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\begin{document}

\begin{center}
\textbf{Functional equations and Lie algebras}

\textit{Emanuela Petracci}

2001/2002

\textbf{Errata and questions by Darij Grinberg - I}

\bigskip
\end{center}

This is a list of errors in Emanuela Petracci's thesis \textquotedblleft
Functional equations and Lie algebras\textquotedblright\ I found while reading
parts of it. The word \textquotedblleft you\textquotedblright\ always refers
to the author of the thesis.

Despite the many errors, the thesis is a masterpiece of algebra. It provides
(among other things) a proof of the Poincar\'{e}-Birkhoff-Witt for
$\mathbb{Q}$-algebras which does not require the ground ring to be a field (so
\textquotedblleft for $\mathbb{Q}$-algebras\textquotedblright\ merely means
that the ground ring is a commutative $\mathbb{Q}$-algebra). Among several
such proofs (all of which are highly nontrivial), the one given in this thesis
is probably the most conceptual one.

\section*{General errors}

\begin{itemize}
\item There seems to be a bug in the style you are using: While there are no
dots after \textquotedblleft Convention\textquotedblright\ and
\textquotedblleft Remark\textquotedblright\ and the likes (for example,
\textquotedblleft Convention 1.1.1\textquotedblright\ and \textquotedblleft
Remark 1.1.1\textquotedblright), there \textit{are} dots after
\textquotedblleft Definition\textquotedblright\ (for instance,
\textquotedblleft Definition. 1.1.1\textquotedblright).

\item The word \textquotedblleft verify\textquotedblright\ is misused as a
synonym for \textquotedblleft satisfy\textquotedblright\ throughout the thesis
(a mistake common of Francophone authors).
\end{itemize}

\section*{Chapter 1}

\begin{itemize}
\item \textbf{Page 9, \S 1.1:} Replace \textquotedblleft
not-zero\textquotedblright\ by \textquotedblleft non-zero\textquotedblright.

\item \textbf{Definition 1.1.1:} Replace \textquotedblleft$m,n\in
M$\textquotedblright\ by \textquotedblleft$m\in M$\textquotedblright.

\item \textbf{Definition 1.1.2:} Replace \textquotedblleft equipped
of\textquotedblright\ by \textquotedblleft equipped with'. (This mistake
occurs in many places throughout the text.)

\item \textbf{Between Definition 1.1.2 and Notation 1.1.1:} Remove the
\textquotedblleft and $\alpha\in\mathbb{K}$\textquotedblright\ part.

\item \textbf{Example 1.1.1:} Replace \textquotedblleft the set formal series
in $s$\textquotedblright\ by \textquotedblleft the set of formal series in
$z$\textquotedblright.

\item \textbf{Example 1.1.2 b):} The last comma in \textquotedblleft$\left\{
v\otimes w;v\in M,w\in N,\right\}  $\textquotedblright\ is misplaced -- it
should be outside the brackets.

\item You do some kind of introduction to superalgebra in Chapter 1. If you
want this to be self-contained, I think a definition of the notion of the
tensor product of two superalgebras would be in place somewhere in \S 1.1: you
use this notion later, and you never define it, although you define much more
basic notions (like Example 1.1.2).

\item \textbf{Example 1.1.2 c):} In \textquotedblleft$T\left(  M\right)
:=\mathbb{K}+\left(  M\otimes M\right)  +\left(  M\otimes M\otimes M\right)
+\cdots$\textquotedblright, you forgot the $M$ addend.

\item \textbf{Between (1.8) and (1.9):} Replace \textquotedblleft%
$X_{1},..,X_{n}\in M$\textquotedblright\ by \textquotedblleft$X_{1}%
,...,X_{n}\in M$\textquotedblright.

\item \textbf{Second absatz of page 12:} Remove the ``of'' from ``Because of
$S\left(  M\right)  $ is a coalgebra''. Also, the ``coalgebra'' here should
probably be a ``cocommutative coalgebra''.

\item \textbf{Remark 1.2.1:} It wouldn't hurt to explicitly remind the reader
here that $X$ denotes the ``constant function'' ($1\mapsto X$, $S^{n}\left(
M\right)  \rightarrow\left\{  0\right\}  $ for $n\neq0$).

\item \textbf{Page 12, one line below remark 1.2.1:} ``\textit{formal vectors
field}'' should be ``\textit{formal vector fields}''.

\item \textbf{Page 12, one line below Definition 1.2.1:} A closing parenthesis
was omitted in \textquotedblleft$P(S\left(  M\right)  $\textquotedblright.

\item \textbf{Proof of Lemma 1.3.1:} This is correct, but I don't understand
why you require $k\geq1$ all the time. Wouldn't $k\geq0$ be completely enough?

\item \textbf{Remark 1.3.1:} Here and in the following, when you write
\textquotedblleft$p\left(  X_{1}+\cdots+X_{n}\right)  $\textquotedblright, you
actually mean $p\left(  X_{1}\right)  +\cdots+p\left(  X_{n}\right)  $ (or,
what is the same, $p\left(  X_{1}\cdots X_{n}\right)  $). This appears so
often in your paper that I am wondering whether it is some standard abuse of
notation, or I am blind?

\item \textbf{Remark 1.3.1:} Replace the \textquotedblleft$:=$%
\textquotedblright\ by \textquotedblleft$=$\textquotedblright\ in
\textquotedblleft$\left(  \operatorname{ad}x\right)  ^{0}\left(  Y\right)
:=Y\in\mathfrak{g}_{x}$\textquotedblright. This is not a definition of
$\left(  \operatorname{ad}x\right)  ^{0}\left(  Y\right)  $; it is already defined.

