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\begin{document}

\begin{center}
\textbf{Counterexample for MathOverflow \#84345}

\textit{Darij Grinberg}

version 19 May 2012

\bigskip
\end{center}

This is not a paper, but just a computation that I am writing up in order to
ensure there are no errors. (As if this could be ensured this way.)

The purpose of the computation is to negatively answer the Concrete Question
in [3].

\begin{center}
\textbf{\S 1. The largest involutive quotient of a Hopf algebra}
\end{center}

First a very easy positive result.

\begin{quote}
\textbf{Definition 1.} Let $H$ be a Hopf algebra over a field $k$. We say that
the Hopf algebra $H$ is \textit{involutive} if $S^{2}=\operatorname*{id}$,
where $S$ denotes the antipode of $H$.

\textbf{Theorem 2.} Let $H$ be a Hopf algebra over a field $k$. Let
$\operatorname*{Inv}H$ denote the $k$-submodule $H\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H$ of $H$
(where $S$ denotes the antipode of $H$). Then:

\textbf{(a)} The $k$-submodule $\operatorname*{Inv}H$ is a Hopf ideal of $H$.

\textbf{(b)} The quotient Hopf algebra $H\diagup\left(  \operatorname*{Inv}%
H\right)  $ is an involutive Hopf algebra.

\textbf{(c)} Whenever $G$ is an involutive Hopf algebra and $f:H\rightarrow G$
is a Hopf algebra homomorphism, we have $f\left(  \operatorname*{Inv}H\right)
=0$, so that the homomorphism $f$ factors through $H\diagup\left(
\operatorname*{Inv}H\right)  $.

\textbf{(d)} If $H$ is a graded Hopf algebra, then $\operatorname*{Inv}H$ is a
homogeneous ideal, so that $H\diagup\left(  \operatorname*{Inv}H\right)  $ is
a graded Hopf algebra canonically.

\textbf{(e)} If $H$ is a connected graded Hopf algebra, then $H\diagup\left(
\operatorname*{Inv}H\right)  $ is a connected graded Hopf algebra.
\end{quote}

Theorem 2 shows that we can consider $H\diagup\left(  \operatorname*{Inv}%
H\right)  $ as the largest involutive quotient Hopf algebra of $H$. While it
is well-known that every commutative and every cocommutative Hopf algebra is
involutive, it is definitely not true that every involutive Hopf algebra is
either commutative or cocommutative; and Theorem 2 shows how to construct
involutive Hopf algebras which are neither commutative nor cocommutative from
"generic" Hopf algebras.

There is also a different way to construct involutive Hopf algebras: namely,
by taking the tensor product of a commutative with a cocommutative Hopf
algebra. However, these are not really the most general possible case; in
particular, they don't help me answering the Concrete Question in [3].

In order to prove the main part of Theorem 2 \textbf{(a)}, we abstract from
the antipode and prove something more general:

\begin{quote}
\textbf{Theorem 3.} Let $H$ be a bialgebra over a field $k$. Let
$t:H\rightarrow H$ be a coalgebra homomorphism. Let $\operatorname*{Inv}%
\nolimits_{t}H$ denote the $k$-submodule $H\cdot\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H$ of $H$. Then,
the $k$-submodule $\operatorname*{Inv}\nolimits_{t}H$ is a biideal of $H$.
\end{quote}

In the following proofs of Theorem 2 and Theorem 3, we will use the standard
notations for bialgebras: namely, we will denote by $\Delta$ the
comultiplication of $H$, and by $\varepsilon$ the counit of $H$.

\textit{Proof of Theorem 3.} In the following, $\operatorname*{id}$ will
always denote $\operatorname*{id}\nolimits_{H}$, whereas the identity map of
$H\otimes H$ will always be written $\operatorname*{id}\nolimits_{H\otimes H}$.

First of all,%
\[
H\cdot\underbrace{\left(  \operatorname*{Inv}\nolimits_{t}H\right)  }%
_{=H\cdot\left(  \left(  t-\operatorname*{id}\right)  \left(  H\right)
\right)  \cdot H}=\underbrace{H\cdot H}_{\subseteq H}\cdot\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H\subseteq
H\cdot\left(  \left(  t-\operatorname*{id}\right)  \left(  H\right)  \right)
\cdot H=\operatorname*{Inv}\nolimits_{t}H,
\]
so that $\operatorname*{Inv}\nolimits_{t}H$ is a left ideal of $H$. Similarly,
$\operatorname*{Inv}\nolimits_{t}H$ is a right ideal of $H$. Thus,
$\operatorname*{Inv}\nolimits_{t}H$ is both a left ideal and a right ideal of
$H$. In other words, $\operatorname*{Inv}\nolimits_{t}H$ is an ideal of $H$.

Note that
\[
\operatorname*{Inv}\nolimits_{t}H=\underbrace{H}_{\ni1}\cdot\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot\underbrace{H}%
_{\ni1}\supseteq1\cdot\left(  \left(  t-\operatorname*{id}\right)  \left(
H\right)  \right)  \cdot1=\left(  t-\operatorname*{id}\right)  \left(
H\right)  .
\]


Furthermore, $t\otimes t=\left(  t\otimes\operatorname*{id}\right)
\circ\left(  \operatorname*{id}\otimes t\right)  $, where $\operatorname*{id}$
denotes $\operatorname*{id}\nolimits_{H}$\ \ \ \ \footnote{\textit{Proof.}
Every $\left(  x,y\right)  \in H\otimes H$ satisfies%
\begin{align*}
\left(  \left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  \right)  \left(  x\otimes y\right)   &
=\left(  t\otimes\operatorname*{id}\right)  \left(  \underbrace{\left(
\operatorname*{id}\otimes t\right)  \left(  x\otimes y\right)  }%
_{=\operatorname*{id}\left(  x\right)  \otimes t\left(  y\right)  }\right)
=\left(  t\otimes\operatorname*{id}\right)  \left(  \operatorname*{id}\left(
x\right)  \otimes t\left(  y\right)  \right) \\
&  =t\left(  \underbrace{\operatorname*{id}\left(  x\right)  }_{=x}\right)
\otimes\underbrace{\operatorname*{id}\left(  t\left(  y\right)  \right)
}_{=t\left(  y\right)  }=t\left(  x\right)  \otimes t\left(  y\right)
=\left(  t\otimes t\right)  \left(  x\otimes y\right)  .
\end{align*}
In other words, the two maps $\left(  t\otimes\operatorname*{id}\right)
\circ\left(  \operatorname*{id}\otimes t\right)  $ and $t\otimes t$ are equal
to each other on every pure tensor. Since these two maps are $k$-linear, this
yields that these two maps are identic (because two $k$-linear maps from a
tensor product must be identic if they are equal to each other on every pure
tensor). In other words, $t\otimes t=\left(  t\otimes\operatorname*{id}%
\right)  \circ\left(  \operatorname*{id}\otimes t\right)  $, qed.}. Also
notice that $\Delta\circ t=\left(  t\otimes t\right)  \circ\Delta$ (since $t$
is a coalgebra homomorphism). Thus,
\[
\Delta\circ t=\underbrace{\left(  t\otimes t\right)  }_{=\left(
t\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
t\right)  }\circ\Delta=\left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  \circ\Delta.
\]
Hence,%
\begin{align*}
\Delta\circ\left(  t-\operatorname*{id}\right)   &  =\underbrace{\Delta\circ
t}_{=\left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  \circ\Delta}-\underbrace{\Delta
\circ\operatorname*{id}}_{=\Delta=\operatorname*{id}\nolimits_{H\otimes
H}\circ\Delta}\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
}k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =\left(  t\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes t\right)  \circ\Delta-\operatorname*{id}\nolimits_{H\otimes H}%
\circ\Delta=\left(  \left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  -\operatorname*{id}\nolimits_{H\otimes
H}\right)  \circ\Delta.
\end{align*}
Since%
\begin{align*}
&  \left(  t\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes t\right)  -\operatorname*{id}\nolimits_{H\otimes H}\\
&  =\left(  t\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes t\right)  -\underbrace{\left(  \operatorname*{id}\otimes t\right)
}_{=\operatorname*{id}\nolimits_{H\otimes H}\circ\left(  \operatorname*{id}%
\otimes t\right)  }+\left(  \operatorname*{id}\otimes t\right)
-\operatorname*{id}\nolimits_{H\otimes H}\\
&  =\left(  t\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}%
\otimes t\right)  -\underbrace{\operatorname*{id}\nolimits_{H\otimes H}%
}_{=\operatorname*{id}\otimes\operatorname*{id}}\circ\left(
\operatorname*{id}\otimes t\right)  +\left(  \operatorname*{id}\otimes
t\right)  -\underbrace{\operatorname*{id}\nolimits_{H\otimes H}}%
_{=\operatorname*{id}\otimes\operatorname*{id}}\\
&  =\underbrace{\left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  -\left(  \operatorname*{id}\otimes
\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes t\right)
}_{\substack{=\left(  t\otimes\operatorname*{id}-\operatorname*{id}%
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
t\right)  \\\text{(since composition of }k\text{-linear maps}\\\text{is
}k\text{-bilinear)}}}+\underbrace{\left(  \operatorname*{id}\otimes t\right)
-\left(  \operatorname*{id}\otimes\operatorname*{id}\right)  }%
_{\substack{=\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right)
\\\text{(since tensoring of }k\text{-linear maps}\\\text{is }%
k\text{-bilinear)}}}\\
&  =\underbrace{\left(  t\otimes\operatorname*{id}-\operatorname*{id}%
\otimes\operatorname*{id}\right)  }_{\substack{=\left(  t-\operatorname*{id}%
\right)  \otimes\operatorname*{id}\\\text{(since tensoring of }k\text{-linear
maps}\\\text{is }k\text{-bilinear)}}}\circ\left(  \operatorname*{id}\otimes
t\right)  -\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right) \\
&  =\left(  \left(  t-\operatorname*{id}\right)  \otimes\operatorname*{id}%
\right)  \circ\left(  \operatorname*{id}\otimes t\right)  -\operatorname*{id}%
\otimes\left(  t-\operatorname*{id}\right)  ,
\end{align*}
this becomes%
\begin{align*}
\Delta\circ\left(  t-\operatorname*{id}\right)   &  =\left(
\underbrace{\left(  t\otimes\operatorname*{id}\right)  \circ\left(
\operatorname*{id}\otimes t\right)  -\operatorname*{id}\nolimits_{H\otimes H}%
}_{=\left(  \left(  t-\operatorname*{id}\right)  \otimes\operatorname*{id}%
\right)  \circ\left(  \operatorname*{id}\otimes t\right)  -\operatorname*{id}%
\otimes\left(  t-\operatorname*{id}\right)  }\right)  \circ\Delta\\
&  =\left(  \left(  \left(  t-\operatorname*{id}\right)  \otimes
\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes t\right)
-\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right)  \right)
\circ\Delta.
\end{align*}
Thus,%
\begin{align*}
\left(  \Delta\circ\left(  t-\operatorname*{id}\right)  \right)  \left(
H\right)   &  =\left(  \left(  \left(  \left(  t-\operatorname*{id}\right)
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
t\right)  -\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right)
\right)  \circ\Delta\right)  \left(  H\right) \\
&  =\left(  \left(  \left(  t-\operatorname*{id}\right)  \otimes
\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes t\right)
-\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right)  \right)
\underbrace{\left(  \Delta\left(  H\right)  \right)  }_{\subseteq H\otimes
H}\\
&  \subseteq\left(  \left(  \left(  t-\operatorname*{id}\right)
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
t\right)  -\operatorname*{id}\otimes\left(  t-\operatorname*{id}\right)
\right)  \left(  H\otimes H\right) \\
&  \subseteq\underbrace{\left(  \left(  \left(  t-\operatorname*{id}\right)
\otimes\operatorname*{id}\right)  \circ\left(  \operatorname*{id}\otimes
t\right)  \right)  \left(  H\otimes H\right)  }_{=\left(  \left(
t-\operatorname*{id}\right)  \otimes\operatorname*{id}\right)  \left(  \left(
\operatorname*{id}\otimes t\right)  \left(  H\otimes H\right)  \right)
}-\underbrace{\left(  \operatorname*{id}\otimes\left(  t-\operatorname*{id}%
\right)  \right)  \left(  H\otimes H\right)  }_{=\operatorname*{id}\left(
H\right)  \otimes\left(  t-\operatorname*{id}\right)  \left(  H\right)  }\\
&  =\left(  \left(  t-\operatorname*{id}\right)  \otimes\operatorname*{id}%
\right)  \underbrace{\left(  \left(  \operatorname*{id}\otimes t\right)
\left(  H\otimes H\right)  \right)  }_{\subseteq H\otimes H}%
-\underbrace{\operatorname*{id}\left(  H\right)  }_{=H}\otimes\left(
t-\operatorname*{id}\right)  \left(  H\right) \\
&  \subseteq\underbrace{\left(  \left(  t-\operatorname*{id}\right)
\otimes\operatorname*{id}\right)  \left(  H\otimes H\right)  }_{=\left(
t-\operatorname*{id}\right)  \left(  H\right)  \otimes\operatorname*{id}%
\left(  H\right)  }-H\otimes\left(  t-\operatorname*{id}\right)  \left(
H\right) \\
&  =\left(  t-\operatorname*{id}\right)  \left(  H\right)  \otimes
\underbrace{\operatorname*{id}\left(  H\right)  }_{=H}-H\otimes\left(
t-\operatorname*{id}\right)  \left(  H\right) \\
&  =\underbrace{\left(  t-\operatorname*{id}\right)  \left(  H\right)
}_{\subseteq\operatorname*{Inv}\nolimits_{t}H}\otimes H-H\otimes
\underbrace{\left(  t-\operatorname*{id}\right)  \left(  H\right)
}_{\substack{\subseteq\operatorname*{Inv}\nolimits_{t}H=-\operatorname*{Inv}%
\nolimits_{t}H\\\text{(since }\operatorname*{Inv}\nolimits_{t}H\text{ is a
}k\text{-module)}}}\\
&  \subseteq\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes
H-H\otimes\left(  -\operatorname*{Inv}\nolimits_{t}H\right)  =\left(
\operatorname*{Inv}\nolimits_{t}H\right)  \otimes H+H\otimes\left(
\operatorname*{Inv}\nolimits_{t}H\right)  .
\end{align*}


