%%%%% This is the source code for my article ``Poisson Lie linear algebra in the graphical language''.  A PDF is available at <http://math.berkeley.edu/~theojf/GraphicalLanguage.pdf>.


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             pdftitle={Poisson Lie linear algebra in the graphical language},
             pdfauthor={Theo Johnson-Freyd},
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\newcommand{\obj}[1][string=green!50!black]{\,\raisebox{-.5ex}{\tikz \draw[#1] (0em,0em) -- (0em,1em);}\,}

\newcommand{\gggg}{\obj[lie]}
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\title{Poisson Lie linear algebra \\ in the graphical language}
\author{Theo Johnson-Freyd}

\begin{document}

\maketitle

\tableofcontents

\section{Introduction}

The two primary purposes of this article are to explain some basic notions from the theory of Lie bialgebras in a graphical language, and to teach the graphical language to readers who know some Lie bialebra theory.\footnote{In fact, another goal of this article was to provide the author practice creating diagrams in the typesetting language \TikZ.  We refer the reader to \cite{TikZ} for a complete description.  If the reader is interested in learning to draw such diagrams, she is welcome to inspect the source code for this document, available at \url{http://math.berkeley.edu/~theojf/GraphicalLanguage.tex}.}

The graphical language, which we will not completely define here, was first introduced in 1971 by Roger Penrose in \cite{Penrose}, and constitutes a merging of two computational conventions used in physics: Einstein's index summation convention, and Feynman's diagrams.  The graphical language has been used to study Lie groups, c.f.\ \cite{Stedman} and \cite{Cvitanovic}.  The language has been expanded quite a lot, most notably to represent constructions in arbitrary tensor categories.  For details, see for example \cite{Selinger} and references therein.

For the purposes of this discussion, it suffices to adopt much closer to Penrose's original meanings.  We pick a finite-dimensional vector space, and denote it, its dual space, and the canonical pairing and unit maps by: 
\[ \begin{tikzpicture}
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (1,1) -- (1,0);
  \draw[lie] (2,0) .. controls +(0,1) and +(0,1) .. (3,0);
  \draw[lie] (4,1) .. controls +(0,-1) and +(0,-1) .. (5,1);
\end{tikzpicture} \]
As we are working in the category of vector spaces, there's nothing interesting to say about crossings (\,\raisebox{-.5ex}{\tikz \draw[blue] (0em,0em) .. controls +(0ex,1ex) and +(0ex,-1ex) .. (1em,1em) (1em,0em) .. controls +(0ex,1ex) and +(0ex,-1ex) .. (0em,1em);}\,) and the like.  We will not discuss infinite-dimensional spaces, but most of these constructions can be made to work in more generality.

It is a matter of taste which direction diagrams ``go.'' In this document we have elected to ``let time go up'', so that morphisms are read from bottom-to-top.  This is largely for ease of typesetting --- Cartesian coordinates are more common than ``matrix'' coordinates --- and because it more closely matches the physicists' conventions from which the notation was derived.

The author learned about the Drinfeld Double and the Chevalley Complex from a graduate seminar on Quantum Groups by Nicolai Reshetikhin.  See \cite{QG} for class notes and references.  In particular, \cite{ES} makes regular use of the graphical language, and treats not only the linear algebra of this article but the non-linear group theory and geometry of Poisson Lie and Quantum groups.

\section{Lie algebras and coalgebras}

A \defn{Lie algebra structure} on our vector space is a trivalent vertex \\
\[ \begin{tikzpicture}
  \node[circle,draw=blue,fill=blue,inner sep=1pt] at (0,0) {};
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
\end{tikzpicture} \]
which should be \defn{antisymmetric}: \[
\begin{tikzpicture}[scale=.7]
  \node[circle,draw=blue,fill=blue,inner sep=1pt] at (0,0) {};
  \draw[lie] (0,0) -- (0,2);
  \draw[lie] (-1,-2) .. controls +(90:1) and +(210:1) .. (0,0);
  \draw[lie] (1,-2) .. controls +(90:1) and +(330:1) .. (0,0);
  \node at (2,0) {$+$};
  \node[circle,draw=blue,fill=blue,inner sep=1pt] at (4,0) {};
  \draw[lie] (4,0) -- (4,2);
  \draw[lie] (3,-2) .. controls +(90:1) and +(330:2) ..  (4,0);
  \draw[lie] (5,-2) .. controls +(90:1) and +(210:2) .. (4,0);
  \node at (6,0) {$=0$};
\end{tikzpicture} \]
and also must satisfy the \defn{Jacobi identity}: \[
\begin{tikzpicture}[scale=1]
  \draw[lie] (1,0) .. controls +(90:.5) and +(210:.5) .. (1.5,1) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (2,0) .. controls +(90:.5) and +(330:.5) .. (1.5,1);
  \draw[lie] (1.5,1) .. controls +(90:.5) and +(210:.5) .. (2,2) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (3,0) .. controls +(90:.5) and +(330:.5) .. (2,2);
  \draw[lie] (2,2) .. controls +(90:.5) and +(270:.5) .. (2,3);
  \node at (3.5,1.5) {$+$};
  \draw[lie] (5,0) .. controls +(90:.5) and +(210:1) .. (4.5,1) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (6,0) .. controls +(90:.5) and +(330:.5) .. (4.5,1);
  \draw[lie] (4.5,1) .. controls +(90:.5) and +(210:.5) .. (5,2) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (4,0) .. controls +(90:.5)  and +(330:1) .. (5,2);
  \draw[lie] (5,2) .. controls +(90:.5) and +(270:.5) .. (5,3);
  \node at (6.5,1.5) {$+$};
  \draw[lie] (9,0) .. controls +(90:.5) and +(210:1) .. (7.5,1) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (7,0) .. controls +(90:.5) and +(330:1) .. (7.5,1);
  \draw[lie] (7.5,1) .. controls +(90:.5) and +(210:.5) .. (8,2) node[circle,draw=blue,fill=blue,inner sep=1pt] {};
  \draw[lie] (8,0) .. controls +(90:.5)  and +(330:1) .. (8,2);
  \draw[lie] (8,2) .. controls +(90:.5) and +(270:.5) .. (8,3);
  \node at (9.5,1.5) {$=0$};
\end{tikzpicture} \]

Let \mmmm\ be another vector space.  We define a \defn{left action} of \gggg\ on \mmmm\ to be a vertex \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[circle,draw=red,fill=red,inner sep=1pt] at (0,0) {};
\end{tikzpicture}}\, satisfying:
\[\begin{tikzpicture}
  \draw[lie] (1,0) .. controls +(90:.5) and +(210:.5) .. (1.5,1) node[dot=blue] {};
  \draw[lie] (2,0) .. controls +(90:.5) and +(330:.5) .. (1.5,1);
  \draw[lie] (1.5,1) .. controls +(90:.5) and +(210:.5) .. (2,2) node[dot=red] {};
  \draw[module] (3,0) .. controls +(90:.5) and +(330:.5) .. (2,2);
  \draw[module] (2,2) .. controls +(90:.5) and +(270:.5) .. (2,3);
  \node at (3.5,1.5) {$=$};
  \draw[module] (6,0) .. controls +(90:.5) and +(330:.5) .. (5.5,1) node[dot=red] {};
  \draw[lie] (5,0) .. controls +(90:.5) and +(210:.5) .. (5.5,1);
  \draw[module] (5.5,1) .. controls +(90:.5) and +(330:.5) .. (5,2) node[dot=red] {};
  \draw[lie] (4,0) .. controls +(90:.5) and +(210:.5) .. (5,2);
  \draw[module] (5,2) .. controls +(90:.5) and +(270:.5) .. (5,3);
  \node at (6.5,1.5) {$-$};
  \draw[module] (9,0) .. controls +(90:.5) and +(330:.5) .. (8.5,1) node[dot=red] {};
  \draw[lie] (7,0) .. controls +(90:.5) and +(210:.5) .. (8.5,1);
  \draw[module] (8.5,1) .. controls +(90:.5) and +(330:.5) .. (8,2) node[dot=red] {};
  \draw[lie] (8,0) .. controls +(90:.5) and +(210:1) .. (8,2);
  \draw[module] (8,2) .. controls +(90:.5) and +(270:.5) .. (8,3);
\end{tikzpicture}\]
A \defn{right action} is a vertex \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[module] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=red] at (0,0) {};
\end{tikzpicture}}\, with defining relation the reflection of the above.  The Jacobi and antisymmetry identities assure that \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=blue] at (0,0) {};
\end{tikzpicture}}\, is both a left and a right action of \gggg\ on itself, called the \defn{adjoint action}.  Moreover, if \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=red] at (0,0) {};
\end{tikzpicture}}\, is a right action, then:
\[  \begin{tikzpicture}
  \node at (-1,0) {$-$};
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (.5,-1) .. controls +(90:.5) and +(210:1) .. (0,0);
  \draw[module] (-.5,-1) .. controls +(90:.5) and +(330:1) .. (0,0);
  \node[dot=red] at (0,0) {};  
\end{tikzpicture} \]
is a left action, and conversely.  In particular, this relates the left and right adjoint actions.

If \gggg\ acts on \mmmm\ from the left, and if \mmmm\ has a dual space \,\raisebox{-.5ex}{\tikz \draw[module] (0em,1em) -- (0em,0em);}\,, then we can define a right action of \gggg\ on \,\raisebox{-.5ex}{\tikz \draw[module] (0em,1em) -- (0em,0em);}\, by:
\[ \begin{tikzpicture}
  \draw[module] (0,1) -- (0,0) node[dot=red] {};
  \draw[module] (0,0) .. controls +(210:.5) and +(90:.5) .. (-.5,-1);
  \draw[lie] (.5,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node at (1,0) {$\defined$};
  \draw[module] (2,0) .. controls +(90:.5) and +(90:1.5) .. (1.5,-1);
  \draw[lie] (2.5,-1) .. controls +(90:.5) and +(210:1) .. (2,0) node[dot=red] {};
  \draw[module] (2,1) .. controls +(270:.5) and +(330:1) .. (2,0);
\end{tikzpicture} \]
Similarly, we can turn a right action into a left action on the dual space (and, of course, switch between left and right actions at the cost of a minus sign).  In this way, we can define the left and right \defn{co-adjoint actions} of \gggg\ on \gggd:
\[ \begin{tikzpicture}
  \draw[lie] (0,1) -- (0,0) node[dot=blue] {};
  \draw[lie] (0,0) .. controls +(210:.5) and +(90:.5) .. (-.5,-1);
  \draw[lie] (.5,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node at (1,0) {$\defined$};
  \draw[lie] (2,0) .. controls +(90:.5) and +(90:1.5) .. (1.5,-1);
  \draw[lie] (2.5,-1) .. controls +(90:.5) and +(210:1) .. (2,0) node[dot=blue] {};
  \draw[lie] (2,1) .. controls +(270:.5) and +(330:1) .. (2,0);
  
  \node at (3.5,0) {and};
  
  \draw[lie] (5,1) -- (5,0) node[dot=blue] {};
  \draw[lie] (5,0) .. controls +(330:.5) and +(90:.5) .. (5.5,-1);
  \draw[lie] (4.5,-1) .. controls +(90:.5) and +(210:.5) .. (5,0);
  \node at (6,0) {$\defined$};
  \draw[lie] (7,0) .. controls +(90:.5) and +(90:1.5) .. (7.5,-1);
  \draw[lie] (6.5,-1) .. controls +(90:.5) and +(330:1) .. (7,0) node[dot=blue] {};
  \draw[lie] (7,1) .. controls +(270:.5) and +(210:1) .. (7,0);
\end{tikzpicture} \]
Just as with the adjoint action, the coadjoint action is antisymmetric in the following sense:
\[ \begin{tikzpicture}
  \draw[lie] (0,1) -- (0,0) node[dot=blue] {};
  \draw[lie] (0,0) .. controls +(210:1.5) and +(90:.5) .. (.5,-1);
  \draw[lie] (-.5,-1) .. controls +(90:.5) and +(330:1.5) .. (0,0);
  \node at (1,0) {$=$}; \node at (1.5,0) {$-$};
  \draw[lie] (2.5,1) -- (2.5,0) node[dot=blue] {};
  \draw[lie] (2.5,0) .. controls +(330:.5) and +(90:.5) .. (3,-1);
  \draw[lie] (2,-1) .. controls +(90:.5) and +(210:.5) .. (2.5,0);
\end{tikzpicture} \]
We remark that one can also easily check the following relationship between the two coadjoint actions:
\[ \begin{tikzpicture}
  \draw[lie] (0,1) -- (0,0) node[dot=blue] {};
  \draw[lie] (0,0) .. controls +(330:.5) and +(90:.5) .. (.5,-1);
  \draw[lie] (-.5,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \node at (1,0) {$\defined$};
  \draw[lie] (1.5,-1) .. controls +(90:1.5) and +(90:.5) .. (2,0) node[dot=blue] {};
  \draw[lie] (2,0) .. controls +(210:1) and +(90:.5) .. (2.5,-1);
  \draw[lie] (2,1) .. controls +(270:.5) and +(330:1) .. (2,0);
\end{tikzpicture} \]

We now dualize all these notions.  A \defn{Lie coalgebra} structure on \gggg\ is a map \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.25]
  \draw[lie] (0,-1) -- (0,0) node[dot=blue] {};
  \draw[lie] (0,0) .. controls +(150:.5) and +(270:.5) .. (-1,1);
  \draw[lie] (0,0) .. controls +(30:.5) and +(270:.5) .. (1,1);
\end{tikzpicture}}\, satisfying the vertical reflections of the antisymmetry and Jacobi relations above.  Equivalently, a Lie coalgebra structure on \gggg\ is a Lie algebra structure on \gggd.  To define such an equivalence, we must choose a dual to the tensor product \,\raisebox{-.5ex}{\begin{tikzpicture} \draw[lie] (0em,0em) -- (0em,1em); \draw[lie] (.5em,0em) -- (.5em,1em); \end{tikzpicture}}\,.  Of course, the dual space is \,\raisebox{-.5ex}{\begin{tikzpicture} \draw[lie] (0em,1em) -- (0em,0em); \draw[lie] (.5em,1em) -- (.5em,0em); \end{tikzpicture}}\,, but whereas \cite{QG} takes the pairing \,\raisebox{0ex}{\begin{tikzpicture} \draw[lie] (1.5em,0em) .. controls +(0em,1em) and +(0em,1em) .. (0em,0em); \draw[lie] (2em,0em) .. controls +(0em,1em) and +(0em,1em) .. (.5em,0em); \end{tikzpicture}}\,, we will take \,\raisebox{0ex}{\begin{tikzpicture} \draw[lie] (1.5em,0em) .. controls +(0em,.5em) and +(0em,.5em) .. (.5em,0em); \draw[lie] (2em,0em) .. controls +(0em,1em) and +(0em,1em) .. (0em,0em); \end{tikzpicture}}\,.  Then we can relate the bracket on \gggd\ to the cobracket on \gggg\ by:
\[
  \begin{tikzpicture}
    \draw[lie] (0,0) node[dot=blue] {} -- (0,1);
    \draw[lie] (-.5,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
    \draw[lie] (.5,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
    \node at (1,0) {$=$};
    \draw[lie] (2,0) node[dot=blue] {} .. controls +(270:.5) and +(270:.5) .. (2.5,0) .. controls +(90:.5) and +(270:.5) .. (2,1);
    \draw[lie] (1.5,-1) .. controls +(90:1) and +(210:.25) .. (1.75,.5) .. controls +(30:.25) and +(30:.5) .. (2,0);
    \draw[lie] (2.5,-1) .. controls +(90:.5) and +(315:.25) .. (1.75,-.25) .. controls +(135:.25) and +(150:.5) .. (2,0);
    \node at (3,0) {$=$};
    \draw[lie] (4,0) node[dot=blue] {} .. controls +(270:.5) and +(270:.5) .. (3.5,0) .. controls +(90:.5) and +(270:.5) .. (4,1);
    \draw[lie] (4.5,-1) .. controls +(90:1) and +(330:.25) .. (4.25,.5) .. controls +(150:.25) and +(150:.5) .. (4,0);
    \draw[lie] (3.5,-1) .. controls +(90:.5) and +(225:.25) .. (4.25,-.25) .. controls +(45:.25) and +(30:.5) .. (4,0);    
\end{tikzpicture} \]

In this way, if \gggg\ is equipped with both a Lie algebra and a Lie coalgebra structure, we can interpret any trivalent vertex with one or two incoming and two or one outgoing edges.

\section{Lie bialgebras and the Drinfeld Double}

Let \gggg\ be equipped with both a Lie algebra structure and a Lie coalgebra structure.  We say that this forms a \defn{Lie bialgebra} if, in addition to the identities demanded in the previous section, the Lie algebra and coalgebra satisfy the following coherence relation:
\[\begin{tikzpicture}

\draw[lie] (0,0) .. controls +(90:.5) and +(210:.5) .. (1,1);
\draw[lie] (2,0) .. controls +(90:.5) and +(330:.5) .. (1,1);
\draw[lie] (1,1) -- (1,2);
\draw[lie] (1,2) .. controls +(150:.5) and +(270:.5) .. (0,3);
\draw[lie] (1,2) .. controls +(30:.5) and +(270:.5) .. (2,3);
\draw (1,1) node[dot=blue] {} (1,2) node[dot=blue] {} ;

\draw[lie] (3,0) .. controls +(90:.5) and +(240:.5) .. (3.5,1.5);
\draw[lie] (3.5,1.5) .. controls +(120:.5) and +(270:.5) .. (3,3);
\draw[lie] (3.5,1.5) -- (4.5,1.5);
\draw[lie] (5,0) .. controls +(90:.5) and +(300:.5) .. (4.5,1.5);
\draw[lie] (4.5,1.5) .. controls +(60:.5) and +(270:.5) .. (5,3);
\draw (3.5,1.5) node[dot=blue] {} (4.5,1.5) node[dot=blue] {} ;

\draw[lie] (6,0) .. controls +(90:.5) and +(240:.5) .. (6.5,1.5);
\draw[lie] (6.5,1.5) .. controls +(120:.5) and +(270:.5) .. (6,3);
\draw[lie] (7.5,1.5) -- (6.5,1.5);
\draw[lie] (8,0) .. controls +(90:.5) and +(300:.5) .. (7.5,1.5);
\draw[lie] (7.5,1.5) .. controls +(60:.5) and +(270:.5) .. (8,3);
\draw (6.5,1.5) node[dot=blue] {} (7.5,1.5) node[dot=blue] {} ;

\draw[lie] (9,0) .. controls +(90:.5) and +(240:.5) .. (9.5,1);
\draw[lie] (9.5,1) .. controls +(120:1) and +(270:1) .. (11,3);
\draw[lie] (9.5,1) -- (10.5,1);
\draw[lie] (11,0) .. controls +(90:.5) and +(300:.5) .. (10.5,1);
\draw[lie] (10.5,1) .. controls +(60:1) and +(270:1) .. (9,3);
\draw (9.5,1) node[dot=blue] {} (10.5,1) node[dot=blue] {} ;

\draw[lie] (12,0) .. controls +(90:.5) and +(240:.5) .. (12.5,1);
\draw[lie] (12.5,1) .. controls +(120:1) and +(270:1) .. (14,3);
\draw[lie] (13.5,1) -- (12.5,1);
\draw[lie] (14,0) .. controls +(90:.5) and +(300:.5) .. (13.5,1);
\draw[lie] (13.5,1) .. controls +(60:1) and +(270:1) .. (12,3);
\draw (12.5,1) node[dot=blue] {} (13.5,1) node[dot=blue] {} ;

\node at (2.5,1.5) {$=$};
\node at (5.5,1.5) {$+$};
\node at (8.5,1.5) {$-$};
\node at (11.5,1.5) {$-$};

\end{tikzpicture}\]
One can easily see that this relation is symmetric, so that \gggd\ is also a Lie bialgebra if \gggg\ is.

If \gggg\ is a Lie coalgebra, then we define the \defn{Drinfeld Double} of \gggg\ to be the vector space $\DDDD = \gggg \oplus \gggd$, with the following bracket and cobracket:

\[\begin{tikzpicture}
\node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (1,0) {$\defined$};
\node[dot=blue] (O1) at (2,0) {};
\draw[lie] (O1) -- +(0,1);
\draw[colie] (O1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (3,0) {$+$};
\node[dot=blue] (O2) at (4,0) {};
\draw[lie] (O2) -- +(0,1);
\draw[lie] (O2) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O2) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (5,0) {$+$};
\node[dot=blue] (O3) at (6,0) {};
\draw[lie] (O3) -- +(0,1);
\draw[colie] (O3) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O3) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (7,0) {$+$};
\node[dot=blue] (O4) at (8,0) {};
\draw[colie] (O4) -- +(0,1);
\draw[lie] (O4) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O4) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (9,0) {$+$};
\node[dot=blue] (O5) at (10,0) {};
\draw[colie] (O5) -- +(0,1);
\draw[colie] (O5) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O5) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (11,0) {$+$};
\node[dot=blue] (O6) at (12,0) {};
\draw[colie] (O6) -- +(0,1);
\draw[lie] (O6) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O6) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);

\node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (0,-3) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\node at (1,-3) {$\defined$};
\node[dot=blue] (T1) at (2,-3) {};
\draw[colie] (T1) -- +(0,-1);
\draw[lie] (T1) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[lie] (T1) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\node at (3,-3) {$-$};
\node[dot=blue] (T2) at (4,-3) {};
\draw[lie] (T2) -- +(0,-1);
\draw[colie] (T2) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[colie] (T2) .. controls +(30:.5) and +(270:.5) .. +(.5,1);

\end{tikzpicture}\]
We should explain the meaning of the above sum.  As a vector space, the double comes with a splitting into a direct sum.  Thus, we can evaluate a tensor by projecting in all ways onto direct summands, evaluating on each projection, and summing.  This is exactly what is expressed above.  

It is immediate that the cobracket \raisebox{-.5ex}{\begin{tikzpicture}[scale=.25] \node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (0,-3) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\end{tikzpicture}} is a Lie coalgebra structure on \DDDD: indeed, it makes \DDDD\ into the direct sum of coalgebras (the coalgebra identities are homogeneous, and so any scalar times a Lie coalgebra structure gives another Lie coalgebra).