\item \textbf{First line of page 14:} You write: \textquotedblleft As a
consequence of the last remark, we can define\textquotedblright. This is
right, but there is no need to use the last remark here. A simpler way to
check that $q\left(  \operatorname*{ad}x\right)  \left(  Y\right)  $ is
well-defined is the following: For every $m\in\mathbb{N}$, let $S_{\leq
m}\left(  \mathfrak{g}\right)  $ define the $\mathbb{K}$-subsupermodule
$\bigoplus\limits_{i=0}^{m}S^{i}\left(  \mathfrak{g}\right)  $ of $S\left(
\mathfrak{g}\right)  $. For every $m\in\mathbb{N}$, let $\mathfrak{g}_{x}%
^{>m}$ denote the $\mathbb{K}$-subsupermodule $\left\{  f\in\mathfrak{g}%
_{x}\ \mid\ f\left(  S_{\leq m}\left(  \mathfrak{g}\right)  \right)
=0\right\}  $ of $\mathfrak{g}_{x}$. Then, it is easy to show that
$x\in\mathfrak{g}_{x}^{>0}$, but every $m\in\mathbb{N}$ satisfies $\left(
\operatorname*{ad}x\right)  \left(  \mathfrak{g}_{x}^{>m}\right)
\subseteq\mathfrak{g}_{x}^{>m+1}$. As a consequence, for every $Y\in
\mathfrak{g}$, every sufficiently high $m\in\mathbb{N}$ satisfies $\left(
\operatorname*{ad}x\right)  ^{m}\left(  Y\right)  =0$, and thus $q\left(
\operatorname*{ad}x\right)  \left(  Y\right)  $ is well-defined.

\item \textbf{Theorem 1.3.1:} The \textquotedblleft$\left(  \dfrac{q\left(
t+u\right)  -q\left(  u\right)  }{t}:\left[  Y,Z\right]  \right)
$\textquotedblright\ should be \newline$\left(  \dfrac{q\left(  t+u\right)
-q\left(  u\right)  }{t}:\left[  Y,Z\right]  \right)  _{x}$ (with an $x$ index).

\item \textbf{Proof of Theorem 1.3.1:} Three typos in the computation:\newline%
- In the first line of the computation, \textquotedblleft$\operatorname{ad}%
x)^{k}$\textquotedblright\ should be \textquotedblleft$\left(
\operatorname{ad}x\right)  ^{k}$\textquotedblright.\newline- In the second
line of the computation, \textquotedblright$\left(  \left(  \operatorname{ad}%
x\right)  ^{k}\right)  \left(  Z\right)  $\textquotedblright\ should be
\textquotedblleft$\left(  \left(  \operatorname{ad}x\right)  ^{k}\left(
Z\right)  \right)  $\textquotedblright.\newline- In the third line of the
computation, \textquotedblleft$\left(  u^{k}:\left[  Y,Z\right]  \right)
$\textquotedblright\ should be \textquotedblleft$\left(  u^{k}:\left[
Y,Z\right]  \right)  _{x}$\textquotedblright.
\end{itemize}

\section*{Chapter 2}

\begin{itemize}
\item \textbf{Page 15, one line below Remark 2.1.2:} You write:
\textquotedblleft$\Phi^{a}:=id\ast\varphi^{a}\equiv\operatorname{Mult}%
\circ\left(  1\otimes\varphi^{a}\right)  \circ\Delta$\textquotedblright. The
$1$ here stands for $id$; maybe it would be better to just call it $id$ (lest
it be confused with the neutral element with respect to convolution).

\item \textbf{Page 16, Lemma 2.1.1:} It might be helpful to explain how
expressions like \textquotedblleft$\varphi^{a}\ast Y$\textquotedblright\ are
to be understood. (As far as I understand, in the expression \textquotedblleft%
$\varphi^{a}\ast Y$\textquotedblright, the terms $\varphi^{a}$ and $Y$ are
understood to mean the maps $S\left(  \mathfrak{g}\right)  \overset{\varphi
^{a}}{\longrightarrow}\mathfrak{g}\overset{\text{inclusion}}{\longrightarrow
}S\left(  \mathfrak{g}\right)  $ and $S\left(  \mathfrak{g}\right)
\overset{Y}{\longrightarrow}\mathfrak{g}\overset{\text{inclusion}%
}{\longrightarrow}S\left(  \mathfrak{g}\right)  $, respectively.)