Now, it is a known (and very easy) fact that if $\mathfrak{A}$ and
$\mathfrak{B}$ are two $k$-algebras, and $\mathfrak{I}$ is an ideal of
$\mathfrak{A}$, then $\mathfrak{I}\otimes\mathfrak{B}$ is an ideal of the
$k$-algebra $\mathfrak{A}\otimes\mathfrak{B}$. Applied to $\mathfrak{A}=H$,
$\mathfrak{B}=H$ and $\mathfrak{I}=\operatorname*{Inv}\nolimits_{t}H$, this
yields that $\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes H$ is
an ideal of $H\otimes H$. Similarly, $H\otimes\left(  \operatorname*{Inv}%
\nolimits_{t}H\right)  $ is an ideal of $H\otimes H$.

So we know that $\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes H$
and $H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  $ are ideals of
$H\otimes H$. Thus, $\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes
H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  $ is an ideal of
$H\otimes H$ (because the sum of two ideals is always an ideal).

Now, $\operatorname*{Inv}\nolimits_{t}H=H\cdot\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H$, so that%
\begin{align*}
\Delta\left(  \operatorname*{Inv}\nolimits_{t}H\right)   &  =\Delta\left(
H\cdot\left(  \left(  t-\operatorname*{id}\right)  \left(  H\right)  \right)
\cdot H\right)  =\underbrace{\left(  \Delta\left(  H\right)  \right)
}_{\subseteq H\otimes H}\cdot\underbrace{\left(  \Delta\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \right)
}_{\substack{=\left(  \Delta\circ\left(  t-\operatorname*{id}\right)  \right)
\left(  H\right)  \\\subseteq\left(  \operatorname*{Inv}\nolimits_{t}H\right)
\otimes H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  }%
}\cdot\underbrace{\left(  \Delta\left(  H\right)  \right)  }_{\subseteq
H\otimes H}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }H\text{ is a bialgebra, so (by the axioms of a bialgebra)}\\
\Delta\text{ is a }k\text{-algebra homomorphism}%
\end{array}
\right) \\
&  \subseteq\left(  H\otimes H\right)  \cdot\underbrace{\left(  \left(
\operatorname*{Inv}\nolimits_{t}H\right)  \otimes H+H\otimes\left(
\operatorname*{Inv}\nolimits_{t}H\right)  \right)  \cdot\left(  H\otimes
H\right)  }_{\substack{\subseteq\left(  \operatorname*{Inv}\nolimits_{t}%
H\right)  \otimes H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)
\\\text{(since }\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes
H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \text{ is an ideal
of }H\otimes H\text{)}}}\\
&  \subseteq\left(  H\otimes H\right)  \cdot\left(  \left(
\operatorname*{Inv}\nolimits_{t}H\right)  \otimes H+H\otimes\left(
\operatorname*{Inv}\nolimits_{t}H\right)  \right)  \subseteq\left(
\operatorname*{Inv}\nolimits_{t}H\right)  \otimes H+H\otimes\left(
\operatorname*{Inv}\nolimits_{t}H\right)
\end{align*}
(since $\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes
H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  $ is an ideal of
$H\otimes H$).

Moreover,
\begin{align*}
\varepsilon\circ\left(  t-\operatorname*{id}\right)   &
=\underbrace{\varepsilon\circ t}_{\substack{=\varepsilon\\\text{(since
}t\text{ is a}\\\text{coalgebra homomorphism)}}}-\underbrace{\varepsilon
\circ\operatorname*{id}}_{=\varepsilon}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
composition of }k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =\varepsilon-\varepsilon=0.
\end{align*}
Now, $\operatorname*{Inv}\nolimits_{t}H=H\cdot\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H$, so that%
\begin{align*}
\varepsilon\left(  \operatorname*{Inv}\nolimits_{t}H\right)   &
=\varepsilon\left(  H\cdot\left(  \left(  t-\operatorname*{id}\right)  \left(
H\right)  \right)  \cdot H\right)  =\left(  \varepsilon\left(  H\right)
\right)  \cdot\underbrace{\left(  \varepsilon\left(  \left(
t-\operatorname*{id}\right)  \left(  H\right)  \right)  \right)  }_{=\left(
\varepsilon\circ\left(  t-\operatorname*{id}\right)  \right)  \left(
H\right)  =0}\cdot\left(  \varepsilon\left(  H\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }H\text{ is a bialgebra, so (by the axioms of a bialgebra)}\\
\varepsilon\text{ is a }k\text{-algebra homomorphism}%
\end{array}
\right) \\
&  =\left(  \varepsilon\left(  H\right)  \right)  \cdot0\cdot\left(
\varepsilon\left(  H\right)  \right)  =0.
\end{align*}


So we have shown that $\Delta\left(  \operatorname*{Inv}\nolimits_{t}H\right)
\subseteq\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes
H+H\otimes\left(  \operatorname*{Inv}\nolimits_{t}H\right)  $ and
$\varepsilon\left(  \operatorname*{Inv}\nolimits_{t}H\right)  =0$. In other
words, $\operatorname*{Inv}\nolimits_{t}H$ is a coideal of $H$.

Since $\operatorname*{Inv}\nolimits_{t}H$ is both an ideal and a coideal of
$H$, it follows that $\operatorname*{Inv}\nolimits_{t}H$ is a biideal of $H$.
This proves Theorem 3.

\textit{Proof of Theorem 2.} \textbf{(a)} It is well-known that the antipode
$S$ is an anti-coalgebra homomorphism. Hence, $S\circ S$ is a coalgebra
homomorphism (because the composition of any two anti-coalgebra homomorphisms
is a coalgebra homomorphism). In other words, $S^{2}$ is a coalgebra
homomorphism (since $S^{2}=S\circ S$). Hence, we can apply Theorem 3 to
$t=S^{2}$. This yields that the $k$-submodule $\operatorname*{Inv}%
\nolimits_{S^{2}}H$ is a biideal of $H$ (where $\operatorname*{Inv}%
\nolimits_{S^{2}}H$ is defined as in Theorem 3). Since%
\begin{align*}
\operatorname*{Inv}\nolimits_{S^{2}}H  &  =H\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot
H\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{Inv}%
\nolimits_{S^{2}}H\right) \\
&  =\operatorname*{Inv}H,
\end{align*}
this yields that $\operatorname*{Inv}H$ is a biideal of $H$.

Now,
\begin{align*}
S\circ\left(  S^{2}-\operatorname*{id}\right)   &  =\underbrace{S\circ S^{2}%
}_{=S^{3}=S^{2}\circ S}-\underbrace{S\circ\operatorname*{id}}%
_{=S=\operatorname*{id}\circ S}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
composition of }k\text{-linear maps is }k\text{-bilinear}\right) \\
&  =S^{2}\circ S-\operatorname*{id}\circ S=\left(  S^{2}-\operatorname*{id}%
\right)  \circ S\ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of
}k\text{-linear maps is }k\text{-bilinear}\right)  ,
\end{align*}
so that%
\[
\left(  S\circ\left(  S^{2}-\operatorname*{id}\right)  \right)  \left(
H\right)  =\left(  \left(  S^{2}-\operatorname*{id}\right)  \circ S\right)
\left(  H\right)  =\left(  S^{2}-\operatorname*{id}\right)  \left(
\underbrace{S\left(  H\right)  }_{\subseteq H}\right)  \subseteq\left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  .
\]
But $\operatorname*{Inv}H=H\cdot\left(  \left(  S^{2}-\operatorname*{id}%
\right)  \left(  H\right)  \right)  \cdot H$ yields%
\begin{align*}
S\left(  \operatorname*{Inv}H\right)   &  =S\left(  H\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H\right)
=\underbrace{\left(  S\left(  H\right)  \right)  }_{\subseteq H}%
\cdot\underbrace{\left(  S\left(  \left(  S^{2}-\operatorname*{id}\right)
\left(  H\right)  \right)  \right)  }_{\substack{=\left(  S\circ\left(
S^{2}-\operatorname*{id}\right)  \right)  \left(  H\right)  \\\subseteq\left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  }}\cdot\underbrace{\left(
S\left(  H\right)  \right)  }_{\subseteq H}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the antipode }S\text{ is an
anti-algebra homomorphism}\right) \\
&  \subseteq H\cdot\left(  \left(  S^{2}-\operatorname*{id}\right)  \left(
H\right)  \right)  \cdot H=\operatorname*{Inv}H.
\end{align*}
Thus, $\operatorname*{Inv}H$ is a biideal of $H$ satisfying $S\left(
\operatorname*{Inv}H\right)  \subseteq\operatorname*{Inv}H$. In other words,
$\operatorname*{Inv}H$ is a Hopf ideal of $H$. This proves Theorem 2
\textbf{(a)}.

\textbf{(b)} For every $y\in H$, let $\overline{y}$ denote the residue class
of $y$ modulo $\operatorname*{Inv}H$.