On the other hand, \raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\end{tikzpicture}} is only a Lie algebra structure if \gggg\ is a Lie bialgebra.  It is immediately antisymmetric, but the Jacobi identity gives trouble:
\[ \begin{tikzpicture}[scale=.75]
\draw (0,0) node[circle,draw=blue,inner sep=2pt] (A1) {} +(.5,1) node[circle,draw=blue,inner sep=2pt] (A2) {};
\draw[drinf] (A1) .. controls +(90:.5) and +(210:.5) .. (A2);
\draw[drinf] (A2) -- +(0,1);
\draw[drinf] (A1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (A1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\draw[drinf] (A2) .. controls +(330:1) and +(90:1) .. +(1,-2);
\path (A1) +(2,.5) node {$+$};
\draw (3,0) node[circle,draw=blue,inner sep=2pt] (B1) {} +(.5,1) node[circle,draw=blue,inner sep=2pt] (B2) {};
\draw[drinf] (B1) .. controls +(90:.5) and +(210:.5) .. (B2);
\draw[drinf] (B2) -- +(0,1);
\draw[drinf] (B1) .. controls +(210:1) and +(90:.5) .. +(.5,-1);
\draw[drinf] (B1) .. controls +(330:.5) and +(90:.5) .. +(1.5,-1);
\draw[drinf] (B2) .. controls +(330:2) and +(90:.5) .. +(-1,-2);
\path (B1) +(2,.5) node {$+$};
\draw (6,0) node[circle,draw=blue,inner sep=2pt] (C1) {} +(.5,1) node[circle,draw=blue,inner sep=2pt] (C2) {};
\draw[drinf] (C1) .. controls +(90:.5) and +(210:.5) .. (C2);
\draw[drinf] (C2) -- +(0,1);
\draw[drinf] (C1) .. controls +(210:1) and +(90:.5) .. +(1.5,-1);
\draw[drinf] (C1) .. controls +(330:1) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (C2) .. controls +(330:1) and +(90:.5) .. +(0,-2);
\path (C1) +(2,.5) node {$=$};

\draw (9,0) node[dot=blue] (D1) {} +(.5,1) node[dot=blue] (D2) {};
\draw[lie] (D1) .. controls +(90:.5) and +(210:.5) .. (D2);
\draw[lie] (D2) -- +(0,1);
\draw[colie] (D1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (D1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\draw[colie] (D2) .. controls +(330:1) and +(90:1) .. +(1,-2);
\path (D1) +(2,.5) node {$+$};
\draw (12,0) node[dot=blue] (E1) {} +(.5,1) node[dot=blue] (E2) {};
\draw[lie] (E1) .. controls +(90:.5) and +(210:.5) .. (E2);
\draw[lie] (E2) -- +(0,1);
\draw[colie] (E1) .. controls +(210:1) and +(90:.5) .. +(.5,-1);
\draw[colie] (E1) .. controls +(330:.5) and +(90:.5) .. +(1.5,-1);
\draw[colie] (E2) .. controls +(330:2) and +(90:1) .. +(-1,-2);
\path (E1) +(2,.5) node {$+$};
\draw (15,0) node[dot=blue] (F1) {} +(.5,1) node[dot=blue] (F2) {};
\draw[lie] (F1) .. controls +(90:.5) and +(210:.5) .. (F2);
\draw[lie] (F2) -- +(0,1);
\draw[colie] (F1) .. controls +(210:1) and +(90:.5) .. +(1.5,-1);
\draw[colie] (F1) .. controls +(330:1) and +(90:.5) .. +(-.5,-1);
\draw[colie] (F2) .. controls +(330:1) and +(90:1) .. +(0,-2);
\path (F1) +(2,.5) node {$+$};

\draw (10,-4) node[dot=blue] (G1) {} +(.5,1) node[dot=blue] (G2) {};
\draw[lie] (G1) .. controls +(90:.5) and +(210:.5) .. (G2);
\draw[lie] (G2) -- +(0,1);
\draw[colie] (G1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (G1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\draw[lie] (G2) .. controls +(330:1) and +(90:1) .. +(1,-2);
\path (G1) +(2,.5) node {$+$};
\draw (13,-4) node[dot=blue] (H1) {} +(.5,1) node[dot=blue] (H2) {};
\draw[lie] (H1) .. controls +(90:.5) and +(210:.5) .. (H2);
\draw[lie] (H2) -- +(0,1);
\draw[lie] (H1) .. controls +(210:1) and +(90:.5) .. +(.5,-1);
\draw[lie] (H1) .. controls +(330:.5) and +(90:.5) .. +(1.5,-1);
\draw[colie] (H2) .. controls +(330:2) and +(90:1) .. +(-1,-2);
\path (H1) +(2,.5) node {$+$};
\draw (16,-4) node[dot=blue] (I1) {} +(.5,1) node[dot=blue] (I2) {};
\draw[lie] (I1) .. controls +(90:.5) and +(210:.5) .. (I2);
\draw[lie] (I2) -- +(0,1);
\draw[lie] (I1) .. controls +(210:1) and +(90:.5) .. +(1.5,-1);
\draw[colie] (I1) .. controls +(330:1) and +(90:.5) .. +(-.5,-1);
\draw[lie] (I2) .. controls +(330:1) and +(90:1) .. +(0,-2);
\path (I1) +(2,.5) node {$+$};
\path (G1) +(-1,.5) node {$+$} ++(-1,-2) node {$+$} +(1em,1em) node[anchor=north west, text width=2.75in] {six more triples with three out of four external edges oriented the same; thus, the interior edge is determined uniquely};

\draw[red,thick,decoration={brace,amplitude=10},decorate] (18.75,2) -- (18.75,-7.5);
\node[anchor=west, text width=1in,text=red] at (19.25,-2.75) {These sums vanish by Jacobi and co-Jacobi.};

\draw (10,-10) node[dot=blue] (J1) {} +(.5,1) node[dot=blue] (J2) {};
\draw[lie] (J1) .. controls +(90:.5) and +(210:.5) .. (J2);
\draw[lie] (J2) -- +(0,1);
\draw[colie] (J1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (J1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\draw[lie] (J2) .. controls +(330:1) and +(90:1) .. +(1,-2);
\path (J1) +(2,.5) node {$+$};
\draw (13,-10) node[circle,draw=blue,inner sep=2pt] (K1) {} +(.5,1) node[circle,draw=blue,inner sep=2pt] (K2) {};
\draw[drinf] (K1) .. controls +(90:.5) and +(210:.5) .. (K2);
\draw[lie] (K2) -- +(0,1);
\draw[colie] (K1) .. controls +(210:1) and +(90:.5) .. +(.5,-1);
\draw[lie] (K1) .. controls +(330:.5) and +(90:.5) .. +(1.5,-1);
\draw[colie] (K2) .. controls +(330:2) and +(90:1) .. +(-1,-2);
\path (K1) +(2,.5) node {$+$};
\draw (16,-10) node[circle,draw=blue,inner sep=2pt] (L1) {} +(.5,1) node[circle,draw=blue,inner sep=2pt] (L2) {};
\draw[drinf] (L1) .. controls +(90:.5) and +(210:.5) .. (L2);
\draw[lie] (L2) -- +(0,1);
\draw[lie] (L1) .. controls +(210:1) and +(90:.5) .. +(1.5,-1);
\draw[colie] (L1) .. controls +(330:1) and +(90:.5) .. +(-.5,-1);
\draw[colie] (L2) .. controls +(330:1) and +(90:1) .. +(0,-2);
\path (L1) +(2,.5) node {$+$};
\path (J1) +(-1,.5) node {$+$} ++(-1,-2) node {$+$} +(1em,1em) node[anchor=north west, text width=2.75in] {five more triples with two external edges of each type; in one diagram in each triple, the interior edge is determined}
 ++(0,-2.5) node[text=red] {$+$} +(1em,1em) node[anchor=north west, text width=2.75in, text=red] {zero diagrams with all external edges going in or all going out};
 
\draw[purple,thick,decoration={brace,amplitude=10},decorate] (8.75,-11) -- (8.75,-8);
\node[text=purple] at (8,-9.5) {$=$};

\fill[purple!10,rounded corners] (7.25,-4.25) rectangle (.75,-14.75);

\draw (1.5,-6.5) node[dot=blue] (Z1) {} +(0,.5) node[dot=blue] (Z2) {};
\draw[lie] (Z1) -- (Z2);
\draw[colie] (Z1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (Z1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\draw[lie] (Z2) .. controls +(30:1) and +(90:.5) .. +(1.5,-1.5);
\draw[lie] (Z2) .. controls +(150:1) and +(270:.5) .. +(.5,1.5);
\path (Z2) +(2,0) node {$-$};
\path (Z2) ++(3,0) node[dot=blue] (Y1) {} +(1,0) node[dot=blue] (Y2) {};
\draw[lie] (Y1) -- (Y2);
\draw[colie] (Y1) .. controls +(240:.5) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (Y1) .. controls +(120:1) and +(270:.5) .. +(.5,1.5);
\draw[colie] (Y2) .. controls +(300:1) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (Y2) .. controls +(60:1) and +(90:.5) .. +(.5,-1.5);
\path (Y1) +(2,0) node {$-$};
\path (Z1) ++(0,-3) node {$-$} ++(1,0) node[dot=blue] (X1) {} +(1,0) node[dot=blue] (X2) {};
\draw[colie] (X1) -- (X2);
\draw[colie] (X1) .. controls +(240:.5) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (X1) .. controls +(120:1) and +(270:.5) .. +(.5,1.5);
\draw[colie] (X2) .. controls +(300:1) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (X2) .. controls +(60:1) and +(90:.5) .. +(.5,-1.5);
\path (X1) +(2,0) node {$+$};
\path (X1) ++(3,0) node[dot=blue] (W1) {} +(1,0) node[dot=blue] (W2) {};
\draw[lie] (W1) -- (W2);
\draw[colie] (W1) .. controls +(240:.5) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (W2) .. controls +(60:1) and +(270:.5) .. +(-.5,1.5);
\draw[colie] (W2) .. controls +(300:1) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (W1) .. controls +(120:2) and +(90:1) .. +(1.5,-1.5);
\path (Z1) ++(0,-6.5) node {$+$} ++(1,0) node[dot=blue] (V1) {} +(1,0) node[dot=blue] (V2) {};
\draw[colie] (V1) -- (V2);
\draw[colie] (V1) .. controls +(240:.5) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (V2) .. controls +(60:1) and +(270:.5) .. +(-.5,1.5);
\draw[colie] (V2) .. controls +(300:1) and +(90:.5) .. +(-.5,-1.5);
\draw[lie] (V1) .. controls +(120:2) and +(90:1) .. +(1.5,-1.5);

\draw[purple,thick,decoration={brace,amplitude=10},decorate] (7.25,-4) -- (7.25,-15);

\draw[purple,thick,decoration={brace,amplitude=10},decorate] (18.75,-8) -- (18.75,-13.5);

\node[text=purple,text width=1in,anchor=west] at (19.25,-10.75) {These sums vanish if and only if \gggg\ is a Lie bialgebra.  See purple rectangle to the left.  There cannot be cancelations between diagrams with miss-matched external edges.};

\end{tikzpicture} \]

We have defined two different (co)brackets on \DDDD: \raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\end{tikzpicture}}, the sum of all canonical actions of \gggg\ and \gggd\ on each other and themselves, and \raisebox{-1ex}{\begin{tikzpicture}[scale=.25] \node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (0,-3) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\end{tikzpicture}}, the difference of cobrackets on \gggg\ and \gggd.  Repeating what we said above, the latter is always a Lie cobracket, and the former is a Lie cobracket if and only if \gggg\ is a Lie bialgebra.  We have not marked a direction on \DDDD\ because it has a canonical pairing \raisebox{-.5ex}{\tikz \draw[drinf] (0em,0em) .. controls +(0em,1em) and +(0em,1em) .. (1em,0em);}.  We have proposed a circle for the first bracket and a square for the second, because the former is rotation-invariant like \raisebox{-.5ex}{\begin{tikzpicture}[scale=.25] \node[dot=blue] (T) at (0,-3) {};
\draw[colie] (T) -- +(0,-1);
\draw[lie] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[lie] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\end{tikzpicture}} whereas the latter is not.

One now can ask if it is possible to turn \DDDD\ into a Lie bialgebra with some combination of these brackets.  It is a direct calculation, in the manner of the check of Jacobi on the previous page, that neither bracket alone along with its rotation makes \DDDD\ into a Lie bialgebra.  However, together the brackets do.  It's standard to think of \raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\end{tikzpicture}} as the ``forward'' bracket and \raisebox{-1ex}{\begin{tikzpicture}[scale=.25] \node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (0,-3) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\end{tikzpicture}} as the ``backward'' cobracket.  The coherence relation reads:

\[\begin{tikzpicture}

\draw (1,1) node[circle,draw=blue,inner sep=2pt] (A1) {} (1,2) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (A2) {} ;
\draw[drinf] (0,0) .. controls +(90:.5) and +(210:.5) .. (A1);
\draw[drinf] (2,0) .. controls +(90:.5) and +(330:.5) .. (A1);
\draw[drinf] (A1) -- (A2);
\draw[drinf] (A2) .. controls +(150:.5) and +(270:.5) .. (0,3);
\draw[drinf] (A2) .. controls +(30:.5) and +(270:.5) .. (2,3);

\draw (3.5,1.5) node[circle,draw=blue,inner sep=2pt] (B1) {} (4.5,1.5) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (B2) {} ;
\draw[drinf] (3,0) .. controls +(90:.5) and +(240:.5) .. (B1);
\draw[drinf] (B1) .. controls +(120:.5) and +(270:.5) .. (3,3);
\draw[drinf] (B1) -- (B2);
\draw[drinf] (5,0) .. controls +(90:.5) and +(300:.5) .. (B2);
\draw[drinf] (B2) .. controls +(60:.5) and +(270:.5) .. (5,3);

\draw (6.5,1.5) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (C1) {} (7.5,1.5) node[circle,draw=blue,inner sep=2pt] (C2) {} ;
\draw[drinf] (6,0) .. controls +(90:.5) and +(240:.5) .. (C1);
\draw[drinf] (C1) .. controls +(120:.5) and +(270:.5) .. (6,3);
\draw[drinf] (C1) -- (C2);
\draw[drinf] (8,0) .. controls +(90:.5) and +(300:.5) .. (C2);
\draw[drinf] (C2) .. controls +(60:.5) and +(270:.5) .. (8,3);

\draw (9.5,1) node[circle,draw=blue,inner sep=2pt] (D1) {} (10.5,1) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (D2) {} ;
\draw[drinf] (9,0) .. controls +(90:.5) and +(240:.5) .. (D1);
\draw[drinf] (D1) .. controls +(120:1) and +(270:1) .. (11,3);
\draw[drinf] (D1) -- (D2);
\draw[drinf] (11,0) .. controls +(90:.5) and +(300:.5) .. (D2);
\draw[drinf] (D2) .. controls +(60:1) and +(270:1) .. (9,3);

\draw (12.5,1) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (E1) {} (13.5,1) node[circle,draw=blue,inner sep=2pt] (E2) {} ;
\draw[drinf] (12,0) .. controls +(90:.5) and +(240:.5) .. (E1);
\draw[drinf] (E1) .. controls +(120:1) and +(270:1) .. (14,3);
\draw[drinf] (E1) -- (E2);
\draw[drinf] (14,0) .. controls +(90:.5) and +(300:.5) .. (E2);
\draw[drinf] (E2) .. controls +(60:1) and +(270:1) .. (12,3);

\node at (2.5,1.5) {$=$};
\node at (5.5,1.5) {$+$};
\node at (8.5,1.5) {$-$};
\node at (11.5,1.5) {$-$};

\end{tikzpicture}\]
To check this, one rewrites the left hand side into a sum of six terms, and then converts each into two to four terms with the Jacobi and bialgebra coherence axioms.  The minus sign in \raisebox{-1ex}{\begin{tikzpicture}[scale=.25] \node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (0,-3) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\end{tikzpicture}} is vital: without it, the coherence relation does not hold.

\section{Lie Algebra Cohomology and the Chevalley Complex}

The Chevalley cohomology of a Lie algebra is present throughout the literature.  For example, in an introductory Lie Groups book (e.g.\ \cite[section 4.4]{LGbook}) it controls the complete reducibility of modules of semisimple Lie groups and leads to Ado's Theorem.  The reader may be interested in the perspective taken in \cite{BV}, which generalizes the Chevalley cohomology to a cohomology theory of Poisson algebras.

Before we define the Chevalley Complex, we establish some notation.  Henceforth, we work only in characteristic zero.  Let \vvvv\ be a vector space.  Then we define a sequence of projections, often called \defn{alternators}:
\[ \begin{tikzpicture}
  \draw[vspace] (4,0) -- (4,2); \draw[green!50!black, fill=green!50!black] (3.9,.9) rectangle (4.1,1.1);
  \node at (4.6,1) {$\defined$};
  \draw[vspace] (5,0) -- (5,2);
  
  \draw[vspace] (7,0) -- (7,2); \draw[vspace] (7.5,0) -- (7.5,2); \draw[green!50!black, fill=green!50!black] (6.9,.9) rectangle (7.6,1.1);
  \node at (8,1) {$\defined$}; \node at (8.75,1) {$\displaystyle \frac12 \Biggl($};
  \draw[vspace] (9.25,0) -- (9.25,2); \draw[vspace] (9.75,0) -- (9.75,2);
  \node at (10.25,1) {$-$};
  \draw[vspace] (10.75,0) .. controls +(0,1) and +(0,-1) .. (11.25,2); \draw[vspace] (11.25,0) .. controls +(0,1) and +(0,-1) .. (10.75,2);
  \node at (11.55,1) {$\displaystyle \Biggr)$}; \end{tikzpicture} \]
  \[ \begin{tikzpicture}
  \draw[vspace] (-.25,-3) -- (-.25,-1); \draw[vspace] (.25,-3) -- (.25,-1); \draw[vspace] (.75,-3) -- (.75,-1); 
  \draw[green!50!black, fill=green!50!black] (-.35,-2.1) rectangle (.85,-1.9);
  \node at (1.35,-2) {$\defined$}; \node at (2,-2) {$\displaystyle \frac16 \Biggl($};
  \draw[vspace] (2.5,-3) .. controls +(0,1) and +(0,-1) .. (2.5,-1);
  \draw[vspace] (3,-3) .. controls +(0,1) and +(0,-1) .. (3,-1);
  \draw[vspace] (3.5,-3) .. controls +(0,1) and +(0,-1) .. (3.5,-1);
  \node at (4,-2) {$-$};
  \draw[vspace] (4.5,-3) .. controls +(0,1) and +(0,-1) .. (4.5,-1);
  \draw[vspace] (5,-3) .. controls +(0,1) and +(0,-1) .. (5.5,-1);
  \draw[vspace] (5.5,-3) .. controls +(0,1) and +(0,-1) .. (5,-1);
  \node at (6,-2) {$+$};
  \draw[vspace] (6.5,-3) .. controls +(0,1) and +(0,-1) .. (7.5,-1);
  \draw[vspace] (7,-3) .. controls +(0,1) and +(0,-1) .. (6.5,-1);
  \draw[vspace] (7.5,-3) .. controls +(0,1) and +(0,-1) .. (7,-1);
  \node at (8,-2) {$-$};
  \draw[vspace] (8.5,-3) .. controls +(0,1) and +(0,-1) .. (9.5,-1);
  \draw[vspace] (9,-3) .. controls +(0,1) and +(0,-1) .. (9,-1);
  \draw[vspace] (9.5,-3) .. controls +(0,1) and +(0,-1) .. (8.5,-1);
  \node at (10,-2) {$+$};
  \draw[vspace] (10.5,-3) .. controls +(0,1) and +(0,-1) .. (11,-1);
  \draw[vspace] (11,-3) .. controls +(0,1) and +(0,-1) .. (11.5,-1);
  \draw[vspace] (11.5,-3) .. controls +(0,1) and +(0,-1) .. (10.5,-1);
  \node at (12,-2) {$-$};
  \draw[vspace] (12.5,-3) .. controls +(0,1) and +(0,-1) .. (13,-1);
  \draw[vspace] (13,-3) .. controls +(0,1) and +(0,-1) .. (12.5,-1);
  \draw[vspace] (13.5,-3) .. controls +(0,1) and +(0,-1) .. (13.5,-1);
  \node at (13.8,-2) {$\displaystyle \Biggr)$};
  
  \node at (15,-2) {etc.};
\end{tikzpicture} \]
In general, the $n$th projection is the signed average of the $n!$ permutations on $n$ strands.  It is easy to check that each alternator is a projection as claimed.  Moreover, the composition of two alternators equals the longer one, in the following sense:
\[ \begin{tikzpicture}
  \draw[vspace] (0,0) -- (0,3); \draw[vspace] (1,0) -- (1,3); \draw[vspace] (1.5,0) -- (1.5,3); \draw[vspace] (2.5,0) -- (2.5,3);
  \path (.5,.5) node {\dots} (.5,1.5) node {\dots} (.5,2.5) node {\dots} (2,.5) node {\dots} (2,1.5) node {\dots} (2,2.5) node {\dots} ;
  \draw[green!50!black, fill=green!50!black] (-.1,.9) rectangle (1.1,1.1);
  \draw[green!50!black, fill=green!50!black] (-.1,1.9) rectangle (2.6,2.1);
  \node at (3.5,1.5) {$=$};
  \draw[vspace] (4.5,0) -- (4.5,3); \draw[vspace] (5.5,0) -- (5.5,3); \draw[vspace] (6,0) -- (6,3); \draw[vspace] (7,0) -- (7,3);
  \path (5,.5) node {\dots}  (5,2.5) node {\dots} (6.5,.5) node {\dots}  (6.5,2.5) node {\dots} ;
  \draw[green!50!black, fill=green!50!black] (4.4,1.4) rectangle (7.1,1.6);
  \node at (8,1.5) {$=$};
  \draw[vspace] (9,0) -- (9,3); \draw[vspace] (10,0) -- (10,3); \draw[vspace] (10.5,0) -- (10.5,3); \draw[vspace] (11.5,0) -- (11.5,3);
  \path (9.5,.5) node {\dots} (9.5,1.5) node {\dots} (9.5,2.5) node {\dots} (11,.5) node {\dots} (11,1.5) node {\dots} (11,2.5) node {\dots} ;
  \draw[green!50!black, fill=green!50!black] (8.9,1.9) rectangle (10.1,2.1);
  \draw[green!50!black, fill=green!50!black] (8.9,.9) rectangle (11.6,1.1);
\end{tikzpicture} \]
Conversely, we can decompose any alternator in terms of a signed average of alternators of one-shorter length:
\[ \begin{tikzpicture}
  \draw[vspace] (0,0) -- (0,2); \draw[vspace] (1,0) -- (1,2); \draw[vspace] (2,.25) .. controls +(180:.25) and +(270:.5) ..  (1.5,1) -- (1.5,2) ;
  \path (.5,.5) node {\dots} (.5,1.5) node {\dots};
  \draw[green!50!black, fill=green!50!black] (-.1,.9) rectangle (1.6,1.1);
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.1,-.1) --  node (n) {} (-.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.1) --  node (n1) {} (1.6,2.1);
  \path (n1) +(0,.4) node {\scriptsize $n+1$};
  \node[anchor=west] at (2,1) {$\displaystyle = \frac1{n+1} \sum_{j=0}^n (-1)^j$};
  \draw[vspace] (5.5,0) -- (5.5,2); \draw[vspace] (7,0) -- (7,2);  
  \path (6.25,.25) node {\dots};
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.1,-.1) --  node (m) {} (5.4,-.1);
  \path (m) +(0,-.35) node {\scriptsize $n$};
  \draw[green!50!black, fill=green!50!black] (5.4,.7) rectangle (7.1,.9);
  \path (5.5,2.2) node {\scriptsize $n$} (5.875,1.8) node {\scriptsize $\dots$} (6.25,2.2) node {\scriptsize $j$} (6.625,1.8) node {\scriptsize $\dots$} (7,2.2) node {\scriptsize $0$};
  \draw[vspace] (8,.25) .. controls +(180:.25) and +(270:.25) .. (7.5,.75) .. controls +(90:1) and +(270:1) .. (6.25,2);  
%  \node[anchor=west, text width=3in] at (8.5,1) {\tiny
%  \verb!\node[anchor=west] at (2,1) {$\displaystyle = \frac1{n+1} \sum_{j=0}^n (-1)^j$};!\\
%  \verb!\draw[vspace] (5.5,0) -- (5.5,2); \draw[vspace] (7,0) -- (7,2);!\\
%  \verb!\path (6.25,.25) node {\dots};!\\
%  \verb!\draw[thick,decoration={brace,amplitude=5},decorate] (7.1,-.1) -- node (m) {} (5.4,-.1);!\\
%  \verb!\path (m) +(0,-.35) node {\scriptsize $n$};!\\
%  \verb!\draw[green!50!black, fill=green!50!black] (5.4,.7) rectangle (7.1,.9);!\\
%  \verb!\path (5.5,2.2) node {\scriptsize $n$} (6.25,2.2) node {\scriptsize $j$}!\\ \verb!      (7,2.2) node {\scriptsize $0$};!\\
%  \verb!\draw[vspace] (8,.25) .. controls +(180:.25) and +(270:.25) .. (7.5,.75)!\\ \verb!      .. controls +(90:1) and +(270:1) .. (6.25,2);!\\
%  \verb!\path (5.875,1.8) node {\scriptsize $\dots$} (6.625,1.8) node {\scriptsize $\dots$};!\\
%  };
\end{tikzpicture} \]
There are similar formulas on the other side.  We will make a habit of identifying projections with their codomains; for example, we will write \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \draw[vspace] (0,0) -- (0,2); \draw[vspace] (.5,0) -- (.5,2); \draw[vspace] (1,0) -- (1,2);   \draw[green!50!black, fill=green!50!black] (-.2,.9) rectangle (1.2,1.1); \end{tikzpicture}}\, for the third wedge-power of \vvvv.