\item \textbf{Proof of Lemma 2.1.1 ii):} I fear I don't understand this proof,
although I suspect the problem is on my side and not on that of the
proof's\footnote{I have troubles understanding the equation \textquotedblleft%
$\Phi_{\mathfrak{g}}^{a}\circ\Psi_{\mathfrak{g}}^{b}=\Phi_{\left(
\mathfrak{g}_{x}\right)  _{y}}^{a}\circ\left(  id\ast\psi\left(
\operatorname*{ad}y\right)  \left(  b\right)  \right)  \mid_{S\left(
\mathfrak{g}\right)  }=\Phi_{\left(  \mathfrak{g}_{x}\right)  _{y}}^{a}%
\circ\psi\left(  \operatorname*{ad}y\right)  \left(  b\right)  ^{L}%
\mid_{S\left(  \mathfrak{g}\right)  }$\textquotedblright. (It is not clear to
me how to interpret the term $\psi\left(  \operatorname*{ad}y\right)  \left(
b\right)  $ -- as an element of $\left(  \mathfrak{g}_{x}\right)  _{y}$
regarded as a constant map $S\left(  \left(  \mathfrak{g}_{x}\right)
_{y}\right)  \rightarrow\left(  \mathfrak{g}_{x}\right)  _{y}$, or as a
(non-constant) map $S\left(  \mathfrak{g}_{x}\right)  \rightarrow
\mathfrak{g}_{x}$ -- in order for both equality signs to be valid.)}.\newline
Anyway, here is a more down-to-earth proof of Lemma 2.1.1 \textit{ii)}%
:\newline\textit{Proof of Lemma 2.1.1 ii):} In the following, we are going to
use the sumfree Sweedler notation for the comultiplication on $S\left(
\mathfrak{g}\right)  $. Also we will assume that all vectors are even, since I
don't want to struggle with the minus signs. I am pretty sure that the general
case can be proven analogously.\newline We start with some straightforward
observations:\newline\textit{Observation 1:} Every $\alpha\in S\left(
\mathfrak{g}\right)  $ and every $Y\in\mathfrak{g}$ satisfy $\left[
x,Y\right]  \left(  \alpha\right)  =\left[  x\left(  \alpha\right)  ,Y\right]
$. (Here, on the left hand side, $Y$ denotes the constant map $Y\in
\mathfrak{g}_{x}$, as usual.)\newline\textit{Proof of Observation 1:} The
constant map $Y\in\mathfrak{g}_{x}=\operatorname*{Hom}\left(  S\left(
\mathfrak{g}\right)  ,\mathfrak{g}\right)  $ maps every $\beta\in S\left(
\mathfrak{g}\right)  $ to $\varepsilon\left(  \beta\right)  Y\in\mathfrak{g}$.
Thus,%
\begin{align*}
\left[  x,Y\right]  \left(  \alpha\right)   &  =\left[  x\left(
\alpha_{\left(  1\right)  }\right)  ,\underbrace{Y\left(  \alpha_{\left(
2\right)  }\right)  }_{=\varepsilon\left(  \alpha_{\left(  2\right)  }\right)
Y}\right]  =\left[  x\left(  \alpha_{\left(  1\right)  }\right)
,\varepsilon\left(  \alpha_{\left(  2\right)  }\right)  Y\right] \\
&  =\left[  x\left(  \underbrace{\alpha_{\left(  1\right)  }\varepsilon\left(
\alpha_{\left(  2\right)  }\right)  }_{=\alpha}\right)  ,Y\right]  =\left[
x\left(  \alpha\right)  ,Y\right]  .
\end{align*}
This proves Observation 1.\newline\textit{Observation 2:} Every $\ell
\in\mathbb{N}$, $b\in\mathfrak{g}$ and $\alpha\in S\left(  \mathfrak{g}%
\right)  $ satisfy%
\[
\left(  \left(  \operatorname*{ad}x\right)  \left(  \left(  \operatorname*{ad}%
x\right)  ^{\ell}\left(  b\right)  \left(  \alpha_{\left(  2\right)  }\right)
\right)  \right)  \left(  \alpha_{\left(  1\right)  }\right)  =\left(
\operatorname*{ad}x\right)  ^{\ell+1}\left(  b\right)  \left(  \alpha\right)
.
\]
\newline\textit{Proof of Observation 2:} We have%
\begin{align*}
&  \left(  \underbrace{\left(  \operatorname*{ad}x\right)  \left(  \left(
\operatorname*{ad}x\right)  ^{\ell}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right)  }_{=\left[  x,\left(  \operatorname*{ad}x\right)
^{\ell}\left(  b\right)  \left(  \alpha_{\left(  2\right)  }\right)  \right]
}\right)  \left(  \alpha_{\left(  1\right)  }\right) \\
&  =\left[  x,\left(  \operatorname*{ad}x\right)  ^{\ell}\left(  b\right)
\left(  \alpha_{\left(  2\right)  }\right)  \right]  \left(  \alpha_{\left(
1\right)  }\right)  =\left[  x\left(  \alpha_{\left(  1\right)  }\right)
,\left(  \operatorname*{ad}x\right)  ^{\ell}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  \right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Observation 1, applied to }\alpha_{\left(  1\right)  }\text{ and
}\left(  \operatorname*{ad}x\right)  ^{\ell}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right) \\
\text{instead of }\alpha\text{ and }Y
\end{array}
\right) \\
&  =\underbrace{\left[  x,\left(  \operatorname*{ad}x\right)  ^{\ell}\left(
b\right)  \right]  }_{=\left(  \operatorname*{ad}x\right)  ^{\ell+1}\left(
b\right)  }\left(  \alpha\right)  =\left(  \operatorname*{ad}x\right)
^{\ell+1}\left(  b\right)  \left(  \alpha\right)  ,
\end{align*}
thus proving Observation 2.\newline\textit{Observation 3:} Every
$q\in\mathbb{N}$, $k\in\mathbb{N}$, $b\in\mathfrak{g}$ and $\alpha\in S\left(
\mathfrak{g}\right)  $ satisfy%
\[
\left(  \left(  \operatorname*{ad}x\right)  ^{q}\left(  \left(
\operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right)  \right)  \left(  \alpha_{\left(  1\right)
}\right)  =\left(  \operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \left(
\alpha\right)  .
\]
\newline\textit{Proof of Observation 3:} We prove Observation 3 by induction
over $q$. The induction base is the case when $q=0$; this case is easy (it
reduces to showing that $\left(  \left(  \operatorname*{ad}x\right)
^{k}\left(  b\right)  \left(  \alpha_{\left(  2\right)  }\right)  \right)
\left(  \alpha_{\left(  1\right)  }\right)  =\left(  \operatorname*{ad}%
x\right)  ^{k}\left(  b\right)  \left(  \alpha\right)  $, but this is clear
since $\left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  $ is a constant map and thus satisfies%
\begin{align*}
\left(  \left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  \right)  \left(  \alpha_{\left(  1\right)
}\right)   &  =\varepsilon\left(  \alpha_{\left(  1\right)  }\right)
\cdot\left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right) \\
&  =\left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\underbrace{\varepsilon\left(  \alpha_{\left(  1\right)  }\right)
\alpha_{\left(  2\right)  }}_{=\alpha}\right)  =\left(  \operatorname*{ad}%
x\right)  ^{k}\left(  b\right)  \left(  \alpha\right)
\end{align*}
). For the induction step, we assume that some $q\in\mathbb{N}$ satisfies%
\begin{equation}
\left(  \left(  \operatorname*{ad}x\right)  ^{q}\left(  \left(
\operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right)  \right)  \left(  \alpha_{\left(  1\right)
}\right)  =\left(  \operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \left(
\alpha\right)  \label{pf1}%
\end{equation}
for all $\alpha\in S\left(  \mathfrak{g}\right)  $, and try to prove that%
\[
\left(  \left(  \operatorname*{ad}x\right)  ^{q+1}\left(  \left(
\operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right)  \right)  \left(  \alpha_{\left(  1\right)
}\right)  =\left(  \operatorname*{ad}x\right)  ^{q+1+k}\left(  b\right)
\left(  \alpha\right)
\]
for all $\alpha\in S\left(  \mathfrak{g}\right)  $. But this follows from%
\begin{align*}
&  \left(  \underbrace{\left(  \operatorname*{ad}x\right)  ^{q+1}\left(
\left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  \right)  }_{\substack{=\left(
\operatorname*{ad}x\right)  \left(  \left(  \operatorname*{ad}x\right)
^{q}\left(  \left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  \right)  \right)  \\=\left[  x,\left(
\operatorname*{ad}x\right)  ^{q}\left(  \left(  \operatorname*{ad}x\right)
^{k}\left(  b\right)  \left(  \alpha_{\left(  2\right)  }\right)  \right)
\right]  }}\right)  \left(  \alpha_{\left(  1\right)  }\right) \\
&  =\left[  x,\left(  \operatorname*{ad}x\right)  ^{q}\left(  \left(
\operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right)  \right]  \left(  \alpha_{\left(  1\right)
}\right) \\
&  =\left[  x\left(  \left(  \alpha_{\left(  1\right)  }\right)  _{\left(
1\right)  }\right)  ,\left(  \left(  \operatorname*{ad}x\right)  ^{q}\left(
\left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\alpha_{\left(  2\right)  }\right)  \right)  \right)  \left(  \left(
\alpha_{\left(  1\right)  }\right)  _{\left(  2\right)  }\right)  \right] \\
&  =\left[  x\left(  \alpha_{\left(  1\right)  }\right)  ,\underbrace{\left(
\left(  \operatorname*{ad}x\right)  ^{q}\left(  \left(  \operatorname*{ad}%
x\right)  ^{k}\left(  b\right)  \left(  \left(  \alpha_{\left(  2\right)
}\right)  _{\left(  2\right)  }\right)  \right)  \right)  \left(  \left(
\alpha_{\left(  2\right)  }\right)  _{\left(  1\right)  }\right)
}_{\substack{=\left(  \operatorname*{ad}x\right)  ^{q+k}\left(  b\right)
\left(  \alpha_{\left(  2\right)  }\right)  \\\text{(by (\ref{pf1}), applied
to }\alpha_{\left(  2\right)  }\text{ instead of }\alpha\text{)}}}\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by coassociativity}\right) \\
&  =\left[  x\left(  \alpha_{\left(  1\right)  }\right)  ,\left(
\operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \left(  \alpha_{\left(
2\right)  }\right)  \right] \\
&  =\underbrace{\left[  x,\left(  \operatorname*{ad}x\right)  ^{q+k}\left(
b\right)  \right]  }_{=\left(  \operatorname*{ad}x\right)  ^{\left(
q+k\right)  +1}\left(  b\right)  =\left(  \operatorname*{ad}x\right)
^{q+1+k}\left(  b\right)  }\left(  \alpha\right)  =\left(  \operatorname*{ad}%
x\right)  ^{q+1+k}\left(  b\right)  \left(  \alpha\right)  .
\end{align*}
Thus, Observation 3 is proven.\newline\textit{Observation 4:} Every
$f\in\operatorname*{Hom}\left(  S\left(  \mathfrak{g}\right)  ,S\left(
\mathfrak{g}\right)  \right)  $, every $Y\in\mathfrak{g}$ and every $\alpha\in
S\left(  \mathfrak{g}\right)  $ satisfy%
\[
\left(  f\ast Y\right)  \left(  \alpha\right)  =f\left(  \alpha\right)  Y
\]
(where the expression $Y$ in \textquotedblleft$f\ast Y$\textquotedblright\ is
regarded as a map $S\left(  \mathfrak{g}\right)  \rightarrow S\left(
\mathfrak{g}\right)  $ by first considering it as a constant map $S\left(
\mathfrak{g}\right)  \rightarrow\mathfrak{g}$ and then composing it with the
inclusion map $\mathfrak{g}\rightarrow S\left(  \mathfrak{g}\right)
$).\newline\textit{Proof of Observation 4:} We have%
\begin{align*}
\left(  f\ast Y\right)  \left(  \alpha\right)   &  =f\left(  \alpha_{\left(
1\right)  }\right)  \underbrace{Y\left(  \alpha_{\left(  2\right)  }\right)
}_{\substack{=Y\varepsilon\left(  \alpha_{\left(  2\right)  }\right)
\\\text{(by the definition}\\\text{of the constant map }Y\text{)}}}=f\left(
\alpha_{\left(  1\right)  }\right)  Y\varepsilon\left(  \alpha_{\left(
2\right)  }\right) \\
&  =f\left(  \underbrace{\alpha_{\left(  1\right)  }\varepsilon\left(
\alpha_{\left(  2\right)  }\right)  }_{=\alpha}\right)  Y=f\left(
\alpha\right)  Y.
\end{align*}
This proves Observation 4.\newline Now let us prove Lemma 2.1.1 \textit{ii)}:
We want to show that%
\[
\Phi^{a}\circ\Psi^{b}=id\ast\left(  \varphi^{a}\ast\psi^{b}-\left(
\dfrac{\varphi\left(  t+u\right)  -\varphi\left(  t\right)  }{u}\psi\left(
u\right)  :\left[  a,b\right]  \right)  _{x}\right)  .
\]
In order to do that, it is enough to show that every $\alpha\in S\left(
\mathfrak{g}\right)  $ satisfies%
\[
\left(  \Phi^{a}\circ\Psi^{b}\right)  \left(  \alpha\right)  =\left(
id\ast\left(  \varphi^{a}\ast\psi^{b}-\left(  \dfrac{\varphi\left(
t+u\right)  -\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[
a,b\right]  \right)  _{x}\right)  \right)  \left(  \alpha\right)  .
\]
Since%
\begin{align*}
&  \left(  \Phi^{a}\circ\Psi^{b}\right)  \left(  \alpha\right) \\
&  =\Phi^{a}\left(  \underbrace{\Psi^{b}\left(  \alpha\right)  }_{=\left(
id\ast\psi^{b}\right)  \left(  \alpha\right)  =\alpha_{\left(  1\right)  }%
\psi^{b}\left(  \alpha_{\left(  2\right)  }\right)  =\left(  \psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right)  ^{L}\left(  \alpha_{\left(
1\right)  }\right)  }\right)  =\Phi^{a}\left(  \left(  \psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right)  ^{L}\left(  \alpha_{\left(
1\right)  }\right)  \right) \\
&  =\left(  \Phi^{a}\circ\left(  \psi^{b}\left(  \alpha_{\left(  2\right)
}\right)  \right)  ^{L}\right)  \left(  \alpha_{\left(  1\right)  }\right) \\
&  =\left(  id\ast\left(  \varphi^{a}\ast\left(  \psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right)  -\left(  \dfrac{\varphi\left(
t+u\right)  -\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right]  \right)  _{x}\right)  \right)
\left(  \alpha_{\left(  1\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since Lemma 2.1.1 \textit{i)} (applied to }Y=\psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \text{) yields}\\
\Phi^{a}\circ\left(  \psi^{b}\left(  \alpha_{\left(  2\right)  }\right)
\right)  ^{L}\\
=id\ast\left(  \varphi^{a}\ast\left(  \psi^{b}\left(  \alpha_{\left(
2\right)  }\right)  \right)  -\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \alpha_{\left(
2\right)  }\right)  \right]  \right)  _{x}\right)
\end{array}
\right) \\
&  =\underbrace{\left(  id\ast\varphi^{a}\ast\left(  \psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right)  \right)  \left(  \alpha_{\left(
1\right)  }\right)  }_{\substack{=\left(  id\ast\varphi^{a}\right)  \left(
\alpha_{\left(  1\right)  }\right)  \cdot\psi^{b}\left(  \alpha_{\left(
2\right)  }\right)  \\\text{(by Observation 4, applied to}\\id\ast\varphi
^{a}\text{, }\psi^{b}\left(  \alpha_{\left(  2\right)  }\right)  \text{ and
}\alpha_{\left(  1\right)  }\\\text{instead of }f\text{, }Y\text{ and }%
\alpha\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\left(  id\ast\left(  \dfrac
{\varphi\left(  t+u\right)  -\varphi\left(  t\right)  }{u}:\left[  a,\psi
^{b}\left(  \alpha_{\left(  2\right)  }\right)  \right]  \right)  _{x}\right)
\left(  \alpha_{\left(  1\right)  }\right)  }_{=\left(  \alpha_{\left(
1\right)  }\right)  _{\left(  1\right)  }\left(  \dfrac{\varphi\left(
t+u\right)  -\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(
\alpha_{\left(  2\right)  }\right)  \right]  \right)  _{x}\left(  \left(
\alpha_{\left(  1\right)  }\right)  _{\left(  2\right)  }\right)  }\\
&  =\underbrace{\left(  id\ast\varphi^{a}\right)  \left(  \alpha_{\left(
1\right)  }\right)  \cdot\psi^{b}\left(  \alpha_{\left(  2\right)  }\right)
}_{=\left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\left(  \alpha_{\left(  1\right)
}\right)  _{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \alpha_{\left(
2\right)  }\right)  \right]  \right)  _{x}\left(  \left(  \alpha_{\left(
1\right)  }\right)  _{\left(  2\right)  }\right)  }_{\substack{=\alpha
_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)  -\varphi\left(
t\right)  }{u}:\left[  a,\psi^{b}\left(  \alpha_{\left(  3\right)  }\right)
\right]  \right)  _{x}\left(  \alpha_{\left(  2\right)  }\right)
\\=\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \left(
\alpha_{\left(  2\right)  }\right)  _{\left(  2\right)  }\right)  \right]
\right)  _{x}\left(  \left(  \alpha_{\left(  2\right)  }\right)  _{\left(
1\right)  }\right)  }}\\
&  =\left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)
-\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \left(
\alpha_{\left(  2\right)  }\right)  _{\left(  2\right)  }\right)  \right]