Note that%
\[
\operatorname*{Inv}H=\underbrace{H}_{\ni1}\cdot\left(  \left(  S^{2}%
-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot\underbrace{H}%
_{\ni1}\supseteq1\cdot\left(  \left(  S^{2}-\operatorname*{id}\right)  \left(
H\right)  \right)  \cdot1=\left(  S^{2}-\operatorname*{id}\right)  \left(
H\right)  .
\]


Let $\overline{S}$ denote the antipode of the quotient Hopf algebra
$H\diagup\left(  \operatorname*{Inv}H\right)  $. By the definition of the
antipode of a quotient Hopf algebra, we have $\overline{S}\left(  \overline
{x}\right)  =\overline{S\left(  x\right)  }$ for every $x\in H$. Thus, every
$y\in H$ satisfies%
\begin{align*}
\overline{S}^{2}\left(  \overline{y}\right)   &  =\overline{S}\left(
\overline{S}\left(  \overline{y}\right)  \right)  =\overline{S}\left(
\overline{S\left(  y\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the formula }\overline{S}\left(
\overline{x}\right)  =\overline{S\left(  x\right)  }\text{ (applied to
}x=y\text{) yields }\overline{S}\left(  \overline{y}\right)  =\overline
{S\left(  y\right)  }\right) \\
&  =\overline{S\left(  S\left(  y\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the formula }\overline{S}\left(
\overline{x}\right)  =\overline{S\left(  x\right)  }\text{ (applied to
}x=S\left(  y\right)  \text{)}\right) \\
&  =\overline{y+\left(  S^{2}-\operatorname*{id}\right)  \left(  y\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since}\\
S\left(  S\left(  y\right)  \right)  =S^{2}\left(  y\right)  =y+S^{2}\left(
y\right)  -\underbrace{y}_{=\operatorname*{id}\left(  y\right)  }%
=y+\underbrace{S^{2}\left(  y\right)  -\operatorname*{id}\left(  y\right)
}_{=\left(  S^{2}-\operatorname*{id}\right)  \left(  y\right)  }=y+\left(
S^{2}-\operatorname*{id}\right)  \left(  y\right)
\end{array}
\right) \\
&  =\overline{y}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  S^{2}-\operatorname*{id}%
\right)  \underbrace{\left(  y\right)  }_{\in H}\in\left(  S^{2}%
-\operatorname*{id}\right)  \left(  H\right)  \subseteq\operatorname*{Inv}%
H\text{, so that }y+\left(  S^{2}-\operatorname*{id}\right)  \left(  y\right)
\equiv y\operatorname{mod}\operatorname*{Inv}H\right)  .
\end{align*}
Thus, every $z\in H\diagup\left(  \operatorname*{Inv}H\right)  $ satisfies
$\overline{S}^{2}\left(  z\right)  =\operatorname*{id}\nolimits_{H\diagup
\left(  \operatorname*{Inv}H\right)  }\left(  z\right)  $%
\ \ \ \ \footnote{\textit{Proof.} Let $z\in H\diagup\left(
\operatorname*{Inv}H\right)  $. Then, there exists some $y\in H$ such that
$z=\overline{y}$. Consider this $y$. Then, $z=\overline{y}$ yields
$\overline{S}^{2}\left(  z\right)  =\overline{S}^{2}\left(  \overline
{y}\right)  =\overline{y}=z=\operatorname*{id}\nolimits_{H\diagup\left(
\operatorname*{Inv}H\right)  }\left(  z\right)  $, qed.}. In other words,
$\overline{S}^{2}=\operatorname*{id}\nolimits_{H\diagup\left(
\operatorname*{Inv}H\right)  }$. This shows that the quotient Hopf algebra
$H\diagup\left(  \operatorname*{Inv}H\right)  $ is involutive. Theorem 2
\textbf{(b)} is proven.

\textbf{(c)} Let $G$ be an involutive Hopf algebra and $f:H\rightarrow G$ be a
Hopf algebra homomorphism. Then, since $f$ is a Hopf algebra homomorphism, we
have $f\circ S=S_{G}\circ f$, where $S_{G}$ denotes the antipode of $G$. Thus,%
\[
f\circ\underbrace{S^{2}}_{=S\circ S}=\underbrace{f\circ S}_{=S_{G}\circ
f}\circ S=S_{G}\circ\underbrace{f\circ S}_{=S_{G}\circ f}=\underbrace{S_{G}%
\circ S_{G}}_{\substack{=S_{G}^{2}=\operatorname*{id}\nolimits_{G}%
\\\text{(since }G\text{ is involutive)}}}\circ f=f,
\]
so that%
\begin{align*}
f\circ\left(  S^{2}-\operatorname*{id}\right)   &  =\underbrace{f\circ S^{2}%
}_{=f}-\underbrace{f\circ\operatorname*{id}}_{=f}\ \ \ \ \ \ \ \ \ \ \left(
\text{since composition of }k\text{-linear maps is }k\text{-bilinear}\right)
\\
&  =f-f=0.
\end{align*}
Hence, $\left(  f\circ\left(  S^{2}-\operatorname*{id}\right)  \right)
\left(  H\right)  =0$. Now, $\operatorname*{Inv}H=H\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H$ yields%
\begin{align*}
f\left(  \operatorname*{Inv}H\right)   &  =f\left(  H\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \cdot H\right)
=\left(  f\left(  H\right)  \right)  \cdot\underbrace{\left(  f\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  H\right)  \right)  \right)
}_{=\left(  f\circ\left(  S^{2}-\operatorname*{id}\right)  \right)  \left(
H\right)  =0}\cdot\left(  f\left(  H\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }k\text{-algebra
homomorphism}\right) \\
&  =\left(  f\left(  H\right)  \right)  \cdot0\cdot\left(  f\left(  H\right)
\right)  =0.
\end{align*}
Hence, the homomorphism $f$ factors through $H\diagup\left(
\operatorname*{Inv}H\right)  $. Theorem 2 \textbf{(c)} is thus proven.

Parts \textbf{(d)} and \textbf{(e)} of Theorem 2 are completely
straightforward and thus left to the reader. The proof of Theorem 2 is thus complete.

We notice that we could generalize Theorems 2 and 3 by replacing "field" by
"commutative ring". These generalizations are proven in exactly the same way
as we verified Theorems 2 and 3, with the only difference that we would have
to use more complicated notations, because we couldn't anymore identify (for
instance) $\left(  \operatorname*{Inv}\nolimits_{t}H\right)  \otimes H$ with a
$k$-submodule of $H\otimes H$ in the proof of Theorem 3 (because if $A$, $B$
and $C$ are three $k$-modules over a commutative ring $k$ such that $A$ is a
submodule of $B$, then $A\otimes C$ is \textit{in general} not a submodule of
$B\otimes C$), so we would have to explicitly work with inclusion maps instead.

\begin{center}
\textbf{\S 2. The Dynkin operator}
\end{center}

We can define a Dynkin operator for an arbitrary graded Hopf algebra, even
though it enjoys most of its interesting properties in less general cases (for
instance, when $H$ is commutative or cocommutative).

\begin{quote}
\textbf{Definition 4.} Let $H$ be a graded Hopf algebra over a field $k$. For
every $n\in\mathbb{N}$, the denote by $H_{n}$ the $n$-th graded component of
$H$.

\textbf{(a)} Let $E:H\rightarrow H$ be the $k$-linear map which sends every
$x\in H_{n}$ to $nx$ for every $n\in\mathbb{N}$. Note that this map $E$ is
easily seen to be a derivation (i. e., it satisfies $E\left(  xy\right)
=E\left(  x\right)  y+xE\left(  y\right)  $ for all $x\in H$ and $y\in H$, or,
equivalently, it satisfies the identity $E\circ\mu=\mu\circ\left(
E\otimes\operatorname*{id}+\operatorname*{id}\otimes E\right)  $ with $\mu$
denoting the multiplication map of $H$) and a coderivation (i. e., it
satisfies $\Delta\circ E=\left(  E\otimes\operatorname*{id}+\operatorname*{id}%
\otimes E\right)  \circ\Delta$ with $\Delta$ denoting the comultiplication map
of $H$).

\textbf{(b)} The \textit{Dynkin operator} of $H$ is defined to mean the
convolution $E\ast S$ of the maps $E:H\rightarrow H$ and $S:H\rightarrow H$
(where $S$ denotes the antipode of $H$, as usual).
\end{quote}

Note that $E\ast S$ is not the only "Dynkin operator" around. One can just as
well call $S\ast E$ a "Dynkin operator" (it is a kind of mirror version of
$E\ast S$), and for a field $k$ of characteristic $0$ one can even
"interpolate" between these two "Dynkin operators" $E\ast S$ and $S\ast E$ by
introducing a "Dynkin operator" $S^{\alpha}\ast E\ast S^{\beta}$ for any two
elements $\alpha$ and $\beta$ of $k$ satisfying $\alpha+\beta=1$. See the
Remark in \S 3 of [2] (where our $E$ is denoted by $D$) for this definition
(which is due to Claudio Procesi).

A well-known theorem states that:

\begin{quote}
\textbf{Theorem 5.} Let $H$ be a graded Hopf algebra over $k$.

\textbf{(a)} If $H$ is cocommutative, then the Dynkin operator $E\ast S$
(defined in Definition 4) satisfies $\left(  E\ast S\right)  \circ\left(
E\ast S\right)  =E\circ\left(  E\ast S\right)  $, $\left(  E\ast S\right)
\mid_{\operatorname*{Prim}H}=E\mid_{\operatorname*{Prim}H}$ and $\left(  E\ast
S\right)  \left(  H\right)  \subseteq\operatorname*{Prim}H$. (Here,
$\operatorname*{Prim}H$ denotes the subspace of $H$ formed by all primitive
elements of $H$.)

\textbf{(b)} If $H$ is commutative, then the Dynkin operator $E\ast S$
(defined in Definition 4) satisfies $\left(  E\ast S\right)  \circ\left(
E\ast S\right)  =E\circ\left(  E\ast S\right)  $, $\operatorname*{Ker}\left(
E\ast S\right)  \subseteq\operatorname*{Ker}E+\left(  H^{+}\right)  ^{2}$ and
$\left(  E\ast S\right)  \left(  \left(  H^{+}\right)  ^{2}\right)  =0$.
(Here, $H^{+}$ denotes the ideal $\operatorname*{Ker}\varepsilon$ of $H$.)
\end{quote}

Our goal in the next section (\S 3) is to prove that $\left(  E\ast S\right)
\circ\left(  E\ast S\right)  =E\circ\left(  E\ast S\right)  $ is not
necessarily true for an\textit{ involutive (but not necessarily commutative or
cocommutative) connected graded Hopf algebra }$H$. (Note that tensor products
of commutative with cocommutative Hopf algebras are easily seen to satisfy
$\left(  E\ast S\right)  \circ\left(  E\ast S\right)  =E\circ\left(  E\ast
S\right)  $, so we need a more subtle counterexample).

\begin{center}
\textbf{\S 3. The counterexample}
\end{center}

To construct a counterexample, we will need the Hopf algebra $\mathbf{H}_{o}$
defined by Foissy in [1], Definition 2. This is the Hopf algebra of ordered
(rooted) forests, where "ordered" means that the set of vertices is totally
ordered (but not necessarily heap-ordered, i. e., a child needs not be greater
than its father). We recall its definition (mostly quoted from [1]):

\begin{quote}
\textbf{Definition 6.} \textbf{(a)} An \textit{ordered forest} means a rooted
forest endowed with a total order on the set of its vertices. We are going to
represent such an ordered forest pictorially by drawing the forest and
decorating every vertex with its position number\footnote{By a
\textit{position number} of an element $a$ in a totally ordered finite set
$T$, I mean the number $n$ such that $a$ is the $n$-th smallest element of
$T$.}. The empty forest is denoted by $1$ (as it will later become the unity
of an algebra).

\textbf{(b)} For every $n\in\mathbb{N}$, let $\mathcal{F}_{o}\left(  n\right)
$ denote the set of all ordered forests with $n$ vertices. The set of all
ordered forests will be called $\mathcal{F}_{o}$; it is the union of the
mutually disjoint sets $\mathcal{F}_{o}\left(  n\right)  $ over all
$n\in\mathbb{N}$.

