]=u$, $wu-quw=1-te^{2\e c}$ (at $\e^2=0$: $[w,u]=\e uw+1-t-2\e tc$), $R=\sum_{m,n}\frac{u^n(b+\e c)^m\otimes c^mw^n}{m![n]_q!}\to \sum_{m,n}\frac{u^n(b+\e c)^m\otimes c^mw^n}{m!n!}\left(1-\frac{\e}{2}\binom{n}{2}\right)$. Also, $\Delta(b,c,u,w)=(b_1+b_2,c_1+c_2,t_2e^{\e c_2}u_1+u_2, e^{\e c_2}w_1+w_2)$ and $S(b,c,u,w)=(-b,-c,-t^{-1}ue^{-\e c},-we^{-\e c})$. Verified {\tt VdVAlgebraAt1-Testing.nb}.

\entry{160801b} $sm_0sl(2)$ formulas, $t=e^b$: $b$ central, $[w,c]=w$, $[c,u]=u$, $wu-uw=1-t$, $R=\sum_{m,n}\frac{u^nb^m\otimes c^mw^n}{m!n!}$ (verified {\tt VdVAlgebraAt0.nb}). Also, $\Delta(b,c,u,w)=(b_1+b_2,c_1+c_2,t_2u_1+u_2, w_1+w_2)$ and $S(b,c,u,w)=(-b,-c,-t^{-1}u,-w)$ (unverified).

\entry{160730} Lessons from Roland: $\bullet$ There is an additional grading, with $ht(b,c,u,w)=(0,0,1,-1)$. $\bullet$ Rescale $u\to\frac{b}{e^b-1}u$. $\bullet$ A simple $R$-matrix for 1-co.

\entry{160628} Figure out duality in $\frakg_1$!

\entry{160622b} $ad(-a_{12})=\{c_1\mapsto u_1w_2$, \hfill $c_2\mapsto -u_1w_2$,\newline
  \null\quad$u_1\mapsto \e u_1c_2$, \hfill ${u_2\mapsto -(b_1-\e c_1)u_2+(b_2-2\e c_2)u_1}$, \newline
  \null\quad$w_1\mapsto -b_1w_2-\e w_1c_2+2\e c_1w_2$, \hfill $w_2\mapsto (b_1-\e c_1)w_2\}$.

{\bf Claim.} Over $\bbQ\llbracket \e, b_i\rrbracket$ the following generate a sub-Lie algebra, sub-meta-monoid, and contains the $a_{ij}$'s:
\[ \{ 1, c_i, u_i, w_i, u_iw_j\} \quad \text{and} \]
\[ \e\{c_ic_j, c_iu_j, c_iw_j, c_iu_jw_k, u_iu_jw_k, u_iw_jw_k, u_iu_jw_kw_l\} \]

\entry{160621} 1-co low algebra ($\e^2=0$): \hfill $a_{12}=I=b_1c_2+u_1w_2\in\frakb_\e^\ast\otimes\frakb_\e$
\newline $[w,c]=w$
\hfill $[b,u]=-\e u$
\hfill $\delta c=0$
\hfill $\delta w=\e(c\wedge w)$
\newline $[b,c]=0$
\hfill $[b,w]=\e w$
\hfill $[c,u]=u$
\hfill $[u,w]=b-\e c$
\newline\null\hfill(verification in \href{http://drorbn.net/AcademicPensieve/2016-06}{pensieve://2016-06})

Also, $ad(-a_{12})=\{u_1\mapsto \e u_1c_2,\ u_2\mapsto -b_1u_2+b_2u_1-\e u_1c_2,\ b_1\mapsto -\e u_1w_2,\ b_2\mapsto \e u_1w_2,\ w_1\mapsto -b_1w_2-\e w_1c_2+\e c_1w_2,\ w_2\mapsto b_1w_2,\ c_1\mapsto u_1w_2,\ c_2\mapsto -u_1w_2\}$.

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{\bf Recycling.}

\end{multicols*}

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