\right)  _{x}\left(  \left(  \alpha_{\left(  2\right)  }\right)  _{\left(
1\right)  }\right)
\end{align*}
and%
\begin{align*}
&  \left(  id\ast\left(  \varphi^{a}\ast\psi^{b}-\left(  \dfrac{\varphi\left(
t+u\right)  -\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[
a,b\right]  \right)  _{x}\right)  \right)  \left(  \alpha\right) \\
&  =\left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)
-\underbrace{\left(  id\ast\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[  a,b\right]
\right)  _{x}\right)  \left(  \alpha\right)  }_{=\alpha_{\left(  1\right)
}\left(  \dfrac{\varphi\left(  t+u\right)  -\varphi\left(  t\right)  }{u}%
\psi\left(  u\right)  :\left[  a,b\right]  \right)  _{x}\left(  \alpha
_{\left(  2\right)  }\right)  }\\
&  =\left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)
-\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[  a,b\right]
\right)  _{x}\left(  \alpha_{\left(  2\right)  }\right)  ,
\end{align*}
this rewrites as%
\begin{align*}
&  \left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)
-\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \left(
\alpha_{\left(  2\right)  }\right)  _{\left(  2\right)  }\right)  \right]
\right)  _{x}\left(  \left(  \alpha_{\left(  2\right)  }\right)  _{\left(
1\right)  }\right) \\
&  =\left(  id\ast\varphi^{a}\ast\psi^{b}\right)  \left(  \alpha\right)
-\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[  a,b\right]
\right)  _{x}\left(  \alpha_{\left(  2\right)  }\right)  .
\end{align*}
Hence, it will be enough to prove that%
\begin{align}
&  \alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}:\left[  a,\psi^{b}\left(  \left(
\alpha_{\left(  2\right)  }\right)  _{\left(  2\right)  }\right)  \right]
\right)  _{x}\left(  \left(  \alpha_{\left(  2\right)  }\right)  _{\left(
1\right)  }\right) \nonumber\\
&  =\alpha_{\left(  1\right)  }\left(  \dfrac{\varphi\left(  t+u\right)
-\varphi\left(  t\right)  }{u}\psi\left(  u\right)  :\left[  a,b\right]
\right)  _{x}\left(  \alpha_{\left(  2\right)  }\right)  . \label{pf2}%
\end{align}
This will clearly be proven if we succeed to show that every $\beta\in
S\left(  \mathfrak{g}\right)  $ satisfies%
\begin{align}
&  \left(  \dfrac{\varphi\left(  t+u\right)  -\varphi\left(  t\right)  }%
{u}:\left[  a,\psi^{b}\left(  \beta_{\left(  2\right)  }\right)  \right]
\right)  _{x}\left(  \beta_{\left(  1\right)  }\right) \nonumber\\
&  =\left(  \dfrac{\varphi\left(  t+u\right)  -\varphi\left(  t\right)  }%
{u}\psi\left(  u\right)  :\left[  a,b\right]  \right)  _{x}\left(
\beta\right)  \label{pf3}%
\end{align}
(because applying (\ref{pf3}) to $\beta=\alpha_{\left(  2\right)  }$ and
multiplying with $\alpha_{\left(  1\right)  }$, we will obtain (\ref{pf2})).
So let us prove (\ref{pf3}).\newline Let us show a somewhat stronger
assertion: let us show that every polynomial $P\in\mathbb{K}\left[
t,u\right]  $ satisfies%
\begin{equation}
\left(  P:\left[  a,\psi^{b}\left(  \beta_{\left(  2\right)  }\right)
\right]  \right)  _{x}\left(  \beta_{\left(  1\right)  }\right)  =\left(
P\cdot\psi\left(  u\right)  :\left[  a,b\right]  \right)  _{x}\left(
\beta\right)  . \label{pf4}%
\end{equation}
Once this equality (\ref{pf4}) is proven, (\ref{pf3}) will immediately follow
(by setting $P=\dfrac{\varphi\left(  t+u\right)  -\varphi\left(  t\right)
}{u}$). So let us prove (\ref{pf4}):\newline Since the equality (\ref{pf4}) is
linear in $\psi$ and $P$ and continuous in $\psi$, we can WLOG assume that
$\psi=z^{k}$ for some $k\in\mathbb{N}$, and that $P=t^{r}u^{q}$ for some
$r\in\mathbb{N}$ and $q\in\mathbb{N}$. Then,%
\begin{align*}
&  \left(  P:\left[  a,\psi^{b}\left(  \beta_{\left(  2\right)  }\right)
\right]  \right)  _{x}\left(  \beta_{\left(  1\right)  }\right) \\
&  =\left(  t^{r}u^{q}:\left[  a,\psi^{b}\left(  \beta_{\left(  2\right)
}\right)  \right]  \right)  _{x}\left(  \beta_{\left(  1\right)  }\right)
=\left[  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)  ,\left(
\operatorname*{ad}x\right)  ^{q}\left(  \psi^{b}\left(  \beta_{\left(
2\right)  }\right)  \right)  \right]  \left(  \beta_{\left(  1\right)
}\right) \\
&  =\left[  \left(  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)
\right)  \left(  \left(  \beta_{\left(  1\right)  }\right)  _{\left(
1\right)  }\right)  ,\left(  \left(  \operatorname*{ad}x\right)  ^{q}\left(
\psi^{b}\left(  \beta_{\left(  2\right)  }\right)  \right)  \right)  \left(
\left(  \beta_{\left(  1\right)  }\right)  _{\left(  2\right)  }\right)
\right] \\
&  =\left[  \left(  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)
\right)  \left(  \beta_{\left(  1\right)  }\right)  ,\left(  \left(
\operatorname*{ad}x\right)  ^{q}\left(  \psi^{b}\left(  \left(  \beta_{\left(
2\right)  }\right)  _{\left(  2\right)  }\right)  \right)  \right)  \left(
\left(  \beta_{\left(  2\right)  }\right)  _{\left(  1\right)  }\right)
\right]
\end{align*}
and%
\begin{align*}
&  \left(  \underbrace{P\cdot\psi\left(  u\right)  }_{=t^{r}u^{q}\cdot
u^{k}=t^{r}u^{q+k}}:\left[  a,b\right]  \right)  _{x}\left(  \beta\right) \\
&  =\left(  t^{r}u^{q+k}:\left[  a,b\right]  \right)  _{x}\left(
\beta\right)  =\left[  \left(  \operatorname*{ad}x\right)  ^{r}\left(
a\right)  ,\left(  \operatorname*{ad}x\right)  ^{q+k}\left(  b\right)
\right]  \left(  \beta\right) \\
&  =\left[  \left(  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)
\right)  \left(  \beta_{\left(  1\right)  }\right)  ,\left(  \left(
\operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \right)  \left(
\beta_{\left(  2\right)  }\right)  \right]  .
\end{align*}
The equality (\ref{pf4}) thus transforms into%
\begin{align}
&  \left[  \left(  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)
\right)  \left(  \beta_{\left(  1\right)  }\right)  ,\left(  \left(
\operatorname*{ad}x\right)  ^{q}\left(  \psi^{b}\left(  \left(  \beta_{\left(
2\right)  }\right)  _{\left(  2\right)  }\right)  \right)  \right)  \left(
\left(  \beta_{\left(  2\right)  }\right)  _{\left(  1\right)  }\right)
\right] \nonumber\\
&  =\left[  \left(  \left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)
\right)  \left(  \beta_{\left(  1\right)  }\right)  ,\left(  \left(
\operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \right)  \left(
\beta_{\left(  2\right)  }\right)  \right]  . \label{pf5}%
\end{align}
It thus remains to prove (\ref{pf5}).\newline Since $\psi^{b}=\underbrace{\psi
}_{=z^{k}}\left(  \operatorname*{ad}x\right)  \left(  b\right)  =\left(
\operatorname*{ad}x\right)  ^{k}\left(  b\right)  $, every $\gamma\in S\left(
\mathfrak{g}\right)  $ satisfies%
\begin{align*}
\left(  \left(  \operatorname*{ad}x\right)  ^{q}\left(  \psi^{b}\left(
\gamma_{\left(  2\right)  }\right)  \right)  \right)  \left(  \gamma_{\left(
1\right)  }\right)   &  =\left(  \left(  \operatorname*{ad}x\right)
^{q}\left(  \left(  \operatorname*{ad}x\right)  ^{k}\left(  b\right)  \left(
\gamma_{\left(  2\right)  }\right)  \right)  \right)  \left(  \gamma_{\left(
1\right)  }\right) \\
&  =\left(  \operatorname*{ad}x\right)  ^{q+k}\left(  b\right)  \left(
\gamma\right)
\end{align*}
(by Observation 3, applied to $\gamma$ instead of $\alpha$). Applying this to
$\gamma=\beta_{\left(  2\right)  }$ and taking the Lie bracket with $\left(
\left(  \operatorname*{ad}x\right)  ^{r}\left(  a\right)  \right)  \left(
\beta_{\left(  1\right)  }\right)  $, we obtain (\ref{pf5}). As explained
above, this completes the proof of Lemma 2.1.1 \textit{ii)}.