\textbf{(c)} Let $k$ be a field. We denote by $\mathbf{H}_{o}$ the free
$k$-vector space with basis $\mathcal{F}_{o}$; it is canonically graded, with
the $n$-th graded component $\mathbf{H}_{o,n}$ being the free $k$-vector space
with basis $\mathcal{F}_{o}\left(  n\right)  $.

\textbf{(d)} We make $\mathbf{H}_{o}$ into a graded $k$-algebra as follows:
For any two ordered forests $a$ and $b$, let $ab$ be the ordered forest whose
underlying forest is the disjoint union of the forests $a$ and $b$, and the
order on which is defined by letting each vertex of $a$ be smaller than each
vertex of $b$ (but the order of the vertices of $a$ is kept from $a$, and the
order of the vertices of $b$ is kept from $b$). The unity of this $k$-algebra
is $1$ (the empty forest).

\textbf{(e)} For every forest $F$, let $V\left(  F\right)  $ denote the set of
the vertices of $F$. If $F$ is a forest and $\mathbf{v}$ is a subset of
$V\left(  F\right)  $, then we write $\mathbf{v}\models V\left(  F\right)  $
(and we say that $\mathbf{v}$ is an \textit{admissible cut} of $F$) if and
only if no element of $\mathbf{v}$ is a descendant\footnote{Descendants mean
direct or indirect descendants. (In particular, every vertex is a descendant
of itself.)} of another element of $\mathbf{v}$ except of itself. If $F$ is a
forest and $\mathbf{v}$ is a subset of $V\left(  F\right)  $ satisfying
$\mathbf{v}\models V\left(  F\right)  $, then we denote by $Lea_{\mathbf{v}}F$
the rooted subforest of $F$ obtained by keeping only the vertices above
$\mathbf{v}$ (where a vertex of $F$ is said to be "above $\mathbf{v}$" if it
is a descendant of an element of $\mathbf{v}$), and we denote by
$Roo_{\mathbf{v}}F$ the rooted subforest of $F$ obtained by keeping only the
other vertices. If $F$ is an ordered forest, then $Lea_{\mathbf{v}}F$ and
$Roo_{\mathbf{v}}F$ canonically become ordered forests.

\textbf{(f)} We now make $\mathbf{H}_{o}$ into a $k$-bialgebra by defining%
\[
\Delta\left(  F\right)  =\sum\limits_{\mathbf{v}\models V\left(  F\right)
}Lea_{\mathbf{v}}F\otimes Roo_{\mathbf{v}}F\ \ \ \ \ \ \ \ \ \ \text{for every
forest }F
\]
and%
\[
\varepsilon\left(  F\right)  =\left\{
\begin{array}
[c]{c}%
1\text{, if }F=1\text{;}\\
0\text{, if }F\neq1
\end{array}
\right.  \ \ \ \ \ \ \ \ \ \ \text{for every forest }F.
\]


\textbf{Theorem 7.} Definition 6 makes $\mathbf{H}_{o}$ into a connected
graded $k$-Hopf algebra.
\end{quote}

For the proof of Theorem 7, see the references in \S 1.2 of [1]. Here are the
$\mathcal{F}_{o}\left(  n\right)  $ for $0\leq n\leq3$ (this table is copied
from [1]):
\begin{align*}
\mathcal{F}_{o}\left(  0\right)   &  =\left\{  1\right\}  ;\\
\mathcal{F}_{o}\left(  1\right)   &  =\left\{
\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 1} \end{picture}\right\}
;\\
\mathcal{F}_{o}\left(  2\right)   &  =\left\{
\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 1} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 2} \end{picture},\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 1} \put(6,3){\tiny 2} \end{picture},\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 2} \put(6,3){\tiny 1} \end{picture}\right\}
;\\
\mathcal{F}_{o}\left(  3\right)   &  =\left\{
\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 1} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 2} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 3} \end{picture},\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 1} \end{picture}\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 2} \put(6,3){\tiny 3} \end{picture},\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 1} \end{picture}\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 3} \put(6,3){\tiny 2} \end{picture},\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 1} \put(6,3){\tiny 3} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 2} \end{picture},\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 2} \end{picture}\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 3} \put(6,3){\tiny 1} \end{picture},\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 1} \put(6,3){\tiny 2} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 3} \end{picture},\begin{picture}(12,5)(0,-1) \put(3,0){\circle*{2}} \put(3,0){\line(0,1){5}} \put(3,5){\circle*{2}} \put(6,-2){\tiny 2} \put(6,3){\tiny 1} \end{picture}\begin{picture}(10,5)(-2,-1) \put(0,0){\circle*{2}} \put(3,-2){\tiny 3} \end{picture},\begin{picture}(20,12)(-5,-1) \put(3,0){\circle*{2}} \put(-0.65,0){$\vee$} \put(6,7){\circle*{2}} \put(0,7){\circle*{2}} \put(5,-2){\tiny 1} \put(9,5){\tiny 3} \put(-5,5){\tiny 2} \end{picture},\begin{picture}(20,12)(-5,-1) \put(3,0){\circle*{2}} \put(-0.65,0){$\vee$} \put(6,7){\circle*{2}} \put(0,7){\circle*{2}} \put(5,-2){\tiny 2} \put(9,5){\tiny 3} \put(-5,5){\tiny 1} \end{picture},\begin{picture}(20,12)(-5,-1) \put(3,0){\circle*{2}} \put(-0.65,0){$\vee$} \put(6,7){\circle*{2}} \put(0,7){\circle*{2}} \put(5,-2){\tiny 3} \put(9,5){\tiny 2} \put(-5,5){\tiny 1} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 1} \put(3,3){\tiny 2} \put(3,9){\tiny 3} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 1} \put(3,3){\tiny 3} \put(3,9){\tiny 2} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 2} \put(3,3){\tiny 1} \put(3,9){\tiny 3} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 2} \put(3,3){\tiny 3} \put(3,9){\tiny 1} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 3} \put(3,3){\tiny 1} \put(3,9){\tiny 2} \end{picture},\begin{picture}(12,12)(-2,-1) \put(0,0){\circle*{2}} \put(0,0){\line(0,1){5}} \put(0,5){\circle*{2}} \put(0,5){\line(0,1){5}} \put(0,10){\circle*{2}} \put(3,-2){\tiny 3} \put(3,3){\tiny 2} \put(3,9){\tiny 1} \end{picture}\right\}
.
\end{align*}


Here are some examples of how multiplication, comultiplication and the
antipode look on $\mathbf{H}_{o}$:

\textit{Multiplication:}%
\begin{align*}%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\cdot%
%TCIMACRO{\TeXButton{1/2,3/2}{\tdtroisun{2}{3}{1}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}%
%EndExpansion
&  =%
%TCIMACRO{\TeXButton{product}{\tddeux{1}{2} \tdtroisun{4}{5}{3}}}%
%BeginExpansion
\tddeux{1}{2} \tdtroisun{4}{5}{3}%
%EndExpansion
;\\%
%TCIMACRO{\TeXButton{1/2,3/2}{\tdtroisun{2}{3}{1}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}%
%EndExpansion
\cdot%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
&  =%
%TCIMACRO{\TeXButton{product}{\tdtroisun{2}{3}{1}\tddeux{4}{5}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}\tddeux{4}{5}%
%EndExpansion
.
\end{align*}


\textit{Comultiplication:}%
\begin{align*}
\Delta\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes1+1\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\otimes1+2%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{3/2,1}{\tddeux{2}{3} \tdun{1}}}%
%BeginExpansion
\tddeux{2}{3} \tdun{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{3/2,1}{\tddeux{2}{3} \tdun{1}}}%
%BeginExpansion
\tddeux{2}{3} \tdun{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{3/2,1}{\tddeux{2}{3} \tdun{1}}}%
%BeginExpansion
\tddeux{2}{3} \tdun{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/2,3/2}{\tdtroisun{2}{3}{1}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/2,3/2}{\tdtroisun{2}{3}{1}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,3/2}{\tdtroisun{2}{3}{1}}}%
%BeginExpansion
\tdtroisun{2}{3}{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/4,4/2,3/3}{\tdquatredeux{2}{3}{4}{1}}}%
%BeginExpansion
\tdquatredeux{2}{3}{4}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/4,4/2,3/3}{\tdquatredeux{2}{3}{4}{1}}}%
%BeginExpansion
\tdquatredeux{2}{3}{4}{1}%
%EndExpansion
\otimes1+1\otimes%
%TCIMACRO{\TeXButton{1/4,4/2,3/3}{\tdquatredeux{2}{3}{4}{1}}}%
%BeginExpansion
\tdquatredeux{2}{3}{4}{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1,3/1}{\tdtroisun{1}{3}{2}}}%
%BeginExpansion
\tdtroisun{1}{3}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/3}{\tddeux{3}{1}}}%
%BeginExpansion
\tddeux{3}{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
;\\
\Delta\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
.
\end{align*}


\textit{Antipode:}\footnote{Computing the antipode can be done by recurrently
solving the equation $\left(  \operatorname*{id}\ast S\right)  \left(
x\right)  =\varepsilon\left(  x\right)  \cdot1$ (or the equation $\left(
S\ast\operatorname*{id}\right)  \left(  x\right)  =\varepsilon\left(
x\right)  \cdot1$), but in some cases one can simplify one's work by using the
fact that the antipode is an anti-algebra homomorphism.}%
\begin{align*}
S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)   &  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\ \ \ \ \ \ \ \ \ \ \left(  \text{no wonder, since }%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\text{ is primitive}\right)  ;\\
S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
;\\
S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
;\\
S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)   &  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{3}{\tdun{3}}}%
%BeginExpansion
\tdun{3}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1} }}%
%BeginExpansion
\tddeux{2}{1}
%EndExpansion%
%TCIMACRO{\TeXButton{3}{\tdun{3}}}%
%BeginExpansion
\tdun{3}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2/3}{\tddeux{3}{2}}}%
%BeginExpansion
\tddeux{3}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
;\\
S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{2}{\tdun{2}}}%
%BeginExpansion
\tdun{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{3}{\tdun{3}}}%
%BeginExpansion
\tdun{3}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{3}{\tdun{3}}}%
%BeginExpansion
\tdun{3}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion%
%TCIMACRO{\TeXButton{3/2}{\tddeux{2}{3}}}%
%BeginExpansion
\tddeux{2}{3}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
.
\end{align*}


Note that, at this place, our notations are slightly ambiguous: For example,
what does $%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{3/5,4/5}{\tdtroisun{5}{3}{4}}}%
%BeginExpansion
\tdtroisun{5}{3}{4}%
%EndExpansion%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
$ mean? Does it mean $\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{3/5,4/5}{\tdtroisun{5}{3}{4}}}%
%BeginExpansion
\tdtroisun{5}{3}{4}%
%EndExpansion
\right)
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
=%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{3/5,4/5}{\tdtroisun{5}{3}{4}}}%
%BeginExpansion
\tdtroisun{5}{3}{4}%
%EndExpansion%
%TCIMACRO{\TeXButton{6/7}{\tddeux{7}{6}}}%
%BeginExpansion
\tddeux{7}{6}%
%EndExpansion
$ or does it mean $%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\left(
%TCIMACRO{\TeXButton{3/5,4/5}{\tdtroisun{5}{3}{4}}}%
%BeginExpansion
\tdtroisun{5}{3}{4}%
%EndExpansion%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion%
%TCIMACRO{\TeXButton{5/7,6/7}{\tdtroisun{7}{5}{6}}}%
%BeginExpansion
\tdtroisun{7}{5}{6}%
%EndExpansion%
%TCIMACRO{\TeXButton{3/4}{\tddeux{4}{3}}}%
%BeginExpansion
\tddeux{4}{3}%
%EndExpansion
$ ? The ambiguity is due to the fact that when we write two trees one after
the other, it is not clear whether we mean to multiply these two trees as
elements of $\mathbf{H}_{o}$ or join these two trees into a forest. One way to
get rid of this ambiguity is never to suppress the product sign (so that when
two trees are written one after the other, it cannot mean multiplication, but
can only mean the forest). This is what we are going to do in the following.
But let us also note that we will use $\times$ instead of $\cdot$ as our
product sign, because $\cdot$ is too similar to a one-vertex tree.