We know let \gggg\ be a Lie algebra with bracket 
\,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=blue] at (0,0) {};
\end{tikzpicture}}\,,
and let \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=red] at (0,0) {};
\end{tikzpicture}}\, be an action of \gggg\ on \mmmm.  Then there are two natural maps, given by the two trivalent vertices, of the form:
\[ \begin{tikzpicture}
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (1,0) -- (1,.5); 
  \draw[colie] (1,.5) -- (1,1); 
  \path (.5,.25) node {\dots} (.5,.75) node {\dots};
  \draw[module] (2,0) -- (2,1);
  \draw[blue, fill=blue] (-.1,.45) rectangle (1.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.1,-.1) -- node (n) {} (-.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \draw[colie] (1.5,1.5) -- (1.5,2); 
  \draw[colie] (1.5,2) -- (1.5,2.5); 
  \path (.5,1.75) node {\dots} (.5,2.25) node {\dots};
  \draw[module] (2,1.5) -- (2,2.5);
  \draw[blue, fill=blue] (-.1,1.95) rectangle (1.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.6) -- node (n1) {} (1.6,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
 
  \draw[purple, fill=purple!30] (-.25,1) rectangle  (2.25,1.5);
\end{tikzpicture} \]
We will let $\Q$ be a linear combination of these two maps, selected so that $\Q^2 = 0$ (we leave to the reader the easy exercise of checking this):
\[ \begin{tikzpicture}
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (1,0) -- (1,.5); 
  \draw[colie] (1,.5) -- (1,1); 
  \path (.5,.25) node {\dots} (.5,.75) node {\dots};
  \draw[module] (2,0) -- (2,1);
  \draw[blue, fill=blue] (-.1,.45) rectangle (1.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.1,-.1) -- node (n) {} (-.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \draw[colie] (1.5,1.5) -- (1.5,2); 
  \draw[colie] (1.5,2) -- (1.5,2.5); 
  \path (.5,1.75) node {\dots} (.5,2.25) node {\dots};
  \draw[module] (2,1.5) -- (2,2.5);
  \draw[blue, fill=blue] (-.1,1.95) rectangle (1.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.6) -- node (n1) {} (1.6,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
 
  \draw[purple, fill=purple!30] (-.25,1) rectangle node[text=black] {$Q$} (2.25,1.5);
  
  \node at (2.5,1.25) {$\defined$};
  
  \draw[colie] (3,0) -- (3,2); 
  \draw[colie] (4,0) -- (4,2); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.1,-.1) -- node (An) {} (2.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (3,2) -- (3,2.5); 
  \draw[colie] (4,2) -- (4,2.5); 
  \draw[colie] (4.5,2) -- (4.5,2.5); 
  \path (3.5,1) node {\dots} (3.5,2.25) node {\dots};
  \draw[blue, fill=blue] (2.9,1.95) rectangle (4.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (2.9,2.6) -- node (An1) {} (4.6,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n+1$};

  \draw[module] (5,0) .. controls +(90:.5) and +(270:.5) .. (4.75,1);
  \draw[module] (4.75,1) .. controls +(30:.25) and +(270:.5) .. (5,2.5);
  \draw[colie] (4.75,1) .. controls +(150:.25) and +(270:.5) .. (4.5,2);
  
  \node[dot=red] at (4.75,1) {};
  
  \node at (5.5,1.25) {$-$};
  \node at (6,1.25) {$\displaystyle \frac{n}{2}$};
  
  \draw[colie] (6.5,0) -- (6.5,.5); 
  \draw[colie] (6.5,.5) -- (6.5,2); 
  \draw[colie] (7.5,0) -- (7.5,.5); 
  \draw[colie] (7.3,.5) -- (7.3,2); 
  \path (7,.25) node {\dots};
  \draw[blue, fill=blue] (6.4,.45) rectangle (7.6,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.6,-.1) -- node (Bn) {} (6.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \path (6.9,1.25) node {\dots};
  \draw[colie] (6.5,2) -- (6.5,2.5); 
  \draw[colie] (7.5,2) -- (7.5,2.5); 
  \draw[colie] (8,2) -- (8,2.5); 
  \path (7,2.25) node {\dots};
  \draw[blue, fill=blue] (6.4,1.95) rectangle (8.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.4,2.6) -- node (Bn1) {} (8.1,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n+1$};
  
  \draw[module] (8.5,0) -- (8.5,2.5);
  
  \draw[colie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.75,1.25) node[dot=blue] {};
  \draw[colie] (7.75,1.25) .. controls +(150:.25) and +(270:.5) .. (7.5,2);
  \draw[colie] (7.75,1.25) .. controls +(30:.25) and +(270:.5) .. (8,2);
\end{tikzpicture} \]
Then $\Q$ is a differential on the \defn{Chevalley complex} $\left(\bigwedge^\bullet \gggd \right)\otimes \mmmm = \bigoplus \,\raisebox{-1ex}{\begin{tikzpicture}[x=.5 cm, y=.25 cm] \draw[colie] (0,0) -- (0,1); \draw[colie] (0,1) -- (0,2);  \draw[colie] (1,0) -- (1,1); \draw[colie] (1,1) -- (1,2);   \draw[blue, fill=blue] (-.2,.9) rectangle (1.2,1.1); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
 \draw[module] (1.5,0) -- (1.5,2);
\end{tikzpicture}}\,$, and defines the \defn{BRST cohomology} of \gggg\ with coefficients in \mmmm.  It is functorial in \mmmm\ in the following sense: if \nnnn\ is another module over \gggg, an {\em intertwiner} is any map \,\raisebox{-1ex}{\begin{tikzpicture}
  \draw[module] (0,-1em) -- (0,0); \draw[nodule] (0,0) -- (0,1em); \node[dot=green!50!black] at (0,0) {};
\end{tikzpicture}}\, such that
\[ \begin{tikzpicture}
  \draw[lie] (0,0) .. controls +(90:.25) and +(210:.5) .. (.5,.5);
  \draw[module] (1,0) .. controls +(90:.25) and +(330:.5) .. (.5,.5);
  \draw[module] (.5,.5) -- (.5,1);
  \draw[nodule] (.5,1) -- (.5,1.5);
  \node[dot=red] at (.5,.5) {};
  \node[dot=green!50!black] at (.5,1) {};
  \node at (1.25,.75) {$=$};
  \draw[lie] (1.5,0) .. controls +(90:.5) and +(210:.5) .. (2,1);
  \draw[nodule] (2.5,.5) .. controls +(90:.25) and +(330:.5) .. (2,1);
  \draw[module] (2.5,0) -- (2.5,.5);
  \draw[nodule] (2,1) -- (2,1.5);
  \node[dot=orange] at (2,1) {};
  \node[dot=green!50!black] at (2.5,.5) {};  
\end{tikzpicture}\,, \]
and the Chevalley complex construction commutes with intertwiners.

If \mmmm\ and \nnnn\ are two modules over \gggg, then we can define an action of \gggg\ on \mmmm\nnnn\ by:
\[ \begin{tikzpicture}
  \draw[lie] (0,0) .. controls +(90:.5) and +(210:.5) .. (.55,1);
  \draw[module] (1,0) .. controls +(90:.5) and +(330:.5) .. (.5,1);
  \draw[nodule] (1.1,0) .. controls +(90:.5) and +(330:.5) .. (.6,1.05);
  \draw[module] (.5,1) -- (.5,2);
  \draw[nodule] (.6,1) -- (.6,2);
  \node[circle,draw=red!50!orange,fill=red!50!orange,inner sep=2pt] at (.555,1.005) {};
  \node at (1.5,1) {$\defined$};
  \draw[lie] (2,0) .. controls +(90:.5) and +(210:.5) .. (2.5,1);
  \draw[module] (3,0) .. controls +(90:.5) and +(330:.5) .. (2.5,1);
  \draw[module] (2.5,1) -- (2.5,2);
  \node[dot=red] at (2.5,1) {};
  \draw[nodule] (3.1,0) .. controls +(90:1) and +(270:1) .. (2.6,2);
  \node at (3.5,1) {$+$};
  \draw[lie] (4,0) .. controls +(90:.5) and +(210:.5) .. (4.6,1);
  \draw[nodule] (5.1,0) .. controls +(90:.5) and +(330:.5) .. (4.6,1);
  \draw[nodule] (4.6,1) -- (4.6,2);
  \node[dot=orange] at (4.6,1) {};
  \draw[module] (5,0) .. controls +(90:1) and +(270:1) .. (4.5,2);
  
%  \node[anchor=west, text width=4in] at (5.5,1) {\tiny
%  \verb@  nodule/.style={draw=orange, postaction={decorate},@\\
%  \verb@    decoration={markings,mark=at position .55 with {\arrow[draw=orange]{>}}}},@\\
%  \mbox{} \\
%  \verb@\draw[lie] (0,0) .. controls +(90:.5) and +(210:.5) .. (.55,1);@\\
%  \verb@\draw[module] (1,0) .. controls +(90:.5) and +(330:.5) .. (.5,1);@\\
%  \verb@\draw[nodule] (1.1,0) .. controls +(90:.5) and +(330:.5) .. (.6,1.05);@\\
%  \verb@\draw[module] (.5,1) -- (.5,2); \draw[nodule] (.6,1) -- (.6,2);@\\
%  \verb@\node[circle,draw=red!50!orange,fill=red!50!orange,inner sep=2pt] at (.555,1.005) {};@\\
%  };
\end{tikzpicture} \]
In this way, if \mmmm\ is a module over \gggg, then we can define an action on \,\raisebox{-1ex}{\begin{tikzpicture}[x=.5 cm, y=.25 cm] \draw[module] (0,0) -- (0,1); \draw[module] (0,1) -- (0,2);  \draw[module] (1,0) -- (1,1); \draw[module] (1,1) -- (1,2);   \draw[red, fill=red] (-.2,.9) rectangle (1.2,1.1); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
\end{tikzpicture}}\, by
\[ \begin{tikzpicture}
  \draw[lie] (0,0) .. controls +(90:.5) and +(210:.5) .. (.5,1);
  \draw[module] (1,.5) .. controls +(90:.25) and +(330:.5) .. (.5,1);
  \draw[module] (.5,1) -- (.5,1.5);
  \draw[module] (1.5,.5) .. controls +(90:.5) and +(270:.5) .. (1,1.5);
  \draw[module] (2.5,.5) .. controls +(90:.5) and +(270:.5) .. (2,1.5);
  \draw[module] (1,0) -- (1,.5);
  \draw[module] (1.5,0) -- (1.5,.5);
  \draw[module] (2.5,0) -- (2.5,.5);
  \draw[module] (.5,1.5) -- (.5,2);
  \draw[module] (1,1.5) -- (1,2);
  \draw[module] (2,1.5) -- (2,2);
  \draw[red, fill=red] (.9,.4) rectangle (2.6,.5);
  \draw[red, fill=red] (.4,1.4) rectangle (2.1,1.5);
  \path (2,.25) node {\dots} (1.75,1) node {\dots} (1.5,1.75) node {\dots};
  \node[dot=red] at (.5,1) {};
  \node at (-.5,1) {$m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (2.6,-.1) -- node (m) {} (.9,-.1);
  \path (m) +(0,-.35) node {\scriptsize $m$};
\end{tikzpicture} \]
The multiplication by $m$ is needed because the condition of being an action is not homogeneous in individual types of vertices.

In particular, the adjoint action defines \gggg\ as a module over itself, and so the map $\Q$ above, with $\mmmm = \,\raisebox{-1ex}{\begin{tikzpicture}[x=.5 cm, y=.25 cm] \draw[lie] (0,0) -- (0,1); \draw[lie] (0,1) -- (0,2);  \draw[lie] (1,0) -- (1,1); \draw[lie] (1,1) -- (1,2);   \draw[blue, fill=blue] (-.2,.85) rectangle (1.2,1.05); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
\end{tikzpicture}}\,$, provides a family of maps
\[ \begin{tikzpicture}
  \draw[colie] (-1,0) -- (-1,.5); 
  \draw[colie] (-1,.5) -- (-1,1); 
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (.5,0) -- (.5,.5); 
  \draw[colie] (.5,.5) -- (.5,1); 
  \path (-.5,.25) node {\dots} (-.5,.75) node {\dots};
  \draw[lie] (1.5,0) -- (1.5,.5);
  \draw[lie] (1.5,.5) -- (1.5,1);
  \draw[lie] (2,0) -- (2,.5);
  \draw[lie] (2,.5) -- (2,1);
  \draw[lie] (3,0) -- (3,.5);
  \draw[lie] (3,.5) -- (3,1);
  \path (2.5,.25) node {\dots} (2.5,.75) node {\dots};  
  \draw[blue, fill=blue] (-1.1,.45) rectangle (0.6,.55);
  \draw[blue, fill=blue] (1.4,.45) rectangle (3.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (.6,-.1) -- node (n) {} (-1.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (3.1,-.1) -- node (m) {} (1.4,-.1);
  \path (m) +(0,-.35) node {\scriptsize $m$};
  
  \draw[colie] (-1,1.5) -- (-1,2); 
  \draw[colie] (-1,2) -- (-1,2.5); 
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (.5,1.5) -- (.5,2); 
  \draw[colie] (.5,2) -- (.5,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \path (-.5,1.75) node {\dots} (-.5,2.25) node {\dots};
  \draw[lie] (1.5,1.5) -- (1.5,2);
  \draw[lie] (2,1.5) -- (2,2);
  \draw[lie] (3,1.5) -- (3,2);
  \draw[lie] (1.5,2) -- (1.5,2.5);
  \draw[lie] (2,2) -- (2,2.5);
  \draw[lie] (3,2) -- (3,2.5);
  \path (2.5,1.75) node {\dots} (2.5,2.25) node {\dots};
  \draw[blue, fill=blue] (-1.1,1.95) rectangle (1.1,2.05);
  \draw[blue, fill=blue] (1.4,1.95) rectangle (3.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-1.1,2.6) -- node (n1) {} (1.1,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.4,2.6) -- node (m1) {} (3.1,2.6);
  \path (m1) +(0,.35) node {\scriptsize $m$};
 
  \draw[purple, fill=purple!30] (-1.25,1) rectangle node[text=black] {$Q$} (3.25,1.5);
  
  \node at (3.75,1.25) {$\defined$}; \node at (4.5,1.25) {$m$};
  
  \draw[colie] (5,0) -- (5,2); 
  \draw[colie] (6,0) -- (6,2); 
  \draw[colie] (6.5,0) -- (6.5,2); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.6,-.1) -- node (An) {} (4.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  \draw[colie] (5,2) -- (5,2.5); 
  \draw[colie] (6,2) -- (6,2.5); 
  \draw[colie] (6.5,2) -- (6.5,2.5); 
  \draw[colie] (7,2) -- (7,2.5); 
  \path (5.5,1) node {\dots} (5.5,2.25) node {\dots};
  \draw[blue, fill=blue] (4.9,1.95) rectangle (7.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.9,2.6) -- node (An1) {} (7.1,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n+1$};

  \draw[lie] (7.5,0) -- (7.5,.5);
  \draw[lie] (8,0) -- (8,.5);
  \draw[lie] (9,0) -- (9,.5);
  \draw[lie] (8,.5) -- (8,2);
  \draw[lie] (9,.5) -- (9,2);
  \draw[lie] (7.5,2) -- (7.5,2.5);
  \draw[lie] (8,2) -- (8,2.5);
  \draw[lie] (9,2) -- (9,2.5);
  \path (8.5,.25) node {\dots} (8.5,1.25) node {\dots} (8.5,2.25) node {\dots};  
  \draw[blue, fill=blue] (7.4,.45) rectangle (9.1,.55);
  \draw[blue, fill=blue] (7.4,1.95) rectangle (9.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (9.1,-.1) -- node (Am) {} (7.4,-.1);
  \path (Am) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.4,2.6) -- node (Am1) {} (9.1,2.6);
  \path (Am1) +(0,.35) node {\scriptsize $m$};
  
  \draw[lie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.25,1.25);
  \draw[lie] (7.25,1.25) .. controls +(30:.25) and +(270:.5) .. (7.5,2);
  \draw[colie] (7.25,1.25) .. controls +(150:.25) and +(270:.5) .. (7,2);
  \node[dot=blue] at (7.25,1.25) {};
  
  \node at (9.5,1.25) {$-$};
  \node at (10,1.25) {$\displaystyle \frac{n}{2}$};

  \draw[colie] (10.5,0) -- (10.5,.5);
  \draw[colie] (10.5,.5) -- (10.5,2);
  \draw[colie] (10.5,2) -- (10.5,2.5);
  \draw[colie] (11.5,0) -- (11.5,.5);
  \draw[colie] (11.5,.5) -- (11.5,2);
  \draw[colie] (11.5,2) -- (11.5,2.5);
  \path (11,.25) node {\dots} (11,1.25) node {\dots} (11,2.25) node {\dots};
  \draw[blue, fill=blue] (10.4,.45) rectangle (12.1,.55);
  \draw[blue, fill=blue] (10.4,1.95) rectangle (12.6,2.05);
  \draw[colie] (12,0) -- (12,.5);
  \draw[colie] (12,2) -- (12,2.5);
  \draw[colie] (12.5,2) -- (12.5,2.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.1,-.1) -- node (Bn) {} (10.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (10.4,2.6) -- node (Bn1) {} (12.6,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n+1$};
  
  \draw[colie] (12,.5) .. controls +(90:.5) and +(270:.5) .. (12.25,1.25);
  \draw[colie] (12.25,1.25) .. controls +(30:.25) and +(270:.5) .. (12.5,2);
  \draw[colie] (12.25,1.25) .. controls +(150:.25) and +(270:.5) .. (12,2);
  \node[dot=blue] at (12.25,1.25) {};
  
  \draw[lie] (13,0) -- (13,1.25);
  \draw[lie] (13,1.25) -- (13,2.5);
  \draw[lie] (13.5,0) -- (13.5,1.25);
  \draw[lie] (13.5,1.25) -- (13.5,2.5);
  \draw[lie] (14.5,0) -- (14.5,1.25);
  \draw[lie] (14.5,1.25) -- (14.5,2.5);
  \draw[blue, fill=blue] (12.9,1.2) rectangle (14.6,1.3);
  \path (14,.5) node {\dots} (14,2) node {\dots};
  \draw[thick,decoration={brace,amplitude=5},decorate] (14.6,-.1) -- node (Bm) {} (12.9,-.1);
  \path (Bm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.9,2.6) -- node (Bm1) {} (14.6,2.6);
  \path (Bm1) +(0,.35) node {\scriptsize $m$};
\end{tikzpicture} \]
What if \gggd\ is also a Lie coalgebra?  Then we can switch all the arrows, getting another interesting map:
\[ \begin{tikzpicture}
  \draw[colie] (-1,0) -- (-1,.5); 
  \draw[colie] (-1,.5) -- (-1,1); 
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (.5,0) -- (.5,.5); 
  \draw[colie] (.5,.5) -- (.5,1); 
  \path (-.5,.25) node {\dots} (-.5,.75) node {\dots};
  \draw[lie] (1.5,0) -- (1.5,.5);
  \draw[lie] (1.5,.5) -- (1.5,1);
  \draw[lie] (2,0) -- (2,.5);
  \draw[lie] (2,.5) -- (2,1);
  \draw[lie] (3,0) -- (3,.5);
  \draw[lie] (3,.5) -- (3,1);
  \path (2.5,.25) node {\dots} (2.5,.75) node {\dots};  
  \draw[blue, fill=blue] (-1.1,.45) rectangle (0.6,.55);
  \draw[blue, fill=blue] (1.4,.45) rectangle (3.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (.6,-.1) -- node (n) {} (-1.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (3.1,-.1) -- node (m) {} (1.4,-.1);
  \path (m) +(0,-.35) node {\scriptsize $m$};
  
  \draw[colie] (-1,1.5) -- (-1,2); 
  \draw[colie] (-1,2) -- (-1,2.5); 
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (.5,1.5) -- (.5,2); 
  \draw[colie] (.5,2) -- (.5,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \path (-.5,1.75) node {\dots} (-.5,2.25) node {\dots};
  \draw[lie] (1.5,1.5) -- (1.5,2);
  \draw[lie] (2,1.5) -- (2,2);
  \draw[lie] (3,1.5) -- (3,2);
  \draw[lie] (1.5,2) -- (1.5,2.5);
  \draw[lie] (2,2) -- (2,2.5);
  \draw[lie] (3,2) -- (3,2.5);
  \path (2.5,1.75) node {\dots} (2.5,2.25) node {\dots};
  \draw[blue, fill=blue] (-1.1,1.95) rectangle (.6,2.05);
  \draw[blue, fill=blue] (.9,1.95) rectangle (3.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-1.1,2.6) -- node (n1) {} (.6,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (.9,2.6) -- node (m1) {} (3.1,2.6);
  \path (m1) +(0,.35) node {\scriptsize $m+1$};
 
  \draw[purple, fill=purple!30] (-1.25,1) rectangle node[text=black] {$R$} (3.25,1.5);
  
  \node at (3.75,1.25) {$\defined$}; \node at (4.5,1.25) {$\displaystyle \frac{m}2$};
  
  \draw[colie] (5,0) -- (5,1.25); 
  \draw[colie] (6,0) -- (6,1.25); 
  \draw[colie] (6.5,0) -- (6.5,1.25); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.6,-.1) -- node (An) {} (4.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  \draw[colie] (5,1.25) -- (5,2.5); 
  \draw[colie] (6,1.25) -- (6,2.5); 
  \draw[colie] (6.5,1.25) -- (6.5,2.5); 
  \draw[lie] (7,2) -- (7,2.5); 
  \path (5.5,.5) node {\dots} (5.5,2) node {\dots};
  \draw[blue, fill=blue] (4.9,1.2) rectangle (6.6,1.3);
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.9,2.6) -- node (An1) {} (6.6,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n$};

  \draw[lie] (7.5,0) -- (7.5,.5);
  \draw[lie] (8,0) -- (8,.5);
  \draw[lie] (9,0) -- (9,.5);
  \draw[lie] (8,.5) -- (8,2);
  \draw[lie] (9,.5) -- (9,2);
  \draw[lie] (7.5,2) -- (7.5,2.5);
  \draw[lie] (8,2) -- (8,2.5);
  \draw[lie] (9,2) -- (9,2.5);
  \path (8.5,.25) node {\dots} (8.5,1.25) node {\dots} (8.5,2.25) node {\dots};  
  \draw[blue, fill=blue] (7.4,.45) rectangle (9.1,.55);
  \draw[blue, fill=blue] (6.9,1.95) rectangle (9.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (9.1,-.1) -- node (Am) {} (7.4,-.1);
  \path (Am) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.9,2.6) -- node (Am1) {} (9.1,2.6);
  \path (Am1) +(0,.35) node {\scriptsize $m+1$};
  
  \draw[lie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.25,1.25);
  \draw[lie] (7.25,1.25) .. controls +(30:.25) and +(270:.5) .. (7.5,2);
  \draw[lie] (7.25,1.25) .. controls +(150:.25) and +(270:.5) .. (7,2);
  \node[dot=blue] at (7.25,1.25) {};
  
  \node at (9.5,1.25) {$-$};
  \node at (10,1.25) {$\displaystyle n$};

  \draw[colie] (10.5,0) -- (10.5,.5);
  \draw[colie] (10.5,.5) -- (10.5,2);
  \draw[colie] (10.5,2) -- (10.5,2.5);
  \draw[colie] (11.5,0) -- (11.5,.5);
  \draw[colie] (11.5,.5) -- (11.5,2);
  \draw[colie] (11.5,2) -- (11.5,2.5);
  \path (11,.25) node {\dots} (11,1.25) node {\dots} (11,2.25) node {\dots};
  \draw[blue, fill=blue] (10.4,.45) rectangle (12.1,.55);
  \draw[blue, fill=blue] (10.4,1.95) rectangle (12.1,2.05);
  \draw[colie] (12,0) -- (12,.5);
  \draw[colie] (12,2) -- (12,2.5);
  \draw[lie] (12.5,2) -- (12.5,2.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.1,-.1) -- node (Bn) {} (10.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (10.4,2.6) -- node (Bn1) {} (12.1,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n$};
  
  \draw[colie] (12,.5) .. controls +(90:.5) and +(270:.5) .. (12.25,1.25);
  \draw[lie] (12.25,1.25) .. controls +(30:.25) and +(270:.5) .. (12.5,2);
  \draw[colie] (12.25,1.25) .. controls +(150:.25) and +(270:.5) .. (12,2);
  \node[dot=blue] at (12.25,1.25) {};
  
  \draw[lie] (13,0) -- (13,2);
  \draw[lie] (13,2) -- (13,2.5);
  \draw[lie] (13.5,0) -- (13.5,2);
  \draw[lie] (13.5,2) -- (13.5,2.5);
  \draw[lie] (14.5,0) -- (14.5,2);
  \draw[lie] (14.5,2) -- (14.5,2.5);
  \draw[blue, fill=blue] (12.4,1.95) rectangle (14.6,2.05);
  \path (14,1) node {\dots} (14,2.25) node {\dots};
  \draw[thick,decoration={brace,amplitude=5},decorate] (14.6,-.1) -- node (Bm) {} (12.9,-.1);
  \path (Bm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.4,2.6) -- node (Bm1) {} (14.6,2.6);
  \path (Bm1) +(0,.35) node {\scriptsize $m+1$};
\end{tikzpicture} \]

Then $\R\Q - \Q\R$ vanishes for every $m,n$ if and only if \gggg\ is a Lie bialgebra.  We sketch the argument, leaving the details to the interested reader as an exercise in manipulating the diagrams of the graphical calculus: the terms in $\R$ and $\Q$ that do not mix \gggg\ with \gggd\ commute; upon expanding the mixed terms using the decomposition of an alternator into a sum of shorter-length alternators,  every summand vanishes except those in which the two vertices are linked directly by an edge; when collected together, these terms sum to the difference of the two sides Lie bialgebra coherence relation, up to composition with alternators.

Thus $\Q$ and $\R$ provide the differentials for a bi-graded complex $\underset{n,m}{\bigoplus} \,\begin{tikzpicture}[x=.5 cm, y=.25 cm,baseline=1ex] 
\draw[colie] (0,0) -- (0,1); \draw[colie] (0,1) -- (0,2);  \draw[colie] (1,0) -- (1,1); \draw[colie] (1,1) -- (1,2);   \draw[blue, fill=blue] (-.2,.9) rectangle (1.2,1.1); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
\draw[lie] (1.8,0) -- (1.8,1); \draw[lie] (1.8,1) -- (1.8,2);  \draw[lie] (2.8,0) -- (2.8,1); \draw[lie] (2.8,1) -- (2.8,2);   \draw[blue, fill=blue] (1.6,.9) rectangle (3,1.1); \path (2.3,.4) node {\tiny \dots} (2.3,1.6) node {\tiny \dots} ;
\path (.5,-.4) node {\tiny $n$} (2.3,-.4) node {\tiny $m$} ;
\end{tikzpicture}\,$.  We will compute the differential $\T_k = \bigoplus_{m+n = k} (-1)^m \Q_{n,m} \oplus (-1)^m \R_{n,m}$ of the total complex\footnote{Any bicomplex has many ``total complexes''.  In our convention, a bicomplex is a (possibly-infinite) grid of the form:
\[ \begin{tikzpicture}[scale=.5]
  \foreach \x in {0,1,2,3} \foreach \y in {0,1,2}
    {\path[draw,->,auto] (\x,-\y) node[dot=black] (N\x\y) {} +(.2,0) -- +(.8,0); \path[draw,->,auto] (N\x\y) +(0,-.2) --  +(0,-.8);}
  \foreach \x in {0,1,2,3} \path[auto] (\x,0) -- node{\tiny $\Q$} +(1,0);
  \foreach \y in {0,1,2} \path[auto,swap] (0,-\y) -- node{\tiny $\R$} +(0,-1);
\end{tikzpicture} \]
where each row and each column is a complex ($\Q^2 = 0 = \R^2$), and such that each square commutes ($\Q\R = \R\Q$).  Then the objects in the total complex are the direct sum of antidiagonals, but the differential depends on a choice of decoration of the edges of the bicomplex with $+$s and $-$s such that each square has an odd number of each sign.  We have chosen the signs $\T_k = \bigoplus_{m+n = k} (-1)^m \Q_{n,m} \oplus (-1)^m \R_{n,m}$ for later convenience.}, and observe that $\T^2 = 0$ if \gggg\ is a Lie bialgebra, thereby showing that $\Q\R = \R\Q$.