\item \textbf{Proof of Theorem 2.1.1:} The first three lines of this proof
don't seem to belong into this proof. Neither does the last line of the
computation. Also, there are some typos:\newline- On the third line of the
computation, \textquotedblleft$\dfrac{\psi\left(  t+u\right)  -\psi\left(
t\right)  }{u}\varphi\left(  u\right)  $\textquotedblright\ should be
$\dfrac{\psi\left(  t+u\right)  -\psi\left(  u\right)  }{t}\varphi\left(
t\right)  $.\newline- On the fourth line of the computation, \textit{I think}
there should be a $\left(  -1\right)  ^{\text{something}}$ term in front of
the second fraction. I am not exactly sure here since I have never been
following the $\left(  -1\right)  $ signs carefully.

\item \textbf{Lemma 2.1.2:} In this lemma (and its proof), \textquotedblleft%
$\mathbb{N}$\textquotedblright\ should be replaced by $\mathbb{N}%
\setminus\left\{  0\right\}  $. (Here I am assuming that $\mathbb{N}$ contains
$0$ in your terminology. This assumption is reinforced by the statement of
Remark 1.3.1.)

\item \textbf{Proof of Lemma 2.1.3:} Replace \textquotedblleft$\left(
w\left(  t,u\right)  ,\left[  \alpha,\beta\right]  \right)  _{x}%
$\textquotedblright\ by \textquotedblleft$\left(  w\left(  t,u\right)
:\left[  \alpha,\beta\right]  \right)  _{x}$\textquotedblright\ (three times).