Let now $R$ be the $k$-subalgebra of $\mathbf{H}_{o}$ generated by the ordered
trees $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
$ and $%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
$. It is very easy to see that $R$ is a connected graded Hopf subalgebra of
$\mathbf{H}_{o}\ \ \ \ $\footnote{More generally: If $S$ is a set of ordered
trees which, for every ordered tree $s\in S$, contains each ordered subtree of
$S$, then the $k$-subalgebra of $\mathbf{H}_{o}$ generated by $S$ is a
connected graded Hopf subalgebra of $\mathbf{H}_{o}$.}. Let $R^{\prime}$ be
the involutive Hopf algebra $R\diagup\left(  \operatorname*{Inv}R\right)  $.
Assume that the field $k$ has characteristic $0$. Then we are going to prove
that $\left(  E\ast S\right)  \circ\left(  E\ast S\right)  \neq E\circ\left(
E\ast S\right)  $ in $R^{\prime}$. This will answer the Concrete Question in
[3] negatively.

We denote by $\Delta,$ $\varepsilon,$ $\mu,$ $\eta$ and $S$ the usual
operations of the Hopf algebra $R$ (in this order: comultiplication, counit,
multiplication map, unit map and antipode). Also, let $\left[  \cdot
,\cdot\right]  $ denote the commutator.

First, a very simple lemma:

\begin{quote}
\textbf{Lemma 8.} Let $H$ be a graded Hopf algebra over $k$. Let $E$ and $S$
be defined as in Theorem 5. Then:

\textbf{(a)} Every graded map $f:H\rightarrow H$ satisfies $E\circ f=f\circ E$.

\textbf{(b)} Let $S$ be the antipode of $H$, let $\varepsilon$ be the counit
map of $H$, and let $\eta$ be the unit map of $H$. Then,%
\[
\left(  E\ast S\right)  \circ\left(  E\ast S\right)  -E\circ\left(  E\ast
S\right)  =\left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)  ,
\]
where $\overline{S}$ denotes the map $S-\eta\circ\varepsilon:H\rightarrow H$.
\end{quote}

\textit{Proof of Lemma 8 (sketched).} Lemma 8 \textbf{(a)} is left to the reader.

\textbf{(b)} By Lemma 8 \textbf{(a)} (applied to $f=E\ast S$, which is easily
seen to be graded), we have $E\circ\left(  E\ast S\right)  =\left(  E\ast
S\right)  \circ E$. Hence,%
\begin{align*}
\left(  E\ast S\right)  \circ\left(  E\ast S\right)  -E\circ\left(  E\ast
S\right)   &  =\left(  E\ast S\right)  \circ\left(  E\ast S\right)  -\left(
E\ast S\right)  \circ E=\left(  E\ast S\right)  \circ\left(  E\ast
S-\underbrace{E}_{=E\ast\left(  \eta\circ\varepsilon\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since composition of }k\text{-linear maps
is }k\text{-bilinear}\right) \\
&  =\left(  E\ast S\right)  \circ\underbrace{\left(  E\ast S-E\ast\left(
\eta\circ\varepsilon\right)  \right)  }_{\substack{=E\ast\left(  S-\eta
\circ\varepsilon\right)  \\\text{(since convolution of }k\text{-linear maps is
}k\text{-bilinear)}}}\\
&  =\left(  E\ast S\right)  \circ\left(  E\ast\underbrace{\left(  S-\eta
\circ\varepsilon\right)  }_{=\overline{S}}\right)  =\left(  E\ast S\right)
\circ\left(  E\ast\overline{S}\right)  ,
\end{align*}
proving Lemma 8 \textbf{(b)}.

Now, let us come back to our $R$. In order to prove that $\left(  E\ast
S\right)  \circ\left(  E\ast S\right)  \neq E\circ\left(  E\ast S\right)  $ in
$R^{\prime}$, it is enough to show that $\left(  E\ast S\right)  \circ\left(
E\ast\overline{S}\right)  \neq0$ in $R^{\prime}$ (by Lemma 8 \textbf{(b)},
applied to $H=R^{\prime}$). We will achieve this by showing that $\left(
\left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)  \right)
\left(  \overline{%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
}\right)  \neq0$ in $R^{\prime}$. This means showing that%
\begin{equation}
\left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \notin\operatorname*{Inv}R. \label{cex.1}%
\end{equation}


We will prove (\ref{cex.1}) by showing something stronger: We will show that%
\begin{equation}
\left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \notin\operatorname*{Inv}R+\left(  R^{+}\right)  ^{3}. \label{cex.2}%
\end{equation}
Here, for every coalgebra $C$, we denote by $C^{+}$ the subspace
$\operatorname*{Ker}\varepsilon$ of $C$. We notice that $R^{+}$ is an ideal of
$R$, so that $\left(  R^{+}\right)  ^{3}$ is an ideal as well. Since $R\otimes
R$ also is a bialgebra, we have a well-defined ideal $\left(  R\otimes
R\right)  ^{+}$ of $R\otimes R$ as well, and it is important to notice that
$\left(  R\otimes R\right)  ^{+}=R^{+}\otimes R+R\otimes R^{+}$, so that every
$n\in\mathbb{N}$ satisfies $\left(  \left(  R\otimes R\right)  ^{+}\right)
^{n}=\sum\limits_{i=0}^{n}\left(  R^{+}\right)  ^{i}\otimes\left(
R^{+}\right)  ^{n-i}$. As a consequence, every $n\in\mathbb{N}$ satisfies
$\mu\left(  \left(  \left(  R\otimes R\right)  ^{+}\right)  ^{n}\right)
\subseteq\left(  R^{+}\right)  ^{n}$. Also, every $n\in\mathbb{N}$ satisfies
$\Delta\left(  \left(  R^{+}\right)  ^{n}\right)  \subseteq\left(  \left(
R\otimes R\right)  ^{+}\right)  ^{n}$ (since $\Delta$ is an algebra
homomorphism and commutes with the counity maps of $R$ and $R\otimes R$),
$S\left(  \left(  R^{+}\right)  ^{n}\right)  \subseteq\left(  R^{+}\right)
^{n}$ (since $S$ is an anti-algebra homomorphism and $S\left(  R^{+}\right)
\subseteq R^{+}$) and $E\left(  \left(  R^{+}\right)  ^{n}\right)
\subseteq\left(  R^{+}\right)  ^{n}$ (very easy to check). As a consequence,
$\left(  E\ast S\right)  \left(  \left(  R^{+}\right)  ^{3}\right)
\subseteq\left(  R^{+}\right)  ^{3}$ (since $E\ast S=\mu\circ\left(  E\otimes
S\right)  \circ\Delta$ by the definition of convolution).

This allows us to work modulo $\left(  R^{+}\right)  ^{3}$ (when we are
working in $R$) and modulo $\left(  \left(  R\otimes R\right)  ^{+}\right)
^{3}$ (when we are working in $R\otimes R$), i. e., to forget all terms which
include more than $2$ (disjoint) trees (because $\left(  R^{+}\right)  ^{3}$
is the $k$-vector subspace of $R$ which is formed by products of more than $2$
disjoint trees). Thus, proving (\ref{cex.2}) will require less computations
than proving (\ref{cex.1}), even though (\ref{cex.2}) is a slightly stronger result.

First of all, we are always going to compute in the first $5$ degrees of $R$.
Let us give bases for these degrees:%
\begin{align*}
R_{0}  &  =\left\langle 1\right\rangle ;\\
R_{1}  &  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right\rangle ;\\
R_{2}  &  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right\rangle \\
R_{3}  &  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right\rangle \\
R_{4}  &  =\left\langle
\begin{array}
[c]{c}%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\end{array}
\right\rangle .
\end{align*}
To be able to prove (\ref{cex.2}), we need to compute a basis for $\left(
\operatorname*{Inv}R+\left(  R^{+}\right)  ^{3}\right)  \cap R_{4}$. Since
everything is graded, we have $\left(  \operatorname*{Inv}R+\left(
R^{+}\right)  ^{3}\right)  \cap R_{4}=\left(  \operatorname*{Inv}R\right)
\cap R_{4}+\left(  R^{+}\right)  ^{3}\cap R_{4}$ (see a more detailed proof of
this further below). It is clear that $\left(  R^{+}\right)  ^{3}\cap R_{4}$
is the $k$-vector subspace of $R$ which is formed by products of more than $2$
disjoint trees, so a basis is obvious:%
\[
\left(  R^{+}\right)  ^{3}\cap R_{4}=\left\langle
\begin{array}
[c]{c}%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\end{array}
\right\rangle .
\]
It remains to compute $\left(  \operatorname*{Inv}R\right)  \cap R_{4}$. For
this we first notice that:

\begin{quote}
\textbf{Lemma 9.} Let $H$ be a Hopf algebra over a field $k$. Then,
$\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $ is a\textit{
}$k$-subalgebra of $H$.
\end{quote}

\textit{Proof of Lemma 9.} Notice that the antipode $S$ is an anti-algebra
homomorphism. Hence, $S\left(  1\right)  =1$, so that $S\left(  S\left(
1\right)  \right)  =S\left(  1\right)  =1$. Hence, $\left(  S^{2}%
-\operatorname*{id}\right)  \left(  1\right)  =\underbrace{S^{2}\left(
1\right)  }_{=S\left(  S\left(  1\right)  \right)  =1}%
-\underbrace{\operatorname*{id}\left(  1\right)  }_{=1}=1-1=0$, so that
$1\in\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $.

Let $a\in\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $ and
$b\in\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $. Then,
$S^{2}\left(  a\right)  -\underbrace{a}_{=\operatorname*{id}\left(  a\right)
}=S^{2}\left(  a\right)  -\operatorname*{id}\left(  a\right)  =\left(
S^{2}-\operatorname*{id}\right)  \left(  a\right)  =0$ (since $a\in
\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $), so that
$S^{2}\left(  a\right)  =a$. Similarly, $S^{2}\left(  b\right)  =b$.

We know that the antipode $S$ is an anti-algebra homomorphism. Hence,
$S\left(  ab\right)  =S\left(  b\right)  S\left(  a\right)  $ and $S\left(
S\left(  b\right)  S\left(  a\right)  \right)  =S\left(  S\left(  a\right)
\right)  S\left(  S\left(  b\right)  \right)  $. Thus,%
\[
S^{2}\left(  ab\right)  =S\left(  \underbrace{S\left(  ab\right)  }_{=S\left(
b\right)  S\left(  a\right)  }\right)  =S\left(  S\left(  b\right)  S\left(
a\right)  \right)  =\underbrace{S\left(  S\left(  a\right)  \right)  }%
_{=S^{2}\left(  a\right)  =a}\underbrace{S\left(  S\left(  b\right)  \right)
}_{=S^{2}\left(  b\right)  =b}=ab,
\]
so that $\left(  S^{2}-\operatorname*{id}\right)  \left(  ab\right)
=\underbrace{S^{2}\left(  ab\right)  }_{=ab}-\underbrace{\operatorname*{id}%
\left(  ab\right)  }_{=ab}=ab-ab=0$. In other words, $ab\in\operatorname*{Ker}%
\left(  S^{2}-\operatorname*{id}\right)  $.

So we have shown that any $a\in\operatorname*{Ker}\left(  S^{2}%
-\operatorname*{id}\right)  $ and $b\in\operatorname*{Ker}\left(
S^{2}-\operatorname*{id}\right)  $ satisfy $ab\in\operatorname*{Ker}\left(
S^{2}-\operatorname*{id}\right)  $. Combined with the fact that $1\in
\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $, this yields
that $\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $ is a
$k$-subalgebra of $H$ (since $\operatorname*{Ker}\left(  S^{2}%
-\operatorname*{id}\right)  $ is clearly a $k$-submodule of $H$). Lemma 9 is proven.