Let \,\begin{tikzpicture}[baseline=-2pt]
  \draw[lie] (0,-1em) -- (0,0); \draw[drinf] (0,0) -- (0,1em); 
\end{tikzpicture}\, and \,\begin{tikzpicture}[baseline=-2pt]
  \draw[colie] (0,-1em) -- (0,0); \draw[drinf] (0,0) -- (0,1em); 
\end{tikzpicture}\, be the canonical injections, and \,\begin{tikzpicture}[baseline=-2pt]
  \draw[drinf] (0,-1em) -- (0,0); \draw[lie] (0,0) -- (0,1em); 
\end{tikzpicture}\, and \,\begin{tikzpicture}[baseline=-2pt]
  \draw[drinf] (0,-1em) -- (0,0); \draw[colie] (0,0) -- (0,1em); 
\end{tikzpicture}\, the canonical surjections, between \gggg, \gggd, and $\DDDD = \gggg \oplus \gggd$.  Moreover, we declare that $\,\begin{tikzpicture}[baseline=-2pt]
  \draw[lie] (0,-1em) -- (0,0); \draw[colie] (0,0.1em) -- (0,1.1em); 
\end{tikzpicture}\, = 0 = \,\begin{tikzpicture}[baseline=-2pt]
  \draw[colie] (0,-1em) -- (0,0); \draw[lie] (0,0.1em) -- (0,1.1em); 
\end{tikzpicture}\,$.  
%Then the alternators are:
%\[  \begin{tikzpicture}
%  \draw[colie] (0,-.5) -- (0,0); 
%  \draw[colie] (.5,-.5) -- (.5,0); 
%  \draw[drinf] (0,0) -- (0,1); \draw[drinf] (.5,0) -- (.5,1);
%  \draw[blue, fill=blue] (-.1,.45) rectangle (.6,.55);
%  \node at (1,.5) {$=$};
%  \draw[colie] (1.5,-.5) -- (1.5,1); 
%  \draw[colie] (2,-.5) -- (2,1); 
%  \draw[blue, fill=blue] (1.4,.45) rectangle (2.1,.55);
%  \node[anchor=south west] at (2.1,0) {, etc., };
%  \draw[colie] (3.5,-.5) -- (3.5,0); \draw[drinf] (3.5,0) -- (3.5,1); \draw[colie] (3.5,1) -- (3.5,1.5);
%  \draw[lie] (4,-.5) -- (4,0); \draw[drinf] (4,0) -- (4,1); \draw[lie] (4,1) -- (4,1.5);
%  \draw[blue, fill=blue] (3.4,.45) rectangle (4.1,.55);
%  \node at (4.5,.5) {$=$};
%  \node at (5,.5) {$\displaystyle \frac12$};
%  \draw[colie] (5.5,-.5) -- (5.5,1.5); \draw[lie] (6,-.5) -- (6,1.5);
%  \node[anchor=south west] at (6,0) {,};
%  \end{tikzpicture}\] \[ \begin{tikzpicture}
%  \node[anchor=south west] at (6,0) {and more generally:};
%  \draw[colie] (10,-.5) -- (10,0); \draw[drinf] (10,0) -- (10,1); \draw[colie] (10,1) -- (10,1.5);
%  \draw[colie] (11,-.5) -- (11,0); \draw[drinf] (11,0) -- (11,1); \draw[colie] (11,1) -- (11,1.5);
%  \draw[thick,decoration={brace,amplitude=5},decorate] (11.1,-.6) -- node (An) {} (9.9,-.6);
%  \path (An) +(0,-.35) node {\scriptsize $n$};
%  \draw[thick,decoration={brace,amplitude=5},decorate] (12.6,-.6) -- node (Am) {} (11.4,-.6);
%  \path (Am) +(0,-.35) node {\scriptsize $m$};  
%  \draw[lie] (11.5,-.5) -- (11.5,0); \draw[drinf] (11.5,0) -- (11.5,1); \draw[lie] (11.5,1) -- (11.5,1.5);
%  \draw[lie] (12.5,-.5) -- (12.5,0); \draw[drinf] (12.5,0) -- (12.5,1); \draw[lie] (12.5,1) -- (12.5,1.5);
%  \path (10.5,0) node {\dots} (10.5,1) node {\dots} (12,0) node {\dots} (12,1) node {\dots};
%  \draw[blue, fill=blue] (9.9,.45) rectangle (12.6,.55);  
%  \node at (13,.5) {$=$};
%  \node at (14,.5) {$\displaystyle \frac{n!\,m!}{(n+m)!}$};
%  \draw[colie] (15,-.5) -- (15,.5);   \draw[colie] (15,.5) -- (15,1.5); 
%  \draw[colie] (16,-.5) -- (16,.5);   \draw[colie] (16,.5) -- (16,1.5); 
%  \draw[lie] (16.5,-.5) -- (16.5,.5);   \draw[lie] (16.5,.5) -- (16.5,1.5); 
%  \draw[lie] (17.5,-.5) -- (17.5,.5);   \draw[lie] (17.5,.5) -- (17.5,1.5); 
%  \path (15.5,0) node {\dots} (15.5,1) node {\dots} (17,0) node {\dots} (17,1) node {\dots};
%  \draw[blue, fill=blue] (14.9,.45) rectangle (16.1,.55);    
%  \draw[blue, fill=blue] (16.4,.45) rectangle (17.6,.55);    
%  \draw[thick,decoration={brace,amplitude=5},decorate] (16.1,-.6) -- node (Bn) {} (14.9,-.6);
%  \path (Bn) +(0,-.35) node {\scriptsize $n$};
%  \draw[thick,decoration={brace,amplitude=5},decorate] (17.6,-.6) -- node (Bm) {} (16.4,-.6);
%  \path (Bm) +(0,-.35) node {\scriptsize $m$};
%  \node[anchor=south west] at (17.6,0) {.};
%\end{tikzpicture} \]
We introduce the following notation, which we define by example:
\[ \begin{tikzpicture}
  \draw[drinf] (-.25,0) -- (-.25,.5) (.25,0) -- (.25,.5) (.75,0) -- (.75,.5);
  \draw[colie] (-.25,1) -- (-.25,1.5); \draw[colie] (.25,1) -- (.25,1.5); \draw[lie] (.75,1) -- (.75,1.5);
  \draw[blue,fill=blue!50] (-.5,.5) rectangle node[text=black] {$\sort^{2,1}$} (1,1);
  \path (1.45,.75) node {$\defined$} (3.5,.75) node {$\oplus$} (4,.75) node {$-$} (6,.75) node {$\oplus$};
  \draw[drinf] (2,0) -- (2,.5) (2.5,0) -- (2.5,.5) (3,0) -- (3,.5);
  \draw[colie] (2,.5) .. controls +(90:.25) and +(270:.25) .. (2,1.5);
  \draw[colie] (2.5,.5) .. controls +(90:.25) and +(270:.25) .. (2.5,1.5);
  \draw[lie] (3,.5) .. controls +(90:.25) and +(270:.25) .. (3,1.5);
  
  \draw[colie] (4.5,.5) .. controls +(90:.25) and +(270:.25) .. (4.5,1.5);
  \draw[colie] (5.5,0) -- (5.5,.5) .. controls +(90:.25) and +(270:.25) .. (5,1.5);
  \draw[lie] (5,0) -- (5,.5) .. controls +(90:.25) and +(270:.25) .. (5.5,1.5);
  \draw[drinf] (5.5,0) -- (5.5,.5) (4.5,0) -- (4.5,.5) (5,0) -- (5,.5);
  
  \draw[drinf] (7.5,0) -- (7.5,.5) (6.5,0) -- (6.5,.5) (7,0) -- (7,.5);
  \draw[colie] (7,.5) .. controls +(90:.25) and +(270:.25) .. (6.5,1.5);
  \draw[colie] (7.5,.5) .. controls +(90:.25) and +(270:.25) .. (7,1.5);
  \draw[lie] (6.5,.5) .. controls +(90:.25) and +(270:.25) .. (7.5,1.5);
\end{tikzpicture} \]
The sign of a term is determined by the number of crossings, just as it is in the alternator projection.  The reader should interpret the $\oplus$s in the above equation as living only on the upper half of the picture, so that $\Sort{n,m}$ is a map $\DDDD^{\otimes n+m} \to \left[ \bigoplus {n +m \choose n} \right] \gggd^{\otimes n} \otimes \gggg^{\otimes m}$, by which we mean a direct sum of $n+m \choose n$ copies of the same vector space.  We will mostly continue to suppress the diagonal and codiagonal (sum) maps $\Delta: \gggg \to \gggg \oplus \gggg$ and $\nabla: \gggg \oplus \gggg \to \gggg$ (so the composition $\Delta \circ \nabla$ is twice the identity operator)\footnote{We remark that in the ``time goes up'' convention we have been adopting, it would make more visual sense to interchange the letters used for these two maps.  However, we will continue with the traditional use of $\Delta$ for ``diagonal''; it is at least the correct first letter.}.  We remark only that using the codiagonal and the canonical injections, one can define the composition of $\Sort{}$ maps: $\Sort{n,k-n} \circ \Sort{m,k-m}$ is zero if $n\neq m$.  Since
\[ \begin{tikzpicture}
  \draw[colie] (0,0) -- (0,.5); \draw[colie] (1,0) -- (1,.5);
  \draw[colie] (0,1) -- (0,1.5); \draw[colie] (1,1) -- (1,1.5);
  \draw[lie] (1.5,0) -- (1.5,.5); \draw[lie] (2.5,0) -- (2.5,.5);
  \draw[lie] (1.5,1) -- (1.5,1.5); \draw[lie] (2.5,1) -- (2.5,1.5);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\sort^{n,m}$} (2.75,1);
  \path (.5,.25) node {\dots}  (2,.25) node {\dots} (.5,1.25) node {\dots}  (2,1.25) node {\dots};
  \node at (3.1,.75) {$=$};
  \draw[colie] (3.5,0) -- (3.5,1.5);
  \draw[colie] (4.5,0) -- (4.5,1.5);
  \draw[lie] (5,0) -- (5,1.5);
  \draw[lie] (6,0) -- (6,1.5);
  \path (4,.25) node {\dots}  (5.5,.25) node {\dots} (4,1.25) node {\dots}  (5.5,1.25) node {\dots};
  \node[anchor=west] at (6.25,.75) {$\mbox{} \oplus 0 \oplus 0 \oplus \dots $,};
\end{tikzpicture} \]
we see that $\nabla \circ \Sort{n,m}$ is a projection onto $\gggd^{\otimes n} \otimes \gggg^{\otimes m}$.

In any case, we see that the dual map to $\Sort{n,m}$ is the map $\sORT{m,n}$ defined by example:
\[ \begin{tikzpicture}
  \draw[drinf] (-.25,2) -- (-.25,2.5) (.25,2) -- (.25,2.5) (.75,2) -- (.75,2.5);
  \draw[colie] (-.25,1) -- (-.25,1.5); \draw[colie] (.25,1) -- (.25,1.5); \draw[lie] (.75,1) -- (.75,1.5);
  \draw[blue,fill=blue!50] (-.5,1.5) rectangle node[text=black] {$\sort_{2,1}$} (1,2);
  \path (1.45,1.75) node {$\defined$} (3.5,1.75) node {$\oplus$} (6,1.75) node {$\oplus$};
  
  \draw[drinf] (2,2) -- (2,2.5) (2.5,2) -- (2.5,2.5) (3,2) -- (3,2.5);
  \draw[colie] (2,1) .. controls +(90:.25) and +(270:.25) .. (2,2);
  \draw[colie] (2.5,1) .. controls +(90:.25) and +(270:.25) .. (2.5,2);
  \draw[lie] (3,1) .. controls +(90:.25) and +(270:.25) .. (3,2);
  
  \node at (4,1.75) {$-$};
  \draw[colie] (4.5,1) .. controls +(90:.25) and +(270:.25) .. (4.5,2);
  \draw[lie] (5.5,1) .. controls +(90:.25) and +(270:.25) .. (5,2) -- (5,2.5);
  \draw[colie] (5,1) .. controls +(90:.25) and +(270:.25) .. (5.5,2) -- (5.5,2.5);
  \draw[drinf] (5.5,2) -- (5.5,2.5) (4.5,2) -- (4.5,2.5) (5,2) -- (5,2.5);
  
  \draw[drinf] (7.5,2) -- (7.5,2.5) (6.5,2) -- (6.5,2.5) (7,2) -- (7,2.5);
  \draw[colie] (6.5,1) .. controls +(90:.25) and +(270:.25) .. (7,2);
  \draw[colie] (7,1) .. controls +(90:.25) and +(270:.25) .. (7.5,2);
  \draw[lie] (7.5,1) .. controls +(90:.25) and +(270:.25) .. (6.5,2);
\end{tikzpicture} \]
Here the $\oplus$s are only on the bottom.  We recall that we have defined the dualization functor to reverse the order of tensor products; we declare that it preserves the order of direct sums.  In algebraic symbols: $(A\otimes B)^* = B^* \otimes A^*$; $(A\oplus B)^* = A^* \oplus B^*$.  Then:
\begin{align*}
  \left( \sum_{n+m = k} \sORT{n,m} \right) \circ \left( \sum_{n+m = k} \Sort{n,m} \right) & = \textrm{identity on $\DDDD^{\otimes k}$} \\
  \Sort{n,m} \circ \sORT{n,m} & = \textrm{identity on $\textstyle \left[ \bigoplus {n+m \choose n} \right] \gggd^{\otimes n} \otimes \gggg^{\otimes m}$}
\end{align*}
As such, ${\color{blue!50!black} \SORT} \defined \left( \bigoplus_{n+m = k} \Sort{n,m} \right)$ and ${\color{blue!50!black} \MIX} \defined \left( \bigoplus_{n+m = k} \sORT{n,m} \right)$ are inverse isomorphisms, corresponding the the binomial expansion $(a+b)^k = \sum_{n=0}^k {k\choose n} a^nb^{k-n}$.  If $\map$ is a map with domain $\gggd^{\otimes n} \otimes \gggg^{\otimes m}$, then we will write
\[ \begin{tikzpicture}
    \draw[drinf] (0,0) -- (0,.5); \draw[drinf] (1,0) -- (1,.5);
  \draw[colie] (0,1) -- (0,1.5); \draw[colie] (1,1) -- (1,1.5);
  \draw[drinf] (1.5,0) -- (1.5,.5); \draw[drinf] (2.5,0) -- (2.5,.5);
  \draw[lie] (1.5,1) -- (1.5,1.5); \draw[lie] (2.5,1) -- (2.5,1.5);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\sort^{n,m}$} (2.75,1);
  \path (.5,.25) node {\dots}  (2,.25) node {\dots} (.5,1.25) node {\dots}  (2,1.25) node {\dots};
  \draw[purple,fill=purple!50] (-.25,1.5) rectangle node[text=black] {$P$} (2.75,2);
  \draw[string=purple] (1.25,2) -- (1.25,2.5);
\end{tikzpicture} \]
for the composition of $\Sort{n,m}$ with the direct sum of $n+m \choose n$ copies of $\map$.  Similarly, if the codomain of $\map$ is $\gggd^{\otimes n} \otimes \gggg^{\otimes m}$, then
\[ \begin{tikzpicture}
    \draw[colie] (0,0) -- (0,.5); \draw[colie] (1,0) -- (1,.5);
  \draw[drinf] (0,1) -- (0,1.5); \draw[drinf] (1,1) -- (1,1.5);
  \draw[lie] (1.5,0) -- (1.5,.5); \draw[lie] (2.5,0) -- (2.5,.5);
  \draw[drinf] (1.5,1) -- (1.5,1.5); \draw[drinf] (2.5,1) -- (2.5,1.5);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\sort_{n,m}$} (2.75,1);
  \path (.5,.25) node {\dots}  (2,.25) node {\dots} (.5,1.25) node {\dots}  (2,1.25) node {\dots};
  \draw[purple,fill=purple!50] (-.25,-.5) rectangle node[text=black] {$P$} (2.75,0);
  \draw[string=purple] (1.25,-1) -- (1.25,-.5);
\end{tikzpicture} \]
will denote the direct sum of $\map$s composed with $\sORT{n,m}$.

We invite the reader to check the following not-too-difficult fact:
\[ \begin{tikzpicture}
  \draw[drinf] (0,-.5) -- (0,.5);  \draw[drinf] (2.5,-.5) -- (2.5,.5);
  \draw[colie] (0,1) -- (0,2); \draw[colie] (1,1) -- (1,2);
  \draw[lie] (1.5,1) -- (1.5,2); \draw[lie] (2.5,1) -- (2.5,2);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\sort^{n,m}$} (2.75,1);
  \path (1.25,-.25) node {\dots} (1.25,.25) node {\dots} (.5,1.5) node {\dots}  (2,1.5) node {\dots} ;
  \draw[blue, fill=blue] (-.1,-.05) rectangle (2.6,.05);
  
  \node at (3.25,.75) {$=$};
  
  \draw[drinf] (4,-.5) -- (4,.5);  \draw[drinf] (6.5,-.5) -- (6.5,.5);
  \draw[colie] (4,1) -- (4,1.5); \draw[colie] (5,1) -- (5,1.5);
  \draw[colie] (4,1.5) -- (4,2); \draw[colie] (5,1.5) -- (5,2);
  \draw[lie] (5.5,1) -- (5.5,1.5); \draw[lie] (6.5,1) -- (6.5,1.5);
  \draw[lie] (5.5,1.5) -- (5.5,2); \draw[lie] (6.5,1.5) -- (6.5,2);
  \draw[blue,fill=blue!50] (3.75,.5) rectangle node[text=black] {$\sort^{n,m}$} (6.75,1);
  \path (5.25,-.25) node {\dots} (5.25,.25) node {\dots}  (4.5,1.25) node {\dots}  (6,1.25) node {\dots} (4.5,1.75) node {\dots}  (6,1.75) node {\dots};
  \draw[blue, fill=blue] (3.9,-.05) rectangle (6.6,.05);
  \draw[blue, fill=blue] (3.9,1.45) rectangle (5.1,1.55) (5.4,1.45) rectangle (6.6,1.55);
\end{tikzpicture} \]
Moreover, if we compose $\Sort{n,m}$ with the codiagonal $\nabla$, then we can erase the lower alternator in the above equation:
\[ \begin{tikzpicture}
  \draw[drinf] (0,-.5) -- (0,.5);  \draw[drinf] (2.5,-.5) -- (2.5,.5);
  \draw[colie] (0,1) -- (0,2); \draw[colie] (1,1) -- (1,2);
  \draw[lie] (1.5,1) -- (1.5,2); \draw[lie] (2.5,1) -- (2.5,2);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\nabla \circ \sort^{n,m}$} (2.75,1);
  \path (1.25,-.25) node {\dots} (1.25,.25) node {\dots} (.5,1.5) node {\dots}  (2,1.5) node {\dots} ;
  \draw[blue, fill=blue] (-.1,-.05) rectangle (2.6,.05);
  
  \node at (3.25,.75) {$=$};
  
  \draw[drinf] (4,-.5) -- (4,.5);  \draw[drinf] (6.5,-.5) -- (6.5,.5);
  \draw[colie] (4,1) -- (4,1.5); \draw[colie] (5,1) -- (5,1.5);
  \draw[colie] (4,1.5) -- (4,2); \draw[colie] (5,1.5) -- (5,2);
  \draw[lie] (5.5,1) -- (5.5,1.5); \draw[lie] (6.5,1) -- (6.5,1.5);
  \draw[lie] (5.5,1.5) -- (5.5,2); \draw[lie] (6.5,1.5) -- (6.5,2);
  \draw[blue,fill=blue!50] (3.75,.5) rectangle node[text=black] {$\nabla \circ \sort^{n,m}$} (6.75,1);
  \path (5.25,0) node {\dots}  (4.5,1.25) node {\dots}  (6,1.25) node {\dots} (4.5,1.75) node {\dots}  (6,1.75) node {\dots};
  \draw[blue, fill=blue] (3.9,1.45) rectangle (5.1,1.55) (5.4,1.45) rectangle (6.6,1.55);
\end{tikzpicture} \]

We define $\nabla \circ {\color{blue!50!black} \SORT} \defined \left( \bigoplus_{n+m = k} \nabla \circ \Sort{n,m} \right)$, even though the different codiagonals in the direct sum have different domains and codomains.  Then $\nabla \circ {\color{blue!50!black} \SORT} $ is the canonical isomorphism of wedge powers $\,\begin{tikzpicture}[x=.5 cm, y=.25 cm,baseline=1ex] 
\draw[drinf] (0,0) -- (0,1); \draw[drinf] (0,1) -- (0,2);  \draw[drinf] (1,0) -- (1,1); \draw[drinf] (1,1) -- (1,2);   \draw[blue, fill=blue] (-.2,.9) rectangle (1.2,1.1); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
\path (.5,-.4) node {\tiny $k$};\end{tikzpicture}\,
\to 
\underset{\tiny n+m = k}{\bigoplus} \,\begin{tikzpicture}[x=.5 cm, y=.25 cm,baseline=1ex] 
\draw[colie] (0,0) -- (0,1); \draw[colie] (0,1) -- (0,2);  \draw[colie] (1,0) -- (1,1); \draw[colie] (1,1) -- (1,2);   \draw[blue, fill=blue] (-.2,.9) rectangle (1.2,1.1); \path (.5,.4) node {\tiny \dots} (.5,1.6) node {\tiny \dots} ;
\draw[lie] (1.8,0) -- (1.8,1); \draw[lie] (1.8,1) -- (1.8,2);  \draw[lie] (2.8,0) -- (2.8,1); \draw[lie] (2.8,1) -- (2.8,2);   \draw[blue, fill=blue] (1.6,.9) rectangle (3,1.1); \path (2.3,.4) node {\tiny \dots} (2.3,1.6) node {\tiny \dots} ;
\path (.5,-.4) node {\tiny $n$} (2.3,-.4) node {\tiny $m$} ;
\end{tikzpicture}\,$.  We will simply call this $\color{blue!50!black} \SORT$, and also confuse $\color{blue!50!black} \MIX$ with ${\color{blue!50!black} \MIX} \circ \Delta$; the mathematicians' careful distinction between $+$ and $\oplus$ doesn't carry too well to the graphical language.  However, we do need to be careful about normalization, to maintain the correct pairing.  We have:
We can write the components of this isomorphism even more simply:
\[\begin{tikzpicture}
  \draw[drinf] (0,-.5) -- (0,.5);  \draw[drinf] (2.5,-.5) -- (2.5,.5);
  \draw[drinf] (1,-.5) -- (1,.5);  \draw[drinf] (1.5,-.5) -- (1.5,.5);
  \draw[colie] (0,1) -- (0,2); \draw[colie] (1,1) -- (1,2);
  \draw[lie] (1.5,1) -- (1.5,2); \draw[lie] (2.5,1) -- (2.5,2);
  \draw[blue,fill=blue!50] (-.25,.5) rectangle node[text=black] {$\nabla \circ \sort^{n,m}$} (2.75,1);
  \path (.5,-.25) node {\dots} (2,-.25) node {\dots} (2,.25) node {\dots} (.5,.25) node {\dots} (.5,1.5) node {\dots}  (2,1.5) node {\dots} ;
  \draw[blue, fill=blue] (-.1,-.05) rectangle (2.6,.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.1) -- node (N) {} (1.1,2.1);
  \path (N) +(0,.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.4,2.1) -- node (M) {} (2.6,2.1);
  \path (M) +(0,.35) node {\scriptsize $m$};
  
  \node at (3.25,.75) {$=$}; \node at (7.6,.75) {$\displaystyle \times {n+ m\choose n}$};

  \draw[drinf] (4,-.5) -- (4,.5);  \draw[drinf] (6.5,-.5) -- (6.5,.5);
  \draw[drinf] (5,-.5) -- (5,.5);  \draw[drinf] (5.5,-.5) -- (5.5,.5);
  \draw[colie] (4,.5) -- (4,2); \draw[colie] (5,.5) -- (5,2);
  \draw[lie] (5.5,.5) -- (5.5,2); \draw[lie] (6.5,.5) -- (6.5,2);
  \path (4.5,-.25) node {\dots} (6,-.25) node {\dots} (6,.25) node {\dots} (4.5,.25) node {\dots} (4.5,1.25) node {\dots}  (6,1.25) node {\dots} ;
  \draw[blue, fill=blue] (3.9,-.05) rectangle (6.6,.05); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (3.9,2.1) -- node (n) {} (5.1,2.1);
  \path (n) +(0,.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (5.4,2.1) -- node (m) {} (6.6,2.1);
  \path (m) +(0,.35) node {\scriptsize $m$};
\end{tikzpicture} \]
whereas we do not add any coefficients to summands in ${\color{blue!50!black} \MIX}$.