\item \textbf{Proof of Lemma 2.1.3:} In the formula, there are two commata
instead of one on the right hand side.

\item \textbf{Proof of Theorem 2.1.2:} You write: \textquotedblleft By theorem
1.2.1, this identity is equivalent to\textquotedblright. In my opinion, what
you are using here is not Theorem 1.2.1, but simply the $\ast$-invertibility
of $id$.

\item \textbf{Remark 2.2.2:} Replace \textquotedblleft$\dfrac{1}{e^{z}-1}%
$\textquotedblright\ by $\dfrac{z}{e^{z}-1}$. Also, replace \textquotedblleft%
$\sum\limits_{k\geq0}^{\infty}$\textquotedblright\ by $\sum\limits_{k\geq0}$.

\item \textbf{Proof of Theorem 2.2.4:} In the first line of this proof,
\textquotedblleft$\omega\left(  ,u\right)  $\textquotedblright\ should be
$\omega\left(  t,u\right)  $.

\item \textbf{Lemma 2.2.3:} You might want to change \textquotedblleft%
$\mathbb{K}\left[  t\right]  /t^{N}$\textquotedblright\ into $\mathbb{K}%
_{0}\left[  t\right]  /t^{N}$. (In fact, you only consider $\varphi
\in\mathbb{K}_{0}\left[  t\right]  /t^{N}$ in the proof. I am not sure whether
this is because the other case is not interesting enough to you, or you can
easily rule it out.)

\item \textbf{Theorem 2.2.5:} Replace \textquotedblleft$t^{n}$%
\textquotedblright\ by $t^{N}$ (I think).

\item \textbf{First line of \S 2.3:} Replace \textquotedblleft
commutating\textquotedblright\ by \textquotedblleft
commuting\textquotedblright.

\item \textbf{(2.16):} The lower arrow of this commutative diagram should be
$F_{\mathfrak{h}}$.

\item \textbf{Proof of Theorem 2.4.1:} A comma is missing in \textquotedblleft%
$x_{1},...x_{n+1}$\textquotedblright.

\item \textbf{Proof of Theorem 2.4.1:} Replace the \textquotedblleft%
$\mathfrak{g}$\textquotedblright\ by an \textquotedblleft$\mathfrak{h}%
$\textquotedblright\ in \textquotedblleft Let $Y:=F_{\mathfrak{g}}\left(
x_{1}\cdots x_{n}\otimes x_{n+1}\right)  $\textquotedblright.

\item \textbf{Proof of Theorem 2.4.1:} Replace every letter \textquotedblleft%
$X$\textquotedblright\ in \textquotedblleft$f_{t,i}:X_{j}\mapsto\left\{
\begin{array}
[c]{c}%
X_{j},\ \ \ j\neq i\\
X_{i}t,\ \ \ j=i
\end{array}
\right.  $\textquotedblright\ by a lowercase \textquotedblleft$x$%
\textquotedblright.

\item \textbf{Proof of Theorem 2.4.1:} Replace \textquotedblleft%
$f_{t,i}\left(  Y\right)  $\textquotedblright\ by \textquotedblleft%
$\widetilde{f}_{t,i}\left(  Y\right)  $\textquotedblright.

\item \textbf{Proof of Theorem 2.4.1:} I don't understand how you obtain
\textquotedblleft$Y=\sum_{i=1}^{n+1}Y_{1,i}$\textquotedblright. However, it is
completely enough to know that \textquotedblleft$Y=\sum_{i\geq0}Y_{1,i}%
$\textquotedblright, and this is obvious.

\item \textbf{Proof of Theorem 2.4.1:} Replace \textquotedblleft$i\in\left\{
i,...,n+1\right\}  $\textquotedblright\ by \textquotedblleft$i\in\left\{
1,...,n+1\right\}  $\textquotedblright.

\item \textbf{Proof of Theorem 2.4.1:} Replace \textquotedblleft brackets of
$n$ elements\textquotedblright\ by \textquotedblleft brackets of $n+1$
elements\textquotedblright.

\item \textbf{Proof of Lemma 2.5.1:} \textquotedblleft using
(2.2)\textquotedblright\ should be \textquotedblleft using
(2.1)\textquotedblright\ in my opinion.