Next we notice that $R_{0}+R_{1}+R_{2}\subseteq\operatorname*{Ker}\left(
S^{2}-\operatorname*{id}\right)  $. This is very easy to check manually, but
actually is a particular case of something general:

\begin{quote}
\textbf{Lemma 10.} Let $H$ be a connected graded Hopf algebra over a field $k$
such that any $a\in H_{1}$ and $b\in H_{1}$ satisfy $ab=ba$. Then,
$H_{0}+H_{1}+H_{2}\subseteq\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}%
\right)  $.
\end{quote}

The proof of Lemma 10 is left to the reader; it can be applied to $H=R$ since
$R_{1}$ is one-dimensional.

Now we notice that $\operatorname*{Inv}R=R\cdot\left(  \left(  S^{2}%
-\operatorname*{id}\right)  \left(  R\right)  \right)  \cdot R$ (by the
definition of $\operatorname*{Inv}R$), so that%
\begin{align}
&  \left(  \operatorname*{Inv}R\right)  \cap R_{4}\nonumber\\
&  =\left(  R\cdot\left(  \left(  S^{2}-\operatorname*{id}\right)  \left(
R\right)  \right)  \cdot R\right)  \cap R_{4}=\sum_{\substack{\left(
i,j,\ell\right)  \in\mathbb{N}^{3};\\i+j+\ell=4}}R_{i}\cdot\left(  \left(
S^{2}-\operatorname*{id}\right)  \left(  R_{j}\right)  \right)  \cdot R_{\ell
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }R\text{ is graded}\right)
\nonumber\\
&  =\sum_{\substack{\left(  i,j,\ell\right)  \in\mathbb{N}^{3};\\i+j+\ell
=4;\\j>2}}R_{i}\cdot\left(  \left(  S^{2}-\operatorname*{id}\right)  \left(
R_{j}\right)  \right)  \cdot R_{\ell}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here we removed all terms with }j\leq2\text{ from the sum,}\\
\text{since }R_{0}+R_{1}+R_{2}\subseteq\operatorname*{Ker}\left(
S^{2}-\operatorname*{id}\right)
\end{array}
\right) \nonumber\\
&  =\left(  S^{2}-\operatorname*{id}\right)  \left(  R_{4}\right)  +\left(
S^{2}-\operatorname*{id}\right)  \left(  R_{3}\right)  \cdot R_{1}+R_{1}%
\cdot\left(  S^{2}-\operatorname*{id}\right)  \left(  R_{3}\right)  .
\label{cex.lm.1}%
\end{align}
We need now to compute $\left(  S^{2}-\operatorname*{id}\right)  \left(
R_{3}\right)  $ and $\left(  S^{2}-\operatorname*{id}\right)  \left(
R_{4}\right)  $.

\textit{Computation of }$\left(  S^{2}-\operatorname*{id}\right)  \left(
R_{3}\right)  $\textit{:} We know that%
\[
R_{3}=\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right\rangle .
\]
Hence, to compute $\left(  S^{2}-\operatorname*{id}\right)  \left(
R_{3}\right)  $, we need to apply $S^{2}-\operatorname*{id}$ to each of these
generators. But since%
\begin{equation}
\text{any product of elements of }R_{0}+R_{1}+R_{2}\text{ lies in
}\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  \label{cex.lm.2}%
\end{equation}
(by $R_{0}+R_{1}+R_{2}\subseteq\operatorname*{Ker}\left(  S^{2}%
-\operatorname*{id}\right)  $ and Lemma 9), it is clear that $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
$ and $%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$ all lie in $\operatorname*{Ker}\left(  S^{2}-\operatorname*{id}\right)  $,
so we only need to apply $S^{2}-\operatorname*{id}$ to $%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
$ and $%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
$ only.

It is easy to check from the recurrent definition of $S$ that%
\begin{align}
S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)   &  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
;\label{cex.S1}\\
S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
. \label{cex.S2}%
\end{align}
Now,%
\begin{align*}
S^{2}\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)   &  =S\left(  S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  \right)  =S\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{cex.S1})}\right) \\
&  =-S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\text{ is an anti-algebra
homomorphism}\right) \\
&  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  -\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{using already-known values of }S\right)
\\
&  =%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
.
\end{align*}
Hence, $\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  =0$. \ \ \ \footnote{This result should not surprise us in the least,
and we could actually have found it without any computation: It is easy to see
that the subalgebra of $R$ generated by ladders ordered from top to bottom (i.
e., trees of the form $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
$, $%
%TCIMACRO{\TeXButton{1/2,2/3,3/4}{\tdquatrecinq{4}{3}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{4}{3}{2}{1}%
%EndExpansion
$, $...$) is a Hopf subalgebra of $R$, and this Hopf subalgebra is
cocommutative, so it satisfies $S^{2}=\operatorname*{id}$. Thus, $\left(
S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  =0$.} \ Besides,%
\begin{align*}
S^{2}\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =S\left(  S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  \right)  =S\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{cex.S2})}\right) \\
&  =-S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\text{ is an anti-algebra
homomorphism}\right) \\
&  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  -\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{using already-known values of }S\right)
\\
&  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  +%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
,
\end{align*}
so that $\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  $.

As a result of this all,%
\begin{equation}
\left(  S^{2}-\operatorname*{id}\right)  \left(  R_{3}\right)  =\left\langle
\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle . \label{cex.S^2-id.3}%
\end{equation}


\textit{Computation of }$\left(  S^{2}-\operatorname*{id}\right)  \left(
R_{4}\right)  $\textit{:} We have%
\[
R_{4}=\left\langle
\begin{array}
[c]{c}%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
,\ \ \\%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\end{array}
\right\rangle .
\]
As in the case of $R_{3}$, we don't have to apply $S^{2}-\operatorname*{id}$
to each of these generators because of (\ref{cex.lm.2}). We only need to apply
$S^{2}-\operatorname*{id}$ to the five elements $%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
$. We notice that whenever $a$ and $b$ are two elements of a Hopf algebra $H$
with $a$ being primitive, then%
\begin{align*}
S^{2}\left(  ab\right)   &  =S\left(  \underbrace{S\left(  ab\right)
}_{=S\left(  b\right)  S\left(  a\right)  }\right)  =S\left(  S\left(
b\right)  S\left(  a\right)  \right)  =\underbrace{S\left(  S\left(  a\right)
\right)  }_{\substack{=a\\\text{(since }a\text{ is primitive, thus }S\left(
a\right)  =-a\text{, and }-a\\\text{is primitive again, so that }S\left(
-a\right)  =a\text{)}}}\underbrace{S\left(  S\left(  b\right)  \right)
}_{=S^{2}\left(  b\right)  }\\
&  =aS^{2}\left(  b\right)  ,
\end{align*}
so that $\left(  S^{2}-\operatorname*{id}\right)  \left(  ab\right)  =a\left(
S^{2}-\operatorname*{id}\right)  \left(  b\right)  $. Applying this to $a=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
$ and $b=%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
$, we obtain%
\[
\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\underbrace{\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  }_{=0}=0.
\]
Similarly,%
\begin{align*}
\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)   &  =0;\\
\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ;\\
\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)   &  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
.
\end{align*}


It is the fifth element, $%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
$, that makes us the most trouble. From%
\[
\Delta\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\]
and the recursive interpretation of $S\ast\operatorname*{id}=\eta
\circ\varepsilon$, we obtain%
\begin{align*}
0  &  =S\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \times1+\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
}\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+\underbrace{S\left(  1\right)  }_{=1}\times%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\\
&  =S\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \times1-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
,
\end{align*}
so that%
\begin{align}
S\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)   &  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
. \label{cex.3}%
\end{align}
Applying $S$ to this again (and using that $S$ is an anti-algebra
homomorphism), we get%
\begin{align*}
&  S^{2}\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right) \\
&  =S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  +S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  -S\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right) \\
&  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  +%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{using the known values }S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  =-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\text{, }S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\text{, }S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\text{,}\\
\text{and the values of }S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  \text{, }S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  \text{ and }S\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \text{ given by (\ref{cex.1}), (\ref{cex.2}) and (\ref{cex.3})}%
\end{array}
\right)  ,
\end{align*}
so that%
\[
\left(  S^{2}-\operatorname*{id}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  .
\]
Hence, altogether,%
\begin{align}
&  \left(  S^{2}-\operatorname*{id}\right)  \left(  R_{4}\right) \nonumber\\
&  =\left\langle
\begin{array}
[c]{c}%
0,\ \ 0,\ \
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \\
\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]
\end{array}
\right\rangle \nonumber\\
&  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle . \label{cex.S^2-id.4}%
\end{align}


Substituting (\ref{cex.S^2-id.3}) and (\ref{cex.S^2-id.4}) (and $R_{1}%
=\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right\rangle $) in (\ref{cex.lm.1}), we obtain%
\begin{align}
&  \left(  \operatorname*{Inv}R\right)  \cap R_{4}\nonumber\\
&  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\left\langle \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle \times\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right\rangle +\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right\rangle \times\left\langle \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle \nonumber\\
&  =\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle \label{cex.Inv}\\
&  =\underbrace{\left\langle
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  ,\ \ \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right\rangle }_{\subseteq\left(  R^{+}\right)  ^{3}\cap R_{4}}+\left\langle
\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle \nonumber\\
&  \subseteq\left(  R^{+}\right)  ^{3}\cap R_{4}+\left\langle \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle ,\nonumber
\end{align}
so that%
\begin{align}
&  \left(  \operatorname*{Inv}R\right)  \cap R_{4}+\left(  R^{+}\right)
^{3}\cap R_{4}\nonumber\\
&  =\left\langle \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle +\left(  R^{+}\right)  ^{3}\cap R_{4}. \label{cex.lm2}%
\end{align}
But since the ideals $\operatorname*{Inv}R$ and $\left(  R^{+}\right)  ^{3}$
are homogeneous, we have $\left(  \operatorname*{Inv}R\right)  \cap
R_{4}+\left(  R^{+}\right)  ^{3}\cap R_{4}=\left(  \left(  \operatorname*{Inv}%
R\right)  +\left(  R^{+}\right)  ^{3}\right)  \cap R_{4}$ (because the sum of
the intersections of two homogeneous ideals with $R_{4}$ equals the
intersection of their sum with $R_{4}$\ \ \ \ \footnote{\textit{Proof
(sketched).} The intersection of a homogeneous ideal with $R_{4}$ is the same
as the projection of this ideal on $R_{4}$. But the sum of the projections of
two ideals on $R_{4}$ clearly equals the projection of their sum on $R_{4}$,
qed.}). Hence, (\ref{cex.lm2}) becomes%
\begin{align}
&  \left(  \left(  \operatorname*{Inv}R\right)  +\left(  R^{+}\right)
^{3}\right)  \cap R_{4}\nonumber\\
&  =\left\langle \left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right\rangle +\left(  R^{+}\right)  ^{3}\cap R_{4}. \label{cex.lm3}%
\end{align}


We now take aim at (\ref{cex.2}) and compute $\left(  E\ast\overline
{S}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  $ modulo $\left(  R^{+}\right)  ^{3}$: Since%
\[
\Delta\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
,
\]
we have (using the fact that $\overline{S}\left(  x\right)  =S\left(
x\right)  $ for every $x\in R^{+}$) the following computation:%
\begin{align}
&  \left(  E\ast\overline{S}\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  =E\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \times\underbrace{\overline{S}\left(  1\right)  }_{=0}%
+\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  }_{=3%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
}\times\underbrace{\overline{S}\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}\times\underbrace{\overline{S}\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=1%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{\overline{S}\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
}+\underbrace{E\left(  1\right)  }_{=0\cdot1}\times\overline{S}\left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  =-3%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \nonumber\\
&  \equiv-3%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\operatorname{mod}\left(  R^{+}\right)  ^{3}. \label{cex.E*S.1}%
\end{align}


We want to apply $E\ast S$ to (\ref{cex.E*S.1}), but for this we need to
compute $\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  $, $\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  $ and $\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  $. This is a routine computation, but we can simplify it using the
following lemma:

\begin{quote}
\textbf{Lemma 11.} Let $H$ be a connected graded Hopf algebra over a field
$k$. We will use the notations of Definition 4. Let $x\in H_{1}$ and $y\in
H^{+}$.