With this, we can now begin to study the total complex $\T =(-1)^m \Q \oplus (-1)^m \R$.  Recall:
\[ \begin{tikzpicture}
  \draw[colie] (-1,0) -- (-1,.5); 
  \draw[colie] (-1,.5) -- (-1,1); 
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (.5,0) -- (.5,.5); 
  \draw[colie] (.5,.5) -- (.5,1); 
  \path (-.5,.25) node {\dots} (-.5,.75) node {\dots};
  \draw[lie] (1.5,0) -- (1.5,.5);
  \draw[lie] (1.5,.5) -- (1.5,1);
  \draw[lie] (2,0) -- (2,.5);
  \draw[lie] (2,.5) -- (2,1);
  \draw[lie] (3,0) -- (3,.5);
  \draw[lie] (3,.5) -- (3,1);
  \path (2.5,.25) node {\dots} (2.5,.75) node {\dots};  
  \draw[blue, fill=blue] (-1.1,.45) rectangle (0.6,.55);
  \draw[blue, fill=blue] (1.4,.45) rectangle (3.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (.6,-.1) -- node (n) {} (-1.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (3.1,-.1) -- node (m) {} (1.4,-.1);
  \path (m) +(0,-.35) node {\scriptsize $m$};
  
  \draw[colie] (-1,1.5) -- (-1,2); 
  \draw[colie] (-1,2) -- (-1,2.5); 
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (.5,1.5) -- (.5,2); 
  \draw[colie] (.5,2) -- (.5,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \path (-.5,1.75) node {\dots} (-.5,2.25) node {\dots};
  \draw[lie] (1.5,1.5) -- (1.5,2);
  \draw[lie] (2,1.5) -- (2,2);
  \draw[lie] (3,1.5) -- (3,2);
  \draw[lie] (1.5,2) -- (1.5,2.5);
  \draw[lie] (2,2) -- (2,2.5);
  \draw[lie] (3,2) -- (3,2.5);
  \path (2.5,1.75) node {\dots} (2.5,2.25) node {\dots};
  \draw[blue, fill=blue] (-1.1,1.95) rectangle (1.1,2.05);
  \draw[blue, fill=blue] (1.4,1.95) rectangle (3.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-1.1,2.6) -- node (n1) {} (1.1,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.4,2.6) -- node (m1) {} (3.1,2.6);
  \path (m1) +(0,.35) node {\scriptsize $m$};
 
  \draw[purple, fill=purple!30] (-1.25,1) rectangle node[text=black] {$Q$} (3.25,1.5);
  
  \node at (3.75,1.25) {$\defined$}; \node at (4.5,1.25) {$m$};
  
  \draw[colie] (5,0) -- (5,2); 
  \draw[colie] (6,0) -- (6,2); 
  \draw[colie] (6.5,0) -- (6.5,2); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.6,-.1) -- node (An) {} (4.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  \draw[colie] (5,2) -- (5,2.5); 
  \draw[colie] (6,2) -- (6,2.5); 
  \draw[colie] (6.5,2) -- (6.5,2.5); 
  \draw[colie] (7,2) -- (7,2.5); 
  \path (5.5,1) node {\dots} (5.5,2.25) node {\dots};
  \draw[blue, fill=blue] (4.9,1.95) rectangle (7.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.9,2.6) -- node (An1) {} (7.1,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n+1$};

  \draw[lie] (7.5,0) -- (7.5,.5);
  \draw[lie] (8,0) -- (8,.5);
  \draw[lie] (9,0) -- (9,.5);
  \draw[lie] (8,.5) -- (8,2);
  \draw[lie] (9,.5) -- (9,2);
  \draw[lie] (7.5,2) -- (7.5,2.5);
  \draw[lie] (8,2) -- (8,2.5);
  \draw[lie] (9,2) -- (9,2.5);
  \path (8.5,.25) node {\dots} (8.5,1.25) node {\dots} (8.5,2.25) node {\dots};  
  \draw[blue, fill=blue] (7.4,.45) rectangle (9.1,.55);
  \draw[blue, fill=blue] (7.4,1.95) rectangle (9.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (9.1,-.1) -- node (Am) {} (7.4,-.1);
  \path (Am) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.4,2.6) -- node (Am1) {} (9.1,2.6);
  \path (Am1) +(0,.35) node {\scriptsize $m$};
  
  \draw[lie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.25,1.25);
  \draw[lie] (7.25,1.25) .. controls +(30:.25) and +(270:.5) .. (7.5,2);
  \draw[colie] (7.25,1.25) .. controls +(150:.25) and +(270:.5) .. (7,2);
  \node[dot=blue] at (7.25,1.25) {};
  
  \node at (9.5,1.25) {$-$};
  \node at (10,1.25) {$\displaystyle \frac{n}{2}$};

  \draw[colie] (10.5,0) -- (10.5,.5);
  \draw[colie] (10.5,.5) -- (10.5,2);
  \draw[colie] (10.5,2) -- (10.5,2.5);
  \draw[colie] (11.5,0) -- (11.5,.5);
  \draw[colie] (11.5,.5) -- (11.5,2);
  \draw[colie] (11.5,2) -- (11.5,2.5);
  \path (11,.25) node {\dots} (11,1.25) node {\dots} (11,2.25) node {\dots};
  \draw[blue, fill=blue] (10.4,.45) rectangle (12.1,.55);
  \draw[blue, fill=blue] (10.4,1.95) rectangle (12.6,2.05);
  \draw[colie] (12,0) -- (12,.5);
  \draw[colie] (12,2) -- (12,2.5);
  \draw[colie] (12.5,2) -- (12.5,2.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.1,-.1) -- node (Bn) {} (10.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (10.4,2.6) -- node (Bn1) {} (12.6,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n+1$};
  
  \draw[colie] (12,.5) .. controls +(90:.5) and +(270:.5) .. (12.25,1.25);
  \draw[colie] (12.25,1.25) .. controls +(30:.25) and +(270:.5) .. (12.5,2);
  \draw[colie] (12.25,1.25) .. controls +(150:.25) and +(270:.5) .. (12,2);
  \node[dot=blue] at (12.25,1.25) {};
  
  \draw[lie] (13,0) -- (13,1.25);
  \draw[lie] (13,1.25) -- (13,2.5);
  \draw[lie] (13.5,0) -- (13.5,1.25);
  \draw[lie] (13.5,1.25) -- (13.5,2.5);
  \draw[lie] (14.5,0) -- (14.5,1.25);
  \draw[lie] (14.5,1.25) -- (14.5,2.5);
  \draw[blue, fill=blue] (12.9,1.2) rectangle (14.6,1.3);
  \path (14,.5) node {\dots} (14,2) node {\dots};
  \draw[thick,decoration={brace,amplitude=5},decorate] (14.6,-.1) -- node (Bm) {} (12.9,-.1);
  \path (Bm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (12.9,2.6) -- node (Bm1) {} (14.6,2.6);
  \path (Bm1) +(0,.35) node {\scriptsize $m$};
\end{tikzpicture} \]
We fix $m+n = k$, consider the composition ${\color{blue!50!black} \MIX} \circ (-1)^m \Q \circ {\color{blue!50!black} \SORT}$, and sum over $0 \leq m \leq k$.  Then:
\[\begin{tikzpicture}
  \draw[drinf] (0,0) node (a) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[drinf] (a) ++(0,2) node (A) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[blue, fill=blue,text=black] (a) +(-.1,.2) rectangle +(1.6,.3) +(.75,.1) node {\dots};
  \draw[blue, fill=blue,text=black] (A) +(-.1,.2) rectangle +(1.6,.3) +(.75,.4) node {\dots};
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,.5) rectangle node {$\SORT$} +(1.75,1);
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,1.5) rectangle node {$\MIX$} +(1.75,2);
  \draw[purple, fill=purple!50,text=black] (a) +(-.25,1) rectangle node (aQ) {$\sum (-1)^m\,Q$} +(1.75,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(1.6,-.1) -- node (ak) {} +(-.1,-.1);
  \path (ak) +(0,-.35) node {\scriptsize $k$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(-.1,2.6) -- node (ak1) {} +(1.6,2.6);
  \path (ak1) +(0,.35) node {\scriptsize $k+1$};
  
  \path (aQ) ++(1.35,0) node (e1) {$=$} ++(.1,0) node[anchor=west] (bEQN) {$\displaystyle \sum_{m=0}^k{k\choose m}(-1)^m\,m$} (bEQN.east) ++(-.3,-1.25) node (b) {};
  
  \draw[drinf] (b) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) ++(.75,0) -- +(0,.5) ++(.75,0) -- +(0,.5) ++(1,0) -- +(0,.5);
  \draw[drinf] (b) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) node (bTOP) {};
  \draw[blue, fill=blue,text=black] (b) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(2.5,0) node {\dots}  ;
  \draw[colie] (b) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (b) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (b) ++(3,.5) -- +(0,1.5);
  \draw[lie] (b) ++(4,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) ++(1.6,-.1) -- node (bn) {} +(-1.2,0);
  \path (bn) +(0,-.35) node {\scriptsize $k-m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) ++(4.1,-.1) -- node (bm) {} +(-1.2,0);
  \path (bm) +(0,-.35) node {\scriptsize $m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) ++(.4,2.6) -- node (bn1) {} +(1.2,0);
  \path (bn1) +(0,.35) node {\scriptsize $k-m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) ++(2.9,2.6) -- node (bm1) {} +(1.2,0);
  \path (bm1) +(0,.35) node {\scriptsize $m-1$};
  \draw[lie] (b) ++(2.25,.5) -- +(0,.75) node[dot=blue] (bDOT) {};
  \draw[colie] (bDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (bDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
  \path (bTOP) ++(.5,-1.25) node {$-$} ++(.1,0) node[anchor=west] (cEQN) {$\displaystyle \sum_{m=0}^k{k\choose m}(-1)^m\,\frac{k-m}2$} (cEQN.east) ++(-.3,-1.25) node (c) {};
  
    \draw[drinf] (c) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) ++(.75,0) -- +(0,.5) ++(.75,0) -- +(0,.5) ++(1,0) -- +(0,.5);
  \draw[drinf] (c) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) node (cTOP) {};
  \draw[blue, fill=blue,text=black] (c) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(2.5,0) node {\dots}  ;
  \draw[colie] (c) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (c) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (c) ++(3,.5) -- +(0,1.5);
  \draw[lie] (c) ++(4,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) ++(1.6,-.1) -- node (cn) {} +(-1.2,0);
  \path (cn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) ++(4.1,-.1) -- node (cm) {} +(-1.2,0);
  \path (cm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) ++(.4,2.6) -- node (cn1) {} +(1.2,0);
  \path (cn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) ++(2.9,2.6) -- node (cm1) {} +(1.2,0);
  \path (cm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (c) ++(2.25,.5) -- +(0,.75) node[dot=blue] (cDOT) {};
  \draw[colie] (cDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[colie] (cDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
  \path (e1) ++(0,-4) node (e2) {$=$} ++(.1,0) node[anchor=west] (dEQN) {$\displaystyle \sum_{m=1}^{k-1}\frac{k!\,(-1)^m\, (-1)^{m-1}}{(m-1)!\,(k-m)!} $} (dEQN.east) ++(-.3,-1.25) node (d) {};
  
    \draw[drinf] (d) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (d) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (dTOP) {};
  \draw[blue, fill=blue,text=black] (d) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (d) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (d) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (d) ++(2,.5) -- +(0,1.5);
  \draw[lie] (d) ++(3,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (d) ++(1.6,-.1) -- node (dn) {} +(-1.2,0);
  \path (dn) +(0,-.35) node {\scriptsize $k-m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (d) ++(3.1,-.1) -- node (dm) {} +(-1.2,0);
  \path (dm) +(0,-.35) node {\scriptsize $m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (d) ++(.4,2.6) -- node (dn1) {} +(1.2,0);
  \path (dn1) +(0,.35) node {\scriptsize $k-m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (d) ++(1.9,2.6) -- node (dm1) {} +(1.2,0);
  \path (dm1) +(0,.35) node {\scriptsize $m-1$};
  \draw[lie] (d) ++(3.75,.5) -- +(0,.75) node[dot=blue] (dDOT) {};
  \draw[colie] (dDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (dDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);

  \path (dTOP) ++(.5,-1.25) node {$-$} ++(.1,0) node[anchor=west] (fEQN) {$\displaystyle \sum_{m=0}^{k-1} \frac{k!\,(-1)^m\,(-1)^m}{2\,m!\,(k-m-1)!}$} (fEQN.east) ++(-.3,-1.25) node (f) {};

  \draw[drinf] (f) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (f) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (fTOP) {};
  \draw[blue, fill=blue,text=black] (f) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (f) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (f) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (f) ++(2,.5) -- +(0,1.5);
  \draw[lie] (f) ++(3,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (f) ++(1.6,-.1) -- node (fn) {} +(-1.2,0);
  \path (fn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (f) ++(3.1,-.1) -- node (fm) {} +(-1.2,0);
  \path (fm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (f) ++(.4,2.6) -- node (fn1) {} +(1.2,0);
  \path (fn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (f) ++(1.9,2.6) -- node (fm1) {} +(1.2,0);
  \path (fm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (f) ++(3.75,.5) -- +(0,.75) node[dot=blue] (fDOT) {};
  \draw[colie] (fDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[colie] (fDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
  \path (e2) ++(0,-4) node (e3) {$=$} ++(.2,0) node[anchor=west] (gEQN) {$\displaystyle -k \sum_{m=0}^{k-1} {k-1 \choose m}$} (gEQN.east) ++(-.3,-1.25) node (g) {};
  
  \draw[drinf] (g) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (g) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (gTOP) {};
  \draw[blue, fill=blue,text=black] (g) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (g) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (g) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (g) ++(2,.5) -- +(0,1.5);
  \draw[lie] (g) ++(3,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (g) ++(1.6,-.1) -- node (gn) {} +(-1.2,0);
  \path (gn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (g) ++(3.1,-.1) -- node (gm) {} +(-1.2,0);
  \path (gm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (g) ++(.4,2.6) -- node (gn1) {} +(1.2,0);
  \path (gn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (g) ++(1.9,2.6) -- node (gm1) {} +(1.2,0);
  \path (gm1) +(0,.35) node {\scriptsize $m$};
  \draw[lie] (g) ++(3.75,.5) -- +(0,.75) node[dot=blue] (gDOT) {};
  \draw[colie] (gDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (gDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);

  \path (gTOP) ++(.5,-1.25) node {$-$} ++(.2,0) node[anchor=west] (hEQN) {$\displaystyle \frac{k}{2} \sum_{m=0}^{k-1} {k-1 \choose m}$} (hEQN.east) ++(-.3,-1.25) node (h) {};

  \draw[drinf] (h) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (h) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (hTOP) {};
  \draw[blue, fill=blue,text=black] (h) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (h) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (h) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (h) ++(2,.5) -- +(0,1.5);
  \draw[lie] (h) ++(3,.5) -- +(0,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (h) ++(1.6,-.1) -- node (hn) {} +(-1.2,0);
  \path (hn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (h) ++(3.1,-.1) -- node (hm) {} +(-1.2,0);
  \path (hm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (h) ++(.4,2.6) -- node (hn1) {} +(1.2,0);
  \path (hn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (h) ++(1.9,2.6) -- node (hm1) {} +(1.2,0);
  \path (hm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (h) ++(3.75,.5) -- +(0,.75) node[dot=blue] (hDOT) {};
  \draw[colie] (hDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[colie] (hDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
    \path (e3) ++(0,-4) node (e4) {$=$} ++(.2,0) node[anchor=west] (iEQN) {$\displaystyle -k \sum_{m=0}^{k-1}$} (iEQN.east) ++(-.1,-1.25) node (i) {};
  
  \draw[drinf] (i) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (i) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (iTOP) {};
  \draw[blue, fill=blue,text=black] (i) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (i) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (i) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (i) ++(2,.5) -- +(0,1.5);
  \draw[lie] (i) ++(3,.5) -- +(0,1.5);
  \draw[blue,fill=blue!50,text=black] (i) ++(.25,.5) rectangle node{$\sort^{k-m-1,m}$} +(3,.5);
  \draw[blue,fill=blue!50,text=black] (i) ++(.25,1.5) rectangle node{$\sort_{k-m-1,m}$} +(3,.5);
  
  \draw[thick,decoration={brace,amplitude=5},decorate] (i) ++(1.6,-.1) -- node (in) {} +(-1.2,0);
  \path (in) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (i) ++(3.1,-.1) -- node (im) {} +(-1.2,0);
  \path (im) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (i) ++(.4,2.6) -- node (in1) {} +(1.2,0);
  \path (in1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (i) ++(1.9,2.6) -- node (im1) {} +(1.2,0);
  \path (im1) +(0,.35) node {\scriptsize $m$};
  \draw[lie] (i) ++(3.75,.5) -- +(0,.75) node[dot=blue] (iDOT) {};
  \draw[colie] (iDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (iDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);

  \path (iTOP) ++(.5,-1.25) node {$-$} ++(.2,0) node[anchor=west] (jEQN) {$\displaystyle \frac{k}{2} \sum_{m=0}^{k-1} $} (jEQN.east) ++(-.1,-1.25) node (j) {};

  \draw[drinf] (j) ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (j) ++(.5,2) -- +(0,.5) ++(1,0) -- +(0,.5) 
  ++(.5,0) -- +(0,.5) ++(1,0) -- +(0,.5)
   ++(.5,0) -- +(0,.5) ++(.5,0) -- +(0,.5) node (jTOP) {};
  \draw[blue, fill=blue,text=black] (j) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[colie] (j) ++(.5,.5) -- +(0,1.5);
  \draw[colie] (j) ++(1.5,.5) -- +(0,1.5);
  \draw[lie] (j) ++(2,.5) -- +(0,1.5);
  \draw[lie] (j) ++(3,.5) -- +(0,1.5);
  \draw[blue,fill=blue!50,text=black] (j) ++(.25,.5) rectangle node{$\sort^{k-m-1}$} +(3,.5);
  \draw[blue,fill=blue!50,text=black] (j) ++(.25,1.5) rectangle node{$\sort_{k-m-1}$} +(3,.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (j) ++(1.6,-.1) -- node (jn) {} +(-1.2,0);
  \path (jn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (j) ++(3.1,-.1) -- node (jm) {} +(-1.2,0);
  \path (jm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (j) ++(.4,2.6) -- node (jn1) {} +(1.2,0);
  \path (jn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (j) ++(1.9,2.6) -- node (jm1) {} +(1.2,0);
  \path (jm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (j) ++(3.75,.5) -- +(0,.75) node[dot=blue] (jDOT) {};
  \draw[colie] (jDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[colie] (jDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
    \path (e4) ++(0,-4) node (e5) {$=$} ++(.2,0) node[anchor=west] (kEQN) {$\displaystyle -k $} (kEQN.east) ++(-.3,-1.25) node (k) {};
  
  \draw[drinf] (k) ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (k) ++(3.5,2)   -- +(0,.5) ++(.5,0) -- +(0,.5) node (kTOP) {};
  \draw[blue, fill=blue,text=black] (k) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(1.6,-.1) -- node (kn) {} +(-1.2,0);
  \path (kn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(3.1,-.1) -- node (km) {} +(-1.2,0);
  \path (km) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(.4,2.6) -- node (kn1) {} +(1.2,0);
  \path (kn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(1.9,2.6) -- node (km1) {} +(1.2,0);
  \path (km1) +(0,.35) node {\scriptsize $m$};
  \draw[lie] (k) ++(3.75,.5) -- +(0,.75) node[dot=blue] (kDOT) {};
  \draw[colie] (kDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (kDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);

  \path (kTOP) ++(.5,-1.25) node {$-$} ++(.2,0) node[anchor=west] (lEQN) {$\displaystyle \frac{k}{2} $} (lEQN.east) ++(-.3,-1.25) node (l) {};

  \draw[drinf] (l) ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (l) ++(3.5,2)   -- +(0,.5) ++(.5,0) -- +(0,.5) node (lTOP) {};
  \draw[blue, fill=blue,text=black] (l) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(1.6,-.1) -- node (ln) {} +(-1.2,0);
  \path (ln) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(3.1,-.1) -- node (lm) {} +(-1.2,0);
  \path (lm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(.4,2.6) -- node (ln1) {} +(1.2,0);
  \path (ln1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(1.9,2.6) -- node (lm1) {} +(1.2,0);
  \path (lm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (l) ++(3.75,.5) -- +(0,.75) node[dot=blue] (lDOT) {};
  \draw[colie] (lDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[colie] (lDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
\end{tikzpicture}\]
Analogously to above, one can compute:
\[\begin{tikzpicture}
  \draw[drinf] (0,0) node (a) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[drinf] (a) ++(0,2) node (A) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[blue, fill=blue,text=black] (a) +(-.1,.2) rectangle +(1.6,.3) +(.75,.1) node {\dots};
  \draw[blue, fill=blue,text=black] (A) +(-.1,.2) rectangle +(1.6,.3) +(.75,.4) node {\dots};
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,.5) rectangle node {$\SORT$} +(1.75,1);
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,1.5) rectangle node {$\MIX$} +(1.75,2);
  \draw[purple, fill=purple!50,text=black] (a) +(-.25,1) rectangle node (aR) {$\sum (-1)^m\,R$} +(1.75,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(1.6,-.1) -- node (ak) {} +(-.1,-.1);
  \path (ak) +(0,-.35) node {\scriptsize $k$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(-.1,2.6) -- node (ak1) {} +(1.6,2.6);
  \path (ak1) +(0,.35) node {\scriptsize $k+1$};
  
  \path (aR) ++(1.35,0) node (e1) {$=$} ++(.2,0) node[anchor=west] (kEQN) {$\displaystyle -\frac{k}2 $} (kEQN.east) ++(-.3,-1.25) node (k) {};
  
  \draw[drinf] (k) ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (k) ++(3.5,2)   -- +(0,.5) ++(.5,0) -- +(0,.5) node (kTOP) {};
  \draw[blue, fill=blue,text=black] (k) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(1.6,-.1) -- node (kn) {} +(-1.2,0);
  \path (kn) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(3.1,-.1) -- node (km) {} +(-1.2,0);
  \path (km) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(.4,2.6) -- node (kn1) {} +(1.2,0);
  \path (kn1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (k) ++(1.9,2.6) -- node (km1) {} +(1.2,0);
  \path (km1) +(0,.35) node {\scriptsize $m$};
  \draw[lie] (k) ++(3.75,.5) -- +(0,.75) node[dot=blue] (kDOT) {};
  \draw[lie] (kDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (kDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);

  \path (kTOP) ++(.5,-1.25) node {$-$} ++(.2,0) node[anchor=west] (lEQN) {$\displaystyle k $} (lEQN.east) ++(-.3,-1.25) node (l) {};

  \draw[drinf] (l) ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.5,0) -- +(0,2.5) ++(1,0) -- +(0,2.5) 
  ++(.75,0) -- +(0,.5);
  \draw[drinf] (l) ++(3.5,2)   -- +(0,.5) ++(.5,0) -- +(0,.5) node (lTOP) {};
  \draw[blue, fill=blue,text=black] (l) +(.4,.2) rectangle +(4.1,.3) +(.4,2.2) rectangle +(4.1,2.3)
  ++(1,1.25) node {\dots} ++(1.5,0) node {\dots}  ;
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(1.6,-.1) -- node (ln) {} +(-1.2,0);
  \path (ln) +(0,-.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(3.1,-.1) -- node (lm) {} +(-1.2,0);
  \path (lm) +(0,-.35) node {\scriptsize $m$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(.4,2.6) -- node (ln1) {} +(1.2,0);
  \path (ln1) +(0,.35) node {\scriptsize $k-m-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (l) ++(1.9,2.6) -- node (lm1) {} +(1.2,0);
  \path (lm1) +(0,.35) node {\scriptsize $m$};
  \draw[colie] (l) ++(3.75,.5) -- +(0,.75) node[dot=blue] (lDOT) {};
  \draw[colie] (lDOT) .. controls +(150:.25) and +(270:.5) .. +(-.25,.75);
  \draw[lie] (lDOT) .. controls +(30:.25) and +(270:.5) .. +(.25,.75);
  
\end{tikzpicture}\]
Then the total complex is:
\[\begin{tikzpicture}
  \draw[drinf] (0,0) node (a) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[drinf] (a) ++(0,2) node (A) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[blue, fill=blue,text=black] (a) +(-.1,.2) rectangle +(1.6,.3) +(.75,.1) node {\dots};
  \draw[blue, fill=blue,text=black] (A) +(-.1,.2) rectangle +(1.6,.3) +(.75,.4) node {\dots};
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,.5) rectangle node {$\SORT$} +(1.75,1);
  \draw[blue, fill=blue!50,text=black] (a) +(-.25,1.5) rectangle node {$\MIX$} +(1.75,2);
  \draw[purple, fill=purple!50,text=black] (a) +(-.25,1) rectangle node (aT) {$T$} +(1.75,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(1.6,-.1) -- node (ak) {} +(-.1,-.1);
  \path (ak) +(0,-.35) node {\scriptsize $k$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (a) +(-.1,2.6) -- node (ak1) {} +(1.6,2.6);
  \path (ak1) +(0,.35) node {\scriptsize $k+1$};
  
  \draw[drinf] (a) ++(3,0) node (b) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[drinf] (b) ++(0,2) node (B) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[blue, fill=blue,text=black] (b) +(-.1,.2) rectangle +(1.6,.3) +(.75,.1) node {\dots};
  \draw[blue, fill=blue,text=black] (B) +(-.1,.2) rectangle +(1.6,.3) +(.75,.4) node {\dots};
  \draw[blue, fill=blue!50,text=black] (b) +(-.25,.5) rectangle node {$\SORT$} +(1.75,1);
  \draw[blue, fill=blue!50,text=black] (b) +(-.25,1.5) rectangle node {$\MIX$} +(1.75,2);
  \draw[purple, fill=purple!50,text=black] (b) +(-.25,1) rectangle node (bQ) {$\sum (-1)^m\,Q$} +(1.75,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) +(1.6,-.1) -- node (bk) {} +(-.1,-.1);
  \path (bk) +(0,-.35) node {\scriptsize $k$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (b) +(-.1,2.6) -- node (bk1) {} +(1.6,2.6);
  \path (bk1) +(0,.35) node {\scriptsize $k+1$};
  