\item \textbf{Proof of Theorem 2.5.1 ii):} In the formula, you write
\textquotedblleft$\left(  \Phi^{a_{1}}\circ\Phi^{g\left(  a_{j}\right)  }%
\circ\Phi^{a_{n}}\right)  \left(  1\right)  $\textquotedblright. This should
be
\[
\left(  \Phi^{a_{1}}\circ\cdots\circ\Phi^{a_{j-1}}\circ\Phi^{g\left(
a_{j}\right)  }\circ\Phi^{a_{j+1}}\circ\cdots\circ\Phi^{a_{n}}\right)  \left(
1\right)  .
\]
(You can leave out the $\Phi^{a_{j-1}}$ and $\Phi^{a_{j+1}}$ terms if you
wish, but at least the $\cdots$ should be there.)

\item \textbf{Proof of Theorem 2.5.1 ii):} You write \textquotedblleft$\left[
g_{2},\Phi_{1}^{a}\right]  =1\otimes\varphi_{1}^{a}\circ\left(  g_{2}%
\otimes1+1\otimes g_{2}-\Delta\circ g_{2}\right)  +\Phi^{g\left(  a\right)  }%
$\textquotedblright. This should be%
\[
\left[  g_{2},\Phi_{1}^{a}\right]  =m\circ\left(  1\otimes\varphi_{1}%
^{a}\right)  \circ\left(  \left(  g_{2}\otimes1+1\otimes g_{2}\right)
\circ\Delta-\Delta\circ g_{2}\right)  +\Phi^{g\left(  a\right)  }.
\]
(Besides I don't understand why you are renaming $id$ as $1$ again, but it's
fine for me.)

\item \textbf{Proof of Lemma 2.5.2:} \textquotedblleft From identity
(1.2)\textquotedblright\ should be \textquotedblleft From identity
(2.2)\textquotedblright.

\item \textbf{Remark 2.5.5:} Replace \textquotedblleft
Dedeking\textquotedblright\ by \textquotedblleft Dedekind\textquotedblright.

\item \textbf{Remark 2.5.5:} There is a useless parenthesis before
\textquotedblleft it is shown that $\beta$ is one-to-one\textquotedblright.
\end{itemize}

\section*{Chapter 3}

\begin{itemize}
\item \textbf{Page 31:} Replace \textquotedblleft if $A$ is a $\mathbb{K}%
$-algebra equipped with a comultiplication $\Delta$\textquotedblright\ by
\textquotedblleft if $A$ is a $\mathbb{K}$-coalgebra with a comultiplication
$\Delta$\textquotedblright.

\item \textbf{(3.1):} The right hand side should be $\left.  \partial\left(
X_{1}\right)  \circ\cdots\circ\partial\left(  X_{n}\right)  \left(  f\right)
\right.  \mid_{0}$ rather than $\left.  \partial\left(  X_{1}\right)
\circ\cdots\circ\partial\left(  X_{n}\right)  \right.  \mid_{0}\left(
f\right)  $.

\item \textbf{Theorem 3.1.1:} I believe \textquotedblleft$-\varphi_{c}\left(
\operatorname*{ad}x\right)  \left(  a\right)  $\textquotedblright\ should be
\textquotedblleft$-\left(  \varphi_{c}\left(  \operatorname*{ad}x\right)
\left(  a\right)  \right)  ^{T}$\textquotedblright.

\item \textbf{Proof of Theorem 3.1.1:} The minus sign in \textquotedblleft%
$-\left(  -1\right)  ^{p\left(  X_{1}\cdots X_{n}\right)  p\left(  a\right)
}\left\langle X_{1}\cdots X_{n},\xi_{c}^{a}\left(  f\right)  \right\rangle
$\textquotedblright\ should be removed. (The minus sign should only appear
later due to the definition of the dual of a representation of a Lie algebra.)

\item \textbf{Remark 3.1.1:} I don't understand what is meant by
\textquotedblleft the evaluation of $\xi_{c}^{a}$ in $X\in\mathfrak{g}_{0}%
$\textquotedblright.
\end{itemize}

\section*{Chapter 4}

\begin{itemize}
\item \textbf{Page 36:} It would be good to clarify if the notions of
\textquotedblleft$\mathbb{K}$-supersymmetric space\textquotedblright\ (or
\textquotedblleft$\mathbb{K}$-super symmetric space\textquotedblright) and
\textquotedblleft$\mathbb{K}$-symmetric space\textquotedblright\ are used
interchangeably. (I think they are, but I am not sure.)

\item \textbf{Page 36, Example 4.0.1:} Doesn't the example i) only work when
$\mathbb{K}=\mathbb{K}_{0}$ ?

\item \textbf{Page 36, proof of Lemma 4.0.2:} Replace \textquotedblleft%
$\Phi_{c}$ et $\Phi_{d}$\textquotedblright\ by \textquotedblleft$\Phi_{c}$ and
$\Phi_{d}$\textquotedblright.

\item \textbf{Page 37, Theorem 4.0.2:} Replace \textquotedblleft A
representations\textquotedblright\ by \textquotedblleft A
representation\textquotedblright.

\item \textbf{Page 40, \S 4.2:} Replace \textquotedblleft of finite
rang\textquotedblright\ by \textquotedblleft of finite rank\textquotedblright.

\item \textbf{Page 40, \S 4.3:} Replace \textquotedblleft we give a an
example\textquotedblright\ by \textquotedblleft we give an
example\textquotedblright.
\end{itemize}

\section*{Chapter 5}

\begin{itemize}
\item \textbf{Page 47, \S 5.1:} Replace \textquotedblleft for any
$X,Y\in\mathfrak{g}$\textquotedblright\ by \textquotedblleft for any $X,Y\in
M$\textquotedblright.

\item \textbf{Page 47, \S 5.1:} Replace \textquotedblleft We say that $\alpha$
is \textit{not-degenerate}\textquotedblright\ by \textquotedblleft If
$M=\mathfrak{g}$ and $N=\mathbb{K}$, then we say that $\alpha$ is
\textit{non-degenerate}\textquotedblright.

\item Generally, replace every appearance of \textquotedblleft
not-degenerate\textquotedblright\ in the text by \textquotedblleft
non-degenerate\textquotedblright.
\end{itemize}

\section*{Chapter 6}

\begin{itemize}
\item \textbf{.}
\end{itemize}

\section*{Appendix}

\begin{itemize}
\item \textbf{.}
\end{itemize}


\end{document}