\textbf{(a)} Then, $\left(  E\ast S\right)  \left(  xy\right)  =\left[
x,\left(  E\ast S\right)  \left(  y\right)  \right]  $.

\textbf{(b)} Besides, $\left(  E\ast S\right)  \left(  yx\right)  =y_{\left(
1\right)  }xS\left(  y_{\left(  2\right)  }\right)  $ using the sumfree
Sweedler notation.
\end{quote}

Note that Lemma 11 \textbf{(a)} is a particular case of Theorem 6 in [2]
(using Theorem 4 in [2]).

\textit{Proof of Lemma 11 (sketched).} We will use the sumfree Sweedler notation.

It is known enough that every element of $H_{1}$ is primitive (due to the
connectedness of $H$). Thus, in particular, $x$ is primitive, so that
$x_{\left(  1\right)  }\otimes x_{\left(  2\right)  }=x\otimes1+1\otimes x$.
Also, $x\in H_{1}$ yields $E\left(  x\right)  =1x=x$.

\textbf{(a)} We have%
\begin{align*}
\left(  E\ast S\right)  \left(  xy\right)   &  =E\left(  \left(  xy\right)
_{\left(  1\right)  }\right)  S\left(  \left(  xy\right)  _{\left(  2\right)
}\right)  =\underbrace{E\left(  x_{\left(  1\right)  }y_{\left(  1\right)
}\right)  }_{\substack{=E\left(  x_{\left(  1\right)  }\right)  y_{\left(
1\right)  }+x_{\left(  1\right)  }E\left(  y_{\left(  1\right)  }\right)
\\\text{(since }E\text{ is a derivation)}}}\underbrace{S\left(  x_{\left(
2\right)  }y_{\left(  2\right)  }\right)  }_{\substack{=S\left(  y_{\left(
2\right)  }\right)  S\left(  x_{\left(  2\right)  }\right)  \\\text{(since
}S\text{ is an anti-algebra homomorphism)}}}\\
&  =\left(  E\left(  x_{\left(  1\right)  }\right)  y_{\left(  1\right)
}+x_{\left(  1\right)  }E\left(  y_{\left(  1\right)  }\right)  \right)
S\left(  y_{\left(  2\right)  }\right)  S\left(  x_{\left(  2\right)  }\right)
\\
&  =E\left(  x_{\left(  1\right)  }\right)  \underbrace{y_{\left(  1\right)
}S\left(  y_{\left(  2\right)  }\right)  }_{=\left(  \operatorname*{id}\ast
S\right)  \left(  y\right)  =\varepsilon\left(  y\right)  1}S\left(
x_{\left(  2\right)  }\right)  +x_{\left(  1\right)  }E\left(  y_{\left(
1\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)  S\left(
x_{\left(  2\right)  }\right) \\
&  =E\left(  x_{\left(  1\right)  }\right)  S\left(  x_{\left(  2\right)
}\right)  \varepsilon\left(  y\right)  1+x_{\left(  1\right)  }E\left(
y_{\left(  1\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)
S\left(  x_{\left(  2\right)  }\right) \\
&  =\underbrace{E\left(  x\right)  }_{=x}\underbrace{S\left(  1\right)  }%
_{=1}\underbrace{\varepsilon\left(  y\right)  }_{\substack{=0\\\text{(since
}y\in H^{+}\text{)}}}1+\underbrace{E\left(  1\right)  }_{=0}S\left(  x\right)
\varepsilon\left(  y\right)  1\\
&  \ \ \ \ \ \ \ \ \ \ +xE\left(  y_{\left(  1\right)  }\right)  S\left(
y_{\left(  2\right)  }\right)  \underbrace{S\left(  1\right)  }_{=1}+1E\left(
y_{\left(  1\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)
\underbrace{S\left(  x\right)  }_{\substack{=-x\\\text{(since }x\text{ is
primitive)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }x_{\left(  1\right)  }\otimes
x_{\left(  2\right)  }=x\otimes1+1\otimes x\right) \\
&  =xE\left(  y_{\left(  1\right)  }\right)  S\left(  y_{\left(  2\right)
}\right)  -E\left(  y_{\left(  1\right)  }\right)  S\left(  y_{\left(
2\right)  }\right)  x=\left[  x,\underbrace{E\left(  y_{\left(  1\right)
}\right)  S\left(  y_{\left(  2\right)  }\right)  }_{=\left(  E\ast S\right)
\left(  y\right)  }\right]  =\left[  x,\left(  E\ast S\right)  \left(
y\right)  \right]  .
\end{align*}
Lemma 11 \textbf{(a)} is thus proven.

\textbf{(b)} We have%
\begin{align*}
\left(  E\ast S\right)  \left(  yx\right)   &  =E\left(  \left(  yx\right)
_{\left(  1\right)  }\right)  S\left(  \left(  yx\right)  _{\left(  2\right)
}\right)  =\underbrace{E\left(  y_{\left(  1\right)  }x_{\left(  1\right)
}\right)  }_{\substack{=E\left(  y_{\left(  1\right)  }\right)  x_{\left(
1\right)  }+y_{\left(  1\right)  }E\left(  x_{\left(  1\right)  }\right)
\\\text{(since }E\text{ is a derivation)}}}\underbrace{S\left(  y_{\left(
2\right)  }x_{\left(  2\right)  }\right)  }_{\substack{=S\left(  x_{\left(
2\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)  \\\text{(since
}S\text{ is an anti-algebra homomorphism)}}}\\
&  =\left(  E\left(  y_{\left(  1\right)  }\right)  x_{\left(  1\right)
}+y_{\left(  1\right)  }E\left(  x_{\left(  1\right)  }\right)  \right)
S\left(  x_{\left(  2\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)
\\
&  =E\left(  y_{\left(  1\right)  }\right)  \underbrace{x_{\left(  1\right)
}S\left(  x_{\left(  2\right)  }\right)  }_{\substack{=\left(
\operatorname*{id}\ast S\right)  \left(  x\right)  =\varepsilon\left(
x\right)  1=0\\\text{(since }x\in H_{1}\subseteq H^{+}\text{)}}}S\left(
y_{\left(  2\right)  }\right)  +y_{\left(  1\right)  }E\left(  x_{\left(
1\right)  }\right)  S\left(  x_{\left(  2\right)  }\right)  S\left(
y_{\left(  2\right)  }\right) \\
&  =y_{\left(  1\right)  }E\left(  x_{\left(  1\right)  }\right)  S\left(
x_{\left(  2\right)  }\right)  S\left(  y_{\left(  2\right)  }\right)
=y_{\left(  1\right)  }\underbrace{E\left(  x\right)  }_{=x}%
\underbrace{S\left(  1\right)  }_{=1}S\left(  y_{\left(  2\right)  }\right)
+y_{\left(  1\right)  }\underbrace{E\left(  1\right)  }_{=0}S\left(  x\right)
S\left(  y_{\left(  2\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }x_{\left(  1\right)  }\otimes
x_{\left(  2\right)  }=x\otimes1+1\otimes x\right) \\
&  =y_{\left(  1\right)  }xS\left(  y_{\left(  2\right)  }\right)  .
\end{align*}
Lemma 11 \textbf{(b)} is thus proven.

Back to work: First, let us compute $\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  $. This is where Lemma 11 doesn't help. Since $\Delta$ is an algebra
homomorphism,%
\begin{align*}
\Delta\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)   &  =\underbrace{\Delta\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}\times\underbrace{\Delta\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\\
&  =\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  \times\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right) \\
&  =%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ +%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\\
&  \ \ \ \ \ \ \ \ \ \ +%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
.
\end{align*}
Hence,%
\begin{align}
&  \left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right) \nonumber\\
&  =\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=4%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\times\underbrace{S\left(  1\right)  }_{=1}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=3%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=3%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=2%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=1%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{\substack{=S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \\=\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  \times\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{E\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=2%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=1%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{\substack{=S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  \\=\left(  -%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  \times\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }}+\underbrace{E\left(  1\right)  }_{=0}\times S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right) \nonumber\\
&  =2\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+2%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\nonumber\\
&  \equiv2\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}. \label{cex.E*S.E*S.1}%
\end{align}


Now, the easy parts: Using the sumfree Sweedler notation,%
\begin{align}
&  \left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right) \nonumber\\
&  =%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
_{\left(  1\right)  }\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
_{\left(  2\right)  }\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 11
\textbf{(b)}, applied to }H=R\text{, }x=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\text{ and }y=%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  =%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\underbrace{S\left(  1\right)  }_{=1}+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}+1\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  =-\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  +\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right]  \times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\equiv-\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}. \label{cex.E*S.E*S.2}%
\end{align}
Also,%
\begin{align*}
\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =\underbrace{E\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  }_{=3%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
}\times\underbrace{S\left(  1\right)  }_{=1}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\right)  }_{=2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}+\underbrace{E\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{=1%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
}\times\underbrace{S\left(
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}+\underbrace{E\left(  1\right)  }_{=0}\times S\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\Delta\left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  =%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\otimes1+%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\otimes%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
+1\otimes%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right) \\
&  =3%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times\left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right) \\
&  =3%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
,
\end{align*}
and now%
\begin{align}
\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)   &  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ \underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  }_{=3%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
}\right] \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma 11 \textbf{(a)}, applied to
}H=R\text{, }x=%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\text{ and }y=%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  =\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\ 3%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
+%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right] \nonumber\\
&  =3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -2\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right]  +\underbrace{\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right]  }_{=0}-\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right] \nonumber\\
&  =3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  -2\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right]  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right] \nonumber\\
&  \equiv3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}. \label{cex.E*S.E*S.4}%
\end{align}


Now, applying $E\ast S$ to (\ref{cex.E*S.1}), we get%
\begin{align}
&  \left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right) \nonumber\\
&  \equiv-3\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{\substack{\equiv-\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}\\\text{(by
(\ref{cex.E*S.E*S.2}))}}}-2\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{\substack{\equiv2\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}.\\\text{(by
(\ref{cex.E*S.E*S.1}))}}}-\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  }_{\substack{\equiv3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}\\\text{(by
(\ref{cex.E*S.E*S.4}))}}}\nonumber\\
&  \equiv-3\left(  -\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  \right)  -2\left(  2\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \right)  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right] \nonumber\\
&  =3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -4\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}. \label{cex.finale}%
\end{align}
This yields%
\[
\left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \equiv3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -4\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  \left(  R^{+}\right)  ^{3}\cap
R_{4}\right)
\]
(because both sides $\left(  \left(  E\ast S\right)  \circ\left(
E\ast\overline{S}\right)  \right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  $ and $3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -4\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  $ lie in $R_{4}$, and thus so does their difference).