  \draw[drinf] (b) ++(3,0) node (c) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[drinf] (c) ++(0,2) node (C) {} -- +(0,.5) ++(1.5,0) -- +(0,.5);
  \draw[blue, fill=blue,text=black] (c) +(-.1,.2) rectangle +(1.6,.3) +(.75,.1) node {\dots};
  \draw[blue, fill=blue,text=black] (C) +(-.1,.2) rectangle +(1.6,.3) +(.75,.4) node {\dots};
  \draw[blue, fill=blue!50,text=black] (c) +(-.25,.5) rectangle node {$\SORT$} +(1.75,1);
  \draw[blue, fill=blue!50,text=black] (c) +(-.25,1.5) rectangle node {$\MIX$} +(1.75,2);
  \draw[purple, fill=purple!50,text=black] (c) +(-.25,1) rectangle node (cR) {$\sum (-1)^m\,R$} +(1.75,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) +(1.6,-.1) -- node (ck) {} +(-.1,-.1);
  \path (ck) +(0,-.35) node {\scriptsize $k$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (c) +(-.1,2.6) -- node (ck1) {} +(1.6,2.6);
  \path (ck1) +(0,.35) node {\scriptsize $k+1$};
  
  \path (aT) ++(1.5,0) node {$=$} ++(3,0) node {$+$} ++(3,0) node (e1) {$=$} ++(.2,0) node[anchor=west] (TT) {$\displaystyle -\frac{k}{2}$};
  
  \draw[drinf] (TT.east) ++(.2,-1.25) node (Tlr) {} -- +(0,2.5) ++(1,0) -- +(0,2.5) ++(.75,0) -- +(0,.5) ++(-.25,2) -- +(0,.5) ++(.5,0) -- +(0,.5);
  \draw[blue,fill=blue] (Tlr) ++(-.1,.2) rectangle +(2.2,.1) ++(0,2) rectangle +(2.2,.1) ;
  \path (Tlr) +(.5,1.25) node {$\dots$};
  \draw[green,fill=green!50] (Tlr) ++(1.25,.5) rectangle +(1,1.5);
  \draw[thick,decoration={brace,amplitude=5},decorate] (Tlr) +(1.1,-.1) -- node (Tk) {} +(-.1,-.1);
  \path (Tk) +(0,-.35) node {\scriptsize $k-1$};
  \draw[thick,decoration={brace,amplitude=5},decorate] (Tlr) +(-.1,2.6) -- node (Tk1) {} +(1.1,2.6);
  \path (Tk1) +(0,.35) node {\scriptsize $k-1$};
\end{tikzpicture}\]
Here \,\tikz[baseline=.5ex] \draw[green,fill=green!50] (0,0) rectangle (.5em,1em);\, is a sum of vertices.  We may as well include an alternator at the top (it will be absorbed by the larger alternator), whence:
\[ \begin{tikzpicture}
  \draw[drinf] (0,0) node (b0) {} -- +(0,.5) ++(-.25,2) -- +(0,.5) ++(.5,0) -- +(0,.5);
  \draw[green,fill=green!50] (b0) ++(-.5,.5) rectangle node (DOT0) {}+(1,1.5);
  
  \foreach \n/\m in {0/1,1/2,2/3,3/4,4/5,5/6} \draw[drinf] (b\n) ++(2,0) node (b\m) {} -- +(0,.5) ++(-.25,2) -- +(0,.5) ++(.5,0) -- +(0,.5) ++(-.25,-.75) node[dot=blue] (DOT\m) {};
  \path (DOT0) +(1,0) node {$=$};
  \foreach \n in {1,2,3,4,5} \path (DOT\n) +(1,0) node {$+$};
  
  \draw[colie] (DOT1) -- ++(-0,-.75);
  \draw[colie] (DOT1) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[lie] (DOT1) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);

  \draw[colie] (DOT2) -- ++(-0,-.75);
  \draw[lie] (DOT2) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[colie] (DOT2) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);
  
  \draw[lie] (DOT3) -- ++(-0,-.75);
  \draw[colie] (DOT3) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[colie] (DOT3) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);
  
  \draw[colie] (DOT4) -- ++(-0,-.75);
  \draw[lie] (DOT4) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[lie] (DOT4) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);
  
  \draw[lie] (DOT5) -- ++(-0,-.75);
  \draw[colie] (DOT5) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[lie] (DOT5) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);
  
  \draw[lie] (DOT6) -- ++(-0,-.75);
  \draw[lie] (DOT6) .. controls +(150:.25) and +(270:.5) .. ++(-.25,.75);
  \draw[colie] (DOT6) .. controls +(30:.25) and +(270:.5) .. ++(.25,.75);
  
  \path (DOT6) +(1,0) node {$=$} ++(2,0) node[circle,draw=blue,inner sep=2pt] (DOT7) {};
  
  \draw[drinf] (DOT7) -- ++(-0,-1.25);
  \draw[drinf] (DOT7) .. controls +(150:.25) and +(270:1) .. ++(-.25,1.25);
  \draw[drinf] (DOT7) .. controls +(30:.25) and +(270:1) .. ++(.25,1.25);

\end{tikzpicture} \]
Thus $\T$ is exactly (the negative of) the Chevalley differential for \DDDD\ with the trivial representation.  In particular, $\T^2 = 0$, and this proves that $\Q\R = \R\Q$ as claimed.  On the other hand, a direct proof that $\Q\R = \R\Q$ implies that $\T^2 = 0$, which is equivalent to \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\end{tikzpicture}}\, being a Lie bracket.  In particular, this provides an alternate proof of the Jacobi identity for  \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3] \node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\end{tikzpicture}}\,.

\section{Coboundary algebras, CYBE, and Triangular structures}

Having developed, via the Chevalley complex, some cohomological language, we can now recast the above definitions, and describe a few more.

Recall that given a Lie algebra \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=blue] at (0,0) {};
\end{tikzpicture}}\, and an action \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=red] at (0,0) {};
\end{tikzpicture}}\, over it, we defined the \defn{Chevalley complex} of \gggg\ with coefficients in \mmmm\ by the differential:
\[ \begin{tikzpicture}
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (1,0) -- (1,.5); 
  \draw[colie] (1,.5) -- (1,1); 
  \path (.5,.25) node {\dots} (.5,.75) node {\dots};
  \draw[module] (2,0) -- (2,1);
  \draw[blue, fill=blue] (-.1,.45) rectangle (1.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.1,-.1) -- node (n) {} (-.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \draw[colie] (1.5,1.5) -- (1.5,2); 
  \draw[colie] (1.5,2) -- (1.5,2.5); 
  \path (.5,1.75) node {\dots} (.5,2.25) node {\dots};
  \draw[module] (2,1.5) -- (2,2.5);
  \draw[blue, fill=blue] (-.1,1.95) rectangle (1.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.6) -- node (n1) {} (1.6,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
 
  \draw[purple, fill=purple!30] (-.25,1) rectangle node[text=black] {$Q$} (2.25,1.5);
  
  \node at (2.5,1.25) {$\defined$};
  
  \draw[colie] (3,0) -- (3,2); 
  \draw[colie] (4,0) -- (4,2); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.1,-.1) -- node (An) {} (2.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (3,2) -- (3,2.5); 
  \draw[colie] (4,2) -- (4,2.5); 
  \draw[colie] (4.5,2) -- (4.5,2.5); 
  \path (3.5,1) node {\dots} (3.5,2.25) node {\dots};
  \draw[blue, fill=blue] (2.9,1.95) rectangle (4.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (2.9,2.6) -- node (An1) {} (4.6,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n+1$};

  \draw[module] (5,0) .. controls +(90:.5) and +(270:.5) .. (4.75,1);
  \draw[module] (4.75,1) .. controls +(30:.25) and +(270:.5) .. (5,2.5);
  \draw[colie] (4.75,1) .. controls +(150:.25) and +(270:.5) .. (4.5,2);
  
  \node[dot=red] at (4.75,1) {};
  
  \node at (5.5,1.25) {$-$};
  \node at (6,1.25) {$\displaystyle \frac{n}{2}$};
  
  \draw[colie] (6.5,0) -- (6.5,.5); 
  \draw[colie] (6.5,.5) -- (6.5,2); 
  \draw[colie] (7.5,0) -- (7.5,.5); 
  \draw[colie] (7.3,.5) -- (7.3,2); 
  \path (7,.25) node {\dots};
  \draw[blue, fill=blue] (6.4,.45) rectangle (7.6,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.6,-.1) -- node (Bn) {} (6.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \path (6.9,1.25) node {\dots};
  \draw[colie] (6.5,2) -- (6.5,2.5); 
  \draw[colie] (7.5,2) -- (7.5,2.5); 
  \draw[colie] (8,2) -- (8,2.5); 
  \path (7,2.25) node {\dots};
  \draw[blue, fill=blue] (6.4,1.95) rectangle (8.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.4,2.6) -- node (Bn1) {} (8.1,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n+1$};
  
  \draw[module] (8.5,0) -- (8.5,2.5);
  
  \draw[colie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.75,1.25) node[dot=blue] {};
  \draw[colie] (7.75,1.25) .. controls +(150:.25) and +(270:.5) .. (7.5,2);
  \draw[colie] (7.75,1.25) .. controls +(30:.25) and +(270:.5) .. (8,2);
\end{tikzpicture} \]
When the action \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=red] at (0,0) {};
\end{tikzpicture}}\, is trivial, the condition that $\Q^2 = 0$ is equivalent to the Jacobi identity.  Given the Jacobi identity, $\Q^2 = 0$ if and only if \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[module] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[module] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[circle,draw=red,fill=red,inner sep=1pt] at (0,0) {};
\end{tikzpicture}}\, satisfies the conditions of a left action.  In particular, \gggg\ acts on itself via \,\raisebox{-1ex}{\begin{tikzpicture}[scale=.3]
  \draw[lie] (0,0) -- (0,1);
  \draw[lie] (-1,-1) .. controls +(90:.5) and +(210:.5) .. (0,0);
  \draw[lie] (1,-1) .. controls +(90:.5) and +(330:.5) .. (0,0);
  \node[dot=blue] at (0,0) {};
\end{tikzpicture}}\,, and so we can consider (and did, above) the Chevalley complex with coefficients in \gggg.  The differential is:
\[ \begin{tikzpicture}
  \draw[colie] (0,0) -- (0,.5); 
  \draw[colie] (0,.5) -- (0,1); 
  \draw[colie] (1,0) -- (1,.5); 
  \draw[colie] (1,.5) -- (1,1); 
  \path (.5,.25) node {\dots} (.5,.75) node {\dots};
  \draw[lie] (2,0) -- (2,1);
  \draw[blue, fill=blue] (-.1,.45) rectangle (1.1,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (1.1,-.1) -- node (n) {} (-.1,-.1);
  \path (n) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (0,1.5) -- (0,2); 
  \draw[colie] (0,2) -- (0,2.5); 
  \draw[colie] (1,1.5) -- (1,2); 
  \draw[colie] (1,2) -- (1,2.5); 
  \draw[colie] (1.5,1.5) -- (1.5,2); 
  \draw[colie] (1.5,2) -- (1.5,2.5); 
  \path (.5,1.75) node {\dots} (.5,2.25) node {\dots};
  \draw[lie] (2,1.5) -- (2,2.5);
  \draw[blue, fill=blue] (-.1,1.95) rectangle (1.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (-.1,2.6) -- node (n1) {} (1.6,2.6);
  \path (n1) +(0,.35) node {\scriptsize $n+1$};
 
  \draw[purple, fill=purple!30] (-.25,1) rectangle node[text=black] {$Q$} (2.25,1.5);
  
  \node at (2.5,1.25) {$\defined$};
  
  \draw[colie] (3,0) -- (3,2); 
  \draw[colie] (4,0) -- (4,2); 
  \draw[thick,decoration={brace,amplitude=5},decorate] (4.1,-.1) -- node (An) {} (2.9,-.1);
  \path (An) +(0,-.35) node {\scriptsize $n$};
  
  \draw[colie] (3,2) -- (3,2.5); 
  \draw[colie] (4,2) -- (4,2.5); 
  \draw[colie] (4.5,2) -- (4.5,2.5); 
  \path (3.5,1) node {\dots} (3.5,2.25) node {\dots};
  \draw[blue, fill=blue] (2.9,1.95) rectangle (4.6,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (2.9,2.6) -- node (An1) {} (4.6,2.6);
  \path (An1) +(0,.35) node {\scriptsize $n+1$};

  \draw[lie] (5,0) .. controls +(90:.5) and +(270:.5) .. (4.75,1);
  \draw[lie] (4.75,1) .. controls +(30:.25) and +(270:.5) .. (5,2.5);
  \draw[colie] (4.75,1) .. controls +(150:.25) and +(270:.5) .. (4.5,2);
  
  \node[dot=blue] at (4.75,1) {};
  
  \node at (5.5,1.25) {$-$};
  \node at (6,1.25) {$\displaystyle \frac{n}{2}$};
  
  \draw[colie] (6.5,0) -- (6.5,.5); 
  \draw[colie] (6.5,.5) -- (6.5,2); 
  \draw[colie] (7.5,0) -- (7.5,.5); 
  \draw[colie] (7.3,.5) -- (7.3,2); 
  \path (7,.25) node {\dots};
  \draw[blue, fill=blue] (6.4,.45) rectangle (7.6,.55);
  \draw[thick,decoration={brace,amplitude=5},decorate] (7.6,-.1) -- node (Bn) {} (6.4,-.1);
  \path (Bn) +(0,-.35) node {\scriptsize $n$};
  \path (6.9,1.25) node {\dots};
  \draw[colie] (6.5,2) -- (6.5,2.5); 
  \draw[colie] (7.5,2) -- (7.5,2.5); 
  \draw[colie] (8,2) -- (8,2.5); 
  \path (7,2.25) node {\dots};
  \draw[blue, fill=blue] (6.4,1.95) rectangle (8.1,2.05);
  \draw[thick,decoration={brace,amplitude=5},decorate] (6.4,2.6) -- node (Bn1) {} (8.1,2.6);
  \path (Bn1) +(0,.35) node {\scriptsize $n+1$};
  
  \draw[lie] (8.5,0) -- (8.5,2.5);
  
  \draw[colie] (7.5,.5) .. controls +(90:.5) and +(270:.5) .. (7.75,1.25) node[dot=blue] {};
  \draw[colie] (7.75,1.25) .. controls +(150:.25) and +(270:.5) .. (7.5,2);
  \draw[colie] (7.75,1.25) .. controls +(30:.25) and +(270:.5) .. (8,2);
\end{tikzpicture} \]

Recall that given any vector space \vvvv, an \defn{element} of \vvvv\ is a map from the ground field to \vvvv:
\[\begin{tikzpicture}
  \draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black, fill=green,inner sep=1pt, shape border rotate=180] {} -- (0,1);
\end{tikzpicture}\]
If 
\,\tikz[baseline=1ex]{\draw[string=green!50!black] (0,0) -- (0,1em) node[dot=purple] (P) {}; \draw[string=green!50!black] (P) -- (0,2em);}\, 
is any complex, then an element 
\,\tikz[baseline=.5ex] \draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black,fill=green,inner sep=1pt, shape border rotate=180] {} -- (0,1em);\,
 is a \defn{cocycle} if 
 $\,\tikz[baseline=1ex]{\draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black,fill=green,inner sep=1pt, shape border rotate=180] {} -- (0,1em) node[dot=purple] (P) {}; \draw[string=green!50!black] (P) -- (0,2em);}\, 
 = 0$.  It is a \defn{coboundary} if 
 $\,\tikz[baseline=.5ex] \draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black,fill=green,inner sep=1pt, shape border rotate=180] {} -- (0,1em);\,
  =
   \,\tikz[baseline=1ex]{\draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black,fill=blue,inner sep=1pt, shape border rotate=180] {} -- (0,1em) node[dot=purple] (P) {}; \draw[string=green!50!black] (P) -- (0,2em);}\,$
for some other element 
\,\tikz[baseline=.5ex] \draw[string=green!50!black] (0,0) node[regular polygon, regular polygon sides=3,draw=green!50!black,fill=blue,inner sep=1pt, shape border rotate=180] {} -- (0,1em);\,.
  Of course, since 
  \,\tikz[baseline=1ex]{\draw[string=green!50!black] (0,0) -- (0,1em) node[dot=purple] (P) {}; \draw[string=green!50!black] (P) -- (0,2em);}\,
   is a complex, 
   $\,\tikz[baseline=1ex]{\draw[string=green!50!black] (0,0) -- (0,.75em) node[dot=purple] (P) {}; \draw[string=green!50!black] (P) -- +(0,.75em) node[dot=purple] (P2) {}; \draw[string=green!50!black] (P2) -- +(0,.75em);}\, 
   = 0$, and so if an element is a coboundary, then it is a cocycle.  The converse holds only if the complex is \defn{exact}; most are not.

We unpack the condition of being a cocycle for the complex $\Q$ above for small $n$.  When $n = 0$:
\[ \begin{tikzpicture}
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] {} -- (0,1);
  \node[anchor=west] (text) at (.25,.5) {is a cocycle if and only if it is central.};
\end{tikzpicture}\]
i.e.\ its bracket with anything is $0$.  (For a general module, the $0$-cocycles are the fixed points.)

When $n = 1$, we are interested in elements \,\tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, such that:
\[ \begin{tikzpicture} \node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {};
   \draw[colie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(-.5,1) -- ++(0,1);
   \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.5,1) node[dot=blue] (O) {};
   \draw[colie] (O) .. controls +(150:.25) and +(270:.5) .. ++(-.5,1);
   \draw[lie] (O) .. controls +(30:.25) and +(270:.5) .. ++(.5,1);
   \draw[blue,fill=blue] (O) ++(-.35,.7) rectangle ++(-.75,.1);
   \path (O) ++(.5,0) node[anchor=west] (m) {$\displaystyle - \,\frac12$};
   \path (m.east) ++(1,-1) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T1) {};
   \draw[lie] (T1) .. controls +(0,.5) and +(0,-1) .. ++(+.75,2);
   \draw[colie] (T1) .. controls +(0,.5) and +(0,-.5) .. ++(-.5,1) node[dot=blue] (O1) {};
   \draw[colie] (O1) .. controls +(150:.25) and +(270:.5) .. ++(-.25,1);
   \draw[colie] (O1) .. controls +(30:.25) and +(270:.5) .. ++(.25,1);
   \path (O1) ++(1,0) node[anchor=west] {$=0$};
\end{tikzpicture}\]
This is equivalent to a map \,\begin{tikzpicture}[baseline=1ex]
  \draw[colie] (0,1em) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50, inner sep=1pt, shape border rotate=270] (T) {} .. controls +(-.5em,0) and +(0,.5em) .. (-1em,0);
  \draw[lie]  (T) .. controls +(-.5em,0) and +(0,-.5em) .. (-1em,2em);
\end{tikzpicture}\, such that:
\[\begin{tikzpicture}
  \draw[lie] (0,0) node[dot=blue] (D1) {} -- ++(0,1);
  \draw[colie] (D1) .. controls +(210:.25) and +(90:.5) .. ++(-.5,-1);
  \draw[colie] (D1) .. controls +(330:.25) and +(180:.25) .. ++(1,-.5) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50, inner sep=1pt, shape border rotate=270] (T1) {};
  \draw[colie] (T1) .. controls +(180:.25) and +(90:.25) .. ++(-.5,-.5);

  \draw[lie] (D1) ++(2,0) node[dot=blue] (D2) {} -- ++(0,1);
  \draw[colie] (D2) .. controls +(330:.25) and +(90:.5) .. ++(.5,-1);
  \draw[colie] (D2) .. controls +(210:.25) and +(180:.25) .. ++(0,-.5) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50, inner sep=1pt, shape border rotate=270] (T2) {};
  \draw[colie] (T2) .. controls +(180:.25) and +(90:.25) .. ++(-.5,-.5);
  
  \path (D1) ++(1,0) node {$+$} ++(2,0) node {$=$};
  
  \draw[lie] (D2) ++(2,0) node[dot=blue] (D3) {} .. controls +(90:.25) and +(180:.25) .. ++(.5,.5) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50, inner sep=1pt, shape border rotate=270] (T3) {};
  \draw[lie] (T3) .. controls +(180:.25) and +(270:.25) .. ++(-.5,.5);
  \draw[colie] (D3) .. controls +(210:.25) and +(90:.5) .. ++(-.5,-1);
  \draw[colie] (D3) .. controls +(330:.25) and +(90:.5) .. ++(.5,-1);
\end{tikzpicture}\]
Such a map is a \defn{derivation} of the Lie algebra.  (This notion also makes sense for any module; the element \,\tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[module]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, is a map from \gggg\ to \mmmm, and such a map is a \defn{derivation} if it is a cocycle.)

Indeed, we can think of the bracket as an element \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[dot=blue] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[colie] (T) -- (0,1.5em);
  \draw[blue, fill=blue] (-1.1em,1.2em) rectangle (.1em,1.3em);
}.  Then it is easy to check that the Jacobi identity implies that the bracket is a $2$-cocycle; however, the Jacobi identity is much stronger than this condition.  There don't seem to be natural words to associate with higher cocycle conditions.\footnote{We will see later that nondegenerate $2$-cocycles in the Chevalley complex for the trivial representation (in the notation of the next paragraph: $m=0$, $n=2$) are in bijection with nondegnerate triangular structures.}

Let us now vary the module \mmmm: we let $\mmmm\, = \,\begin{tikzpicture}[x=.5 cm, y=.25 cm,baseline=1ex] 
\draw[lie] (1.8,0) -- (1.8,1); \draw[lie] (1.8,1) -- (1.8,2);  \draw[lie] (2.8,0) -- (2.8,1); \draw[lie] (2.8,1) -- (2.8,2);   \draw[blue, fill=blue] (1.6,.9) rectangle (3,1.1); \path (2.3,.4) node {\tiny \dots} (2.3,1.6) node {\tiny \dots} ;
\path (2.3,-.4) node {\tiny $m$} ;
\end{tikzpicture}\, = \gggg^{\wedge m}$.  When $m = n =0$, the only elements are trivial, and the cocycle condition is trivially satisfied.  When $m=0$, the $1$-cocycles are linear maps on \gggg\ that are zero on the derived subalgebra; i.e.\ they are linear maps on the abelianization of \gggg.  We have explored already the case when $n=0,1$ and $m=1$.