Thus, $\left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  $ is \textit{not} in $\left(  \left(  \operatorname*{Inv}R\right)
+\left(  R^{+}\right)  ^{3}\right)  \cap R_{4}$ (according to (\ref{cex.lm3}%
)). Hence, $\left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline
{S}\right)  \right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  $ lies in $R_{4}$ (because the maps $E\ast S$ and $E\ast\overline{S}$
are graded) but not in $\left(  \left(  \operatorname*{Inv}R\right)  +\left(
R^{+}\right)  ^{3}\right)  \cap R_{4}$. Consequently, $\left(  \left(  E\ast
S\right)  \circ\left(  E\ast\overline{S}\right)  \right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \notin\operatorname*{Inv}R+\left(  R^{+}\right)  ^{3}$. This proves
(\ref{cex.2}). Thus, we have shown that $\left(  E\ast S\right)  \circ\left(
E\ast S\right)  \neq E\circ\left(  E\ast S\right)  $ in $R^{\prime}$.

\begin{center}
\textbf{\S 4. A shortcut}
\end{center}

The proof that $\left(  E\ast S\right)  \circ\left(  E\ast S\right)  \neq
E\circ\left(  E\ast S\right)  $ in $R^{\prime}$ is complete, but let us give
an alternative shortcut from (\ref{cex.E*S.1}) to (\ref{cex.finale}), without
requiring that much computation:

We will prove that%
\begin{equation}
\left(  E\ast S\right)  \left(  a\times b\right)  \equiv\deg b\cdot\left[
a,b\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}%
\ \ \ \ \ \ \ \ \ \ \text{for any two ordered trees }a\text{ and }b\text{.}
\label{shortcut.1}%
\end{equation}
(Here, $\deg b$ means the number of vertices of the tree $b$, or,
equivalently, the degree of $b$ in the graded Hopf algebra $R$.)

\textit{Proof of (\ref{shortcut.1}).} Let $t$ be an ordered tree. We first
notice that, from the definition of $\Delta$, it is pretty much clear that%
\[
\Delta\left(  t\right)  \equiv t\otimes1+1\otimes t\operatorname{mod}%
R^{+}\otimes R^{+}.
\]
Applying $\operatorname*{id}\otimes S$ to this congruence, we get%
\[
\left(  \operatorname*{id}\otimes S\right)  \left(  \Delta\left(  t\right)
\right)  \equiv t\otimes\underbrace{S\left(  1\right)  }_{=1}+1\otimes
S\left(  t\right)  =t\otimes1+1\otimes S\left(  t\right)  \operatorname{mod}%
\left(  \operatorname*{id}\otimes S\right)  \left(  R^{+}\otimes R^{+}\right)
.
\]
Since $\left(  \operatorname*{id}\otimes S\right)  \left(  R^{+}\otimes
R^{+}\right)  =R^{+}\otimes\underbrace{S\left(  R^{+}\right)  }_{\subseteq
R^{+}}\subseteq R^{+}\otimes R^{+}$, this becomes
\[
\left(  \operatorname*{id}\otimes S\right)  \left(  \Delta\left(  t\right)
\right)  \equiv t\otimes1+1\otimes S\left(  t\right)  \operatorname{mod}%
R^{+}\otimes R^{+}.
\]
Since $\mu\left(  R^{+}\otimes R^{+}\right)  =\left(  R^{+}\right)  ^{2}$, we
can apply $\mu$ to this congruence, and obtain
\[
\mu\left(  \left(  \operatorname*{id}\otimes S\right)  \left(  \Delta\left(
t\right)  \right)  \right)  \equiv t+S\left(  t\right)  \operatorname{mod}%
\left(  R^{+}\right)  ^{2}.
\]
Since (by the axioms of a Hopf algebra) we have $\mu\left(  \left(
\operatorname*{id}\otimes S\right)  \left(  \Delta\left(  t\right)  \right)
\right)  =\varepsilon\left(  t\right)  =0$, this becomes $0\equiv t+S\left(
t\right)  \operatorname{mod}\left(  R^{+}\right)  ^{2}$, so that $S\left(
t\right)  \equiv-t\operatorname{mod}\left(  R^{+}\right)  ^{2}$. We have thus
proven that%
\begin{equation}
S\left(  t\right)  \equiv-t\operatorname{mod}\left(  R^{+}\right)
^{2}\ \ \ \ \ \ \ \ \ \ \text{for every ordered tree }t. \label{shortcut.2}%
\end{equation}


Now, let $a$ and $b$ be two ordered trees. Then,%
\[
\Delta\left(  a\times b\right)  \equiv\left(  a\times b\right)  \otimes
1+1\otimes\left(  a\times b\right)  +a\otimes b+b\otimes a\operatorname{mod}%
\left(  \left(  R^{+}\right)  ^{2}\otimes R^{+}+R^{+}\otimes\left(
R^{+}\right)  ^{2}\right)
\]
(this is, again, easy to see from the definition of $\Delta$: for every
$\mathbf{v}\models V\left(  a\times b\right)  $, we have either $\mathbf{v}%
\in\left\{  V\left(  a\times b\right)  ,V\left(  a\right)  ,V\left(  b\right)
,\varnothing\right\}  $ or $Lea_{\mathbf{v}}F\otimes Roo_{\mathbf{v}}%
F\in\left(  R^{+}\right)  ^{2}\otimes R^{+}+R^{+}\otimes\left(  R^{+}\right)
^{2}$). Thus,%
\begin{align*}
\left(  E\otimes S\right)  \left(  \Delta\left(  a\times b\right)  \right)
&  \equiv\underbrace{E\left(  a\times b\right)  }_{\substack{=\deg\left(
a\times b\right)  \cdot a\times b\\=\left(  \deg a+\deg b\right)  \cdot
a\times b}}\otimes\underbrace{S\left(  1\right)  }_{=1}+\underbrace{E\left(
1\right)  }_{=0}\otimes S\left(  a\times b\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{E\left(  a\right)  }_{=\deg a\cdot
a}\otimes S\left(  b\right)  +\underbrace{E\left(  b\right)  }_{=\deg b\cdot
b}\otimes S\left(  a\right) \\
&  \equiv\left(  \deg a+\deg b\right)  \cdot\left(  a\times b\right)
\otimes1+\deg a\cdot a\otimes S\left(  b\right)  +\deg b\cdot b\otimes
S\left(  a\right) \\
&  \ \ \ \ \ \ \ \ \ \ \operatorname{mod}\left(  E\otimes S\right)  \left(
\left(  R^{+}\right)  ^{2}\otimes R^{+}+R^{+}\otimes\left(  R^{+}\right)
^{2}\right)  .
\end{align*}
Since $\left(  E\otimes S\right)  \left(  \left(  R^{+}\right)  ^{2}\otimes
R^{+}+R^{+}\otimes\left(  R^{+}\right)  ^{2}\right)  \subseteq\left(
R^{+}\right)  ^{2}\otimes R^{+}+R^{+}\otimes\left(  R^{+}\right)  ^{2}$ (very
easily seen), this becomes%
\begin{align*}
&  \left(  E\otimes S\right)  \left(  \Delta\left(  a\times b\right)  \right)
\\
&  \equiv\left(  \deg a+\deg b\right)  \cdot\left(  a\times b\right)
\otimes1+\deg a\cdot\underbrace{a\otimes S\left(  b\right)  }%
_{\substack{\equiv a\otimes\left(  -b\right)  \operatorname{mod}R^{+}%
\otimes\left(  R^{+}\right)  ^{2}\\\text{(since }a\in R^{+}\text{ and
}S\left(  b\right)  \equiv-b\operatorname{mod}\left(  R^{+}\right)  ^{2}\text{
(by (\ref{shortcut.2})))}}}\\
&  \ \ \ \ \ \ \ \ \ \ +\deg b\cdot\underbrace{b\otimes S\left(  a\right)
}_{\substack{\equiv b\otimes\left(  -a\right)  \operatorname{mod}R^{+}%
\otimes\left(  R^{+}\right)  ^{2}\\\text{(since }b\in R^{+}\text{ and
}S\left(  a\right)  \equiv-a\operatorname{mod}\left(  R^{+}\right)  ^{2}\text{
(by (\ref{shortcut.2})))}}}\\
&  \equiv\left(  \deg a+\deg b\right)  \cdot\left(  a\times b\right)
\otimes1+\deg a\cdot a\otimes\left(  -b\right)  +\deg b\cdot b\otimes\left(
-a\right) \\
&  \ \ \ \ \ \ \ \ \ \ \operatorname{mod}\left(  \left(  R^{+}\right)
^{2}\otimes R^{+}+R^{+}\otimes\left(  R^{+}\right)  ^{2}\right)  .
\end{align*}
Applying $\mu$ to this equation, and using $\mu\left(  \left(  R^{+}\right)
^{2}\otimes R^{+}+R^{+}\otimes\left(  R^{+}\right)  ^{2}\right)  =\left(
R^{+}\right)  ^{3}$, we obtain%
\begin{align*}
\mu\left(  \left(  E\otimes S\right)  \left(  \Delta\left(  a\times b\right)
\right)  \right)   &  \equiv\left(  \deg a+\deg b\right)  \cdot a\times b+\deg
a\cdot a\times\left(  -b\right)  +\deg b\cdot b\times\left(  -a\right) \\
&  =\deg b\cdot\left[  a,b\right]  \operatorname{mod}\left(  R^{+}\right)
^{3}.
\end{align*}
Since $\mu\left(  \left(  E\otimes S\right)  \left(  \Delta\left(  a\times
b\right)  \right)  \right)  =\left(  E\ast S\right)  \left(  a\times b\right)
$, this proves (\ref{shortcut.1}).

Now, applying $E\ast S$ to (\ref{cex.E*S.1}) (and using that $\left(  E\ast
S\right)  \left(  \left(  R^{+}\right)  ^{3}\right)  \subseteq\left(
R^{+}\right)  ^{3}$), we obtain%
\begin{align*}
&  \left(  \left(  E\ast S\right)  \circ\left(  E\ast\overline{S}\right)
\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/4,4/3}{\tdquatrecinq{3}{4}{2}{1}}}%
%BeginExpansion
\tdquatrecinq{3}{4}{2}{1}%
%EndExpansion
\right)  \\
&  \equiv\left(  E\ast S\right)  \left(  -3%
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
-2%
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
-%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  \\
&  =-3\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right)  }_{\substack{\equiv1\left[
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}\\\text{(after
(\ref{shortcut.1}))}}}-2\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right)  }_{\substack{\equiv2\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}\\\text{(after
(\ref{shortcut.1}))}}}-\underbrace{\left(  E\ast S\right)  \left(
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\times%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right)  }_{\substack{\equiv3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}\\\text{(after
(\ref{shortcut.1}))}}}\\
&  \equiv-3\underbrace{\left[
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
,%
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
\right]  }_{=-\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  }-4\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \\
&  =3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/2,2/3}{\tdtroisdeux{3}{2}{1}}}%
%BeginExpansion
\tdtroisdeux{3}{2}{1}%
%EndExpansion
\right]  -4\left[
%TCIMACRO{\TeXButton{1/2}{\tddeux{2}{1}}}%
%BeginExpansion
\tddeux{2}{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{2/1}{\tddeux{1}{2}}}%
%BeginExpansion
\tddeux{1}{2}%
%EndExpansion
\right]  -3\left[
%TCIMACRO{\TeXButton{1}{\tdun{1}}}%
%BeginExpansion
\tdun{1}%
%EndExpansion
,\
%TCIMACRO{\TeXButton{1/3,3/2}{\tdtroisdeux{2}{3}{1}}}%
%BeginExpansion
\tdtroisdeux{2}{3}{1}%
%EndExpansion
\right]  \operatorname{mod}\left(  R^{+}\right)  ^{3}.
\end{align*}
Hence, (\ref{cex.finale}) is proven once again.

\begin{center}
\textbf{References}
\end{center}

[1] Loic Foissy, \textit{Ordered forests and parking functions},
arXiv:1007.1547v3.\newline\texttt{http://arxiv.org/abs/1007.1547v3}

[2] Fr\'{e}d\'{e}ric Patras, Christophe Reutenauer, \textit{On Dynkin and
Klyachko idempotents in graded bialgebras}.\newline%
\texttt{http://www-irma.u-strasbg.fr/annexes/publications/pdf/01029.pdf}

[3] Darij Grinberg, \textit{MathOverflow question \#84345}.\newline%
\texttt{http://mathoverflow.net/questions/84345}


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