When $m=2$, the $0$-cocycles are strange.  They are antisymmetric elements \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, such that for any element \,\tikz[baseline=.5ex] \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue,fill=blue!50,inner sep=1pt, shape border rotate=180] {} -- (0,1em);\,, the element \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:-.25em) .. (-.5em,.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  \draw[lie] (-1em,0) node[regular polygon, regular polygon sides=4,draw=blue,fill=blue!50,inner sep=1pt, shape border rotate=180] {} .. controls +(0,.25em) and +(210:-.25em) .. (-.5em,.5em);
  \draw[lie] (-.5em,.5em) node[dot=blue] {} -- (-.5em,1em);
}\, is antisymmetric.  We will now explore the cocycle condition when $m=2$ and $n=1$.  We have an element \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
  \draw[blue, fill=blue] (-.1em,1.2em) rectangle (1.1em,1.3em);
} satisfying:
\[2 \begin{tikzpicture}[baseline=(b)]
   \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-1) .. ++(-1,2);
   \draw[lie] (T) -- ++(0,1) node[dot=blue] (b) {};
   \draw[colie] (b) .. controls +(150:.25) and +(270:.5) .. ++(-.5,1);
   \draw[lie] (b) .. controls +(30:.25) and +(270:.5) .. ++(.5,1);
   \draw[lie] (T) .. controls +(0,.5) and +(0,-1) .. ++(1,2);
   \draw[blue, fill=blue] (T) ++(-1.1,1.7) rectangle ++(.7,.1);
   \draw[blue, fill=blue] (T) ++(1.1,1.7) rectangle ++(-.7,.1);   
 \end{tikzpicture}
 - \frac12 \;\;
 \begin{tikzpicture}[baseline=(b)]
   \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-.5) .. ++(-.75,1) node[dot=blue] (b) {};
   \draw[colie] (b) .. controls +(150:.25) and +(270:.5) .. ++(-.25,1);
   \draw[colie] (b) .. controls +(30:.25) and +(270:.5) .. ++(.25,1);
   \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.5,2);
   \draw[lie] (T) .. controls +(0,.5) and +(0,-1) .. ++(1,2);
 \end{tikzpicture}
 = 0
\]
Consider the element \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
} as a map \tikz[baseline=.5ex]{
  \draw[lie] (0,0) -- (0,.75em) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt] (T) {};
  \draw[lie]  (T) .. controls +(150:.25em) and +(0,-.25em) .. (-.5em,1.5em);
  \draw[lie]  (T) .. controls +(30:.25em) and +(0,-.25em) .. (.5em,1.5em);
}.  Then the cocycle condition reads:
\[\begin{tikzpicture}[scale=.5]

\draw[lie] (0,0) .. controls +(90:.5) and +(210:.5) .. (1,1);
\draw[lie] (2,0) .. controls +(90:.5) and +(330:.5) .. (1,1);
\draw[lie] (1,1) -- (1,2);
\draw[lie] (1,2) .. controls +(150:.5) and +(270:.5) .. (0,3);
\draw[lie] (1,2) .. controls +(30:.5) and +(270:.5) .. (2,3);
\draw (1,1) node[dot=blue] {} (1,2) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt] {} ;

\draw[lie] (3,0) .. controls +(90:.5) and +(240:.5) .. (3.5,1.5);
\draw[lie] (3.5,1.5) .. controls +(120:.5) and +(270:.5) .. (3,3);
\draw[lie] (3.5,1.5) -- (4.5,1.5);
\draw[lie] (5,0) .. controls +(90:.5) and +(300:.5) .. (4.5,1.5);
\draw[lie] (4.5,1.5) .. controls +(60:.5) and +(270:.5) .. (5,3);
\draw (3.5,1.5) node[dot=blue] {} (4.5,1.5) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt,shape border rotate=270] {} ;

\draw[lie] (6,0) .. controls +(90:.5) and +(240:.5) .. (6.5,1.5);
\draw[lie] (6.5,1.5) .. controls +(120:.5) and +(270:.5) .. (6,3);
\draw[lie] (7.5,1.5) -- (6.5,1.5);
\draw[lie] (8,0) .. controls +(90:.5) and +(300:.5) .. (7.5,1.5);
\draw[lie] (7.5,1.5) .. controls +(60:.5) and +(270:.5) .. (8,3);
\draw (6.5,1.5) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt,shape border rotate=90] {} (7.5,1.5) node[dot=blue] {} ;

\draw[lie] (9,0) .. controls +(90:.5) and +(240:.5) .. (9.5,1);
\draw[lie] (9.5,1) .. controls +(120:1) and +(270:1) .. (11,3);
\draw[lie] (9.5,1) -- (10.5,1);
\draw[lie] (11,0) .. controls +(90:.5) and +(300:.5) .. (10.5,1);
\draw[lie] (10.5,1) .. controls +(60:1) and +(270:1) .. (9,3);
\draw (9.5,1) node[dot=blue] {} (10.5,1) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt,shape border rotate=270] {} ;

\draw[lie] (12,0) .. controls +(90:.5) and +(240:.5) .. (12.5,1);
\draw[lie] (12.5,1) .. controls +(120:1) and +(270:1) .. (14,3);
\draw[lie] (13.5,1) -- (12.5,1);
\draw[lie] (14,0) .. controls +(90:.5) and +(300:.5) .. (13.5,1);
\draw[lie] (13.5,1) .. controls +(60:1) and +(270:1) .. (12,3);
\draw (12.5,1) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt,shape border rotate=90] {} (13.5,1) node[dot=blue] {} ;

\node at (2.5,1.5) {$=$};
\node at (5.5,1.5) {$+$};
\node at (8.5,1.5) {$-$};
\node at (11.5,1.5) {$-$};

\end{tikzpicture}\]
In particular, if \gggg\ is a Lie bialgebra, then the cobracket \tikz[baseline=.5ex]{
  \draw[lie] (0,0) -- (0,.75em) node[dot=blue] (T) {};
  \draw[lie]  (T) .. controls +(150:.25em) and +(0,-.25em) .. (-.5em,1.5em);
  \draw[lie]  (T) .. controls +(30:.25em) and +(0,-.25em) .. (.5em,1.5em);
} is a $1$-cocycle.  We remark that a $1$-cocycle \tikz[baseline=.5ex]{
  \draw[lie] (0,0) -- (0,.75em) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt] (T) {};
  \draw[lie]  (T) .. controls +(150:.25em) and +(0,-.25em) .. (-.5em,1.5em);
  \draw[lie]  (T) .. controls +(30:.25em) and +(0,-.25em) .. (.5em,1.5em);
} need not define a Lie bialgebra structure on \gggg, because it may not satisfy the coJacobi identity.  We write this identity in terms of elements of our Chevalley complex --- the coJacobi identity asserts that the following element vanishes:
\[ \begin{tikzpicture}
   \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (P) {} .. controls +(0,.25) and +(0,-.5) .. ++(-1,1);
   \draw[lie] (P) -- ++(0,1);
   \draw[lie] (P) .. controls +(0,.25) and +(0,-.25) .. ++(.5,1);
   \draw[lie] (P) .. controls +(0,.25) and +(0,-.5) .. ++(1,1);
   \node at (1.5,.5) {$\defined$};
   
   \draw[colie] (2.5,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T1) {} .. controls +(0,.25) and +(0,-.5) .. ++(-.5,1);
   \draw[lie] (3.5,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T2) {} .. controls +(0,.25) and +(0,-.5) .. ++(.5,1);
   \draw[lie] (T1) .. controls +(0,.5) and +(0,.5) .. (T2);
   \draw[lie] (T2) -- ++(0,1);
   \draw[lie] (T1) .. controls +(0,.5) and +(0,-.5) .. ++(.5,1);
   
   \draw[blue,fill=blue] (T2) +(-.65,.75) rectangle +(.6,.85);
\end{tikzpicture}\]
As we said above, that   \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
}   vanishes does not follow from the fact that  \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
} is a $1$-cocycle.  But the cocycle condition does imply that    \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
}   is a $1$-cocycle --- checking this is a long but straightforward calculation which we do not repeat here, because it is not part of our primary story.  Then it may occur that   \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
}  is a $1$-coboundary.  We say that the Lie algebra \gggg\ along with a $1$-cocycle \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=3,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
} is a \defn{Lie quasibialgebra} if   \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
}   defined above is a $1$-coboundary.  We will not develop the theory of quasibialgebras here; see for example \cite{ES}.

Let \gggg\ be a Lie bialgebra.  Then the cobracket \tikz[baseline=.5ex]{
  \draw[lie] (0,0) -- (0,.75em) node[dot=blue] (T) {};
  \draw[lie]  (T) .. controls +(150:.25em) and +(0,-.25em) .. (-.5em,1.5em);
  \draw[lie]  (T) .. controls +(30:.25em) and +(0,-.25em) .. (.5em,1.5em);
} is a $1$-cocycle, as we have described.  As in the previous paragraph, any time we find a cocycle, we should ask if it is a coboundary.  We say that the Lie bialgebra \gggg\ is \defn{coboundary} if the cobracket is a $1$-coboundary, i.e.\ if there is an element \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, such that $\tikz[baseline=.5ex]{
  \draw[lie] (0,0) -- (0,.75em) node[dot=blue] (T) {};
  \draw[lie]  (T) .. controls +(150:.25em) and +(0,-.25em) .. (-.5em,1.5em);
  \draw[lie]  (T) .. controls +(30:.25em) and +(0,-.25em) .. (.5em,1.5em);
} = 
\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (-.5em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (-1em,0)  .. controls +(0,.25em) and +(210:.25em) .. (-.5em,.75em);
  \draw[lie] (-.5em,.75em) node[dot=blue] {} -- (-.5em,1.5em);
}\, + 
\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (.5em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1.5em);
  \draw[lie] (-1em,0)  .. controls +(0,.25em) and +(210:.25em) .. (.5em,.75em);
  \draw[lie] (.5em,.75em) node[dot=blue] {} -- (.5em,1.5em);
}\,
$, and a choice of such an element is a \defn{coboundary structure} on \gggg, and a \defn{coboundary Lie bialgebra} is a Lie bialgebra equipped with a choice of coboundary structure.  We remark that a coboundary structure for a given Lie bialgebra is determined only up to a choice of $0$-cocycle.  A Lie bialgebra homomorphism of coboundary Lie bialgebras is \defn{coboundary} if it preserves the coboundary structure.

Let \gggg\ be a Lie algebra and let $\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.25em);
}\, = \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.25em);
  \draw[blue,fill=blue] (-.6em,.8em) rectangle (.6em,.9em);
}\,$ be an antisymmetric element; then $\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  \draw[colie] (-.5em,1em) -- (-.5em,1.5em);
  \draw[lie] (.5em,1em) -- (.5em,1.5em);
  \draw[lie] (0em,1em) -- (0em,1.5em);
  \draw[purple,fill=purple!50] (-.75em,.75em) rectangle (.75em,1em);
}\, = 
\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (0em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (-.5em,1.5em)  .. controls +(0,-.25em) and +(210:.25em) .. (0em,.75em);
  \draw[lie] (0em,.75em) node[dot=blue] {} -- (0em,1.5em);
}\, + 
\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (.5em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (0em,1.5em);
  \draw[lie] (-.5em,1.5em)  .. controls +(0,-.25em) and +(210:.25em) .. (.5em,.75em);
  \draw[lie] (.5em,.75em) node[dot=blue] {} -- (.5em,1.5em);
}\,
$ is a $1$-coboundary and hence a $1$-cocycle, where \,\tikz[baseline=1.25ex]  \draw[purple,fill=purple!50] (-.75em,.75em) rectangle (.5em,1em);\, is the differential $\Q$ that we have used throughout these notes.  Then it is not necessarily the case that \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  \draw[colie] (-.5em,1em) -- (-.5em,1.5em);
  \draw[lie] (.5em,1em) -- (.5em,1.5em);
  \draw[lie] (0em,1em) -- (0em,1.5em);
  \draw[purple,fill=purple!50] (-.75em,.75em) rectangle (.75em,1em);
}\, is a Lie bialgebra structure on \gggg: it may not satisfy the coJacobi identity.  However, it is a theorem of Drinfeld that \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  \draw[colie] (-.5em,1em) -- (-.5em,1.5em);
  \draw[lie] (.5em,1em) -- (.5em,1.5em);
  \draw[lie] (0em,1em) -- (0em,1.5em);
  \draw[purple,fill=purple!50] (-.75em,.75em) rectangle (.75em,1em);
}\, satisfies the coJacobi identity if and only if the element
\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (-.5em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (-1em,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {}  .. controls +(0,.25em) and +(210:.25em) .. (-.5em,.75em);
  \draw[lie] (-.5em,.75em) node[dot=blue] {} -- (-.5em,1.5em);
  \draw[lie]  (S) .. controls +(0,.5em) and +(0,-.5em) .. (-1.5em,1.5em);
  \draw[blue,fill=blue] (-1.6em,1.3em) rectangle (.6em,1.4em);
}\,
 is a $0$-cocycle.  The proof is straightforward and we leave it to the reader.\footnote{In \cite{ES}, the proof is described as a ``rather long direct computation using the Jacobi identity''.  In fact, it is completely straightforward in the graphical language we have been using, but like the proof that  \tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=5,draw=blue, fill=blue!50,inner sep=2pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (1em,1.5em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (T) -- (0,1.5em);
} is a cocycle, it is not deeply enlightening.}  We write $\CYB$ for the quadratic map $\gggg^{\wedge 2} \to \gggg^{\wedge 3}$ given by $\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.25em);
}\, \mapsto \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.25em) and +(330:.25em) .. (-.5em,.75em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1.5em);
  \draw[lie] (-1em,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {}  .. controls +(0,.25em) and +(210:.25em) .. (-.5em,.75em);
  \draw[lie] (-.5em,.75em) node[dot=blue] {} -- (-.5em,1.5em);
  \draw[lie]  (S) .. controls +(0,.5em) and +(0,-.5em) .. (-1.5em,1.5em);
  \draw[blue,fill=blue] (-1.6em,1.3em) rectangle (.6em,1.4em);
}\,$.  Then $\CYB$ extends to a map $\CYB: \gggg^{\otimes 2}  \to \gggg^{\otimes 3}$ by
\[ \CYB: \, \begin{tikzpicture}[baseline]
  \draw[lie] (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-.5) .. ++(-.25,.75);
  \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,.75);
\end{tikzpicture} \mapsto
  \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (-.25,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (.25,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.25) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (.75,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
  \end{tikzpicture}
\]

The map $\CYB$ is called the \defn{classical Yang-Baxter map}, and the quadratic equation $\CYB\!\left(\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\right) = 0$ the \defn{classical Yang-Baxter equation}.  Certainly any antisymmetric solution to the classical Yang-Baxter equation gives rise to a Lie bialgebra.  We say that a coboundary Lie bialgebra is \defn{triangular} if the coboundary structure satisfies the classical Yang-Baxter equation; a \defn{triangular structure} on a Lie bialgebra is a coboundary structure that is triangular, and a \defn{triangular structure} on a Lie algebra is an antisymmetric solution to the classical Yang-Baxter equation.  We say that any (possibly non-antisymmetric) solution to the classical Yang-Baxter equation is an \defn{$r$-matrix}; thus a triangular structure is precisely an antisymmetric $r$-matrix.  Following \cite{ES}, we remark that the image of a coboundary structure under a Lie algebra homomorphism is not necessarily a coboundary structure, but that the image of an $r$-matrix necessarily is an $r$-matrix.

Recall that an arbitrary element \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, is \defn{nondegenerate} if the map \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[lie]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\, is invertible; by finite-dimensionality, the adjoint map \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(-.25em,0) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(-.25em,0) and +(0,.5em) .. (-.5em,-.25em);
}\, is then also invertible.  We say that a triangular Lie algebra is \defn{nondegenerate} if the triangular structure is nondegenerate; we have seen that any triangular Lie algebra has a nondegenerate triangular subalgebra.  

We will now describe nondegenerate triangular structures in terms of the cohomology we have developed.  Let \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, be  nondegenerate and antisymmetric: that the map \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[lie]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\, is invertible and  $  \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(-.25em,0) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(-.25em,0) and +(0,.5em) .. (-.5em,-.25em);
}\,   = - \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[lie]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\,$.  Define the element \,\tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[colie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, such that \,\tikz[baseline=.5ex]{
  \draw[colie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[colie]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\, is the inverse of \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[lie]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\,, i.e.\ $\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,1em) and +(0,1em) .. (1em,0em) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {};
  \draw[colie] (S) .. controls +(0,.5em) and +(0,-.5em) .. (1.5em,1em);
}\, = \tikz[baseline=.5ex] \draw[colie] (-.5em,1em) .. controls +(0,-.5em) and +(0,-.5em) .. (1.5em,1em);$; then \,\tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[colie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\,      is also antisymmetric.  Then:
\[    \begin{tikzpicture}[baseline=(X)]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (.25,.25) node[dot=blue] (B)  {} -- (.25,.75);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.25) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. (-.25,.75);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. (.75,.75);
    \draw[blue,fill=blue] (-.35,.7) rectangle (.85,.8);
    \path (-.5,1.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A1) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A2) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A3) {};
    \draw[colie] (A1) .. controls +(.25,0) and +(0,.25) .. ++(.25,-.5);
    \draw[colie] (A2) .. controls +(.25,0) and +(0,.25) .. ++(.25,-.5);
    \draw[colie] (A3) .. controls +(.25,0) and +(0,.25) .. ++(.25,-.5);
    \draw[colie] (A1) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5);
    \draw[colie] (A2) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5);
    \draw[colie] (A3) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5);
    \coordinate (X) at (0,.75);
  \end{tikzpicture}
  \, = \,
  \begin{tikzpicture}[baseline=(X)]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw (.25,.25) node[dot=blue] (B)  {} ;
    \draw[lie] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.25) ..  (B);

    \draw[blue,fill=blue] (-.35,1.2) rectangle (.85,1.3);
    \path (-.5,.75) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A1) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A2) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A3) {};

    \draw[lie] (S) .. controls +(0,.5) and +(.25,0) .. (A1);
    \draw[lie] (B) .. controls +(0,.25) and +(.25,0) .. (A2);
    \draw[lie] (T) .. controls +(0,.5) and +(.5,0) .. (A3);

    \draw[colie] (A1) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5) coordinate (a1);
    \draw[colie] (A2) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5) coordinate (a2);
    \draw[colie] (A3) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5) coordinate (a3);
    
    \draw[colie] (a1) -- ++(0,.5);  \draw[colie] (a2) -- ++(0,.5);  \draw[colie] (a3) -- ++(0,.5);
    \coordinate (X) at (0,.75);
  \end{tikzpicture}
  = - \,
  \begin{tikzpicture}[baseline=(X)]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw (.25,.25) node[dot=blue] (B)  {} ;
    \draw[lie] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.25) ..  (B);

    \draw[blue,fill=blue] (-.35,1.2) rectangle (.85,1.3);
    \path (-.5,.75) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A1) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A2) {} ++(1,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A3) {};

    \draw[lie] (S) .. controls +(0,.5) and +(.25,0) .. (A1);
    \draw[lie] (B) .. controls +(0,.25) and +(.25,0) .. (A2);
    \draw[lie] (T) .. controls +(0,.5) and +(-.25,0) .. (A3);

    \draw[colie] (A1) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5) coordinate (a1);
    \draw[colie] (A2) .. controls +(.25,0) and +(0,-.25) .. ++(.25,.5) coordinate (a2);
    \draw[colie] (A3) .. controls +(-.25,0) and +(0,-.25) .. ++(-.25,.5) coordinate (a3);
    
    \draw[colie] (a1) -- ++(0,.5);  \draw[colie] (a2) -- ++(0,.5);  \draw[colie] (a3) -- ++(0,.5);
    \coordinate (X) at (0,.75);
  \end{tikzpicture}
  \, = -
  \begin{tikzpicture}[baseline=(X)]
    \path (-.25,1.25) coordinate (a1) ++(.5,0) coordinate (a2) ++(.5,0) coordinate (a3) (0,.75) coordinate (X)  (.25,.25) node[dot=blue] (B) {}  (0,.75) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A) {} ;
    \draw[colie] (a1) -- ++(0,.5);  \draw[colie] (a2) -- ++(0,.5);  \draw[colie] (a3) -- ++(0,.5);
    \draw[blue,fill=blue] (-.35,1.2) rectangle (.85,1.3);    
    
    \draw[lie] (B) .. controls +(0,.25) and +(.25,0) .. (A);    
    \draw[lie] (a2) .. controls +(0,-.5) and +(.25,0) .. (A);    
    \draw[lie] (a1) .. controls +(0,-.5) and +(210:.5) ..  (B);
    \draw[lie] (a3) .. controls +(0,-.5) and +(330:.5) ..  (B);
  \end{tikzpicture}
  \, = - \!\!\!\!\!
  \begin{tikzpicture}[baseline=(X)]
    \path (-.25,1.25) coordinate (a1) ++(.5,0) coordinate (a2) ++(.5,0) coordinate (a3) (0,.75) coordinate (X)  (-.25,.25) node[dot=blue] (B) {}  (-.5,.75) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A) {} ;
    \draw[colie] (a1) -- ++(0,.5);  \draw[colie] (a2) -- ++(0,.5);  \draw[colie] (a3) -- ++(0,.5);
    \draw[blue,fill=blue] (-.35,1.2) rectangle (.85,1.3);    
    
    \draw[lie] (B) .. controls +(0,.25) and +(.25,0) .. (A);    
    \draw[lie] (a1) .. controls +(0,-.5) and +(.25,0) .. (A);    
    \draw[lie] (a3) .. controls +(0,-2) and +(210:1) ..  (B);
    \draw[lie] (a2) .. controls +(0,-.5) and +(330:.5) ..  (B);
  \end{tikzpicture}
  = - 
  \begin{tikzpicture}[baseline=(X)]
    \path (-.25,1.25) coordinate (a1) ++(.5,0) coordinate (a2) ++(.5,0) coordinate (a3) (0,.75) coordinate (X) ;
    \draw[colie] (a1) -- ++(0,.5);  \draw[colie] (a2) -- ++(0,.5);  \draw[colie] (a3) -- ++(0,.5);
    \draw[blue,fill=blue] (-.35,1.2) rectangle (.85,1.3);    
    \path (.25,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (A) {} (.5,.5) node[dot=blue] (B) {};
    \draw[lie] (a1) .. controls +(0,-.5) and +(0,.5) .. (A);
    \draw[lie] (a2) .. controls +(0,-.5) and +(150:.25) .. (B);
    \draw[lie] (a3) .. controls +(0,-.5) and +(30:.25) .. (B);
    \draw[colie] (A) .. controls +(0,.25) and +(0,-.25) .. (B);
  \end{tikzpicture}
\]
Thus \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, is a nondegenerate triangular structure on a Lie algebra \gggg\ if and only if its inverse \,\tikz[baseline=.5ex]{
  \draw[colie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=blue!50,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[colie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, is a $2$-cocycle.\footnote{This and the previous assertions are given geometric meaning and geometric proofs in \cite{ES}.}  Then the (possible degenerate) triangular structures on a Lie algebra are in bijection with the pairs consisting of a Lie subalgebra and a nondegenerate $2$-cocycle (with coefficients in the trivial representation) on it.

\section{Quasitriangularity}

Let \gggg\ be a Lie algebra.  Recall that a possibly non-antisymmetric element \tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
} is an $r$-matrix if it a solution to the classical Yang-Baxter equation:
\[ \CYB\!\left( \begin{tikzpicture}[baseline]
  \draw[lie] (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-.5) .. ++(-.25,.75);
  \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,.75);
\end{tikzpicture}\right) \defined
  \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (-.25,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (.25,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.25) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[lie] (.75,.25) node[dot=blue] (B)  {} -- ++(0,.5);
    \draw[lie] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[lie] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[lie] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[lie] (T) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
  \end{tikzpicture}
  = 0
\]
As is clear from the diagrams, if \tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
} is an $r$-matrix, then so is its reflection \tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
}.
The function $\CYB$ is homogeneous quadratic, so any scalar multiple of an $r$-matrix is an $r$-matrix, but the sum of two $r$-matrices is not necessarily an $r$-matrix.  In particular,
  $ \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
    \draw[blue,fill=blue] (-.6em,.8em) rectangle (.6em,.9em);
  } = \frac12 \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } - \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$
is not necessarily an $r$-matrix, and so does not make \gggg\ into a triangular Lie bialgebra.  

Recall that $\tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
    \draw[blue,fill=blue] (-.6em,.8em) rectangle (.6em,.9em);
  }$ makes \gggg\ into a coboundary algebra if and only if $\CYB\!\left(\tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
    \draw[blue,fill=blue] (-.6em,.8em) rectangle (.6em,.9em);
  }\right)$ is invariant (a $0$-cocycle).  Since $\CYB\!\left( \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right) = 0 = \CYB\! \left( \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$, it's clear that $\CYB\!\left(\frac12\! \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } - \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)\right) = -\CYB\!\left(\frac12\! \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } + \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)\right)$, and by the Jacobi identity this is definitely invariant provided that the symmetric element $\frac12 \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } + \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$ is invariant.  (There is nothing special about being symmetric: if \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=red,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } is invariant, then so is $\CYB\!\left(\tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=red,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$.)  This is a sufficiently natural condition that we give it a name: a \defn{quasitriangular structure} on a Lie algebra \gggg\ is an $r$-matrix \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } such that $\frac12 \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } + \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$ is invariant under the action by \gggg.  In particular, if \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } is a quasitriangular structure, then $ \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
    \draw[blue,fill=blue] (-.6em,.8em) rectangle (.6em,.9em);
  } = \frac12 \left(
  \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
  } - \tikz[baseline=.5ex]{
    \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.5em,.5em) and +(0,-.5em) .. (-.5em,1em);
    \draw[lie]  (T) .. controls +(-.5em,.5em) and +(0,-.5em) .. (.5em,1em);
  }\right)$ makes \gggg\ into a coboundary Lie bialgebra; a coboundary Lie bialgebra with a choice of quasitriangular $r$-matrix that gives rise to the cobracket is a \defn{quasitriangular Lie bialgebra}.  Although the image of a quasitriangular structure under a Lie algebra homomorphism is an $r$-matrix, it is not necessarily quasitriangular --- its symmetrization need not be invariant --- and does not necessarily define a coboundary structure.

Triangular Lie bialgebras are few and far between, but quasitriangular Lie bialgebras are of great importance.  We will now describe a quasitriangular structure for the double of any Lie bialgebra \gggg.  Recall that as a vector space, the double is $\DDDD = \gggg \oplus \gggd$, with bracket and cobracket given by:
\[\begin{tikzpicture}[scale=.5]
\node[circle,draw=blue,inner sep=2pt] (O) at (0,0) {};
\draw[drinf] (O) -- +(0,1);
\draw[drinf] (O) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[drinf] (O) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (1,0) {$\defined$};
\node[dot=blue] (O1) at (2,0) {};
\draw[lie] (O1) -- +(0,1);
\draw[colie] (O1) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O1) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (3,0) {$+$};
\node[dot=blue] (O2) at (4,0) {};
\draw[lie] (O2) -- +(0,1);
\draw[lie] (O2) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O2) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (5,0) {$+$};
\node[dot=blue] (O3) at (6,0) {};
\draw[lie] (O3) -- +(0,1);
\draw[colie] (O3) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O3) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (7,0) {$+$};
\node[dot=blue] (O4) at (8,0) {};
\draw[colie] (O4) -- +(0,1);
\draw[lie] (O4) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[colie] (O4) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (9,0) {$+$};
\node[dot=blue] (O5) at (10,0) {};
\draw[colie] (O5) -- +(0,1);
\draw[colie] (O5) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O5) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);
\node at (11,0) {$+$};
\node[dot=blue] (O6) at (12,0) {};
\draw[colie] (O6) -- +(0,1);
\draw[lie] (O6) .. controls +(210:.5) and +(90:.5) .. +(-.5,-1);
\draw[lie] (O6) .. controls +(330:.5) and +(90:.5) .. +(.5,-1);

\node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (T) at (16,0) {};
\draw[drinf] (T) -- +(0,-1);
\draw[drinf] (T) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[drinf] (T) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\node at (17,0) {$\defined$};
\node[dot=blue] (T1) at (18,0) {};
\draw[colie] (T1) -- +(0,-1);
\draw[lie] (T1) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[lie] (T1) .. controls +(30:.5) and +(270:.5) .. +(.5,1);
\node at (19,0) {$-$};
\node[dot=blue] (T2) at (20,0) {};
\draw[lie] (T2) -- +(0,-1);
\draw[colie] (T2) .. controls +(150:.5) and +(270:.5) .. +(-.5,1);
\draw[colie] (T2) .. controls +(30:.5) and +(270:.5) .. +(.5,1);

\end{tikzpicture}\]
Indeed, let $\tikz[baseline=.5ex]{
  \draw[drinf] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[drinf]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
} \defined \tikz[baseline=.5ex] \draw[lie] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);$.  Then its symmetrization $\frac12 \left(\tikz[baseline=.5ex] \draw[colie] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em); + \tikz[baseline=.5ex] \draw[lie] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\right) = \frac12 \tikz[baseline=.5ex] \draw[drinf] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);$ is the inverse of the canonical invariant pairing $\tikz[baseline=.5ex] \draw[drinf] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); \defined \tikz[baseline=.5ex] \draw[lie] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[colie] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em);$, and so is invariant.  We check the classical Yang-Baxter equation:
\begin{align*}
  \CYB\!\left( \begin{tikzpicture}[baseline]
  \draw[drinf] (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-.5) .. ++(-.25,.75);
  \draw[drinf] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,.75);
\end{tikzpicture}\right) & = 
  \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw (-.25,.25) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
    \draw[drinf] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[drinf] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[drinf] (S) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
    \draw[drinf] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[drinf] (.25,.25) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
    \draw[drinf] (S) .. controls +(90:.25) and +(210:.25) ..  (B);
    \draw[drinf] (T) .. controls +(90:.25) and +(330:.25) ..  (B);
    \draw[drinf] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[drinf] (T) .. controls +(0,.5) and +(0,-.5) .. ++(.25,1);
  \end{tikzpicture}
  +
   \begin{tikzpicture}[baseline]
    \draw (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (S) {}  (.5,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ;
    \draw[drinf] (.75,.25) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
    \draw[drinf] (S) .. controls +(90:.25) and +(210:.5) ..  (B);
    \draw[drinf] (T) .. controls +(90:.25) and +(330:.5) ..  (B);
    \draw[drinf] (S) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
    \draw[drinf] (T) .. controls +(0,.5) and +(0,-.5) .. ++(-.25,1);
  \end{tikzpicture} \\
  & = \!\!\!\! \begin{tikzpicture}[baseline]
        \draw (-.25,0) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
        \draw[colie] (.25,.5) .. controls +(0,-1) and +(210:1) .. (B);
        \draw[colie] (.75,.5) .. controls +(0,-.5) and +(330:.5) .. (B);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline]
        \draw (.25,0) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
        \draw[lie] (-.25,.5) .. controls +(0,-.5) and +(210:.5) .. (B);
        \draw[colie] (.75,.5) .. controls +(0,-.5) and +(330:.5) .. (B);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline]
        \draw (.75,0) node[circle,draw=blue,inner sep=2pt] (B)  {}; \draw[drinf] (B) -- ++(0,.5);
        \draw[lie] (-.25,.5) .. controls +(0,-.5) and +(210:.5) .. (B);
        \draw[lie] (.25,.5) .. controls +(0,-1) and +(330:1) .. (B);
  \end{tikzpicture} \\
  & = \!\!\!\! \begin{tikzpicture}[baseline]
        \draw (-.25,0) node[dot=blue] (B)  {}; \draw[colie] (B) -- ++(0,.5);
        \draw[colie] (.25,.5) .. controls +(0,-1) and +(210:1) .. (B);
        \draw[colie] (.75,.5) .. controls +(0,-.5) and +(330:.5) .. (B);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline]
        \draw (.25,0) node[dot=blue] (B)  {}; \draw[lie] (B) -- ++(0,.5);
        \draw[lie] (-.25,.5) .. controls +(0,-.5) and +(210:.5) .. (B);
        \draw[colie] (.75,.5) .. controls +(0,-.5) and +(330:.5) .. (B);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline]
        \draw (.25,0) node[dot=blue] (B)  {}; \draw[colie] (B) -- ++(0,.5);
        \draw[lie] (-.25,.5) .. controls +(0,-.5) and +(210:.5) .. (B);
        \draw[colie] (.75,.5) .. controls +(0,-.5) and +(330:.5) .. (B);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline]
        \draw (.75,0) node[dot=blue] (B)  {}; \draw[lie] (B) -- ++(0,.5);
        \draw[lie] (-.25,.5) .. controls +(0,-.5) and +(210:.5) .. (B);
        \draw[lie] (.25,.5) .. controls +(0,-1) and +(330:1) .. (B);
  \end{tikzpicture} \!\!\!\! \\
  & = 0 + 0 = 0
\end{align*}
by antisymmetry.  Moreover:
\begin{align*}
  \begin{tikzpicture}[baseline=(B)]
    \draw[drinf] (0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5) and +(0,-.5) .. (-.25,.25);
    \draw[drinf] (T) .. controls +(0,.5) and +(0,-.5) .. (.25,.25);
    \draw[purple,fill=purple!50] (-1,.25) rectangle node (Q) {$Q$} (.5,.75);
    \draw[drinf] (-.25,.75) -- ++(0,.25);
    \draw[drinf] (.25,.75) -- ++(0,.25);
    \draw[drinf] (-.75,.75) -- ++(0,.25);
    \coordinate (B) at (0,.5);
  \end{tikzpicture} & = 
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[circle,draw=blue,inner sep=2pt] (B) {};
    \draw[drinf] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[drinf] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(.75,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[circle,draw=blue,inner sep=2pt] (B) {};
    \draw[drinf] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.75,.5);
    \draw[drinf] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.25,.5);
  \end{tikzpicture} \\
  & = 
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[colie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[colie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(.75,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[colie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(.75,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[colie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(.75,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.75,.5);
    \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.25,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.75,.5);
    \draw[colie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.25,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[colie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.75,.5);
    \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.25,.5);
  \end{tikzpicture}
  \\ & = 
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[colie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(.75,.5);
  \end{tikzpicture}
  +
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.75,.5);
    \draw[colie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.25,.5);
  \end{tikzpicture}
  + 0 + 0
  \\ & =   
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[colie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.75,.5);
  \end{tikzpicture}
  -
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[dot=blue] (B) {};
    \draw[colie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[colie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[lie] (B) .. controls +(270:.5) and +(270:1) .. ++(-.75,.5);
  \end{tikzpicture}
  \\ & =
  \begin{tikzpicture}[baseline=(B)]
    \draw (.75,0) node[regular polygon, regular polygon sides=4,draw=blue,inner sep=2pt] (B) {};
    \draw[drinf] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
    \draw[drinf] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
    \draw[drinf] (B) .. controls +(270:.5) and +(270:1) .. ++(-.75,.5);
  \end{tikzpicture}
\end{align*}
Therefore \tikz[baseline=.5ex] \draw[lie] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em); is a quasitriangular structure for \DDDD.  We will now prove that in some sense the double is a universal construction of quasitriangular Lie bialgebras.

Let \gggg\ be a Lie algebra and let \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, be an $r$-matrix, and consider the map \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(-.25em,0) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(-.25em,0) and +(0,.5em) .. (-.5em,-.25em);
}\,.  It is immediately clear that the image \obj[string=purple] of this map is a Lie subalgebra of \gggg.  Indeed, it follows directly from the classical Yang-Baxter equation that:
\[ \begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T1) {} +(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T2) {} ++(0,.5) node[dot=blue] (B) {} ++(0,-.25) coordinate (X);
  \draw[string=purple] (T1) .. controls +(180:.25) and +(210:.5) .. (B);
  \draw[string=purple] (T2) .. controls +(180:.25) and +(330:.5) .. (B);
  \draw[lie] (T1) .. controls +(180:.25) and +(90:.5) .. ++(-.25,-.5);
  \draw[lie] (T2) .. controls +(180:.25) and +(90:.5) .. ++(-.25,-.5);
  \draw[lie] (B) -- ++(0,.5);
\end{tikzpicture}
=
\begin{tikzpicture}[baseline=(B.base)]
  \path (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T1) {} ++(-.5,.5) node[dot=blue] (B) {} ++(.25,.5) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T2) {};
  \draw[lie] (T1) .. controls +(180:.25) and +(330:.5) .. (B);
  \draw[lie] (T2) .. controls +(180:.25) and +(90:.5) .. (B);
  \draw[lie] (B) .. controls +(210:.25) and +(90:.5) .. ++(-.25,-1);
  \draw[lie] (T1) .. controls +(180:.25) and +(90:.5) .. ++(-.25,-.5);
  \draw[string=purple] (T2) .. controls +(180:.25) and +(270:.5) .. ++(-.25,.5);
\end{tikzpicture}
-
\begin{tikzpicture}[baseline=(B.base)]
  \path (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T1) {} ++(.5,.5) node[dot=blue] (B) {} ++(.25,.5) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T2) {};
  \draw[lie] (T1) .. controls +(0:.25) and +(210:.5) .. (B);
  \draw[lie] (T2) .. controls +(180:.25) and +(90:.5) .. (B);
  \draw[lie] (B) .. controls +(330:.25) and +(90:.5) .. ++(.25,-1);
  \draw[lie] (T1) .. controls +(0:.25) and +(90:.5) .. ++(.25,-.5);
  \draw[string=purple] (T2) .. controls +(180:.25) and +(270:.5) .. ++(-.25,.5);
\end{tikzpicture}
\]
If furthermore \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, is a quasitriangular structure on \gggg, then the classical Yang-Baxter equation and the definition of the cobracket assure that:
\[ \begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180]  (T) {} ++(-.25,.5) node[dot=blue] (B) {} ++(0,-.25) coordinate (X);
  \draw[string=purple] (T) .. controls +(180:.25) and +(270:.5) .. (B);
  \draw[lie] (T) .. controls +(180:.25) and +(90:.5) .. ++(-.25,-.5);
  \draw[lie] (B) .. controls +(150:.25) and +(270:.25) .. ++(-.25,.5);
  \draw[lie] (B) .. controls +(30:.25) and +(270:.25) .. ++(.25,.5);
\end{tikzpicture} =
 \begin{tikzpicture}[baseline=(X.base)]
   \path (0,0) node[dot=blue] (B) {} ++(0,.5) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180]  (T1) {} ++(.5,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180]  (T2) {} ++(0,-.25) coordinate (X);
   \draw[lie] (B) -- ++(0,-.5);
   \draw[lie] (T1) .. controls +(180:.25) and +(150:.25) .. (B);
   \draw[lie] (T2) .. controls +(180:.25) and +(30:.25) .. (B);
   \draw[string=purple] (T1) ..  controls +(180:.25) and +(270:.5) .. ++(-.25,.5);
   \draw[string=purple] (T2) ..  controls +(180:.25) and +(270:.5) .. ++(-.25,.5);
 \end{tikzpicture} 
 \]
This proves that \obj[string=purple] is a Lie subbialgebra of \gggg, and that   \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[string=purple]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\, is a Lie bialgebra homomorphism (in fact, it is an embedding, as can be seen by counting dimensions).

Let \obj[draw=purple,double,double distance=2pt] be the Drinfeld double of \obj[string=purple].  Then the embeddings \,\tikz[baseline=.5ex]{ \draw[string=purple] (0,-.25em) -- (0,.5em); \draw[lie] (0,.5em) -- (0,1.25em); }\, and \,\tikz[baseline=.5ex]{
  \draw[lie] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(.25em,0) and +(0,-.5em) .. (.5em,1.25em);
  \draw[string=purple]  (T) .. controls +(.25em,0) and +(0,.5em) .. (.5em,-.25em);
}\, sum to a map \,\tikz[baseline=.5ex]{ \draw[draw=purple,double,double distance=2pt] (0,-.25em) -- (0,.5em); \draw[lie] (0,.5em) -- (0,1.25em); }\,.  For it to be a Lie algebra homomorphism, we need only to check that
\[ 
\begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) coordinate (X) +(0,.25) coordinate (n) ++(0,-.25) node[circle,draw=purple,inner sep=2pt] (B) {} +(-.25,-.5) coordinate (a);
  \draw[purple,double,double distance=2pt] (B) -- (n);
  \draw[lie] (n) -- ++(0,.5);
  \draw[string=purple] (B) .. controls +(330:.25) and +(90:.25) .. ++(.25,-.5);
  \draw[string=purple] (a) .. controls +(90:.25) and +(210:.25) .. (B);
\end{tikzpicture}
\defined
\begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) coordinate (X) +(0,.25) coordinate (n) ++(0,-.25) node[dot=purple] (B) {} +(-.25,-.5) coordinate (a);
  \draw[string=purple] (B) -- (n);
  \draw[lie] (n) -- ++(0,.5);
  \draw[string=purple] (B) .. controls +(330:.25) and +(90:.25) .. ++(.25,-.5);
  \draw[string=purple] (a) .. controls +(90:.25) and +(210:.25) .. (B);
\end{tikzpicture}
+
\begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) coordinate (X) +(-.25,.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ++(0,-.25) node[dot=purple] (B) {} +(-.25,-.5) coordinate (a);
  \draw[string=purple] (T) .. controls +(.25,0) and +(0,.5) .. (B);
  \draw[lie] (T) .. controls +(.25,0) and +(0,-.5) .. ++(.25,.5);
  \draw[string=purple] (B) .. controls +(330:.25) and +(90:.25) .. ++(.25,-.5);
  \draw[string=purple] (a) .. controls +(90:.25) and +(210:.25) .. (B);
\end{tikzpicture}
 \overset{?}= 
\begin{tikzpicture}[baseline=(X.base)]
  \path (0,0) coordinate (X) +(0,-.25) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} ++(0,.25)  node[dot=blue] (B) {} ++(-.25,-.5) coordinate (a) ++(0,-.5) coordinate (a2);
  \draw[lie] (B) -- ++(0,.5);
  \draw[string=purple] (a2) -- (a);
  \draw[lie] (a) .. controls +(90:.25) and +(210:.25) .. (B);
  \draw[lie] (T) .. controls +(.25,0) and +(330:.25) .. (B);
  \draw[string=purple] (T) .. controls +(.25,0) and +(0,.5) .. ++(.25,-.5);
\end{tikzpicture}
 \]
If we append \,\tikz[baseline=.5ex]{
  \draw[string=purple] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(-.25em,0) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(-.25em,0) and +(0,.5em) .. (-.5em,-.25em);
}\, to the lower-left string, we get precisely the classical Yang-Baxter equation.  But since \,\tikz[baseline=.5ex]{
  \draw[string=purple] (0,.5em) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(-.25em,0) and +(0,-.5em) .. (-.5em,1.25em);
  \draw[lie]  (T) .. controls +(-.25em,0) and +(0,.5em) .. (-.5em,-.25em);
}\, is onto, this suffices to verify the equation.  We have shown that \,\tikz[baseline=.5ex]{ \draw[draw=purple,double,double distance=2pt] (0,-.25em) -- (0,.5em); \draw[lie] (0,.5em) -- (0,1.25em); }\, is a Lie algebra homomorphism.  Moreover, it is clear that the quasitriangular structure \,\tikz[baseline=.5ex] \draw[string=purple] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,  on \obj[draw=purple,double,double distance=2pt]  maps to \,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, under this homomorphism.  Thus we have shown that inside every quasitriangular Lie bialgebra there is a subalgebra whose double maps to the algebra whose double maps to the Lie bialgebra, such that this map is a bijection on the subalgebra and such that the quasitriangular structure is the image of the canonical one.  
% Conversely, we get nothing new upon iterating this construction.  The subalgebra of the double of a Lie bialgebra \gggg\ that corresponds to the subalgebra \obj[string=purple] in the previous construction is precisely the algebra \gggg\ itself.  Moreover, t

\section{Standard structure}

So far our entire story has applied to arbitrary Lie (bi)algebras.  We will conclude with a description of the standard bialgebra structure on a simple Lie algebra; we will see that a simple Lie algebra is almost the double of its Borel subalgebra.

Let \gggg\ be a semisimple Lie algebra.  Then the Killing form \,\tikz[baseline=.5ex]{ \draw[lie] (-.5em,0em) .. controls +(0,1em) and +(-.25em,-.25em) .. (0em,1em); \draw[lie] (.5em,0em) .. controls +(0,1em) and +(.25em,-.25em) .. (0em,1em) node[dot=blue] {}; }\, is a nondegenerate pairing.  We choose a Cartan subalgebra and system of positive roots.  Then the Killing form pairs the Cartan with itself, so we will draw it without a direction: \obj[orange].  We now choose a system of positive roots, inducing upper- and lower-triangular subalgebras, which pair with eachother under the Killing form.  We write \obj[string=red] for the upper-triangular subalgebra, and identity the lower-triangular subalgebra with its dual \obj[costring=red].  Then we have $\gggg = \obj[string=red] \oplus \obj[orange] \oplus \obj[costring=red]$, and $\,\tikz[baseline=.5ex]{ \draw[lie] (-.5em,0em) .. controls +(0,1em) and +(-.25em,-.25em) .. (0em,1em); \draw[lie] (.5em,0em) .. controls +(0,1em) and +(.25em,-.25em) .. (0em,1em) node[dot=blue] {}; }\, = \tikz[baseline=.5ex] \draw[string=red] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[orange] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[costring=red] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em);\,$.  Let us compare this to the Double construction: $\obj[purple,double,double distance=2pt] = \obj[string=purple] \oplus \obj[costring=purple]$ with pairing $\,\tikz[baseline=.5ex] \draw[purple,double,double distance=2pt] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); = \tikz[baseline=.5ex] \draw[string=purple] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[costring=purple] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em);\,$ and quasitriangular $r$-matrix \,\tikz[baseline=.5ex] \draw[string=purple] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,.  This suggests that for the semisimple Lie algebra \gggg\ we consider the element $\,\tikz[baseline=.5ex]{
  \draw[lie] (0,0) node[regular polygon, regular polygon sides=4,draw=blue, fill=green,inner sep=1pt, shape border rotate=180] (T) {} .. controls +(0,.5em) and +(0,-.5em) .. (-.5em,1em);
  \draw[lie]  (T) .. controls +(0,.5em) and +(0,-.5em) .. (.5em,1em);
}\, = \,\tikz[baseline=.5ex] \draw[string=red] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\, + \frac12 \,\tikz[baseline=.5ex] \draw[orange] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,$.  Then it is straightforward to check that this does indeed define an $r$-matrix.  It is quasitriangular trivially: its symmetrization is the \defn{Casimir} \,\tikz[baseline=.5ex]{ \draw[colie] (-.5em,1em) .. controls +(0,-1em) and +(-.25em,0em) .. (0em,0em); \draw[colie] (.5em,1em) .. controls +(0,-1em) and +(.25em,0em) .. (0em,0em) node[dot=blue] {}; }\,.  The $r$-matrix $\,\tikz[baseline=.5ex] \draw[string=red] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\, + \frac12 \,\tikz[baseline=.5ex] \draw[orange] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,$ defines on a semisimple Lie algebra its \defn{standard structure} as a quasitriangular Lie bialgebra.  The standard structure depends on a choice of Cartan and system of positive roots; of course, any two such choices are related by an automorphism of \gggg.

We conclude by describing the standard construction of the standard structure.  Consider the Borel Lie algebra $\obj[string=purple] \defined \obj[string=red] \oplus \obj[orange]$, a subalgebra of \gggg.  The Killing form on \gggg\ pairs \obj[string=purple] invariantly with $\obj[costring=purple] \defined \obj[costring=red] \oplus \obj[orange]$, and it's not too hard to see that the bracket on \obj[costring=purple] pulls back to a bialgebra structure on \obj[string=purple] (namely, it definitely pulls back to a cobracket, and one can check the cocycle identity by hand).  Thus we may construct the double \obj[purple,double,double distance=2pt] of \obj[string=purple].  We add arrows to the Cartan parts to tell them apart: we have $\obj[purple,double,double distance=2pt] = \obj[string=red] \oplus \obj[string=orange] \oplus \obj[costring=orange] \oplus \obj[costring=red]$, and the nondegenerate invariant pairing is $\,\tikz[baseline=.5ex] \draw[string=red] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[costring=red] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[string=orange] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em); + \tikz[baseline=.5ex] \draw[costring=orange] (-.5em,0em) .. controls +(0,1em) and +(0,1em) .. (.5em,0em);\,$.  The reader is invited to check that the map $\,\tikz[baseline=.5ex]{ \draw[draw=purple,double,double distance=2pt] (0,-.25em) -- (0,.5em); \draw[lie] (0,.5em) -- (0,1.25em); }\, = 
\,\tikz[baseline=.5ex]{ \draw[string=red] (0,-.25em) -- (0,.5em); \draw[string=red] (0,.5em) -- (0,1.25em); }\, + 
\,\tikz[baseline=.5ex]{ \draw[costring=red] (0,-.25em) -- (0,.5em); \draw[costring=red] (0,.5em) -- (0,1.25em); }\, + \frac12\left(
\,\tikz[baseline=.5ex]{ \draw[string=orange] (0,-.25em) -- (0,.5em); \draw[orange] (0,.5em) -- (0,1.25em); }\, + \,\tikz[baseline=.5ex]{ \draw[costring=orange] (0,-.25em) -- (0,.5em); \draw[orange] (0,.5em) -- (0,1.25em); }\,\right)
$ that averages the two Cartan components is a Lie algebra homomorphism. 
Indeed, the kernel of this map is the image of the Cartan \obj[orange] under the difference map $\,\tikz[baseline=.5ex]{ \draw[orange] (0,-.25em) -- (0,.5em); \draw[string=orange] (0,.5em) -- (0,1.25em); }\, - \,\tikz[baseline=.5ex]{ \draw[orange] (0,-.25em) -- (0,.5em); \draw[string=orange] (0,.5em) -- (0,1.25em); }\,$, and this image is central in \obj[purple,double,double distance=2pt].  The projection \,\tikz[baseline=.5ex]{ \draw[draw=purple,double,double distance=2pt] (0,-.25em) -- (0,.5em); \draw[lie] (0,.5em) -- (0,1.25em); }\, 
 sends the quasitriangular structure $\,\tikz[baseline=.5ex] \draw[string=red] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\, +  \,\tikz[baseline=.5ex] \draw[string=orange] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,$ on \obj[purple,double,double distance=2pt] to the quasitriangular structure $\,\tikz[baseline=.5ex] \draw[string=red] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\, + \frac12 \,\tikz[baseline=.5ex] \draw[orange] (-.5em,1em) .. controls +(0,-1em) and +(0,-1em) .. (.5em,1em);\,$ defined previously.\footnote{In \cite{ES} the authors present a slightly modified definition: rather than averaging the Cartans, they sum them, requiring extra $2$s in their presentation.}

Finally, a straightforward calculation shows that the cobracket when restricted to the Cartan \obj[orange] is trivial --- this is a strong reminder that a choice is involved in constructing the bialgebra structure, since any given element of a semisimple Lie algebra can be chosen to lie in a Cartan.  The simplest semisimple Lie algebra is $\gggg = \mathfrak{sl}_2$, for which the spaces \obj[string=red], \obj[orange], and \obj[costring=red] are all one-dimensional.  It is then a triviality to see that the cobracket defines on \gggd\ the structure of the three-dimensional Heisenberg algebra.  For a general semisimple Lie algebra one can compute the cobracket easily provided one has computed the coefficients of the bracket in terms of the root basis.  Unfortunately, the graphical notation for linear algebra we have been promoting in this paper is not well suited for computations within a basis --- this is entirely the point of the notation.


\begin{thebibliography}{99}

\bibitem{BV} K. Bhaskara and K. Viswanath.  {\em Poisson algebras and Poisson manifolds}.  Longman Scientific and Technical, Essex, UK, 1988.

\bibitem{Cvitanovic} P. Cvitanovi\'c.  {\em Group Theory: Birdtracks, Lie�s, and Exceptional Groups}.  Version 8.9.  Princeton University Press, Princeton, NJ, 2008, available at \url{http://www.nbi.dk/GroupTheory/}.

\bibitem{ES} P. Etingof and O. Schiffmann.  {\em Lectures on quantum groups}. Second edition. Lectures in Mathematical Physics. International Press, Somerville, MA, 2002.

\bibitem{LGbook} M. Haiman.  {\em Lie Groups}, graduate seminar, UC Berkeley, Fall 2008.  Edited lecture notes by T. Johnson-Freyd available at \url{http://math.berkeley.edu/~theojf/LieGroupsBook.pdf}.

\bibitem{Penrose} R. Penrose.  Applications of negative dimensional tensors. In D.J.A.Welsh, editor, \emph{Combinatorial Mathematics and its Applications}, pages 221--244. Academic Press, New York, 1971.

\bibitem{QG} N. Reshetikhin.  {\em Quantum Groups}, graduate seminar, UC Berkeley, Spring 2009.  Lecture notes by T. Johnson-Freyd available at \url{http://math.berkeley.edu/~theojf/QuantumGroups.pdf}.

\bibitem{Selinger} P. Selinger. \emph{A survey of graphical languages for monoidal categories}, preprint (2008), available at \url{http://www.mscs.dal.ca/~selinger/papers/graphical.pdf}.

\bibitem{Stedman} G. Stedman.  {\em Diagram techniques in group theory}.   Cambridge University Press, Cambridge, UK, 1990.

\bibitem{TikZ} T.\ Tantau. \emph{\TikZ\  \& PGF: Manual for Version 2.00}, 2007, available at \url{http://www.ctan.org/tex-archive/graphics/pgf/base/doc/generic/pgf/pgfmanual.pdf}.


\end{thebibliography}


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